690 reputation
616
bio website
location United Kingdom
age
visits member for 3 years, 1 month
seen yesterday

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.


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 ...
Jan
15
comment Calculation of spherical Bessel functions - meaning of $\left(\frac{1}{x}\frac{d}{dx}\right)^{l}$
Thanks. So, in this case, you'd simply take the derivative, then multiply by 1/x, then take the derivative again and then multiply by 1/x? Afraid I'm not too familiar with operators.
Dec
31
comment Don't understand the integral over the square of the Dirac delta function
To me, that seems even clearer (in so far as anything involving this bizarre function can be clear). Apologies twistor59, I've now accepted this answer.
Dec
31
comment Don't understand the integral over the square of the Dirac delta function
That's plausible enough for me. Out of interest, why isn't this a proof?
Dec
30
comment Don't understand the integral over the square of the Dirac delta function
This is a little advanced for me as I'm not familiar with Heaviside step functions. Twistor59's answer is more my level though I'm still trying to think it through.
May
7
comment Zero divergence of energy-momentum tensor and gravitational energy
apologies for being a bit slow on the uptake. This stuff would be a lot easier if (a) I was smarter and (b) I had a degree in physics. I confess I've only just today found out what a continuity equation is! I've also just read a reasonably understandable on-line piece by Weiss and Baez titled, "Is energy conserved in general relativity". Their answer is an unambiguous, "In special cases, yes. In general — it depends on what you mean by 'energy', and what you mean by 'conserved'." Glad that's sorted out.
May
6
comment Zero divergence of energy-momentum tensor and gravitational energy
Is it possible to describe, in absolute baby steps why that implication is true?
May
6
comment Zero divergence of energy-momentum tensor and gravitational energy
thanks. Any chance of more slowly and even simplier, with no reference to Lagrangians? I have a rudimentary understanding of the divergence theorem, connection coefficients and covariant derivatives. My understanding of divergence is at the level of sources and sinks of a liquid sloshing about in a container - if no liquid is entering or leaving the container, the divergence equals zero. I'm trying to relate that simple picture to $\nabla_{\mu}T^{\mu\nu}=0$ and why that implies the conservation of energy in flat but not in curved spacetime.