Defining Emitter vs Observer for Schwarzschild Metric

[MarkovMarakov]

Let's say we are working with the Schwarzschild metric and we have an emitter of light falling into a Schwarzschild black hole.

Suppose we define the quantity $u=t- v$ where $dv/dr= 1/(1-r_{s}/r)$ where $r_s$ is the Schwarzschild radius. What is the $u$ as observed by the emitter? I just need a *definition of $u_e$*. I have problems identifying the quantities as measured by an observer at large $r$ and that of the emitter. Would I be right at least to say that $t_{e}=\tau$ the proper time? Many thanks.

_____

In fact, I've been told that

$$du_o/d\tau=du_e/d\tau$$

Why is it?

 

[Dale]

Let's say we are working with the Schwarzschild metric and we have an emitter of light falling into a Schwarzschild black hole.

Suppose we define the quantity $$u=t- v$$ where $$dv/dr= 1/(1-r_{s}/r)$$ where $r_s$ is the Schwarzschild radius. What is the $u$ as observed by the emitter? I just need a *definition of $u$*.

I don't understand. You already gave the definition of u as u=t-v.

 

[MarkovMarakov]

@DaleSpam: You are right. I have missed out the subscript $e$. I meant $u$ as measured by the emitter.

 

[Dale]

Well, v is just some function of r and so u is just some function of t and r. I don't see how it is something measured by anyone.

Please fix your LaTeX.

 

[MarkovMarakov]

@DaleSpam: LaTex fixed. I suppose $u_e$ would be in terms of $t_e,r_e$ measured by the emitter, no? But then what is its explicit form?

 

[Dale]

Well, what coordinate system do you want to use for the emitter? Then you would just have to substitute in the transforms for the emitter's coordinates.

 

[MarkovMarakov]

@DaleSpam: the proper time is that of the emitter. I am still not sure how to do the substitution though. (Sorry about my daftness. :( )