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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.
Jul
23
comment Gradient is covariant or contravariant?
Sorry, I don't understand what you're asking me to do.
Jul
22
comment Gradient is covariant or contravariant?
The Cartesian bases are the same whether they're covariant or contravariant, so I don't see how you can draw a conclusion from that.
Jul
21
answered Gradient is covariant or contravariant?
Jul
21
comment Metric tensor in special and general relativity
It's common in differential geometry to identify vectors with directional derivatives. It's kind of a definition of last resort: if your manifold were a vector space, you could take $\partial \vec x/\partial x^\alpha$ and get a vector because $\vec x$ is an element of a vector space. When positions are no longer elements of vector spaces, that notion breaks down. The directional derivatives themselves still obey the vector space structure, though.
Jul
5
revised Tensors in special relativity
some subject-verb agreement
Jul
5
answered Tensors in special relativity
Jul
3
comment E&M and geometry - a historical perspective
You might also find geometric calculus, based upon clifford algebra, interesting here. It manages to take those free space equations for the EM field and marry them into one equation.
Jun
29
comment “Vectors” (i.e. 1-tensors) their definition and motivation for relativity
Perhaps they mean that $p'$ cannot have that form in every basis, which would be true.
Jun
29
comment “Vectors” (i.e. 1-tensors) their definition and motivation for relativity
I don't understand "the question that motivated this post..." Is $p'$ supposed to be the result of some transformation? Is it merely supposed to be some other vector-like quantity that may or may not obey the correct transformation laws?
Jun
26
comment What invariant counting process derives the Minkowski metric?
This answer doesn't strike me as in analogy to Larry's question, though: why not consider an object of unit spacetime interval and proceed exactly as in Larry's explanation for Euclidean space?
Jun
26
comment What invariant counting process derives the Minkowski metric?
So, you presuppose the rotation transformation, and then it follows that the Euclidean metric is the one that is kept invariant under that rotation transformation. Do you think this process would turn out differently by presupposing the Lorentz transformation instead?
Jun
26
comment What invariant counting process derives the Minkowski metric?
I don't understand. How does each observer measuring the same length derive the Euclidean metric? How are the copies of each coordinate system different from one another? Are they merely rotated with respect to each other?
Jun
18
comment How big or small is a reference frame in Relativity?
Re: train. Yes, relativity tells us that's purely a geometric phenomenon. Imagine two people standing in line with two fenceposts. One person faces the line of fenceposts and perceives that one post is closer in the direction he's facing than the other. The other person faces perpendicular to the posts and perceives that both posts are equally far ahead of him. That's trivial, of course: you know that the two people, though facing different directions, are perceiving the same physical system. Think of the observers in your question as merely facing different directions through spacetime.