Calculating Time to Singularity in a Schwarzschild Black Hole

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Homework Help Overview

The discussion revolves around calculating the time elapsed for an observer falling into a Schwarzschild black hole, starting from rest at a specific radial coordinate. The participants explore the implications of the Schwarzschild metric and the equations of motion relevant to the scenario.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of the Schwarzschild metric and the Euler Lagrange equations to derive equations of motion. There are attempts to substitute energy terms into the metric and questions about the implications of boundary conditions at infinity versus the starting point at r = 10M.

Discussion Status

Multiple approaches to the problem are being explored, with some participants expressing confusion over the substitutions and assumptions made regarding energy and the behavior of the observer at different radial distances. Guidance has been offered regarding the use of the metric and the significance of boundary conditions, but no consensus has been reached on the correct approach.

Contextual Notes

There is a noted confusion regarding the initial conditions, particularly the velocity of the observer at the starting point compared to at infinity. Participants are also grappling with the implications of their substitutions and the resulting integrals.

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An observer falls radially into a Schwarzschild black hole of mass [itex]M[/itex]. She starts from rest (i.e. [itex]\frac{dr}{d \tau} = 0[/itex]) at [itex]r = 10M[/itex]. How much time elapses on her clock before she hits the singularity?

Is my first step to take the metric equation
[itex]ds^2 = - ( 1 - \frac{2M}{r} ) dt^2 + \frac{dr^2}{(1 - \frac{2M}{r})} + r^2 ( d \theta^2 + \sin^2{\theta} d \phi^2 )[/itex]
and find the Euler Lagrange equations of motion?

Or is this barking up the wrong tree?
 
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just did this in class today and after substituting E=(1-2M/r)dt/dT and d(phi) and d(theta)=0 into the metric and after some algebra and assuming E=1 so dr/dT=0 at infinity we got:
(dr/dT)^2=2M/r and therefore
T=integral(sqrt(2M/r) dr) from R0 to R1

please let me know if this is what your looking for or if there is a better easier way
 
VanOosten said:
just did this in class today and after substituting E=(1-2M/r)dt/dT and d(phi) and d(theta)=0 into the metric and after some algebra and assuming E=1 so dr/dT=0 at infinity we got:
(dr/dT)^2=2M/r and therefore
T=integral(sqrt(2M/r) dr) from R0 to R1

please let me know if this is what your looking for or if there is a better easier way

ok. I'm a bit confused about the E substitution, I get a metric eqn of

[itex]ds^2 = - (1 - \frac{2M}{r}) dt^2 + \frac{dr^2}{1- \frac{2M}{r}}[/itex]

Now if I didvide through by [itex]d \tau^2[/itex] I get:

[itex]\frac{ds^2}{d \tau^2} = - ( 1- \frac{2M}{r} ) ( \frac{dt}{d \tau})^2 + \frac{1}{1- \frac{2M}{r}} ( \frac{dr}{d \tau} )^2[/itex]

How do I substitute E into that? Do I need to multiply through by [itex]1- \frac{2M}{r}[/itex] first?

Doing this, I get

[itex](1-\frac{2M}{r})(\frac{ds}{d \tau})^2 = - E^2 + (\frac{dr}{d \tau})^2[/itex]
Why do we know to find E by looking at dr/dT at infinity?

Also, I assum you changed [itex]ds= c d \tau = d \tau[/itex]
Thank you.
 
Last edited:
substitute (1-2M/r)(dt/dT)^2=(E^2)/(1-2M/r) to get
(dr/dT)^2=E^2-(1-2M/r)

and we look at r at infinity because at this limit dr/dT=0 which tells us E must be 1
also yes i assumed ds = dT
hope that helps
 
VanOosten said:
substitute (1-2M/r)(dt/dT)^2=(E^2)/(1-2M/r) to get
(dr/dT)^2=E^2-(1-2M/r)

and we look at r at infinity because at this limit dr/dT=0 which tells us E must be 1
also yes i assumed ds = dT
hope that helps

Yes but surely we don't want dr/dT=0 at r=infinity. The question tells us that dr/dT=0 at r=10M.
If you follow this through you get E^2=4/5.
This then leads to the following (much more complicated) integral that I'm sure must be wrong!
[itex]\tau = \int_{10M}^0 ( \frac{2M}{r} - \frac{1}{5} ) ^{\frac{1}{2}} dr[/itex]
 
bump.
 

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