# Invariance of inner product under Poincare transformation

The Poincare transformation reads, $$x\rightarrow x^\prime=\Lambda x +a$$ The scalar product is preserved under Lorentz transformation. However I do not see how it is preserved under the more general Poincare transformation, \begin{align} x^T\eta x\rightarrow {x^{\prime}}^T\eta x^{\prime}&=\left(\Lambda x +a\right)^T\eta\left(\Lambda x +a\right)\\ &=\left(x^T\Lambda^T+a^T\right)\eta\left(\Lambda x +a\right)\\ &=x^T\Lambda^T\eta\Lambda x+a^T\eta\Lambda x+x^T\Lambda^T\eta a+a^T\eta a\\ &=x^T\eta x+a^T\eta\Lambda x+x^T\Lambda^T\eta a+a^T\eta a \end{align} I don't know what to do with the rest of the terms. I will also have to show that the scalar product $$u^\mu A_{\mu},$$ where $$u^\mu$$ is the four velocity and $$A_{\mu}$$ is the vector gauge potential, is invariant under Poincare transformations. Basically I am trying piece by piece to show that the charged particle in a Electromagnetic field is invariant under Poincare transformations, $$\int \left(-m +eA_{\mu}u^\mu\right)d\tau-\int d^4x\frac{1}{4}\left(F_{\mu\nu}F^{\mu\nu}\right)$$

The quantity $$x^{2} = x^{\mu}x_{\mu}$$ is Lorentz invariant but not Poincar'{e} as you have clearly shown. However, the four velocity is defined by the derivative $$u^{\mu} = \frac{dx^{\mu}}{d\tau}$$ and hence, under $$x' = \Lambda x + a$$, we have $$u' = \Lambda u$$. Now, it is easy to see why $$u^{\mu}A_{\mu}$$ is Poincare invariant using $$u'=\Lambda u$$ and $$A'_{\mu}(x') = \Lambda_{\mu}^{\nu}A_{\nu}(x)$$.
• I don't understand this: why is $A_{\mu}(x^\prime)=\Lambda^{\,\,\,\nu}_\mu A_\nu(x)$? Shouldn't this be something like: $A_\mu^\prime(x^\prime)=\Lambda^{\,\,\,\nu}_\mu A_\nu(\Lambda x+a)+a_\mu$? Sorry, but I am a very confused soul! Please explain this a bit further. – Faber Bosch Nov 5 '20 at 9:23
• Sorry, I think I made a typo earlier. It's now corrected. No, there won't be any extra $a_{\mu}$ term. For example, think of a scalar field $\phi(x)$. If one performs a translation $x'=x+a$ then we have $\phi'(x')=\phi(x)$ or $\phi'(x) = \phi(x) - a^{\mu}\partial_{\mu}\phi(x)$ (if $a$ is infiniitesimal). The same idea applies for the vector field $A_{\mu}(x)$ but since it is a vector under Lorentz transformation, it picks up a $\Lambda$ term. – Viraj Meruliya Nov 5 '20 at 13:28
• Okay! I see. The key thing to understand is then $A_{\mu}$ is a vector under Lorentz transformations and not the whole Poincare transformation. I was working under a wrong impression then. – Faber Bosch Nov 5 '20 at 14:37