Recent content by Kyuubi

Part a is fairly straightforward and can be found in many text books. I just want to ask if part b is a well posed question for an undergrad solid state course. Over this semester, my instructor has asked many badly posed questions in my opinion. Questions that outright make no sense or are...

It's been answered for me. Indeed the number of polarizations goes down. I suppose then that EM modes retain their 2 polarizations regardless of dimension (?)

When going from 3 to 2 dimensions, I am unsure about how the number of polarizations will be affected. I know the following though: The 1/8 factor becomes a 1/4 since we are now integrating over the positive quadrant in 2d rather than the positive octant in 3d. The ##4\pi n^2## becomes a ##2\pi...

I want to verify some inspection I'm making at this problem. Because of the infinite barrier at ##x=0##, we expect the wave function to take the value 0 there to preserve continuity. As such, we can make the conclusion that the wave function will just be a sine term in the [0,a] region. But...

Now if I'm given a ##\phi(k)##, and I'm asked to find ##\langle p \rangle##, ##\langle p^2 \rangle##, etc. Am I justified to say that ##\langle p \rangle = \hbar \langle k \rangle## and that ##\langle p^2 \rangle = \hbar^2 \langle k^2 \rangle## ?

I tried with powers as well, but I did it in the form n^n. I just felt like taking powers of 2 would be a bit arbitrary. I want the entire expression to only depend on n. Anyways looking for numbers that satisfy n = digits_in(n^n), we get 1 (as expected), but we also simply get 8 and 9. 8^8 =...

This may have already been found by many people but I discovered the pattern on my own out of curiosity with some coding. There are only 4 natural numbers whose factorial contains the same number of digits as the number itself. That is to say n = digits_in(n!). The trivial case is obviously...

Thank you so much! This is exactly what I needed, and I got the right answer. I see now that I shouldn't plug in the momentum into the Energy conservation, but rather I should see the resultant change in energy from the applied momentum and use that instead.

Sorry for the double reply, but I didn't address your other point. I'm not saying that Energy and momentum are conserved before and after the impulse is delivered. I'm saying if we think of the system right *after* the impulse is delivered, then conservation holds. Say if the impulse is...

Ok tbf I had two methods to reach the orbital speed. This was one of them, but like you are saying it feels weird. My other reason for my velocity term is that the potential energy function corresponds to the centripetal force ##F(r) = -kr## which can be equated to ##\frac{mv^2}{r}##. Plugging...

I've already solved the orbital speed by equating the kinetic and potential energy in the circle orbit case. $$\frac{1}{2}mv^2 = \frac{1}{2}ka^2.$$And so $$v^2 = \frac{k}{m}a^2$$Now when the impulse is added, the particle will obviously change course. If we set our reference point in time just...

Surely that's not too much of a bad thing no?:bugeye:

Helloz. I want to become an astrophysicist and was wondering if there topics in math that are HEAVILY used in astrophysics so that I can start focusing on them from early on. I understand that A LOT of different aspects of math are used in almost every field of physics, but I want to know which...

I've found the inductance and capacitance per unit length in a long coaxial cable. I even clearly see that if I multiply the two, I can get the speed of light. How do I begin to find the current wave and its speed though?

Sorry I'm new to this. Is there a way to delete a thread or close it? I think I've solved my problem.