Calculating Time to Singularity in a Schwarzschild Black Hole

In summary, an observer falls radially into a Schwarzschild black hole of mass M. She starts from rest (i.e. \frac{dr}{d \tau} = 0) at r = 10M. She hits the singularity after a time t.
  • #1
latentcorpse
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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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  • #2
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
 
  • #3
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:
  • #4
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
 
  • #5
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]
 
  • #6
bump.
 

1. What is a Schwarzschild Black Hole?

A Schwarzschild black hole is a type of black hole that is defined by its "event horizon", the point of no return for anything that enters its gravitational field. It is named after the German physicist Karl Schwarzschild, who first described this type of black hole in 1916.

2. How is a Schwarzschild Black Hole formed?

A Schwarzschild black hole is formed when a massive star dies and its core collapses under its own gravity. As the core collapses, it becomes infinitely dense and its escape velocity exceeds the speed of light, creating a black hole.

3. What is the size of a Schwarzschild Black Hole?

The size of a Schwarzschild black hole is determined by its event horizon, which is directly related to its mass. The event horizon of a black hole with the mass of the sun would be around 3 kilometers in diameter, while a black hole with the mass of a million stars would have an event horizon around 3 million kilometers in diameter.

4. Can anything escape from a Schwarzschild Black Hole?

Once something crosses the event horizon of a Schwarzschild black hole, it cannot escape. This includes light, which is why black holes are invisible. However, objects that are far enough away from the black hole can still escape its gravitational pull.

5. What is the significance of the Schwarzschild radius?

The Schwarzschild radius is the distance from the center of a black hole to its event horizon. It is significant because it marks the point of no return for anything that enters a black hole's gravitational field. It is also used to calculate the size and mass of a black hole.

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