2
$\begingroup$

Can one explain the relativistic energy transformation formula:

$$E = \gamma\ E',$$

where the primed frame has a velocity $v$ relative to the unprimed frame, in terms of relativistic time dilation and the quantum relation $E=h\nu$?

I imagine a pair of observers, A and B, initially at rest, each with an identical quantum system with oscillation period $T$.

Now A stays at rest whereas B is boosted to velocity $v$.

Just as in the "twin paradox" the two observers are no longer identical: B has experienced a boost whereas A has not. Both observers should agree on the fact that B has more energy than A.

From A's perspective B has extra kinetic energy by virtue of his velocity $v$. Relativistically A should use the energy transformation formula above.

But we should also be able to argue that B has more energy from B's perspective as well.

From B's perspective he is stationary and A has velocity $-v$. Therefore, due to relativistic time dilation, B sees A's oscillation period $T$ increased to $\gamma\ T$.

Thus B finds that his quantum oscillator will perform a factor of $\gamma\ T/T=\gamma$ more oscillations in the same period as A's quantum system.

Thus B sees that the frequency of his quantum system has increased by a factor of $\gamma$ over the frequency of A's system.

As we have the quantum relation, $E=h\nu$, this implies that B observes that the energy of his quantum system is a factor of $\gamma$ larger than the energy of A's stationary system.

Thus observer B too, using his frame of reference, can confirm that his system has more energy than observer A's system.

Is this reasoning correct?

$\endgroup$
1
  • $\begingroup$ the object $\left(E,\vec{p}\right)$ transforms as 4-vector, so Lorentz transformation rules are applied on it, thus I'm not sure from where you get your first equity of how energy transforms. $\endgroup$
    – TMS
    Commented Jul 24, 2013 at 22:51

1 Answer 1

1
$\begingroup$

No, your reasoning is not correct. As is indicated by the relativistic transformation of energy, energy is completely relative. There is no absolute measurement of energy, and therefor observers A and B will not agree that B is more energetic. B is only more energetic from the standpoint of A, which is the standpoint you chose to analyse the problem. (You can see this relativity of energy in real live through the relativistic Doppler Effect.) Furthermore, the analogy with the twin paradox fails because the solution to the twin paradox was the deaccaleration of the travelling twin when encountering the distant star, this deacceleration invalidated the reference frame of the travelling twin. In your question, there is no such invalidation, and therefor both observers will be correct, even though they draw diffeerent conclusions. Finally you seem to use time dilation to argue about B from the perspective of B. You cannot do this, since B cannot measure his own velocity or experience his own time dillation. Time dillation can only b used to argue about the observations of others. However, even though your reasoning is partially flawed, it is possible to derive the lorentz transformation for energy using time dilation and $E=h\nu$.

Let's look at this from a different perspective:

Imagine a pair of observers A and B, moving relative to each other with a velocity $v$, each observing their own oscillation period as $T_A = T_B = T$. This must be true because the only thing differentiating observers A and B is their relative velocity, and neither observer can observe his own velocity.

Observer A will now argue that observer B will see time pass more slowly (by a factor $\gamma$). Therefor observer B will measure the oscillation period of observer A as $T_A' = T/\gamma$. This will lead him to conclude that $E_A' = h\nu_A' = h*{1\over{T_A'}} = h*{\gamma\over{T}} = \gamma*h*1/\nu=\gamma E$. Which is exactly the lorentz transformation for energy. However, it is very important to realize that observer B can use the exact same reasoning to conclude that A must measure a higher energy for B; $E_B'=...=\gamma E$

In other words, both observers will measure a higher energy for the other observer, and both will argue that the other observer will measure their own energy to be higher. And all of this is perfectly valid.

Of course, the situation is completely different if you introduce an actualy force that causes B to accelarate. In this case work is being done and B will measure his own period to be different from $T$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.