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Aug
15
comment Why the notation of Lorentz Transform has to be like this?
Sounds like you're asking, "it's not a tensor, so why do we write it like a tensor?" Maybe you should be asking, "is it actually a tensor?"
Jul
24
comment 9-point stencil “equivalent” for advection equation
Can you write the particular equation you are trying to solve?
Jul
7
comment Vector product in a 4-dimensional Minkowski spacetime
I take "event" here to mean a particular location and moment--a position in spacetime, in other words. Of course, one should always keep in mind that positions in spacetime do not obey the transformation laws of four-vectors.
Jul
4
comment Is the stress energy tensor continuous at the interface between an object and vacuum?
I think a better title for this question would be, "Is the stress energy tensor continuous?" or "Is the stress energy tensor continuous at the interface between an object and vacuum?"
Jun
17
comment Clarification on meaning of scalar in math and scalar in physics
My position is that you should not take the physics "definition" literally, but it is instead a sloppy rephrasing of the mathematics definition.
Jun
17
answered Determining the Lorenz gauge condition
Jun
17
comment Clarification on meaning of scalar in math and scalar in physics
The physicist could call it a scalar, but he would be imprecise. Volume forms are textbook examples of pseudoscalar quantities.
Jun
17
comment Clarification on meaning of scalar in math and scalar in physics
Physicists just emphasize aspects of the definition that are most useful to them. I don't see two conflicting definitions here. I see one definition that some people tend to talk about rather formally and others more imprecisely.
Jun
17
comment Clarification on meaning of scalar in math and scalar in physics
I think considering more than those transformations is simply overkill. By considering general changes of basis, your experiment only gains information about whether the quantity lives in the framework of pseudo-Riemannian geometry at all.
Jun
17
answered Clarification on meaning of scalar in math and scalar in physics
Jun
9
comment Tricks for evaluating tensor contractions with Levi-Civita symbol
Yes, the algebra of Dirac matrices is a clifford algebra (as is the algebra of Pauli marices for 3d spin). - Physicists tend to think of the the Dirac and Pauli algebras as having to do with matrices, but in mathematics, the Dirac and Pauli matrices are considered a representation--a way to build matrices that obey the desired commutation relation. - In this answer, everything I deduce only relies on the commutation relations; making the gammas into matrices is just a convenience. - By nature, the clifford algebra is very similar to exterior algebra and differential forms, as well.
Jun
9
revised Tricks for evaluating tensor contractions with Levi-Civita symbol
fixed more sign issues from the sign of the metric
Jun
9
answered Tricks for evaluating tensor contractions with Levi-Civita symbol
May
27
answered Resolving General relativity and Newtonian mechanics to a computer
May
25
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.
May
24
answered Where does this relation between gamma, energy and mass come from?
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).