Search Results

Once you know the phase velocity $\omega$, you know that along the $x$ axis the red dot is described by $x(t) = \omega t$. Then you just apply the chain rule. The value of $y(x,t)$ at the red dot is a …

How about the study of minimal surfaces (physical applications in soap films etc.)? In fact one might argue the Lagrangian formulation of minimal surfaces (the problem of Plateau) is one of the oldest …

For shapes of liquid drops, it is probably not driven by Ricci flow. Fluid interfaces with surface tension is better modeled by mean curvature, going back to Young and Laplace; and there is a lot of …

Looking at Figure 3 and Step B of the proof of the theorem, it looks like the $C_{\tau_k}$ should be of the form $\partial \tilde{J}^-_{\theta_k}(Q_k)$. I am also pretty sure that he chose the symbol …

Maybe derive the Polyakov formula? Like KConrad says, whether it can be understood on an undergraduate level depends a lot on your presentation and the level of the undergraduates. But the basic idea …

To make my comment more concrete: if you define $$\mathcal{G}_0 = \phi $$ and $$ \mathcal{G}_{n+1} = \Box \mathcal{G}_n - \frac{1}{n+1} \partial (\partial\cdot \mathcal{G}_n) + \frac{1}{(n+1)(2n+1)} …

Q1 The topology on $\mathbb{R}^4$ is the usual one. This is the general case for Lorentzian geometry: the topology is the one defined by the charts in your atlas. Q2 Given a fixed Lorentz transformati …

Somewhat related to the ergodic hypothesis mentioned in another answer is the assumption that generic non-linearities leads to thermalization and equipartition of energy. To be more precise, start wit …

The one thing which I think you are missing is that Unlike the Riemann curvature, the spin coefficients are not tensors (they are not co- or contra-variant). In particular, the spin coefficient …

Spin coefficients are indeed invariant under coordinate changes. What changing coordinates does is that it changes the coordinate representation of the tetrad vectors (in terms of the coordinate vect …

Multiply your equation by $u$ and integrate over $\mathbb{R}^3$ in azimuthal coordinates (so the volume form is $r~\mathrm{d} r~\mathrm{d}z ~\mathrm{d}\theta$ you get $$ 0 = 2\pi \iint_{r,z} (u_{rr} …

You should re-read the immediately previous section on Petrov Type II first. There he did the computation is a tiny bit more detail. Then you will realize that since we are dealing with a fourth ord …

Yes, there is one such example: $u \equiv 0$. The answer above is not facetious! That $u$ is in fact the only example (modulo measure zero modifications). By Titchmarsh's theorem, if $u\in L^2(\ …

I am somewhat doubtful that the question as posed as any sort of reasonable answer. (Also, I don't really see how the Lorentzian metric even enter into the problem.) (a) ANY hyperbolic PDE in (1+3) d …

Hi, I am adding another answer because this suggests a rather different approach then what I have outlined before, and this is targeted at the fact you are willing to grant smooth with compact support …