# Connection coefficients in Earth orbit

1. Sep 17, 2015

### TerryW

I've run into a couple of problems with exercise 12.6 in MTW (attached).

At this stage, all I'm asking is for someone to give me answers to the following questions:

1. What are the Rα'β'γ'δ' in the primed co-ordinate system? and

2. What are the non-zero connection coefficients in the primed co-ordinate system?

I started with an Earth bound co-ordinate system fixed so that the x-axis runs from the centre of the earth through the prime meridian at the equator and worked to get the co-ordinate transformations between (x' , y') and (x , y) and used -2GM/R2 for Rα1010 and GM/R2 for R2020. The results for R1'0'1'0' and R2'0'2'0' didn't turn out as I think they should.

For my connection coefficients I ended up with

Γ1'0'0' = +ω2x' ,

Γ2'0'0' = -2ω2y'

Γ1'0'2' = -ω

Γ2'0'1' = +ω

Which didn't give a good result for the geodesic equation for y'.

I used the connection coefficients to calculate R1'0'1'0' and Rα2'0'2'0' and ended up with 2ω2 and -ω2 respectively which also doesn't look quite right.

Can anyone put me on the right track here?

Regards

TerryW

#### Attached Files:

• ###### Geodesic Deviation above the Earth.pdf
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1.2 MB
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2. Sep 17, 2015

### Staff: Mentor

Moved to homework since this is a textbook problem.

3. Sep 17, 2015

### Staff: Mentor

What transformations did you come up with?

Where did you get these from? They don't look right. Remember that the Riemann tensor describes tidal gravity, not "acceleration due to gravity". (Also, I'm not sure the problem is asking you to take the Earth's gravity into account. See below.)

What did you start with? If you started with the connection coefficients in the unprimed frame being all zero, are you aware that this is only true if we ignore the Earth's gravity? That may indeed be what the problem is asking you to do (I'm not sure), but if so, the Riemann tensor components will also all be zero in the unprimed frame; all of the actual interesting stuff comes from the coordinate transformations.

4. Sep 18, 2015

### TerryW

I've written out my transformations and attached them, along with a page from MTW containing the Riemann tensor I used (with the Z and X axes interchanged)

I must say I have tried hard to find examples of Riemann tensor components and Connection Coefficients for this situation on the Internet with no success at all.

Regards

Terry W

#### Attached Files:

• ###### My transformation and Riemann components.pdf
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5. Sep 18, 2015

### Staff: Mentor

Ok, you had a typo in the OP, you had $R^2$ where it should be $R^3$ in the denominators of the Riemann tensor coefficients, that was what confused me about those. With the typo corrected, those are the right Riemann tensor components in the vicinity of the Earth, if we are including Earth's gravity. But if we are including Earth's gravity, then things get a lot more complicated because space is no longer Euclidean. Your coordinate transformation appears to assume that space is Euclidean, which would mean we are ignoring Earth's gravity and working in an SR inertial frame centered on an idealized "Earth" that has no mass (so "Earth" is really just an identifier for a region of space). If that is the case, then the Riemann tensor components in the unprimed frame are all zero. If it's not, then your coordinate transformation needs to start from Schwarzschild coordinates, not ordinary Cartesian coordinates, in the unprimed frame.

Even if we assume we are ignoring Earth's gravity (which is the easier way to proceed), your coordinate transformation doesn't look right. First, since both coordinate systems are supposed to be Cartesian (x, y, z), you don't need to go through the intermediate step of translating into radius and angular coordinates; that just introduces confusion. You should be able to write down, at a given instant of time, a direct transformation between the x, y, z of the unprimed (non-rotating) coordinates and the x', y', z' of the primed (rotating) coordinates.

Second, note that I said "a given instant of time". The primed coordinates are attached to an object moving in a circle, so the coordinate transformation should have factors in it describing that motion--such as the angular velocity $\omega$. No such factor appears in what you wrote down.

6. Sep 18, 2015

### TerryW

Peter,

Thanks for your detailed reply. It prompted me to take a fresh look at the problem and I've spotted something I'd overlooked earlier.

Some of the work in this chapter of MTW is looking at Galilean frames which have the particular characteristic that Γj00 = ∂φ/∂xj, all other Γαβγ = 0 and Rj0k0 = Γj00,k.

This then leads to Γ100,1 = R1010 and Γ200,2 = R2020.

As φ = -GM/R (I need to check out the minus sign, but MTW says that φ = - U (the Newtonian potential) so I think it is OK), then with φ = -GM/(x2+y2)½ then R1010 and R2020 are the same, so the transformation is simple, leading to R1'0'1'0' and R2'0'2'0' being the same as R1010 and R2020 respectively. ( A simple transformation as the problem says!)

(I realise now that the Riemann components I'd been using earlier were incorrect because they represent the components in a local Lorentz frame which isn't the same as my frame fixed at the centre of the earth.)

I feel confident about my transformations at this stage. As you say, they represent a given instance of time, but once you differentiate wrt to time, then d/dt sinθ = cosθdθ/dt, where dθ/dt = ω.

All I need to do now is go through my workings on the results for Γα'β'γ' to see if I can square them up with the values for R1'0'1'0' and R2'0'2'0'.

I'm off on holiday tomorrow so I won't be able to do too much over the next ten days, but I'll drop a post to you when I've reworked everything - hopefully clearing up the issues.

Regards

TerryW

7. Oct 3, 2015

### TerryW

Hi Peter,

I've made progress while on holiday and I'm a step nearer a consistent set of answers to this problem.

I noticed that I made a transcription error in my 'Transformations' attachment so I've attached a corrected version.

I had a further look at the transformations for R1010 and R2020 and found that they were a bit more complicated, but when I worked through them, the results I got for R1'0'1'0' and R2'0'2'0' were -2GM/R3 and GM/R3 respectively, ie they have the same form as the Riemann tensor components derived in Chapter 1 (which I included in my original attachment). I'm pretty sure that these are correct because they represent the accelerations one would observe between two test particles floating inside an orbiting space station, which is in effect, the same situation as this problem - I also found a paragraph in Chapter 1 which is pretty conclusive on the matter.

I then tried to derive the connection coefficients Γ1'0'0' and Γ2'0'0' using formulae given in the text and which I have derived myself in an earlier exercise. I was hoping the result would give Γ1'0'0' = R1'0'1'0'x and Γ2'0'0' = R2'0'2'0'y, but my results are wide of the mark.

My workings run to several pages but I would really appreciate it if you could take a look at what I have done to see if you can see where it is all going wrong. Would you be willing to do this?

Regards

TerryW

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8. Oct 3, 2015

### Staff: Mentor

I can, but it won't be for several days. I don't currently have access to my copy of MTW, and I would like to review it while I'm looking at what you've done.

9. Oct 3, 2015

### TerryW

Thanks for this. It will take me a couple of days to write up my workings into a presentable form anyway.

By the way, although I do have my original copy of MTW (bought in the early 70s), I also recently came across a PDF on the internet which is a scan of the entire book. I'll see if I can find the link again for you - I find it useful to have on my iPad and for printing out the odd page when I need it.

Regards

Terry

10. Oct 3, 2015

### Staff: Mentor

I don't think PF rules would permit posting the link here since it's still a copyrighted work.

11. Oct 5, 2015

### TerryW

Hi Peter,

Here is my write up for the derivation of the connection coefficients.

Many thanks for agreeing to take a look.

Regards

TerryW

#### Attached Files:

• ###### Finding Connection Coeffients.pdf
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969.7 KB
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12. Oct 5, 2015

### TSny

Terry, you have gotten me interested in this problem! I have worked on it off and on for several days and managed to get some results. I might very well have made some errors (including conceptual ones). I hope you and Peter don’t mind if I enter the discussion.

First, I agree with your results for the nonzero components of the Riemann tensor.

However, I don’t agree with your argument for showing that $R^{1’}_{0’ 1’ 0’}x^{1’} = \Gamma^{1’}_{0’ 0’}$. On the first page of the last set of notes you posted, the geodesic deviation equation doesn’t look correct. The second derivative with respect to lambda should be the covariant second derivative rather than the ordinary second derivative. This will involve some $\Gamma$’s. See box 12.1 for a similar calculation. But in Box 12.1 the calculation is done in a Galilean coordinate system while in our problem we are working in the non-Galilean system (primed frame) of the orbiting laboratory.

In your expression for the geodesic equation, you have only $\Gamma^{\alpha}_{0’ 0’}$ appearing. But I believe there will be other terms of the form $\Gamma^{\alpha}_{k’ 0’} \frac{dx^{k’}}{d \lambda}$ . However, the solution to the problem will only require using the geodesic deviation equation. The geodesic equation will not be needed.

When you derive the expressions for the $\Gamma^{k’}_{0’ 0’}$ connection coefficients in the rest of your notes, they will only be needed at the origin of the primed frame when using the geodesic deviation equation. So you can set $x’ = y’ = z’ = 0$. You can see that all of the $\Gamma^{k’}_{0’ 0’}$ terms are zero at the origin. However, some of the $\Gamma^{j’}_{0’ k’}$ are nonzero at the origin and they will occur in the geodesic deviation equation.

13. Oct 5, 2015

### TerryW

Hi TSny,

Thanks for your thoughts. You are more than welcome to join in.

That's a relief - if you had worked through it in fifteen minutes and reached a consistent answer that I'd missed, I'd feel a bit crushed!

Regards

TerryW

14. Oct 6, 2015

### TSny

Suppose you release an object at rest a small distance out along the $x^{1’}$ axis.

The geodesic equation would imply that just after release $$\frac{d^2x^{1’}}{d{ t’}^2}+ \Gamma^{1’}_{0’ 0’} = 0$$ where $\Gamma^{1’}_{0’ 0’}$ is to be valuated at the location of the object.

The geodesic deviation equation will imply that just after release $$\frac{d^2x^{1’}}{d{ t}’^2}+ \left(\Gamma^{1’}_{2' 0’}\Gamma^{2’}_{1’ 0’} + R^{1’}_{0’ 1’ 0’}\right)x^{1'} = 0$$ where the gamma factors come from the covariant derivatives. Here the gammas and R factors are to be evaluated at the origin of the primed frame.

You can check that the two equations are in agreement if you use your results for the values of the $\Gamma$'s and R’s.

15. Oct 6, 2015

### TerryW

Hi TSny,

Many thanks for your help with this. Much appreciated.

I agree that my attempt to equate the Riemann components and the Connection Coefficients was a mistake. I can see where I got it from and I just got it wrong!

There were also a couple of things in the question that I missed. One is the quite clear statement that the primed frame is not Galilean, the other is that the question, as you quite rightly point out, asks only for an evaluation of the Connection Coefficients at the origin of the primed frame. (RTFQ!)

I had already evaluated Γ1'0'2' ( = -ω) and Γ2'0'1' ( = ω) and I have used these, in conjunction with Γ1'0'0' and Γ2'0'0' to derive R1'0'1'0', R2'0'2'0' and R1'0'2'0'. They all check out fine!

I'm a bit puzzled by your equation d2x'/dt'2 + (Γ1'0'2'Γ2'0'1' + R1'0'1'0')x1' = 0. Surely the equation is just d2x'/dt'2 + R1'0'1'0'x1' = 0, the geodesic separation equation, isn't it?

The other equation is d2x2'/dt'2 + R2'0'2'0'x2' = 0

These equations would seem to indicate that the bag of rubbish released at a distance y' out in the direction of motion would see the bag slowly approach the space station but if the bag is placed at a distance x' perpendicular to the direction of motion, it will gradually move away from the space station. I have a bit of a problem validating this analysis by examining the Keplerian orbits though. If the rubbish bag is released at a distance x' perpendicular to the direction of motion, assuming it has the same tangential velocity as the space station, it will be travelling too fast for a circular orbit (if it is further from earth) and will therefore move in an elliptical orbit, and although it may move away initially, it will not do so indefinitely as it's elliptical orbit will limit it's maximum distance from the space station. If it is released at a distance y' out in the direction of motion, as it has the same velocity, it will just continue in a circular orbit ahead of the space station. These are just qualitative ideas at the moment, I'm still working on the maths but I would appreciate any thoughts you have.

Regards

TerryW

16. Oct 6, 2015

### TSny

To get the Riemann components in the primed frame, you don't need to derive them from the connection coefficients. As hinted in the problem statement the primed Riemann coefficients can be obtained by a "trivial transformation of tensorial components". I actually had to spend some time thinking about that, but I believe it's true that it is easy to get the primed components from the unprimed components with virtually no calculation. I'll let you think about that if you want.

No. This is a good exercise in carrying out covariant differentiation. To avoid possible confusion, I think it is a good idea to use the notation $n^k$ for the spatial components of the deviation vector that measures the deviation between the "fiducial" geodesic and the neighboring geodesic of the object of interest (or "test particle"). The fidcuial geodesic in our problem is the geodesic of the center of the freely falling space laboratory. That is, it is the geodesic along which the origin of the primed coordinate system is moving. So, the "tail" of $n^k$ will always be at the origin of the primed system and the "head" of $n^k$ will be at the location of the test particle.

The covariant derivative as written in the middle of Box 12.1 is: $$\frac{Dn^k}{dt} = \frac{dn^k}{dt} + \Gamma^k_{\beta \mu} n^{\beta}\frac{dx^\mu}{dt}$$ For convenience, I will drop all primes since it is understood that we are working now exclusively in the primed frame. In the above equation, the gammas are evaluated on the fiducial geodesic at the location of the tail of $n^k$ and, hence, at the origin of the primed coordinate system. This is why we only need the gammas at the origin. The $\frac{dx^\mu}{dt}$ factor is the time derivative of $\mu$th coordinate of the point on the fiducial geodesic corresponding to the tail of $n^k$. But since we are working in the primed system, we always have $x^k = 0$ and $x^0 = t$. Thus, see what you get for $\frac{Dn^1}{dt}$ and $\frac{Dn^2}{dt}$ keeping in mind that most of the gammas are zero. Then you can go on to try to get the second covariant derivatives.

Yes, if the bag is released from rest at a small distance along the y' axis, then it should essentially stay at that point in the primed frame. Once you get the correct differential equation for $n^2$ you will see that this is the case.

And, yes, if the bag is released from rest at a positive value of x', then it should initially start moving farther away from the origin. This will be validated by the correct differential equation for $n^1$.

17. Oct 8, 2015

### TerryW

Hi again TSny,

You really are doing sterling work on my behalf!

I know this is the case and I'd done that. I just thought it would be fun to see if deriving Riemann from the Connection Coeffs and their derivatives would work and it did.

Yes, I can see how - just put θ = 0 in my transformation! I did this originally but I was worried about the fact that I seemed to be choosing a rotating frame, but as the process does not involve any derivatives of the Aj'k terms, no problems arise.

I've worked this through and got to:
$$\frac{d^2n^{1}}{dt^2} = \Gamma^1_ { 02}\Gamma^2_ { 01} n^1 = R^1_ { 010}n^1$$

and
$$\frac{d^2n^{2}}{dt^2} = \Gamma^2_ { 01}\Gamma^1_ { 02} n^1 = R^2_ { 020}n^2$$

Using my results for $\Gamma^2_ { 01}$ and $\Gamma^1_{02}$, I get
$$\frac{d^2n^{1}}{dt^2} = 3\omega^2n^1$$ and
$$\frac{d^2n^{2}}{dt^2} =0$$

So the bag placed further out on the x' axis moves away and there is no acceleration along the y' axis.

So this agrees with you and my earlier Keplerian consideration but it doesn't fit with MTW page 29, which in §1.6 states that in orbit around the earth, the two particles will move towards each other.

Is this one of the legendary 2 mistakes? I know the usual argument is that there is a tiny angle between the gravitational force vectors acting on the two particles which results in a tiny force of attraction and I can see how this would manifest itself if the spaceship was falling radially inward towards the centre of the earth. But in orbital motion, the radial gravitational forces are providing the inward accelerations which maintain circular orbits which will have exactly the same period of motion - no change in separation.

Do you think I've cracked this now (with your much appreciated help of course)? And how do I give you a big 'like"?

Regards

TerryW

18. Oct 8, 2015

### TSny

OK. ( I'm sure you meant to type "+" instead of "=" in front of the last terms on the right in the first two equations.) The above equations hold at an instant when the bag is at rest near the origin in the primed frame. So, they tell us how the bag will initially begin to move just after being released at rest. However, when the bag is moving (as in the problem where the bag is jettisoned from the origin), there are additional Coriolis terms that involve the first time derivatives of $n_1$ or $n_2$.

I'm not familiar with the "2 legendary mistakes". It would be interesting to know what they are. But, it does seem to me that they made a gaffe here. Certainly, as you say, two orbiting test particles of negligible mass on the same circular orbit would maintain constant separation.

19. Oct 9, 2015

### TerryW

Hi TSny

Yes, it is a typo.

Looking back at some of my original post, I actually did have the answer (for Γ100 and Γ200) but then tried to reconcile these answers with MTW §1.6, which we now agree is wrong.

However, I think I learned quite a lot from following your guidance so thanks again for your help.

I can recall seeing this somewhere in the past but I haven't been able to find the reference again today. I believe it was one of the authors who said it but, tantalisingly, didn't reveal what the mistakes were.

Anyway, on to Chapter 14, (I've done all the problems on Ch 13!)

Regards

TerryW

20. Oct 9, 2015

### TSny

OK. Sounds good. Enjoy chapter 14. The book is truly amazing. Seems like people either love it or hate it. Yesterday, I ran across this interesting article related to the initial publication of the book: http://www.jstor.org/stable/10.1086/664983