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location United Kingdom
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visits member for 3 years, 4 months
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Manual worker. Live in England. Formal scientific education "peaked" many years ago when I failed maths and chemistry A-level and scraped a bare pass in physics. Still enjoy pottering about in maths and physics.


5h
accepted Geodesic deviation equation - why does the ordinary second derivative give the correct answer?
Oct
20
comment Geodesic deviation equation - why does the ordinary second derivative give the correct answer?
So, can I assume that as only one of the coordinate curves (hope that's the correct term) is a geodesic (ie the $\phi=constant$ one), then the answer has nothing to do with Riemann normal coordinates? Also, I didn't realise that if $\frac{D\zeta^{\mu}}{d\lambda}=0$ then $\frac{D\zeta_{\mu}}{d\lambda} =0$. I've seen a similar problem to this online but with one of the particles moving along the equator ($\theta =\pi/2$), meaning the connection coefficients “naturally” disappear and the absolute 2nd derivative equals the ordinary 2nd derivative, which makes things a lot simpler.
Oct
19
revised Geodesic deviation equation - why does the ordinary second derivative give the correct answer?
added 159 characters in body
Oct
19
awarded  Promoter
Oct
19
revised Geodesic deviation equation - why does the ordinary second derivative give the correct answer?
added 335 characters in body
Oct
17
comment Geodesic deviation equation - why does the ordinary second derivative give the correct answer?
I think so. Isn't that what Schutz calls the "local flatness" theorem? What I don't understand is that I correctly calculated the answer using the ordinary spherical $\theta,\phi$ coordinates. I can't see how the connection coefficients vanish using these.
Oct
17
asked Geodesic deviation equation - why does the ordinary second derivative give the correct answer?
Oct
1
answered Geodesic deviation on a unit sphere
Sep
24
awarded  Autobiographer
Sep
4
revised Geodesic deviation on a unit sphere
added 841 characters in body; edited title
Sep
3
comment Geodesic deviation on a unit sphere
Whoops. I've now amended the question to show my results after summing over the indices.
Sep
3
revised Geodesic deviation on a unit sphere
I forgot to sum over the indices when calculating the absolute derivative, which I've now done.
Sep
3
asked Geodesic deviation on a unit sphere
Aug
29
comment Which of these two textbook equations of geodesic deviation is correct?
I know this is ultra longwinded, but I think I see it now.
Aug
29
comment Which of these two textbook equations of geodesic deviation is correct?
So if, for eg, we had $x,y,z$ coordinates, the two terms would multiply out to give a symmetric matrix $$\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{d\lambda}=\left(\begin{array}{cc‌​c} \frac{dx}{d\lambda}\frac{dx}{d\lambda} & \frac{dx}{d\lambda}\frac{dy}{d\lambda} & \frac{dx}{d\lambda}\frac{dz}{d\lambda}\\ \frac{dy}{d\lambda}\frac{dx}{d\lambda} & \frac{dy}{d\lambda}\frac{dy}{d\lambda} & \frac{dy}{d\lambda}\frac{dz}{d\lambda}\\ \frac{dz}{d\lambda}\frac{dx}{d\lambda} & \frac{dz}{d\lambda}\frac{dy}{d\lambda} & \frac{dz}{d\lambda}\frac{dz}{d\lambda} \end{array}\right)$$ aka a symmetric tensor.
Aug
29
comment Which of these two textbook equations of geodesic deviation is correct?
@jerryschirmer - sorry to come back to this, but I woke up in the middle of last night and realised I don't understand why$$\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{d\lambda}$$ is a symmetric tensor. My understanding of a ST is at the level of a symmetric matrix. Why should the tensor product of $\frac{dx^{\beta}}{d\lambda}$ and $\frac{dx^{\gamma}}{d\lambda}$ give a symmetric matrix? Can't they each be a set of any four numbers? Thanks
Aug
28
comment Which of these two textbook equations of geodesic deviation is correct?
Thanks. I need to study that.
Aug
28
accepted Which of these two textbook equations of geodesic deviation is correct?
Aug
28
comment Which of these two textbook equations of geodesic deviation is correct?
I found it difficult to understand the stuff about oppositely signed Riemann tensors, but thanks for letting me know that Lambourne's equation is incorrect. Looking through various examples of the geodesic deviation equations online, they all seem to show the right-hand side $\xi$ index as equal to the second or third lower index on the Riemann tensor. I guess that corresponds to whether the Riemann tensor is positively or negatively signed.
Aug
28
comment Which of these two textbook equations of geodesic deviation is correct?
I'm struggling to take this in, but Lambourne gives $$R_{\phantom{l}ijk}^{l}=\frac{\partial\Gamma_{ik}^{l}}{\partial x^{j}}-\frac{\partial\Gamma_{ij}^{l}}{\partial x^{k}}+\Gamma_{ik}^{m}\Gamma_{mj}^{l}-\Gamma_{ij}^{m}\Gamma_{mk}^{l}.$$