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location United Kingdom
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visits member for 3 years, 2 months
seen 6 mins ago

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.


6h
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
19h
comment Which of these two textbook equations of geodesic deviation is correct?
Thanks. I need to study that.
20h
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.
22h
comment Which of these two textbook equations of geodesic deviation is correct?
I'm struggling to take this in, but Lambourne gives $$R_{\phantom{l}ijk}^{l}=\frac{\partial\Gamma_{ik}^{l}}{\partial x^{j}}-\frac{\partial\Gamma_{ij}^{l}}{\partial x^{k}}+\Gamma_{ik}^{m}\Gamma_{mj}^{l}-\Gamma_{ij}^{m}\Gamma_{mk}^{l}.$$
22h
comment Textbook disagreement on geodesic deviation on a 2-sphere
Thank you, though I'm struggling to understand the answers.
1d
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.
1d
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.
Jul
13
comment Derivation of freely falling frame in Schwarzschild spacetime
Took me ages to work through the algebra, but I then ground to a halt when you said $r$ (presumably you meant $\bar{r}$ ) can be treated as a constant in the new metric coefficients? What's the justification for that? Also, what's the problem with my guess at subsituting $ar=-GM/r$ into the metric? Does that only work for a weak gravitational field? Also, it's still not obvious how the above transformation is linked to something freely falling. Thanks.
May
29
comment Kronecker delta confusion
The penny has dropped. Thanks
May
29
comment Kronecker delta confusion
Sorry, still can't see it. Don't both equation refer to multiplying a metric by its inverse? If so, why two different answers. Please don't worry about making your answers too simple!
Jan
22
comment Is this interpretation of $\psi=\frac{1}{\sqrt{\pi a^{3}}}e^{-r/a}$ correct?
The $\pi$ 's cancelled, so I didn't mention them. Nor did I mention that I meant the complex square of the wave function. If $x=\infty$ then $P=1$, and if $x=0.00000000009$ then $P=0.66077$. I now know my interpretation was right, but my maths was a bit ragged. Thanks.
Jan
22
comment Is this interpretation of $\psi=\frac{1}{\sqrt{\pi a^{3}}}e^{-r/a}$ correct?
I cut a few corners in my question and have now been found out. My original calculation of the probability $P$ of finding the electron in a sphere radius $x$ was along the lines of $$P=\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{x}\left|\frac{1}{\sqrt{\pi a^{3}}}e^{-r/a}\right|^{2}\left(r^{2}\sin\theta drd\theta d\phi\right)=\frac{4\pi}{\pi a^{3}}\int_{0}^{x}r^{2}e^{-2r/a}dr.$$ continued ...