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network profile
| bio | website | |
|---|---|---|
| location | United Kingdom | |
| age | ||
| visits | member for | 1 year, 11 months |
| seen | May 10 at 17:51 | |
| stats | profile views | 108 |
I live in Essex, England.
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Jan 22 |
accepted | Is this interpretation of $\psi=\frac{1}{\sqrt{\pi a^{3}}}e^{-r/a}$ correct? |
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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. |
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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 ... |
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Jan 21 |
asked | Is this interpretation of $\psi=\frac{1}{\sqrt{\pi a^{3}}}e^{-r/a}$ correct? |
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Jan 15 |
accepted | Calculation of spherical Bessel functions - meaning of $\left(\frac{1}{x}\frac{d}{dx}\right)^{l}$ |
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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. |
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Jan 15 |
asked | Calculation of spherical Bessel functions - meaning of $\left(\frac{1}{x}\frac{d}{dx}\right)^{l}$ |
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Dec 31 |
accepted | Don't understand the integral over the square of the Dirac delta function |
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Dec 31 |
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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. |
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Dec 31 |
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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? |
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Dec 30 |
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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. |
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Dec 30 |
asked | Don't understand the integral over the square of the Dirac delta function |
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Dec 24 |
awarded | Critic |
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Jun 17 |
awarded | Yearling |
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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. |
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May 7 |
accepted | Zero divergence of energy-momentum tensor and gravitational energy |
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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? |
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May 6 |
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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. |
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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? |
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May 5 |
comment |
Zero divergence of energy-momentum tensor and gravitational energy in some way from the rhs, hence my confusion. I didn't know that (as you say) “the curvature of spacetime is ONLY dictated by the density of the non-gravitational energy density T or EMT”. So, thankfully, all the calculations I've struggled through re Schwarzschild and cosmology are still valid because only the EMT (and not gravitational energy) curves spacetime? I'm still puzzled about why zero divergence 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 |
