# Inner product of four-vectors in special relativity

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.

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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.

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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.

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+1 This is the correct answer. – Killercam Jul 24 '12 at 17:36