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Units of E=mc²

I was wondering if someone could explain the units used in this equation. I have never seen an explanation which includes the units. It would seem that the equation would not hold if the units changed. User:jimaginator

This relation will be true if the units are consistent. For example using all SI units (E in Joules, m Kilograms and c in metres/second). But other units would be valid, for example natural units (E and m in MeV, c=1). 137.138.46.155 07:37, 23 August 2006 (UTC)

Units of E=mc² again

E=MC2

THE e IN e=mc2 MEANS energy..................

THE = IN e=mc2 MEANS equals..................

THE m IN e=mc2 MEANS mass..................

THE c IN e=mc2 MEANS cillibus.................. I DONT KNOW IF I SPELT THE WORD CORECT. BUT BASICLY cillibus MEANS the speed Of light..................

THE 2 IN e=mc2 MEANS squared. (TIMES BY IT SELF)

e=mc2 = ENERGY = MASS CILLIBUS TIMES BY ITS SELF

...............................................................................................

jimaginator

I had the same question. For the unit of mass, it seems like electronvolt (eV) is normally used in particle physics.[http:// Units: E] c is a constant the speed of light(m/s). But when calculating the energy of a nucler fission of a chunk of plutonium for example, eV is not suitable to use because the unit is too small. According this website E=mc^2 The Basics, Kg is used in that instance. Both are SI units and are convertible, so I guess that explains it.--Nc622 16:58, 5 Nov 2004 (UTC)

Thanks for your response. The E=mc2 Basics website states that it's Energy in Joules, mass in kg, and speed in meters per second. So far so good. Now Wiki says a Joule is: "One joule is the work required to exert a force of one newton for a distance of one metre". ---So now we have: 1N x 1m = 1kg x (m/s) x (m/s) A newton is a SI derived unit defined as the amount of force required to accelerate a mass of one kilogram at a rate of one metre per second squared. ---So now we have: 1kg x m x s^-2 = 1kg x (m/s) x (m/s). So, both sides end up being defined with kg,m,s; the basis of the SI units in the first place. It all seems rather circular in structure to me. If the units are picked to have some inherent relationship in the first place, anything can be used. And I suppose the squaring of the right side is where the really big energy quantity comes from, but somehow it's not all that satisfying. I guess it's the nature of the concept of energy in the first place, since this is more abstract than a meter which we can pace out, or a second which we have all have a subjective feel for. I had always felt that the simplicity of the equation was due to something simple in the fabric of the universe. When I first heard the equation in junior high, it was a wow moment, how could it be so simple? But really it isn't. What would the equation look like in other measurement systems? I suspect not so elegant. I wonder if Einstein was using SI, or even if he was thinking in terms of any particular system at all at first. Jimaginator 14 Nov 2004 UTC

The equation would look exactly the same in any other measurement system. I'm having trouble envisioning a thought process that could possibly lead you to believe otherwise. Of course the units are picked because of their inherent relationship; that's why the equation is meaningful! E is any unit of energy, m is any unit of mass, and c is any unit of velocity: so long as the three units are consistently defined, the equation will look exactly the same.

Further, eV is most emphatically not a unit of mass. An electron-volt is a unit of energy. The figures you see quoted in particle physics texts are in units of eV/c^2 which, not coincidentally, is a direct consequence of E=mc^2.

Mote 04:43, 9 Jan 2005 (UTC)

Well, sorry about the old thought processes, but the equation being based on circular, intertwined units is the crux of the matter. If we use pounds, parsecs/hour, and BTUs or whatever, I believe the equation WILL look terrible. If Einstein did not invent the SI units, then the interelationship between the units somehow pre-cognitively anticipates the future equation. Why even bother with the ^2 element at all? Pre-define the interconnection of the units right, and the equation could be E=mc! Still not satisfied here Jimaginator 19:30, Mar 17, 2005 (UTC)

Think of it this way: energy equals a constant times mass multiplied by the square of the speed of light. These are all abstract concepts. This relationship between the abstract concepts of energy, mass, and the speed of light is the simple truth which struck you so in junior high. SI units are chosen so that the constant equals one. If you use imperial units or made up units, the abstract relation still holds. There will be a different constant in the equation, but energy always equals a constant times mass multiplied by the square of the speed of light. There is no way to measure speed such that energy equals mass times the speed of light. Speed is always a measure of distance traveled per unit of time. There is no way to define the "interrelationship" of units so that energy is not proportional to mass, or so that energy is not proportional to the square of the speed of light.

Many of the questions above illustrate confusion about how units and formulas work. Let's think for a minute about a simple, intuitive formula: the area of a rectangle as a function of its width and height:

$A=W\times H$

The important thing to understand is that when you plug in units, they are subject to the same mathematics as the numbers. If you plug in W and H in meters, you get A in meters times meters, or m². If you plug in W and H in feet, you get A in feet times feet, or square feet. If you plug in W in parsecs and H in nanometers, you get A in nanometer-parsecs, which is not a convenient unit of area, but is a legitimate one.

If you say, "What if W is in feet, H is in meters, and A is in acres?" the question doesn't make sense; it's like asking, "What if W is 3, H is 7, and A is 18?". The units of the three variables are no more independent than their values; specifying two of them tells you what the third is.

E=mc² works the same way. If m is in kilograms, and c is in meters/second (SI units), then E is in kg-m²/s². That's the SI units for energy, and it has another name: the joule. If m is in grams, and c is in cm/s (cgs units), then E is in g-cm²/s², which is also called an erg (the cgs unit for energy). If m is in slugs and c is in furlongs/fortnight, then E is in slug-furlong²/fortnight², which is an inconvenient but legitimate unit of energy. (Note, of course, that the numerical value of c is different in these different units, just like your car's speed is different in MPH and km/h.)

In summary, physical formulas are not tied to a particular system of units, but the formula tells you the relationship between the units, just as it does for the numerical quantities. They have to be in agreement, with the same units on both sides of the equals sign. -- Coneslayer 2005 June 30 21:12 (UTC)

Thanks one and all for your responses. I believe I am starting to get the abstract nature of the relationship and the changing constant. But still I wonder how that ^2 got in there. Why isn't it ^2.039874 or something? I can't help but think of Pi's transcendental nature (OK, maybe that's a bad example, but the relationship of the circumference of a circle to it's diameter does seem to be locked into the universe, as is the speed of light. Why isn't Pi 3.00000...?) E=mc^2 still seems amazingly elegant, which I suppose is a testament to Einstein's genius. Anyway, thanks very much for the increased understanding Jimaginator 18:33, 22 September 2005 (UTC)

For a better understanding, see the same article in the french wikipedia.--24.37.172.194 04:20, 23 August 2005 (UTC)

I see what jimaginator is saying though, if E is expressed in joules ( kg x m^2 / s^2) and c^2 is expressed in m/s then the only real "mystery" is the signficants of the speed of light or electromagnetic radiation. I.e. kg x (m/s)^2 = kg x (m/s)^2. With the constant being 3 X 10^8. So, what's the deal with that number for speed (is it really saying that at the speed of light, matter becomes energy?) Also just below (E = m), the whole point of the equation is to relate energy to mass, if you define energy as mass then you defeat the purpose of the whole equation, no? Energy, as define in joules, is the amount of force it takes to accelarate 1kg of mass 1meter per second, every second. To me that feels like a real unit of energy; not kg = kg, I guess you could say 1 joule = .00000003kg, and so would that mean, in a vaccuum, if .00000003 kg of whatever, moving at or very near the speed of light, smacked into 1kg of something it would accellarate it to nearly 1m/s. But what is that saying? and why, when the "whatever" accellarates to the speed of light, does it start acting like a wave and not a particle? Niubrad 07:07, 9 October 2006 (UTC)

E = m

If we measure E and m in units of mass (kilograms, grams, etc...), the formula becomes E = m, which is much more simpler than the usual one and much more easier to understand. The c² is just a factor of conversion to adjust the mass measure in grams to the E measure in joules: grams can not simply be equal to joules. 24.202.163.194 17:31, 30 December 2005 (UTC)

That is 100% true and, simltaneously, completely non-enlightening to non-expert users. -- SCZenz 00:35, 14 January 2006 (UTC)

I agree with you. This little text was an answer to a contributor question, at the beginning of the section, who writes: 'It would seem that the equation would not hold if the units changed.' [[1]]
To be completely enlightening to non-expert users, it would take an entirely new article. I think there is one in the french part of wikipedia, or at least a section of article, but I did not verify recently. On the other hand, as it is used in the world of physicists, it should be present in wikipedia. It is our job to do so. --Aïki 01:12, 14 January 2006 (UTC)

This might be stupid but when using E=MC^2 i just use tons of TNT which is 21,480,764,310 tons of TNT for one gram of mass i think --Longchuan 03:01, 04/28/2006

Equation and the bomb

I tried to make more accurate the connection between this equation and the bomb, which seems to often erroneously imply that the uranium itself can serve as the "mass" end of this equation (it refers to the binding energy of the nucleus, much smaller than the mass of the atoms themselves). But I'm not a physicist. If someone with more expertise here could look over what I've done and see if it is correct, I'd be very happy. --Fastfission 01:52, 15 Jan 2005 (UTC)

I'm currently reading Greene's The Elegant Universe, which states: "The world grasped the devastating power arising from the conversion of less than 1 percent of two pounds of uranium into energy at Hiroshima" (51) - which would seem to suggest that the uranium is what's converted. ~ Booyabazooka 01:57, 4 May 2005 (UTC)

A piece of uranium emits radiations. If you weight it at a time t and later at a time t + dt, you will have less weight at the time t + dt than at the time t. Thus we can say that mass contain energy, or, that energy (radiations) convey mass. 24.202.163.194 18:16, 30 December 2005 (UTC)

Fastfission is along the right lines here with binding energy. The Uranium splits into two smaller elements, which are nucleon for nucleon, the same total relative atomic mass (a nucleon being a proton or neutron and having a mass of 1 unit). However, overall mass is still lost in the process. The mass lost is the m in the equation, and the resultant E is termed the binding energy. It's this energy which blows stuff up. --Nathan (Talk) 04:11, 8 January 2006 (UTC)

Matter to energy

The mass which is converted into energy is not the fissionable material itself (i.e. the uranium or the plutonium), but related to the strong nuclear force of the bonds which holds the large nucleus all together.

Actually, then, it seems that a fission reaction doesn't convert any matter into energy at all; it just releases a lot of energy that was formerly "contained", correct? If incorrect, can someone explain this in some detail? The popular belief is that fission and fusion convert some amount of mass into energy. Thanks. Tempshill 17:33, 16 May 2005 (UTC)

By matter, do you mean the protons and the neutrons?
What we call the mass, is what we obtain by weighing the object and by deducing the mass from calculation. 24.202.163.194 18:44, 30 December 2005 (UTC)

Popular belief is correct: Fission product.

Yes, mass is converted into energy - however it is much easier to think of this concept if you understand that truly mass *is* energy. Energy *is* mass. A good example to drive this home is a frozen dinner - take a dinner out of the freezer, put it in the microwave and heat it up. The dinner will GAIN MASS. This mass comes in the form of heat, and yes I do mean energy as well, but if you weigh the dinner on a very sensitive scale, you will see that the dinner weighs more than it did before you heated it. Fresheneesz 07:46, 20 November 2005 (UTC)

Erm, mass isn't energy. If it was we could measure it in joules. The fact that we have to multiply it by a constant (with units) means that it can be converted into energy, not that it is energy. I know units are arbitrary, but this is the best way I can think of to describe it. --Nathan (Talk) 04:18, 8 January 2006 (UTC)

See above the subsection 1.1 E = m.--Aïki 02:38, 2 February 2006 (UTC)

Maybe it's easier to think that energy has mass, the energy released is actually the energy 'trapped' in the material which gave it extra mass. In fact 'intrinsic mass' (maybe see mass for more on this) of fundamental particles is pretty negligible when considering everyday mass which almost all results from binding energy within nucleons. 137.138.46.155 07:45, 23 August 2006 (UTC)

---

See the article on rest mass to explain why this does not account for the modern understanding of mass.

The equation also proves one of the fundamental limiting factors of space travel - the inability for anything with mass to exceed (or even reach) the speed of light. As the speed of light squared is always constant, the two variables in the equation are energy on one side, and mass on the other.

Therefore, when energy (in this case velocity) increases, mass must increase. Conversely, the more mass an object has, the more energy it requires to accelerate even further. The speed of light (c = 299,792,458 meters per second) is the point where the energy required to reach it is infinite, and attaining such a speed becomes impossible.

Roadrunner 04:08, 4 Jun 2005 (UTC)

Alternative view on E=mc^2

This is undoubtedly rubbish but would like to have it confirmed. Sure the 'c' is only valid as a function of time. So could time be an alternative form of energy and mass? It would answer a few of questions that seem to float around. E.g. the universe doesn't have enough mass - but it does have lots of time. In the instant after big bang a huge amount of mass, energy and time were created - which implies a strong relationship. What is at the centre of an atom/proton/quark - perhaps concentrated time. What is gravity and why does it distort space/time? - gravity is a function of mass which correlates to time and therefore attracts other time as it tries to return a singular point of time. The universe is only infinite for as much time as is available for it to exist. What's beyond the universe is no time. This is not the linear time that we experience day to day but time the essence. -unsigned

I can confirm "This is [...] rubbish". WAS 4.250 20:03, 9 October 2005 (UTC)

Time, space, mass, energy, the particle-like nature of reality (e.g. interactions occur at definite places in space-time), and the wave-like nature of reality (e.g. between interactions particles can pop into and out of existence and lack a definite location or momentum) are all one indivisible THING. None can exist without all of them existing. The expansion of the univrse since the big bang is NOT matter flying into space; but space-time being CREATED between the particles (that's why the expansion is not limited to being below the speed of light). As you read this, space-time is being created between the particles that make up your body (and everything else). WAS 4.250 20:03, 9 October 2005 (UTC)

'Sure the 'c' is only valid as a function of time.'
The 'c' being a constant, it cannot vary at all, as a function of time or anything else! 24.202.163.194 19:23, 30 December 2005 (UTC)

'So could time be an alternative form of energy and mass? '

Not in special relativity! Where did you get that nonsense? 24.202.163.194 19:28, 30 December 2005 (UTC)

Can e=mc2 be plotted as a function of time? And if so what would the plot look like? Zim

Celeritas

I have written a new article on Celeritas which I feel should be linked to from this article but can't really figure out how. Any input would be welcomed. Majts 00:03, 12 October 2005 (UTC)

I added "Celeritas is said to be the c in E=mc²." to the See also section. WAS 4.250 00:51, 12 October 2005 (UTC)

thanks, that works, although I took the liberty of making a change to the wording Majts 01:12, 12 October 2005 (UTC)

Energy to mass?

Is this even theoretically possible? Would a giant discharge of energy produce a miniscule amount of matter? JD79 04:31, 5 November 2005 (UTC)

Well... yes... and it was first done in 1932. You can see the picture here [2]

Cultural Significance

Unfortunately I don't have the time to do it, but may I suggest perhaps a section to the article stipulating the cultural significance of the famous equation aswell as notable references that allude to it in various forms of popular culture.

Equation is an approximation

I think it should be noted that the equation E=mc^2 is an approximation good only when v<<c . My edit in this respect was reverted as being *wrong* - while it is quite clear (and is shown later down the page) that it most certainly *is* an approximation. Fresheneesz 07:04, 21 November 2005 (UTC)

Provide a source and we'll talk. WAS 4.250 11:57, 21 November 2005 (UTC)

You aren't confusing m with rest mass, are you? In relativity that's usually denoted m0. - 172.135.179.230 17:03, 21 November 2005 (UTC)

Well, actually m is the standard way to write mass - and it usually means rest mass. m0 is used to differentiate rest mass from "relativistic mass" - a term that many physicsists (including einstein) think is a "wrong" way to think about it. The point is that in the equation E=mc^2 - m is REST mass. And in any case, the page *should* indicate that m is relativistic mass if it is such.

But you're right, if m indicated relativistic mass - then it would be correct - but it would be very misleading for those who are used to m being rest mass, namely almost every physicist out there. - As for a source, how bout THIS PAGE. WAS 4.250 - have you even read the article? Do you understand the meaning of the relativistic part of this page? It is OBVIOUS that E=mc^2 is an approximation, and I don't appreciate this constant nagging for a source when the source is on the same damn page. Just because I don't provide an ISBN number doesn't mean that my information is incorrect. - Not to mention, I find it infinitely irksome that you don't even explain your stance - you just want a fucking source. Try *THINKING* for once. E=mc^2 only works when m is relativistic mass. And given that that is NOT the case (as of yet the page never mentions that that is the case), the equation must be an approximation. I believe I already explained this on your talk page WAS - so TALK to me. Fresheneesz 18:28, 21 November 2005 (UTC)

I plan on either noting that m is relativistic mass - or changing the equation back to an approximation. Please, anyone whos interested discuss. Fresheneesz 01:45, 22 November 2005 (UTC)

Noting that the m refered to here OF COURSE refers to the actual mass rather than the Newtonian mistaken concept of mass is appropriate. "Relativistic" mass is the real mass - the only mass - when equating mass to energy. Equating the Newtonian concept of mass to energy is ridiculous. As velocity increases the mass increases. The REAL mass. Not some funny special definition mass. Relativistic mass is that REAL mass. WAS 4.250 02:07, 22 November 2005 (UTC)

If you missed the entire conversation at this point... mass is not used to describe "real" relativistic mass. Most people would say rest mass is more "real" because it is constant, and it is the mass that the mass itself would measure if it measured itself. Do you not agree that relativistic mass has conventionally gone out of use? Fresheneesz 17:32, 26 November 2005 (UTC)

If we take the symbol m as the relativistic mass, then we should use m0 (or else) to denote the mass when the mass of the object is measure from an inertial frame of reference in which the object his at rest. Thus, if we talk of the Einstein famous formula, it should then be write E = m0c². But if we write E = mc², with m as the relativistic mass, it can be rewrite E = ym0c², and E = m0c² becomes a special case where v = 0, and where y = 1. With y = 1, we then have E = ym0c² reduce to E = m0c². 24.202.163.194 20:08, 30 December 2005 (UTC)

about Fresheneesz and physics article edits from my user talk page

Hey WAS, thanks so much for reverting those ridiculous edits on the Entropy page. Is there some way to prevent this kind of ridiculous editing in the future? The same guy totally messed up the Arrow of Time article (I reverted that) - he seems to mean well but... These are nontrivial subjects that should probably not have fundamental changes made to them by individuals without formal training in the subject matter. Heck, I don't even remember enough of my physics classes to edit this stuff without looking it up in a book! - JustinWick 18:01, 20 November 2005 (UTC)

Wikipedia:Vandalism in progress WAS 4.250 12:03, 21 November 2005 (UTC)

Hi, its fine you think my edits are ridiculous - but did you guys actually go and *look* it up in a book before you reverted my changes? E=mc^2 is most definately an approximation - have you heard of relativity? I'm editing these pages to match current theory, and current knowlege - as is the entire point of wikipedia. I am most definately *NOT* trying to match text books, or in any way parallel the idiotic way our society goes about teaching us science. E=mc^2 only works when the velocity of an object is near 0. What is so hard about understanding that? Fresheneesz 20:40, 20 November 2005 (UTC)

You say "E=mc^2 only works when the velocity of an object is near 0." Provide a source and we'll talk. WAS 4.250 12:03, 21 November 2005 (UTC)

Max Plank apparently said so: [3], tell me - is the m in that equation relatvistic mass? Because if it is, that needs to be explained on that page and a link to relativistic mass given. Fresheneesz 18:37, 21 November 2005 (UTC)

There are good sources and there are bad sources. The author of the piece you referenced doesn't know what he's talking about. Please read this. Changes to the article to reflect knowledge contained in it but not in the wikipedia article MIGHT be appropriate, and probably ARE, if done in the right way. I myself am much better at facts than the best way to present those facts. Maybe the talk page of the article would be a good place to suggest additions to the article to present what you discover at the source I just suggested you read. But make sure you understand something before you jump in and change stuff. Saying you don't believe in entropy and making changes to the entropy article seems, well, vandal-like... WAS 4.250 21:30, 21 November 2005 (UTC)

I didn't say I don't understand enropy. I said I don't understand it fully - and i doubt many chemists do either. Its a difficult subject. I edit things I *do* understand, and I am most definately correct in the relationship between free energy and entropy.

- secondly, your constant crying for sources is getting ridiculous - a source is not what makes something correct. I would urge you to think about what *is* and what *isn't* correct before you revert an edit. When I see edits I don't understand - I ignore them, because I can't say whether its right or wrong. I'll only corect something in cases where i'm sure i'm right.

- I think your statement "A little knowlege is a dangerous thing" is a horrible outlook, and completely contrary to the ideas on which wikipedia is based. The point of Wikipedia is not to put people through the rigamarole we went through in school, but to teach the full depth of a concept - with all its knowlege in full view.

- all that said, If you want me to take the m in E=mc^2 as relativistic mass - I will note that on the page myself. If you want me to take m as rest mass (as it should be - agreed upon by most physicists - see the page for rest mass, its the same page as relativistic mass), then I will reinstate my approximation claim. So please choose what you want. Fresheneesz 01:39, 22 November 2005 (UTC)

Fresheneesz, you have run afoul of several issues at once. (1) The simplest is the question of a proper understanding and statement of the physics. Assuming good faith and civil discussion, there's a fair chance we can sort that out to everyone's satisfaction. (2) A little knowledge is a dangerous thing; however, it has never been established how much knowledge is safe. The Wikipedia approach to resolving questions draws heavily on the concept of no original research. Pragmatically, this often comes down to citing sources. For example, suppose I have just completed a rock-solid proof of the Riemann Hypothesis, at home in my study. Further suppose twenty trustworthy mathematician editors here vouch for it. Can I put that on Wikipedia? The answer is no; it must first appear in a reputable journal. Or suppose the two of us witness the landing of an interstellar craft, chat with its occupants, and obtain unique tangible items from them that prove it happened. Can that incident be reported in Wikipedia? Again, no; not until it has been published elsewhere. So WAS 4.250 is merely following standard protocol by asking you to cite sources. (3) A century has passed since Einstein published the equation, but Wikipedia is very young. There is a growing awareness of a need for better quality control, and poorly written articles like this are a symptom. The unfortunate truth is that an excellent article requires a rare combination: subject knowledge, expressive skill, free time, adequate motivation, editor consensus, and constant vigilance. And to obtain editor consensus, you may need outstanding political skills coupled with extraordinary patience. Don't expect Wikipedia to be something it is not. --KSmrq^T 16:00, 22 November 2005 (UTC)

I think a fresheneesz is needed in every wiki-page Fresheneesz, I believe there is a place for your type of edits on wikipedia. Ist, WAS, you dont own this page, or wikipedia. You dont make the rules. Asking for sources is ok, but what is the source for the color blue, and could you at least source the need for sources (as not to be a hypocrite)?. KSmrq states that information must appear in a reputable journal and that is just not the case, if everything that goes onto wikipedia had to appear in a reputable journal then wikipedia would be much much smaller, and the whole nature of wikipedia would change. Exempla gratia: Legend of Zelda. It's a balance, no? And, without fresheneesz coments on this discussion page, I would have less trust in wikipeda content. Let the debate continue. Niubrad 08:10, 9 October 2006 (UTC)

See Internet troll. WAS 4.250 01:49, 22 November 2005 (UTC)

By the way, there is a Wikipedia talk:WikiProject Physics and Wikipedia talk:WikiProject Mathematics where you can ask for comment on this kind of stuff. Oleg Alexandrov (talk) 06:01, 22 November 2005 (UTC)

e=mc^2 is not an approximation

Hi Fresheneesz. I came here from Oleg's house. E=mc^2 is not an approximation within its domain of applicability. Perhaps the reason that you think it is an approximation is because this article shows you an approximation where E is approximately mc2 for low v. That's true: in the case that the kinetic energy is negligible, the total energy is approximately equal to mc2. However, the equation E=mc2 is meant to apply to the case of rest energy (not total energy), or else to total energy with the m understood to be relativistic mass. The former case is more standard, but in neither case is the equation an approximation. -lethe ^talk 08:26, 22 November 2005 (UTC)

Thank you for the clarification. I think some clarification of this sort should appear in the article. Perhaps noting that E is "rest energy" ? Fresheneesz 08:41, 22 November 2005 (UTC)

I was going to add some text, but this article is in bad shape: explaining the technical details of the equation should not go into the "background" section, nor should it go into the "approximation" section. I couldn't figure out a clean place to add something. This article needs a section devoted to the equation, its derivation, and its meaning. When I finish everything on my to-do list (haha). -lethe ^talk 10:24, 22 November 2005 (UTC)

I'd happily stand corrected if adequate sources were provided. I hate this arguing. But the subject in question is the meaning of the "m" in Einstein's "E=mc squared" and the sources provided AGREE that the equation when written ("outdated") meant relativistic mass. That it is also true when velocity equals zero is a subset. Please read this. Dumbing it down for students doesn't change the meaning of the equation. Further discussion at Talk:Relativistic mass. WAS 4.250 17:54, 22 November 2005 (UTC)

Arguing? Who's arguing? I explained something to Fresheneez, and he thanked me. You probably could have done this yourself, instead of reverting with "cite sources". Do you know see how this can be confusing for the beginner? And you won't be corrected by any sources, because you are not incorrect. The equation is not an approximation.

Anyway, I seem to have another issue that I want to talk about with you. Is m in the equation rest mass or relativistic mass? You want to claim relativistic mass (in which case the formula is exact for all v) . I said above that rest mass is standard (in which case the formula is exact only for v=0, and wrong for other v), so we do appear to have a bona fide disagreement on our hands. Furthermore, I see that you've editted several pages to bring them into agreement with your position on this matter. I am inclined to revert, but let's discuss first, shall we?

Firstly, are we talking about the historical usage of the equation E=mc2? The more modern equation is m²=(E/c²)²−(p/c)². In this equation, there can be no doubt that m is rest mass (it's the Casimir invariant of the Poincaré group, aka the magnitude of the 4-momentum). The simpler formula can be viewed as a special case of this one for p=0. But that doesn't mean it has to be so viewed. It can also be viewed with the relativistic mass in mind. While it is not done today, perhaps it was in the past. I argue that not even that is true.

From the physics FAQ:

"The first record of the relationship of mass and energy explicitly in the form E = mc² was written by Einstein in a review of relativity in 1907. If this formula is taken to include kinetic energy, then it is only valid for relativistic mass, but it can also be taken as valid in the rest frame for invariant mass. Einstein's conventions and interpretations were sometimes ambivalent and varied a little over the years; however an examination of his papers and books on relativity shows that he almost never used relativistic mass himself. Whenever the symbol m for mass appears in his equations it is always invariant mass. He did not introduce the notion that the mass of a body increases with velocity--just that it increases with energy content. The equation E = mc2 was only meant to be applied in the rest frame of the particle. Perhaps Einstein's only definite reference to mass increasing with kinetic energy is in his "autobiographical notes"."

Apparently the notion of relativistic mass was introduced by the chemist Lewis much later. Here are some more quotes from the FAQ:

"It is not good to introduce the concept of the mass M = m/(1-v2/c2)1/2 of a body for which no clear definition can be given. It is better to introduce no other mass than `the rest mass' m. Instead of introducing M, it is better to mention the expression for the momentum and energy of a body in motion." -A. Einstein

"The concept of `relativistic mass' is subject to misunderstanding. That's why we don't use it. First, it applies the name mass--belonging to the magnitude of a four-vector--to a very different concept, the time component of a four-vector. Second, it makes increase of energy of an object with velocity or momentum appear to be connected with some change in internal structure of the object. In reality, the increase of energy with velocity originates not in the object but in the geometric properties of space-time itself." -Tayler and Wheeler

Relativistic mass is used in a lot of pop science books, for example books by Hawking or Kaku or the like. There are also some scientists who advocate their use in their work (for example the link you provided by Flores). However, it is hard to disagree with the fact that today, m always means rest mass. It is also likely that in Einstein's time in the equation E=mc², m also means rest mass. Your friend Flores is in a minority, and for Wikipedia to espouse his view is to breach NPOV. We should mention as often as is reasonable, that m usually means rest mass, but there are some who use it to mean relativistic mass, and what the difference is, and who uses what.

Those are my arguments. I disagree with the following edits of yours:

• this edit
  • Unilateral "since mass increases with velocity" may be OK in relativistic mass, but is no good in mass.
• this one
  • It is not only for teaching purposes. mass is rest mass in all kinds of research contexts. Really. Most practicing physicists in many theoretical branches don't ever consider the notion of relativistic mass. It is nearly impossible to do field theory, for example, with relativistic mass instead of rest mass.
• more
  • It is not inaccurate to call things with zero rest mass "massless". It is not NPOV to change all occurrences of the word "massless" to the phrase "with zero mass". "Massless" is standard usage among physicists, we are not here to redefine the language of the field, merely to display it. Let's be absolutely clear that the word massless refers to rest mass, but then use the term.
• [4]
  • This gives NPOV preferential treatment to relativistic mass: "since mass depends on velocity..." I think rest mass should be given primary standing, but failing that, there should be at least parity.
• [5]
  • I disagree with your source Flores, and don't think he should be taken as canon for wikipedia.
• [6]
  • equality of gravitational and inertial mass is not a theory. It's a hypothesis (synonymous with assumtion?) this edit also includes an enormous deletion of text with the edit summary "correct mistakes". I haven't followed the history of that article, so I can't say what's going on there, but I'm nervous.
• [7]
  • In this one you claim to be "removing POV", and you delete the suggestion that relativistic mass is used in popular science books, as well as several equations showing the benefits of relativistic mass. WTF?
• [8]
  • Another massive deletion of text without even an edit summary. What is going on here?

I feel that I have to revert all these edits. -lethe ^talk 19:26, 22 November 2005 (UTC)

I think Lethe has covered anything I could add. I agree, especially on your edits on the page mass - your edits there most defintately need discussion before they can be permenantly done. I hope that, although my actions have been rather distructive (to people's free time) I hope good change will come out of it when discussing the standard use of the word mass. Fresheneesz 20:57, 22 November 2005 (UTC)

A physicist's view

OK, I can't even read the mess above without hurting my head, so here area couple of facts to help you:

1. Physicists rarely use the "relativistic mass"—usually m just denotes the rest mass.
2. E=mc² is incomplete. The full version, as it is usually written out, is

   $E^{2}=p^{2}c^{2}+m^{2}c^{4}$.

3. E = mc² is thus valid only in the reference frame where the object is stationary.

Yes, you can do this other ways, but they are much less common in modern physics. -- SCZenz 00:53, 23 November 2005 (UTC)

That being said, if the article is presenting Einstein's original treatment, and is clear about it, that's not a problem. But it definitely is different than how the material would be presented today. -- SCZenz 00:56, 23 November 2005 (UTC)

I agree 100% with SCZenz, including the head-hurting part, and I'm glad someone else said all this so I don't have to. (go Bears!) Melchoir 01:04, 23 November 2005 (UTC)

Blame the professors. I think introductory, undergraduate relativity is often taught so that students learn the formula $M=m{\sqrt {1+v^{2}/c^{2}}}$ with $m$ the rest mass and M the relativistic mass. Yes, real physicists never use this formula, yes, Einstein may have thought it was a bad idea to word it this way ... yet, none-the-less, this is a part of the undergrad curriculum, and some fraction of WP readers will expect to see this formula when they go searching for it. I can only suggest that it needs to be explained in some mind-numbing way, i.e. that its just a whiz-bang variant of SCZenz's formula 2. Or, I duuno, maybe we can just banish it from WP, based on principles. linas 04:48, 23 November 2005 (UTC)

I agree that this is what's taught, and what some will expect to see. I therefore think it has to be visible in this article too. However, certain users seem to be having a giant POV war over equivalent formulations of these equations, which is absolutely absurd. I think this needs to be rewritten by someone who understands this and doesn't have an axe to grind. I'll put it on my own list unless someone else wants to do it. -- SCZenz 04:58, 23 November 2005 (UTC)

Sorry if I made your heads hurt with the long rant, guys. I just saw a whole long list of edits across multiple articles, including massive text deletions, all of which seemed to be replacing rest mass with relativistic mass based on some internet source and sort of lost it. Thanks for your help, it looks fine now. -lethe ^talk 22:16, 25 November 2005 (UTC)

I didn't help, but I'm glad to hear that it's cleared up! Melchoir 00:23, 27 November 2005 (UTC)

In E=mc2, E is the total energy when the object is at rest in a reference frame. In that case, there is no kinetic energy. The only energy is thus the mass. That is why we often call the mass, the rest energy. At rest, means that v=o. At v=o, m=mo. In others words, at v=o, mass and relativistic mass have (or are?) the same quantity. So there is no sense to argue if it is mass OR relativistic mass, in this special case. When v is not zero but very small compared to c, the Newton's system is applied, i.e. we use E=½mv2, where the total energy is considered to be only kinetic, because the energy due to the mass, in this case, is insignifiant compare to the kinetic energy. 24.202.163.194 06:53, 24 December 2005 (UTC)

That sounds more or less what the article says now, I think. -- SCZenz 07:06, 24 December 2005 (UTC)

There is a super discussion of this whole issue in Physics Today, June 1989, p.31 by the Russian physicist Lev Okun. Based on his article, Einstein did not originate E=mc^2, Henri Poincare did in Arch Neerland, 5, 252 (1900). Einstein introduced ΔE0=Δmc^2 in his second Special Relativity paper Ann. Phys. (Leipzig) 18, 639 (1905). Okun, in his article, suggests modern physics should dispense with different forms of mass; there is only one mass and that does not vary with velocity. I think there is merit to Okun's suggestion; perhaps the article might be rewritten with an intention to correct many popular misconceptions about Einstein's famous equation and dispense with the mass variants propagated by Wolfgang Pauli's book "The Theory of Relativity".--Beanmf 23:11, 28 December 2005 (UTC)

All energy

Can all matter with mass be converted into massless energy? Could all mass in the universe (theoretically) be converted as such? Is it viable to consider a universe with just energy and no mass? --moxon 02:29, 18 December 2005 (UTC)

The question assumes mass and energy are different. Special relativity asserts that they are equivalent. Moving into general relativity, energy is the same as mass in its gravitational effects. Or consider that a photon, for example, has no rest mass, but is never at rest; so a photon always has mass and momentum. We can smack a photon into a target and measure the impact, the same as for a "massive" particle like an electron.

In future, it's better to ask such questions at the reference desk. --KSmrq^T 05:52, 18 December 2005 (UTC)

OK thanks --moxon 14:40, 18 December 2005 (UTC)

"Relativistic mass" or velocity-dependent mass is still prevalent in popular books and some research papers:

1. The God Particle: If the Universe Is the Answer, What Is the Question? by Leon Lederman, Nobel Laureate, Houghton Mifflin (1993) Pg 205

2. In search of the ultimate building blocks by Gerard't Hooft, Nobel Laureate, (Cambridge University Press, Cambridge, 1997), p.17

3. A Different Universe: Reinventing Physics from the Bottom Down by Robert Laughlin, Nobel Laureate, Basic Books, (2005) Pg 125

4. Perfectly Reasonable Deviations from the Beaten Track by Richard Feynman, Nobel Laureate, Basic Books, (2005) "It was successful, the necessary consequential phenomena (like mass changing with velocity) were ultimately observed experimentally." Pg 283.

5. Facts and Mysteries in Elementary Particle Physics by Martinus J.G. Veltman, Nobel Laureate, World Scientific Publishing Company (2003) Pg 137

6. Einstein's Legacy: The unity of space and time by Julian Schwinger, Nobel Laureate, Dover, (2002) Pg 84

7. The Road to Reality by Roger Penrose, Jonathan Cape, London (2004)

8. A Brief History of Time by Stephen Hawking (1988) "Because of the equivalence of energy and mass, the energy which an object has due to its motion will add to its mass. In other words, it will make it harder to increase its speed."

9. Einstein's Cosmos by Michio Kaku (2004) "For example, Einstein could show that the mass of an object increased the faster it moved. (Its mass would in fact become infinite if you hit the speed of light—which is impossible, which proves the unattainability of the speed of light.) This meant that the energy of motion was somehow being transformed into increasing the mass of the object. Thus, matter and energy are interchangeable!"

10. Understanding Relativity by Leo Sartori (1996) "Whether or not to speak of velocity-dependent mass is largely a matter of taste. Although it is currently unfashionable to do so, Einstein did and we shall as well."

11. John Roche (2005) "What is mass?" European Journal of Physics 26 (2), 225?42.

12. John W. Luetzelschwab (2003) "Apparatus to measure relativistic mass increase", American Journal of Physics, 71(9), 878.

13. Gerald Gabrielse (1995) "Relativistic mass increase at slow speeds", American Journal of Physics, 63(6), 568.

Inertia and energy content

In the Newton's system, when you applie a force F to an object of mass M you obtain an acceleration A, no matter if the object is in motion or not.

In the Einstein's system, that makes a difference, because if the object is in motion, he as then energy in form of kinetic energy in addition of his mass. This energy add inertia to the object, so that if you applie that same force F to the object, the acceleration resulting will be less then A. But this is not detectable at the speeds we are used to: that is why we don't bother whit it, and that's why we continue to use the Newton's laws at those speeds. It takes speeds near the speed of light, beginning about 2/3 of c, to be significantly detectable, and measurable.

This kinetic energy behave like mass, so we say that the mass of the object increase when the energy increase. It is not because of the mass itself increase, it is because energy, kinetic or any other form of energy, is also mass. That's why too, it would be better to use the term inertia instead of mass when we speak of the increase of mass due to the presence of energy, like in this following sentence: the inertia of the object increase when we add energy to the object. Also: mass and energy are inertia. I think it would be less confusing. 24.202.163.194 05:19, 26 December 2005 (UTC)

I don't like to say that mass increases; I prefer to say that

$F={\frac {dp}{dt}}$ and $p=\gamma mv$

and leave it at that. The article currently does both the "mass changes with velocity" and the "mass is a constant" bits. I'm not sure how the language with inertia would be clearer. But of course this is a wiki; please make changes that you think are appropriate and we'll see how they look. -- SCZenz 05:49, 26 December 2005 (UTC)

Special relativity insists that we cannot know if an object is in motion or not, and that the laws of physics should be stated in such a way that they are independent of the motion. The effects of forces are not as Newton predicts, but we can adopt viewpoints where the cause is time dilation, length contraction, or mass increase. As SCZenz says, it's more neutral and often better to lump all the possibilities together as momentum change. --KSmrq^T 06:30, 26 December 2005 (UTC)

3 systems of concepts

It seems to me that there is 3 systems of concepts related to mass:
1- Newtonian system where the mass stay always the same;
2- Relativistic mass system where the mass is changing with is content in energy, wich one give raise to the concept of rest mass;
3- The energy-momentum 4-vector system where the mass is an invariant as seen from various reference frames of inertia, but not necessary invariant in a collision, and in wich the mass is also called rest energy. But even in this last one, physicists still use the term rest mass, increasing the confusion.

I think that when we use the word mass, we have to know, or specifie if it is not clear, in wich system we are, and adjust all our wording to that system. It would be less confusing, the theory of relativity being enough confusing in itself without the neeed of adding to it.

What do you think of this view?
24.202.163.194 17:33, 27 December 2005 (UTC)

I think too much specificity could make some articles confusing. I also don't think that 1 and 3 are really very different; in fact, 1 is just a special case of 3. -- SCZenz 18:22, 27 December 2005 (UTC)

It is true that 3 use 1 for low velocity, but it is only the equations of 1 that they use. They don't adopt the conceptual system of 1 who is clearly different. For example, in 1 the inertia consist only in mass while in 3 inertia depend on mass and energy. For 3, a cloud of photons has inertia because it has kinetic energy, but not for 1 for which each photon having no mass, a photon or a cloud of photons could not be forced to change direction.

Specifity: a word, or expression, can change of meaning when we change the context in which it is use. So, when we utilise a word, or expression, we have to look in which context it is use. We can not use, for example, the expression 'relativistic mass' in 1, neither in 3: in both, this expression has absolutely no meaning at all!
That's why we have to use words and expressions that fits with the context, and a context that have the right words and expressions.
There is never too much specifity in science. 24.202.163.194 19:42, 27 December 2005 (UTC)

Yes, but there can be too much specificity in an encyclopedia. We want to avoid confusing people with details that aren't relevant, and we want to use common terms in the majority of sources when possible. Unfortunately, in this case, undergrad-level/popular sources (which we should be taking as a cue for our terminology) don't all agree. -- SCZenz 19:45, 27 December 2005 (UTC)

Origin of E=mc²

Contrary to the first sentence in the article, Einstein was not the originator of E=mc². Instead it was Henri Poincaré, 1900 (5 years before Einstein) in Arch. Neerland. 5, 252. "a pulse of light with energy E has mass m = E/c²", Lev Okun in Phyiscs Today, June 1989, p.33. Lev Okun makes a reasoned suggestion that we use a single (velocity-invariant) mass definition and substitute E0=mc² as the standard formula for today's use. The article could use a revision that clarified the source of confusion seen elsewhere in this discussion, including exactly how Einstein's relativity theories helped clarify issues concerning mass and energy.

Does the Okun article also point out that the equation only takes on its present importance in the broader context of special relativity, not just a special case of light pulses? Since special relativity had yet to be proposed in 1900, Poincaré could not have been saying what Einstein said in 1905. --KSmrq^T 05:26, 29 December 2005 (UTC)

Yes, that is precisely his point; thank you.--192.154.91.225 15:59, 29 December 2005 (UTC)

There are many misleading statements in Okun's paper:

1. Poincaŕe's equation of 1900,p = J/c^2, arose out of a comprehensive study of electromagnetism. p represents the effective mass density and J the energy density. Poincaŕe's relationship concerned the energy density of a fictitious fluid of radiation, and made no statement about matter itself.

2. Frank Wilczek, Nobel Laureate, explains that the way Einstein wrote it was M = e/c^2 and the original paper had a title that was a question, which was, "Does the inertia of a body depend on its energy content?" So right from the beginning Einstein was thinking about the question of could you explain mass in terms of energy. It turned out that the realization of that vision, the understanding of how not only a little bit of mass but most of the mass, 90 percent or 95 percent of the mass of matter as we know it, comes from energy. We build it up out of massless gluons and almost massless quarks, producing mass from pure energy. That's the deeper vision.

(Okun is wrong, and advocates that Einstein's true formula is "E0 = mc2")

I hate to say it, but...

I do not think that these rewrites have made this article easier to read. I think they confuse the issues, and in many cases are written in a non-encyclopedaic style. (One issue: the introduction is now too short, explaining nothing.) I suggest the users who are making extensive edits sit down, read over what's here now, and consider if it's what someone who looked the equation up in an encyclopedia would expect to find, or what an ordinary user with relatively minimal science education would benefit from reading. There's no reason that a topic like this can't be accessible. -- SCZenz 03:06, 30 December 2005 (UTC)

Each recent edit has degraded the article further. Clearly the writer has a limited command of the English language, as evidenced by linguistic errors. Nor is there clarity, nor is there helpful structure. I'd suggest SCZenz should revert to a better, earlier version and discuss further objections here if necessary. --KSmrq^T 05:22, 30 December 2005 (UTC)

I don't like simply reverting away good-faith edits. Instead, I'd like to incorporate the new material as much as possible, so I plan to do a more detailed merge/rewrite that incorporates both the old structure and the new information. That will take time, so I won't do it for some days anyway; if the editor who added the new material has a chance to consider some of my questions above in the meantime, that would be very helpful. -- SCZenz 06:10, 30 December 2005 (UTC)

I'm happy to see that others contributors had finally react to what I have edited. Here is what I have to say on this subject for the moment.
First, I am very aware of what you have said both of you, which does not mean that I agree on everything you said.
Second, wikipedia is a collective work, and a progressive one. If an information is wrong, please supressed it. If you find linguistic errors, please correct it. If a structure does not please you, please try an other one. Revertion is regression!
Third, even someone who goes at the university to study special relativity find it hard to understand. An ordinary reader cannot hope better just reading one article of wikipedia on the subject, even many articles. Secondly, to understand something, whatever it is, you must start by getting the right information first.
Fourth, what is a encyclopedaic style? Where can we find the criterias?
Thank you for your comments. I hope that others contributors will do the same.
24.202.163.194 16:53, 30 December 2005 (UTC)

Some stylistic ideas, off the top of my head, are:

1. It's too choppy. The intro should contain more information, including the equation's relevance to Einstein and Relativity as well as some of the idea of what it means. The major sections should all be at least one big paragraph and preferably more.
2. The most important information should be first. Since the background section actually explains the equation in understandable language, with concrete examples, it should be near the top. Instead it's moved downward.
3. The historical discussion, including the origin, should be in one coherent section. It should aknowledge that the equation is nearly universally associated with Einstein, but also include that some (one?) source(s) say that the equation really came from the other guy in 1900. This section should occur after the equation is explained.
4. Also, I'd like to say again that although physics is hard, I think this topic is understandable by people with a pretty basic education (say, High School level Chemistry, but no physics). Parts which aren't understandable in this manner should be near the bottom, and should still be as simple as possible. -- SCZenz 23:01, 30 December 2005 (UTC)

There is no (as far as I know) one place that say you should write like this or that, and I don't think there should be anyway as editors should have the fexilibity to adjust their apporach according to needs. However, like art, there is an ability for readers to read an article and then say whether they feel the language to be encyclopedic or not. One can have two articles that says the same thing in different ways, and readers would come out feeling completely differently about them. On the other hand, the ideas listed above by SCZene is a good guide to which all editors (including me) could use to base their writings on.

I disagree with you on the difficulty of speical relativity, but then I would seeing I've been a Mathematical Physics student. But that is actually neither here or there as SCZene point out, this topic (E=mc^2) should be (is) understandable by most people. Having made the last two point, I have to regretably say that I find the article in its current state confusing and hard to understand, even given someone that otherwise understand the concepts present in the article.

That is not to diminish the great help you're providing in improving this article. If it weren't for people like you, this article would not be able to improve. Like you said, we can always improve the writing once we get the right information in. -- KTC 01:09, 31 December 2005 (UTC)

There's also Wikipedia:Manual of style, but that's rather long. -- SCZenz

Thank you for the information. I just finish to read it all, ...almost. 24.202.163.194 03:46, 31 December 2005 (UTC)

It is never a good idea to revert good-faith edits without comment, nor did I suggest doing so. Please, do discuss efforts and intentions here on the talk page.

New editors of physics-related articles are encouraged to visit Wikipedia:WikiProject Physics and its talk page. All editors will want to read the Wikipedia:Manual of Style, with physics editors taking particular interest in Wikipedia:Manual of Style (mathematics) and Help:Formula.

This is an important article in need of substantial improvement. Chances are it will take extended collaboration and/or one really good editor to bring it to featured article quality.

Einstein himself wrote clearly and simply for the general public to explain relativity, and it might be helpful to read some of those efforts, such as Relativity : the Special and General Theory. Also consult the numerous links at "Albert Einstein" and "special relativity".

The opening sentences of an article are often the most challenging to write well. One helpful strategy is to first concentrate effort on the body of the article, then come back and see how best to give a brief overview and introduction for the general public.

Also, do not hesitate to ask for technical or editorial assistance at the village pump. --KSmrq^T 06:33, 1 January 2006 (UTC)

Except for minor things, I agree with all what you say. (For Einstein, note that he was alone to write is text. If he had putted it on wikipedia first, I don't know what would have happened to that text!?) --Aïki 16:03, 13 January 2006 (UTC)

New structure: first try

If you don't like this new structure, we can always go back to the precedent one or try another new one. What do you decide? --24.202.163.194 22:52, 31 December 2005 (UTC)

Bomb again

This article seems to imply that this equation led to the atomic bomb being developed. I don't think this is true at all -- the developments in nuclear fission were not as a result of pursuing this equation at all. It is true that the equation is often associated with the bomb (in a misleading way, usually), but that doesn't actually mean it was important in the bomb's physical development. --Fastfission 03:39, 8 January 2006 (UTC)

Agree, the equation has totally no practical implication or practical direction to the society till nowadays. Neither airplane nor A-Bomb is created becasue of the existence of the equation. May God show us the usefulness of the equation in our following days! armor 08:47, 11 January 2006 (UTC)

Apology for the word «body» in the Einstein statement

I edited this, thinking I was signed in. I changed the first sentence of "Background and consequences" to give our possesive pronoun an antecedent, at approximately 1951EST.--Josh Rocchio 00:53, 11 January 2006 (UTC)

To Josh Rocchio and 69.143.249.21 about the word «body» in the Einstein statement. I have verified with the original text in the links at the bottom of the article, and you are right. I was wrong. Thus, it is to me to apologise! So, I do apologise. Thank you to have indicated me this error.
--Aïki 01:45, 11 January 2006 (UTC)

No prob, it was the grammar that irked me more than the translation. Can't well have an its without a something.----Josh Rocchio 04:42, 11 January 2006 (UTC)69.143.249.21 04:41, 11 January 2006 (UTC)

Ack...both 69.yadda.yaddas were me.--Josh Rocchio 04:42, 11 January 2006 (UTC)

The first sentence in the first section is physically wrong

A body at rest in a reference frame, according to Newtonian mechanics, usually has some form of Potential Energy. You cannot say it has no energy at all. --atou 01:01, 13 January 2006 (UTC)

You are completely free to define the 0 of potential energy so can chose any body to have no energy. 137.138.46.155 07:54, 23 August 2006 (UTC)

Can somebody rewrite this article?

Going through this article I found pretty many errors in physics. Such as confusion in energy, rest mass, relativistic mass, and some convoluted sentences. I think this article need not to be a special relativity thesis, but at least this should be clear and correct. --atou 01:40, 13 January 2006 (UTC)

Yes, I've stopped paying attention since I rewrote it last. The rate at which people make helpful contributions that screw things up is mind-boggling; I just noticed a section which used to be correct, with words added to the equations so that it was now wrong. I'll rewrite the whole thing again soon. -- SCZenz 15:17, 13 January 2006 (UTC)

Equation not original to Einstein?

In his book How the World Was One: Beyond the Global Village, Arthur C. Clarke mentions that the equation was known to the famous physicist Heaviside; that's older than the other alternate origin.

Most standard works agree with this article in attributing it to Einstein. But they don't deal with claims that it was older. We do need clarification.

--GwydionM 15:31, 15 January 2006 (UTC)

• The notion that energy and mass were in some way interchangeable appears in many papers in late-19th century physics, and was considered a major part of what would now be called searches for the "theory of everything" — attempts to treat all physical phenomena as aspects of the same thing. Some sort of deep connection between mass and energy is virtually implied by the first law of thermodynamics, for example. However the formulation which came out of special relativity was unique and far more specific than these earlier attempts. Kragh talks about this in some detail in the first chapter of his Quantum Generations book (1999). I think some more work on the history section would be worthwhile, both to give more ample discussion to the earlier idea of mass-energy equivalence, as well as a discussion about why Einstein's approach was eventually seen as specifically different. --Fastfission 04:29, 9 March 2006 (UTC)

For this one, I suggest looking into the Wikipedia article for Olinto de Pretto. As usual, popular science simplifies and romantisizes the process of scientific inquiry, and attributes a discovery purely to one guy as if he just came up with it out of the blue like some magic Buddha. --129.10.116.232 16:44, 22 May 2006 (UTC)

inertia perpendicular to motion

what I am missing in the article, is the difference in apparant inertia perpendicular to and parallel to the motion of a particle? I don't feel very confident to write about that without looking into the stuff again first... &#151; Xiutwel (talk) 20:19, 15 January 2006 (UTC)

How did he get the equation?

How did he get the equation? Could there be some explaination on that? I mean did he know the ratio of energy to mass was going to be the speed of light squared to begin with or was it E=mx; x being the unknown. Is it an coincidence that the mass to energy ratio is the speed of light squared or is there some speical reason? If c² is only a conversion factor from kilograms to joules, and a kilogram was originally defined as "the mass of one litre of pure water at a temperature of 3.98 degrees Celsius and standard atmospheric pressure" and then later defined as "precisely the mass of a particular standard mass created to approximate the original definition." and a joule is "the absolute minimum amount of energy required (on the surface of Earth) to lift a one kilogram object up by a height of 10 centimetres." then how did he get the equation? Since the measurement of both above are defined by humans before they had any knowledge of mass to energy relationship. Unless there is some scientific reason for joules and kg that I am not getting. Or...does the speed of light squared/the speed of light have something to do with the amount of energy used to move an given mass? If so, someone should explain it. Pseudoanonymous 02:22, 14 February 2006 (UTC)

The equation arose as a consequence of taking existing laws of physics and modifying them to work in a spacetime setting. In particular, all equations should be invariant under Lorentz transformations, which are the permissible changes between inertial frames. This familiar form of the equation is a limited version of the general rule. Maxwell's equations for electromagnetism, discovered years earlier, were already invariant and a helpful guide. In fact, Maxwell needed to add an electromagnetism term to his equations, just as Einstein later added and combined terms for the rest of classical physics. Gravity was still a problem in special relativity, one which was to force Einstein to curve spacetime for general relativity theory. Not only does energy act as inertial mass, it also acts as gravitational mass, as embodied in Einstein's stress-energy tensor. The upshot of all this is that Einstein could not choose some arbitrary formula and expect it to satisfy the conditions required to make a consistent theory, no more than an elementary school student could look at 3+x=5 and choose anything other than x=2. --KSmrq^T 04:16, 14 February 2006 (UTC)

I just entered the proof, and if you know a little bit of calculus, it's relatively easy to understand. --AmaanMitha

Wouldn't derivation be a better title? You don't really prove physics equations. Zarniwoot 22:54, 13 June 2006 (UTC)

It should be noted that the derivation is of my own work, and I really have no idea how Einstein figured it out. I remember seeing in some physics video that he figured it out with his little thought experiments, but as I can't remember the source on that, don't take my word for it. Again, the derivation may not be the same one Einstein used; I have no idea what he did, and the purpose of the derivation is so that somebody wondering why it is true can see why. Also, the derivation shows that (total energy) = (relativistic mass)c^2, and while that is true, a lot of people on this page say that Einstein meant his equation to be in terms of the change in internal energy and rest mass, not total energy and relativistic mass. Sorry for any confusion. --AmaanMitha 03:32, 14 June 2006 (UTC)

The modern viewpoint

It is unfortunate that at present this article presents an out of date and deprecated viewpoint in several places. This is a viewpoint inconsistent with better articles on Wikipedia. It is now accepted that the term mass should be reserved for rest mass, which is a 4-scalar, and there is no point at all in using the second term relativistic mass for total energy divided by c². There is an interesting study published last year [9] that shows how over the last few decades the viewpoint of most experts has filtered down into textbooks with 20 current textbooks never using the term mass for anything but rest mass, and only 6 (one of which is Physics for poets) using the term mass in other ways. It is interesting that Einstein himself appears to have only used the concept of relativistic mass early in his work, and later was an active member of the consensus that reserved the term mass for rest mass. It is perhaps surprising that it took over half a century for most textbooks to catch up, and a little longer for this article.

The article can be made better by being careful only to use the term mass for rest mass except when reporting how the term was used historically. E = mc² is still a perfectly good equation for describing the relationship between rest mass and the energy associated with rest mass, but E² = m²c⁴ + p²c² is a much more useful equation that relates directly to 4-vectors and 4-scalars. 80.0.184.11 01:34, 26 February 2006 (UTC)

I agree that relativistic mass is a completely unused concept in modern physics, but I think there is strong reason to keep the subsection, for historical reasons. In the first couple sections, I think it would actually be best to gloss over the difference, since these are the sections the "poets" would be interested in. As for using E² = m²c⁴ + p²c², that would require introducing the concept of momentum, as well as explaining why it is not simply equal to mv, but equal to $mv/{\sqrt {1-v^{2}}}$.

It would be interesting to see a 4-vector treatment of Special Rel, but this really isn't the page for it, since this is probably the one page on Relativity that would be most visited by people who have no interest in seeing that better treatment. Short and sweet is the way to go on this article, without leaving gaping holes. As the man said, "Everything should be made as simple as possible, but not one bit simpler."

The writing may not be the best (especially in "History and Consequences), but I think the concepts used are good. I'm rewriting a little to clear up a few points.--- MOBle 21:56, 26 February 2006 (UTC)

C² clarification requested

The speed of light "C" is a velocity, just as 9.81 meters per second is a velocity. This presumably means that "C²" is a rate of acceleration, just as 9.81 meters per second² is a rate of acceration (the approximate acceleration from earth's gravity at the earth's surface). I believe it would help to explain this in the main article. This also raises a question. If matter were physically accelerated at this rate, would it turn into energy? I realize matter is energy, but would the matter change form? I also realize that matter cannot travel at a velocity greater than the speed of light, but it could presumably be accelerated at this rate for a very short period of time. Kmorford 09:07, 26 March 2006 (UTC)

I think you've made a mistake. The units of velocity squared are meters squared per second squared in SI, while the units of acceleration are meters (to the first power) per second squared. They are not the same, not do I know of any simple relation. -lethe ^{talk +} 10:07, 26 March 2006 (UTC)

I see your point. So, if C² is not a rate of acceleration, is there any "real world" analogy to explain what C² is, or is it just a very large number (conversion factor) which makes the equation ballance? Could it be considerd an accelerating rate of acceleration? If so, would matter change form if subjected to that accelerating rate of acceleration? Kmorford 17:55, 26 March 2006 (UTC)

Forget acceleration. Any unit of acceleration must have dimensions of distance over squared time. The way c (note: always written lowercase) works in special relativity is as a factor putting distance and time on a common footing. That is, imagine that we had only measured things in a two-dimensional world, using meters as our unit of distance for x and y displacements. Then, separately, we began measuring vertical displacements using kilometers. Under these circumstances we would have a problem computing three-dimensional total displacement, which ordinarily would be a simple application of the Pythagoras theorem:

$d^{2}=x^{2}+y^{2}+z^{2}.\,\!$

We must correct for z being measured in kilometers:

$d^{2}{\mbox{(in meters, squared)}}=x^{2}{\mbox{(in meters, squared)}}+y^{2}{\mbox{(in meters, squared)}}+{\frac {z^{2}{\mbox{(in kilometers, squared)}}}{1000^{2}}}.$

Just so, to measure displacements in space-time we must use c to adjust time units. Advanced physics usually takes the obvious step of measuring time in "natural" units, comparable to the distance units and the mass units, so that c disappears from the equations. (There is no need to write a factor of "1".) --KSmrq^T 18:52, 26 March 2006 (UTC)

Makes sense. Thanks for the clarification. Kmorford 22:38, 26 March 2006 (UTC)

Non-gravitational forces and spacetime

The article says:

It is not a coincidence that mass bends spacetime, while the charges of the other three fundamental forces do not.

I keep on staring at this and not understanding it. If there exist two electric charges at a certain distance, the electric field contains energy. And by E=mc², the system containing the electric field will therefore have mass proportional to the energy in the electric field (in addition to the non-electric-related mass of the particles involved). And because all mass is the same, that mass will bend spacetime. So the sentence above is not correct - an electric field will bend spacetime - and so, presumably, will the other forces. Just slightly.

What did I miss? --Alvestrand 20:04, 27 March 2006 (UTC)

You're not wrong. Charge creates field, and field contains mass/energy, and that bends spacetime. But you're supposed to think of the mass/energy as the source of curvature, and the fact that the mass/energy has some charge as its own source is incidental. Maybe this could be worded a little better. -lethe ^{talk +} 20:24, 27 March 2006 (UTC)

Association with relativity

This equauation is undeniably associated with relativity, although whether it's derived depends on your pedagogical approach to the subject. In any case, relativity should certainly be mentioned in the introduction somehow; what should we say? -- SCZenz 17:42, 9 April 2006 (UTC)

That is a good point (alough I think the sentence was factual correct). Maybe something like: The formula was first published in a slightly different formulation by Albert Einstein in 1905 in one of his famous articles. He derived it as a consequence of the special theory of relativity which he had proposed the same year.? Zarniwoot 21:41, 22 April 2006 (UTC)

Just thinking...

Using the equation, if you turn 1kg of a substance into pure energy at 100% efficiency, you will create 89,875,517,873,681,764 joules of energy. This amount of energy is the amount required to lift that same 1 kilogram of a substance on earth to a height of 9,161,622,617,092,942 kilometres. Now you cannot do that on earth, as this distance is the same distance that the would travel if it had gone to the moon and back 11,916,689,798 times, and then travelled a further 251,754 kilometres, but it is still a long way.

Derivation

Surely I'm not the only one who objects to the big ugly "Derivation" section plunked down in the middle of the article, with no formatting, no discussion, and no concern for the structure of the article. Even a physics text would do better, and this article has a wider audience than that. Nevertheless, I have only commented out the section, so that anyone who wants to spin off a separate article can cut and paste, or so that anyone who can see how to do something useful with it has the (very) raw material to mold. --KSmrq^T 00:00, 14 June 2006 (UTC)

I sort of agree. How about sticking it in an appendix at the end, the way many journals do with ugly math derivations? Sbharris 02:12, 14 June 2006 (UTC)

Well, don't bite the newbies. It would perhaps fit better into a new article and it needs a reference. Zarniwoot 02:44, 14 June 2006 (UTC)

I am a new member of wikipedia, even though I've been using it as a reference for ages. There is no reference, as I derived the equation myself when I couldn't find it online. After figuring it out, I decided that it probably should be up somewhere in case somebody else wanted to see it, and since wikipedia attracts such a large crowd, I figured it was probably the best place. If you would like me to create a new article, I don't mind. I don't really know what you mean by molding the material into something useful, as the purpose of the derivation is to be useful in and of itself. I put it here because I thought it would be too small of an article otherwise (as I doubt very many people would look up the derivation, and it just being a smaller part of the article made more sense to me). Personally, I think the appendix idea is pretty good. --AmaanMitha 03:01, 14 June 2006 (UTC)

And another quick question for anybody who knows his/her math...I'm a rising junior in high school, so as I haven't actually taken calculus yet (but I have studied it), I wasn't sure if I could use values of v instead of m as my limits since the derivation I put up here has dp in terms of dm, not dv. I solved for dp in terms of dv also, but the derivation that proceeds from that point is longer and more complex, so I just put up the shorter one. If I was wrong about that, then I can put up the longer proof if it is agreeable with everybody else. --AmaanMitha 03:10, 14 June 2006 (UTC)

To answer your calculus question, when you do a definite integration (one with limits), if you want the answer in terms of one "nice" formula in terms of one variable, you need to have the same variable as integral limits, as appears in the argument (the integrand). To do that, you can substitute variables in either the integrand, or the limits (as you did at first). However, if you never intend to express one in terms of the other, you can just ignore the whole problem, so long as you remember what your associated limits are for the equation you get. Thus, your integral of dm is simply Δm, but don't forget that one m in that difference is the m at "v = V" and the other m is the m at v=0. In your proof, you don't really have to give those explicitly in terms of v, so you can not bother with the change of variables

The derivation is a bit circular. It assumes the relativistic mass formula, but this formula is usually derived from or along with E = mc². Therefore, all we really have here is a derivation that E=mc² is compatible with m = m₀/√1–v²/c², which should come as no surprise, but is also not a derivation. I favor removal. -lethe ^{talk +} 03:17, 14 June 2006 (UTC)

Did the transformation for mass/relativistic mass come about from $E=mc^{2}$? Because I was under the impression that it was one of Lorentz's transformations, which would have come before Einstein's paper. Not sure, so a clarification would be nice. --AmaanMitha 04:13, 14 June 2006 (UTC)

Actually, you're right. You can arrive at that formula from the Lorentz transformation of the momentum four vector. -lethe ^{talk +} 13:50, 14 June 2006 (UTC)

I have replaced the derivation with a more careful but also more sophisticated one. I have long thought that this article needs such a derivation, but now we have to decide what derivation we want (if any at all. it is rather long, do others agree that we should have one?). My derivation is more sophisticated, making use of Minkowski vectors, but I like the idea of a derivation which uses only high school calculus, so that might be preferable. But I think Amaan's derivation would have to be cleaned up if we were going to keep it. One of the problems with Amaan's derivation is that I seem to think that F = dp/dt only holds in the rest frame. I'm looking for feedback from others. -lethe ^{talk +} 13:58, 14 June 2006 (UTC)

I have no idea what the new derivation means, so I'm not going to comment on it. And, if you use relativistic momentum instead of rest momentum, wouldn't F = dp/dt still be true in any reference frame? And if your p and t are relativistic p and t, then shouldn't the gammas multiply to one and leave you with the same amount of force as if it were v<<c? --AmaanMitha 22:16, 14 June 2006 (UTC)

The problem is that force transforms weird under Lorentz transformations. So in an arbitrary reference frame, you get

$\mathbf {F} +{\frac {\gamma ^{2}}{\gamma +1}}(\mathbf {F} \cdot \mathbf {v} )={\frac {d\mathbf {p} }{d\tau }}$

which only reduces to F = dp/dτ in the rest frame. Of course if you use the Minkowski vectors, then the equation holds in all frames, but then F isn't really force any more. For your second point, are you suggesting that

${\frac {d\mathbf {p} }{dt}}={\frac {d(m\gamma c)}{d(\gamma \tau )}}={\frac {\gamma }{\gamma }}{\frac {d(mc)}{d\tau }}?$

Because that is not correct. If that's not what you meant, then please clarify. -lethe ^{talk +} 22:48, 14 June 2006 (UTC)

That's not what I was thinking (I know you can't pull the gammas out), but I tried out what I was thinking and I've realized that it was wrong anyways. Also, I didn't know how force transformed, and I don't really understand what a Minkowski vector is. --AmaanMitha 03:57, 15 June 2006 (UTC)

Spatial vectors include three spatial components. So a spatial position vector looks like (x,y,z), a spatial velocity vector looks like (v_x,v_y,v_z). Spatial vectors are well-behaved under rotations. Apply a rotation, you get a new spatial vector which looks about the same, except for some sines and cosines thrown in there to tell you how big your rotation was. Spatial vectors don't behave very nicely under Lorentz transformations. To figure out what happens to your (x,y,z) coordinates under a Lorentz transformation, you also have to know the the time coordinate of the boost transformation. Thus the right kind of vector includes also a time coordinate and looks like (t,x,y,z). This is the Minkowski vector, also called a four vector since it has four components. There is a Minkowski vector for velocity which includes c as its time component. The time component of force is power, the time component of momentum is energy, the time component of electromagnetic potential is voltage, etc etc. Lorentz transformations always intermix spatial parts with time parts, which you will see if you figure out how the Lorentz transformation gives you $m=m_{0}/{\sqrt {1-v^{2}/c^{2}}}$. The Minkowski vector is just the collection of these two parts. -lethe ^{talk +} 08:33, 15 June 2006 (UTC)

If you guys do want to put in a "high school calculus" version of the proof, I have two. Both use the equation W = int_0^v v dp. One of them solves for dp in terms of dm (that's the shorter one and the one I originally put up), the other solves for dp in terms of dv (which is longer and more complex, but I think more accurate). I also found a mistake in my derivation; it should only go down to W=mc^2-m_0c^2. The rest of the derivation was flawed, only because I showed that E_0=m_0c^2 the wrong way, and I'm not sure how to prove that...I think that's what Einstein actually showed, so if I just assume it, then my derivation would only continue off of Einstein's and show that the formula holds true in relativistic situations. --AmaanMitha 22:25, 14 June 2006 (UTC)

Well, please post them, and let's have a look. Is one of the two derivations the one you've already put in the article? Like I said, I have some problems with that derivation, but perhaps they can be addressed. -lethe ^{talk +} 22:48, 14 June 2006 (UTC)

One of the proofs was the one I put up; that's the shorter one in which I find dp in terms of dm rather than dv. I just went back to an older copy of the page and lifted the code out and I'm placing it here, although I am making a couple of changes based on stuff I have realized since I put my derivation up.

In this derivation $W$ represents Work (or Kinetic Energy), $m$ is relativistic mass (meaning it is a function of v, which is why I used two measures of v as my limits for the integrals), $m_{0}$ is rest mass, $c$ is the speed of light in a vacuum (Einstein's upper limit for velocity in his Special Theory of Relativity), $v$ is the velocity of the object, $E_{0}$ is rest energy, and $E$ is total energy. The purpose of this derivation is to show that $E=mc^{2}$ is true whether E and m are at relativistic velocities or v<<c.

We begin with the formula that yields the work that a force F produces while accelerating any mass on a infinitesimal distance ds.

$W=\int _{0}^{v}F\ ds$

We also know

$F={\frac {d}{dt}}mv$

Therefore,

$W=\int _{0}^{v}{\frac {d}{dt}}mv\ ds$

$W=\int _{0}^{v}{\frac {ds}{dt}}\ d(mv)$

$W=\int _{0}^{v}v\ d(mv)$

This much is common to both derivations; it only provides a different method of calculating work.

From here, we begin the shorter derivation.

We know from the product rule that

$\ d(mv)=v(dm)+m(dv)$

$d(mv)=(v+m{\frac {dv}{dm}})dm$

Now we must find ${\frac {dv}{dm}}$

According to the Lorentz transformation for mass ($m$ is relativistic mass, $m_{0}$ is rest mass),

$m={\frac {m_{0}}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}$

where $c$ is the speed of light in a vacuum (Einstein's upper limit on velocity in his Special Theory of Relativity

${\frac {dm}{dv}}={\frac {m_{0}v}{c^{2}({\sqrt {1-{\frac {v^{2}}{c^{2}}}}})^{3}}}$

${\frac {dm}{dv}}=({\frac {m_{0}}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}})({\frac {\frac {v}{c^{2}}}{1-{\frac {v^{2}}{c^{2}}}}})$

${\frac {dm}{dv}}=m{\frac {\frac {v}{c^{2}}}{{\frac {c^{2}}{c^{2}}}-{\frac {v^{2}}{c^{2}}}}}$

${\frac {dm}{dv}}=m{\frac {v}{c^{2}-v^{2}}}$

To find ${\frac {dv}{dm}}$ we will simply find the reciprocal of ${\frac {dm}{dv}}$

${\frac {dv}{dm}}={\frac {c^{2}-v^{2}}{mv}}$

$d(mv)=(v+m({\frac {c^{2}-v^{2}}{mv}})dm$

$d(mv)=(v+{\frac {c^{2}-v^{2}}{v}})dm$

$W=\int _{0}^{v}v\ d(mv)$

$W=\int _{0}^{v}v(v+{\frac {c^{2}-v^{2}}{v}})\ dm$

$W=\int _{0}^{v}v^{2}+c^{2}-v^{2}\ dm$

$W=\int _{0}^{v}c^{2}\ dm$

$\Delta W=\Delta E=\Delta mc^{2}\,\!$

Therefore, the work done by the force acting on ds is the previous expression. ΔE is the change in energy due to kinetic energy, and Δm is the change in the object's relativistic mass associated with that energy.

This is very close to the form in which Einstein first introduced this equation in 1905, although his original derivation was concerned with rest mass change due to an increase in internal energy in a body, and not relativistic mass change due to a body's kinetic energy, which is the route used above, and which Einstein explored in 1907. Note that modern physicists often prefer not to use the relativistic mass term $m$ (see mass in special relativity) even though it allows Einstein's $E=mc^{2}$ equation to remain correct, even when $E$ is the total energy. In general, however, Einstein did not use $m$ as above and below, and his equation when used with ordinary mass $m_{0}$ is not correct, except in the special case of objects at rest, where $E=E_{0}$, the rest energy.

However, continuing to use the notation of $m$ as relativistic mass:

$W=c^{2}\left(m-m_{0}\right)$

$\,\!W=mc^{2}-m_{0}c^{2}$

At rest, $\,\!E_{0}=m_{0}c^{2}$, which is what Einstein showed.

$E=W+E_{0}\,\!$

$\,\!E=mc^{2}-m_{0}c^{2}+m_{0}c^{2}$

$E=mc^{2}\,\!$

Therefore, $E=mc^{2}$ holds true as Einstein said for an object at rest (when E is rest energy and m is rest mass), or as shown through this derivation, for an object moving at relativistic velocities (when E is relativistic energy and m is relativistic mass).

I'll try to get the longer one up later, might not be able to get it up until the weekend. --AmaanMitha 04:20, 15 June 2006 (UTC)

For AmaanMitha: One point about Wikipedia formatting style that you seem to have overlooked is that equations should be indented. Thus we use

---

The roots of the quadratic equation

$ax^{2}+bx+c=0\,\!$

are the two values

$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}},\,\!$

assuming a is not zero.

---

not

---

The roots of the quadratic equation

$ax^{2}+bx+c=0\,\!$

are the two values

$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}},\,\!$

assuming a is not zero.

---

For exciting bedtime reading, try Help:Formula and WP:MSM. --KSmrq^T 06:09, 15 June 2006 (UTC)

Ok, sorry about the mistake. I'll be sure to indent my equations next time. --AmaanMitha 17:35, 15 June 2006 (UTC)

A derivation of Δm = E/c^2 without calculus

So long as we’re doing a high school algebra presentation of E= mc^2, how about one that doesn’t need either calculus or anything else but a couple of conservation laws and one fact from Maxwell’s theory of light? And doesn't use force?

We should note that Einstein’s original presentation of this equation in 1905 makes use of the Lorentz transformation, plus his own use of it to transform the energy of a light beam from one reference frame to another. By means of examining the energy of a moving object emitting light beams in opposite directions to balance momentum, he was able to come up with the idea that the inertia of body which emits light of energy E, varies as if its m had decreased by E/c^2. But his derivation is tricky and opaque. The math is good, but you don’t understand much more of the guts of why you get the result you do after reading it, than you did in the first place.

Let’s see if we can do better. Paricularly, we'd like to avoid kinematics and force integrals.

Consider a perfectly uniform cylindrical spaceship floating in space, with length L. The ship is so perfectly built that the center of mass of the ship is at the exact center of it, in all axes. If we make the ship long and skinny so we can neglect all but the Y axis, there’s equal mass on either side of the Y axis. We pick this line to be our inertial reference frame, in which the ship is at rest. We choose our coordinate center to be at the center of the ship, with the Y axis cutting our semi- one dimensional ship in half.

Now, Huey and Louie are twins of identical mass, situated at each end of the ship, and each armed with a wad of chewing gum. Still the center of mass doesn’t change, and the ship is at rest. Now Huey throws his gum at Louie. While the gum is in flight (let us say to the right), the whole ship moves to the left, in reaction, due to conservation of momentum. If the gum with mass m has velocity V, then the ship with mass M moves in the other direction with much smaller velocity v (we assume the gum mass is so small that M doesn’t change enough to worry about when m is lost from it). And mV = Mv. The gum is in flight for t = L/V, and before it sticks, the ship moves for a distance D = v*t = L (v/V). Since V is much larger than v, this isn’t far. Once the gum sticks, the ship stops.

Now for the interesting part. The center of mass doesn’t change in all of this, nor does it move off our reference Y line. HOWEVER, the Y line no longer perfectly bisects the ship, because the extra gum stuck at the right end of the ship now makes that half of the ship more massive, so the amounts of mass on both sides of the Y axis remain the same (the important part) even though the ship is unsymmetrical insofar as our original axis. What happens is that the mass of ship which crosses over the Y line to the left, while the gum is in flight to the right, exactly equals the mass of the gum which now adds to mass on the right. And there it stops when the gum stops.

And further, this is true no matter how fast or slowly you throw the gum! You can’t trick this. If you throw the gum faster, the ship moves faster, but for a shorter time, and less of it crosses over. In the end, no matter how you get the gum from one end of the ship to the other, fast or slow, the mass of the ship that crosses the line before the ship stops, equals the mass of the gum that crosses the line before IT stops.

To prove it, let us examine the distance D the ship moves before it stops. How much ship does this represent? We note that the mass of ship which crosses the line, ΔM, is intuitively = (D/L)M = D(M/L). But now D, the distance the ship moves, is t*v = L (v/V). So ΔM = D(M/L) = L (v/V)(M/L) = M (v/V) = (Mv)/V. We've eliminated L. But since we know from conservation of momentum that Mv = mV, we can now also eliminate two more variables, so ΔM = mV/V = m. So ΔM, the amount of ship crossing over the line before it stops, is equal to the mass of the gum, m, as was to be proved.

Now for the photon! Instead of gum, Louie fires a photon at Huey. If we can find out how much mass of the ship cross the center of mass line, due to conservation of momentum, we can find the “equivalent” mass which is missing from one end of the ship, and now stuck on the other, as a result of this photon translocation. This is not quite the mass of the photon per se, but it’s surely the mass that disappears from one end of the ship when the photon leaves it, and reappears on the other end when the photon is absorbed! It's the mass associated with translocation of the energy of the photon. You pick your viewpoint. But the mass does move, because the ship moves (the photon has momentum, so ship must move), yet the center of mass of the whole shebang can’t move. And again, the ship mass which moves over the line, tells us quantitatively how much mass crossed over, in the other direction due to the photon. Let's find it quantitatively, and see how much mass that photon carried over!

The proof is exactly the same as the gum example, up to the point that ΔM = M (v/V), except that at this point now we now obviously want to replace gum V with the photon c, so ΔM = M (v/c). The conservation of momentum equation we need now only requires one new fact, which is that Mv = E/c, since E/c replaces the gum mV as the photon momentum (we need that fact from Maxwell). Putting E/c in for Mv, we get:

ΔM = M(v/c) = Mv/c = E/c^2.

Since this is the amount of rocket that cross the line of center of mass before it stops (one side slightly longer from the line than the other), this new mass which has transfered over the line is now how much more one half of the rocket would weigh than the other, if the photon had transferred no mass from one side to the other, in compensation. Of course, since both halves of the rocket on either side of the line have the same mass, ΔM = E/c^2 is the mass which has been transferred by the photon.

It follows that if we open the end of the rocket before the photon hits, and just let the photon leave before it sticks and transfers its mass, THEN the whole rocket doesn’t get a transfer of E/c^2 from once side to the other, but actually gets less massive by ΔM = E/c^2 as a result of losing the photon with energy E. And there we are. That's the mass associated with the energy of the photon, when considered WRT to the system of the rocket. Sbharris 07:29, 16 June 2006 (UTC)

When I started out reading this, I was highly skeptical, because I don't think you can get the result without Newton's second law. However, I see that what you've derived is not the full velocity dependent equation E = mγc² for massive bodies, but rather just the equivalent rest mass for a photon E = mc². Even photons need to have some information in place of Newton, but it's hidden in your explanation E = pc for EM radiation follows from the dynamics set out in Maxwell's equations. I guess Maxwell's equations are more sophisticated than Newton's laws, but perhaps the reader is also more willing to take them for granted, for that very reason. It's a little dishonest to say you're going to derive the equation with just high school algebra and then invoke Maxwell's equations, but I suppose it can't be helped. Somewhere along the line, a dynamical equation must be invoked. Anyway, I think it's a cute explanation. Does it matter if we use the relativistic momentum for the ship? -lethe ^{talk +} 11:21, 16 June 2006 (UTC)

I don’t think the 2nd law is really necessary to do anything, although perhaps it is needed to underlie a breezy assumption which I’ll make in a moment. If the mass ΔM changes by E/c^2 every time you fire out a photon, then you can fire them in pairs in opposite directions to keep you in the in the COM frame by the previous argument, until you’re down to nothing. Then Eo = Mo c^2. The whole energy Eo of all those photons (which we define as what E we get when we do this) is the rest energy equivalent (Eo) to that ship you just turned into photons, and it is Mo c^2. That’s Einstein’s equation. It applies not just to photons, but resting ships you turned into photons by the process described, and thus any rest mass.

If you want the relativistic kinematic form of kinetic energy KE in terms of total energy E = Eo + KE, THEN you need the axiom (here's the assumption) that ALL the KE in relativity appears or should appear as a ΔM c^2 associated with motion (rather than a ΔM from firing out photons). Perhaps that’s the fly in the ointment, but it seems reasonable (maybe there’s a way to prove it from conservation of momentum only, but I haven’t got the time now to see). In any case, that assumption gives you

KE = ΔM(motion) = (γ-1) Mo c^2

and that plus the definition of total E above, gives you E = γ Mo c^2.

Does the derivation I gave yesterday work with full relativistic momentum, you ask? Yes. It’s just a matter of sticking in all the gammas where you see length or mass quantities in the proof. These all cancel out for the length quantities, since lengths cancel in the proof. You’re left with the γ for the mass, so the conservation of momentum equation is now E/c = γMv and from the other part of the proof the only γ left is the one in front of the M so ΔM = γMv/c. Which still leaves you with ΔM = E/c^2. Sbharris 18:14, 16 June 2006 (UTC)

I don’t think the 2nd law is really necessary to do anything, although perhaps it is needed to underlie a breezy assumption which I’ll make in a moment. If the mass ΔM changes by E/c^2 every time you fire out a photon You have apparently already used a dynamical law for photons to arrive at this result. -lethe ^{talk +} 19:02, 16 June 2006 (UTC)

Well, yes, but as discussed, it was the one from Maxwell: p = E/c. So the point is tht Einstein could in theory have derived the subject of this article Eo = Mo*c^2 (at least up to there) without recourse to any relativistic theory at all. Or at least no NEW theory of his own-- as we all know by now, there's a lot of relativity implied in Maxwell if you know here to look for it [Einstein found it just this way] so using a Maxwell result is sneaking SR through the back door before it's fully called SR. Still, historically, in theory it could have been done like this.Sbharris 20:13, 16 June 2006 (UTC)

Right. I'm just thinking from an axiomatic point of view, it's nice to have a derivation from first principles, and that's what I've put in the article, using calculus and Minkowski space. It would be cool if there were a high school level derivation, and what you've done almost gets it, but it relies on Maxwell's theory as a first principle, which is not accessible to a high schooler. My preference for axiomatic treatments may be a bit idiosyncratic, and probably this kind of argument would be very nice for a lot of purposes. -lethe ^{talk +} 20:22, 16 June 2006 (UTC)

It's a great thought experiment and it works wonderfully, but I don't like assuming p=E/c just because Maxwell discovered that, and if that could also be proven , I think it would be a great addition. --AmaanMitha 21:20, 17 June 2006

Rest energy to kinetic??

The energy is only defined up to an additive constant, so it is conceivable that we could define the total energy of a free particle to be given simply by the kinetic energy T = mc²(γ – 1) which differs from E by a constant, which is afterall the case in nonrelativistic mechanics. To see that the rest energy must be included, the law of conservation of momentum (which will serve as the relativistic replacement for Newton's third law) must be invoked, which dictates that the quantity mγc² = mc² + T be conserved and allows that rest energy can be converted into kinetic energy and vice versa, a phenomenon that is observed in many experiments.

I don't see that the last bit follows as a result of the formula. All that formula means is that total energy of a moving body (given by this formula) is a sum of rest energy plus kinetic energy T. Still not proven by that alone is that rest energy can be converted to kinetic energy. Yes, if you take mγ to be relativistic MASS, then the increase in that mass due to kinetic energy is ΔM = T/c^2. But we've gone through all that-- that just uses a definition of mass which is the same as total energy (with a scale factor) so of course the difference in this quantity when due to nothing but T, will end up T/c^2. That's how the quantity was defined in the first place. Steve 22:27, 27 June 2006 (UTC)

I can't understand your complaint. The law of conservation of momentum does ensure that kinetic energy can be converted into rest energy (and vice versa). What's the problem? -lethe ^{talk +} 22:38, 27 June 2006 (UTC)

For multi-particle systems where p = 0, yes. But that's true basically because in multiparticle systems there's a minimum total system kinetic energy which is frame-independent (you can always increase it by switching inertial frames, but you can never go lower than a certain minimum value--- the frame where this least total KE occurs defines the center-of-mass frame). This residual system KE show up as system invariant mass, so it's already "rest mass" in the sense that the system COM is at rest, and so is the system, if bounded. This KE can indeed be "converted" freely to some kind of extra "rest" mass if you insist on measuring rest mass of a system as the sum of rest masses of particles in it. But that's a no-good way to measure system mass, so we're stuck. In essense I'm saying that I'm not impressed that two particles m headed for each other with KE = T end up sticking with total mass 2m+(T/c^2). Since that was the invariant mass of the system in the first place, and the mass if would have had if you'd let the particles enter a can and bounce around inside while you weighed it, without the particles ever sticking to anything (let alone each other). System mass never changes, whether you let the particles stick or not. So this "rest mass" which is supposedly "created" this way is entirely a figment of the imagination, resulting from the (bad) and pre-relativistic idea that extra "rest mass" only "counts" if it shows up as an extra quantity when rest mass of a system is gotten by summing from the inertial frames of all separate system masses, and then compared with that mass that is measured for the whole bound system at the end. But you can get that kind of "extra rest mass" from a can in which nothing is at rest. So it isn't REALLY rest mass. There really isn't any such thing as "rest mass" except for single leptons. Everything else is system mass, and it includes a lot of KE in all systems. As I've noted, "resting" hadrons seem to be mostly KE.

I still can't understand what your complaint is, but instead of standing still, I'll just respond to some points in your paragraph. The law of conservation of momentum holds in all frames, not just the COM frame. It has absolutely nothing to do with the fact that there is a minimum to the energy in the COM frame. Like I say, it's true in all frames. The conversion of kinetic energy into rest energy may seem meaningless if you only talk about the system invariant mass which includes both. But lots of people like to talk about rest masses of particles as well, in which case the ability to convert rest energy into rest mass is not a meaningless fact. -lethe ^{talk +} 00:28, 28 June 2006 (UTC)

Anyway, for single particles in a single reference frame, however, we're stuck with our old simple result that KE = (γ-1)mc^2. But we're stuck there, because it's now a moving system. But that's not an increase in rest anything. No results for rest mass are possible, because nothing is at rest, and if you catch up with that moving particle to see what it's rest mass is, it's still mc^2. If one switches inertial frames during an interaction, that's cheating. Any particle has more momentum outside its rest frame, but that increase in momentum is not an increase in "mass," unless "mass" is defined trivially, as relativistic-E/c^2. But as we've discussed, there are much better definitions of mass, namely invariant mass, which is a separately conserved quantity, and needs a name. Steve 00:08, 28 June 2006 (UTC)

It's probably true that we can't derive E = mγc² for a single noninteracting particle. We need some dynamics, some interactions. The law of conservation of momentum (and Newton's third law) become trivial for that system. So? -lethe ^{talk +} 00:28, 28 June 2006 (UTC)

Could you perhaps sum up what your complaint is, in a sentence or two, because I can't figure it out. Is it that the proof that E = mγc² because we should only talk about total invariant system mass for interacting particles, rather than individual rest masses? -lethe ^{talk +} 01:41, 28 June 2006 (UTC)

Well it looks pretty simple lethe, he's arguing that since rest mass is defined based on a system, that system will never have more kinetic energy than it every did, unless an outside force is acted on it. He's also suggesting that "rest mass" is not a proper term for a system of particles.

Rest mass is the proper term however, in the very general case of a system of particles (which we have to assume any "elementary particle" also is) the rest mass is the energy measured in the frame of the center of mass. Thus you're wrong about rest mass, but you're right about the fact that we can't convert rest mass into kinetic energy. That would amount to creating a reactionless drive. The law of conservation of momentum in fact ensures that this cannot happen. Half the system must spray in one direction, and the other half in the other - preserving the center of mass, and preserving the rest energy of the system. Fresheneesz 19:30, 29 June 2006 (UTC)

No, I have no problem with "rest mass" for a system of particles. We use it as a term all the time, since ordinary objects "at rest" are systems of moving particles, from hadrons to tin cans.

In such systems the mass doesn't change. You can decide for yourself that you have new particles in there made from kinetic energy, and that's interesting, but it doesn't change the system. If you fire two electrons into a tin can with enough energy to make a new proton and antiproton, then yes you've made two new particles out of kinetic energy. They don't last long. And they themselves are just smaller cans for their consituent quarks, ie, smaller tin cans for your energy. When they anihilate, the pions and electrons and gammas (finally) bounce around inside the can until finally you get your two electrons back, in a thermalized photon gas which has a mass (in your system) of exactly the kinetic energy you fired in on the electrons. So new particles have been created and destroyed in all kinds of ways, and now you're down to "heat." Made out of your kinetic energy. System "rest" mass doesn't change during all of this. And your particles temporarily made from kinetic energy as "rest mass" have now dissolved again back into to photons and your "new" particle rest mass is gone again. I suppose Lethe finds this more exciting than I do. Particles with "rest mass" pop into reality and then go back again, all the time. The vacuum seeths when them. What stays constant are system properties. Total energy, total charge, total momentum, to some extent some additional quantum numbers like lepton or baryon numbers. But those sum across particles and antiparticles.Steve 20:26, 29 June 2006 (UTC)

Explanation

Could someone please add an explanation of how the equation works, especially the bit about C^2? I understand the result of the equation, but I do not understand the exact math involved. For instance, why is the speed of light squared used to relate mass and energy, and what are the resulting units? It seems to me that if I multiply 1 pound by the speed of light in miles per second, squared, I get 34701131432.066 (lbs.)(mi.^2)/sec^2, which is not like any unit of energy I've ever seen. --scienceman 21:31, 10 July 2006 (UTC)

Assuming that you are using a "pound" as mass rather than weight (force), it may not be a common or useful unit of energy, but it is a valid unit of energy (mass × length² / time², much like the joule=kg×m²/s²). A very similar discussion occurred near the top of this talk page. -- Coneslayer 22:02, 10 July 2006 (UTC)

Yep. I personally vote for dram-furlongs^2/fortnight^2 as our basic energy unit. The drams to come from a dram shop not too many furlongs and fortnights away. SBHarris 22:28, 10 July 2006 (UTC)

Why not E=c²m ?

Conventionally, a linear relationship between two variables, x and y, is written y = ax (where a is the constant of proportionality), not y = xa. In the relation E = mc², the energy E and the mass m are the variables and c² is the constant of proportionality. So conventionally it should be written E = c²m. Why is it not written like that? Einstein is not to blame, because he wrote ΔL = m/c². Who was responsible for this unconventional formulation E = mc², and what can be said to his defence ? Bo Jacoby 13:22, 13 July 2006 (UTC)

Perhaps by analogy with kinetic energy, E=½mv². -- Coneslayer 15:04, 13 July 2006 (UTC)

Yeah and it looks nicer with the squared at the end, most simple equations containing squares are written like this.137.138.46.155 08:00, 23 August 2006 (UTC)

actually mathematically the variable with the superscript is put on the end because, yes, it is more visually pleasing. although y=ax, y=xa2 in this case.

Relativistic mass

Whole article needs rewriting to emphasize right from the start, instead of mentioning it halfway through, that the m is the relativistic mass, not a rest mass. A crank is going apeshit crazy over this at various photon articles. --Michael C. Price ^talk 18:24, 9 August 2006 (UTC)

It can be a rest mass. Then it gives you the rest energy. Which isn't very interesting for a particle, but IS interesting for a system at rest, since it includes all kinds of system internal energies (rest mass = relativistic mass for systems in the system "rest" frame = COM frame). SBHarris 07:52, 23 August 2006 (UTC)

why is it kilos?

Okay, how comes that this formula is using kilos, not any other scale of weight?

You measure the earth from pole to pole, then divide that with a large number. then you devide this number by 10 and get a decimeter. You then make a cube of decimeter sticks and fill it with water that is under normal preassure and is 4 degress C. This weight is now a kilo. And somehow this e=mc2 , wich seem kinda simple, is using this strange scale.

Am I just confuzing myself? —Preceding unsigned comment added by 217.209.215.117 (talk • contribs)

The formula is not tied to any particular units. The units you use for two of the quantities will determine the units of the third. See my longer explanation near the top of this page. -- Coneslayer 16:40, 17 August 2006 (UTC)

Because the speed of light is in metres per second and energy in Joules. All the SI units are derived to be internally consistent. There is no coincidence here, that is just the way the units are derived, the definition of Joules depends on metres, seconds and kilograms so it all works out nice. 137.138.46.155 08:04, 23 August 2006 (UTC)

Practical example

Annual consumer power usage in Hungary is about 216 Petajoules. This is equal to the energy equivalent of 2.5 kilos of matter based on this article. I'd also put such examples in there. Perhaps US ones.

Lede change

Proposed: -Ste|vertigo 17:33, 27 August 2006 (UTC)

In physics E=mc2 is an important and well-known equation, which defines a relationship between energy (E, in whatever form), relativistic mass (m), and (relativistic) time —represented by the square of the speed of light (the constant c) in a vacuum. Hence c² is the conversion factor required to formally convert from units of mass to units of energy, i.e. the energy per unit mass. In unit-specific terms, E (joules) = m (kilograms) multiplied by (299792458 m/s)2.

The speed of light is not a "time." Nor is the square of it. SBHarris 20:41, 30 August 2006 (UTC)

Nor does it say the speed of light is "a time." It says that the formula stated a relationship between energy, mass and time, where the speed of light is simply used as a measure thereof.continued...--Ste|vertigo 01:57, 31 August 2006 (UTC)

But that is completely wrong. The formula does NOT state a relationship between energy, mass and time. The speed of light is not being used as a measure of "time". The speed of light is a constant and does not vary. You can pick its value to be anything you like, including 1. In that case, the equation only notes the relationship between energy and mass: E = m. Time does not enter and is not a factor, or a variable in the equation. Period. SBHarris 17:53, 31 August 2006 (UTC)

This doesn't make a whole lot of sense, I agree. However, I'd like to note that it's perfectly reasonable to make big chances to the article without discussion; this is a Wiki, after all. Of course, in this case changing it back was the right thing to do. -- SCZenz 23:00, 30 August 2006 (UTC)

There are two schools of thought: be bold and taciturn (like calling a speed "relativistic time") without explanation, and be prepared to be reverted with no more explanation than YOU gave. Or argue your case in the change notes (or even better for something this drastic) on the TALK page. Either one is okay, I suppose. I've done it both ways myself. If you're lazy about justification, you can lose your work quickly to the reflexive reverter, so you don't really save the time you thought you did in not defending your idea. SBHarris 23:24, 30 August 2006 (UTC)

See WP:BOLD. Useful changes, no matter how large, do not require explanation in advance. Reflexive reversion is bad; however, evaluating an edit and reverting if it doesn't help the article is good. -- SCZenz 23:27, 30 August 2006 (UTC)

I agree that drastic changes should be discussed first, which is why I commented here a few days ago. Nobody responded, so I made the change on the article. -Ste|vertigo 01:57, 31 August 2006 (UTC)

cont.

FYI : In The Meaning of Relativity Einstein explained:

"The theory of relativity is often criticised for giving without justification, a central theoretical role to the propagation of light, in that it founds the concept of time upon the law of propagation of light. The situation, however, is somewhat as follows. In order to give physical significance to the concept of time, processes of some kind are required which enable relations to be established between different places. It is immaterial what kind of processes one chooses for such a definition of time. It is advantageous however for the theory to choose only those processes concerning which we know something certain. This holds for the propagation of light in vacuo in a higher degree than for any other process which could be considered, thanks to the investigations of Maxwell and H. A. Lorentz. "[10]

-Ste|vertigo 01:57, 31 August 2006 (UTC)

I, for one, really like the idea of bold edits. However, I think that when people are willing to discuss edits like these, the discussion should take place first. I find the wording very confusing. The passage Stevertigo refers to above seems to be talking about causality, and the relation of the units we use for space with the units we use for time (via c). Noting that clarity (as opposed to an oblique reference to deep facets of GR) is paramount in the intro, I don't see what's wrong with the following:

In physics E=mc² is an important and well-known equation, which states a directly proportional relationship between energy (E, in whatever form), relativistic mass (m), and the speed of light (the constant c) in a vacuum.

Additionally, I think the wording about the "basis for" GR isn't the best. It would be useful to note that the Newtonian notion of mass is best replaced by the notion of mass-energy. It might be reasonable to note that this mass-energy is what causes curvature of spacetime.

The equation was first published in a slightly different formulation by Albert Einstein in one of his ground-breaking articles of 1905. He derived it as a consequence of the special theory of relativity which he had proposed the same year, which defines "time" as relative to the observer. This revolutionary equation unifies the concepts of mass and energy.

Hence c² is the conversion factor required to formally convert from units of mass to units of energy, i.e. the energy per unit mass.

In unit-specific terms, E (joules) = m (kilograms) multiplied by (299792458 m/s)².

I feel strongly about the "time" issue. It seems like others agree, and I think it should be changed back to the older form. This is certainly the less intellectually adventurous route, but this article is probably one of the most prominent faces of Relativity in Wikipedia. As such, I think it should have a very clear and direct intro. MOBle 02:51, 31 August 2006 (UTC)

I wonder if those gleefully citing WP:BOLD have actually read it completely. An article like this is explicitly not the place to make big bold changes. Everyone and his dog thinks they know just the thing to make this article better. We can excuse the dog, but humans should know better, or be told. Don't believe me? Then go through the history of the article, and all the discussions on the talk page. Decide for yourself how many edits have been "improvements", and whether the huge number of drive-by shootings have achieved steady forward progress. Does that mean there is no room for improvement? Not at all. But it's arrogantly foolish for anyone to make major edits — especially to the intro — without prior consultation.--KSmrq^T 03:44, 31 August 2006 (UTC)

Gosh, not only do I know what WP:BOLD says in its entirety, I also happen to know a lot about the subject of this article. This article is not hot shit as it is; it needs work, and contributions are good in whatever form people want to make them. It is also not edited particularly frequently for a Wikipedia aritcle; a few changes every so often would not be a big deal. Finally, this is a wiki, and we do sometimes make changes by, well, making changes first and discussing them later. I know most of the edits make things worse, but bad edits are dealt with easily enough. -- SCZenz 06:24, 31 August 2006 (UTC)

Reply to Mobile

Mobile, Thank you for your detail. 1) You add that the article requires "a very clear and direct intro" because of its prominence, but fail to explain why the current one is less clear, other than that its more "intellectually adventurous". You wrote "I dont see whats wrong with" (the intro with without the mention of time)... and you "think it should be changed back to the older form." I may agree about the second section you quoted, but not about the first.

Professor Einstein, in the above referred section, is not talking about causality, as you have interpreted (and thus appear to have based your whole criticism). He was talking about the usage of $c^{2}$ as a measurement throughout the whole of his formulation of special (and thus also general) relativity. His explanation of this usage in his theory even smacks of a little ire at having to answer the question ("the situation, however, is somewhat as follows") just as he was probably a little disappointed at the fact that learned people had even years later failed to make the distinction between his usage of light and time. He said "it is immaterial what kind of processes one chooses for such a definition of time", which makes it fairly clear that time is the object, and that light is just the "immaterial" physical measurement thereof. He states (perhaps sarcastically?) that "it is advantageous" for a theory to be based on "what we know to be certain," which might translate to something like: 'theories ought to be based on (known) constants, dont you think?' And he concludes by stating the fact that "the propagation of light in vacuo" qualfies more than "any other [measurement]" (and it so remains).

I understand there is a difference between the eloquent and the mechanical description of a formula, and there are purists on both sides. I am neither. In this case, as in the case of numerous other important concepts, I would suggest that the accomplishment that the formula represents demands eloquence, and not merely a technical definition - particularly one which mistates the purpose of the relationship as one referring to light and not time. Regards, -Ste|vertigo 06:30, 31 August 2006 (UTC)

While I haven't read the rest of Einstein's statement, I still disagree with your interpretation of the text you've quoted. That isn't important, though. What matters is clarity of the leader in an important article like this. I have two points to make about the "time" thing:

1) The statement that "E=mc² is an important and well-known equation, which states a directly proportional relationship between energy (E, in whatever form), relativistic mass (m), and the speed of light (the constant c) in a vacuum" is very clearly true, and serves the very direct purpose of defining what E is, what m is, and what c is for someone who just knows the letters, and has no understanding of the formula's meaning. This is a crucial objective.

2) Your insertion of time points to an attempt to bring a deeper understanding of the equation to the article. This is good, in its place. However, it's not what my 13 year-old nephew will want to see when he wonders what $E=mc^{2}$ means. Seeing a fleeting reference to "time" there will just confuse him. Perhaps you could flesh out this idea, and put somewhere other than the very first sentence.

I think that I'll move your statement to later in the leader, as a compromise, and that'll be the last I do on this issue. Try to give the statement a little meaningful content, rather than just the nebulous and confusing suggestion it is now. MOBle 18:56, 31 August 2006 (UTC)

lede...

1. "Einstein's statement [on the usage of $c^{2}$] isn't important" - I disagree. I think what Einstein said related to $E=mc^{2}$ is important to an article titled "$E=mc^{2}$."
2. "What [your] 13 year-old nephew will want to see" is not really as clear to me. IMHO I think he would benefit from a description which actually gets to the heart of relativity - that time is not a constant, as was previously assumed. Perhaps thats why $c^{2}$ seems preferable, as it is necessary in the actual equasion. But, as Albert pointed out, that wasnt the point, and in that context, I think its fair to state what the point is, rather than just leave that to an interpretation section. Regards, -Ste|vertigo 20:52, 31 August 2006 (UTC)

1. I didn't say (or mean to say) that Einstein's statement was unimportant; I said that our disagreement on its meaning is unimportant. I do sincerely disagree with your reading of the passage. Having read a little more of the text from which you took that passage, I am even more convinced that Einstein was talking about causality -- light being the most quickly propagating "cause" known; therefore, its speed being of fundamental importance. Again, though, our difference of opinion is not important.
2. Certainly my nephew will benefit from any understanding he is able to gain on any subject. (I.e., education is a good thing.) If the article goes right over his head, and only serves to confuse him -- that's a different matter. Someone who has only gotten as far as the first sentence is probably in prime position to be confused. MOBle 21:18, 31 August 2006 (UTC)

1. I think you misread the section.
2. I think you underestimate your 13 year old nephew's powers of understanding. Time is a concept I have certainly thought about since childhood. I have no reason to assume others of young age are any less inquisitive. -Ste|vertigo 21:45, 31 August 2006 (UTC)

OK, I will agree with that for this article defining the formula terms first is best (that $c$ should be mentioned before time). But then my concern is that the conversion of mass to energy not supercede the tie to relativity, which itself has something to do with space and time, in which energy and mass are based.

The concept of light-speed is abstract to people's experience, isnt it? OTOH we all grow up developing some notion of energy, mass, and time. Thus its important IMHO to state upfront how 1) SR rests on "light" (EM) to shed light on the nature of time and space 2) hence these are fundamentally unified in $c$ (compare $c^{4}$) 3) hence mass and energy are fundamentally unified in $E=mc^{2}$ 4) hence GR. All of these can be done in a couple sentences! :) Regards, -Ste|vertigo 03:16, 1 September 2006 (UTC)

In physics E=mc2 is an important and well-known equation, which states an equivalence between energy (E) and relativistic mass (m), in direct proportion to the square of the speed of light in a vacuum (c2). The equation was first published (in a slightly different formulation) in 1905 by Albert Einstein, in what are known as his Annus Mirabilis ("Miracle Year") Papers. With these papers, he showed that a unified four-dimensional model of space and time ("spacetime") could accurately describe observable phenomena in a way that was consistent with Galileo's Principle of Relativity, but also accounted for the constant speed of light. His special theory of relativity ultimately showed that the traditional (Euclidean-Galilean) assumption that time and distance were absolutes was incorrect, and, as a consequence, mass and energy were different only in form. Thus c² is the conversion factor required to formally convert from units of mass to units of energy, i.e. the energy per unit mass. In unit-specific terms, E (joules) = m (kilograms) multiplied by (299792458 m/s)2.

I like this much better. I made a couple small changes, and one to the sentence about light. (Relativity didn't really account for E&M radiation.) I'm happy with this formulation. MOBle 17:28, 1 September 2006 (UTC)

Cool! -Ste|vertigo 23:34, 1 September 2006 (UTC)

E=MC3!!!

E=MC2 should be E=MC3 because if it is 2 then that means everything is in 2D, although space is in 3 dimentions. -David Knott

why does the squared make it mean that everything is in 2D? - 9/19/06

It doesn't. The statement was nonsense. -- SCZenz 02:22, 20 September 2006 (UTC)

Error in Low-Speed Approximation Section

The article misleadingly implies that the classical physics equation "E=1/2mv^2" was Isaac Newton's. This is incorrect. Gottfried Leibniz originally proposed the "vis-viva" equation E=mv^2 and it was later refined to the correct form of E=1/2mv^2 by Coriolis & Poncelet. Refer to the article on "Conservation of Energy" -JT

User:Thljcl's edits

User:Thljcl has made several edits to this article. Based on his questionable edits in kinetic energy, and his failure to discuss the matter when asked to, he appears to be a problem user. I do not, however, have the knowledge to comment on his edits here. Any thoughts? -- Meni Rosenfeld (talk) 09:56, 21 September 2006 (UTC)

Which One?

I'm not going to claim to know much on the subject; I'm merely able to spot discrepancies. This article says 'It is a little known piece of trivia that Einstein originally wrote the equation in the form Δm = L/c² (with an "L", instead of an "E", representing energy, the E being utilised elsewhere in the demonstration to represent energy too).'

The article Celeritas says 'In the nineteenth century, an upper-case V was commonly used to describe the speed of light. This was the notation used by Einstein in his 1905 papers, thus his most famous equation was originally written as m=L/V² (E having being used elsewhere for a different energy)'

Which of these is correct?--Jcvamp 23:32, 5 October 2006 (UTC)

Static vs. Expanding Universe

I was looking over some of the assumptions made when Einstein derived his original equations and I found an oddity. He assumed that the universe was static and that time was infinite in both directions, and from this created the concept of a rest mass with zero momentum, as seen in the formula F * p = 0 stating: In the particle's rest frame, the momentum is (mc,0) and so for the force four-vector to be orthogonal, its time component must be zero in the rest frame as well, so F = (0,F).

Since we've known the universe to be expanding since Hubble's discovery in 1929, and we've already observed a violation in cosmological isotropy from the WMAP observations of cosmic microwave background radiation, should the equation be rederived to reflect the change in our knowledge? It seems to me that there's an unaccounted for momentum vector that needs to be added into the equation.

I could be very wrong, but I'd like to know anyways.--Mad Morlock 05:58, 8 December 2006 (UTC)

This is an article about Special Relativity. In SR, it is assumed that spacetime is flat and static. Specifically, matter and energy moving around in it don't warp spacetime. This idea has to wait for General Relativity. The Universe can only expand when you use the full GR theory.

However, this is all still relevant because "locally" anything described by GR can be described by SR. That is, if you take just a small chunk of GR's spacetime (small lengths in space and time compared to the curvature of spacetime), you get something that looks exactly like a small chunk of SR's spacetime. It is in this small chunk that the usual Special Relativity results (like E=mc^2) apply. This may seem obvious and useless, but it's really pretty deep, and crucial to getting a consistent theory of General Relativity.

This is a good question. However, it's considered very bad form to put your own questions or your own ideas in an article; use the talk page first, then find reputable outside references. Remember: WP:NOR and WP:CITE are crucial to the quality of Wikipedia. --MOBle 17:57, 9 December 2006 (UTC)

Whats the problem with using original reasoning if you're using a system of logic to refute the original assumptions made in deriving the equation?

Or should this page be kept as a historical reference with a link attached to the page refering to a more uptodate theory that includes the expansion of spacetime? --Mad Morlock 20:47, 9 December 2006 (UTC)

The problem is that what seems logical to you may contain a flaw in the logic; or you may not take into account all of the facts (which I claim is the problem here); etc. The standard we use to decide on this site is Wikipedia policy. That policy prohibits original research (WP:NOR), and requires the use of relevant sources (WP:CITE). Now, in this case, my challenge to you would be to find a reputable, peer-reviewed source that says E=mc^2 is unreliable in an expanding Universe. By contrast, I could point to basically any modern treatment of General Relativity for a source saying that it is valid in curved spacetime (locally).

For a little more technical detail, a textbook on General Relativity will say something like the following:

In General Relativity, spacetime is made up of a Lorentzian manifold with three space dimensions and one time dimension.

That, plus a little more gobbledygook to explain those terms, is physicist-speak for "locally, Special Relativity is valid". The equation is validly derived in Special Relativity, so it is valid locally in General Relativity.

This article is about the equation useful for things like particle interactions (nuclear reactions, for example). These things are local in the sense that size scales of nuclear reactions are huge compared to size scales on which the Universe is curved. The exception to that is at times and places where we have no accepted theory of Physics: the Big Bang; singularities in black holes; maybe a few other places we don't understand yet.

There is no new page that needs to be created. This page is not just an "historical reference"; it is as up-to-date as I expect it will be possible to get for many years to come. I'd be happy to be proven wrong, but I'll have to see it in a peer-reviewed journal before I believe it. --MOBle 07:19, 10 December 2006 (UTC)