What do the three indices in \Gamma ^a_{bc} represent in the unit sphere?

[ehrenfest]

Homework Statement

I am confused about this question and probably the connection in general. When we only have two independent coordinates here what do the three indices in \Gamma ^a_{bc} stand for?

I found the metric for this space, but every formula for Gamma has indices a,b,c and I do not know what to plug in for them?

Homework Equations

The Attempt at a Solution

 

[Dick]

a,b and c stand for either phi or theta. I can say this without even seeing your attachment.

 

[ehrenfest]

So, when it asks me to calculate the coefficients, i need to calculate for

(a,b,c) = (theta, theta, theta)
(a,b,c) = (theta, theta, phi)
(a,b,c) = (theta, phi, theta)
(a,b,c) = (theta, phi, phi)

and then four more with a = phi ?

 

[Dick]

Yes. but you know case (a,b,c)=(a,c,b) right? And didn't you just do an exercise about diagonal metrics that would come in handy here?

 

[ehrenfest]

So, to do the next part, should I just plug the connection into 3.41 and integrate from 0 to 2 pi?

 

[Dick]

Your attachments still haven't been approved, but you should plug the connection into the parallel transport equations, if that's what you mean. And then solve them.

 

[ehrenfest]

Which equations do you mean? I know the connection and I think that I know the initial vector, so which equation relates these and the rotation angle to the final vector?

 

[Dick]

Which equations do you mean? I know the connection and I think that I know the initial vector, so which equation relates these and the rotation angle to the final vector?

Look up the 'parallel transport' operation mentioned in the problem.

 

[ehrenfest]

What to you mean look up the parallel transport operation? I already looked it up and I even posted that entire section from my book. Did you mean look it up elsewhere? Because I have tried that also.

I just want help figuring out which equations to plug the connection and the initial vector into and get the final vector.

 

[Dick]

Equation (3.42). Start trying to use it and then people can start trying to help you. Did you read what you posted? It's a differential equation for the rate of change of the transported vector components in terms of the tangent vector to the curve and the connection coefficients. You have to solve this differential equation.

 

[ehrenfest]

Here is my attempt:

\Gamma^{\phi}_{\phi \theta} = \Gamma^{\phi}_{\theta \phi} = \cot \theta

\Gamma^{\theta}_{\phi \phi} = - \sin \theta \cos \theta

The rest of the connection vanishes.

I plug this into the equation along with the initial position and the theta equation vanishes and the phi equation is

\frac{ dv^\phi}{du} = - \theta_0 \cot \theta \frac{dx^\phi}{du}

So, is the next step finding dx^phi/du with some parametrization?

 

[Dick]

The theta equation does not vanish. And why theta_0? Shouldn't that be v^theta? And use the obvious parametrization. Use phi as the path parameter.

 

[ehrenfest]

You're right. I was using the initial conditions for v^theta and v^phi. That is why I thought the theta equation vanishes.

So, the two equations are:

\frac{ dv^\theta}{du} = \sin \theta \cos \phi v^{\theta} \frac{d\phi}{du}\frac{ dv^\phi}{du} = - v^{\theta} \cot \theta \frac{d\phi}{du}

So when I replace u with theta as my parameter, why don't both of these equation vanish since they both would have a d\phi/d\theta and theta and phi are supposed to be independent?

 

[Dick]

You're right. I was using the initial conditions for v^theta and v^phi. That is why I thought the theta equation vanishes.

So, the two equations are:

\frac{ dv^\theta}{du} = \sin \theta \cos \phi v^{\theta} \frac{d\phi}{du}


\frac{ dv^\phi}{du} = - v^{\theta} \cot \theta \frac{d\phi}{du}

So when I replace u with theta as my parameter, why don't both of these equation vanish since they both would have a d\phi/d\theta and theta and phi are supposed to be independent?

That's why theta would not be a good choice of parameter. Your path is along constant theta with varying phi. Use phi. You also have some phi's and theta's in the wrong place in the first equation.

 

[ehrenfest]

You're right, the equations should be:

\frac{ dv^\theta}{du} = \sin \theta \cos \theta v^{\phi} \frac{d\phi}{du}
\frac{ dv^\phi}{du} = - v^{\theta} \cot \theta \frac{d\phi}{du}

And when I apply the phi parametrization (I don't know what I was thinking with the theta parametrization), I get\frac{ dv^\theta}{d\phi} = \sin \theta \cos \theta v^{\phi}
\frac{ dv^\phi}{d\phi} = - v^{\theta} \cot \theta

So if we integrate both of these equations w.r.t. phi, we get:

\Delta v^\theta = \sin \theta_0 \cos \theta_0 \int_{0}^{2\pi}v^{\phi}d\phi
\Delta v^\phi = - \cot\theta_0 \int_{0}^{2\pi} v^{\theta} d\phi

How do I evaluate the integrals on the RHS?

 

[Dick]

You can't just integrate them like that. You have two ODE's wrt to two different functions. You want to get a single ODE in terms of a single function to start out. Hint: differentiate the second one wrt phi so you get a second order ODE and then substitute the first one into eliminate v_theta.

 

[ehrenfest]

I get

\frac{d^2v^{\phi}}{d\theta} = - \cos \theta v^{\phi}

Since \theta = \theta_0 is a constant all along this parametrization, this DE has solutions:

v^{\phi} = A e^{i \cos \theta_0 \phi} + B e^{-i \cos \theta_0 \phi}

So, now I need to plug this into the other DE, find A and B, and plug in 2 pi for phi to get the final value of v^phi and v^theta, right?

EDIT: the first equation should be \frac{d^2v^{\phi}}{d\phi^2} = - \cos^2 \theta v^{\phi}

 

[Dick]

Basically right. Except I get a cos^2(theta) in your ODE. And life might be a little simpler if you use the sin,cos form of the solutions instead of complex exponentials, but it's up to you.

 

[ehrenfest]

I am having trouble finding A and B.

I get v^{\phi}_{initial} = A + B from the starting point.

The other equation should come from the fact that the vector is parallel to the circle phi = 0. Doesn't that condition just imply that the vector has only a phi component?
Should I use the unit size of the vector to get the other equation instead?

 

[Dick]

The other condition is the initial values of the derivatives of v_phi and v_theta. You aren't free to choose them arbitrarily. Your transport equations specify initial values of derivatives in terms of initial values of the v vectors. I don't know what makes you think v will have only a phi part.

 

[ehrenfest]

So you are saying I should use

\frac{ dv^\phi}{d\phi} = - v^{\theta} \cot \theta
i \cos \theta_0 A e^{i \cos \theta_0 \phi} + -i \cos \theta_0 B e^{-i \cos \theta_0 \phi} = - v^{\theta} \cot \theta_0

to get
i \cos \theta_0 A -i \cos \theta_0 B = - v^{\theta} \cot \theta_0
i A + -i B = - v^{\theta_0} \csc \theta_0
as my second equation, correct?

Seems weird that these coefficients are imaginary.

 

[Dick]

Why don't you just use the form v^{\phi} = A \cos {\cos \theta_0 \phi} + B \sin {\cos \theta_0 \phi} if the i's are making you nervous? And I think I've already pointed out that it looks to me like cos(theta) should be cos(theta)^2. And why don't you just take the initial values of v to be something like v^phi=1 and v^theta=0 at phi=0?

 

[Dick]

The original problem calls for something like (1,0) as the initial condition. But it does say to take it as a unit vector. That's not a unit vector. You'll have to adjust the first component to make it a unit vector.

 

[ehrenfest]

Why don't you just use the form v^{\phi} = A \cos {\cos \theta_0 \phi} + B \sin {\cos \theta_0 \phi} if the i's are making you nervous? And I think I've already pointed out that it looks to me like cos(theta) should be cos(theta)^2.

I thought you meant that

\frac{d^2v^{\phi}}{d\theta^2} = - \cos^2 \theta v^{\phi}

instead of

\frac{d^2v^{\phi}}{d\theta^2} = - \cos \theta v^{\phi} ?

My other equations are right, aren't they?

 

[ehrenfest]

Let's just use exponential form (just for fun):
Are these the correct equations that I should use to get A and B:

v^{\phi}_{initial} = A + B
i A + -i B = - v^{\theta_0} \csc \theta_0

Can we actually calculate the components of the initial vector just from the sentence "A vector v of unit length..."

 

[Dick]

I thought you meant that

\frac{d^2v^{\phi}}{d\theta^2} = - \cos^2 \theta v^{\phi}

instead of

\frac{d^2v^{\phi}}{d\theta^2} = - \cos \theta v^{\phi} ?

My other equations are right, aren't they?

I did. But it's dphi right? And certainly not all of them.

 

[Dick]

Let's just use exponential form (just for fun):
Are these the correct equations that I should use to get A and B:

v^{\phi}_{initial} = A + B
i A + -i B = - v^{\theta_0} \csc \theta_0

Can we actually calculate the components of the initial vector just from the sentence "A vector v of unit length..."

Yes you can. It has unit length and it's pointed in the phi direction. And if you are having fun with exponentials, then those equations just say A=B with an initial condition of v^theta(0)=0. And stop differentiating theta. It's a constant.

 

[ehrenfest]

But it's dphi right?

Fixed it.

 

[ehrenfest]

Yes you can. It has unit length and it's pointed in the phi direction. And if you are having fun with exponentials, then those equations just say A=B with an initial condition of v^theta(0)=0. And stop differentiating theta. It's a constant.

So we get that v^{\phi}_{initial} = v^0 = A/2 = B/2, correct?

Where did I differentiate theta?

You said earlier that the initial problem called for something "like" (v^theta_initial, v^phi_initial) = (0,1) but don't those two above conditions require that this and only this for (v^theta_initial, v^phi_initial)?

 

[Dick]

Yes, you are only asked that the initial conditions for v make it a unit vector (but you will find it remains a unit vector as it's transported - if we ever finish this). I was wondering where the csc(theta) came from but maybe it's just a typo. Oh, yeah and how many more times do I have to tell you it's cos^2(theta) in the argument of the exp? This is the third.

 

[ehrenfest]

You are saying that v^{\phi} = A e^{i \cos \theta_0 \phi} + B e^{-i \cos \theta_0 \phi} should be v^{\phi} = A e^{i \cos ^2\theta_0 \phi} + B e^{-i \cos ^2\theta_0 \phi} ?

I disagree because it is a solution to

\frac{d^2v^{\phi}}{d\phi^2} = - \cos^2 \theta v^{\phi}
Also, in post 21 (could you tell me what post I differentiated theta in?) the first equation is just the parallel transport equation for phi and I just plugged in the expression for v^phi.

So, if we collect everything up to know we get that A = B =2, do you agree?

 

[Dick]

Ok, right. I agree with you. My mistake. And I think I was mistaking a typo for a differentiation. Don't worry about it. I get that A=B=v^phi(0)/2.

 

[ehrenfest]

Yes you can. It has unit length and it's pointed in the phi direction. And if you are having fun with exponentials, then those equations just say A=B with an initial condition of v^theta(0)=0. And stop differentiating theta. It's a constant.

How do you know that v^theta(0)=0?

 

[Dick]

It really doesn't matter what initial conditions you take. But if you want to follow the instructions of the problem to the letter, then (I've just reread it) it says to take the initial unit vector parallel to the phi=0 circle. I guess this would mean we actually should have had the initial conditions to be v^phi(0)=0 and v^theta(0)=1.

 

[ehrenfest]

It really doesn't matter what initial conditions you take. But if you want to follow the instructions of the problem to the letter, then (I've just reread it) it says to take the initial unit vector parallel to the phi=0 circle. I guess this would mean we actually should have had the initial conditions to be v^phi(0)=0 and v^theta(0)=1.

I would think that it does matter what the initial conditions are but we'll see. So, with v^phi(0)=0 and v^theta(0)=1, I get:

A = -B = i/2 csc theta_0

 

[Dick]

Looks ok to me.

 

[ehrenfest]

Plugging in 2 pi for theta gives v^phi (phi = 2pi) = 0, right? So, now I need to solve the DE for v^theta and hopefully get v^theta (phi = 2pi) = 1, right?

 

[Dick]

Nooo. I get v^phi=constant*sin(cos(theta)*phi). Putting in phi=2pi doesn't give you zero for a general value of theta. And you should expect this - the problem asks you to show that after transport the direction v is not generally the same, but the length is the same.

 

[Dick]

Sure, but it's e^(i*cos(theta)*2pi), not e^(i*2pi).

 

[ehrenfest]

Sorry, I deleted a post in between there.

Anyway, e^(i*cos theta 2pi) = (e^(i*2pi))^cos(theta) = 1^{cos theta} = 1, right?

 

[Dick]

Wrong. That rule only works for positive real numbers. i=exp(i*2pi*(1/4))=exp(i*2pi)^(1/4)=1^(1/4)=1.

What went wrong there? It's true that i and 1 are both fourth roots of 1 but they aren't equal.

 

[ehrenfest]

Wrong. That rule only works for positive real numbers. i=exp(i*2pi*(1/4))=exp(i*2pi)^(1/4)=1^(1/4)=1.

What went wrong there? It's true that i and 1 are both fourth roots of 1 but they aren't equal.

Interesting!

I would need to see a counterexample for negative real numbers, though.

Yes, so then I get:

v^{\phi}(\phi = 2 \pi) = \sin ( \cos (\theta) 2 \pi) \csc \theta .

I am not really sure how to simplify that.

So, now I need to solve the DE for v^theta and hopefully get v^theta (phi = 2pi) = sqrt( 1- sin^2 ( cos (theta) 2 pi) csc^2 theta ), right?

 

[Dick]

Something like that. Just find v^phi and v^theta and put them into the metric. BTW ((-1)^2)^(1/2)=1 which is NOT EQUAL to (-1)^(2*(1/2))=(-1)^1=-1. In the real case you can allow the exponents to be negative, but not the base (or whatever you call the -1 part).

 

[ehrenfest]

OK. I plugged the expression for v^phi into the DE for v^theta and integrated and plugged in 2 pi for the phi parameter and got:

v^{\theta}( \phi = 2 \pi) = \cos^2 \theta \sin (\cos (\theta) 2 \pi)

When you say plug it into the metric, do you mean find g_ab v^a v_b?

 

[Dick]

I don't get the cos(theta)^2 and do I get that its a cos function of the argument. Check your derivation.

 

[ehrenfest]

Do you get:

v^{\theta}( \phi = 2 \pi) = -\cos (\cos (\theta) 2 \pi)

No way, that actually does give 1 when we plug into the metric. I think that completes the problem.

 

[Dick]

If you plug phi=0 in you get v^phi(0)=0, v^theta(0)=(-1) (we wanted +1). I think you have another sign error lurking around. But, yeah, that's basically it, once you correct your derivation of v^theta.