# Gr vs. newtonian orbits

1. May 14, 2007

### charybdis

i am a physics enthusiast studying general relativity on my own with the book: "Exploring Black Holes: Introduction to General Relativity" by Edwin F. Taylor and John Archibald Wheeler. I am having the hardest time with the fourth chapter of the book and especially the 4th question in that chapter (i seem to have a thing with 4's). I have no knowledge of tensors, but that is not required for the book. The answer that i get for the first part of the problem is confusing and i want to verify that what i have as an answer is either right or wrong and then proceed from there.
the problem states:
Under what circumstances are circular orbits predicted by Newton indistinguishable from circular orbits predicted by Einstein?
A. Find the Newtonian expression similar to:r=((L/m)^2/2)[1+/-(1-(12M^2)/((L/m)^2))^(1/2)] for the radius of a stable circular orbit, starting with V(r)/m=-M/r+(L/m)^2/(2r^2).
B. Recast for the general relativistic prediction of r for stable orbits in the form:r=rnewt(1+q)

Where r¬Newt is the radius of the orbit predicted by Newton and q is a small quantity. This expression neglects differences between the Newtonian and relativistic values of L/m (angular momentum) when expressed in the same units. Use the approximation:
(1+d)^n ≈1+nd provided |d|<<1 and |nd|<<1.
To derive a simple algebraic expression for q in terms of M and rNewt.

as a little aside how do you type equations into this the equation editor i have apparently does not work for html.
thank you.

2. May 14, 2007

### Chris Hillman

As luck would have it...

One of the few currently popular gtr books which I typically don't have easily available to me

Try left clicking with your mouse on any formula you see at PF to see the latex pseudocode used to format it.

You should have an effective potential which looks something like
$$V_{\rm GTR}(r) = \left( 1- \frac{2m}{r} \right) \; \left( 1 + \frac{L^2}{r^2} \right) = 1 \, - \, \frac{2m}{r} \, + \, \frac{L^2}{r^2} \, - \, \frac{2 m \, L^2}{r^3}$$
where the last term is the gtr correction to the corresponding Newtonian effective potential
$$V_{\rm Newton} (r) = 1 - \frac{2m}{r} + \frac{L^2}{r^2}$$
where $2m/r$ is the gravitational term and $L^2/r^2$ is the centrifugal term. In both cases, m is the mass of the central object, L is the (specific) angular momentum of the test particle, and r is a radial coordinate. (It looks like Taylor & Wheeler might have tossed in the mass of the test particle, but this should drop out of the problem.)

For both potentials, sketch the graph for typical values of m, L. The place (value of Schwarschild radial coordinate r) where the graph has a local minimum is the location of the stable circular orbit for a given L. So differentiate wrt r, set the result equal to zero, and solve for r, obtaining
$$r_{\rm GTR} = \frac{L}{2m} \, \left( L + \sqrt{L^2-4 m^2} \right), \; \; r_{\rm Newt} = \frac{L^2}{m}$$

Now you want to compare the two radii $r_{\rm GTR}, \; r_{\rm Newton}$ for a given m, L. Are you familiar with Taylor series? Or better yet, asymptotic series? (Hint: we want to consider stable circular orbits far from the horizon, so take the case L much larger than m.)

This is a very good problem, because it should cause you to think hard about the operational meaning of a "radial coordinate". In the Newtonian case, r is the usual euclidean radial coordinate, but in the Schwarzschild vacuum case, r is Schwarzschild radial coordinate (do you recall its geometric significance?), so you should be worried that we are trying to compare apples and oranges. Are we?

(So how come everyone is posting about Schwarzschild geodesics all of a sudden? Did something happen in the popular press that I don't know about?)

Last edited: May 14, 2007
3. May 14, 2007

### smallphi

To find the radius of the circular orbit, set to zero the derivative with respect to r of the effective potential. I got

$$r = \frac{(L/m)^2}{M}$$

The same result can be obtained by more elementary methods by solving

centripetal acceleration = acceleration by gravity

and expressing the speed in terms of angular momentum L = mvr. Don't forget that the book uses geometrized units in which G =1.

4. May 14, 2007

### pervect

Staff Emeritus
I'm a little confused by what you are confused about and what answers you say you are getting.

So far all I've seen is a problem statement, but not much indication of what work you've done with it or why you are confused.

I've also seen a lot of helpful posts by others in this thread, but I've also seen some minor differences in notation and (I think) some typos, which might not be helping your state of confusion any.

The bottom line is that I'm not sure how much you've been helped, and I'm not sure why you're confused or what you're confused about.

5. May 18, 2007

### charybdis

First, thank you for such a quick response. A couple of things;
1. Chris Hillman: according to my book the equation you gave:
$$V_{\rm GTR}(r) = \left( 1- \frac{2m}{r} \right) \; \left( 1 + \frac{L^2}{r^2} \right) = 1 \, - \, \frac{2m}{r} \, + \, \frac{L^2}{r^2} \, - \, \frac{2 m \, L^2}{r^3}$$
Is the book for the square of the effective potential. The equation for the effective potential is:
$$V_{\rm GTR}(r) = - \frac{2M}{r} + \frac{L^2}{2r^2}$$
2. Pervect: my confusion with part A, is resolved. But, my confusion with part B remains. My confusion with part B is how do I derive an expression for q, in terms of M and r¬newt. The book says to start with the equation:
$$r= \frac {L^2}{2M} (1 \pm \sqrt{1- \frac{12M^2}{L^2}0)$$

And put in the equation:
$$r=r_{\rm newt}(1+q)$$
I guess I put the rnewt equation I derived earlier into rnewt, but then, how do I get q?
I don’t know where to start is all I think the problem is.
All I ask right now is to be pointed in the right direction.

6. May 18, 2007

### pervect

Staff Emeritus
OK, so you know that
$$r_{newt} = \frac{\tilde{L}^2}{M}$$

where $$\tilde{L} = L/m$$, i.e. the tilde means we want the specific angular momentum, the angular momentum per unit mass of the orbiting body, which is just velocity * orbital distance.

And you are happy with that and how it was derived from Newton's law?

The first thing to note is that you need to use the plus sign in the GR expression for r. The minus sign will put the orbit inside the body for any non-black hole, and will be an unstable orbit for any black hole.

See for instance http://www.fourmilab.ch/gravitation/orbits/

The next thing you need to know is that $$\frac{12 M^2}{\tilde{L}^2}$$ is "small". Consider the Earth's orbit around the sun, for instance

The mass of the sun in geometric units is 1.5 km. The specific angular momentum of the earth is the orbital velocity of the earth times the semi-major axis.

http://hyperphysics.phy-astr.gsu.edu/hbase/solar/soldata2.html

puts the orbital velocity of the earth at about 30 km/sec, that's 10^-4 c. And the orbital radius is about 1.5*10^8 km. So$$\tilde{L}$$ = v*r = 1.5*10^4 km in geometric units. Thus, because of the large orbital radius L >> M.

This means you know M^2/L^2 is small, and you can do a taylor series expansion.

$$\sqrt{1 - \epsilon} \approx 1 - \frac{\epsilon}{2}$$

with $\epsilon = \frac{12 M^2}{L^2}$

Last edited: May 18, 2007
7. May 18, 2007

### Chris Hillman

Oops!

Oops, I did mean to write $V^2$. Thanks for pointing this out.

Let me try again: (25.16) in Misner, Thorne, & Wheeler, Gravitation (the first "modern" GTR textbook, the book from which your book--- same Wheeler, BTW--- and all more recent modern textbooks derive) states
$$\dot{r}^2 = E^2 - V(r)^2,$$
$$V(r)^2 = \left( 1- \frac{2m}{r} \right) \, \left( -\varepsilon + \frac{L^2}{r^2} \right)$$
where $\varepsilon=-1, \, 0, \, 1$ for timelike, null, spacelike orbits respectively. This expression--- which is standard, trust me--- is also given as (11.10) in Schutz, A First Course in General Relativity, as (10.7.12) in de Felice and Clarke, Relativity on Curved Manifolds, and as the second equation on p. 314 of Landau and Lif****z, The Classical Theory of Fields.

(EDIT: that's pretty funny, PF autobleeped the name of a distinguished physicist. His name is also sometimes transliterated Lifschitz, but the publishers of the translation left out the Germanic "s".)

But Carroll, Spacetime and Geometry: An Introduction to General Relativity, gives in (5.66) the potential as the square, as does Clarke, Elementary General Relativity in (21), Chapter 4. So while the method of effective potentials is completely standard, the quantity called "the effective potential" is not!

Note that in mathphys, a "potential" is just something which when "differentiated" in some fashion gives a quantity of interest, such as velocity vector field or electromagnetic field. So there's no convention stating that potential must have given units, unlike potential energy which of course has the units of energy.

Dunno if this helps answer your question, which unfortunately it seems we are having difficulty understanding.

Last edited: May 18, 2007
8. May 19, 2007

### charybdis

my question has been answered. thank you both very much