Why did we need relativity to derive $E=mc^2$?

Okay, so the way I understand one of the "derivations" of $E=mc^2$ is roughly as follows:

• We observe a light bulb floating in space. It appears motionless. It gives off a brief flash of light. We measure the frequency of this light, and conclude that the light bulb lost some amount of energy $E_1$.
• Next, we imagine flying by this light bulb at 90% the speed of light (or something similar), and again observe the flash. Due to the Relativistic Doppler Effect, the frequency of the light is decreased, and so we observe that the bulb only lost a smaller amount of energy $E_2$.
• Since the total amount of energy lost must be the same in both reference frames, we conclude that the bulb must have lost kinetic energy in the moving reference frame. Kinetic energy is $\frac{1}{2}mv^2$. Since $v$ did not decrease, $m$ must have decreased. Thus, when bodies lose energy, they also lose mass. QED.

So obviously it's a little more complicated than that, but my question is: why do we even need special relativity? Can't we replace "relativistic doppler effect" with "classical doppler effect" and still get a similar result? The exact relationship between $E$ and $m$ will change, but shouldn't the insight that a relationship exists be accessible even with only classical physics?

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Please formulate the argument properly: there are two flashes of light in opposite directions. The energy lost is not the same in the two frames if you are moving at 90% of the speed of light. The answer is that in the Galilean relativity the argument wouldn't work because the light would have no momentum, or else there would be an ether, and the ether would pick out a rest frame. –  Ron Maimon Jul 4 '12 at 16:09
In Einstein's original derivation, two light rays leave the bulb in opposite directions. That is very important because it means the bulb isn't kicked by the momentum of light. I showed here that the first order classical doopler effect (the v/c term) cancels out as a consequence (science20.com/curious_inquirer/blog/was_einstein_wrong-90405). According to Tony Rothman, E=mc^2 isn't a relativistic result. (physicsworld.com/cws/article/news/2011/aug/23/…). –  MadScientist Jul 4 '12 at 18:23
YouTube video is simply wrong. He got to the correct result because k.e. has been canceled out. –  SS-3.1415926535897932384626433 Jul 4 '12 at 18:27
@Shekhar, If it is emitting in all directions and the bulb is perfectly spherical then the same result applies because for each ray of light leaving one side there is a ray of light leaving the other. Anyway, these thought experiments should always be made as simple as possible. Emitting in all directions is a needless complication (given Einstein's simpler derivation.) –  MadScientist Jul 4 '12 at 18:37
@Physikslover: So what. The assumption that C is the same before and after is fine. Please see my answer for a way of making it manifest. –  Ron Maimon Jul 8 '12 at 18:01

These thought experiments should always be made as simple as possible. Emitting in all directions is a needless complication. Consider Einstein's original derivation of E=mc^2 with only two light rays pointing in opposite directions (as Ron said above, this is a necessary assumption- it means the momentum of the light rays won't kick the bulb in one direction). You will find that, if you only put in the first term of the Relativistic Doopler Effect (the v/c term that -alone- is the classical doopler effect), it cancels out. Only the v^2/c^2 and higher terms remain. Using only the classical doopler effect produces no result. EDIT: The +/-v/c terms produces no net energy from the point of view of the moving frame because adding the energies results in the cancellation of +/-v/c terms (see Einstein's original paper), but these terms do produce a net change in the momentum of light in the right direction from the point of view of the moving system (See * below for how this net momentum is used in Rohrlich's derivation of E=mc^2)

On the cancellation of the v/c term: (http://www.science20.com/curious_inquirer/blog/was_einstein_wrong-90405)

On Einstein's original paper:(http://www.science20.com/curious_inquirer/most_famous_equation_physics_and_its_derivation-89948)

The Original Paper on E=mc^2:( http://www.fourmilab.ch/etexts/einstein/E_mc2/www/)

*BUT! There is a derivation of E=mc^2 using the classical doopler effect:(http://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence#Alternative_version)

[The last comment was EDITED due to Alfred Centauri's comment below.].

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I don't think it's correct to say that Rohrlich's derivation is manifestly non-relativistic. –  Alfred Centauri Jul 4 '12 at 19:25
@Alfred Centauri, Why not? –  MadScientist Jul 4 '12 at 19:30
He smuggled in the constancy of the speed of light. From the Wiki article: "The momentum of the light is its energy divided by c". His argument begs the question. –  Alfred Centauri Jul 4 '12 at 19:36
BB1, if one drops the constancy of the speed of light (a postulate of SR), then the classical doppler effect does not produce a result. Rohrlich essentially is using the relativistic doppler effect when he divides the change in energy, due to the classical doppler shift, by the constant c to get the change in momentum. –  Alfred Centauri Jul 4 '12 at 20:31
In your link you say "Because the H-E terms measure the change in energy of the box due to the relative motion of the two frames only, the additive constant C representing any other energy left over (such as the internal molecular energies of the box etc) is constant" why shouldn't the internal energy of the body and therefore C change? –  Physiks lover Jul 5 '12 at 23:04

The argument is Einstein is a little annoying to read, here's a simpler version (I put it on Wikipedia under mass-energy equivalence a long time ago. It also appears somewhere on Terry Tao's blog, and it's the right way to make the argument).

Consider a mass M which is stationary. It emits two identical pulses of light in opposite directions, each of equal energy E/2. After the emmission it still isn't moving.

Now consider the frame where the mass is moving with velocity v to the right (so the frame is moving to the left compared to the original frame). It emits two pulses of light in opposite direction, but now the light moving to the right is blueshifted and the light moving to the left is redshifted. The frequency shift is by the doppler shift formula, the shift factor is v/c, and the momentum is E/2c. So the right moving light has more momentum by Ev/2c^2 and left moving light has less momentum by Ev/2c^2, so the total momentum of the light is

Ev/2c^2

to the right. This means that the body has it's momentum reduced by this amount, but the velocity didn't change, so the mass went down. The amount of the mass decrease, to conserve momentum, must be so that the new momentum of the body is:

P = (m - E/c^2) v

so the mass is decreased by the outgoing energy. Einstein uses the energy of the body, not the momentum, but Einstein's argument is equivalent to the one above by 4-vector nature of energy momentum. So Einstein's argument is fine, and there should be no controversy.

Let's consider the same experiment using sound in a nonrelativistic theory of a mass in air. A body emits two pulses of sound to the left and to the right with total energy E/2. Now we shift frames to where the body and the air are moving with velocity v. But now we can't conclude anything--- the air is moving, there is no relativity, the background the body is moving in isn't invariant to boosts.

But consider the momentum balance in the moving frame anyway: the sound moving to the right is blueshifted, the sound moving to the left is redshifted by v/c, but there is no imbalance in momentum, because in Galilean transformations, if the momentum of two things is balanced in one frame, it is balanced in every frame.

Einstein's argument requires the principle of relativity, but he used a small v limit where all the equations are nonrelativistic. The only relativistic equation you use in the whole thing is the relation between energy and momentum of light E=pc. That doesn't mean it's a nonrelativistic argument, because the principle of relativity is invoked multiple times, and the only reason the momentum doesn't balance in the moving frame is because the light is relativistic.

I should point out that Einstein already knew about photons, and for sure was considering two photon emission to come up with this. From the photon principle, he knew that frequency and energy have the same transformation. Otherwise it would be harder to know what the change in momentum energy under blueshifting/redshifting will be.

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Do you think you could stick with the original derivation of Einstein since that's what the OP is referring to? –  Physiks lover Jul 8 '12 at 19:59
@Physikslover: Aargh--- that means I have to read the paper! Ok, I shouldn't be lazy. The argument I give is equivalent, since the physical situation is identical, I just did the analysis myself instead of following Einstein's suboptimal 1905 notation conventions. The issue with the constant C is annoying, because Einstein didn't fix it by 4-vector considerations, but by hamiltonian considerations (or perhaps thermodynamics) and I hate trying to guess what someone long dead was thinking, but I'll try. –  Ron Maimon Jul 8 '12 at 20:21
One can speculate that Einstein used energy of the outgoing light and not momentum conservation because he hadn't yet become comfortable with the notion that photons carry momentum proportional to the inverse wavelength. This only came in 1909 or so when he published a paper showing that a photon carrying energy must carry momentum, since both are 4-vectors. This simplifies the 1905 argument to the one I give, but Einstein might have been reluctant to use momentum for photons, and so used energy. Remember that in 1905, even Einstein wasn't 100% comfortable with relativity, it took a few years. –  Ron Maimon Jul 8 '12 at 20:24
Thanks Ron, that's precisely the point I was attempting to get across. At the very moment that the argument introduces a particle where the momentum and energy are proportional, the inquiring mind should pause and ask "WTF??? Where did that come from? What are the implications of the existence of such a particle?" Of course, the implication is non-other than the existence of an invariant speed. –  Alfred Centauri Jul 9 '12 at 0:26
@AlfredCentauri: The momentum and energy are already proportional for classical electromagnetic waves. This is not what I am saying--- I am saying that Einstein was further using photon concepts in secret, and publishing equivalent light-pulse translations of these photon arguments. –  Ron Maimon Jul 9 '12 at 2:09

If I understand your question correctly, you're asking if there is an argument using light for some kind of mass - energy equivalence in the context of non-relativistic mechanics.

I think you might be forgetting that the classical theory of light, Maxwell's equations, are relativistically covariant and it was this tension between Maxwell's theory and non-relativistic mechanics that led Einstein to his relativistic mechanics.

One way to see the problem is to consider that the Galilean transformation is recovered in the limit as the invariant speed $c \rightarrow \infty$. So, from this "Galilean" angle, non-zero mass represents infinite energy.

What's going on here? The speed of light is observationally finite so we have that in a non-relativistic context, the speed of light is not invariant. If you go with that and, as Ron suggested, you actually use two oppositely directed pulses of light, the sum of the energy of the two light pulses is invariant. So, there's no change in kinetic energy required to balance the books, no need for a mass - energy relation.

But, observationally, the speed of light is invariant so we have a conflict and a need for a new mechanics to do this problem.

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Okay, so if I understand correctly: in classical physics, if the bulb emits light in all directions, then the total amount of energy lost is invariant, even in other frames of reference (so, classical Doppler effect has no impact on the total amount of energy lost due to the light pulse). It is only with special relativity, and a relativistic Doppler effect, that the light pulse carries away different amounts of energy in different reference frames. Is that correct? –  Ord Jul 4 '12 at 18:13
Also, just to clarify, I recognize the need for special relativity in general. I just wasn't sold on why it is needed to discover the mass-energy relation. –  Ord Jul 4 '12 at 18:14
@Ord Yes, that's the problem. I have highlighted that in my answer. –  SS-3.1415926535897932384626433 Jul 4 '12 at 18:16
@Ord: the total amount of energy lost is not invariant, but the fact that energy is conserved IS. –  Jerry Schirmer Jul 4 '12 at 19:07
@JerrySchirmer ... so ... if the total amount of energy lost is not invariant in different reference frames with a classical Doppler effect, why do we need special relativity to discover the relationship between energy and mass? –  Ord Jul 4 '12 at 20:45

1. Classical physics can't answer such problems if you forcefully apply your point 2 (classical doppler effect can't cause lesser energy loss). With classical physics, we start like this: m is constant, so there must be change in v. If not, we need to in-invent physics. Mind it, particle mass is constant in classical physics and has higher priority than v.

2. Special Theory of Relativity wasn't introduced to create E=mc^2. It was introduced because classical physics failed to explain relativistic phenomena due to wrong understanding of space and time scales. New "Spacetime" scale was introduced to measure everything with new way. And, with new scale, we found E=mc^2.

Remember: Relativistic Doppler effect needs time dilation for understanding (which requires Special Theory of Relativity). In that YouTube video, mass is motion dependent which is also from Special Theory of Relativity.

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-1: This doesn't answer the question, it is a rambling collection of unrelated thoughts. –  Ron Maimon Jul 8 '12 at 6:47
@Ron The first point is the answer. And, if you can see the bolder part, that's the key.. –  SS-3.1415926535897932384626433 Jul 8 '12 at 7:23
I see, that's kind of right. But the issue is where exactly does the argument fail in classical mechanics--- two beams of light leave a source in opposite directions, you boost, naively there is a change in the emitted momentum, so you conclude that E=mc^2. The answer is just that the classical doppler shift doesn't lead the momentum in the outgoing waves to be imbalanced, but you bury this in unnecessary extraneous verbiage, and you never clearly say it. I will admit it has a little more of the correct answer in it than I thought initially. –  Ron Maimon Jul 8 '12 at 8:02