Search results for query: *

The del operator is not a vector that crosses with vectors, although it resembles the property of vectors. I am sorry I am not an expert who can explain clearly to you about it, but here is a derive of the curl in other...

The general formula for cylindrical coordinate is as follows: $$\vec{\nabla} \times \vec{V} = \frac{1}{r}\begin{vmatrix} \hat{r}\ & r\hat{\theta} & \hat{z} \\ \partial/\partial r & \partial/\partial \theta & \partial/\partial z \\ V_{r} & V_{\theta} & V_{z} \end{vmatrix} $$ Because ##V_r## and...

I am sorry. I thought the approximation only refers to ##\frac{d^2\psi}{d\xi^2} \approx \xi^2\psi##; with regards to the approximation ##\frac{d^2\psi}{d\xi^2} \approx \xi^2\psi## , the following ##\psi = Ae^{-\frac{\xi^2}{2}}+ Be^{\frac{\xi^2}{2}}## is exact and is not an approximation.

By the way, it would be very nice if you could explain what asymptotic form is in the second picture. I don't understand the transition from ##\psi = Ae^{\frac{-\xi^2}{2}}+Be^{\frac{\xi^2}{2}}## to ##\psi = h(\xi)e^{\frac{-\xi^2}{2}}##

The solution is same as I stated. Actually this is from a proof on my textbook on quantum mechanics, where the proof is about solving the simple harmonic oscillator. ##\psi(x)## refers to the wave function; ##\xi## refers to the constant ##\sqrt{\frac{m\omega}{\hbar}}x##

This is a very simple question: I would like to solve for ##\psi## in this equation $$\frac{d^{2}\psi}{d\xi^2} =\xi^2\psi$$ I so apply ##y=c_{1}e^{-kx}+c_{2}e^{kx}## and ##\psi## should be equal to ##\psi=c_{1}e^{-\xi^2}+c_{2}e^{\xi^2}##, because ##(D^2-\xi^2)\psi=0##. However the answer is...

At least the transform covered in the Boas book...

Yes, I have found the online version of your book. I will work through it later because I have just finished the Fourier series and transform. The problem of Hong Kong's educational system is not at the university level, but at the secondary school level. At secondary school, except for a few...

By the way, where did you guys learn these from during your ug level study? Did you guys actually take a math course on complex analysis?

$$(\frac {\partial \psi} {\partial t})^* = (\frac {\partial \psi_{1}} {\partial t} + i \frac {\partial \psi_2} {\partial t})^* $$ $$=(\frac {\partial \psi_{1}} {\partial t})^* + ( i \frac {\partial \psi_2} {\partial t})^*$$ $$=\frac {\partial \psi_{1}} {\partial t} - i\frac {\partial \psi_2}...

That's easy. $$ (e^{i\theta_1}\cdot e^{i\theta_2})^* = (e^{i(\theta_1 +\theta_2)})^*$$ $$= e^{-i(\theta_1 +\theta_2)}$$ $$=e^{-i\theta_1}\cdot e^{-i\theta_2}$$ $$=e^{i\theta_1*} \cdot e^{i\theta_2 *}$$ What really surprises me is the complex conjugate on the partial derivatives.

Thanks for your hints. That's much much better. Honestly I don't know what it means by learning complex number. I have taken a course(maths for physics student I, you can say) where I was taught the Euler's formula, the complex equation for natural logarithm, sine and cosine. But obviously I...

I have asked one of my frds from math department and obtained the following proof. Forgive me for uploading only two piece of paper because I would like to spare my effort in typing latex. . I do not understand why taking the complex conjugate seems so intuitive and obvious to you guys.

Good idea. You know my school's education sucks. It even doesn't teach Fourier Series and Fourier transform, which I am self-learning during this winter break. I am using the Mathematical methods for physical science of Boas, which my school uses for teaching ODEs.

How do you take the complex conjugate of a function? I know how to take a complex conjugate of a complex number ##z##. For example, for ##z= 1 + 2i##, its conjugate is ##z^* = 1-2i##. Forgive me but my complex number knowledge stops there.

So how does it apply to the wave function case? Maybe I shouldn't generalize that ##\frac {\partial \Psi^*} {\partial t} = - \frac {\partial \Psi} {\partial t}##, but that happens in my textbook...

It is a rather simple question: In my textbook it writes something like: $$\frac {\partial \Psi} {\partial t}= \frac{i\hbar}{2m}\frac {\partial^2 \Psi} {\partial x^2}- \frac{i}{\hbar}V\Psi$$ $$\frac {\partial \Psi^*} {\partial t}= -\frac{i\hbar}{2m}\frac {\partial^2 \Psi^*} {\partial...

By the way, the poet is David J. Griffiths :)

A very silly question: when performing the integration wrt to dx, we keep the variable in ##\Psi## constant, right? Sorry but I am not very familiar with such type of multivariable integration.

I have added back that ##x##. So here is what I think: $$\frac{d\langle x \rangle}{dt} = \frac{i \hbar}{2m}\int_{-\infty}^{+\infty} x\frac {\partial } {\partial x} (\Psi^{*}\frac{\partial \Psi} {\partial x}-\frac{\partial \Psi^{*}} {\partial x}\Psi)dx $$ $$= -\int_{-\infty}^{+\infty}...

Yes, I have seen a few times, for example in electrodynamics. But I chose to ignore the proofs at that time. Now I feel obliged to understand it...

So I am confused about a proof in which the formula for expected value of velocity, ##\frac{d\langle x \rangle}{dt} ##, is derived. Firstly, because the expected value of the position of wave function is $$\langle x \rangle =\int_{-\infty}^{+\infty} x|\Psi(x,t)|^2 dx$$Therefore...

Thanks. This is a nice interpretation that is not mentioned in the book.

Thanks. No joke but this thing has bothered me for the whole afternoon... :)

The answer in the textbook writes: $$ f(x) = \frac{1}{4} +\frac{1}{\pi}(\frac{\cos(x)}{1}-\frac{\cos(3x)}{3}+\frac{\cos(5x)}{5} \dots) + \frac{1}{\pi}(\frac{\sin(x)}{1}-\frac{2\sin(2x)}{2}+\frac{\sin(3x)}{3} + \frac{\sin(5x)}{5}\dots)$$ I am ok with the two trigonometric series in the answer...

The outer layer is magnetized and radiates H field, which alters the H field inside the sphere, am I right?

However I have come up with another way for d) and f): For d), $$ \nabla \cdot \vec H = - \nabla \cdot \vec M$$, $$ H^{\perp}_{l} - H^{\perp}_{s} = -(M^{\perp}_{l} - M^{\perp}_{s})$$ Given that: ##M^{\perp}_{l} = \vec M_{l} \cdot \hat r## For ##\vec M_{l}##, because ## \vec B_{l} = \mu_o (\vec...

The solution is finally released:

Alright. But your article is a bit complicated and may take some time...

Yes, ##\vec B = \mu \vec H## is what I meant. I use this equation because there is no linear material in this inner magnetized sphere. Hence, we can assume that ##\mu## in ##\vec B = \mu \vec H## can be assumed to be ##\mu_o##. Why is it incorrect for this region? By the way, I have come up...

But isn't it what I am doing for part f? $$ \nabla \cdot \vec H = -\nabla \cdot \vec M$$ $$ \oint \vec H \cdot d\vec a = -\oint \vec M \cdot d\vec a$$

The question is as follows: Solutions given only contain part a) to c), which is as follows: So I now try to attemp d), e) and f). d) The magnetic field of a uniformly magnetized sphere is: $$ \vec B =\frac{2}{3}\mu_{o} \vec M = \mu_{o}\vec H$$ $$\frac{2}{3}\vec M = \vec H$$ The perpendicular...

To be honest my school doesn't allow laziness; if you are lazy, your grade will be heavily dragged down and you will end up with 3rd honours. In that case it would be better to quit the university. Competition here is huge, as in any other megacities. So people work very hard and you end up...

The direction of the magnetic potential, ##\vec A##, must be in the direction of the current, which is in ##\hat z## direction in cylindrical coordinates. It is obvious that the potential only varies with ##s##. Therefore, $$\vec A = A(s) \hat z$$ Therefore, $$\nabla \times \vec A = \vec B$$...

Even if I supply a work into the heat pump so that the the amount of heat is pumped back into the hot resevoir, fuel is burnt eventually.

You must put in some work to convert it back to higher temperature. I think in this sense, you are burning more and more fuel for this conversion. So high grade energy must go to low grade energy.

So my professor says that the implication of 2nd law of thermodynamics is that high quality energy will be degraded into low quality energy. By high quality energy he means something like coal or fuel. By low quality energy he means something like heat entering the cold resevoir in a heat...

You are totally correct and smart! Thanks. Here is the solution I have worked out: The potetial for the interior of the sphere in spherical coordinates is: $$V_{in} = \sum_{l=0}^\infty A_l r^l P_l(cos\theta)$$ By the boundary condition that: $$V_{in} = V_{out}$$ and $$\frac{\partial...

The Laplace's equation ##\nabla ^{2} V =0##, has the following solution in spherical coordinates: $$V = \sum_{l=0}^{\infty} (A_l r^l + \frac{B_l}{r^{l+1})P_{l}(cos\theta) $$. The potential boundary condition can be calculated by letting ##V_{in} = V_{out}##. But I don't know as well how you...

This is an example of Griffith's book on bound charge, and the following is the solution to this example. We choose the z-axis to conincide with the direction of polarization. By $$\sigma_b \equiv \mathbf P \cdot \hat {\mathbf n} $$ and $$\rho_b \equiv - \nabla \cdot \mathbf P$$ we can...

I suppose we can treat the electron cloud as a point charge, just like what we do for centre of mass? If that's not the case, why would the author draw two points inside the sphere?

The question is like this: The solution is like this: However, according to the equation for ##E_{dip}## , what I think is that it should be: $$E=\frac {1}{4 \pi \epsilon_o} \frac {qd}{d^3} \hat {\mathbf z} $$, where I take the centre of the sphere in figure 2 as the centre of the...

My poor English, that's what I want to say.

I think ##\vec r## refers to the direction of the radial distance of the potential at a general point from the centre of the coordinate system, whereas ##\vec r^{'}## is the direction from the centre of the coordinate system to the infinitesmal charge; the angle between them is ##\alpha##, which...

Then what is ##r##?

The diagram of the problem should look something like this: ,which is just the normal spherical coordinate. To calculate the potential far away, we use the multipole expansion. ##I_o## in the expansion is ok, because ##(r^{'})^{0} = 1##. However, I am wondering how I should calculate...

Nice explanation!

This is hard but I will make a guess. I think for even numbers of ##l## the Legendre polynomials are tossed. The ##V_{out}## is given by: $$\sum_{l=0}^{\infty}\frac{B_l}{r^{l+1}}P_l(cos\theta)$$ For the first few ##l##s, $$\frac{B_1}{r^2} + \frac{B_2}{r^3}cos\theta +...

Thanks. Your explanation is super clear.