Recent content by qbert

looks like you're having problems with the implicit sums. when in doubt write it out. T_{ij}=\mu S_{ij} + \lambda \delta_{ij} \sum_k \delta S_{kk} You then hit by a \delta_{ij} and sum over both i and j \sum_{ij} \delta_{ij} T_{ij}=\mu \sum_{ij} \delta_{ij} S_{ij} + \lambda...

It's just the chain rule: \frac{d}{dt} f(y(t), z(t)) = \frac{\partial f}{\partial y} \frac{dy}{dt}+\frac{\partial f}{\partial z} \frac{dz}{dt} Where your function has the form P = P(y(t), t) <-- has two t dependent terms To make this clearer let y(t) just as is and put z(t)=t, So that...

I don't know about two conditions. but, transversality (div A=0) says not that the sum of all of the Fourier components is 0, but instead that for each k {\bf k} \cdot {\bf A}_{\bf k}(t) = 0. just brainstorming: If your original A was real you also have a reality condition that...

what? your definitions make no sense. ``dx = dy \wedge dz'' is literally nonsensical.

Both are simple chain rule consequences, i'll illustrate with the divergence since it's quicker \nabla \cdot {\bf F}(r) = \sum_i \frac{\partial F_i(r)}{\partial x^i} = \sum_i \frac{\partial F_i(r)}{\partial r} \frac{\partial r}{\partial x^i} now \frac{\partial r}{\partial...

short answer: under the time ordering you can collect all of the U(τ) terms together and they cancel.

There is a pretty standard representation of the delta-function here \lim_{\eta \rightarrow 0} \frac{\eta}{x^2 + \eta^2} = \pi \delta(x). See, for example, the wikipedia article on the delta function.

There is no \widehat{\theta} direction, it is wiped out when you take the inner product in {\bf E} \cdot d{\bf A}, and that is my point. You have to do the inner product when you're doing flux integrals. It's only the piece of electric field which pierces the surface element which contributes...

that's what I'm worried about. the right answer for the wrong reason. do you see what's different between what I'm saying and what you are?

I'm not sure what you are doing. {\bf E}=(rsin(\theta))^2(sin(\theta)\hat{r}+cos(\theta)\hat{\theta}) and d{\bf A} = r^2 \sin \theta d \theta d\phi \widehat{r}, tell you {\bf E} \cdot d{\bf A} = r^4 sin^4(\theta) d\theta d\phi .

I think the approach should be something like: S = S(U,V) => dS = dS/dU dU + dS/dV dV now use what you know about "monatomic ideal gasses" and the fact dU =-P dV to show the terms in dS cancel.

Now supposing you got E right you need d{\bf A} = r^2 \sin \theta d \theta d\phi \widehat{r}, and that product is the inner product.

essentially it is a restatement of: xdx = 1/2 d(x^2)

You use the fact that the wavefunction is continuous at each connecting point, (-a/2), a/2, to write C and D in terms of A. Then you integrate the piecewise-function squared over the whole interval to tell you what A should be. [also this might be a solution to the finite square well but...

Second is right technique, although you've mathed wrongly. for three states: 0, 1, 2 U = avg energy = E(0) P(0) + E(1) P(1) + E(2) P(2) For E(0)=0, E(1)=E(2) =ε, and P(energy) = exp(-energy/kT)/Z P(0) = 1/Z, P(1)=P(2)=exp(-ε/kT)/Z