685 reputation
315
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location United Kingdom
age
visits member for 2 years, 10 months
seen Apr 17 at 18:19

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.


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.
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
21
asked Is this interpretation of $\psi=\frac{1}{\sqrt{\pi a^{3}}}e^{-r/a}$ correct?
Jan
15
accepted Calculation of spherical Bessel functions - meaning of $\left(\frac{1}{x}\frac{d}{dx}\right)^{l}$
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.
Jan
15
asked Calculation of spherical Bessel functions - meaning of $\left(\frac{1}{x}\frac{d}{dx}\right)^{l}$
Dec
31
accepted Don't understand the integral over the square of the Dirac delta function
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?