0
$\begingroup$

In Weinberg book The quantum theory of fields he says that one condition for the $S$ matrix to be Lorentz invariant invariant is that the interactions therm takes the form

$$V(t)=\int dx^3H(x,t)$$

Such that $H(x,t)$ transform as a scalar under Lorentz transformation.

But under the lorentz transfrom $$ x' = \frac{x - ut}{\sqrt{1 - \frac{u^{2}}{c^{2}}}}, \quad t' = \frac{t - \frac{u}{c^{2}}x}{\sqrt{1 - \frac{u^{2}}{c^{2}}}} . $$ Since $H(x,t)$ is a scalar and $dx'=\gamma dx$ we have $$V'(t)=\gamma V(t)$$

where $$\gamma =\sqrt{1 - \frac{u^{2}}{c^{2}}}$$ But $V(t)$ should transform as the $0$ component of a momentum four vector that is , take the four vector $p=(p^0.p^1,p^2,p^3)$ with $p^0=V(t)$ we should have

$$V'(t)=\gamma V(t) -\gamma \frac{u}{c^{2}}p^1$$

My question is why we have $V'(t)=\gamma V(t)$?

$\endgroup$
0
$\begingroup$

The reason is that the S matrix is written like $$ S=\langle f|{\cal T}\exp\left(-\frac{i}{\hbar}\int dt'V(t')\right)|i\rangle $$ being $\cal T$ the time-ordering operator. This makes the integration with respect to time Lorentz-invariant.

$\endgroup$
4
  • $\begingroup$ What i am asking is why $V(t)$ does not transform as energy $\endgroup$ Nov 18 '19 at 12:42
  • $\begingroup$ In this case $p_1=0$, as you have $p_{\mu}=(V(t),0,0,0)$. This particular choice makes the trick. $\endgroup$
    – Jon
    Nov 18 '19 at 12:55
  • $\begingroup$ Why should we have $p_1=0$ ? $\endgroup$ Nov 18 '19 at 13:50
  • $\begingroup$ Because you are in the rest frame. Note that, if you are not, the $S-$matrix will grant you to be there because you are just an unitary transformation away. $\endgroup$
    – Jon
    Nov 18 '19 at 14:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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