I consider it worth learning just for the simplicity it introduces into problems that otherwise would've been extremely tedious or difficult.
It's not uncommon that one will have to manipulate a tensor expression to try to get to a simpler result. GA's generality and identities make this...
It sounds like you're imagining a box with a finite extent and getting confused because the magnetic field should be weaker on one side of the box or the other. What you should be imagining is that in the limit that the box takes on an infinitesimal extent, the magnetic field strength is the...
I admit I'm not well versed in stuff with manifolds and such, but isn't a diffeomorphism basically a way of mapping positions on one manifold to positions on another? If so, then ##f## is just the Jacobian of the mapping, and it is inherently dependent on position within the manifold. It's...
Regarding Schrodinger's equation: the imaginary there is not merely used for convenience. Quantum mechanics works on a system of probability that is fundamentally different from what we usually think of--it is, essentially, probability on a plane. I've believed for a while that one can convert...
The stress energy tensor is a linear operator on a vector. Feeding it different vectors (or evaluating the resultant vector by its covariant or contravariant components) doesn't change that.
For example, the stress-energy tensor of an ideal fluid is
$$\underline T(a) = (\rho + p)(a \cdot...
I personally have no need of dyadics--most linear operators can be directly expressed as functions of the vector they act on. For instance, a reflection operator is
$$\underline N(a) = a - 2(a \cdot n)n$$
for any unit vector ##n##. Ultimately, that's what linear operators are--functions of...
One approach is to use multivectors and wedge products.
What are bivectors? Oriented planes, the same way vectors can be taken to represent oriented lines with magnitudes. If you have two basis vectors ##e_x, e_y##, then the bivector ##e_x \wedge e_y## represents the unit bivector of the...
Let there be two vectors ##e_0, e_1## such that ##e_0 \cdot e_0 = -1## and ##e_1 \cdot e_1 = 1##, as well as ##e_0 \cdot e_1 = 0##. This is an orthonormal basis for a 1+1 Minkowski space.
Isotropy of this space allows us to freely change the basis. Let ##{e_0}' = ge_0 + he_1## and ##{e_1}'...
Subspaces can be viewed as geometric objects containing the origin: the point at the origin, a line through the origin, a plane through the origin, etc. Each of these constitutes a subspace of the overall vector space.
It works just fine? :P
In all seriousness, this topic of passive transformations is one I'm still tackling myself. I've usually just gone along with passive-active equivalence and always used only active transformations.
Hmm. I guess, in the passive transformation, you can "convert" to an active one by considering the point with coordinates ##{x'}^\mu## but on the original coordinate chart ##O##, and then go about your business as with the active transformation?
Alternatively, can you use a point with the...
I'm not so sure. An observer can use whatever coordinate system they want.
How is a relabeling of events in spacetime distinguishable from remapping points?
Let me be concrete with this. Let there be a scalar field ##\phi(x^0, x^1)##. A 1+1 spacetime is sufficient to get the point...
Agreed, usually we just say the point ##x## is expressed in terms of different coordinates and basis vectors. But that doesn't stop that there exists a point with the same coordinates as ##x## originally had, but evaluated on the transformed basis vectors.
You're probing at the problem...
stevendaryl's statement is correct. If the spacetime is flat, there exists a plain old cartesian coordinate system that can describe everything everywhere because all the tangent spaces are the same.
You make sweeping, general statements that on their face are incorrect. You do not clarify...