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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.


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.
Aug
28
comment Textbook disagreement on geodesic deviation on a 2-sphere
Maybe I should have simply asked which of the two textbook equations of geodesic deviation is correct. Instead I tried to show effort by attempting an answer myself in the form of calculating the geodesic deviation on the surface of a unit 2-sphere. I wasn't confident about these calculation, hence my request for someone to take a look at them. I thus seem to have fallen into the “check my work” trap. As a self-studier, I can't say I feel encouraged by the negative responses to my question. Anyway, I'll try again with a rephrased question.
Aug
27
comment Textbook disagreement on geodesic deviation on a 2-sphere
Is my question controversial? It's been tagged as "Homework", but I'm in my sixth decade and haven't formally studied physics for forty years. I'm just curious as to whether I'm on the right track. Thanks.
Jun
5
comment Minkowski metric and definition of coordinate differentials?
Yes, I can see it now. Thanks.
Jun
5
comment Minkowski metric and definition of coordinate differentials?
Of course! I should have realised.
Dec
19
comment Riemann curvature tensor symmetries confusion
@StanLiou: Thanks. Final question. Can the Ricci tensor therefore be defined using other index permutations that don't involve the Riemann tensor having the same 1 and 2 or 3 and 4 indices as the metric, ie $R_{\mu\nu}=g^{\sigma\rho}R_{\sigma\mu\nu\rho}$ , $R_{\mu\nu}=g^{\sigma\rho}R_{\mu\sigma\rho\nu}$ , $R_{\mu\nu}=g^{\sigma\rho}R_{\mu\sigma\nu\rho}$ ?
Dec
18
comment Riemann curvature tensor symmetries confusion
Oh, showing my huge ignorance, I thought you could just sort of “cancel” any upper and lower index. I didn't know about symmetric/anti-symmetric. Does that mean that $g^{\alpha\beta}R_{\alpha\beta\gamma\mu}=g^{\alpha\beta}R_{\gamma\mu\alpha\beta}‌​=0$ ? Would that also mean (swapping the metric tensor indices) that $g^{\beta\alpha}R_{\alpha\beta\gamma\mu}=g^{\beta\alpha}R_{\gamma\mu\alpha\beta}‌​=0$ ? So, if I avoid the Riemann tensor having the same 1 and 2 or 3 and 4 indices as the metric, can I state the Ricci tensor as (for example) $R_{\mu\nu}=g^{\sigma\rho}R_{\sigma\mu\rho\nu}$ ?
Sep
9
comment Index raising and lowering - how does it work?
I've no idea what this means.
Sep
9
comment Index raising and lowering - how does it work?
I'm afraid I don't understand where the “integrability condition” comes from. However, I can live with that as (I think) I get the gist that all gradients are covectors but not all covectors are gradients, and therefore $g_{\mu\nu}V^{\nu}=\partial_{\mu}\phi$ is only sometimes true. How about going the other way? Does $g_{\mu\nu}V^{\nu}$ always give a tangent vector to a parametrised curve, and how would you show that? Apologies if you've already shown this and I've just got lost in the maths.
Jul
13
comment Derivation of freely falling frame in Schwarzschild spacetime
“The transformation to $\bar{r}$ then allows to write the metric directly in the standard form, in a mathematically rigorous way.” Yes, I can see that now. Is there a specific name for the above transformation? One more thing. I confess I could look at $r=\bar{r}\left(1+\frac{r_{\text{s}}}{4\bar{r}}\right)^{2}$ for many years and not think, “ah yes, that describes a local, freely falling frame”. Is there an obvious way of showing that the radial coordinate $\bar{r}$ describes such a frame? Thank you for your patience.
Jul
13
comment Derivation of freely falling frame in Schwarzschild spacetime
But if we treat $r$ as a constant why do we need to go to all the trouble of introducing $\bar{r}$ ? Why not just say $\left(1-\frac{2GM}{c^{2}r}\right)=a$ , $\left(1-\frac{2GM}{c^{2}r}\right)^{-1}=b$ and $r^{2}=c$ , where $a$, $b$ and $c$ are constants, and get to the Minkowski metric that way? I'm probably missing the obvious here.