Recent content by plasmoid

Makes sense. Now I have |B|. To get the components of \vec{B}, I can use 0 = A_{x} B_{x} + A_{y}B_{y}+ A_{z}B_{z} and 0 = C_{x} B_{x} + C_{y}B_{y}+ C_{z}B_{z}, and B^{2} = B_{x}^{2} + B_{y}^{2} + B_{z}^{2} . Does that sound right?

I have a vector equation: \vec{A} \times \vec{B} = \vec{C} . \vec{A} and \vec{C} are known, and \vec{B} must be determined. However, upon trying to use Cramer's rule to solve the system of three equations, I find that the determinant we need is zero. I know now that I need to choose a...

Yes, but using the Levi-Civita symbol will essentially mean writing out the components, and I was trying to prove it without doing that...

How is del (u . u) = |u|^2 (del) ?

Yes I have, but that doesn't give me the 1/2 in the first term ...

Can someone help me prove the identity \ u \times (\nabla \times u) = \nabla(u^2 /2) - (u.\nabla)u without having to write it out in components?

If \int_{-\infty}^{\infty} f(k) e^{i\mathbf{k}.\mathbf{r}} d\mathbf{k} =0 then is \ f(k)=0 ? Is it correct to say that this is an expansion in an orthonormal basis, \ e^{ik.r} , and so linear independence demands that f(k) be zero for all k?

After you click 'preview', refresh the page - it should now show you what you typed. This is a known issue on these forums.

Thanks guys ... I had started on the "divide one equation by the other" path, but for some reason did not carry it to it's conclusion. @HallsofIvy, the first equation actually does depend on l; I guess you mistook the l in the numerator for 1. Thanks anyway :)

I have the equations \frac{l}{u^{2}} \frac{du}{dx}=constant and \frac{1}{u} \frac{dl}{dx}=constant. By "eyeball", I can say the solution is l \propto x^{n} \ and \ u \propto x^{n-1}. I can't see how I could arrive at these solutions 'properly', if you know what I mean

Here's my definition of temporal average - <f(t,x)> = \stackrel{lim}{T\rightarrow\infty} \frac{1}{T} \int_{t_{0}}^{t_{0}+T} f(t,x) dt And it's all making even less sense to me now. With this definition, <f(t,x)> is...

That's it actually. But all the math texts I saw move differentiation past the integral sign only when the two involve independent variables. Is there a rigorous justification for doing it when both the differentiation and integration involve the same variable? Edit - I just can't get the latex...

I've been studying Turbulence, and there's a lot of averaging of differential equations involved. The books I've seen remark offhandedly that differentiation and averaging commute for eg. < \frac{df}{dt} > = \frac{d<f>}{dt} Here < > is temporal averaging. If...