How to Calculate Christoffel Symbols in Spherical Coordinates?

[gabbagabbahey]

Your $\LaTeX$ contains a few typos, but yes, that's correct.

Also, it is worth writing $\dot{\phi}$ and $\dot{r}$ in spherical coordinates;

$$\dot{r}=b_x\sin\theta\cos\phi+ b_y\sin\theta\sin\phi+b_z\cos\theta$$

$$\dot{\phi}=\frac{b_y\cos\phi-b_x\sin\phi}{r\sin\theta}$$

These will come in handy when calculating $\ddot{\theta}$ (Since $\dot{\theta}$ will be a function of $r$, $\theta$ and $\phi$; knowing these derivatives allows you to find $\ddot{\theta}$ quickly, using the chain rule)

 

[Phrak]

I recall that a straight line in cartesian coordinates is $$\ \frac{d^2 x^i}{d \lambda^2} = 0 \ ,$$
where the curve is parametric in lambda. The parameter can be time.

$$\frac{d^2 x^i}{d t^2} = 0$$

 

[gabbagabbahey]

I recall that a straight line in cartesian coordinates is $$\ \frac{d^2 x^i}{d \lambda^2} = 0 \ ,$$
where the curve is parametric in lambda. The parameter can be time.

$$\frac{d^2 x^i}{d t^2} = 0$$

This is true; which is why the solution is of the form $x^i=a_i+b_it$ in Cartesians.

 

[latentcorpse]

ok. so finally $\ddot{\theta}$

$$\theta=\cos^{-1} \left( \frac{z}{r} \right)$$

$$\dot{\theta} = - \frac{1}{\sqrt{1 - \left( \frac{z}{r} \right)^2 }} \left( \dot{z} r^{-1} - z r^{-2} \dot{r} \right)$$

$$\dot{\theta}=-\frac{1}{\sqrt{r^4-z^2r^2}} \left(\dot{z}r-z \dot{r} \right)$$

is that ok for $\dot{\theta}$?

 

[gabbagabbahey]

Looks fine so far; now express it in terms of spherical coordinates and $b_x$, $b_y$ and $b_z$...

 

[latentcorpse]

$$\dot{\theta}=-\frac{1}{r^2 \sqrt{1-\cos^2{\theta}}} \left( \dot{z}r - z \dot{r} \right) = -\frac{1}{r^2 \sin{\theta}} \left( b_z r - \left( a_z + b_z t \right) \left(b_x \sin{\theta} \cos{\phi} + b_y \sin{\theta} \sin{\phi} + b_z \cos{\theta} \right)$$

looking ok?

 

[gabbagabbahey]

I'd use $z=r\cos\theta$ instead of $z=a_z+b_zt$...

 

[latentcorpse]

$$\dot{\theta}=-\frac{1}{r \sin{\theta}} \left(b_z - \cos{\theta} \right) \left( b_x \sin{\theta} \cos{\phi} + b_y \sin{\theta} \sin{\phi} + b_z \cos{\theta} \right)$$
that's ok?

should i take the derivative of the above or should i take the derivative of something shorter like:

$$\dot{\theta}=-\frac{1}{r \sin{\theta}} \left( \dot{z} r - z \dot{r} \right)$$?

this gives

$$\ddot{\theta}=\frac{ \left( \ddot{z} r + \dot{z} \dot{r} - \dot{z} \dot{r} - z \ddot{r} \right) r \sin{\theta} - \left( \dot{r} \sin{\theta} + r \dot{\theta} \cos{\theta} \right) \left( \dot{z} r - z \dot{r} \right) }{r^2 \sin^2{\theta}}$$

$$\ddot{\theta}=\frac{-zr \ddot{r} \sin{\theta}- \left( \dot{r} \sin{\theta} + r \dot{\theta} \cos{\theta} \right) \left( \dot{z} r - z \dot{r} \right)}{r^2 \sin^2{\theta}}$$

on the right lines?

 

[gabbagabbahey]

$$\dot{\theta}=-\frac{1}{r \sin{\theta}} \left(b_z - \cos{\theta} \right) \left( b_x \sin{\theta} \cos{\phi} + b_y \sin{\theta} \sin{\phi} + b_z \cos{\theta} \right)$$
that's ok?

Do $b_z$ and $\cos\theta$ have the same units? If not, the appearance of the factor $b_z-\cos\theta$ should be a dead giveaway that you've made an error somewhere...(It's always a good idea to check the units of your expression at each step of a complicated calculation, it will help cut down on the number of error you make)

 

[latentcorpse]

you said i was fine at $$\dot{\theta}=-\frac{1}{\sqrt{r^4-z^2r^2}} \left(\dot{z}r-z \dot{r} \right)$$

so i need to change the coords here

$$\dot{\theta}=-\frac{1}{\sqrt{r^4-r^4 \cos^2{\theta}}} \left(b_z r - r \cos{\theta} \dot{r} \right) = -\frac{1}{r \sin{\theta}} \left( b_z - \cos{\theta} \dot{r} \right)$$
ok that should be better

now

$$\ddot{\theta}=- \left( \frac{ \left( - \ddot{r} \cos{\theta} + \dot{r} \dot{\theta} \sin{\theta} \right) r \sin{\theta} - \left( \dot{r} \sin{\theta} + r \dot{\theta} \cos{\theta} \right) \left(b_z - \dot{r} \cos{\theta} \right)}{r^2 \sin^2{\theta}} \right)$$

is that ok? it doesn't look like it's going to get much simpler : i can bring the minus at the front in and cancel the $r \sin{\theta}$ in the first term i guess...

 

[gabbagabbahey]

you said i was fine at $$\dot{\theta}=-\frac{1}{\sqrt{r^4-z^2r^2}} \left(\dot{z}r-z \dot{r} \right)$$

so i need to change the coords here

$$\dot{\theta}=-\frac{1}{\sqrt{r^4-r^4 \cos^2{\theta}}} \left(b_z r - r \cos{\theta} \dot{r} \right) = -\frac{1}{r \sin{\theta}} \left( b_z - \cos{\theta} \dot{r} \right)$$
ok that should be better

Why not substitute in your expression for $\dot{r}$ here?

 

[latentcorpse]

i get

$$\dot{\theta}=-\frac{1}{r \sin{\theta}} \left(b_z-\cos{\theta} \dot{r} \right)$$

$$\dot{\theta}=-\frac{1}{r \sin{\theta}} \left( b_z - b_x \sin{\theta} \cos{\theta} \cos{\phi} - b_y \sin{\theta} \cos{\theta} \sin{\phi} - b_z \cos^2{\theta} \right)$$

$$\dot{\theta}=-\frac{1}{r \sin{\theta}} \left( b_z \sin^2{\theta} - b_x \sin{\theta} \cos{\theta} \cos{\phi} - b_y \sin{\theta} \cos{\theta} \sin{\phi} \right)$$

$$\dot{\theta}=-\frac{1}{r} \left(b_z \sin{\theta} - b_x \cos{\theta} \cos{\phi} - b_y \cos{\theta} \sin{\phi} \right)$$
now do i take the time derivative of this?

that gives, by product rule:

$$\ddot{\theta}=r^{-2} \dot{r} \left( b_z \sin{\theta} - b_x \cos{\theta} \cos{\phi} - b_y \cos{\theta} \sin{\phi} \right) - \frac{1}{r} \left( b_z \dot{\theta} \cos{\theta} + b_x \dot{\theta} \sin{\theta} \cos{\phi} + b_x \dot{\phi} \cos{\theta} \sin{\phi} + b_y \dot{\theta} \sin{\theta} \sin{\phi} - b_y \dot{\phi} \cos{\theta} \cos{\phi} \right)$$

$$\ddot{\theta} = \frac{\dot{r}}{r^2} \left(b_z \sin{\theta} - b_x \cos{\theta} \cos{\phi} - b_y \cos{\theta} \sin{\phi} \right) - \frac{1}{r} \left( \dot{\theta} \dot{r} + \dot{\phi} b_x \cos{\theta} \sin{\phi} - \dot{\phi} b_y \cos{\theta} \cos{\phi} \right)$$

is this looking ok? i can't seem to simplify it any further though.

 

[gabbagabbahey]

Looks fine to me...now realize that the first term is just

$$\frac{\dot{r}}{r^2} \left(b_z \sin{\theta} - b_x \cos{\theta} \cos{\phi} - b_y \cos{\theta} \sin{\phi} \right)=-\frac{\dot{r}\dot{\theta}}{r}$$

And

$$-\frac{1}{r}\left(\dot{\phi} b_x \cos{\theta} \sin{\phi} - \dot{\phi} b_y \cos{\theta} \cos{\phi} \right)=\sin\theta\cos\theta\dot{\phi}^2$$

So,

$$\ddot{\theta}=-\frac{2}{r}\dot{r}\dot{\theta}+ \sin\theta\cos\theta\dot{\phi}^2$$

Compare that to your second Geodesic equation...

Edit: Is there a reason you've interpreted $$\frac{dx^i}{dt}\frac{dx^i}{dt}$$ as $$\frac{d^2x^i}{dt^2}$$ when calculating your Geodesic equations in post#22?

 

[latentcorpse]

should the geodesic eqns be

$$\frac{d^2 r}{dt^2}-r \frac{d \theta}{dt} \frac{d \theta}{dt} - r \sin^2{\theta} \frac{d \phi}{dt} \frac{d \phi}{dt}=0$$
$$\frac{d^2 \theta}{dt^2} + \frac{2}{r} \frac{dr}{dt} \frac{d \theta}{dt} - \sin{\theta} \cos{\theta} \frac{d \phi}{dt} \frac{d \phi}{dt}=0$$
$$\frac{d^2 \phi}{dt^2} + \frac{2}{r} \frac{dr}{dt} \frac{d \phi}{dt} + 2 \cot{\theta} \frac{d \theta}{dt} \frac{d \phi}{dt}=0$$?

 

[gabbagabbahey]

Yup.

 

[latentcorpse]

okay so I am testing the equations now:
looking at the first one i have

$$\ddot{r}-r \dot{\theta}^2 - r \sin^2{\theta} \dot{\phi}^2$$
$$=\frac{b^2}{r}-\frac{ \left( b_x x + b_y y + b_z z \right)^2}{r^3} + b_z \sin{\theta} - b_x \cos{\theta} \cos{\phi} - b_y \cos{\theta} \sin{\phi} + \frac{2}{r} \left( b_x b_y \cos^2{\phi} + b_y^2 \sin{\phi} \cos{\phi} - b_x^2 \sin{\phi} \cos{\phi} -b_x b_y \sin^2{\phi} \right)$$

i can't get any constructive cancellation after this line though...

 

[gabbagabbahey]

It might be easier if you just use the chain rule to find $\ddot{r}$ from this equation $\dot{r}=b_x\sin\theta\cos\phi+ b_y\sin\theta\sin\phi+b_z\cos\theta$...that way everything is in Spherical coords.

 

[latentcorpse]

so $$\ddot{r}= \dot{\theta} \left( b_x \cos{\theta} \cos{\phi} + b_y \cos{\theta} \sin{\phi} - b_z \sin{\theta} \right) + \dot{\phi} \left( b_y \sin{\theta} \cos{\phi} - b_x \sin{\theta} \sin{\phi} \right)$$

$$\ddot{r}=r \dot{\theta}^2 + \dot{\phi} \left( b_y \sin{\theta} \cos{\phi} - b_x \sin{\theta} \sin{\phi} \right)$$

$$\ddot{r}=r \dot{\theta}^2 + r \sin^2{\theta} \dot{\phi}^2$$
so geodesic eqns 1 and 2 are obviously satisfied.

the third one is giving me a bit of grief though:

$$\ddot{\phi} + \frac{2}{r} \dot{r} \dot{\phi} + 2 \cot{\theta} \dot{\theta} \dot{\phi}$$
$$=\frac{-2 \left(b_xb_y \cos^2{\phi} + b_y^2 \sin{\phi} \cos{\phi} - b_x^2 \sin{\phi} \cos{\phi} - b_x b_y \sin^2{\phi} \right)}{r^2 \sin^2{\theta}}$$
$$+\frac{2 \left(b_x b_y \sin{\theta} \cos^2{\phi} - b_x^2 \sin{\theta} \cos{\theta} \cos{\phi} + b_y^2 \sin{\theta} \cos{\phi} \sin{\phi} - b_x b_y \sin{\theta} \sin^2{\phi} + b_y b_z \cos{\theta} \cos{\phi} - b_x b_z \sin{\phi} \cos{\theta} \right) }{r^2 \sin{\theta}}$$
$$-\frac{2 \cos{\theta}}{r^2 \sin^2{\theta}} \left(b_y b_z \sin{\theta} \cos{\phi} - b_x b_y \cos{\theta} \cos^2{\phi} - b_y^2 \cos{\theta} \cos{\phi} \sin{\phi} - b_x b_z \sin{\theta} \sin{\phi} - b_x^2 \cos{\theta} \sin{\phi} \cos{\phi} + b_x b_y \cos{\theta} \sin{\phi} \cos{\phi} \right)$$

which is proving hard to simplify. in particular the second term has a denominator that is different from that of the the first two terms - should i multiply through by $\frac{\sin{\theta}}{\sin{\theta}}$?

 

[gabbagabbahey]

It shouldn't be too hard to simplify; just collect terms with $b_x^2$ in them, and terms with $b_xb_y$ etc..

 

[latentcorpse]

that goes to

$$\frac{2 b_x b_y}{r^2 \sin^2{\theta}} \left( - \cos^2{\phi} + \sin^2{\phi} + \sin^2{\theta} \cos^2{\phi} - \sin^2{\theta} \sin^2{\phi} + \cos^2{\theta} \cos^2{\phi} - \cos^2{\theta} \sin{\phi} \cos{\phi} \right)$$
$$+\frac{2b_x b_z}{r^2 \sin^2{\theta}} \left(- \sin{\theta} \cos{\theta} \sin{\phi} + \sin{\theta} \sin{\phi} \right) = \frac{2 b_y b_z}{r^2 \sin^2{\theta}} \left( \sin{\theta} \cos{\theta} \cos{\phi} - \sin{\theta} \cos{\phi} \right)$$
$$+\frac{2 b_y^2}{r^2 \sin^2{\theta}} \left( - \sin{\phi} \cos{\phi} + \sin^2{\theta} \sin{\phi} \cos{\phi} + \cos^2{\theta} \sin{\phi} \cos{\phi} \right) + \frac{2 b_x^2}{r^2 \sin^2{\theta}} \left( \sin{\phi} \cos{\phi} - \sin^2{\theta} \cos{\theta} \cos{\phi} + \cos^2{\theta} \sin{\phi} \cos{\phi} \right)$$

looks like I've made a mistake somewhere but i can't see where.

 

[gabbagabbahey]

The quickest way to do this is probably to start with this expression (from post #60)

$$\ddot{\phi}=\frac{-2r^2 \sin^2{\theta} \left(b_x \cos{\phi} + b_y \sin{\phi} \right) \left( b_y \cos{\phi} - b_x \sin{\phi} \right)}{r^4 \sin^4{\theta}}=\frac{-2\dot{\phi} \left(b_x \cos{\phi} + b_y \sin{\phi} \right) }{r\sin{\theta}}$$

 

[latentcorpse]

got it finally! thank you so much for your help!

what did you mean in post 23 when you asked if there could be any other solutions?
how would i go about answering that?

 

[gabbagabbahey]

Wald discusses a certain uniqueness theorem right after equation 3.3.5...does that help you here?

 

[latentcorpse]

well the idea behind it would suggest that, no, there can be no other solutions.

however, the theorem says that solutions are only unique if we define a point p in hte manifold M and a tangent vector, $T^a \in V_p$. In the question, neither of these were specified so perhaps that does leave scope for alternative solutions?

 

[gabbagabbahey]

Realize that any potential solution $\textbf{r}(t)$ can be expanded in a Taylor series as

$$\textbf{r}(t)=\textbf{a}+\textbf{b}t+\textbf{c}t^2+\ldots$$

If $\textbf{a}$ and $\textbf{b}$ are specified, the uniqueness theorem tells you there is a unique solution...but you just showed that $\textbf{r}(t)=\textbf{a}+\textbf{b}t$ satisfies the geodesic equation for all $\textbf{a}$ and $\textbf{b}$...therfore ____?

 

[latentcorpse]

therefore straight lines are the unique solution as the values of a and b in the taylor expansion are precisely those of the coefficients a and b in the straight line eqn given in Cartesian coordinates, is that ok?

how did you manage to get round the fact that the point p and the tangent weren't specified though?

 

[gabbagabbahey]

Doesn't $\textbf{a}$ specify a point on the geodesic (where $t=0$) ?...And the tangent vector is___?

 

[latentcorpse]

the vector $\vec{b}$. great. thanks.