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I Gravity due to pressure

  1. Mar 10, 2017 #1
    How do I calculate the gravitational mass of a cylinder of compressed gas, including the effects of pressure? By gravitational mass, I mean what I would measure on an ideal mass balance.
    (I know that the pressure is negligibly small in a realistic container, but I want to have a conceptual understanding.)

    My understanding is that the time time component of the Ricci curvature is $$R_{00}=\frac{1}{2}\left(\rho_E+P_x+P_y+P_z\right)$$
    so pressure should have an analogous contribution to energy on gravity. But I've never seen it applied to any sort of ordinary objects so I'm having a hard time connecting it to reality.

    Suppose the gas has a energy of ##E_g##, a pressure of ##P##, and a volume ##V##. The cylinder has an energy ##E_c## and a wall tension of ##-P## due to the confinement of the gas and a surface area ##A##.

    If we ignore pressure, then the gravitational mass is just ##(E_g+E_c)/c^2##.

    Is the pressure contribution then just ##3PV/c^2##? So the total gravitational mass is ##(E_g+E_c+3PV)/c^2##?

    Is the wall tension irrelevant because the wall has 0 volume?
     
  2. jcsd
  3. Mar 10, 2017 #2

    PeterDonis

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    For a static container of compressed gas, the gravitational effect of pressure in the gas is exactly cancelled by the gravitational effect of tension (which is negative pressure) in the walls of the container. So the gravitational mass is just the ordinary mass (density times volume) of the gas.
     
  4. Mar 10, 2017 #3
    I thought that might be the case, but I don't understand how the math works that way, since the volume of the gas is much greater than the volume of the container walls.
     
  5. Mar 10, 2017 #4
    I'll admit that I knew nothing about this to begin with, but I think I understand why the math works this way. It shouldn't have anything to do with the volume. As long as the cylinder is in equilibrium (i.e. the gas isn't breaking the walls due to pressure), then the summation of forces at every point will equal zero. As @PeterDonis pointed out: there is an equal and opposite force for every point where pressure is acting. So the equation would be...
    ΣF=0
    P + (-T) = 0
    P = T
    Same deal for the normal force cancelling out the force caused by gravity when you have an object resting on the Earth.
     
  6. Mar 10, 2017 #5

    PeterDonis

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    Comeback City's response is basically correct (but see my follow-up post to him). For a more technical answer, look up "Komar Mass"; that is the actual calculation you would do in GR to find the total mass of an isolated static object by "adding up" the contributions from all the stress-energy in it.
     
  7. Mar 10, 2017 #6

    PeterDonis

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    Yes, but force is not the same as pressure. The force balance turns out to lead to a relationship between the pressure in the gas and the tension in the walls of the container, but that relationship is not equality. For a spherical container, the relationship is

    $$
    \sigma = \frac{p r}{2t}
    $$

    where ##p## is the gas pressure, ##r## is the inner radius of the sphere, and ##t## is its thickness, which is assumed to be much smaller than ##r## (the usual constraint is ##r / t > 10##). Note that as ##t## gets smaller, ##\sigma## gets larger in relation to ##p##, and in the limit of a zero thickness wall, ##t## [Edit: actually ##\sigma##] increases without bound. That is a key part of the resolution to the issue @Khashishi raises.
     
    Last edited: Mar 11, 2017
  8. Mar 10, 2017 #7
    What does σ actually represent in this?
    Shouldn't σ increase without bound, not t?
     
  9. Mar 11, 2017 #8

    PeterDonis

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    The tension in the container wall.

    Sorry, yes, ##\sigma## increases without bound as ##t \rightarrow 0##. I have edited my previous post to correct this.
     
  10. Mar 11, 2017 #9
    So for a spherical vessel, the tension integrated over a wall of thickness t is ##4\pi r^2 \sigma t = 2 \pi P r^3##. But, the tension is expressed in two directions (tangential to the wall) so this value should be doubled (##4 \pi P r^3##).
    The pressure of the gas integrated over a sphere is ##PV = \frac{4\pi}{3} P r^3##. But the pressure is isotropic so we multiply by 3, and we get the same result. It makes sense now. Thank you.

    For a cylindrical vessel with thin walls, we can cap the top and bottom with hemispheres. We already know the tension in the hemispheres will cancel the pressure in the hemispheres as shown above, but what about the cylindrical portion? Wikipedia (https://en.wikipedia.org/wiki/Cylinder_stress) gives a circumferential stress of ##\sigma_\theta = \frac{Pr}{t}## and the axial stress as approximately ##\sigma_z = \frac{Pr}{2t}##. The total tension (ignoring the top and bottom) should be ##2 \pi r l t (\sigma_\theta+\sigma_z)## or ##3 \pi P r^2 l##. Meanwhile, the integral of the trace of the gas pressure is ##3 \pi r^2 l P##. Unsurprisingly, it works out for this case, too.
     
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