Geodesic Derivation Question


by Hypnotoad
Tags: derivation, geodesic
Hypnotoad
Hypnotoad is offline
#1
Dec16-04, 08:00 PM
P: 35
I'm studying for my math physics final tomorrow and I'm going through a derivation done in our book, but I'm stuck on this one step. The derivation is of the geodesic equation using variational calculus (this is done in the Arfken and Weber book, on page 156 if you have it). Anyways, I follow the derivation up to this point:

[tex]\frac{1}{2}\int{(\frac{dq^i}{ds}\frac{dq^j}{ds}\partial{g_{ij}}{q^k}-\frac{d}{ds}(g_{ik}\frac{dq^i}{ds}+g_kj\frac{dq^j}{ds}))\delta q^kds}=0[/tex]

since there is an independant variation with the [tex]\delta q^k[/tex], the rest of the integral is zero:

[tex]\frac{1}{2}( \frac{dq^i}{ds} \frac{dq^j}{ds} \frac{\partial{g_{ij}}}{\partial{q^k}} - \frac{d}{ds} (g_{ik} \frac{dq^i}{ds} + g_kj \frac{dq^j}{ds}))=0[/tex]

He then goes on to expand the derivatives of g in the last term:

[tex]\frac{dg_{ik}}{ds}=\frac{\partial{g_{ik}}}{\partial{q^j}}\frac{dq^j}{ds }[/tex]

[tex]\frac{dg_{kj}}{ds}=\frac{\partial{g_{kj}}}{\partial{q^i}}\frac{dq^i}{ds }[/tex]

The next step is the one I don't understand. He gets:

[tex]\frac{1}{2} \frac{dq^i}{ds} \frac{dq^j}{ds} (\frac{\partial {g_{ij}}} {\partial {q^k}} - \frac{\partial {g_{ik}}}{\partial {q^j}} - \frac{\partial {g_{jk}}}{\partial {q^i}})-g_{ik} \frac{d^2q^i}{ds^2}=0[/tex]


I don't know where this last term is coming from. This seems a lot like something my professor would ask on the test, so I would like to understand what is going on but I don't have time to talk to him before the test tomorrow. Can anyone explain where that second derivative is coming from?
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Dr Transport
Dr Transport is offline
#2
Dec16-04, 08:29 PM
Sci Advisor
PF Gold
P: 1,458
[tex] \frac{d}{ds} (g_{ik} \frac{dq^i}{ds} + g_{kj} \frac{dq^j}{ds}) =


\frac{dg_{ik}}{ds}\frac{dq^{i}}{ds} + \frac{dg_{kj}}{ds}\frac{dq^{j}}{ds} + 2g_{ik}\frac{d^{2}q^{i}}{ds^{2}}

[/tex]

you just use the chain rule on the second term and change [tex] j [/tex] to [tex] i [/tex].
Hypnotoad
Hypnotoad is offline
#3
Dec16-04, 08:42 PM
P: 35
Quote Quote by Dr Transport
[tex] \frac{d}{ds} (g_{ik} \frac{dq^i}{ds} + g_{kj} \frac{dq^j}{ds}) =


\frac{dg_{ik}}{ds}\frac{dq^{i}}{ds} + \frac{dg_{kj}}{ds}\frac{dq^{j}}{ds} + 2g_{ik}\frac{d^{2}q^{i}}{ds^{2}}

[/tex]

you just use the chain rule on the second term and change [tex] j [/tex] to [tex] i [/tex].

Thanks. I can't beleive I missed that.

dextercioby
dextercioby is offline
#4
Dec16-04, 08:42 PM
Sci Advisor
HW Helper
P: 11,863

Geodesic Derivation Question


Quote Quote by Hypnotoad
I'm studying for my math physics final tomorrow and I'm going through a derivation done in our book, but I'm stuck on this one step. The derivation is of the geodesic equation using variational calculus (this is done in the Arfken and Weber book, on page 156 if you have it). Anyways, I follow the derivation up to this point:

[tex]\frac{1}{2}\int{(\frac{dq^i}{ds}\frac{dq^j}{ds}\partial{g_{ij}}{q^k}-\frac{d}{ds}(g_{ik}\frac{dq^i}{ds}+g_kj\frac{dq^j}{ds}))\delta q^kds}=0[/tex]

since there is an independant variation with the [tex]\delta q^k[/tex], the rest of the integral is zero:

[tex]\frac{1}{2}( \frac{dq^i}{ds} \frac{dq^j}{ds} \frac{\partial{g_{ij}}}{\partial{q^k}} - \frac{d}{ds} (g_{ik} \frac{dq^i}{ds} + g_kj \frac{dq^j}{ds}))=0[/tex]

He then goes on to expand the derivatives of g in the last term:

[tex]\frac{dg_{ik}}{ds}=\frac{\partial{g_{ik}}}{\partial{q^j}}\frac{dq^j}{ds }[/tex]

[tex]\frac{dg_{kj}}{ds}=\frac{\partial{g_{kj}}}{\partial{q^i}}\frac{dq^i}{ds }[/tex]

The next step is the one I don't understand. He gets:

[tex]\frac{1}{2} \frac{dq^i}{ds} \frac{dq^j}{ds} (\frac{\partial {g_{ij}}} {\partial {q^k}} - \frac{\partial {g_{ik}}}{\partial {q^j}} - \frac{\partial {g_{jk}}}{\partial {q^i}})-g_{ik} \frac{d^2q^i}{ds^2}=0[/tex]


I don't know where this last term is coming from. This seems a lot like something my professor would ask on the test, so I would like to understand what is going on but I don't have time to talk to him before the test tomorrow. Can anyone explain where that second derivative is coming from?
To me,it's all very clear (i know,i've worked with tensors a lot...).In the second equation (actually the first to come out of the action's variation wrt to parameters),he substitutes the two derivatives of the metric.The number of terms is 5,before and after the substitution.Next,he uses the fact that the metric is symmetric wrt to indices and gets read of one tems involvin second order derivatives of the metric,but at the price of multiplying the remaining one with 2 (those two terms were equal).So from there,comes the last term,the one without 2.
As for the first three,he just factors the product of derivatives (wrt to the parameter) along the curves.So he should be getting beside that 1/2 which stays (no equal terms this time) another three terms in the bracket multiplied with the factor.

Daniel.

PS.I guess u haven't shuffled,swhitched,raised and lowered too many suffices in your life. At the math faculty,there are sure as hell less than at the phyiscs faculty.


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