I Show Maxwell's Eqns. on a Cauchy Surface (Wald Ch. 10 Pr.2)

ergospherical
Science Advisor
Homework Helper
Education Advisor
Insights Author
Messages
1,097
Reaction score
1,384
This problem is Wald Ch. 10 Pr. 2.; it asks us to show that ##D_a E^a = 4\pi \rho## and ##D_a B^a = 0## on a spacelike Cauchy surface ##\Sigma## (with normal vector ##n^a##) of a globally hyperbolic spacetime ##(M, g_{ab})##. Using the expression ##E_a = F_{ab} n^b## for the electric field gives ##D_a E^a = {h^a}_b {h_a}^c \nabla_c E^b##. I replace ##{h^a}_b {h_a}^c = (\delta^a_b + n^a n_b)(\delta_a^c + n_a n^c) = \delta_b^c+ n_b n^c = {h_b}^c##, thereby obtaining\begin{align*}
D_a E^a &= {h_b}^c \nabla_c (F^{bd} n_d) = \nabla_b (F^{bd} n_d) + n_b n^c \nabla_c (F^{bd} n_d) \\
&= n_d \nabla_b F^{bd}+ F^{bd} \nabla_b n_d + n^c n_b n_d \nabla_c F^{bd} + n_b n^c F^{bd} \nabla_c n_d
\end{align*}The third term vanishes because ##n_b n_d \nabla_c F^{bd} = n_{(b} n_{d)} \nabla_c F^{[bd]} = 0##. Also, since ##n^a## is orthogonal to ##\Sigma##, the condition ##n_{[b} \nabla_c n_{d]}## holds i.e. ##n_b \nabla_c n_d## is totally antisymmetric, and the sum of the second and fourth terms is ##F^{bd} \nabla_b n_d + n_b n^c F^{bd} \nabla_c n_d = F^{bd} \nabla_b n_d - n_c n^c F^{bd} \nabla_b n_d = 2F^{bd} \nabla_b n_d##. Using Maxwell's equation ##\nabla^a F_{ab} = -4\pi j_b## on the first term gives\begin{align*}
D_a E^a &= -4\pi n_d j^d + 2F^{bd} \nabla_b n_d \\
&= 4\pi \rho + 2F^{bd} \nabla_b n_d
\end{align*}Does ##2F^{bd} \nabla_b n_d = 0##, i.e. is ##\nabla_b n_d## symmetric? I can't see why this should be so. (I thought about ##0 = \nabla_b(-1) = \nabla_b(n_a n^a) = n^a \nabla_b n_a##, but this doesn't seem to help).
 
Physics news on Phys.org
ergospherical said:
Does ##2F^{bd} \nabla_b n_d = 0##, i.e. is ##\nabla_b n_d## symmetric? I can't see why this should be so. (I thought about ##0 = \nabla_b(-1) = \nabla_b(n_a n^a) = n^a \nabla_b n_a##, but this doesn't seem to help).
This is up to a sign the second fundamental form, which is symmetric. With mathematical notations you can do the following. Let ##n## be the normal, and ##X,Y## tangent to the surface. Then differentiating ##g(n,Y)=0## you get ##g(\nabla _X n,Y)+g(n,\nabla _XY)=0##. So your expression is the negative of ##g(n,\nabla _XY)##, which is symmetric because ##g(n,\nabla _XY)=g(n,\nabla _YX)+g(n,[X,Y])##. The last term is zero because the comutator is also tangent to the surface.
 
  • Like
  • Love
Likes George Jones, vanhees71, ergospherical and 1 other person
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Does the speed of light change in a gravitational field depending on whether the direction of travel is parallel to the field, or perpendicular to the field? And is it the same in both directions at each orientation? This question could be answered experimentally to some degree of accuracy. Experiment design: Place two identical clocks A and B on the circumference of a wheel at opposite ends of the diameter of length L. The wheel is positioned upright, i.e., perpendicular to the ground...
According to the General Theory of Relativity, time does not pass on a black hole, which means that processes they don't work either. As the object becomes heavier, the speed of matter falling on it for an observer on Earth will first increase, and then slow down, due to the effect of time dilation. And then it will stop altogether. As a result, we will not get a black hole, since the critical mass will not be reached. Although the object will continue to attract matter, it will not be a...
Back
Top