
#19
Oct2909, 10:44 AM

HW Helper
P: 5,004

You seem to have used [itex]g^{ab}=g_{ab}[/itex] instead of using the inverse matrix you calculated...
[tex]\Gamma^2{}_{12}=\frac{1}{2} g^{22} \left( \frac{\partial g_{22}}{\partial r} + \frac{\partial g_{12}}{\partial \theta}  \frac{\partial g_{12}}{\partial \theta} \right) = \frac{1}{2} \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \right) = \frac{1}{r}[/tex] 



#20
Oct2909, 11:05 AM

P: 1,431

lol. i'm an idiot sometimes...thanks.
the next bit asks me to write out the copmonents of the geodesic equaiton in this coordinate system and verify the solutions correspond ot straight lines in Cartesian coordinates. so i guess the eqn tehy're referring to is 3.3.5. [tex]\frac{d^2 x^{\mu}}{dt^2} + \sum_{\sigma, \nu} \Gamma^{\mu}{}_{\sigma \nu} \frac{d x^{\sigma}}{dt} \frac{dx^{\nu}}{dt}=0[/tex] am i required here to solve 17 different differential equations? even if the Christoffelsymbol is zero there will still be that first term equal to zero so none of the equaitons are going to be trivial. or do i just solve for one term of the Christoffel symbol and show the solutino is a straight line? 



#21
Oct2909, 11:17 AM

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P: 5,004

Once you sum over [itex]\sigma[/itex] and [itex]\nu[/itex] you will only have 3 ODEs (one for each value of [itex]\mu[/itex])...




#22
Oct2909, 11:26 AM

P: 1,431

ok so i get my three equations as:
[tex]\frac{d^2 r}{dt^2}r \frac{d^2 \theta}{dt^2}  r \sin^2{\theta} \frac{d^2 \phi}{dt^2}=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^2 \phi}{dt^2}=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] do they look correct? how on earth do i solve them? 



#23
Oct2909, 11:34 AM

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P: 5,004

Assuming you meant [itex]\frac{d^2 \theta}{dt^2} +\frac{2}{r} \frac{dr}{dt} \frac{d \theta}{dt}  \sin{\theta} \cos{\theta} \frac{d^2 \phi}{dt^2}=0[/itex] for the 2nd ODE, then yes those look right....Now, do you really need to solve them, in order to verify that the solutions are straight lines in Cartesian coordinates?...What is the general (parametrized in terms of [itex]t[/itex]) form of a straight line in Cartesian coords?...What is it when you convert to Spherical coords?...Does that form satisfy the ODEs? Could there be any other solutions?




#24
Oct2909, 11:47 AM

P: 1,431

in cartesian isnt it just going to be
[itex]\vec{r}(t)=\vec{a}+\vec{b}t[/itex] with [itex]\vec{a}, \vec{b} \in \mathbb{R}^3[/itex]. i'm not sure how to convert that to sphericals though? have i used the wrong form above? 



#25
Oct2909, 11:51 AM

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P: 5,004

No, the general parametrized form of a line in 3D, is [itex]ax(t)+by(t)+cz(t)=0[/itex]




#26
Oct2909, 11:52 AM

P: 1,431

[tex]\frac{xx_0}{a}=\frac{yy_0}{b}=\frac{zz_0}{c}[/tex]?




#28
Oct2909, 03:23 PM

P: 1,431

ok. so if that's the cartesian form i need to convert it to spherical polars:
[tex]ar(t) \sin{\theta (t) } \cos{\phi (t) } + br(t) \sin{\theta (t)} \sin{\phi (t)} + c r(t) \cos{\theta (t)}=0[/tex] but i don't see where to substitute that into the geodesic eqn. or am i wanting to show [tex] \left( \frac{d^2 x^{\mu}}{dt^2} + \sum_{\sigma, \nu} \Gamma^{\mu}{}_{\sigma \nu} \frac{d x^{\sigma}}{dt} \frac{d x^{\nu}}{dt} \right) \left( ar(t) \sin{\theta (t)} \cos{\phi (t)} + br(t) \sin{\theta (t)} \sin{\phi (t)} + c r(t) \cos{\theta (t)} \right) = 0[/tex]? 



#29
Oct2909, 03:37 PM

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P: 5,004





#30
Oct3009, 08:04 AM

P: 1,431

sorry. i don't understand what's going on here.
am i substituting that equation into the geodesic equations? if so what is r theta and phi? or am i just multiplying the geodesic eqn with the straight line eqn? 



#31
Oct3009, 08:42 AM

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P: 5,004

You want to show that [itex]r(t)[/itex], [itex]\theta(t)[/itex] and [itex]\phi(t)[/itex] satisfy each of your 3 ODEs, so long as they also satisfy your straight line equation..




#32
Oct3009, 09:50 AM

P: 1,431

ok. so i did the double derivative with time by hand (is there a way to write a simple maple code to do this for me? i tried but couldn't get it to work) and got:
[tex]a \ddot{r} \sin{\theta} \cos{\phi} + a \dot{r} \dot{\theta} \cos{\theta} \cos{\phi}  a \dot{r} \dot{\phi} \sin{\theta} \sin{\phi}[/tex] [tex] + a \dot{r} \dot{\theta} \cos{\theta} \cos{\phi} + a r \ddot{\theta} \cos{\theta} \cos{\phi}[/tex] [tex]  a r \dot{\theta}^2 \sin{\theta} \cos{\phi}  a r \dot{\theta} \dot{\phi} \cos{\theta} \sin{\phi}  a \dot{r} \dot{\phi} \sin{\theta} \sin{\phi}  a r \ddot{\phi} \sin{\theta} \sin{\phi}  a r \dot{\theta} \dot{\phi} \cos{\theta} \sin{\phi}[/tex] [tex]  a r \dot{\phi}^2 \sin{\theta} \cos{\phi} + b \ddot{r} \sin{\theta} \sin{\phi} + b \dot{r} \dot{\theta} \cos{\theta} \sin{\phi} + b \dot{r} \dot{\phi} \sin{\theta} \cos{\phi} + b \dot{r} \dot{\theta} \cos{\theta} \sin{\phi} + b r \ddot{\theta} \cos{\theta} \sin{\phi}[/tex] [tex]  b r \dot{\theta}^2 \sin{\theta} \sin{\phi} + b r \dot{\theta} \dot{\phi} \cos{\theta} \cos{\phi} + b \dot{r} \dot{\phi} \sin{\theta} \cos{\phi} + b r \ddot{\phi} \sin{\theta} \cos{\phi} + b r \dot{\theta} \dot{\phi} \cos{\theta} \cos{\phi}[/tex] [tex]  b r \dot{\phi}^2 \sin{\theta} \ubs{\phi} + c \ddot{r} \cos{\theta{  c \dot{r} \dot{\theta} \sin{\theta}  c \dot{r} \dot{\theta} \sin{\theta}  c r \ddot{\theta} \sin{\theta}  cr \dot{\theta}^2 \cos{\theta} = 0[/tex] 



#33
Oct3009, 09:53 AM

P: 1,431

hmm. i don't know why that all posted on one line? can you read it by just clicking on the code?
anyway, i can group some terms together and what not but i'm still not sure what im doing...surely i want to take the second derivative of r, theta and phi not of the straight line eqn? 



#34
Oct3009, 01:34 PM

P: 1,431

sorry, i sorted out the LaTeX in post 32. see my previous 2 posts.




#35
Oct3109, 08:42 AM

P: 1,431

basically i dont understand why differentitating the straight line eqn wrt time twice helps us to show they also satisfy the geodesic eqn???



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