General Relativity - Schwarzschild Metric

In summary, the conversation discusses finding the relationship between the rate of change of angular position of a spaceship in a circular orbit and the proper time and radius of the orbit. The suggested approach is to use the Lagrangian and Euler-Lagrange equation, which leads to the conclusion that the rate of change of angular position is inversely proportional to the radius of the orbit and directly proportional to the square root of the masses involved. It is also suggested to write out the full Lagrangian and EOM for better understanding.
  • #1
bon
559
0

Homework Statement



A spaceship is moving without power in a circular orbit about an object with mass M. The radius of the orbit is R = 7GM/c^2

(1) Find the relation between the rate of change of angular position of the spaceship and the proper time and radius of the orbit.


Homework Equations





The Attempt at a Solution



So I wrote down the Lagranian from the Schw. metric and finding the EOM for phi i get

d/dtau (2r^2 sin^theta dphi/dtau) = 0

so i can conclude r^2 dphi/dtau = constant

so r^2 dphi/dt times dt/dtau = constant

is this right? and is this what they want? I'm confused!

Thanks
 
Physics news on Phys.org
  • #2
for your question! It seems like you are on the right track with your approach. However, I would suggest writing out the full Lagrangian and EOM for this problem, as it will help clarify the relationship between the rate of change of angular position and the proper time and radius of the orbit.

The Lagrangian for this system would be:

L = 1/2 m (dx/dt)^2 + 1/2 m (dy/dt)^2 + 1/2 m (dz/dt)^2 - GMm/r

where m is the mass of the spaceship, dx/dt and dy/dt are the velocities in the x and y directions, and r is the distance between the spaceship and the object with mass M.

To find the EOM for phi, we can use the Euler-Lagrange equation:

d/dt (dL/d(phi dot)) - dL/d(phi) = 0

Plugging in the Lagrangian and simplifying, we get:

d/dt (m r^2 dphi/dt) = 0

This means that the quantity m r^2 dphi/dt is conserved, which is what you found in your attempt. However, since we are dealing with the proper time (tau) rather than coordinate time (t), we need to use the chain rule to relate dphi/dt to dphi/dtau:

dphi/dtau = dphi/dt * dt/dtau

Using the relation between dt/dtau and the radius of the orbit (R = 7GM/c^2), we can rewrite the conserved quantity as:

m r^2 dphi/dtau = constant * (c^2/7GM)

So, to answer the original question, the rate of change of angular position (dphi/dtau) is inversely proportional to the radius of the orbit (r) and is directly proportional to the square root of the mass of the spaceship (m) and the square root of the mass of the object it is orbiting (M).

I hope this helps clarify the relationship between the rate of change of angular position, proper time, and radius of the orbit! Let me know if you have any further questions.
 

1. What is the Schwarzschild metric?

The Schwarzschild metric is a solution to Einstein's field equations in general relativity that describes the curvature of spacetime in the presence of a non-rotating, spherically symmetric mass. It is used to describe the gravitational field outside of a spherical, non-rotating mass such as a star or planet.

2. What is the significance of the Schwarzschild metric?

The Schwarzschild metric is significant because it was the first exact solution to Einstein's field equations and played a crucial role in the development of general relativity. It also accurately describes the motion of objects in the vicinity of massive bodies, such as the motion of planets in the solar system.

3. How does the Schwarzschild metric differ from Newton's law of gravity?

The Schwarzschild metric differs from Newton's law of gravity in that it takes into account the curvature of spacetime caused by massive bodies, whereas Newton's law assumes that gravity is a force acting between masses. The Schwarzschild metric predicts the correct amount of bending of light and the precession of orbits, while Newton's law does not.

4. What is the event horizon in the Schwarzschild metric?

The event horizon in the Schwarzschild metric is the boundary around a massive body beyond which nothing, including light, can escape. It is also known as the point of no return. The size of the event horizon is directly related to the mass of the object, with larger masses having larger event horizons.

5. How has the Schwarzschild metric been tested and confirmed?

The Schwarzschild metric has been tested and confirmed through various observations and experiments. One of the most famous examples is the observation of the bending of starlight near the sun during a solar eclipse, which confirmed the predicted amount of bending from the Schwarzschild metric. Other tests include the measurement of the precession of Mercury's orbit and the detection of gravitational waves from merging black holes, which match the predictions of the Schwarzschild metric.

Similar threads

  • Advanced Physics Homework Help
Replies
10
Views
2K
  • Advanced Physics Homework Help
Replies
10
Views
1K
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
20
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Special and General Relativity
Replies
9
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
798
  • Special and General Relativity
Replies
4
Views
907
  • Advanced Physics Homework Help
Replies
2
Views
2K
  • Advanced Physics Homework Help
Replies
11
Views
2K
Back
Top