Starting with finding the probability of getting one of the states will make finding the other trivial, as the sum of their probabilities would be 1.
Some confusion came because I never represented the states ##|\pm \textbf{z}\rangle## as a superposition of other states, but I guess you would...
It seems a semester of vector calculus had me fixated on the definition of a conservative field. Scalars are in place for gravitational and electric fields to distinct them from each other (or some better wording than what I said).
I seem to have forgotten that g was associated with a vector field, thank you. So it is written as ##\textbf{F}=m\textbf{g}= -m\nabla u##.
I have never seen this expression. So are you saying ##u= \lim_{m \to 0} \frac{U}{m}##(U as potential energy)? Could you point me to where I can find this...
If we have a conservative vector field, then we can describe it as ##\textbf{F}=\nabla\phi## where ##\phi## is some potential.
This here is the derivation of Newtons law of gravity:
Where ##\nabla u## is the gravitational potential. If we were to ignore it as a gravitational field, why is it...
I should have clarified. I know ##dS= r^2sin\theta d\theta d\phi##. I want to be able to "interpret" it as ##d\textbf S=(dr e_r,rd\theta e_\theta,r\sin\theta d\phi e_\phi)##, or some equivalent form (I know that is not how you interpret it as it is not a Cartesian vector), so that I can evaluate...
That was a typo.
Evaluating the right hand side: ##\int_0^aB_n\sqrt{\frac{2}{a}}^2sin^2(\frac{n\pi x}{a})dx = B_n\frac{(2\pi n - sin(2\pi n))}{2\pi n}##. If calculated right, will let me solve for ##B_n##
I guess I could get the coefficients by using ##\int_0^a \psi(x,0)\psi_ndx=0## because the functions would be orthogonal.
If this is true, still how would I relate the coefficients to the energy and energy probability?
It was said that :
##\psi(x, 0) = \sum_{n = 1}^\infty a_n \psi_n(x)##
So expanding the first three n terms: ##Ax(a−x)= B\psi_1+C\psi_2+D\psi_3##. Where I am assuming ##\psi_n=\sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a})##.
This is the relation that I do not know. How is ##E_n=\frac{n^2...
I'm not sure how to find the probability of being in an energy state. Are we using the real valued eigenfunction for n=1 then finding its probability distribution?
"Calculate the probability that the measurement of the energy would yield the value ##E_n##".
If it helps,the energy level question I was referring to is "in particular, calculate the numerical values of the probability to measure ##E_1,E_2## or ##E_3##"
I worded that poorly. This is getting into another part of the question. The question wants me to find the energy expectation for ##E_n##. I know that ##\langle E\rangle=\int_{\infty}^\infty \psi*\hat H\psi dx##. So would ##\langle E_n\rangle=\int_{\infty}^\infty \psi*\hat H\psi dx## where:
I...
So for the wave function to be an eigenfunction, does it have to satisfy the Schrodinger equation?
I see. For a later part of the question, it wants me to measure its first three energy levels. So would I now assume the general solution (sinusoidal) of the Schrodinger equation?
Let ##S_t## be a uniformly expanding hemisphere described by ##x^2+y^2+z^2=(vt)^2, (z\ge0)##
I assume by verify they just want me to calculate this for the surface. I guess that ##\textbf{v}=(x/t,y/t,z/t)## because ##v=\frac{\sqrt{x^2+y^2+z^2}}{t}##. The three terms in the parentheses evaluate...
Some questions:
Why is this even a valid wave function? I thought that a wave function had to approach zero as x goes to +/- infinity in all of space. Unless all of space just means the bounds of the square well.
Since we have no complex components. I am guessing that the ##\psi *=\psi##.
If...
I'm not entirely sure, but I guess it's because we did some ODE's in Calc 2 and they see fit that we start ODE's as well as Calc 3.
No ours starts off at level 300 for some reason. My uni is weird in that there are some courses that are listed as a level x course, but are usually taken a year...
How many hours a week would you say you spent studying? I have a pretty good work ethic during the week(though I do tend to get a bit lazy during the weekends).
So this is what my course load is going to look like for my fall of second year as a physics major.
math 251= Calculus 3
phys 211= Intermediate mechanics
phys 201= just some weekly seminar that "expose students majoring in any Physics program to opportunities available with a physics degree "...
Yes sorry I changed it, ##d-w## is from another problem stuck in my head.
So I carried out ##E=\frac{1}{4\pi\epsilon_0}\frac{Q}{L}\int_0^L \frac{(w-x)}{(R^2+(w-x)^2)^{3/2}}dx##
Giving me...
My bad, ##z## represents an arbitrary location, ##Q## is the charge of the ring and ##R## is the radius of the ring.
Yes that's what bothers me, my answer doesn't depend on a distance.
I think my mistake lies some where from the transition from...
I understand your derivation and I can see that it shows oscillation. I'm more wanting to know what's going on microscopically.
##\frac{d^2}{dt^2}x=-kx##. But why? Why does this oscillation occur? Why is there a restoring force? (those are rhetorical questions) All that can be explained to an...
I uploaded a diagram of the problem.
I treated this as many thin rings and integrated it over the length. I placed my origin as in the same place as the uploaded picture.
Finding the electric field due to one small ring:
##\vec r =\langle w-x, 0, 0 \rangle## where ##x## is the distance of the...
And you can fundamentally explain that can't you? So the springs are made of some material who's molecules like to be positioned the way they are, and when displaced will exert a force the opposite way to go back to their original position, probably due to how they bind geometrically (I'm not...
I know that's not possible obviously, but you took that out of context. Wouldn't it be reasonable to say that this inductor creates an electric field to oppose the decrease in current, and in doing so the electric field never decreases. (I uploaded an image of my reasoning).
So why does the voltage of an RC circuit with no battery drop to zero over time?
I'm talking about an LC circuit with no battery. To my understanding, a fully charged capacitor discharges and over time the fringing fields of the capacitor decrease because the original charge imbalance on the...
What I'm thinking about is a capacitor and a resistor for example. With no battery. Initially at ##t=0## the capacitor is fully charged and has some potential across it's plates. Over time, the current falls of as ##I=\frac{emf}{R}e^{\frac{-t}{RC}}## therefore the potential of the capacitor...
I mean of how a capacitor is initially fully charged, and over time the electric field of the capacitor weakens because the charge buildups cancel each other out. So then over time the potential and the current goes to zero.
During a period, there is current going into the capacitor but that doesn't mean an increase in potential because it matters what the net charge over the capacitor plates are. So that doesn't make sense to me.
Yes I mean I understand but I don't know where it came from. What I want to know is...
Why would the capacitor the capacitor gain voltage if it was initially charged with no battery?
What I was trying to get at was what drives the current in the opposite direction? There has to be some electric or magnetic field that causes it to reverse. Near the end of one period, the current...
I understand Faraday's law and about induced electric fields created by a changing magnetic fields, etc.
But what causes the current to oscillate in an LC circuit, with no battery? If you picture that there is current going into an inductor, and that current is decreasing over time, then you...
Wow that actually makes so much sense. My biggest issue is once I got the second derivative, I got this large function which seemed annoying to evaluate with x=5 (given that this question could be on an exam). Thank you.
I got the first part of it down, $$L=\int_1^5 \sqrt{1+(\frac{1}{x^2})}dx$$
I just want to know if it's right to make your ##f(x)=\sqrt{1+\frac{1}{x^2}}## then compute it's second derivative and find it's max value, for the trapezoidal error formula.
It's the same thing I think.
##B_{wire}=\frac{\mu_{0}I}{2\pi r}=\frac{\mu_0}{4\pi}\frac{2I}{r}##
I just leave the 2 because of ##\frac{\mu_0}{4\pi}## being a defined constant
It comes from the approximation of the length of the wire being much larger than a distance ##z##. So I can cancel the ##z^2## term in the denominator
##B_{wire}=\frac{\mu_0}{4\pi}\frac{LI}{z\sqrt{z^2+(\frac{L}{2})^2}}\longrightarrow B_{wire}\approx\frac{\mu_0}{4\pi}\frac{2I}{z}## (if the...
So I have
##\vec F_{near}=I_{coil}L(\frac{\mu_0}{4\pi}\frac{2I_{wire}}{z}) \langle 0,0,1 \rangle## (attraction)
##\vec F_{far}=I_{coil}L(\frac{\mu_0}{4\pi}\frac{2I_{wire}}{z+w}) \langle 0,0,-1 \rangle## (repulsion)
where ##I_{wire}=a-bt##
and ##I_{coil}=...
Yes that's my bad, I do know of this, I was just leaving the general expression for now where ##z## is an arbitrary distance, I probably should have used another variable.
So now I have ##I_{wire}=-\frac{d\phi}{dt}=-\frac{d}{dt}(BA)=-\frac{dB}{dt}Lw##
Plugging in, I get...
Yes that makes sense, that's my bad.
If that's not true then wouldn't the net magnetic force on the coil be 0? (I'm not sure what you mean exactly, but if you're saying that the magnetic field is weaker at z+w, then that is something that I considered.
I am actually stuck again,
So you know ##F_m=I\Delta\vec l\times \vec B= I_{coil}L(\frac{\mu_0}{4\pi}\frac{2I_{wire}}{z})## and ##I_{wire}= a-bt##
This force is attractive to the closest length of the coil, and repulsive to the further length of the wire. The magnetic field has no effect on the...
I don't understand why a force would be acting on this rectangular coil at all. The magnetic field of the wire would only induce a force on the coil, if the coil had a current flowing through it. At first I would think that the electric field from the varying magnetic field would induce such...
I will first calculate the magnetic flux of the coil in motion.
$$\frac {d\phi}{dt} = -\frac {dB_{loop}}{dt}A = -\frac{d}{dt}(\frac{\mu_o}{4\pi}\frac {2\pi NR^2I}{(R^2+z^2)^{\frac{3}{2}}})A$$differentiating in terms of ##z##, we get $$\frac {d\phi}{dt} =(\frac{\mu_o}{4\pi}\frac {6\pi^2...