Blackbody temperature in special relativity?

Question

This is from Gravitation Foundations and Frontiers - Padmanabhan

Consider an observer moving with a velocity v through a radiation bath of temperature $T_0$. The observer will see an anisotropy in the radiation field with the effective temperature in a direction making an angle θ with the direction of motion being $T(\theta) = T_0 [\gamma(1 + v \cos \theta)]^{−1}$ with a plus sign

But this answer https://physics.stackexchange.com/a/491096/356911 gives $T(\theta) = T_0 [\gamma(1 - v \cos \theta)]^{−1}$ with a minus sign

What confuses me is Radiative Processes in Astrophysics - George B. Rybicki, Alan P. Lightman - Wiley-VCH (1985), which uses the plus sign at question 4.13

but on its solution:

I want to know if the $\theta \to \theta+\pi$ is necessary, thanks!

Answer 1

It looks like there are two conventions. In the reference frame of the observer you have two angles $\theta_{look}$ and $\theta_k$, where $k$ refers to the wave vector $\vec k$.

They are related by:

$$ \theta_k = \theta_{look} + \pi $$

so that:

$$ \cos\theta_k = -\cos\theta_{look} $$

So when you just say $\theta$, you need to know which one you're talking about.

Choose $\theta=0$, if $T(\theta) > T_0$ you got the right combination of $\pm$ and definition of $\theta$.

If not, switch ONE.

Answer 2

It depends exactly how you define $\theta$ and which frames the observer and the blackbody are defined to be in.

I am guessing that in the first example $T$ is the temperature observed by an observer moving with speed $v$ with respect to a stationary blackbody with temperature $T_0$. The angle $\theta$ then would have to be the angle between the wave vector of the radiation and the velocity vector of the observer.

This equation appears to be correct, since when $\theta=0$ it would mean the observer looks in the direction opposite to their relative motion and $T<T_0$.

The equation in the stack exchange answer you reference is the observed temperature when the angle $\theta$ is the angle between the line joining a moving source and observer and the source motion. i.e. It is the angle between the source velocity and the received radiation wave vector. When $\theta = 0$ in this case, the blackbody is heading straight for the observer and $T > T_0$.