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Feb
6
awarded  Enlightened
Feb
6
awarded  Nice Answer
Feb
4
revised What exactly is the connection between gauge transformations and symmetry groups?
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Feb
1
comment Can Quantum Field Theory be right even though it doesn't include gravity?
It depends on what you mean by "right." If you mean "complete," then of course there is a sense in which our current understanding of QFT is incomplete because we haven't nailed down how to use it to accurately describe gravity. I'd say, however, that rarely would a physicist say that a model is wrong simply because it only has a limited domain of validity (as indicated by ACuriousMind's answer below).
Jan
24
awarded  Necromancer
Jan
22
comment What is “mass” in particle physics?
@Strilanc I figured, and I don't think that question ever received an answer that was particularly satisfying either. I tried adding some thoughts at the time in the form of an "answer" which you might find useful, but they're certainly not definitive by any means.
Jan
18
awarded  Yearling
Jan
13
comment Derivation of hyperbolic motion in Special Relativity
@MaddeAnerson This may be helpful: physics.stackexchange.com/questions/66839/…
Dec
29
awarded  Enlightened
Dec
29
awarded  Nice Answer
Dec
29
awarded  Nice Answer
Dec
24
awarded  Enlightened
Dec
24
awarded  Nice Answer
Dec
23
comment Equation for the equipotential lines?
@H.R. The dipole formula I wrote is already for an appropriate limit of opposite charges. If the same limit is taken for like charges, then one simply gets the potential for a point charge. I encourage you to explicitly work all of this out yourself, and then you'll be equipped to answer these questions for yourself!
Dec
23
comment Equation for the equipotential lines?
@H.R. I'm not sure I understand what you mean. What does it mean for a dipole to have a charge?
Dec
23
comment Equation for the equipotential lines?
@H.R. Sure if you insist, but I'd prefer to call it an idealization rather than an approximation since one can rigorously take the limit I mention and get that precise expression -- no approximations actually necessary :)
Dec
23
comment Equation for the equipotential lines?
@H.R. No. A pure dipole is defined as the limit in which the charges become infinitely close while keeping the dipole moment fixed (by requiring the charges to increase appropriately). In this limit, one finds the expression I wrote. Try doing the computation for charges $\pm q/\epsilon$ with positions $\pm \epsilon a \hat{\mathbf z}$. Expand the potential to first non-vanishing order in $\epsilon$, use the definition of dipole moment, and everything will work out.
Dec
23
comment Equation for the equipotential lines?
@H.R. The pure dipole potential applies for all $\mathbf x \neq 0$, not just positions with large norms.
Dec
23
awarded  Enlightened
Dec
23
awarded  Nice Answer