How to Calculate Christoffel Symbols in Spherical Coordinates?

AI Thread Summary
The discussion focuses on calculating the Christoffel symbols in spherical coordinates, starting from the metric of Euclidean \(\mathbb{R}^3\). Participants clarify the relationship between the metric \(ds^2\) and the components \(g_{ab}\), leading to the correct formulation of the metric tensor. They derive the Christoffel symbols using Wald's formula, addressing confusion about the indices and the necessity of using the inverse metric. The conversation also touches on the symmetry of the Christoffel symbols and how to express them concisely, ultimately leading to the formulation of geodesic equations in spherical coordinates. The participants explore the implications of these equations for verifying straight-line solutions in Cartesian coordinates.
  • #51
Shouldn't you have \phi=\tan^{-1}(y/x)?
 
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  • #52
ok so

\ddot{\phi}=\frac{\ddot{y} x^{-1} -2 \dot{y} x^{-2} \dot{x} - y x^{-2} \dot{x} +2y x^{-3} \dot{x}^2}{1+ \left( \frac{y}{x} \right)^2} - \frac{2y}{x} \frac{ \left( \dot{y} x^{-1} - y x^{-2} \dot{x} \right)^2}{\left( 1+ \left( \frac{y}{x} \right)^2 \right)^2}
 
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  • #53
That doesn't look quite right to me...I'd start by simplifying \dot{\phi} as much as possible, and then converting it to spherical coordinates...what do you get when you do that?
 
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  • #54
\dot{\phi}=\frac{1}{1+ \left( \frac{y}{x} \right)^2 } \left( \dot{y} x^{-1} - y x^{-2} \dot{x}} \right)=\frac{1}{1+ \left( \frac{a_y+b_yt}{a_x+b_xt} \right)^2} \left( \frac{b_y}{a_x+b_xt}-\frac{b_x(a_y+b_yt)}{(a_x+b_xt)^2} \right)
that ok?
 
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  • #55
latentcorpse said:
\dot{\phi}=\frac{1}{1+ \left( \frac{y}{x} \right)^2 } \left( \dot{y} x^{-1} - y x^{-2} \dot{x}} \right)
I'd multiply both the numerator and denominator by x^2, sub in \dot{x}=b_x and \dot{y}=b_y and then write x and y in terms of r, \theta and \phi...
 
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  • #56
\dot{\phi} = \frac{r \sin{\theta}}{1+r^2 \sin^2{\theta} \sin^2{\phi}} \left( b_y \cos{\phi} - b_x \sin{\phi} \right)

is that going to help? shouldn't i take the second derivative before i switch to spherical coordinates?
 
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  • #57
latentcorpse said:
\dot{\phi} = \frac{r \sin{\theta}}{1+r^2 \sin^2{\theta} \sin^2{\phi}} \left( b_y \cos{\phi} - b_x \sin{\phi} \right)

is that going to help? shouldn't i take the second derivative before i switch to spherical coordinates?

It would help, if you did it properly:wink:

Try again, you have an error...and don't be afraid to simplify your result as much as possible.
 
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  • #58
\dot{\phi}=\frac{1}{1+ \left( \frac{y}{x} \right)^2} \left( \dot{y} x^{-1} - y x^{-2} \dot{x} \right) = \frac{1}{1+y^2} \left( \dot{y} x - y \dot{x} \right)

\ddot{\phi} = \frac{ \left( \ddot{y} x + \dot{y} \dot{x} - \dot{y} \dot{x} - y \ddot{x} \right) \left( 1+ y^2 \right) - 2 y \dot{y} \left( \dot{y} x - y \dot{x} \right) }{ \left(1+y^2 \right)^2 } note the first term in the numerator will dissappear because \ddot{y}=\ddot{x}=0 and \dot{y}\dot{x}-\dot{x}\dot{y}=0.

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

i can't see how to simplify that further...
 
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  • #59
When you multiply 1+\left(\frac{y}{x}\right)^2 by x^2, should you get x^2+y^2?
 
  • #60
\dot{\phi}=\frac{\dot{y}x-y \dot{x}}{x^2+y^2}

\ddot{\phi}=\frac{ \left( \ddot{y} x + \dot{y} \dot{x} - \dot{y} \dot{x} - y \ddot{x} \right) \left( x^2+y^2 \right) - \left( 2 x \dot{x} + 2 y \dot{y} \right) \left( \dot{y} x - y \dot{x} \right)}{ \left( x^2+y^2 \right)^2}

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

\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}}

\ddot{\phi} = \frac{-2 \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)}{r^2 \sin^2{\theta}}

hopefully that looks better? i was hoping the numerator would simplify nicer though.
 
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  • #61
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)
 
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  • #62
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
 
  • #63
Phrak said:
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.
 
  • #64
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}?
 
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  • #65
Looks fine so far; now express it in terms of spherical coordinates and b_x, b_y and b_z...
 
  • #66
\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?
 
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  • #67
I'd use z=r\cos\theta instead of z=a_z+b_zt...
 
  • #68
\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?
 
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  • #69
latentcorpse said:
\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)
 
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  • #70
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...
 
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  • #71
latentcorpse said:
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?
 
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  • #72
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.
 
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  • #73
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...:wink:

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?
 
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  • #74
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?
 
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  • #75
Yup.
 
  • #76
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...
 
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  • #77
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.
 
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  • #78
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}}?
 
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  • #79
It shouldn't be too hard to simplify; just collect terms with b_x^2 in them, and terms with b_xb_y etc..
 
  • #80
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.
 
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  • #81
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}}
 
  • #82
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?
 
  • #83
Wald discusses a certain uniqueness theorem right after equation 3.3.5...does that help you here?:wink:
 
  • #84
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?
 
  • #85
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 ____?
 
  • #86
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?
 
  • #87
Doesn't \textbf{a} specify a point on the geodesic (where t=0) ?...And the tangent vector is___?
 
  • #88
the vector \vec{b}. great. thanks.
 
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