Curved Space-time and Relative Velocity

[Dale]

I am on a mobile device and cannot open the attachment. Can you put it in LaTeX? If not I will look at it in a couple of days when I return.

 

[Anamitra]

In my last [Thread #90]I got solutions of the type:
T(theta)=A cos[phi/root2] + B sin[phi/root2] ------------------------- (1)
And
T(phi)= C cos[phi/root2] + D sin [phi/root2] ---------------- (2)
[I have used T for the tensor instead of A]

To keep the norm constant we should have:
R^2 [ T(theta)^ 2+ [sin(theta)]^2 T(phi)^2]=const
We denote sin (theta) by K
R^2[T(theta)^2+k^T(phi)^2]= const
=>[T(theta)^2+k^T(phi)^2]= const
We may make the norm constant by the following choice
A^2+K^2 C^2 = B^2+K^2 D^2
K^2 CD=-AB
This may be achieved by choosing
A=B
C=-D
K^2 C^2=A^2
We have,
T(theta)/T(phi)=A[Cos (phi/root2) +Sin (phi/root2)]/C[Cos(phi/root2)-Sin (phi/root2)/]
For theta=0 , we have,
T(theta)/T(phi)=A/B
For theta =360 degrees
T(theta)/T(phi) obtains a different value
If norm is kept constant the vector is changing its orientation as it comes back to its original location.
Otherwise of course we may have variations, if the norm is allowed to change. We may write B=-A and D=-C to relate the constants. This gives the same value of T(theta)/T(phi) for theta=0 and theta=360 . At the same time the sign of each individual expression for T(theta) and T(phi) remains unaltered.

 

[Dale]

Hi Anamitra, I was able to look at your work this evening. The first page is correct. Starting on the second page you had a few mistakes.

First, you have a system of two first order differential equations, so there are only two unspecified constants of integration, not four. The correct form for the solution of this system is:
$$A^\theta=C_1 \; cos\left(\frac{\phi}{\sqrt{2}}\right)+\frac{C_2}{\sqrt{2}} \; sin\left(\frac{\phi}{\sqrt{2}}\right)$$
$$A^\phi=C_2 \; cos\left(\frac{\phi}{\sqrt{2}}\right)-\sqrt{2} \, C_1 \; sin\left(\frac{\phi}{\sqrt{2}}\right)$$

Then in the next section you got a little confused. When we set $\phi=0$ we get:
$$A^\theta(0)=C_1$$
$$A^\phi(0)=C_2$$

Plugging that back into the above we clearly see that
$$A^\theta(2 \pi) \ne A^\theta(0)$$
$$A^\phi(2 \pi) \ne A^\phi(0)$$
therefore the vector does not return to itself when parallel transported around a smooth curve, contrary to your above assertions.

Overall, you did fine setting up the equations, you just solved them wrong.

 

[Anamitra]

Yes you have got the constants so well by keeping the norm constant. In fact I have just now tested the norm by using the relation:

norm=R^2[A(theta)^2 + sin(theta)^2 A(phi)^square]
For your equations[solutions] the norm happens to be constant.In fact I have said this in my previous thread -that the vector changes its orientation if the the norm is kept constant. But I chose a different relation between the constants.


But in the case of light, if it is treated as a four vector how do we calculate its time component?[ I am asking this for my own understanding]

A Subsidiary Point

If along the line of latitude we make small adjustments of curvature to change the values of the metric coeffocients at some point or some lenghths (which could be small enough )we could generate a parallel transport that does not change the orientation of the vector after a round trip.In such a case the differential equations would change in different parts of the trip.The idea is to produce solutions which are periodic functions of 2*pi.But the inner part of the surface remains the same.If we make the enclosed surface very small to prevent this manoevre we are coming very close to flat space-time

 

[Dale]

Yes you have got the constants so well by keeping the norm constant. In fact I have just now tested the norm by using the relation:

norm=R^2[A(theta)^2 + sin(theta)^2 A(phi)^square]
For your equations[solutions] the norm happens to be constant.In fact I have said this in my previous thread -that the vector changes its orientation if the the norm is kept constant. But I chose a different relation between the constants.

Two quick points:

1) It is not simply a matter of choosing a different relation between the constants, you had too many constants. Any system of two first order differential equations has two constants, not four.

2) You are not free to choose a "different relation between the constants". If the norm does not remain constant then the dot product is not preserved and therefore it is not parallel transport.

With this counter-example using a smooth curve, do you now understand and agree that parallel transport is path-dependent? Do you have a better feeling for the geometric idea of curvature and the effect it has on parallel transport?

 

[Anamitra]

It is true that a relation between the constants is not sufficient--we need to have only two constants and not four.What physical condition did you use for that?We can assign arbitrary values to A(theta) and A(phi). But it is difficult to predict the initial value of the derivatives.Did you find the constants by the invariance of the dot product or you have you used some other physical condition? I am very much interested in knowing that for my own understanding.

Another point to be addressed is my Subsidiary Point in thread #94.

 

[Dale]

It is true that a relation between the constants is not sufficient--we need to have only two constants and not four.What physical condition did you use for that?We can assign arbitrary values to A(theta) and A(phi). But it is difficult to predict the initial value of the derivatives.Did you find the constants by the invariance of the dot product or you have you used some other physical condition?

The value of the two-dimensional vector A at anyone point along the curve (e.g. phi=0) provides the physical condition needed to determine the two constants of integration. There is no need to "predict the initial value of the derivatives". The invariance of the dot product falls out naturally from the definition of parallel transport and does not need to be added in by hand later, i.e. it is already in the equations that you correctly set up on the first page.

Regarding your subsidiary point: Even if you find an example of a closed path that maps the vectors back onto themselves that still does not make the result of parallel transport independent of the path in general.

 

[Anamitra]

Regarding your subsidiary point: Even if you find an example of a closed path that maps the vectors back onto themselves that still does not make the result of parallel transport independent of the path in general.

So parallel transport does not give us a report of curvature for every instance!

Next issue: How do we calculate the time component of the velocity of light if it is to be treated as a four vector?

 

[nismaratwork]

So parallel transport does not give us a report of curvature for every instance!

Next issue: How do we calculate the time component of the velocity of light if it is to be treated as a four vector?

The time component of light is 0, or the equal of its space components, but with an opposite sign to cancel. Whatever gets you a null-geodesic works in my view, but that's just my opinion.

 

[Dale]

Anamitra, I don't think we are ready to pursue any next issues yet. We still need to come to a resolution on the many previous issues:

Do you now understand how parallel transport is path dependent in curved spaces? Essentially, do you understand that your "final conclusions" post was wrong and (more importantly) why it was wrong?

Do you understand how different chains of inertial frames could lead to different results? Do you understand how the dot product could be preserved even though parallel transport is path-dependent?

On a less important note, are you still stuck on smooth paths or do you understand how a finite number of sharp bends is acceptable?

 

[Anamitra]

A local inertial frame is not unique. There are an infinite number of such frames. If you go on different chains of inertial states you may still wind up with different final inertial frames and different final vectors. A chain of locally inertial frames is not sufficient to establish uniqueness. The argument is not as strong as you seem to believe.

We may have an infinite number of inertial frames at each point and therefore we may have several such chains connecting a pair of initial and final point on the space time surface. For each such chain dA(mu)/d(xi)=0.For a "particular inertial frame" at the initial point we may choose several inertial paths connecting the initial point to the final point.We end up with the same tensor finally.If we change the inertial frame at the initial point we adjust the "inertial paths". There should be no problem in such a procedure.

Of course parallel transport is causing a lot of problem in defining Relative velocity if we leave aside the aspect of the chain of inertial frames.
Now let us come to a relevant issue.If somebody sees a moving car at some distance in front of him in curved space-time he should have some idea/report of the motion.If there are problems in defining relative motion in curved spacetime[I am assuming this for argument's sake] it does not mean that relative motion is meaningless.I tried to highlight this in Query 1 of thread #1.The incapability of the mathematical apparatus in defining a physical quantity[in case such an incapability exists] does not imply the non-existence of the physical quantity itself.

 

[Anamitra]

An Important Point

For flat space-time the equation for parallel transport is given by:

dA(mu)/dx(i)=0 as we move along the path[The christoffel tensors are equal to zero]

So the components should not change individually .But this not true for spherical or polar coordinates. Though the vector in the final state remains parallel to its initial position the components do change for a path that is not closed.
"Gravity" by James Hatley,page 456,Figure 20.3
[Chapter 20,Section 20.4:The covariant derivative]

 

[Dale]

We may have an infinite number of inertial frames at each point and therefore we may have several such chains connecting a pair of initial and final point on the space time surface. For each such chain dA(mu)/d(xi)=0.For a "particular inertial frame" at the initial point we may choose several inertial paths connecting the initial point to the final point.We end up with the same tensor finally.

Not in general, no. If you cannot accept this then you need to do some homework problems from your favorite GR textbook. Wald's problems may not be of the "practical" sort that you need.

I don't see any reason to move to other points when you are still stuck on the basics. If we cannot agree on the simple and obvious mathematical fact of the non-uniqueness of parallel transport in curved spaces then any other discussion will be pointless.

 

[Dale]

An Important Point

For flat space-time the equation for parallel transport is given by:

dA(mu)/dx(i)=0 as we move along the path[The christoffel tensors are equal to zero]

No, the equation for parallel transport remains the same in flat spacetime. Some coordinate systems in flat spacetime have non-zero Christoffel symbols as you point out, therefore you cannot simply drop them.

However, in flat spacetime it is always possible to perform a global coordinate transform to a standard Minkowski inertial frame where the Christoffel symbols are all 0 and the parallel transport equation simplifies as you propose. In this simplification you immediately see that the parallel transport equation becomes independent of the path in a Minkowski inertial frame, and therefore (since the covariant derivative is a tensor operation) it is true in any coordinate system in flat spacetime.

[nitpick]The Christoffel symbols are not tensors. The correct terminology is "Christoffel symbols".[/nitpick]

 

[Anamitra]

Regarding #103: I have sufficient difficulty in accepting what DaleSpam has to say. I have no hesitation in working out any homework problem he suggests. I have done this before.Nevertheless I would like to clarify my stand on this issue once more.
A freely falling lift is an inertial frame in the gravitational field of the earth.Now we may assign different velocities to it without spoiling the inertial nature of the frame. This may be accepted in a general way. We consider two transformations from the same metric leading to the Minkowski matrix[1 -1 -1 -1] in a local way.From special relativity we know that they must be moving with uniform speed with respect to each other. Now we divide our path from A to B into small intervals (A ,A1),(A1,A2)...(A[n-1],B)
In each interval we choose a frame with the same velocity V. The intervals being very small we choose for every interval V=V+delta_V approximately.So we have several coordinate systems which are not in relative motion.We may view them as rectangular coordinate systems in consideration of the Minkowski matrix[1 -1-1-1]Then we move our vector through these intervals.It remains constant since dA(mu)/dx(i)=0 for each interval.The vector remains unchanged at the end point B. For any other path we repeat the same manoeuvre starting with the same velocity at A.
In case there is some mistake in my method it has to be pointed out in a specific way. Of course I am ready to work out any practice problem suggested.No harm in doing that.

On Parallel transport:For parallel transport we have simply come to the conclusion that for a round trip the orientation of the vector may change or it may not change even for a curved surface.So the relationship between parallel transport and the curvature is not as strong or conclusive as we might be tempted to think of.

A fundamental issue has been raised in thread #101 in the second paragraph. It has not been addressed.

Regarding Thread #102: Yes ,there has been a mistake

 

[Dale]

Regarding #103: I have sufficient difficulty in accepting what DaleSpam has to say. I have no hesitation in working out any homework problem he suggests. I have done this before.Nevertheless I would like to clarify my stand on this issue once more.
A freely falling lift is an inertial frame in the gravitational field of the earth.Now we may assign different velocities to it without spoiling the inertial nature of the frame. This may be accepted in a general way. We consider two transformations from the same metric leading to the Minkowski matrix[1 -1 -1 -1] in a local way.From special relativity we know that they must be moving with uniform speed with respect to each other. Now we divide our path from A to B into small intervals (A ,A1),(A1,A2)...(A[n-1],B)
In each interval we choose a frame with the same velocity V. The intervals being very small we choose for every interval V=V+delta_V approximately.So we have several coordinate systems which are not in relative motion.We may view them as rectangular coordinate systems in consideration of the Minkowski matrix[1 -1-1-1]Then we move our vector through these intervals.It remains constant since dA(mu)/dx(i)=0 for each interval.The vector remains unchanged at the end point B. For any other path we repeat the same manoeuvre starting with the same velocity at A.
In case there is some mistake in my method it has to be pointed out in a specific way. Of course I am ready to work out any practice problem suggested.No harm in doing that.

OK, then using the Schwarzschild metric (in units where c=1, G=1, and M=1/2):
$$ds^{2} = <br /> -\left(1 - \frac{1}{r} \right) dt^2 + \frac{dr^2}{\displaystyle{1-\frac{1}{r}}} + r^2 \left(d\theta^2 + \sin^2\theta \, d\phi^2\right)$$

Start at $$t=0$$, $$r=2$$, $$\theta=45^{\circ}$$, $$\phi=0^{\circ}$$ and parallel transport an arbitrary vector to $$\theta=45^{\circ}$$, $$\phi=180^{\circ}$$ along:
a) the 45º colatitude line
b) the 0º and 180º longitude line

Work both cases using the standard parallel transport equation, and also try to work both using your "inertial frame" method.

 

[Anamitra]

Parallel Transport through Inertial Frames
We have the Schwarzschild metric for m=1/2

ds^2=[1-1/r]dt^2-[1-1/r]^(-1) dr^2 - r^2[d(theta)^2+[sin(theta)]^2 d(phi)^2]

In the above metric we hold r constant making r=R
We keep time variable . In the final stage we will consider dt->0 so that time becomes constant and we are back in a loop.
We have for the 45 degree latitude line:
ds^2=[1-1/R]dt^2 - R^2[d(theta)^2+1/2 d(phi)^2]
[Though theta is constant for a latitude we consider small variations in theta for better workability of the metric. Of course sin(theta) does not change appreciably. It may be taken as a constant]

We use the transformations:

T=root[(1-1/R)t]

x=R theta

y=R(1/root2)phi

The last two transformations apply locally.
Regarding time: It represents physical time [having the dimension of time]
Regarding x: The x-axis is parallel to the vector e(phi)
Regarding y: The y-axis is parallel to the vector e(theta)
[important to note that we have a z-axis along e(r) ]
Our metric now:
ds^2=dT^2 - dx^2-dy^2
Movement along the latitude:
We consider an infinitesimal movement of the coordinate frame along the line of latitude.In each new position the new [x,y,z] triad corresponds to the new position of [e(phi),e(theta),e(r)]There are two effects:
1) An infinitesimal translation that leaves the direction of the axes unchanged.
x=x+x'

y=y+y'

z=z+z'
The metric now:
ds^2=dT^2 - dx'^2-dy'^2-dz'^2
2) A infinitesimal rotation which may be expressed by a matrix consisting of the Eulerian angles.
We leave the translation unchanged but we apply the inverse transformation to cancel the effect of rotation.

Form of the metric:
ds^2=dT^2 - dx'^2-dy'^2-dz'^2

We carry on this process as we move along the line of latitude
It is to be noted that the axes do not change their orientation as we move from one frame to another. The frames are not in relative motion.

Parallel Transport Equation in each frame:

dA(mu)/dA(nu)=0

The components do not change as we pass in the same frame. They do not change as we pass from one frame to another.
The vector does not change its orientation at any instant.

Movement along the meridian: Similar arguments may be applied in the case of a meridian. We simply move the [x,y,z]triad along the meridian instead of the latitude. Each small movement can be decomposed into a translation and a rotation. We reverse the effect of rotation keeping the translation in tact.
Net Effect:
1) The metric has the diagonal form[1-1-1-1]
2) In each situation the axes remain parallel to the previous situation[respectively]
3) There is no relative motion between the frames.
4) In each frame we have
5) As we pass from frame to frame the components remain unchanged

 

[Anamitra]

A Suggested Experiment:

Two persons A and B are standing at two distinct points in curved spacetime where the metrics have different values[especially in relation to g(0,0)].B performs an experiment at his own location by sending a light ray(say gamma rays) through a cloud chamber and measures the length of the track and also records the time interval

Speed=[length of track]/time interval measured by B

This should be something less than "c" considering the the slowing down of the light ray by the medium

Let the observed speed be c/2

Observation of A:The clock of A runs at a different rate from that of B.Let us assume that the intervals measured by A are 100 times shorter than that of BSpeed measured by A=[Length of Track]/time interval of A
=[[Length of Track]/time interval of B] * 100
=50 * c
W are getting this result because A and B are non-local points.

Observation in the physical world cannot be suppressed by by the incapability of the mathematical apparatus to cope up with the situation[in case such an incapability exists, for the sake of argument]

[Instead of a light ray we could use a fast moving electron (or an alpha particle)also,say one that moves at 0.99c]

 

[Dale]

Please finish working the problems. Then we can discuss the results which should help you understand the next parts of the discussion.

 

[Anamitra]

Regarding DaleSpam's Problems:
Four equations for parallel transport have been considered for each case,the latitude and the longitude.
1)Changes of A(theta) A(phi) A(r) and A(t) for changes in phi have been considered for theta=const and r=constant
2)Changes of A(theta) A(phi) A(r) and A(t)[for changes theta)have been considered for phi=const and r=constant

The initial values of A(r) and A(t) are zero. But even with that the equations yield non-zero values of A(r) in general. A(t) is not showing any problem. The vector seems to lift out of the tangent plane to satisfy the equations. That needs an explanation.

[The work has been shown in the attachment]

 

[Dale]

Hi Anamitra,

I have looked at the first part, transport along the 45º lattitude line. My results are pretty close to yours, with I suspect only a single small mistake that had minimal impact.

In the third differential equation you are missing a sin² term:
$$\frac{dA^r}{d\phi}-(r-2M) A^{\phi} sin^2(\theta)=0$$

This then leads to a couple of factors of 2 (at 45º) in the solution:
$$A^r=\frac{r-2M}{2k}\left( C_1 sin(k\phi) - C_2 cos(k\phi) \right)-2 r C_3 cos^2(\theta)$$

The errors don't affect the form of the solution, just a couple of small details, so I don't think they are a big deal.

Regarding the lifting from the tangent plane, this is obvious and expected. In fact, if it didn't happen we would immediately know that there is a problem. Consider a vector pointing due north at the equator in the flat space limit of M=0, this vector is purely tangential at the equator and purely radial when transported to the poles.

I will get to the second half in a bit.

 

[Dale]

On the second half I have a different expression fro the second equation. Specifically, I have:

$$\frac{dA^{\phi}}{d\theta}+A^{\phi}cot(\theta)=0$$

You have an extra term in there. Unfortunately, this one makes a big difference. I don't get even close to your solutions.

 

[Anamitra]

Yes ,I made a mistake with a Christoffel symbol.And I am repeating the thing.In any case cot(theta) causes some problem for theta=0 [and theta=pi]. How to get over it?I mean to say, one must write a separate equation for the neighborhood of theta=0[and pi].The values are enormously large here.
.

 

[Anamitra]

The revised calculations have been uploaded.The effect of the infinite discontinuity reflects itself in the problem.A(phi) is undefined at theta=0 and theta=pi.It tends to an unbounded value as theta tends to 0 or pi.

 

[Dale]

Your revised section is correct. Also, the cot and csc functions are correct. These don't have anything particularly to do with curved spacetime or parallel transport, but simply are features of spherical coordinates where near the poles a small change in $\theta$ leads to a large change in $A^{\phi}$. You can set M=0 and work in a flat spacetime as we did before to see that this is correct.

So, let us now review the recent results. Using the definition of parallel transport you have parallel transported a vector from one location to another location in curved spacetime using two different paths and obtained two different results.

Do you now agree with that and understand that parallel transport is indeed path dependent in a curved spacetime?

If you agree with that then we can investigate in a little more detail the "inertial frames" approach.

 

[Anamitra]

$$\frac{dA^{\phi}}{d\theta}+A^{\phi}cot(\theta)=0$$

If you the solve the equation in thread#112[ https://www.physicsforums.com/showpost.php?p=2865831&postcount=112] you get:

A(phi)=c* Cosec (theta)

Now this blows up in the neighborhood of theta =0 or theta =pi.

A small change in theta produces a large change in A(phi)----that part is ,perhaps,OK. But the value of A(phi) getting infinitely large at the poles[in the actual sense of its value] cannot be entertained.The concept of parallel transport is simply undefined at the poles.

 

[Dale]

If you the solve the equation in thread#112[ https://www.physicsforums.com/showpost.php?p=2865831&postcount=112] you get:

A(phi)=c* Cosec (theta)

Now this blows up in the neighborhood of theta =0 or theta =pi.

Yes, as it should in spherical coordinates.

A small change in theta produces a large change in A(phi)----that part is OK. But the value of A(phi) getting infinitely large at the poles[in the actual sense of its value] cannot be entertained.The concept of parallel transport is simply undefined at the poles.

Parallel transport is fine. It is just spherical coordinates where phi is undefined at the poles. This has nothing to do with parallel transport or curved spacetime, this is just "business as usual" in spherical coordinates. Your objections in both .pdf files are not about parallel transport, but spherical coordinates. I get the impression that you have not worked in spherical coordinates very much, or you would have seen this type of behavior previously.

However, if you want to modify the path so that it avoids the poles then feel free to do so. I am glad to work with any alternative path you choose (other than the latitude line since we already did that one). Do you have a suggestion?

 

[Anamitra]

For the Swarzschild Sphere parallel transport is working well with the exclusion of the poles which admit themselves to multiple values for phi.

Now let us look into the fact that an ordinary sphere is commonly used to illustrate the concept of parallel transport in the texts. You did the same thing the following thread:
https://www.physicsforums.com/showpost.php?p=2758350&postcount=25

If you write the three equations [for A(r),A(phi) and A(theta)] for a line of latitude and solve them you get functions like Sin(phi) and cos(phi) in the solutions which are periodic functions of 2pi .

So the vector returns to its original orientation in this case!
[File has been uploaded for consideration]

 

[Dale]

I will be on a mobile device for the next couple of days. Please use LaTeX directly in the forum. There is no need to keep posting .pdf files.

For the Swarzschild Sphere parallel transport is working well with the exclusion of the poles which admit themselves to multiple values for phi.

What do you mean by "working well"? Do you mean that you finally understand parallel transport and how it is path dependent?

If so then we can begin with the other issues.

 

[George Jones]

Now let us look into the fact that an ordinary sphere is commonly used to illustrate the concept of parallel transport in the texts. You did the same thing the following thread:
https://www.physicsforums.com/showpost.php?p=2758350&postcount=25

If you write the three equations [for A(r),A(phi) and A(theta)] for a line of latitude and solve them you get functions like Sin(phi) and cos(phi) in the solutions which are periodic functions of 2pi .

So the vector returns to its original orientation in this case!

Does it?

My daughter was constantly pestering me while I was working on this, so what follows could be full of mistakes.

Tangent spaces of an "ordinary sphere'' are 2-dimensional, and vectors in these tangent spaces don't have $r$ components. Therefore, set $r=1$ and $A^{r}=0$. Your three equations then become two equations,

$$\begin{equation*}<br /> \begin{split}<br /> \frac{dA^{\phi }}{d\phi }+\cot \theta A^{\theta } &=0 \\<br /> \frac{dA^{\theta }}{d\phi }-\sin \theta \cos \theta A^{\phi } &=0.<br /> \end{split}<br /> \end{equation*}$$

This set of equations can be solved elegantly using 2x2 matrices, but I'll solve it using a different method. Differentiating the first equation gives

$$\frac{d^{2}A^{\phi }}{d\phi ^{2}}+\cot \theta \frac{dA^{\theta }}{d\phi }=0,$$

and using the second equation in this gives

$$\frac{d^{2}A^{\phi }}{d\phi ^{2}}+\cos ^{2}\theta A^{\phi }=0.$$

Since $\theta$ is constant along a line of longitude, this is just a harmonic oscillator equation with solution

$$A^{\phi }\left( \phi \right) =c_{1}\sin \left( \phi \cos \theta \right) +c_{2}\cos \left( \phi \cos \theta \right)$$

Then,

$$\begin{equation*}<br /> \begin{split}<br /> \frac{dA^{\phi }}{d\phi }\left( \phi \right) &= \cos \theta \left[ c_{1}\cos \left( \phi \cos \theta \right) -c_{2}\sin \left( \phi \cos \theta \right) \right] \\<br /> &=-\cot \theta A^{\phi }<br /> \end{split}<br /> \end{equation*}$$

and, from the first equation,

$$A^{\theta }\left( \phi \right) =\sin \theta \left[ c_{1}\cos \left( \phi \cos \theta \right) -c_{2}\sin \left( \phi \cos \theta \right) \right].$$

Consequently,

$$\begin{equation*}<br /> \begin{split}<br /> A^{\phi }\left( 0\right) &= c_{2}\cos \left( \phi \cos \theta \right) \\<br /> A^{\theta }\left( 0\right) &= c_{1}\sin \theta \cos \left( \phi \cos \theta \right) <br /> \end{split}<br /> \end{equation*}$$

and

$$\begin{equation*}<br /> \begin{split}<br /> A^{\phi }\left( 2\pi \right) &= c_{1}\sin \left( 2\pi \cos \theta \right) +c_{2}\cos \left( 2\pi \cos \theta \right) \\<br /> A^{\theta }\left( 2\pi \right) &= \sin \theta \left[ c_{1}\cos \left( 2\pi \cos \theta \right) -c_{2}\sin \left( 2\pi \cos \theta \right) \right].<br /> \end{split}<br /> \end{equation*}$$

As a simple example, take $c_{1}=1/\sin \theta$ and $c_{2}=0$, so that

$$\begin{equation*}<br /> \begin{split}<br /> A^{\phi }\left( 0\right) &= 0 \\<br /> A^{\theta }\left( 0\right) &= 1<br /> \end{split}<br /> \end{equation*}$$

and

$$\begin{equation*}<br /> \begin{split}<br /> A^{\phi }\left( 2\pi \right) &= \sin \left( 2\pi \cos \theta \right) /\sin \theta \\<br /> A^{\theta }\left( 2\pi \right) &= \cos \left( 2\pi \cos \theta \right).<br /> \end{split}<br /> \end{equation*}$$

What does this mean?