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1d
comment Technical question about 2-forms
If you want to do contractions in an index-free manner, and still you may want to try clifford algebra. For instance, differential forms would require you to write $\partial^\mu F_{\mu \nu}$ as $\star d (\star F)$. Clifford algebra would just denote it $\partial \cdot F$ instead.
2d
answered Where does this relation between gamma, energy and mass come from?
May
21
comment Scalar functions and manifolds
Sorry, I've reworded the edit. Hopefully that makes the answer clear.
May
21
revised Scalar functions and manifolds
changed the wording; the scalar fields agree for the same points on M
May
21
comment Scalar functions and manifolds
Let me know if that edit is to your liking.
May
21
revised Scalar functions and manifolds
added a section on the definition of a scalar field
May
21
answered Scalar functions and manifolds
May
21
comment Scalar functions and manifolds
Are you sure he said a scalar function maps $M \mapsto \mathbb R^n$? Because one might expect that to be $M \mapsto \mathbb R$ instead.
May
18
comment What are Einstein constraint equations?
Additionally, books on relativity from a computational physics standpoint also tend to discuss the constraints: see, for example, Baumgarte and Shapiro, or Alcubierre.
May
6
answered Typical Mathematical Questions about the Lorentz-Transformation
May
6
comment Magnetic monopoles and special relativity
@LandosAdam That the electric and magnetic fields can be unified is true regardless of whether there are magnetic monopoles or not. If there can be magnetic monpoles, then the fields may take on different values, but they are still unified. - In my answer, I talk about adding a monopole source term in 3d; that doesn't change that the magnetic field is still a vector field in 3d. The same logic applies for special relativity. Adding a "magnetic four-current' doesn't change that the EM field is unified in 3+1 special relativity.
May
6
comment Magnetic monopoles and special relativity
@ChrisWhite Not at all; a magnetic four-current is just a (four)-trivector source term, compared to the electric four-current that is a (four)-vector source term. The electric and magnetic four-currents can be transformed into each other through Hodge duality, just as the electric and magnetic fields can be (since bivectors, like the Faraday bivector, dualize to other bivectors). - Are you confusing the two sides of the equation? Adding magnetic monopoles in no way changes that the Faraday bivector...is a bivector (and thus can be represented with an antisymmetric matrix).
May
6
answered Magnetic monopoles and special relativity
May
3
answered What is the meaning of EM field having curl?
May
2
awarded  Nice Answer
May
2
comment What is the uncertainty principle?
@StanLiou Sorry, I don't really see the difference. I described (on a very broad and hopefully accessible level) the notion of standard deviation of some function or distribution. I used the word "wave" to emphasize the connection with wavefunctions in Hilbert space, yes, but this is applicable to any square integrable function. Even in phase space formalism, one ultimately arrives at a statement of uncertainty that involves a product of standard deviations, does one not? So I'm not appreciating any fundamental difference here.
May
2
comment What is the uncertainty principle?
Sorry, what do you mean by, "this is only a peculiarity of the Hilbert space formulation"?
May
1
answered What is the uncertainty principle?
May
1
comment Rotation in the x-t plane
Yes, the Minkowski distance. - The hyperbola drawn there is appropriate for spacelike objects. Imagine the $tx$-plane cut into quarters by the diagonals (the lightlike lines). Each quarter corresponds to a distinct region: the quarter containing $+ct$ is the causal future, for instance--the set of points that can be influenced by an object at the origin. A point on the hyperbola in the causal future will stay on that hyperbola piece no matter what Lorentz transformations are applied, and cannot reach the lightlike lines on either side (since the hyperbola never intersects them).
May
1
answered Rotation in the x-t plane