# E=mc^2 and Special Relativity

1. Jan 29, 2007

### NanakiXIII

While searching the net I've come across numerous derivations for the equation of $$E=mc^2$$, the most understandable of which I found to be the photon-in-the-box momentum derivation. However, this derivation and other ones don't necessarily seem to be based on Special Relativity. So my question is this: is $$E=mc^2$$ a relativistic concept per se?

2. Jan 29, 2007

### arcnets

I think the answer is yes.
Here's a derivation which I rather like.

You start from the invariance of light speed:
(ct)^2 - x^2 = invariant.
Now you multiply by m and divide by t, getting
(mc)^2 - (mv)^2 = invariant (since v = x/t).
Now you look at two frames, one of which is the rest frame (v = 0):
(m0 c)^2 = (mc)^2 - (mv)^2.
Now you solve that for m and get relativistic mass:
m = m0 / sqrt(1 - v^2/c^2).
Now, Energy is defined as the integral
E = Integral F dx.
Now, Newton says F = dp/dt, so
E = Integral dp dx/dt.
Now dx/dt = v, so
E = Integral v dp.
Now p = mv, so
p = m0 v / sqrt(...)
Now we solve this for v, getting
v = p / sqrt(m0^2 + (p/c)^2)
So, obviousliy m = sqrt(m0^2 + (p/c)^2)
...and if you do the Integral v(p) dp, you get E = mc^2.

So, what is this based on?
1) Invariant light speed
2) Definition of Energy
3) Newton's Law.

3. Jan 29, 2007

### nakurusil

This is not a valid "proof", I don't know where you got it frpm but I flagged it once before as a "sleigh of hand" . Do you know why?

4. Jan 29, 2007

### cesiumfrog

Nanaki, I'd be surprised if such a derivation didn't implicitly assume the speed of light is the same in multiple frames; got a link?

5. Jan 29, 2007

### arcnets

You are fast! Where do you see an error?

6. Jan 29, 2007

### cesiumfrog

This is the second time I've quoted myself..

7. Jan 29, 2007

### NanakiXIII

This is the photon-in-the-box derivation I referred to:

However, I've also in my Googling seen page titles such as "Non-Relativistic Derivation of E=mc^2".

8. Jan 29, 2007

### arcnets

cesiumfrog, I understand that you see my error in my derivation of relativistic mass but not in the rest of the "proof".

I admit there is a weak spot there.
(ct)^2 - x^2 = invariant.
Let's define the 4-vector X = (ct; ix). Then the invariant is the square length of this vector. So a Lorentz transform looks like a rotation in this basis.
Now, we define 4-momentum P = mX/t. Then P = (mc; imv).
Now, if we ASSUME that a Lorentz transform acts like a rotation of 4-momentum also, then
(mc)^2 - (mv)^2 = invariant.

Better?

9. Jan 29, 2007

### Fredrik

Staff Emeritus
I wrote something here and then removed it because it contained a blunder. Maybe I'll fix it and update this thread in a while.

Last edited: Jan 29, 2007
10. Jan 29, 2007

### nakurusil

Still bad, there is nothing that enables you to multiply the invariant (ct)^2 - x^2 by m and declare it another "invariant" after the fact. You cannot divide the invariant (ct)^2 - x^2 by t and declare it invariant after the fact as you did earlier. By this logic you would be getting c^2-v^2=invariant (NOT).

Now, the correct way to do this is to start with the relativistic enery-momentum:

$$E=\gamma*mc^2$$
$$p=\gamma*mv$$

Then:

$$E^2-(pc)^2=m^2c^4$$ (1)
where m=proper mass and $$\gamma=1/\sqrt(1-v^2/c^2)$$

We can rewrite (1) as :

$$E=\sqrt((pc)^2+m^2c^4))$$ (2)
which is the most general form .
In the proper frame of the object $$p=0$$ so (2) becomes:

$$E=mc^2$$ (3)

And Nanaki, you are welcome.

.

11. Jan 29, 2007

### cesiumfrog

That's an interesting question.

A nuclear physicist can take a large (stationary) sample of material, release some of the potential energy, and experimentally measure that the mass has decreased. So indeed, it's hard to attribute this to any relativistic concept.

In the case of your linked derivation: it simply uses (after providing motivation) the definition that mass is the ratio of momentum to velocity, combined with the momentum of light derived by classical electrodynamics. But since classical electrodynamics is the original relativistic theory, I can't call this an independent derivation.

12. Jan 29, 2007

### nakurusil

Yes, it is , it required relativity in order to derive it.

13. Jan 29, 2007

### rbj

i think people might, for pedagogical reasons, wonder how you got these equations to start with. are they axioms?

14. Jan 29, 2007

### nakurusil

Yes, they are "by definition" the components of the energy-momentum 4-vector

$$(p,E/c)=\gamma(mv,mc)$$

15. Jan 29, 2007

### bernhard.rothenstein

mass

I think that we should make a net distinction between how we arive at a given equation and how we deriveit.
If you consider the experiment performed say by Bertozzi, then he puts on a graph the expeimental results (energy as a function of v^2) and finds out that they fit if we use a formula which can be derived using relativistic concepts.
If we want to derive it we should consider the same experiment (say the intercation between a photon and a tardyon) from two inertial reference frames in relative motion and so special relativity becomes involved and we can say that it is a relativistic effect.
If it is "per se" the my answer is no.
sine ira et studio

16. Feb 3, 2007

### cesiumfrog

What do you think this adds to the scenario?? You're still using the same energy-momentum relation from classical electrodynamics, so your conclusion still isn't independent from special relativity. At least the original scenario didn't start out by writing that mass and energy are interchangeable.

Maybe it really just is a relativistic result, and for example the missing atomic masses arise simply from the colour field being composed of (virtual) gluons propagating at the speed of light?

17. Feb 3, 2007

### country boy

I came across your interesting discussion by accident and wondered if I could suggest another way of getting E=mc^2 that might be helpful in answering the question about it being a relativistic concept. The following is a version of Einstein's photon-in-a-box that doesn't require a box.

Start with two atoms, A and B, with masses Ma and Mb, that are at rest and separated by a distance D. Atom A emits a photon toward B with energy E and momentum E/c, reducing the mass of A from Ma to ma. From conservation of momentum, the emission imparts a velocity to A of Va=-E/ma*c. After a time t atom A has moved to position Xa=Va*t=-E*t/ma*c.

At time T=D/c the photon is absorbed by atom B and increases its mass from Mb to mb. By similar reasoning, the position of atom B after a time t is
Xb=D+E*(t-D/c)/mb*c.

The position of the center of mass before the emission is Ma*(0)+Mb*D and after the emission is ma*Xa+mb*Xb (both divided by total mass). Because this is an isolated system, the center of mass cannot have changed, so these two quantities are equal. Therefore,
Mb*D = -E*t/c+mb*D+E*t/c-E*D/c^2 = mb*D-E*D/c^2.

The increase in mass of atom B and decrease in mass of atom A are then
M = (mb-Mb) = (Ma-ma) = E/c^2, giving the relation between mass and energy E=Mc^2.

This derivation seems to make no use of relativity and, in fact, can be made without knowing about relativity. However, the method of equating a change in rest mass (a mechanical concept) with the energy of a photon (a radiation concept) is at the basis of special relativity.

18. Feb 3, 2007

### country boy

Wow cesiumfrog. Two question marks? This is similar to the way Einstein derived it, and he certainly added to the scenario. The original question had to do with attempting to understand the basis for the relation, and I just though that an alternative approach might help.

19. Feb 3, 2007

### nakurusil

Do you realise that the moment you said "momentum E/c" you have already used the conclusion in your demonstration?

In general $$E^2=(pc)^2+(mc^2)^2$$
The above is the generalisation of $$E=mc^2$$
For massless particles, it reduces to $$E=pc$$, which is exactly what you are attempting to use in your proof.

20. Feb 3, 2007

### country boy

For a photon, momentum = E/c. There is no mention of mass in this statement. The problem we are addressing is how to relate energy to mass and the photon-in-a-box makes the connection. Of course, if you write down the correct expressions you are already assuming the correct conclusion.