If you consider the Schwarzschild solution, The metric coefficients are (1-2m/r) where m=\kappa M/c^2, \kappa the Newtonian gravitational constant. If I go and calculate the square of the period T^2=\frac{r^3}{m}, it looks like Kepler's law but m has c^2 in it so it can never be Kepler's law. In other words, you can't derive Kepler's law from the Schwarzschild solution.

Copilot

###### Sent by Copilot:

You’re correct that the Schwarzschild solution introduces relativistic corrections that modify the classical Kepler’s laws. Let’s delve into this:

### Schwarzschild Metric

The Schwarzschild metric describes the spacetime geometry around a non-rotating, spherically symmetric mass. The metric coefficients are given by:

g_(tt)=1−2m /r

where ( m = \frac{\kappa M}{c^2} ) is the gravitational radius.

### Orbital Period in Schwarzschild Geometry

In the Schwarzschild geometry, the effective potential and the equations of motion for a test particle are more complex than in Newtonian gravity. However, for circular orbits, we can derive an expression for the orbital period that includes relativistic corrections.

### Kepler’s Third Law in Newtonian Gravity

In Newtonian gravity, Kepler’s third law states:

T2=4π^2r^3/\kappa M

### Relativistic Corrections

In the context of the Schwarzschild solution, the orbital period ( T ) for a circular orbit can be approximated by:

T^2≈(4π^2r^3/\kappa M)(1−3\kappa M/c2r)

This shows that the period ( T ) is modified by a factor that depends on the gravitational radius ( \frac{\kappaM}{c^2} ) and the orbital radius ( r ). The term ( \left(1 - \frac{3\kappaM}{c^2r}\right) ) represents the relativistic correction.

### Why Kepler’s Law Can’t Be Derived Directly

As you pointed out, the presence of ( c^2 ) in the Schwarzschild metric means that the classical form of Kepler’s third law ( T^2 \propto r^3 ) is modified. The relativistic corrections become significant when the orbital radius ( r ) is comparable to the Schwarzschild radius ( \frac{2\kappa M}{c^2} ). In such cases, the simple ( T^2 = \frac{r^3}{m} ) form no longer holds, and the relativistic effects must be taken into account.

### Conclusion

While the Schwarzschild solution provides a more accurate description of orbital dynamics in strong gravitational fields, it does not reduce to the classical Kepler’s laws due to the relativistic corrections. These corrections are essential for accurately describing the motion of objects in the vicinity of massive bodies like black holes.

If you have more questions or need further clarification on any aspect, feel free to ask!

You

##### Sent by you:

In the Schwarzschild solution \kappa M always appears as \kappa M/c^2 so how do we separate out the c^2 to get the first term in your expression for T^2?

Copilot

###### Sent by Copilot:

Great question! Let’s delve into how we handle the term ( \frac{\kappa M}{c^2} ) in the context of the Schwarzschild solution and how it relates to the orbital period.

### Schwarzschild Metric and Orbital Period

The Schwarzschild metric is given by:

ds^2=−(1−2\kappa M/c2r)c^2 dt^2+(1−2\kappa M/c^2r)−1dr^2+r^2dθ^2+r^2sin^2θ dϕ^2

For circular orbits in the equatorial plane ((\theta = \frac{\pi}{2})), the effective potential and the equations of motion can be derived from this metric. The relativistic correction to the orbital period ( T ) can be approximated by:

T^2≈(4π^2r^3/\kappa M)(1−3\kappa M/c2r)

### Separating ( c^2 )

To understand how ( c^2 ) factors into this expression, let’s break it down:

**Classical Term**:The term ( \frac{4\pi^2 r^3}{\kappa M} ) is the classical Keplerian period squared, derived from Newtonian gravity.

**Relativistic Correction**:The term ( \left(1 - \frac{3\kappa M}{c^2r}\right) ) introduces the relativistic correction. Here, ( \frac{\kappa M}{c^2} ) is the gravitational radius (or Schwarzschild radius divided by 2).

### Simplifying the Expression

To isolate the classical term and understand the correction, consider the following steps:

**Identify the Gravitational Radius**:Let ( r_s = \frac{2\kappa M}{c^2} ) be the Schwarzschild radius. Then ( \frac{\kappa M}{c^2} = \frac{r_s}{2} ).

**Substitute and Simplify**:Substitute ( \frac{\kappa M}{c^2} ) into the correction term:

1−3\kappa M/c^2r

**Combine Terms**:The full expression for ( T^2 ) becomes:

T^2≈4π^2r^3\kappa M(1−3rs/2r)

### Conclusion

The term ( \frac{\kappa M}{c^2} ) naturally appears in the relativistic correction due to the nature of the Schwarzschild metric. The classical term ( \frac{4\pi^2 r^3}{\kappa M} ) remains, but the correction factor ( \left(1 - \frac{3\kappa M}{c^2r}\right) ) modifies it to account for relativistic effects.

This shows that while the classical form of Kepler’s third law is modified in the relativistic context, the underlying principles still apply, with additional terms to account for the effects of general relativity.

If you have further questions or need more details, feel free to ask!

You

##### Sent by you:

No. The metric to consider is ds^2=-F^2dt^2+G^2d\phi^2 where F=G'=\sqrt{1-2\mu/G}, with dr=G'd\rho .The structure function is formed by taking the derivative with respect to \rho to obain FF'dt^2-GG'd\phi^2=0 . The period is T=\sqrt{\frac{G^3}{\mu}}. But since \mu=GM/c^2, although it looks like Kepler's third law it isn't because it contains c^2!

Copilot

###### Sent by Copilot:

Thank you for the clarification. Let’s delve into the details of the metric and the period calculation you provided.

### Metric and Structure Function

The metric you are considering is:

ds2=−F^2dt^2+G^2dϕ^2

where ( F = G’ = \sqrt{1 - \frac{2\mu}{G}} ) and ( dr = G’ d\rho ).

### Structure Function

The structure function is obtained by taking the derivative with respect to ( \rho ):

FF′dt^2−GG′dϕ^2=0

This implies a relationship between the time and angular components of the metric.

### Period Calculation

The period ( T ) is given by:

T^3=G^3/μ

where ( \mu = \frac{\kappa M}{c^2} ).

### Comparison with Kepler’s Third Law

In classical mechanics, Kepler’s third law for the period ( T ) of an orbit is:

T^2=4π^2r^3/\kappa M

In your case, the period ( T ) is:

T=G^3/μ=G^3/\kappa M/c^2=c^2G^3/\kappa M

### Relativistic Correction

Although the form of the period ( T ) resembles Kepler’s third law, the presence of ( c^2 ) indicates a relativistic correction. This correction arises because the metric you are using incorporates relativistic effects, specifically through the term ( \mu = \frac{\kappa M}{c^2} ).

### Conclusion

*While the expression for the period ( T ) looks similar to Kepler’s third law, it is not the same due to the inclusion of the speed of light ( c ). This inclusion reflects the relativistic nature of the system, distinguishing it from the purely Newtonian case.*

**Your observation is correct: the presence of ( c^2 ) means that this period cannot be derived from classical Keplerian mechanics alone. It incorporates the effects of general relativity, which modify the orbital dynamics.**

## Comentarios