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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, $$\int \left(-m +eA_{\mu}u^\mu\right)d\tau-\int d^4x\frac{1}{4}\left(F_{\mu\nu}F^{\mu\nu}\right)$$

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  • $\begingroup$ It's not. Like the inner product of two position vectors under Galielean transformation. You should look at displacement, or any derivative of the position, not at the position itself $\endgroup$
    – basics
    Commented Sep 9, 2022 at 20:14

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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 $A'_{\mu}(x') = \Lambda_{\mu}^{\nu}A_{\nu}(x)$.

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  • $\begingroup$ 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. $\endgroup$ Commented Nov 5, 2020 at 9:23
  • $\begingroup$ 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. $\endgroup$ Commented Nov 5, 2020 at 13:28
  • $\begingroup$ 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. $\endgroup$ Commented Nov 5, 2020 at 14:37
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Lets look at the displacement vector with the components explicitly written out as $\Delta s = (x^\mu - 0^\mu)$. Under a full Poincare transformation, this coordinate difference would transform as $\Delta s^{´} = (\Lambda x^\mu+a^\mu - (\Lambda 0^\mu+a^\mu))$, and the invariance of the inner product for 4 - vectors would then follow from invariance under homogenous Lorentz transformations.

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