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Strength of materials and special relativity

  1. Jun 21, 2005 #1
    While mowing the grass, I thought of the following problem (to take my mind off of the heat!)
    Consider a vertical column of water in a container. The pressure exerted by the water on the sides of the container is rho x g x h, where rho is the density of the water, h is the height of the column above the specific vertical location being considered. Remember that the pressure doesn't depend on the horizontal size. Now, if the strength of the container material (measured in pressure units) is too small (less than the pressure at that point), the container breaks. An observer sitting at rest w.r.t. the column observes the container is not breaking.

    Another observer is moving rapidly, perpendicular to the column (horizontally). The width of the container is contracted and the mass is increased according to most explanations. Also, the walls become thinner. Therefore, the density of the the liquid increases by approx. (gamma)^2. Yet, the container cannot break because the other observer doesn't see it.

    Is the strength of the material greater in the improper frame? If so, why? One might say the intermolecular distances contract, but only in the horizontal direction, and the strength that's important is the shear strength between different heights; that spacing hasn't changed, has it? Why doesn't the thinning of the walls make the wall weaker?

    Or maybe the density doesn't change? If not, why not? You still have the problem of thin walls.

    Explanations, please? Hope this tickles a bit.

  2. jcsd
  3. Jun 22, 2005 #2


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    I don't think I've ever seen this sort of problem discussed in a SR textbook. Certain aspects of it are discussed in GR textbooks, but not all. An ideal fluid can be represented by a stress-energy tensor

    [tex]T^{ab} = P g^{ab} + (\rho + P) u^a\times u^b[/tex]

    where [tex]u^a[/tex] is the 4-velocity of a fluid element (zero in its rest frame).

    In the bulk of the fluid we can say

    [tex]\nabla_u T^{uv} = 0[/tex]

    The container can also be represented by a stress-energy tensor. What isn't terribly clear to me is how to write the boundary conditions at the fluid-container bondary.

    Common sense suggests that we just solve for the stresses in the container in the rest frame, then transform the stress-energy tensor of the container, and forget about writing the hydrodynamic equations. But this would sort of sidestep the problem, I think.

    On the other hand, the side-stepped problem is a bit interesting in its own right, and it's certainly a LOT easier to talk about.

    It turns out that forces in the direction of motion are transformed by the identity relationship, while forces perpendicular to the direction of motion are decreased by a factor of gamma. This is not particularly intuitive, but results from the fact that the four force

    F4 = dP/dtau = (dE/dtau, dPx/dtau, dPy/dtau, dpz/dtau)

    transforms as a 4-vector. Note that when v=0, F*v=0 and dE/dtau=0, so that the 4-force is equivalent to the 3-force in the rest frame of the object to which the force is being applied.

    So we start with F4 = (0,Fx,Fy,Fz)

    and then do the Lorentz transform for a boost in the x-direction to get


    Here gamma is a number larger than 1, equal to 1/sqrt(1-(v/c)^2)

    Now we need to convert the 4-forces back to three-forces. We do this by noting that any force component Fa has

    Fa = dPa/dt = (dPa/dtau)*(dtau/dt), and since (dtau/dt) = 1/gamma, we divide all the components of the 4-force by gamma to get the three force.

    This means we wind up with

    Fx, Fy/gamma, Fz/gamma

    for the relativistically transformed force.

    I guess I need to describe the container walls and the forces in them. Let's take the simplest case, a container with a circular cross section. Then the forces in the walls will be pure tension forces - there won't be any shear forces. With thin walls, the tension forces will be the same at the inner and outer radius of the circular container.
    [end clarify]

    Now let's apply the above transformation to these tension forces in the wall of the circular container.

    This means that the thinner walls of the container that are oriented perpendicularly to the direction of motion have less force on them, by a factor of gamma and the same force/unit area.

    The walls of the container oriented in the same direction as the force have the same force, not a larger force (and the same area, so the force/area is the same).

    This should be what's predicted by the tensor transformation laws for this case - otherwise there is an error in my reasoning somewhere. It's getting late, I'll think about it more from this angle tomorrow.
    Last edited: Jun 22, 2005
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