Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Reference) "Feynman lectures on Physics Vol.3 , p.7-4 ."

With four vectors $x_{\mu} = (t,x,y,z)\ , \ p_{\mu} = (E,p_{x},p_{y},p_{z})$

the inner product of these two four vectors is scalar invariant and equals to $Et - \overrightarrow{p} \overrightarrow{x}$ . Alright.

But I cannot understand why $p_{\mu}x_{\mu}$ is just $Et$ in the rest frame as Feynman writes in the book above.

share|cite|improve this question
up vote 1 down vote accepted

The value of $xp=x^\mu p_\mu=\eta^{\mu\nu}x_\mu p_\nu$ (with the sum over repreted indices) is invariant, i.e. if $A$ and $A'$ denote two frames, then the number given by the product $xy$ is the same as $x'y'$.

In one frame $A$, the expression $xp$ might evaluate to $Et-\vec x \vec p$, in another frame $A'$ where $x_0'\equiv t',\vec p' = 0$ (the rest frame) you'll have $x'p'=E't'-\vec x' \vec p'=E't'$. So the product can look like something which only contain energy and time this way.

By the invariance discussed in the first sentece, i.e. $xp=x'p'$, you have $Et-\vec x \vec p=E't'$. Not more, not less. In general $E't'\ne Et$, so probably it's not written explicitly that the energy and time variable on the other page really are these quantities in another frame.

share|cite|improve this answer

Because $\vec{p}=0$ in the rest frame!

So $p_\mu x_\mu = Et − \vec{p}\vec{x}=Et-0=Et$.

The $E$ and $t$ may of course be different in different frames.

share|cite|improve this answer
+1 This is the correct answer. – Killercam Jul 24 '12 at 17:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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