Obtaining the Orbit equation (Effective Potential) from the Newtonian metric

[Arman777]

We can write the Newtonian metric in the form of

$$ds^2 = -(1 - 2M/r)dt^2 + (1+2M/r)[dr^2 + r^2d\Omega^2]$$

In order to obtain the orbit equation I have written the constant of motion,

$$e = (1 - 2M/r)(\frac{dt}{d\tau})$$

and

$$l = r^2sin^2(\theta)(\frac{d\phi}{d\tau})$$

I can divide the metric to $$d \tau^2$$ and write

$$(\frac{ds}{d\tau})^2 = -(1 - 2M/r)(\frac{dt}{d\tau})^2 + (1+2M/r)[(\frac{dr}{d\tau})^2 + r^2((\frac{d\theta}{d\tau})^2 + r^2sin^2(\theta)(\frac{d\phi}{d\tau})^2]$$

by using $e$ and $l$, and from the initial conditions we know that $(\frac{d\theta}{d\tau}) = 0$ for $\theta = \pi /2$

$$-1 = -(1-2M/r)^{-1}e^2 + (1+2M/r)[(u^r)^2 + l^2/r^2]$$

where $(u^r)^2 = (\frac{dr}{d\tau})^2$

I want to format this equation in such a way that I would obtain similar to

$$ e^2 = \frac{1}{2} (u^r)^2 + \text{terms}$$

where some terms will give the effective potential energy.

Here is a version of it done for the Schwarzschild Metric

 

[pervect]

We can write the Newtonian metric in the form of

$$ds^2 = -(1 - 2M/r)dt^2 + (1+2M/r)[dr^2 + r^2d\Omega^2]$$

Surely, that's the Schwarzschild metric?

In order to obtain the orbit equation I have written the constant of motion,

$$e = (1 - 2M/r)(\frac{dt}{d\tau})^2$$

and

$$l = r^2sin^2(\theta)(\frac{d\phi}{d\tau})^2$$

I believe that's also incorrect - it doesn't match my text.

$$\tilde{E} = (1-2M/r) \frac{dt}{d\tau} \quad \tilde{L} = r^2 \frac{d\phi}{d\tau}$$

Here $\tilde{E}$ is energy/unit mass, and $\tilde{L}$ is angular momentum per unit mass.

See for instance https://www.fourmilab.ch/gravitation/orbits/, or the textbook from which it is sourced, MTW's "Gravitation".

I'm also not quite sure of your question - if you fix these errors, you'll get motion in the Schwarzschild geometry, but you seem to be asking about a "Newtonian" something. So I wouldn't be surprised if you're struggling, but I don't know what you're looking for.

 

[PeterDonis]

Surely, that's the Schwarzschild metric?

It's the Schwarzschild metric in isotropic coordinates in the "near Newtonian" limit, where we ignore terms of second or higher order in $M / r$ (which is assumed to be small).

 

[PeterDonis]

We can write the Newtonian metric

Yes, we can, but why do you need to? This metric is just an approximation, but the exact metric in question, the Schwarzschild metric, can be solved exactly for orbit equations and effective potential, as you note in your post. So why not just do that?

 

[Arman777]

Surely, that's the Schwarzschild metric?

I believe that's also incorrect - it doesn't match my text.

$$\tilde{E} = (1-2M/r) \frac{dt}{d\tau} \quad \tilde{L} = r^2 \frac{d\phi}{d\tau}$$

Here $\tilde{E}$ is energy/unit mass, and $\tilde{L}$ is angular momentum per unit mass.

See for instance https://www.fourmilab.ch/gravitation/orbits/, or the textbook from which it is sourced, MTW's "Gravitation".

I'm also not quite sure of your question - if you fix these errors, you'll get motion in the Schwarzschild geometry, but you seem to be asking about a "Newtonian" something. So I wouldn't be surprised if you're struggling, but I don't know what you're looking for.

I have fixed some of them. I am trying to find the Newtonian effective potential from the given Newtonian metric.

 

[Arman777]

Yes, we can, but why do you need to? This metric is just an approximation, but the exact metric in question, the Schwarzschild metric, can be solved exactly for orbit equations and effective potential, as you note in your post. So why not just do that?

It's kind o homework...I need to derive the photons and timelike particle's orbit in the Newtonian metric. I tried to calculate the timelike particles' orbit using the method that I did in post 1, but I am stuck.

I mean I need to follow the original steps..

 

[romsofia]

http://sites.science.oregonstate.edu/physics/coursewikis/GGR/book/ggr/oNewton.html

Is this what you're looking for?

 

[Arman777]

http://sites.science.oregonstate.edu/physics/coursewikis/GGR/book/ggr/oNewton.html

Is this what you're looking for?

Nope. I need to derive it by using Newtonian metric

 

[vanhees71]

You get the most simple form of the EoM in terms of arbitrary affine parameters using the Euler-Lagrange equations of the Lagrangian
$$L=\frac{1}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}.$$
Then automatically $L=\text{const}$ along the world line. For massive particles you can choose this constant 1 (then the affine parameter is proper time). For massless "particles" $L=0$, and you must live with an arbitrary affine parameter.

 

[romsofia]

Can you tell us where you're getting stuck exactly? It's literally the same procedure as you would for schwartzschild metric since you're using Killing Vectors, if you aren't comfortable with using them (which I believe you should be since you've identified them), then start here: http://sites.science.oregonstate.edu/physics/coursewikis/GGR/book/ggr/symmetries.html

Go through the examples, then hit up:
http://sites.science.oregonstate.edu/physics/coursewikis/GGR/book/ggr/orbits.html
http://sites.science.oregonstate.edu/physics/coursewikis/GGR/book/ggr/onull.html

If you've understood all this, then we'll need to know where exactly you're getting stuck since it really isn't obvious.

EDIT: Didn't see Vanhees reply, but that is a different way to derive the EOM (more efficient), but I prefer this way since it's more intuitive for me. But, don't feel like you have to use my way or his way, since you're young, try both!

 

[pervect]

An alternative to Killing vectors for trajectories of massive objects (which must have timelike worldlines) is the slightly oversimplified idea of the principle of maximal aging, which happens to be one of my favorite approaches.

This ties into vanhees71's post as
$$g_{\mu\nu} \dot{x}^\mu \dot{x}^\nu$$

is the square of the proper time, with possible sign differfences depending on sign conventions. The Lagrangian for the case of a timelike worldline of a free particle is proportional to the proper time of the worldline. The proportionality constant is negative, so the principle of least action (most negative Lagrangian) becomes the principle of maximal aging, the maximum proper time.

There are some subtle points I've glossed over, involving why vanheese71 used the square of the proper time, and I did not. To gloss over and oversimplify, it turns out the math is a lot easier by taking the square of the proper time as the Lagrangian, and this is acceptable for an affinely paramaterized geodesic.

Energy and angular momentum conservation come into play when the equations of motion of the free particle have the specialized formulations $dE/d\tau=0$ and $dL/d\tau=0$, which applies in the case we are considering, the Schwarzschild case.

I'm afraid that the multiplicity of possible approaches may confuse the OP, but I'm not sure what the best way to handle it is. It's a question of whether it's better to stick with an approach from a textbook and make it work, or to take possibly a different approach that may be more appealing and fit the OP's background better.

Maximal aging, more precisely extremal aging, is appealing for the relatively modest background knowledge requirments, but it may not match the OP's textbook reference.

E.F. Taylor has a webpage with some resources for the "action" approach to physics at https://www.eftaylor.com/leastaction.html. "Exploring black holes", of which Taylor is a co-author, uses the principle of maximal aging, though I don't own a copy.

 

[Ibix]

though I don't own a copy.

Free for download from Taylor's website if you want one. Fair warning - I think it's a scanned copy, so it's a big file.

 

[Arman777]

I solved the problem. Guys the solution is really simple.

$$-1 = -(1-2M/r)^{-1}e^2 + (1+2M/r)[(u^r)^2 + l^2/r^2]$$

just turn $(1-2M/r)^{-1}$ into $(1+2M/r)$ and proceed