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

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Discussion Overview

The discussion revolves around exercise 21.26 from MTW, focusing on the derivation of junction conditions for a thin shell of dust surrounded by vacuum. Participants explore the mathematical relationships and physical interpretations of the equations related to the 4-accelerations of the shell and the forces acting on it.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses difficulty in deriving the equation ##\mathbf{a}^+ + \mathbf{a}^- = 0##, suggesting it relates to the total non-gravitational force on the shell element.
  • Another participant proposes adding the second and third equations to explore their implications.
  • A participant notes that adding these equations yields ##\mathbf{a}^\pm = \pm 2\pi\sigma\mathbf{n}##, drawing a parallel to electrostatics and discussing the symmetry of forces on the accelerometers.
  • There is a question about deriving equation (c) and the use of Gaussian Normal Coordinates (GNC) in the derivation process.
  • A participant describes their approach to deriving equation (d) from the book's equations and discusses the implications of the vacuum condition on the shell's accelerations.
  • Another participant mentions the need to show that certain quantities are the same on both faces of the shell to complete the derivation of equation (b).
  • A later reply suggests that the solution can be found in another resource, indicating that it may require further debugging.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and approaches to the problem, with no consensus reached on the derivation of the equations, particularly equation (b). Multiple competing views and methods are presented without resolution.

Contextual Notes

Participants note the limitations of their approaches, including unresolved mathematical steps and dependencies on specific coordinate choices. The discussion reflects the complexity of the problem and the need for careful consideration of the conditions involved.

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?
 
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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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