I MTW exercise 21.26: junction conditions for a thin shell of dust

JimWhoKnew
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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.
 
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JimWhoKnew said:
The second ( ##\mathbf{a}^+ +\mathbf{a}^- =0## ) is where I get stuck.
What do you get if you add the second and third equations?
 
PeterDonis said:
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?
 
JimWhoKnew said:
##\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.

JimWhoKnew said:
Can we show it from the equations?
How did you derive equation (c)?
 
PeterDonis said:
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.
 
The solution can be found in "Problem Book in Relativity and Gravitation" (Lightman, Press, Price & Teukolsky). Needs a little debugging.
 
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