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

by kaleidoscope
Tags: energy, momentum
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kaleidoscope
#1
Aug7-10, 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:

E2 = (pc)2 + (mc2)2
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zhermes
#2
Aug7-10, 02:18 PM
P: 1,261
Start with einstein's equation:
[tex]
E = \gamma mc^2
[/tex]
Plug in for gamma
Altabeh
#3
Aug7-10, 02:27 PM
P: 665
Quote Quote by kaleidoscope View Post
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:

E2 = (pc)2 + (mc2)2
The simplest way is just dependent on how much you know Algebra. zhermes hinted at the starting point. If you had any problem proving it, let us know.

AB

bcrowell
#4
Aug7-10, 02:30 PM
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How do you derive E2= (pc)2 + (mc2)2

http://www.lightandmatter.com/html_b...tml#Section4.2
Petr Mugver
#5
Aug7-10, 03:23 PM
P: 280
Quote Quote by kaleidoscope View Post
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:

E2 = (pc)2 + (mc2)2
You can't actually "derive" it, you have to make some postulates. For a free particle of mass m, the equations of motion are derived frome the action

[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 v-L=\frac{mc^2}{\sqrt^{1-(v^2/c^2)}}[/tex]

And now the relation you want to prove is easy to verify.
Altabeh
#6
Aug7-10, 03:36 PM
P: 665
Quote Quote by Petr Mugver View Post

[tex](d\tau)^2=dx^\mu dx_\mu[/tex]

is the proper time of the particle, that you can write

.
Just a minor error: Following the Riemannian line-element for a general spacetime, this must have been

[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
Petr Mugver
#7
Aug7-10, 03:42 PM
P: 280
Quote Quote by Altabeh View Post
Just a minor error: Following the Riemannian line-element for a general spacetime, this must have been

[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
True. I usually put c=1, and when I have to put them back I always forget some!
jtbell
#8
Aug7-10, 10:56 PM
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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.
bcrowell
#9
Aug8-10, 01:08 PM
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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 four-vector exists that is a generalization of the momentum three-vector 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 four-vector 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 self-consistency.

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 mass-energy equivalence. A good example is Einstein's 1905 paper "Does the inertia of a body depend on its energy content?," where he does a thought-experiment 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.
Altabeh
#10
Aug8-10, 04:35 PM
P: 665
Quote Quote by bcrowell View Post
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 four-vector exists that is a generalization of the momentum three-vector 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 four-vector 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 self-consistency.
The bold-faced requirement is not actually what the relativistic energy equation [itex]E^2=p^2+m^2[/itex] offers because here energy is taken to the power two and its square root won't satisfy an additive property between quantities of energy nature. But I agree that conservation law always calls for an additive relation between all kinds of energies.

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.
Of course when one says that energy is (approximately) conserved in general relativity, such derivation in the zone of SR isn't automatically conserved if we all agree that the energy-momentum tensor is zero for a non-gravitational metric, despite the usual application of the logic "SR is a special case of GR so that whatever is satisfied by the latter holds true for SR too. This is all because of the mechanism we use in GR to derive conservation laws which is essentially based upon the Landau-Lifgarbagez pseudo-tensor and unfortunately this cannot be made zero when gravity disappears. Here the conservation of the energy-momentum tensor density results in the appearance of an extra term -Landau-Lifgarbagez pseudo-tensor- added to this tensor density under the operation of ordinary differentiation that stands for the validity of your guess.

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
DaTario
#11
Aug17-10, 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
starthaus
#12
Aug18-10, 12:17 AM
P: 1,568
Quote Quote by DaTario View Post
Is there a derivation which only takes into account the LT's and Newton's Second Law ?

Best wishes

DaTario
no, there is no such thing
starthaus
#13
Aug18-10, 12:21 AM
P: 1,568
Quote Quote by jtbell View Post
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.
I think this is the stanfdard one since :

[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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