MTW 21.26: deriving junction condition a+ + a− = 0 for dust shell

[JimWhoKnew]

TL;DR

see below

I need help with exercise 21.26 in MTW. The question goes like this:

For a thin shell of dust surrounded by vacuum ( $[T^{in}]=0$ , $\mathbf{t}=0$ ), derive the following equations$$\frac{d\sigma}{d\tau}=-\sigma^b{}_{|b}\;\; ,\tag{21.175a}$$$$\mathbf{a}^+ +\mathbf{a}^- =0 \;\; ,\tag{21.175b}$$$$\mathbf{a}^+ -\mathbf{a}^- =4\pi\sigma\mathbf{n} \;\; ,\tag{21.175c}$$$$\mathbf{\gamma}=8\pi\sigma\left(\mathbf{u}\otimes\mathbf{u}+\frac12 \mathbf{g}\right) \;\; .\tag{21.175d}$$Here $\mathbf{a}^+$ and $\mathbf{a}^-$ are the 4-accelerations as measured by accelerometers that are fastened onto the outer and inner sides of the shell, and $\mathbf{g}$ is the 3-metric of the shell.

It's not hard to derive equations a, c & d. The second ( $\mathbf{a}^+ +\mathbf{a}^- =0$ ) is where I get stuck. Intuitively, I understand it as proportional to the "total non-gravitational force" exerted on the shell element between the accelerometers, and therefore it should vanish. But I fail to spot how it can be derived from the equations in section 21.13 and exercise 21.25.

Please help.

 

[PeterDonis]

The second ( $\mathbf{a}^+ +\mathbf{a}^- =0$ ) is where I get stuck.

What do you get if you add the second and third equations?

 

[JimWhoKnew]

What do you get if you add the second and third equations?

$\mathbf{a}^\pm=\pm 2\pi\sigma\mathbf{n}$ , which looks like the well known result in electrostatics. If the accelerometers were not fastened to the shell, they would have followed a geodesic and measure 0. So the shell exerts a force on them which is symmetric on both its faces. But can we show it from the equations?

 

[PeterDonis]

$\mathbf{a}^\pm=\pm 2\pi\sigma\mathbf{n}$ , which looks like the well known result in electrostatics. If the accelerometers were not fastened to the shell, they would have followed a geodesic and measure 0. So the shell exerts a force on them which is symmetric on both its faces.

Yes, that's the physical interpretation of the result.

Can we show it from the equations?

How did you derive equation (c)?

 

[JimWhoKnew]

How did you derive equation (c)?

The book uses Gaussian Normal Coordinates (GNC) in this section, so I do too.
First I derived equation (d) out of equation (21.168b). The equation numbering follows that of the book.
Then I used the condition that the shell is surrounded by vacuum (see OP) in equation (21.173) to get $a_j=0$ (Greek letters are used for 4D, roman for 3D). In GNC this means $a^j=0$ also, so $~\hat{\mathbf{a}}=\hat{\mathbf{n}}$ . From the identity$$\left(u^\nu u_\mu n^\mu\right)_{;\nu}=0$$I get$$\mathbf{a}\cdot\mathbf{n}=\mathbf{K}\left(\mathbf{u},\mathbf{u}\right)\;\; .\tag{1}$$Using this in equation (d) yields (c).
In GNC we have$$K_{ij}=-\frac12 g_{ij,n} \;\; ,\tag{2}$$so following (1), to get equation (b) I have to argue that $\left| u^i u^j g_{ij,n}\right|$ are the same on both faces of the shell in vacuum. That's where I get stuck.

Alternatively, following post #3, If I could show $~-\frac12 u^i u^j g_{ij,n}=\pm2\pi\sigma~$ at the faces, then the derivation of (b) will be completed. But I don't know how to show that either.

 

[JimWhoKnew]

The solution can be found in "Problem Book in Relativity and Gravitation" (Lightman, Press, Price & Teukolsky). Needs a little debugging.