How do we get E=mc2?

by fawk3s
Tags: emc2
 P: 342 Hi Somewhat usual question for you guys I guess. How do we get the formula E=mc2? I saw in a video where it said Einstein combined these equations Mv=E/c t=L/c x=vt Mx=mL and got EL/c2=mL => E/c2=m But I dont quite follow. What is Mv? What is Mx? And what time does L/c represent? How exactly are they combined? Always interested me. Thanks in advance, fawk3s
 P: 342 I actually found this: http://fotonowy.pl/index.php?main_page=page&id=6 So I think I get it now. :)
 P: 1,345 How do we get E=mc2? It looks like they assumed some of the results in their derivation which is circular logic. Perhaps a better derivation is: http://en.wikipedia.org/wiki/Mass%E2...nce#Background
P: 1,060
 Quote by Feldoh It looks like they assumed some of the results in their derivation which is circular logic.
Now as you say it, I agree. That derivation already assumes the result from the beginning.

In fact after assuming
$$p=\frac{E}{c}$$
one could directly use $p=mc$ to get the result.

However, I don't see a derivation in Wikipedia either?!
 HW Helper PF Gold P: 1,976 I've written up a derivation that I think is pretty concise. go to www.shadycrypt.com Click on the E=mc2 link at the top.
Emeritus
PF Gold
P: 5,598
Here is the argument that Einstein originally published: http://fourmilab.ch/etexts/einstein/E_mc2/www/

Here is a different argument: http://www.lightandmatter.com/html_b...tml#Section1.3

 Quote by Feldoh It looks like they assumed some of the results in their derivation which is circular logic.
No, the derivation on the fotonowy.pl page is not circular. This is another well known proof, originating with Einstein. One delicate issue in it is that in the original form of this thought experiment, the box is implicitly assumed to be perfectly rigid. This is a flaw, but it can be fixed: http://galileo.phys.virginia.edu/cla...nd_energy.html

 Quote by Feldoh Perhaps a better derivation is: http://en.wikipedia.org/wiki/Mass%E2...nce#Background
That isn't a derivation. They point out that it's a special case of the relativistic energy-momentum relation, but they haven't proved the energy-momentum relation.
P: 1,060
 Quote by bcrowell No, the derivation on the fotonowy.pl page is not circular.
It's not even circular. It already assumes E=mc^2 from the very beginning. After writing p=E/c you do not need a lengthy derivation, but just use p=mc to derive the final result. So the two equations are one step apart from being equivalent.
Emeritus
PF Gold
P: 5,598
 Quote by Gerenuk It's not even circular. It already assumes E=mc^2 from the very beginning. After writing p=E/c you do not need a lengthy derivation
No, p=E/c for electromagnetic waves follows directly from Maxwell's equations, so that had been known for 30 or 40 years before Einstein published SR in 1905. Here is an explanation: http://www.lightandmatter.com/html_b...ch01/ch01.html (see subsection 1.5.7).

 Quote by Gerenuk but just use p=mc to derive the final result. So the two equations are one step apart from being equivalent.
No, this is incorrect. You can't just plug v=c in to p=mv and expect it to be correct for a photon. p=mv is a nonrelativistic equation, which can't be expected to hold for light.

Considering that the argument given at fotonowy.pl is due to Einstein, I really don't think you're going to find obvious logical fallacies in it.
P: 1,060
I need to check the links you proposed.

Can you point me to a link where they show the prove how to derive p=mv from Maxwell?
(I'm not surprised it works, since Maxwell is already relativistic?!)

However:
 Quote by bcrowell No, this is incorrect. You can't just plug v=c in to p=mv and expect it to be correct for a photon. p=mv is a nonrelativistic equation, which can't be expected to hold for light.
p=mv is always true. Also in relativity, since judging by the proof they use the relativistic mass. They use E=p/c, so E is the total energy. They derive E=mc^2 with the same variable E, so m must be the relativistic mass. In that case p=mv in both classical and relativistic theory.
Emeritus
PF Gold
P: 5,598
 Quote by Gerenuk I need to check the links you proposed. Can you point me to a link where they show the prove how to derive p=mv from Maxwell?
No, because it's not true. But I did provide a link that shows p=E/c for an electromagnetic wave.

 Quote by Gerenuk p=mv is always true. Also in relativity, since judging by the proof they use the relativistic mass. They use E=p/c, so E is the total energy. They derive E=mc^2 with the same variable E, so m must be the relativistic mass. In that case p=mv in both classical and relativistic theory.
You've misunderstood the content of E=mc2.

[EDIT] Actually the link I gave above only proves that p is nonzero for an electromagnetic wave (which is inconsistent with the classical relation p=mv, since m=0 for light). By linearity and units we must have $p=kE/c$, where k is a unitless constant. For the proof that k=1, see this link: http://www.lightandmatter.com/html_b...ml#Section11.6 (subsection 11.6.2).
P: 1,060
 Quote by bcrowell No, because it's not true. But I did provide a link that shows p=E/c for an electromagnetic wave.
That link states this equation follows from Maxwell's equation, but there isn't even a single Maxwell equation. There might be a vague hint in the text, but there is no derivation.

Oh, but if you mean the link you posted later... I still have to go through it...

 Quote by bcrowell You've misunderstood the content of E=mc2.
You have to be more specific and tell where my argumentation is wrong, if you believe you know it.
P: 1,060
 Quote by bcrowell By linearity and units we must have $p=kE/c$, where k is a unitless constant.
Where is the proof for linearity then? Why are normal particles not linear?
P: 72
 Quote by Gerenuk Where is the proof for linearity then? Why are normal particles not linear?
Take the Maxwell's equations in vacuum. Transform them to a wave equation for B. Try the solution in the form $$B=B_ocos(kx-\omega t)$$.

The normal particles have nonzero rest masses, so their energy is proportional to $$p^2$$ rather then to $$p$$.