Graduate Massless Particle Action under Conformal Killing Vector Transformation

[ergospherical]

For a massless particle let\begin{align*}
S[x,e] = \dfrac{1}{2} \int d\lambda e^{-1} \dot{x}^{\mu} \dot{x}^{\nu} g_{\mu \nu}(x)
\end{align*}Let $\xi$ be a conformal Killing vector of $ds^2$, then under a transformation $x^{\mu} \rightarrow x^{\mu} + \alpha \xi^{\mu}$ and $e \rightarrow e + \dfrac{1}{4} \alpha e g^{\mu \nu} (L_{\xi} g)_{\mu \nu}$ the action changes (to first order in $\alpha$) as\begin{align*}
S[x,e] &\rightarrow \dfrac{1}{2} \int d\lambda e^{-1} \left(1- \dfrac{1}{4} \alpha g^{\rho \sigma} (L_{\xi} g)_{\rho \sigma} \right)(\dot{x}^{\mu} + \alpha \dot{\xi}^{\mu})(\dot{x}^{\nu} + \alpha \dot{\xi}^{\nu}) g_{\mu \nu}(x) \\ \\

&= S[x,e] + \dfrac{1}{2} \int d\lambda e^{-1} \left( \alpha [\dot{x}^{\mu} \dot{\xi}^{\nu} + \dot{x}^{\nu} \dot{\xi}^{\mu}] - \dfrac{1}{4}\alpha g^{\rho \sigma} (L_{\xi} g)_{\rho \sigma} \dot{x}^{\mu} \dot{x}^{\nu} \right) g_{\mu \nu}(x)
\end{align*}Since $\xi$ is a conformal Killing vector it satisfies $(L_{\xi} g)_{\mu \nu} = \Omega^2 g_{\mu \nu}$ therefore $\dfrac{1}{4} g^{\rho \sigma} (L_{\xi} g)_{\rho \sigma} = \dfrac{1}{4} \Omega^2 {\delta^{\rho}}_{\rho} = \Omega^2$ so\begin{align*}

S[x,e] \rightarrow S[x,e] + \dfrac{1}{2} \int d\lambda e^{-1} \alpha \left( [\dot{x}^{\mu} \dot{\xi}^{\nu} + \dot{x}^{\nu} \dot{\xi}^{\mu}] - \Omega^2 \dot{x}^{\mu} \dot{x}^{\nu} \right)

\end{align*}This transformation should leave $S$ invariant, so how does one show that the remaining integral vanishes? Is it a case of integrating by parts and neglecting a boundary term, in which case, which one?

 

[Ibix]

You seem to have dropped a $g_{\mu\nu}$ from your last expression. Putting it back in, the last term becomes $\Omega^2\dot x^\mu\dot x^\nu g_{\mu\nu}$ which is zero for a massless particle.

The term in square brackets becomes $2\dot x_\mu\dot\xi^\mu$. Noting that $\frac d{d\lambda}(\dot x_\mu\xi^\mu)=\dot x_\mu\dot\xi^\mu+\ddot x_\mu\xi^\mu$ and that $\dot x_\mu\xi^\mu=\mathrm{const}$ it reduces to $-2\ddot x_\mu\xi^\mu$, although I'm not sure that helps.

 

[ergospherical]

You seem to have dropped a $g_{\mu\nu}$ from your last expression. Putting it back in, the last term becomes $\Omega^2\dot x^\mu\dot x^\nu g_{\mu\nu}$ which is zero for a massless particle.

One thing I don't understand is that if $\dot{x}^\mu\dot x^\nu g_{\mu\nu}=0$ then the integrand of $S[x,e]$ is also identically zero hence also $S[x,e]$, which seems wrong. Maybe you are supposed to treat it as the action of a massive particle then at the end take $m \rightarrow 0$?

 

[Ibix]

Yeah, you're right. Never mind.

But $\frac 12\int d\lambda e^{-1}\alpha\Omega^2\dot x^\mu\dot x^\nu g_{\mu\nu}$ would appear to be $\alpha\Omega^2S[x,e]$. Is that useful?

 

[ergospherical]

I don't know haha, not sure what to do with $\dot{x}^{\mu} \dot{\xi}^{\nu} + \dot{x}^{\nu} \dot{\xi}^{\mu}$

 

[vanhees71]

Isn't the upshot of this idea to finally end up with the parametrization independent formulation of the action functional for massive particles (i.e., time-like geodesics)
$$S=-\int \mathrm{d} \lambda \sqrt{g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}}?$$

 

[haushofer]

Isn't the upshot of this idea to finally end up with the parametrization independent formulation of the action functional for massive particles (i.e., time-like geodesics)
$$S=-\int \mathrm{d} \lambda \sqrt{g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}}?$$

Yes, by using the algebraic equation of motion of the einbein. But here TS asks about the explicit invariance of the world line action under general coordinate transformations (target space) in the guise of conformal transformations. I'd say you need to use the equations of motion for e; how else would the particle know it's massless and exhibits conformal symmetry?

 

[PeterDonis]

@Alexandru123 please use the LaTeX Guide (there's a link to it at the bottom left of the post window) to get the correct LaTeX code for formatting math.

 

[Alexandru123]

Make $d\xi/d\lambda = (d\xi^{\mu}/dx^{\mu})*dx^{\mu}/d\lambda$. Which will allow you to put the 2 $\dot{x}$ in evidence. ( this is the chain rule with $x^{\mu}$ until I find out to write out formulas in latex here)
Then you'll get $2* \nabla \cdot \xi - \Omega^2$ in the integrand.
There is one last theorem where $(1/2) \nabla \cdot \xi = \Omega^2$ by tracing $L_{\xi} g= \Omega^2 g$. However here is where I get stuck.

 

[PeterDonis]

@Alexandru123 if you are trying to write a divergence in LaTeX, i.e., $\nabla \cdot \xi$, use "\nabla \cdot" as the operator.