# Showing the Poincare invariance of a term

I know that this is a simple question! But I would like to know the details. How we can show that the term

$$A_\mu(x)\dot{x}^\mu$$

is global and local Poincare invariant? Where $A_\mu(x)$ is supposed to be a four-covector and This term is a part of reparametrization-invariant square-root Lagrangian $\mathcal{S}=\int d\tau[-mc\sqrt{-(\dot x^\mu)^2}+\frac{e}{c}A_\mu\dot x^\mu]$ where $x^\mu$ are parametric equations of the physical trajectory $x_i(t)$. We assume $\frac{dt}{d\tau}>0$ and $A^\mu$ is electromagnetic four-vector potential.

• Maybe I'll be ashamed of this comment, but the statement does not look obvious. Could you give a reference? And what is $\dot{x^\mu}$? This does not even look Lorentz-invariant. Commented Dec 7, 2013 at 15:44
• Dear @akhmeteli, $\dot x^\mu \equiv \frac{d x^\mu}{d\tau}$. In fact $x^\mu$ are parametrized variables. You can assume $A^\mu$ as electromagnetic four-potential and $x^\mu$ as parametrized coordinates. Here we define Poincare transformation as $x^\prime=\Lambda x+a$. Thanks for your help. :) Commented Dec 7, 2013 at 22:50
• Well, I thought the dot was for time derivative; that was quite confusing:-) Is your expression for some particle trajectory? Or what parametrization is used? Why don't you disclose all you know about this problem? Commented Dec 8, 2013 at 1:00
• @Valid Nikoofard: Now it looks almost reasonable:-). So do I get it right that actually you want to prove Poincare-invariance of $∫dτA_μ\dot{x}^\mu$, rather than $A_μ\dot{x}^\mu$? Commented Dec 8, 2013 at 15:17
• @VahidNikoofard: The question might become clearer to the readers if you explicitly define your notation, especially what physical time quantities $t$ and $\tau$ are. Commented Dec 8, 2013 at 16:29

So I assume you actually need to prove Poincare invariance of $\int d\tau A_\mu\dot{x}^\mu$ for a particle trajectory, rather than invariance of $A_\mu\dot{x}^\mu$, but the former expression is equal to $\int_a^b dx^\mu A_\mu$, where $a$ and $b$ are the initial and the final points of the trajectory, and Poincare invariance is indeed almost obvious for this expression.

• Dear @akhmeteli, thanks you VERY much for your answer! :) Commented Dec 9, 2013 at 11:59

If you're familiar with differential forms, then akhmeteli's answer is great, especially if you want to generalize to curved geometries.

Let's try to be notationally and mathematically precise without using forms and be as explicit as possible.

Let a vector potential $A = (A^\mu) = (A^0, \mathbf A)$ be given. Consider a parametrized path $x(\lambda) = (x^\mu(\lambda)) = (x^0(\lambda), \mathbf x(\lambda))$ with $\dot x^0(\lambda)>0$ in some inertial frame $F$ with coordinates $x^\mu$. I'm using the symbol $\lambda$ as the parameter along the path to emphasize that it need not be time or proper time in general. Also, an overdot will always denote the derivative with respect to the argument of the function in question.

Then, in a new inertial frame $F'$ whose coordinates $x'^\mu$ are related to the $F$ coordinates by a Poincare transformation, \begin{align} x'^\mu = \Lambda^\mu_{\phantom\mu\nu}x^\nu + a^\mu, \end{align} the vector potential and parametrized curve transform as follows: \begin{align} A'_\mu(x) &= \Lambda_\mu^{\phantom\mu\nu} A_\nu(\Lambda^{-1}(x-a)) \\ x'^\mu(\lambda) &= \Lambda^\mu_{\phantom\mu\nu} x^\nu(\lambda) + a^\mu \end{align} Now, the action you wrote down is that of a charged particle moving in an externally-applied vector potential $A_\mu$. In the notation I'm using here, the second term would be written as follows for an arbitrary parametrization: \begin{align} I[A, x;\lambda_1,\lambda_2]=\frac{e}{c}\int_{\lambda_1}^{\lambda_2} d\lambda \, A_\mu(x(\lambda)) \dot x^\mu(\lambda) \end{align} Now, for convenience, suppress the explicit $\lambda_1, \lambda_2$ arguments on $I$. So how do we show that $I[A,x]$ is Poincare-invariant? Namely, how do we show that \begin{align} I[A',x'] = I[A,x]? \end{align} Well, we compute; \begin{align} \frac{c}{e}I[A',x'] &= \int d\lambda\, A'_\mu(x'(\lambda)) \dot x'^\mu(\lambda) \\ &= \int d\lambda\, \Lambda_\mu^{\phantom\mu\alpha}A_\alpha\Big(\Lambda^{-1}\big[(\Lambda x(\lambda)+a)-a\big]\Big)\cdot\Lambda^\mu_{\phantom\mu\beta}\dot x^\beta(\lambda) \\ &= \int d\lambda\,\Lambda_\mu^{\phantom\mu\alpha}\Lambda^\mu_{\phantom\mu\beta}A_\alpha(x(\lambda))\dot x^\beta(\lambda) \end{align} but recall that \begin{align} \Lambda_\mu^{\phantom\mu\alpha}\Lambda^\mu_{\phantom\mu\beta} &= \delta^\alpha_\beta \end{align} so we get \begin{align} \frac{c}{e}I[A',x'] &= \int d\lambda\,A_\alpha(x(\lambda))\dot x^\alpha(\lambda) \\ &= \frac{c}{e}I[A,x] \end{align} as desired.

Interesting Aside. The charged particle Lagrangian is actually also reparametrization-invariant, namely given any sufficiently smooth, invertible function $f:(s_1, s_2)\to (\lambda_1, \lambda_2)$, we have \begin{align} I[A,x;\lambda_1, \lambda_2] = I[A, x\circ f,s_1, s_2] \end{align} I strongly encourage you to try to prove this; it's very instructive.

• Dear @joshphysics, why does parametrized curve transform in this way? Is there no translation? Commented Dec 9, 2013 at 11:59
• @VahidNikoofard That was an error, fixed! Commented Dec 9, 2013 at 16:35