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I'm an ex-twistor theorist who worked on the original twistor programme until 1984, when I decided that I wasn't clever enough to be a mathematical physicist ! Have worked in industry since then and am now trying to catch up on what has been happening in physics.


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awarded  Yearling
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awarded  Nice Question
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awarded  Revival
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awarded  Nice Answer
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Feb
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awarded  quantum-field-theory
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Nov
7
comment How algebraic geometry and motives appears in physics?
Related: physics.stackexchange.com/q/29424
Nov
4
comment single-particle wavepackets in QFT and position measurement
Also relevant to the discussion of wavefunctions in QFT is the question of the causal behaviour of the "naive" wavefunction (not the Schroedinger wavefunctional that Trimok is referring to). This is discussed here
Nov
2
awarded  Nice Answer
Nov
1
comment Euclidean black hole extrinsic curvature
Not sure how to answer your question fully, but since the horizon is two dimensional, there are two normals with respect to which extrinsic curvatures can be defined. If $n$ is one of the normals, the ext. curvature for that direction is $\mathcal{L}_ng_{\mu\nu}$ and so vanishes if $n$ is a Killing vector for your metric. In your Euclidean case this Killing property is what you'd need to show presumably.
Oct
21
comment Diffeomorphism Invariance of General Relativity
To specify the action of the active diff. you have to be extremely precise with the notation, see for example here and the reference to the paper by Lusanna.
Oct
21
comment Diffeomorphism Invariance of General Relativity
@JLA If you take your example of changing to polar coordinates, this is a passive diffeomorphism - the metric components have changed, but the metric tensor $\eta_{\mu\nu}e^{\mu}\otimes e^{\nu}$ hasn't. With an active diffeomorphism, the metric tensor itself changes, so a solution of the wave equation doesn't (necessarily) get mapped to a solution. In the Euler-Lagrange equations for the wave equation, the metric tensor is still fixed, i.e. part of the background.
Oct
20
comment Diffeomorphism Invariance of General Relativity
Incidentally, there are possibilities to make any theory diffeomorphism invariant by parametrization see here for example.
Oct
20
comment Diffeomorphism Invariance of General Relativity
If I take the wave equation $\Box \psi=0$ for example, this is not diff. invariant because it really means $\eta^{\mu\nu}\nabla_{\mu}\nabla_{\nu}\psi=0$, and if I apply a diffeo $\phi$, it messes up $\eta_{\mu\nu}$, which is part of the background. In other words, the wave equation is not "solving for" $\eta_{\mu\nu}$. However with Einstein's equations, all the variables you have which define the physics - $g_{\mu\nu}$ are the things you're solving for - there are no fixed background quantities.
Oct
14
comment What is the weight equation through general relativity?
@user1551 the condition $1-\frac{2GM}{c^2r}<0$ would mean that you're at a point inside the Schwarzschild radius i.e. inside the horizon of a black hole. So, $M$ can exceed $\frac{c^2r}{2G}$, but it just means that a black hole is present.
Oct
3
comment Isn't gravity non-local and non-causal?
They're indeed not identical in all cases, and the apparent horizon can be just as hard to work with as the event horizon. I thought it worth mentioning because the apparent horizon has featured a few times in recent papers.
Oct
3
comment Is “A Brief History of Time” still up to date?
Nuclear conflagration invoked by last sentence in 3...2...1....