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Jan
31
answered Why do the Einstein field equations (EFE) involve the Ricci curvature tensor instead of Riemann curvature tensor?
Jan
20
answered What is the mathematical nature of the stress-momentum-energy tensor?
Jan
16
comment Is a metric tensor field the same thing as $ds² = -dt² + dx²+ dy² + dz²$?
@StanShunpike Multilinear maps are functions. They're just linear in each argument.
Jan
1
awarded  Nice Answer
Dec
10
revised Basic question in electromagnetic duality
fixed some sines and cosines
Dec
6
awarded  Enlightened
Dec
6
awarded  Nice Answer
Nov
27
revised Do there exist functions $\phi$ and $A$ such that $\vec E$ satisfies the Helmholtz Theorem $\vec E = -\nabla \phi + \nabla \times \vec A$?
fixed some math about the symbol for reals
Nov
26
comment What do people actually mean by “rolling without slipping”?
@NeuroFuzzy Interesting. I suspect this comment section isn't the place for this discussion, though. Perhaps you could ask a question about rolling on curved surfaces, present your work, and ask for references that might verify or refute your idea? I'd be happy to follow that question if you link to it here.
Nov
26
comment What do people actually mean by “rolling without slipping”?
@NeuroFuzzy Sure, if you define the overall travel of the wheel with respect to the center of the wheel. If instead you measure the overall distance traveled by whatever point is in contact with the ground, I think the bold statement still holds. Regardless, travel of a wheel along a curved surface is more involved, and I think it's good you point that out.
Nov
7
comment Can Gauss' and Ampere's Laws be written in terms of the divergence of an energy four-vector?
Are you not familiar with the special relativistic form of the EM fields, the faraday bivector $F$? Of how Maxwell's equations in vacuum can be written $\nabla \wedge F = -\mu_0 J$ and $\nabla \cdot F = 0$?
Oct
29
comment Is $E^2=(mc^2)^2+(pc)^2$ correct, or is $E=mc^2$ the correct one?
@user929304 It's a purely geometrical phenomenon. It'd be like rotating a vector and asking it to trace out an ellipse instead of a circle. In Minkowski space, boosting a four-velocity traces out a hyperbola. Hyperbolas have asymptotes--lightlike rays, in this case.
Oct
19
awarded  Yearling
Sep
25
comment Four-momentum, four-velocity, energy
@Kennan You just wrote $\eta_{ab} = e_{\hat a} \cdot e_{\hat b}$. Clearly the $p^{\hat 0}$ term comes from $p^{\hat 0} e_{\hat 0} \cdot e_{\hat 0} = p^{\hat 0} \eta_{\hat 0 \hat 0}$. What is the value of that metric component, $\eta_{\hat 0 \hat 0}$?
Sep
25
comment Four-momentum, four-velocity, energy
What's the signature of your metric?
Aug
29
answered When can I use $\wedge$ instead of curl?
Aug
23
comment Why does the second Weyl scalar describe electromagnetic radiation?
I could interpret that to mean something like telling you about the total mass causing the curvature (see the Schwarzschild case, for instance, which is characterized only by the mass), the way the Coulomb field is characterized by only the total charge of the point charge.
Aug
9
answered Is relative velocity invariant under special relativity?
Jul
24
comment Gradient is covariant or contravariant?
$\nabla \varphi = \sum_i g^i \partial_i \varphi$, where each $g^i$ is a cotangent basis vector. $g^i \cdot g_i = 1$ always, while $g^i \cdot g^j = g^{ij}$ gives you metrical components.
Jul
24
comment Gradient is covariant or contravariant?
Obviously you can expand any such quantity in terms of either the tangent basis or the cotangent basis, but expanding the quantity $\nabla \varphi$ that I defined above would give you metrical terms if you expanded in terms of the tangent basis--it is naturally written in terms of the cotangent basis.