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| 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 |
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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 |
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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 15 |
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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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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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May 7 |
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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 6 |
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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 |
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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 |
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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 |
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May 5 |
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Zero divergence of energy-momentum tensor and gravitational energy Oh dear. I think I've got this seriously wrong. Please bear with me. I thought the rhs of the field equations (ie the EMT) described the total energy-momentum of a system (which I think is correct). I thought the lhs was a description of the curvature of spacetime caused by the rhs (which I think also is correct). But I never thought of the lhs as describing gravitational energy. (Thinking about it, both sides must of course have the same units, so if the rhs describes energy so must the lhs). I thought that the “additional source of energy”, ie gravitational energy, was missing |
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May 5 |
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Zero divergence of energy-momentum tensor and gravitational energy thanks. I did try to read the link but it was a little over my head (my level - I've no idea what Hamiltonians or Lagrangians are!). I've been trying to understand the Schwarzshild metric and relativistic cosmology both of which are based on a particular energy-momentum tensor (zero and perfect fluid). I still don't see how that can be valid if the EMT doesn't include gravitational energy. I sense you answering my question in your final paragraph ("very-long-distance effective description") but, sorry, don't understand. Please don't worry about making your answer too simple! |
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Apr 26 |
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Confusion regarding geodesic of thrown ball - curved or Cartesian coordinates? Thank you, though I find your first sentence a little ambiguous. Are you saying it's a valid approximation to plot the ball's path using Cartesian coordinates because (1) it's moving relatively slowly and/or (2) is in a weak gravitational field and/or (3) is not being thrown very far? So if I was throwing the ball at a velocity near the speed of light on the surface of a neutron star, for example, I couldn't plot its path using Cartesian coordinates but would need to take into account spacetime curvature and use curved coordinates - whatever they are? |
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Apr 25 |
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Why do objects follow geodesics in spacetime? Thanks for the replies, though I'm afraid they are way over my head. I still don't get it. |
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Apr 24 |
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Why do objects follow geodesics in spacetime? @David Zaslavsky Both, but of course I can't promise I'll understand the mathematical one. |
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Apr 8 |
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Hubble time, the age of the Universe and expansion rate thanks very much. |
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Apr 7 |
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How does the critical density decide the fate of the Universe? of course! Funny how some things are obvious when they're pointed out. I was getting the strangest results trying to feed my equation into the WolframAlpha differential equation calculator. Thanks. |
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Apr 6 |
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How does the critical density decide the fate of the Universe? thanks. Ryden uses parametric equations to get her big bang to big crunch graph of a matter only universe on p87. Just out of interest, do you know why I can't set the first term on the rhs of the Friedmann equation to a constant $C$, $k=+1$, $c=1$ to obtain $$\left[\frac{1}{R}\frac{dR}{dt}\right]^{2}=C-\frac{1}{R^{2}}$$ $$\frac{dR}{dt}=\left(R^{2}-1\right)^{1/2}$$ but when I try to solve this, I don't get the nice Big Bang to Big Crunch graph that Ryden does? |
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Apr 3 |
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Very basic question about empty universe Thanks for that. I looked at the paper - superluminal galaxies are not travelling faster than light in our or any other observer's inertial frame so don't contradict STR. That is amazing. |
