690 reputation
616
bio website
location United Kingdom
age
visits member for 3 years, 1 month
seen Jul 26 at 16:16

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.


Sep
9
comment Index raising and lowering - how does it work?
I've no idea what this means.
Sep
9
accepted Index raising and lowering - how does it work?
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.
Sep
9
asked Index raising and lowering - how does it work?
Aug
28
awarded  Popular Question
Aug
5
answered Understanding Tensors
Jul
13
accepted Derivation of freely falling frame in Schwarzschild spacetime
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.
Jul
12
asked Derivation of freely falling frame in Schwarzschild spacetime
Jun
17
awarded  Yearling
Jun
5
accepted Einstein-de Sitter - confusing constant of integration
Jun
3
asked Einstein-de Sitter - confusing constant of integration
May
29
comment Kronecker delta confusion
The penny has dropped. Thanks
May
29
accepted Kronecker delta confusion
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!
May
29
asked Kronecker delta confusion
Jan
22
accepted Is this interpretation of $\psi=\frac{1}{\sqrt{\pi a^{3}}}e^{-r/a}$ correct?
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.