# Massless Particle Action under Conformal Killing Vector Transformation

• A
• ergospherical
In summary: Also, 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.In summary, the conversation discusses the transformation of the action for a massless particle under a conformal Killing vector, and how this transformation should leave the action invariant. It is shown that the remaining integral vanishes due to the masslessness of the particle, but there are still some unresolved questions regarding the explicit invariance of the world line action under general coordinate transformations.
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?

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
Ibix said:
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##?

Last edited:
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}##

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)
$$S=-\int \mathrm{d} \lambda \sqrt{g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}}?$$

vanhees71 said:
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?

vanhees71
@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.

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

Alexandru123

## 1. What is a massless particle?

A massless particle is a particle that has no rest mass, meaning it travels at the speed of light and has no resistance to acceleration. Examples of massless particles include photons, gluons, and gravitons.

## 2. What is a conformal Killing vector transformation?

A conformal Killing vector transformation is a type of symmetry transformation that preserves the angles between curves on a manifold. It is used to study the behavior of massless particles in curved spacetime.

## 3. How does a conformal Killing vector transformation affect the action of a massless particle?

A conformal Killing vector transformation changes the form of the action of a massless particle, but does not change its physical content. This transformation is used to simplify the equations of motion for massless particles in curved spacetime.

## 4. What is the significance of studying massless particle action under conformal Killing vector transformation?

Studying massless particle action under conformal Killing vector transformation allows us to understand the behavior of massless particles in curved spacetime, which is important for understanding phenomena such as black holes and the expansion of the universe. It also has applications in theoretical physics and cosmology.

## 5. Are there any experimental implications of massless particle action under conformal Killing vector transformation?

While there are currently no direct experimental implications, studying massless particle action under conformal Killing vector transformation can lead to a better understanding of the fundamental laws of physics and potentially inform future experiments.

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