I got a job.
It's just a part-time one, the details of it don't really matter. Fact is, I got accepted to UC Berkeley, so I'm in Sydney with a couple of months to waste. "Haha, I'll get some money from a job!" I said. "It won't be that bad!" I said. Just two weeks in, and I feel like floating down the river styx, whispering prayers of a dead man. Every minute, every second is an eternity.
In other slightly less depressing news, I think I solved a long-standing problem I've had. The problem is pretty easy to state. Suppose I have a magnetic field that looks like:
$$B(x,y,z,t) = t\hat{z}$$
what's the electric field like? I've had quite a long history with this actually. When I first thought of the problem, I thought I could figure it out using the integral form of Maxwell's equations. (this one: $\int E\cdot d\vec{l} = -\frac{d\phi}{dt}$, where the line integral of the electric field is given by the change in magnetic flux through that surface). It didn't really work out though, because I just ended up with a bunch of line integrals telling me that the electric field should swirl 'anticlockwise' everywhere (when looking down the $z$ axis), but not actually getting me the electric field.
I asked senpai M about it, and he said something wise along the lines "Maybe the integral formulation ain't good enough, try the differential formulation. And remember, boundary conditions!!" and I was all like "er ok that's cool" (I didn't know a thing about the differential formulation).
Anyways, many months passed, and since I was learning a bit of vector calculus (about time) (I hope that's what it's called, the one with all those upside down triangle things) I thought I'd try solving it again. So I popped out my Maxwell's equations from google:
$$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}\\
\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}\\
\nabla \cdot \mathbf{B} = 0\\
\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}$$
With the given situation, we know $\mathbf{B}$ for all $t$, and let's just assume there ain't no charges in sight either. By subbing in these conditions we get the equations:
$$\nabla \cdot \mathbf{E} = 0\\
\nabla \times \mathbf{E} = -\hat{z}\\
\frac{\partial \mathbf{E}}{\partial t} = 0$$
Notice that the last equation tells us that the electric field is static (because there ain't no curl in the magnetic field). Finally, we can crack these nuts into all their components, and we get 4 equations that $\mathbf{E}$ has to satisfy:
$$\frac{\partial \mathbf{E}_x}{\partial x} + \frac{\partial \mathbf{E}_y}{\partial y} +\frac{\partial \mathbf{E}_z}{\partial z} = 0 \\
\frac{\partial \mathbf{E}_z}{\partial y} = \frac{\partial \mathbf{E}_y}{\partial z}\\
\frac{\partial \mathbf{E}_z}{\partial x} = \frac{\partial \mathbf{E}_x}{\partial z}\\
\frac{\partial \mathbf{E}_y}{\partial x} - \frac{\partial \mathbf{E}_x}{\partial y} = -1$$
The thing about this solution that for some reason I never anticipated is that there are like heaps and heaps of solutions. For some reason I thought given a $\mathbf{B}$ field, then the corresponding $\mathbf{E}$ field is unique, but it depends on the initial configuration of the $\mathbf{E}$ field. Whatever the initial $\mathbf{E}$ field is (as long as it sticks to the four rules above), that's what the $\mathbf{E}$ field will be forever.
Probably. I'm about 66% sure (I only just learnt upside down triangle, the chance I made a mistake is not negligible). Anyway, if this is right, then one potential answer is:
$$\mathbf{E} = -x\hat{y}$$
which seems alright to me.
OYEA P.S. I remembered what was bugging me. If the $\mathbf{E}$ field is like stationary, then so is the energy stored in it. But somehow the magnetic field just grows and grows with time. And like, isn't the energy stored in a magnetic field something like something times $B^2$? Then energy shoots up as time goes on, does energy conservation just go out the window?
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I've forgotten why I started this blog now. I might as well use it as a place to remember stuff. I thought up a problem the other day (which is really in the same vein as a whole bunch of other problems that I've done) that seems hard but who knows.
Adventures of Melon boy, part 2:
Melon boy is back at it again! This time, he lives in an $n$-dimensional space. A set of coordinates $(x_1, x_2, x_3, ... x_n)$ define a point Melon boy can travel to, provided that $(\forall i)(0 \leq x_i \leq m)$ where $m$ is the 'size' of the grid. Each point is either a melon (with a 50% chance) or a pumpkin (with a 50% chance). If Melon boy is currently at a point $(x_1, x_2, x_3, ... x_n)$, he can travel to a point $(y_1, y_2, y_3, ... y_n)$ if the condition: $(\exists j)(0 \leq j \leq n)$ such that $[(|y_j - x_j| = 1) \wedge (\forall k) (k \neq j \Rightarrow y_k = x_k)]$ holds. Melon boy can travel from square $x$ to square $y$ for $0$ time if the squares are both melons or both pumpkins, else he travels for $1$ time. Melon boy starts at square $(\forall i) (x_i = 0)$. What is the shortest time it takes him to get to square $(\forall i) (x_i = m)$?
I'd rate the probability of that making sense (even to myself later on) as 33%. So, here, have a diagram:
A case for $n = 2$ and $m = 6$ (2 dimensional, 6 big square). The squares are either red (pumpkin) or blue (melon). The green numbers written on each square is the time it would take melon boy to get to that square if he took the most efficient path. In this case, melon boy can move up right down and left. In this case, the time to get to the upper corner is $t = 4$.
That's all for now folks. Stay tuned. Hopefully my soul doesn't escape any further than it already has.
