Showing that Energy-momentum relation is invariant

In summary, the particle has momentum p and energy E in the frame S, and has momentum p' and energy E' in the frame S' if S' is moving at speed v.
  • #1
Rubber Ducky
14
0

Homework Statement


[/B]
A particle of mass [itex]m[/itex] is moving in the [itex]+x[/itex]-direction with speed [itex]u[/itex] and has momentum [itex]p[/itex] and energy [itex]E[/itex] in the frame [itex]S[/itex].

(a) If [itex]S'[/itex] is moving at speed [itex]v[/itex], find the momentum [itex]p'[/itex] and energy [itex]E'[/itex] in the [itex]S'[/itex] frame.
(b) Note that [itex]E' \neq E[/itex] and [itex]p' \neq p[/itex], but show that [itex](E')^2-(p')^2c^2=E^2-p^2c^2[/itex]

Homework Equations


[/B]
[itex]E=p^2c^2+m^2c^4[/itex]
[itex]E'=\gamma mc^2=\frac{mc^2}{\sqrt{1-\frac{(u')^2}{c^2}}}[/itex]
[itex]p'=\gamma mu'=\frac{mu'}{\sqrt{1-\frac{(u')^2}{c^2}}}[/itex]
[itex]u'=\frac{u-v}{1-\frac{uv}{c^2}}[/itex]

The Attempt at a Solution



My prof says we need to use (a) to derive (b). It seems like it'd be easier a different way, but whatever. My strategy has been to show that [itex](E')^2-(p')^2c^2=m^2c^4[/itex], which would complete the derivation.

I tried playing around with some algebra and got [itex](E')^2=\frac{m^2c^4(c^2-uv)^2}{(c^2-u^2)(c^2-v^2)}[/itex] and [itex](p')^2=\frac{m^2(u-v)^2}{(1-\frac{uv}{c^2})^2(c^2-u^2)(c^2-v^2)}[/itex]

I got a little overwhelmed with the algebra at this point. I'd like to know if this is right so far, at least, before trying anything else.
 
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  • #2
For part (a), you should be able to show that
\begin{eqnarray*}
E' &= \gamma(E - \beta pc) \\
p'c &= \gamma(pc - \beta E)
\end{eqnarray*} where ##\beta = v/c## and ##\gamma = 1/\sqrt{1-\beta^2}##.

Your expression for ##E'^2## looks okay. Divide both the numerator and denominator by ##c^4## and take the square root. You should be able to show the first equation above. Just make sure you keep track of the three ##\gamma##'s you have in this problem.
 
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  • #3
vela said:
For part (a), you should be able to show that
\begin{eqnarray*}
E' &= \gamma(E - \beta pc) \\
p'c &= \gamma(pc - \beta E)
\end{eqnarray*} where ##\beta = v/c## and ##\gamma = 1/\sqrt{1-\beta^2}##.

Your expression for ##E'^2## looks okay. Divide both the numerator and denominator by ##c^4## and take the square root. You should be able to show the first equation above. Just make sure you keep track of the three ##\gamma##'s you have in this problem.
Thanks very much for the help. It worked like a charm - the last part of the derivation that I ended up using was pretty neat. My only worry is that my prof will wonder how I thought to express ##E'## and ##cp'## in that way.
 
  • #4
Look up the Lorentz transformation. :smile: You might already be familiar with it applied to x and t. It applies to all four-vectors.
 
  • #5
vela said:
Look up the Lorentz transformation. :) You might already be familiar with it applied to x and t. It applies to all four-vectors.
Interesting - we've covered the Lorentz transformation, but not four-vectors. My class gives a brief overview of SR, but maybe I'll see it in future.

I have another question that I'm editing into the OP now - It's somewhat related, but I didn't want to flood the section with threads.

EDIT: Woops, guess it's been too long to edit. See you in the next thread XD
 

Related to Showing that Energy-momentum relation is invariant

1. What is the energy-momentum relation?

The energy-momentum relation is a fundamental equation in physics that describes the relationship between energy and momentum for a particle or system. It states that the total energy of a particle is equal to its mass multiplied by the square of its velocity, plus its potential energy.

2. Why is it important to show that the energy-momentum relation is invariant?

It is important to show that the energy-momentum relation is invariant because it demonstrates that the laws of physics are the same for all observers, regardless of their relative motion. This principle, known as the principle of relativity, is a cornerstone of modern physics and has far-reaching implications for our understanding of the universe.

3. How can we show that the energy-momentum relation is invariant?

The energy-momentum relation can be shown to be invariant by using mathematical equations and principles of special relativity, such as the Lorentz transformations. These transformations allow us to relate the energy and momentum of a particle as measured by one observer to the energy and momentum measured by another observer in a different frame of reference.

4. What evidence supports the invariance of the energy-momentum relation?

Experimental data from a variety of sources, such as particle accelerators and high-energy collisions, have consistently shown that the energy-momentum relation holds true regardless of the observer's frame of reference. This provides strong evidence for the invariance of the relation and supports the validity of the principle of relativity.

5. What are the implications of the invariance of the energy-momentum relation?

The invariance of the energy-momentum relation has several important implications in physics. It allows us to make accurate predictions about the behavior of particles and systems, and it also provides a foundation for the development of theories such as special relativity and quantum mechanics. Furthermore, the invariance of this relation has led to significant advancements in fields such as particle physics, cosmology, and astrophysics.

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