
#1
Aug710, 01:14 PM

P: 67

I know you need to use some spacetime geometry to solve some conservation equations but what is the simplest way you derive the following equation about energy and momentum:
E^{2} = (pc)^{2} + (mc^{2})^{2}




#2
Aug710, 02:18 PM

P: 1,262

Start with einstein's equation:
[tex] E = \gamma mc^2 [/tex] Plug in for gamma 



#3
Aug710, 02:27 PM

P: 665

AB 



#4
Aug710, 02:30 PM

Emeritus
Sci Advisor
PF Gold
P: 5,500

How do you derive E2= (pc)2 + (mc2)2 



#5
Aug710, 03:23 PM

P: 280

[tex]S=mc\int d\tau[/tex] where [tex](d\tau)^2=dx^\mu dx_\mu[/tex] is the proper time of the particle, that you can write [tex]d\tau=\sqrt^{1(v^2/c^2)}dt[/tex] So you have a lagrangian [tex]L=mc\sqrt^{1(v^2/c^2)}[/tex] from which you derive momentum and energy in the usual way: [tex]p_i=\frac{\partial L}{\partial v_i}=\frac{mv_i}{\sqrt^{1(v^2/c^2)}}[/tex] [tex]E=p\cdot vL=\frac{mc^2}{\sqrt^{1(v^2/c^2)}}[/tex] And now the relation you want to prove is easy to verify. 



#6
Aug710, 03:36 PM

P: 665

[tex](cd\tau)^2=dx^\mu dx_\mu[/tex] because the LHS is just the square of the differential of proper length, [tex]ds^2.[/tex] AB 



#7
Aug710, 03:42 PM

P: 280





#8
Aug710, 10:56 PM

Mentor
P: 11,255

Another way: assume the equations for energy and momentum,
[tex]E = \frac{mc^2}{\sqrt{1  v^2/c^2}}[/tex] [tex]p = \frac{mv}{\sqrt{1  v^2/c^2}}[/tex] and solve them together algebraically to eliminate v. 



#9
Aug810, 01:08 PM

Emeritus
Sci Advisor
PF Gold
P: 5,500

I agree with Petr that the answer to this really depends on what you're willing to assume.
Suppose you want these things: (1) A fourvector exists that is a generalization of the momentum threevector from Newtonian mechanics. (2) The relationship between the relativistic and Newtonian versions satisfies the correspondence principle. (3) The quantity is additive (because we hope to have a conservation law). Then I think it's quite difficult to come up with any other definition for the momentum fourvector than the standard one, and then the [itex]E^2=p^2+m^2[/itex] identity follows immediately. But this isn't anything like a proof of uniqueness or selfconsistency. Petr's derivation is likewise very natural, but it depends on the assumption of a certain form for the action, and there's no guarantee that the results it outputs obey the correspondence principle or result in a conservation law. Those properties have to be checked mathematically and experimentally. There's a variety of very persuasive physical arguments that once you've accepted SR's description of spacetime, you have to believe in massenergy equivalence. A good example is Einstein's 1905 paper "Does the inertia of a body depend on its energy content?," where he does a thoughtexperiment involving an object emitting rays of light in opposite directions. He only says there that "The fact that the energy withdrawn from the body becomes energy of radiation evidently makes no difference[...]," without giving any real justification, but it's not hard to come up with arguments to that effect (see, e.g., http://www.lightandmatter.com/html_b...tml#Section4.2 , at "the same must be true for other forms of energy"). No amount of mathematical manipulation can substitute for this kind of physical reasoning, or for empirical verification. 



#10
Aug810, 04:35 PM

P: 665

In general the only alternative way to prove Einstein's relation is just what jtbell or zhermes said and one cannot expect this to happen within GR by considering any special restrictive conditions by which GR drops down to embrace SR. AB 



#11
Aug1710, 11:24 PM

P: 555

Is there a derivation which only takes into account the LT's and Newton's Second Law ?
Best wishes DaTario 



#12
Aug1810, 12:17 AM

P: 1,568





#13
Aug1810, 12:21 AM

P: 1,568

[tex]E=\gamma mc^2[/tex] [tex]p=\gamma mv[/tex] by definition. As a twist to your suggested proof, the one that gets used most often relies on calculating: [tex]E^2(pc)^2[/tex] and reducing the answer to [tex](mc^2)^2[/tex] 


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