- #1

- 683

- 406

- TL;DR Summary
- I've read that speed of light for a Rindler observer is not constant.

I've read that speed of light for a Rindler observer is not constant. I wasn't sure why and I tried to do this as an exercise for myself, I think I came with the correct answer, but I'm not sure, is the following argument correct?

Let's describe the (flat) space-time using Rindler coordinates ##\tau, \rho## and let's assume a (non-inertial) observer moving by the worldline ##\rho=\rho_0##. I want to compute the speed of light for that observer.

First of all, to define the velocity we need to talk about distances and time. In SR the time measured by any observer is the "length" of its worldline, ##s##. Therefore because our observer's worldline is defined as ##d\rho=0## the Rindler metric gives us ##ds=\rho_0 d\tau##. Now the distance that this observed sees is the "negative length" between two points at equal time, i.e., equal ##\tau##. So again the Rindler metric gives ##dl=-ds=d\rho##. Finally, the velocity is the change in space divided by the change on time, so

$$v=\frac{1}{\rho_0}\frac{d\rho}{d\tau}$$

Because a lightray has ##ds=0## we find that

$$v=\frac{1}{\rho_0}\frac{d\rho}{d\tau}=\frac{\rho}{\rho_0}$$

Which depends on the ##\rho## coordinate.

Let's describe the (flat) space-time using Rindler coordinates ##\tau, \rho## and let's assume a (non-inertial) observer moving by the worldline ##\rho=\rho_0##. I want to compute the speed of light for that observer.

First of all, to define the velocity we need to talk about distances and time. In SR the time measured by any observer is the "length" of its worldline, ##s##. Therefore because our observer's worldline is defined as ##d\rho=0## the Rindler metric gives us ##ds=\rho_0 d\tau##. Now the distance that this observed sees is the "negative length" between two points at equal time, i.e., equal ##\tau##. So again the Rindler metric gives ##dl=-ds=d\rho##. Finally, the velocity is the change in space divided by the change on time, so

$$v=\frac{1}{\rho_0}\frac{d\rho}{d\tau}$$

Because a lightray has ##ds=0## we find that

$$v=\frac{1}{\rho_0}\frac{d\rho}{d\tau}=\frac{\rho}{\rho_0}$$

Which depends on the ##\rho## coordinate.

Last edited: