Conserving Quantity in Schwarzschild Metric

In summary, the conversation revolved around the concept of conserved quantities in the Schwarzschild metric. It was mentioned that along a geodesic, ##K=k^uV_u## is constant, where ##V^u## is the tangent vector to an affinely parameterized geodesic. The example of the Schwarzschild metric was given, where ##K^u=(1,0,0,0)## and ##V^u=(\dot{t},\dot{r},\dot{\theta},\dot{\phi})##, leading to the conclusion that ##K= (1-\frac{2GM}{r})\dot{t}## is a conserved quantity. The question was
  • #1
binbagsss
1,254
11

Homework Statement



Conserved quantity Schwarzschild metric.

Homework Equations

The Attempt at a Solution


[/B]
##\partial_u=\delta^u_i=k^u## is the KVF ##i=1,2,3##

We have that along a geodesic ##K=k^uV_u## is constant , where ##V^u ## is the tangent vector to some affinely parameterised geodesic.

For example if we take the Schwarzschild metric, ##K^u=(1,0,0,0)## , ##V^u=(\dot{t},\dot{r},\dot{\theta},\dot{\phi})## so we get ##K= (1-\frac{2GM}{r})\dot{t}## is conserved, for example.

where dot denotes a derivative with respect to some affine parameter ##s##

QUESTION:

What here is to say that ##s## is an affine parameter?
I.e- why is ##V^u=(\dot{t},\dot{r},\dot{\theta},\dot{\phi})## the tangent vector to an affinely parameterised geodesic?
 
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  • #2
You assumed it to be an affine parameter and then wonder why it is affine?

If it is not an affine parameter ##k^u V_u## will not be a conserved quantity.
 
  • #3
Moderator's note: moved to homework.
 
  • #4
Orodruin said:
You assumed it to be an affine parameter and then wonder why it is affine?

If it is not an affine parameter ##k^u V_u## will not be a conserved quantity.

ahh apologies got it yes, it is in the proof showing that ##k^u V_u## will not be a conserved quantity is a conserved quantity.
 

1. What is the Schwarzschild metric?

The Schwarzschild metric is a mathematical formula that describes the curvature of space-time in the vicinity of a non-rotating, spherically symmetric object, such as a black hole. It is a key concept in Einstein's theory of general relativity and is used to describe the gravitational field around a massive object.

2. How does conservation of quantity apply to the Schwarzschild metric?

In the Schwarzschild metric, there are two quantities that are conserved: energy and angular momentum. This means that the total amount of energy and angular momentum of a particle moving in the gravitational field of a massive object will remain constant throughout its trajectory.

3. Why is it important to conserve quantity in the Schwarzschild metric?

Conservation of quantity in the Schwarzschild metric is important because it allows us to accurately predict the behavior of particles in the gravitational field of a massive object. Without this conservation, the laws of physics would not hold and our understanding of the universe would be incomplete.

4. Are there any exceptions to the conservation of quantity in the Schwarzschild metric?

Yes, there are some exceptions to the conservation of quantity in the Schwarzschild metric. For example, if a particle falls into a rotating black hole, its energy and angular momentum can change. This is due to the effects of frame dragging, which is caused by the rotation of the black hole.

5. How does the Schwarzschild metric affect the concept of time dilation?

The Schwarzschild metric predicts that time will pass more slowly near a massive object, such as a black hole. This is known as gravitational time dilation and is a result of the curvature of space-time. The closer a particle is to the massive object, the slower time will appear to pass for that particle.

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