Components of Christoffel symbol

In summary, we discussed the metric of Euclidean \mathbb{R}^3 in spherical coordinates and how to calculate the Christoffel components \Gamma^{\sigma}{}_{\mu \nu} in this coordinate system using the formula \Gamma^{\sigma}{}_{\mu \nu}=\frac{1}{2} \sum_{\rho} g^{\sigma \rho} \left( \frac{\partial g_{\nu \rho}}{\partial x^{\mu}} + \frac{\partial g_{\mu \rho}}{\partial x^{\nu}} - \frac{\partial g_{\mu \nu}}{\partial x^{\sigma}}
  • #71
latentcorpse said:
you said i was fine at [tex]\dot{\theta}=-\frac{1}{\sqrt{r^4-z^2r^2}} \left(\dot{z}r-z \dot{r} \right)[/tex]

so i need to change the coords here

[tex]\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)[/tex]
ok that should be better

Why not substitute in your expression for [itex]\dot{r}[/itex] here?
 
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  • #72
i get

[tex]\dot{\theta}=-\frac{1}{r \sin{\theta}} \left(b_z-\cos{\theta} \dot{r} \right)[/tex]

[tex]\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)[/tex]

[tex]\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)[/tex]

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

that gives, by product rule:

[tex]\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)[/tex]

[tex]\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)[/tex]

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

[tex]\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}[/tex]

And

[tex]-\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[/tex]

So,

[tex]\ddot{\theta}=-\frac{2}{r}\dot{r}\dot{\theta}+ \sin\theta\cos\theta\dot{\phi}^2[/tex]

Compare that to your second Geodesic equation...:wink:

Edit: Is there a reason you've interpreted [tex]\frac{dx^i}{dt}\frac{dx^i}{dt}[/tex] as [tex]\frac{d^2x^i}{dt^2}[/tex] when calculating your Geodesic equations in post#22?
 
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  • #74
should the geodesic eqns be

[tex]\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[/tex]
[tex]\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[/tex]
[tex]\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[/tex]?
 
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  • #75
Yup.
 
  • #76
okay so I am testing the equations now:
looking at the first one i have

[tex]\ddot{r}-r \dot{\theta}^2 - r \sin^2{\theta} \dot{\phi}^2[/tex]
[tex]=\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)[/tex]

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 [itex]\ddot{r}[/itex] from this equation [itex]\dot{r}=b_x\sin\theta\cos\phi+ b_y\sin\theta\sin\phi+b_z\cos\theta[/itex]...that way everything is in Spherical coords.
 
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  • #78
so [tex]\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)[/tex]

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

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

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

[tex]\ddot{\phi} + \frac{2}{r} \dot{r} \dot{\phi} + 2 \cot{\theta} \dot{\theta} \dot{\phi}[/tex]
[tex]=\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}}[/tex]
[tex]+\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}}[/tex]
[tex]-\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)[/tex]

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 [itex]\frac{\sin{\theta}}{\sin{\theta}}[/itex]?
 
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  • #79
It shouldn't be too hard to simplify; just collect terms with [itex]b_x^2[/itex] in them, and terms with [itex]b_xb_y[/itex] etc..
 
  • #80
that goes to

[tex]\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)[/tex]
[tex]+\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)[/tex]
[tex]+\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)[/tex]

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)

[tex]\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}}[/tex]
 
  • #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, [itex]T^a \in V_p[/itex]. In the question, neither of these were specified so perhaps that does leave scope for alternative solutions?
 
  • #85
Realize that any potential solution [itex]\textbf{r}(t)[/itex] can be expanded in a Taylor series as

[tex]\textbf{r}(t)=\textbf{a}+\textbf{b}t+\textbf{c}t^2+\ldots[/tex]

If [itex]\textbf{a}[/itex] and [itex]\textbf{b}[/itex] are specified, the uniqueness theorem tells you there is a unique solution...but you just showed that [itex]\textbf{r}(t)=\textbf{a}+\textbf{b}t[/itex] satisfies the geodesic equation for all [itex]\textbf{a}[/itex] and [itex]\textbf{b}[/itex]...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 [itex]\textbf{a}[/itex] specify a point on the geodesic (where [itex]t=0[/itex]) ?...And the tangent vector is___?
 
  • #88
the vector [itex]\vec{b}[/itex]. great. thanks.
 
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<h2>1. What are the components of the Christoffel symbol?</h2><p>The components of the Christoffel symbol are mathematical quantities used in differential geometry to describe the curvature of a manifold. There are three components for each index, resulting in a total of nine components.</p><h2>2. How are the components of the Christoffel symbol calculated?</h2><p>The components of the Christoffel symbol can be calculated using the metric tensor and its derivatives. Specifically, they are calculated using the inverse metric tensor and the Christoffel symbols of the second kind.</p><h2>3. What is the significance of the Christoffel symbol in physics?</h2><p>In physics, the Christoffel symbol is used to calculate the geodesic equation, which describes the path of a free particle in curved spacetime. It is also used in the Einstein field equations to describe the curvature of spacetime in general relativity.</p><h2>4. Can the Christoffel symbol be used in other areas of mathematics?</h2><p>Yes, the Christoffel symbol can also be used in other areas of mathematics such as differential equations and differential geometry. It is a fundamental tool for understanding the curvature of a manifold and is applicable in various fields of mathematics.</p><h2>5. Are there any practical applications of the Christoffel symbol?</h2><p>Yes, the Christoffel symbol has practical applications in fields such as physics, engineering, and computer graphics. It is used to model curved surfaces and to calculate the shortest path between two points on a curved surface.</p>

1. What are the components of the Christoffel symbol?

The components of the Christoffel symbol are mathematical quantities used in differential geometry to describe the curvature of a manifold. There are three components for each index, resulting in a total of nine components.

2. How are the components of the Christoffel symbol calculated?

The components of the Christoffel symbol can be calculated using the metric tensor and its derivatives. Specifically, they are calculated using the inverse metric tensor and the Christoffel symbols of the second kind.

3. What is the significance of the Christoffel symbol in physics?

In physics, the Christoffel symbol is used to calculate the geodesic equation, which describes the path of a free particle in curved spacetime. It is also used in the Einstein field equations to describe the curvature of spacetime in general relativity.

4. Can the Christoffel symbol be used in other areas of mathematics?

Yes, the Christoffel symbol can also be used in other areas of mathematics such as differential equations and differential geometry. It is a fundamental tool for understanding the curvature of a manifold and is applicable in various fields of mathematics.

5. Are there any practical applications of the Christoffel symbol?

Yes, the Christoffel symbol has practical applications in fields such as physics, engineering, and computer graphics. It is used to model curved surfaces and to calculate the shortest path between two points on a curved surface.

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