Why no absolute derivative in this example of geodesic deviation?

[peter46464]

On the surface of a unit sphere two cars are on the equator moving north with velocity $v$. Their initial separation on the equator is $d$. I've used the equation of geodesic deviation $$\frac{D^{2}\xi^{\mu}}{D\lambda^{2}}+R_{\phantom{\mu}\beta\alpha\gamma}^{\mu}\xi^{\alpha}\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{d\lambda}=0.$$to find their separation $s=\xi^{\phi}$ after time $t$. I used $\lambda=t$ to give $$\frac{d{}^{2}\xi^{\mu}}{dt{}^{2}}+R_{\phantom{\mu}\beta\alpha\gamma}^{\mu}\xi^{\alpha}\frac{dx^{\beta}}{dt}\frac{dx^{\gamma}}{dt}=0,$$expanded out the Riemann components for a unit sphere to eventually get $$\xi^{\phi}=d\cos\left(vt\right),$$which is correct. So far, so good. However, my problem is that in order to do the calculation I assumed the absolute second derivative $$\frac{D^{2}\xi^{\mu}}{D\lambda^{2}}$$ can be replaced the the ordinary second derivative $$\frac{d{}^{2}\xi^{\mu}}{dt{}^{2}}.$$ This assumption works, but I don't understand why/how I can get away with it. In other words, why don't I need to calculate the absolute derivative using $$\frac{DV^{\alpha}}{D\lambda}=\frac{dV^{\alpha}}{d\lambda}+ V^{\gamma}\Gamma_{\gamma\beta}^{\alpha}\frac{dx^{\beta}}{d\lambda}.$$
Can anyone explain why the ordinary second derivative works in this calculation? Thanks.

 

[WannabeNewton]

Could you post the details of your calculation? Attach it as a document or image if that's more convenient.

 

[peter46464]

Thanks, I've attached a pdf showing the calculation.

 

[TrickyDicky]

You are just setting the problem in such a way that allows you to use geodesic normal local coordinates, that gives you the vanishing of the Christoffel symbols at the point you compute the distance, and thus you may use the ordinary derivative.

 

[peter46464]

What are geodesic normal local coordinates? Are spherical coordinates always geodesic normal local coordinates? I've tried Googling the term, and got something about "applying the exponential map to the tangent space at p". I'm OK with "tangent space" but have no idea what "exponential map" means.

 

[TrickyDicky]

What are geodesic normal local coordinates?

http://en.wikipedia.org/wiki/Normal_coordinates
http://en.wikipedia.org/wiki/Fermi_coordinates

Are spherical coordinates always geodesic normal local coordinates?

No. It depends on the symmetries of the problem/space.

I've tried Googling the term, and got something about "applying the exponential map to the tangent space at p". I'm OK with "tangent space" but have no idea what "exponential map" means.

http://en.wikipedia.org/wiki/Exponential_map see the Riemannian geometry part, under Properties.

When you set up the problem so that the distance traveled by the car along the geodesic is a linear function of t, you are introducing a unit time(therefore you can use an orthonormal basis).

 

[TrickyDicky]

Similarly in GR's geodesic deviation problems it is often possible to simplify the problem by working on linear inertial frames so that the Christoffel symbols vanish at a point or along a geodesic.

 

[peter46464]

Thanks, I need to think about this.

I still can't see why the Christoffel symbols vanish along geodesics on the surface of a sphere. Surely the Christoffel symbols will only vanish if we're using a flat metric?

 

[TrickyDicky]

Thanks, I need to think about this.

I still can't see why the Christoffel symbols vanish along geodesics on the surface of a sphere. Surely the Christoffel symbols will only vanish if we're using a flat metric?

It's a common misconception. It is the Riemann curvature(second derivatives of the metric) that will only vanish in a flat manifold. The Christoffel symbols(first derivatives of the metric) can be made to vanish at a point by the use of the appropriate coordinates(just like they can appear in a flat space with non-cartesian coordinates), and by the exponential map in the neighborhood of the point along the geodesic.
It is not always guaranteed at all points of the curve, if this is possible they are called Fermi coordinates .

 

[peter46464]

Sorry to labour this, but how do you get from the Christoffel symbols for a unit sphere $$\Gamma_{\theta\phi}^{\phi}=\Gamma_{\phi\theta}^{\phi}=\frac{\cos\theta}{\mathbf{\mathbf{\sin\theta}}},\Gamma_{\phi\phi}^{\theta}= \sin\theta\cos\theta$$ to Christoffel symbols that vanish?

I know that the Christoffel symbols are given by:
$$\Gamma_{jk}^{i}=\frac{1}{2}g^{il}\left(\frac{\partial g_{lk}}{\partial x^{j}}+\frac{\partial g_{jl}}{\partial x^{k}}-\frac{\partial g_{jk}}{\partial x^{l}}\right).$$How do I get those partial derivatives to vanish? Or, in other words, what is it about being on a geodesic that makes the Christoffel symbols equal zero?

 

[TrickyDicky]

Sorry to labour this, but how do you get from the Christoffel symbols for a unit sphere $$\Gamma_{\theta\phi}^{\phi}=\Gamma_{\phi\theta}^{\phi}=\frac{\cos\theta}{\mathbf{\mathbf{\sin\theta}}},\Gamma_{\phi\phi}^{\theta}= \sin\theta\cos\theta$$ to Christoffel symbols that vanish?

I know that the Christoffel symbols are given by:
$$\Gamma_{jk}^{i}=\frac{1}{2}g^{il}\left(\frac{\partial g_{lk}}{\partial x^{j}}+\frac{\partial g_{jl}}p{\partial x^{k}}-\frac{\partial g_{jk}}{\partial x^{l}}\right).$$How do I get those partial derivatives to vanish? Or, in other words, what is it about being on a geodesic that makes the Christoffel symbols equal zero?

No, sorry I might have confused you in a previous answer, you need normalized coordinates, spherical coordinates are just orthogonal, that's what you did when you reparametrized the geodesic deviation equation with t.

 

[peter46464]

Thanks. I've been away for a few days, but still can't see this. In my example how are the new (normalized) coordinates related to the old ($\theta,\phi$) coordinates? Also, the original line element is $$ds{}^{2}=d\theta^{2}+\sin^{2}\theta d\phi^{2}.$$How would I find the new line element using the new coordinates? (I think/hope that if I can see the line element, I might have a chance of seeing how the connections coefficients vanish.)

 

[peter46464]

But surely (except at the equator) the distance between the geodesics is not measured along geodesics but along circles of constant $\theta$. Therefore I can't be using geodesic coordinates?

 

[PeterDonis]

the distance between the geodesics is not measured along geodesics

Yes, it is. At least, the concept of "distance between geodesics" that enters into the equation of geodesic deviation is.

 

[TSny]

Here's how I see it. Corrections welcomed!

I think the confusion is related to the definition of the deviation vector $\xi^{\mu}$. In a general space, a family of geodesics can be described by some functions $x^{\mu}(\lambda; p)$ where p denotes a particular geodesic and $\lambda$ can be taken as arc length along the geodesic. Two neighboring geodesics would be $x^{\mu}(\lambda; p)$ and $x^{\mu}(\lambda; p+dp)$.

For geodesics on the unit sphere that run perpendicular to the equator, you can use the spherical coordinate $\theta$ as arc length $\lambda$ and the spherical coordinate $\phi$ to pick out different geodesics. So, two neighboring geodesics would be $x^{\mu}(\theta;\phi)$ and $x^{\mu}(\theta;\phi + d\phi)$. The deviation vector between the two neighboring geodesics is defined as $$\xi^{\mu} = \frac{\partial x^{\mu}(\lambda;p)}{\partial p} = \frac{\partial x^{\mu}(\theta;\phi)}{\partial \phi} $$ For example, see http://en.wikipedia.org/wiki/Geodesic_deviation (with modified notation).

Moreover, if you use spherical coordinates on the unit sphere, then the coordinates $x^{\mu}$ are of course just the spherical coordinates $\theta$ and $\phi$. Thus, the deviation vector reduces to something fairly trivial: $\xi^{\theta} = \frac{\partial \theta}{\partial \phi} = 0$ and $\xi^{\phi} = \frac{\partial \phi}{\partial \phi} = 1$.

This describes the geodesic deviation vector and there is no need to solve the geodesic deviation equation. However, you can check that this deviation vector does satisfy the geodesic deviation equation. But you will need to keep the $\Gamma$ terms in the covariant derivatives.

Note that the deviation vector does not equal the distance of separation between two neighboring geodesics. The deviation vector is defined so that the coordinate difference between two points on neighboring geodesics (for the same $\lambda$) is $\delta x^{\mu} = \xi^{\mu} dp$. This is easily checked for our example where $\delta x^{\theta} = \xi^{\theta} d\phi = 0 \cdot d\phi = 0$ and $\delta x^{\phi} = \xi^{\phi} d\phi = 1 \cdot d\phi = d\phi$.

To get the distance $ds$ between two neighboring geodesics, use the metric $$ds^2 = g_{\mu \nu} \delta x^{\mu} \delta x^{\nu} = g_{\phi \phi} d\phi^2 = \sin^2\theta \; d\phi^2$$

 

[peter46464]

My physics is struggling here, so I'm going to resort to a classic “argument by authority”. My example of geodesic deviation on a sphere is similar to the CalTech Exercise 24.10 here:

http://www.pma.caltech.edu/Courses/ph136/yr2004/0424.1.K.pdf

The solution (Answer 23.10) is given here:

http://www.pma.caltech.edu/Courses/ph136/yr2011/ps23_09.pdf

In the CalTech problem the reference geodesic is not along constant $\phi$ geodesics, but along the equator ($\theta=\pi/2$). The solution certainly does involve using the equation of geodesic deviation, but because the connection coefficients vanish along the $\theta=\pi/2$ geodesic the absolute second derivative becomes the ordinary second derivative (CalTech's equation 67 in the solution). The answer to the CalTech problem is $\xi^{\phi}=0$ and $\xi^{\theta}=b\cos\phi$, where $b$ is the initial separation. Note the similarity to my answer ($\xi^{\theta}=0$, $\xi^{\phi}=d\sin\theta$) for constant $\phi$ geodesics, where $d$ is my initial separation. Wheeling out the very heavy guns, the same (CalTech) answer is given in Misner, Thorne and Wheeler in the Figure 10 text, page 31.

The most convincing answer I've had to the question why the absolute second derivative equals the ordinary second derivative in my problem is Void's answer here:

http://physics.stackexchange.com/qu...oes-the-ordinary-second-derivative-give-the-c

though I can't honestly say I can confidently follow all of this explanation.

 

[PeterDonis]

To get the distance $ds$ between two neighboring geodesics, use the metric

$ds^2 = g_{\mu \nu} \delta x^{\mu} \delta x^{\nu} = g_{\phi \phi} d \phi^2 = \sin^2 \theta d \phi^2$​

Applying the term "distance" to this can be misleading, because (except at $\theta = \pi / 2$) it isn't evaluated along a geodesic curve. If, for example, we were to set up a local Euclidean coordinate chart centered on a point on the reference geodesic, a curve of constant $\theta$, which is what this is being evaluated along, would appear curved, not straight.

 

[TrickyDicky]

The most convincing answer I've had to the question why the absolute second derivative equals the ordinary second derivative in my problem is Void's answer here:

http://physics.stackexchange.com/qu...oes-the-ordinary-second-derivative-give-the-c

though I can't honestly say I can confidently follow all of this explanation.

I forgot about this thread.
So the part I guess I should have stressed but it seemed obvious is that the sphere is of course a surface of constant curvature, which means that the deviation of the geodesics is also constant. So this fact is what allows you to make the assumption about normalizing the spherical coordinates with a unit time and also computing a correct answer without using the Christoffel symbols.

 

[TSny]

In the attachment of post #4, the geodesics are lines of constant $\phi$. Along these geodesics the $\Gamma$’s are not zero (with the usual $g_{\theta \theta} = 1$ and $g_{\phi \phi} = \sin ^2 \theta$ ). As you move along one of these geodesics, $\frac{D^{2}\xi^{\mu}}{D\lambda^{2}} \neq \frac{d^{2}\xi^{\mu}}{d\lambda^{2}}$. Also, as mentioned in my previous post, the deviation vector $\xi^{\mu}$ for this case does not represent the distance between neighboring geodesics.

In the link given in post #16, the equator is now taken as the “fiducial” geodesic. Along the equator the $\Gamma$’s are all zero (but the derivatives with respect to $\theta$ of the $\Gamma$’s evaluated on the equator are not all zero.) In this case you can show that $\frac{D^{2}\xi^{\mu}}{D\lambda^{2}} = \frac{d^{2}\xi^{\mu}}{d\lambda^{2}}$. Also, for this case, you can show that the deviation vector does represent the distance between neighboring geodesics (at least up to a constant factor).

The geodesic deviation equation applies to two neighboring geodesics. In the attachment to post #4, the two cars are apparently not infinitesimally close to each other. So, technically, the geodesic equation does not apply directly to the geodesics of the two cars. But since the sphere has the same intrinsic geometry at all points, it is not hard to see how to go from the distance between neighboring geodesics to the distance between finitely separated geodesics.

 

[TSny]

Applying the term "distance" to this can be misleading, because (except at $\theta = \pi / 2$) it isn't evaluated along a geodesic curve. If, for example, we were to set up a local Euclidean coordinate chart centered on a point on the reference geodesic, a curve of constant $\theta$, which is what this is being evaluated along, would appear curved, not straight.

Yes. But for two neighboring points whose coordinates differ by an infinitesimal amount, $\delta x^{\mu}$, the distance between the points can be determined from $ds^2 = g_{\mu \nu} \delta x^{\mu} \delta x^{\nu}$. You don't need to worry about whether the distance is measured "along a geodesic" here.

For points separated by a finite distance, then the "distance between the points" needs to be interpreted. If you take this to mean the shortest distance on the sphere between the points, then it would be the distance along a geodesic connecting the points. But, of course, you could also measure distance along any curve between the points.

For two points widely separated on a curve of constant $\small \theta$, you could take the distance between the points as measured along the curve of constant $\small \theta$. This might be a convenient way to define the distance even though it does not correspond to the shortest distance between the points.

Suppose you have two geodesics such that each geodesic corresponds to a constant value of $\small \phi$ (i.e., two longitude lines) and you select a point on each geodesic corresponding to the same value of $\small \theta$ (i.e., two points on the same latitude). It might be convenient to define "the distance between the two longitude geodesics at that latitude" as the distance between the two points measured along the latitude line (rather than measured along a geodesic). [Of course, for some applications (e.g., airline flights!), this might not be a convenient definition.]

 

[peter46464]

Thanks to all for your help. Who would have thought two cars on a sphere could be so complicated.

Can I now assume (a) there's a consensus that Void's "constant curvature" answer (http://physics.stackexchange.com/qu...oes-the-ordinary-second-derivative-give-the-c) is correct, and (b) that the answer does not involve Riemann normal coordinates (because ξ^ϕ is measured along non-geodesic curves)?

Finally, can anyone provide a reasonably straightforward definition of "constant curvature" that might further my understanding of this problem but doesn't zoom off into the higher reaches of differential geometry? At my basic level, can I simply assume that constant curvature means that (as on the sphere) the equations of the components of the curvature tensor are constant?

 

[TrickyDicky]

Can I now assume (a) there's a consensus that Void's "constant curvature" answer (http://physics.stackexchange.com/qu...oes-the-ordinary-second-derivative-give-the-c) is correct, and (b) that the answer does not involve Riemann normal coordinates (because ξ^ϕ is measured along non-geodesic curves)?

There seems to be some basic misunderstanding here, the problem only involves computing a distance from certain given distance; as commented above the only reason it can be turned into a geodesic deviation type of problem and use the GD equation is because the sphere has constant curvature. You are not measuring anything, you are simply capable of normalizing your orthogonal basis to obtain a correct answer because of the particular properties of the space the problem is set in. But in a general manifold these assumptions wouldn't have allowed you to obtain the correct answer, you would have needed the Christoffels and you might not have been able to solve it from the given data.
Riemann normal coordinates can be used locally on any Riemannian manifold, they are an example of a situation where the Christoffel symbols vanish, that doesn't mean they are required to solve geodesic deviation problems. In this particular space setting, they, or more exactly the introduction of the orthonormal basis allow you to compute correctly the distance after time t, from the initial distance and the GD equation with regular derivative, but again you are not measuring anything.
See http://en.wikipedia.org/wiki/Jacobi_field it treats the sphere case.

Finally, can anyone provide a reasonably straightforward definition of "constant curvature" that might further my understanding of this problem but doesn't zoom off into the higher reaches of differential geometry?

It's just the same curvature at all points of the space.
http://en.wikipedia.org/wiki/Constant_curvature

 

[TrickyDicky]

(because ξ^ϕ is measured along non-geodesic curves)

The distance ξ^ϕ computed is always the shortest just like in the equator and it would correspond to the distance along great circles.

Edit: This distance is (as discussed by peterdonis and TSny) not actually ξ^ϕ, which is an infinitesimal distance, just trying not to confuse things further.

 

[peter46464]

There seems to be some basic misunderstanding here

Yep, as far as physics is concerned “basic misunderstanding” is my default position. However, in my defence I would say that to the uninitiated this is a genuinely confusing problem. The cars' separation $s$ is given by
$$s=\Delta\phi=d\cos\left(vt\right)=d\sin\theta$$and is calculated trignometrically here:

http://physics.stackexchange.com/questions/107421/geodesic-devation-on-a-two-sphere

I have used the geometric deviation equation to reach the exact same solution, yet it now turns out that along geodesics of constant $\phi$
$$\Delta\phi\neq\xi^{\phi}.$$To me that is subtle, especially given that the CalTech problem (post #16) at first glance looks pretty similar to this one. I can't claim to understand all the details of the solution, but I'm happy with getting the gist of it. This has been a convoluted (and thoroughly enjoyable) journey.

 

[TrickyDicky]

Yep, as far as physics is concerned “basic misunderstanding” is my default position. However, in my defence I would say that to the uninitiated this is a genuinely confusing problem. The cars' separation $s$ is given by
$$s=\Delta\phi=d\cos\left(vt\right)=d\sin\theta$$and is calculated trignometrically here:

http://physics.stackexchange.com/questions/107421/geodesic-devation-on-a-two-sphere

I have used the geometric deviation equation to reach the exact same solution, yet it now turns out that along geodesics of constant $\phi$
$$\Delta\phi\neq\xi^{\phi}.$$To me that is subtle, especially given that the CalTech problem (post #16) at first glance looks pretty similar to this one. I can't claim to understand all the details of the solution, but I'm happy with getting the gist of it. This has been a convoluted (and thoroughly enjoyable) journey.

I'm a layman so you should take what I write wth a grain of salt.
I'd say your second expression is an equality for the infinitesimal distance ds.