This account is temporarily suspended to cool down. The suspension period ends on Aug 24 at 17:32.
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479185
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location New York City
age 40
visits member for 2 years, 11 months
seen 4 hours ago

I do not participate on this site any longer, except to respond to comments regarding my own text, if that text is unavailable in another form. I do not accept the political moderation atmosphere here, it is not compatible with open science.


Jun
28
comment On-shell symmetry from a path integral point of view
@drake: I understand what you are saying, I agree with you on the physics, and the content you gave was fine, although not written the way or with the meaning I intended at all. I am sorry for sniping at you, I got annoyed. Thanks for making the answer nice, and good luck.
Jun
28
comment On-shell symmetry from a path integral point of view
@drake: I just noticed when I did the example that I was missing the other O's, it was dimensionally wrong too. I give up on edits, the editor is impossibly slow (every keystroke takes 10 seconds to register, and the time gets longer and longer). The point is: equation of motion for the $\phi$ field, times the other operators, is only nonzero at the collisions, where the contribution is the variation of these operators under the variation of $\phi$, I did it from memory, I made a botchy notation mess, but I really understand this, and it was annoying that the fixes were changing the meaning.
Jun
28
comment On-shell symmetry from a path integral point of view
@drake: Yes! You are right, of course, stupid me. I just noticed this too, I am sorry.
Jun
28
revised On-shell symmetry from a path integral point of view
more fixes
Jun
28
revised On-shell symmetry from a path integral point of view
more fixes
Jun
28
comment On-shell symmetry from a path integral point of view
But I understood where the confusion came from--- you thought the $\phi_k$'s were other fundamental fields in the Lagrangian--- not necessarily (and the relation would be trivial then). The $\phi_k$'s are just arbitrary composite local operators, and the right hand side has the variational derivative. Since it is not clear to you yet (and you know your stuff) I changed the notation, and fixed the "dimensional inconsistency" (which just means you didn't like it that I factored out the stupid delta function without changing the notation for $\delta \phi$, ok fixed)
Jun
28
revised On-shell symmetry from a path integral point of view
fix
Jun
28
reviewed Approve suggested edit on On-shell symmetry from a path integral point of view
Jun
27
awarded  Necromancer
Jun
26
revised On-shell symmetry from a path integral point of view
i's and signs
Jun
26
comment On-shell symmetry from a path integral point of view
@drake: Reading the corrections--- I want to make sure the main point of this is clear: the equation of motion obeys a ward identity with other insertions, so that the product of the phi equation of motion times an operator is equal to zero away from the collision, and is equal to the variation of that operator under a shift in phi at the point of the collision (times a delta function). So the product stays identically zero when the fields are independent variables, or when you have a composite operator that doesn't vary when you vary phi, which happens to be the case for the SUSY closures.
Jun
26
comment How does gravitational lensing account for Einstein's Cross?
@JoeHobbit: Unfortunately, I don't have a copy--- it's described in words--- draw two saddles and two sources and link them to the origin and infinity in the unique way with flow-lines.
Jun
26
revised On-shell symmetry from a path integral point of view
final fixes
Jun
26
revised On-shell symmetry from a path integral point of view
remove brackets
Jun
26
revised On-shell symmetry from a path integral point of view
rolled back to a previous revision
Jun
25
comment The necessity of the B field
@LarryHarson: I was talking about the 1905 paper, where he just knows that E and B exist. But the derivation can be made using only the fact that q is independent of frame. If E were a gradient of a scalar, so that there were no B, then q would have to transform like mass under boost. If E were the gradient of a metric tensor, like in GR, again, q would have to transform. This how Purcell gets the general form, although I don't like it so much. There are only a discrete set of tensorial covariant expressions that match the low energy limit, and it's not wrong to select one based on experience.
Jun
25
revised If we know the universe is made up of a relativistic ether, why wouldn't gravity just be a pressure gradient of the ether?
fix stupidities
Jun
24
comment The necessity of the B field
@LarryHarson: Under the assumption of covariance, the special case at low velocities implies the general thing: once you know the behavior at slow speeds, you find the behavior at high speeds by boosting. How do you find the correct covariant behavior quickly? You find a convariant expression which reduces to the correct nonrelativistic limit, in this case it's uniquely determined, from the tensor form of F, which follows from the invariance of charge under boosts. This is how you derive the second equation from the special case of the first, and it is how Einstein did so.
Jun
22
comment How does electricity flow in conductor when potential difference is applied?
@SatwikPasani: Because the current is constant, and the voltage adjusts itself quickly so that the heat production from the current matches the potential drop--- it's the only self-consistent solution, and it's set up when you close the circuit at the speed of light. The distribution of charges doesn't change the voltage difference across the battery, because this is determined by the chemical reactions in the battery, this is the initial driver to set up the voltage differences in the wire.
Jun
21
awarded  Nice Answer