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location United Kingdom
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visits member for 3 years, 6 months
seen Dec 7 at 18:05

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.


Nov
22
comment Derivation of the Riemann tensor confusion
After spending ages carefully typing it out in LyX, I couldn't resist showing it off here!
Nov
22
comment Derivation of the Riemann tensor confusion
I've just found this for the covariant derivative for a general tensor:$\nabla_{k}T_{j_{1}\ldots j_{s}}^{i_{1}\ldots i_{r}}=\partial_{k}T_{j_{1}\ldots j_{s}}^{i_{1}\ldots i_{r}}+\Gamma_{lk}^{i_{1}}T_{j_{1}\ldots j_{s}}^{li_{2}\ldots i_{r}}+\ldots+\Gamma_{lk}^{i_{r}}T_{j_{1}\ldots j_{s}}^{i_{1}\ldots i_{r-1}l}-\Gamma_{j_{1}k}^{l}T_{lj_{2}\ldots j_{s}}^{i_{1}\ldots i_{r}}\ldots-\Gamma_{j_{s}k}^{l}T_{j_{1}\ldots j_{s-1}l}^{i_{1}\ldots i_{r}}.$
Nov
22
comment Derivation of the Riemann tensor confusion
Carroll gives it on p58 here: arxiv.org/pdf/gr-qc/9712019.pdf. I notice that the upper index on the negative Gammas and the lower indices on the positive Gammas are constant, whilst the lower indices on the negative Gammas and the upper index on the positive Gammas are not.
Nov
21
comment Derivation of the Riemann tensor confusion
OK. It was that particular detailed equation for a general tensor I was interested in, but I'll take a look around.
Nov
21
comment Derivation of the Riemann tensor confusion
Thanks. Does $T^{d...a_{n}}$ mean $T^{da_{2}...a_{n}}$ etc? Do you have a reference for the derivation of this equation?
Nov
21
comment Derivation of the Riemann tensor confusion
Right. So the mechanism is that I start with a four component vector $\vec{V}$. Take the covariant derivative of this using $\frac{\partial V^{\alpha}}{\partial x^{\beta}}+V^{\gamma}\Gamma_{\gamma\beta}^{\alpha}$. Summing over the repeated $\gamma$ index then gives me a $4\times4$ matrix aka a rank-2 tensor $V_{;\beta}^{\alpha}$. Is that correct?
Nov
21
comment Derivation of the Riemann tensor confusion
The semi-colon threw me.
Nov
21
comment Derivation of the Riemann tensor confusion
Oh. I didn't realise you treat $\lambda_{a;b}$ as having two lower indices. That does tend to put things in a new light.
Nov
17
comment Is this covariant derivative identity true?
Excellent answer. I can see it now. Thanks.
Nov
17
comment Is this covariant derivative identity true?
@ValterMoretti Any hints as to how I can show it's true?
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
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.
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.
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
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 $$R_{\phantom{l}ijk}^{l}=\frac{\partial\Gamma_{ik}^{l}}{\partial x^{j}}-\frac{\partial\Gamma_{ij}^{l}}{\partial x^{k}}+\Gamma_{ik}^{m}\Gamma_{mj}^{l}-\Gamma_{ij}^{m}\Gamma_{mk}^{l}.$$
Aug
28
comment Textbook disagreement on geodesic deviation on a 2-sphere
Thank you, though I'm struggling to understand the answers.