I Can Rindler coordinates work when a=0 without looking ugly?

  • Thread starter Anwyl
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Rindler coordinates result in T=0/0 when a=0. Is there a clean way to fix that?
Rindler coordinates are nice, but they fall apart when a=0, where ##T=\frac{sinh(at)}{a}=\frac{0}{0}##. Is there a good way to fix that?

Intuitively I'd want to do out the taylor expansion, divide by a, then collapse it back to... something...

$$T=\frac {\sinh(at)} {a}=\frac{ \sum_{n=0}^\infty {\frac {(at)^{2n+1}}{ (2n+1)!}}} {a} = \sum_{n=0}^\infty {\frac {a^{2n}t^{2n+1}} {(2n+1)!}} = ???$$

Is there any simplification of that taylor expansion to something sane that doesn't result in 0/0?

Is there another approach which might work better?
 
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Is there another approach which might work better?
If ##a = 0## there is no such thing as "Rindler coordinates"; you just have Minkowski coordinates, since you have objects at rest in the coordinates moving inertially (zero proper acceleration). That's what the 0/0 issue is really telling you.
 
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Intuitively I'd want to do out the taylor expansion, divide by a, then collapse it back to... something...
For ##n = 0## your expression with ##a = 0## is ##a^0 t^1 / 1! = t##. For ##n > 0## your expression gives zero if ##a = 0##. What does that tell you?
 
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Summary: Rindler coordinates result in T=0/0 when a=0. Is there a clean way to fix that?

Is there another approach which might work better?
Just take the limit as a goes to 0.
 
For ##n = 0## your expression with ##a = 0## is ##a^0 t^1 / 1! = t##. For ##n > 0## your expression gives zero if ##a = 0##. What does that tell you?
If ##a = 0## it gives ##T = t##, which is the desired result, since it should look just like flat Minkowski coordinates in that case.
I'm just hoping for a way of writing it which is clearly ##T = t## when ##a = 0## and ##T = \frac{sinh(at)}{a}## otherwise, without the gap in the domain of the function, or doing something like
##T = \begin{cases}t & a=0 \\ \frac{sinh(at)}{a} & a \neq 0\end{cases}##
 
24,303
5,989
If ##a = 0## it gives ##T = t##, which is the desired result, since it should look just like flat Minkowski coordinates in that case.
Exactly.

I'm just hoping for a way of writing it which is clearly #T = t## when ##a = 0## and ##T = \frac{sinh(at)}{a}## otherwise
I'm confused. Isn't that what you just agreed that your Taylor expansion shows?

If you want another way of looking at it, take the suggestion @Dale made and figure out what the limit of ##\frac{sinh(at)}{a}##, considered as a function of ##a##, is when ##a \rightarrow 0##. An easy way to evaluate the limit is L'Hopital's rule.
 
I'm confused. Isn't that what you just agreed that your Taylor expansion shows?
Yeah, I think I'm just trying to oversimplify the equation. It works fine with the sum there, but I'm always suspicious when working with infinite sums that I'll do something subtly wrong, and it feels like there should be some elegant continuous function there.
 

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