@ACuriousMind well i am kind of just picturing two $\mathbb{R}^2$ and im just picturing associating these points according to the transformation statement and im not really visualizing any sort of motion in this situation.
or i dont even know what it would mean to have motion in this situation
It just means that x'=0 is a straight line in the (x, t) co-ordinate system
@Relativisticcucumber
So it means Lorentz transform maps some constant velocity trajectory to the time-axis, which is like "switching to the PoV of a constant velocity frame"
but you don't need to write the power series in that weird way - since $\mathrm{ad}_X$ is linear, you can just represent it as a matrix (in terms of acting on the basis $\Lambda_i$ of $\mathfrak{g}$)
then just use the computer algebra system of your choice to compute the power series for $\frac{1 - \mathrm{e}^{-\mathrm{ad}_X}}{\mathrm{ad}_X}$ as a series in matrices
unless you have a specific reason to believe there should be a simpler way, computations are sometimes just tedious
the $e^{-i\Sigma \theta_i \Lambda_i}$ are $g \in G$. taking all $\theta_i = 0$ gives the identity id $\in G$. Fix $i$. So, the directional derivatives at the identity can be found by taking all $\theta_j = 0$ for $i \neq j$. Then, the exponential derivative simplifies greatly (the whole power series term is now gone) because there is no "translation term". Namely, the directional derivatives at the identity look like $-i\theta_i \Lambda_i$.
Sure, that's the special case when $X(0) = 0$ in the formula for the derivative of the exponential
the term that "shifts" this to some other $X(0)$ is precisely the $\frac{1 - \mathrm{e}^{-\mathrm{ad}_X}}{\mathrm{ad}_X}$ term you seem to be trying to avoid :P
but the pushforward $g \cdot (v \in T_{\text{id}}G)$ if all i cared about is mapping from the directional derivatives at the identity to the directional derivatives at an arbitrary $g \in G$ does the same thing?
when you asked about "directional derivatives" without this context, I assumed you mean operator that compute direction derivatives, i.e. the $\frac{\partial}{\partial \theta_i}$
these are just the tangent vectors, and so they are related by the pushforward, etc. etc.
if i laxen this condition, because really i would like to compute the directional derivative at every single point and hence an arbitrary point, what you said previously still doesn't apply?
you can solve the whole thing of course while leaving the value of $X(0) = \sum_i \phi_i \Lambda_i$ arbitrary
just like when we compute the derivative of a function $f$ we also have the formula for the derivative "at point x" as $f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$ but no one forces you to plug in a specific value for $x$, you can just leave it arbitrary
hm so to my understanding the basis of a tangent space $T_gG$ are the directional derivatives at $g \in G$. and the pushforward map will send the basis of $T_eG$ to the basis for $T_gG$. So it will send the set of directional derivatives at the identity to the set of directional derivatives at $g$
well i am trying to minimize the linear entropy of a traced-out-of-the-environment arbitrary-hamiltonian-time evolved state and this reduces to computing the directional derivatives of $\exp(-i\Sigma \theta_i \Lambda_i)$ at least to my understanding
Suppose I have a parameterized quantum state: $\rho(\theta) = U(\theta) \rho U^\dagger(\theta)$. I am curious to know whether the following holds:
$\frac{\partial \text{Tr}_A (\rho(\theta))}{\partial \theta} = \text{Tr}_A \Big(\frac{\partial \rho(\theta)}{\partial \theta} \Big)$, where $\text{Tr}...
@RyderRude I always recommend students to learn Cartesian tensors first, because in that context it is much less confusing, much more clear what is happening, and at the same time, vastly useful because it would simplify a lot of the proofs that appears in vector calculus. Then, once you are used to the notation and indices, it will be easier to move to the curvilinear stuff.
In a transition where the particle number in the $E_n$'th energy level raises by 1 you get a $e^{-iE_n t}$ factor, and if it decreases by 1 you get a $e^{+iE_n t}$, now the $-E$ solution makes perfect sense and everything in QFT becomes consistent
@bolbteppa I'm just saying that when we define the position space wavefunction using the $\langle 0|\phi(x) |\psi\rangle$ scheme, there are no positive frequency exponentials in the evolution of $\psi(x)$
You're using about 3 or so distinct ideas in writing that down, trace any of it backwards and you'll end up facing this $-E$ thing, once you even write down the Poincare group you're faced with the question of its irreducible representations which immediately throws the Casimir $P^2=m^2$ in your face and you're faced with negative energies
I thought of a physical reason but I'm not sure it's correct regarding the formalism. If it is true that $$g_{\mu\nu} = \mathbf{e_{\mu}}\mathbf{e_{\nu}}$$ then it is equivalent to saying that no basis vector has a projection on a time-like direction? Meaning $g_{00}$ represents roughly, time dilation only, and the off diagonal $g_{m0}$ represent length contraction?
@Slereah But I kind of think what you're saying has a physical intuition too, that we must unlink the time components from the spatial ones 'cause that's taken for granted in CM
Also $\mathbf{e}_{\mu} = \frac{\partial \mathbf{r}}{\partial x^{\mu}}$, where $\mathbf{r} = t \hat{t} + x \hat{i} + y \hat{j} + z \hat{k} = x^{\mu} \hat{e}_{\mu} $ is the position vector in space-time
I think it's because the gravitational field is static i.e. does not depend on time so it should be invariant under $t \to - t$ and if you include the $g_{m0} dx^m dx^0$ terms the metric will change under this transformation
I do remember reading somewhere about this time reversal, they also mentioned a difference between a static and a stationary metric. Honestly I didn't quite understand.
In $ds^2 = g_{00} dt^2 + 2 g_{0i} dx^0 dx^i + g_{ij} dx^j dx^j$, since the gravitational field is static we see none of the $g$'s depend on $t$ so the gravitational field is invariant under $t \to - t$, but the line element $ds^2$, i.e. squared differential length in space-time, is not invariant under $t \to - t$, instead we get $ds'^2 = g_{00} dt^2 - 2 g_{0i} dt dx^i + g_{ij} dx^i dx^j$, unless of course $g_{0i} = 0$ in which case $ds'^2 = ds^2$ as should be the case since $ds^2$
is physically a squared line element in the space-time with $g$ as its gravitational field
Okay, that's helpful. But I am not 100% clear on what this invariance under $t \rightarrow -t$ represents, is it a kind of a time dependent curvature of space?
Yes, I understand. But translating to some physical concept... what are we requiring, that a world line has the same length forward and backwards in time?
Because $g_{\mu}(t,x,y,z) = g_{\mu \nu}(x,y,z)$ holds (the assumption that $g_{\mu \nu}$ is static), we trivially have $g_{\mu}(t,x,y,z) = g_{\mu \nu}(-t,x,y,z) = g_{\mu \nu}(x,y,z)$, this trivial statement should be reflected in the expression for any given (squared) line element $ds^2$ in such a space-time where the gravitational field possesses this trivial symmetry
Hey everyone. I'm trying to find out exactly how current flows in an ideal wire—assuming that there even is a solution to Maxwell's equations describing that.
It seems like since the E field is zero, its curl also has to be zero, meaning that the B field can't change. Let's assume that the B field is zero. Its curl also has to be zero, and so no current can ever flow. Which... obviously doesn't sound right.
Wait, I think I've got the answer. All the current has to flow at the surface. Both the E field and the B field will remain zero inside the wire, but nonzero outside the wire. I think they're both of bounded magnitude, too.
