That's really cool. I know nothing about coding so I can't help you. I'm just wondering if you've compared your result with the known exact result via the image charge method. It would be cool to see the quality of this lattice method.
As far as I understand, such a charge distribution (a) always exists; and (b) is unique. This is a boundary value problem for Poisson's equation and existence and uniqueness of solutions (given completely specified boundary conditions) is a theorem.
I believe what you have here is tensor notation but with the tensor sign left out to reduce clutter. By the tensor A \otimes B of two matrices A,B, we mean at every entry of A, insert a copy of B multiplied by that entry in A. For example, i'll work out the first term in that Hamiltonian. I'll...
\langle N \rangle is the Fermi-Dirac distribution, which is derived on that wikipedia page. So, you can perform the derivative yourself and verify the second equality.
The first equality can be derived as follows. First,
\displaystyle \langle ( \Delta N )^2 \rangle = \langle (N - \langle N...
I'm assuming you know what to do without the absorbing wall (i.e. how to derive the appropriate diffusion equation and show that the solution is Gaussian of some particular width in the continuum limit etc. etc. etc.) The diffusion equation is linear and has a unique solution given complete and...
You are correct that the angle between Q and P is not the same as the angle between Q' and P. However, the length of Q is also not the same as the length of Q'. The cross product depends on both of these things. It might not be immediately obvious why these two changes exactly compensate for one...
"Linear" in "Linear Algebra" means "closed under addition". In physics a "linear system" is one that satisfies the superposition principle, which is just the physics way of saying closed under addition. This means that if S_1 and S_2 are two possible states of the system (i.e. two possible...
The 100 in sin(100t) has units of radians per second. When you multiply 100\ \text{rad/s} with t seconds, you get an angle in radians. Then you can take the sine of that. The sine of something that has units of seconds is meaningless.
This is the reason why you should NEVER EVER EVER solve...
I may be misunderstanding something, but I think the second term (with the double sum in i and j) should be multiplied by 2. Either that or the sum in j should be over j \neq i rather than j<i.
Anyway, you are correct to say that the blue term needs to be expanded further. Just try writing...
Let X_n=\mathbb{R} and X= \prod_{n \in \mathbb{N}} X_n = \mathbb{R}^{\mathbb{N}}. Suppose U_{n \in \mathbb{N}} is a countable neighborhood basis for the point \vec{0} = (0,0, \ldots ) \in X.
See if you can think of an open neighborhood of \vec{0} that does not contain any one of the U_n's...
The solution says IF aA+bB=0, THEN a=b=0. That is what it means for the vectors A and B to be linearly independent. Vectors in a basis must be linearly independent.
Absolutely! Check out https://www.physicsforums.com/showthread.php?t=105050. It's kind of a long thread about the proof of the orthogonality of Legendre polynomials. I attached a pdf there (mine is the fifth message) showing such a proof which relies solely on the fact that Legendre polynomials...
You don't actually get 2 \pi at \omega = 0, you get infinity. You get 2 \pi \delta ( \omega ) evaluated at \omega = 0, which is infinite or not well-defined at least.
Yes, that is correct. You should get 0 when \omega \neq 2 \pi k for any integer k.
Do you know how to calculate the...
\displaystyle \sum_{n=-\infty}^{\infty} e^{-jn \omega} = \sum_{n=0}^{\infty} e^{-jn \omega} + \sum_{n=-\infty}^{0} e^{-jn \omega} -1 = \sum_{n=0}^{\infty} e^{-jn \omega} + \sum_{n=0}^{\infty} e^{jn \omega} - 1.
When the sum converges, each of the last two sums is a geometric series that you...
You're using conservation of energy. Energy is not the only thing conserved in elastic collisions. In fact, the thing I'm thinking of is conserved even in inelastic collisions.
P(X>10) = P(X<0) in this specific case. It is due to the fact that the normal distribution is symmetric about the mean. Therefore, P(X> \mu + \epsilon) = P(X< \mu - \epsilon ) for arbitrary real \epsilon. In this case, P(X>10) = P(X> \mu + 5) = P(X < \mu - 5) = P(X < 0).
Correct.
P(X>10) = P(X<0) in this specific case. It is due to the fact that the normal distribution is symmetric about the mean. Therefore, P(X> \mu + \epsilon) = P(X< \mu - \epsilon ) for arbitrary real \epsilon. In this case, P(X>10) = P(X> \mu + 5) = P(X < \mu - 5) = P(X < 0).
1) You are correct. The released energy is electromagnetic radiation. The frequency of the radiation depends on the strength of the static magnetic field in the MRI as well as the duration of the pulsed magnetic field. In case you don't know what these two things are, here is a youtube video of...
The mistake happens in the third line when you convert to polar coordinates. You are no longer integrating over the same region. In the Cartesian case (second line) the integration region is a semi-infinite square with top-right corner at coordinates (0.5,0.5). In the polar case (third line) the...
Certain actions depend only on the topology of the underlying space on which the action is defined (as an integral on a smooth space or a sum on a lattice). This may happen even though the way the action is presented may look like it depends on more than topology. For example, the...
In Newtonian mechanics, F=ma will hold. Actually, that's tautological because Newtonian mechanics is defined by Newton's laws of motion and F=ma is Newton's second law.
Anyway, often the tricky part is to figure out what F is. In the E-frame, or any other non-inertial frame, you just have to...
All the possible fictitious forces pertain to this problem, which is partly what makes going to the E-frame difficult. There is a linear fictitious force parallel to the incline just because the disk as a whole is accelerating down the incline. There are also the three rotational fictitious...
Well, the unit vectors of a frame can always be made constant with respect to an observer in that frame so their constancy alone can't determine whether or not the frame is inertial. I won't bore you with inertial reference frames since that's probably on wikipedia or something and the problem...
So, it's all a matter of where you define your origin to be. As drawn, \vec{r} is with respect to the center of the disk. If the center of the disk is the origin, then \vec{r} = \frac{R}{2} \hat{e}_r.
If you say \vec{r} = R \hat{n}_2 + \frac{R}{2} \hat{e}_r, then you have defined your origin...
Did you forget to attach the picture? Maybe I'm just stupid and don't know how to find it. I would want to make sure I understand the question correctly. The integral in your "attempt" goes from -\phi to +\phi. That makes me think that the rod is only a semicircle when \phi = \pi / 2. I'm...
Looks good to me. So, when you combine the x and y components of the field using the Pythagorean theorem, you are effectively calculating the magnitude of the electric field. I don't think you need to do that, although you certainly can do. I would think it sufficient to say the x and y...