# Lorentz transformation of Interaction therm in Weinberg book

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 $$V'(t)=\gamma V(t)$$?

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.
• What i am asking is why $V(t)$ does not transform as energy Nov 18 '19 at 12:42
• In this case $p_1=0$, as you have $p_{\mu}=(V(t),0,0,0)$. This particular choice makes the trick.
• Why should we have $p_1=0$ ? Nov 18 '19 at 13:50
• 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.