694 reputation
620
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location United Kingdom
age
visits member for 3 years, 5 months
seen 17 hours 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.


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?
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.
Dec
30
asked Don't understand the integral over the square of the Dirac delta function
Dec
24
awarded  Critic
Jun
17
awarded  Yearling
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
7
accepted Zero divergence of energy-momentum tensor and gravitational energy
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.
May
5
comment Zero divergence of energy-momentum tensor and gravitational energy
of the rhs (the EMT) doesn't imply the conservation of energy and momentum. If you put the Einstein tensor on the rhs (ie, the lhs equals zero), would the zero divergence of the rhs then imply total conservation of energy and momentum?