# Invariance of inner product under Poincaré 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, $$\int \left(-m +eA_{\mu}u^\mu\right)d\tau-\int d^4x\frac{1}{4}\left(F_{\mu\nu}F^{\mu\nu}\right)$$

• 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 Commented Sep 9, 2022 at 20:14

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. Commented Nov 5, 2020 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. Commented Nov 5, 2020 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. Commented Nov 5, 2020 at 14:37
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.