Recent content by jdwood983

The 1/k_BT comes from the normalization constant A that you have in your first post. If you are requiring that to be the answer then I suppose that z is a speed? If that's the case, then f(E)=A\exp\left[-\frac{E}{k_BT}\right]=\sqrt{\frac{b}{\pi k_BT}}\exp\left[-\frac{bz^2}{k_BT}\right]...

I think your problem is two-fold: You're not using the correct formula for the average value You're putting in E=bz^2 when you don't really need to Both points are explained a little more below \langle E\rangle=\frac{1}{k_BT}\int_0^\infty E\exp\left[-\frac{E}{k_BT}\right]dE then using...

Try integration by parts: \int u\,dv=uv-\int v\,du where u=E and v=-k_BT\exp\left[-E/k_BT\right]

You are correct about the boundary conditions, but there still can be a contribution from the radial component (you'll see why/how soon). Clearly there is no z dependence so from separation of variables we can write V(r,\theta)=R(r)\Theta(\theta) (where 0\leq\theta\leq\beta) so that we get...

Looks to me like cylindrical Laplacian is necessary for this problem. What are your boundary conditions??

This seems a little longer than it should be: \mathrm{Tr}(XY) = \sum_{a'}\langle a'|XY|a'\rangle=\sum_{a',a''}\langle a'|X|a''\rangle\langle a''|Y|a'\rangle = \sum_{a',a''}\langle a''|Y|a'\rangle\langle a'|X|a''\rangle=\sum_{a''}\langle a''|YX|a''\rangle = \mathrm{Tr}(YX) I think your biggest...

If you have expanded Equation 15.5.6 using the substitution given, try separating the resulting equation into the real and imaginary components. It looks to me like Equation I in 15.5.7 is the imaginary part and Equation II of 15.5.7 is the real part after substituting for \sin[x].

Did you try using the relation \sin[x]=\frac{\exp[ix]-\exp[-ix]}{2i} in Equation 15.5.6 and expanding to see what you get? It looks like you should get the result they obtain if you work at it using that relation.

Because you have U(t) defined as U(t)=u\left[x(t),t\right] in the problem. If U(t) is only a function of t, then \left(\mathbf{u}\cdot\nabla\right)u=0 and dU/dt=\partial U/\partial t, which is not what you want for this problem. Glad I can help.

This isn't the total derivative, this is: \frac{dU}{dt}=\frac{\partial u}{\partial t}+\frac{dx}{dt}\frac{\partial u}{\partial x}+\frac{dy}{dt}\frac{\partial u}{\partial y}+\frac{dz}{dt}\frac{\partial u}{\partial z} Note that the last three terms look like: \frac{dx}{dt}\frac{\partial...

It seems to me that a rotating fluid creates a force and thus pressure. I imagine that to keep pressure either constant or zero, the velocity should be zero as well. I could be wrong though.

Woohoo, I was right! It just wasn't the power I was thinking of (you already pointed out the 3/2 power should be as is due to the 5/2 not working). I'm glad you were able to figure the solution out. I imagine that there's plenty more typos in that text, and am very surprised that the text made...

I started thinking that it might be for this function only after I typed my response. I would think, then, that there must be a typo in the problem. What the error is, I can't be sure; perhaps the power should be something else?

What do you know about the total derivative?

I'm not sure that it is a valid equality in the first place. Imagine our function is a little bit more simple: z(x,y)=3xy-y^2+x^2 Then z_{xx}=2, z_{yy}=-2 \rightarrow z_{xx}\cdot z_{yy}=-4 compared to z_{yx}=3\rightarrow z_{yx}^2=9 so these two are not equal. Unless, that is, this...