Help Needed for Schwarzschild Solution Model - Romeo

In summary, Romeo is looking for help with a problem with general relativity. He has 13 books on the subject and is stuck. He has been told that a solution avoids tensor calculus, but he is unsure how to find the time for a body to fall from a given distance. He has been told that the Lagrangian may be adapted to a form of the equations of motion, and that the equation for the body's potential energy may be found. However, he is not sure if the expression for the Lagrangian is correct. From MTW, it is possible to find the body's potential energy as a function of the body's position and energy. However, the book does not provide the full derivation for this equation
  • #1
Romeo
13
0
Consider a model in which we let a body fall (radially) into a star, being the simplest example of the schwarzschild solution, in which the angular parts of the solution may be ignored, so that we consider:
[tex]
ds^2 = -(1- GM/r) dt^2 + (1- GM/r)^{-1}dr^2.
[/tex]

I have been told that this may be adapted to a Lagrangian of form:

[tex]

L(r, \dot{r}) = -(1- GM/r) (\frac{dt}{d\lambda})^2 + (1- GM/r)^{-1}(\frac{dr}{d\lambda})^2,

[/tex]

and then solved to find a general time for a body to 'fall' from rest, from a general distance R. How, I don't know (...I love supervisors).

I now have 13 books out upon general relativity, and am royally stuck. It is important that a solution avoids tensor calculus- i am aware that this is possible.

Any help with this would be incredibly appreciated.

Regards

Romeo.
 
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  • #2
Romeo said:
Consider a model in which we let a body fall (radially) into a star, being the simplest example of the schwarzschild solution, in which the angular parts of the solution may be ignored, so that we consider:
[tex]
ds^2 = -(1- GM/r) dt^2 + (1- GM/r)^{-1}dr^2.
[/tex]

I have been told that this may be adapted to a Lagrangian of form:

[tex]

L(r, \dot{r}) = -(1- GM/r) (\frac{dt}{d\lambda})^2 + (1- GM/r)^{-1}(\frac{dr}{d\lambda})^2[/tex]

[/tex]

and then solved to find a general time for a body to 'fall' from rest, from a general distance R. How, I don't know (...I love supervisors).

I now have 13 books out upon general relativity, and am royally stuck. It is important that a solution avoids tensor calculus- i am aware that this is possible.

Any help with this would be incredibly appreciated.

Regards

Romeo.

I think most of what you need should be in the Lagrangian - you shouldn't need any GR since you have the Lagrangian. But I think perhaps your intepretation of what variables the Lagrangain is a funciton of is a bit off, and after looking at the problem even more I think that maybe the Lagrangian itself is a bit off. So you should ask where this Lagrangian came from.

You have a system where [tex]\lambda[/tex] is the "time" variable, and you have two functions

[tex]r(\lambda)[/tex], [tex]t(\lambda)[/tex]

Letting [tex]\dot{r} = \frac{dr}{d\lambda}[/tex] and [tex]\dot{t}=\frac{dt}{d\lambda}[/tex]

we can write

[tex]L(r,\dot{r},t,\dot{t}) = -(1-GM/r)\dot{t}^2+\frac{\dot{r}^2}{1-GM/r}[/tex]

which gives by Lagrange's equations a couple of equations, one of which is

[tex]\frac{d}{dt} (2\frac{\dot{r}}{1-GM/r}) =0[/tex]

This implies that

[tex]\frac{\dot{r}}{1-GM/r} = K[/tex] where K is some constant.


Unfortunately, if we compare this to MTW pg 656, eq 25.16a, or some other source, we cannot veryify this expression, which therefore seems suspect. It's possible that the expression for the Lagrangian is correct, but not if [tex]\lambda[/tex] is supposed to be "proper time". But my guess is that the Lagrangain isn't complete. It really doesn't look like it has the right form.

From MTW, for an infalling body, with [tex]\lambda[/tex] being the proper time, we should have

[tex](\frac{dr}{d\lambda})^2 = E^2 - (1-2GM/r)[/tex]

[I've taken various liberties in converting MTW's equations into a comparable form]

for the correct expression, here E is a constant which depends on the energy of the infalling particle, and E=1 when the particle falls from infinity.

In the "fall from infinity" case, [tex](\frac{dr}{d\lambda})^2 = 2GM/r[/tex]
 
Last edited:
  • #3
Romeo, the book Exploring Black Holes covers all the major scenarios for a Schwarzschild black hole, including the one you mention (on pg. 3-31). It doesn't always give the full derivation, but gives lots of hints. (There's an answer book for it, that's mentioned in the acknowledgments but I haven't seen it.) No tensor calculus in the book. You might want to add this one to your library.
 
  • #4
Pervect- very kind, thank you. It's annoying that MTW is the only text i couldn't get my hands on (on-loan), so have a plethora of others (although Weinberg and Schultz have been useful).

Zanket- Who authors that text? (I'm going to feel very silly if that's the MTW title... )

Best

Romeo
 
  • #5
Anything that talkes about the "effective potential" should enable you to find dr/dtau as a function of r.

http://www.fourmilab.ch/gravitation/orbits/

has the effective potential equations written down, and a neat Java applet. In your case, L is zero (no angular momentum). It also has the auxillary equations that you need to find t and theta (but since L=0, theta is a constant in your case).
 
  • #6
Romeo said:
Zanket- Who authors that text? (I'm going to feel very silly if that's the MTW title... )

The W from MTW and Edwin F. Taylor. For more info look here.
 
  • #7
Zanket's choice is excellent. Taylor & Wheeler's book is very insightful. Easy to follow [as evidenced by my completing all the exercises].
 

Related to Help Needed for Schwarzschild Solution Model - Romeo

1. What is the Schwarzschild solution model?

The Schwarzschild solution model is a mathematical solution to Einstein's field equations of general relativity that describes the gravitational field outside a spherical, non-rotating mass. It is often used to model mass distributions, such as planets and stars.

2. How does the Schwarzschild solution model relate to Romeo's situation?

In Romeo's case, the Schwarzschild solution model can be used to calculate the gravitational field around his body, as he is a spherical, non-rotating mass. This can help determine the amount of force acting on his body, which can then be used to assess his physical condition and potential medical needs.

3. What information is needed to apply the Schwarzschild solution model to Romeo?

To apply the Schwarzschild solution model to Romeo, we would need to know his mass and his distance from the center of the Earth. This information can then be used to calculate the gravitational force acting on him.

4. How accurate is the Schwarzschild solution model?

The Schwarzschild solution model is considered to be highly accurate in most cases, as it takes into account the curvature of space and time caused by a massive object. However, it may not be accurate in extreme cases, such as near a black hole, where other factors come into play.

5. Are there any limitations to the Schwarzschild solution model?

One limitation of the Schwarzschild solution model is that it assumes a non-rotating, spherically symmetric mass. This means it may not accurately describe the gravitational field around irregularly shaped or rotating objects. Additionally, it does not take into account other factors such as the influence of other nearby masses.

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