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I Stress as momentum flux

  1. Feb 14, 2017 #1
    I'm trtying to get a better understanding of the spatial part of the energy-momentum tensor, and although similar questions have been asked here, I think the point I do not fully grasp has not been covered so far.
    The stress tensor can be considered as "momentum flux density" tensor.
    If I consider a small test volume V, I can calculate the total flux of momentum into this test volume as
    [tex] \int_{\partial V} \sigma dA= \dot{p} [/tex]
    so this is the momentum change (i.e. the force) on the volume.

    Now consider a material in homogeneous uniaxial stress, [itex]\sigma_{11}\ne 0[/itex], all other stress components are zero.
    For each small test volume, the total force on the volume vanishes as it should.

    The question I have is the basically on the meaning of "momentum flux". If I consider a test surface [itex]A[/itex] orthoginal to the 1-direction somewhere in my volume, The momentum flux through this surface can be calculated (similar to above, but now the surface is not closed)
    [tex] \int_{A} \sigma dA [/tex]
    which is non-zero (and is just a surface traction vector). This tells me that there is a momentum flux through the surface. (In other words, the material on the other side of the surface exerts a force) If I now reverse the surface (flip its normal vector), I get the same result, but with a minus sign. So through each surface A, there is momentum flowing from the left to the right and from the right to the left - basically, this seems to be Newtons third law. (Force exerted from the part of the material on the left and right side of the surface has to be equal and opposite in a static situation.)

    First of all, I'd like to ask whether my understanding so far is correct.

    Second, I find it puzzling that on the one hand, the momentum flux through any surface (if I add up the flux from left to right and right to left) is zero (the same is true for any test volume using my first equation above), but that we can still speak of a non-zero momentum flux density. Is there any more intuitve way to grasp this or should I just stick to the maths (which seems to make sense)?
     
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  3. Feb 14, 2017 #2

    Paul Colby

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    I'm just learning continuum mechanics for the first time and I find the following observation has been helpful. In Cartesian coordinates a stress tensor with equal diagonal components correspond to an isotropic pressure in a medium. For pressure to cause acceleration (strain in a solid or wind in a gas) one needs the pressure to have a divergence. Looking at the continuum equations of motion this is seen to be a universal requirement for a stress field to cause an acceleration of the medium. Having a divergence is what is required for the surface integrals you are considering to be non-zero. I hope this mechanical view is helpful for you. I found continuum mechanics quite baffling until this occurred to me.
     
  4. Feb 14, 2017 #3
    This makes sense to me, because I could transform my first surface integral to a volume integral over the divergence of the stress tensor.
    Still not totally sure whether this clears up my problem with the momentum flux since this is related to the tensor, not to its divergence.
     
  5. Feb 14, 2017 #4

    Paul Colby

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    Maybe it's more helpful than I realize. No divergence no momentum flux.
     
  6. Feb 14, 2017 #5

    Orodruin

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    This is true for any current, not only momentum currents. The flux of a quantity through a surface with one direction of the surface normal is the negative of the flux through the same surface with the normal in the opposite direction. It just tells you that what flows into the region the normal is pointing to is the same as what flows out of the region the normal is pointing from.

    Edit: The fallacy is to think that the fluxes affect the same region. It is a bit like thinking that objects cannot accelerate because Newton's third law tells you that there is an equal but opposite force. The point is that the fluxes describe the flux out of particular (and distinct regions) and the flux out of one region is equal to the flux into another.
     
  7. Feb 14, 2017 #6

    Paul Colby

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    So in the pressure example ##T_{xx}## is the momentum flux across an ##yz## area element yet there is no acceleration in this case. What is meant by "momentum flux". There would seem to be no net momentum flow across the area element.
     
  8. Feb 14, 2017 #7

    Paul Colby

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    This is a really good question you are asking. For charge current the area normal dotted with the current vector is the charge moving across the area element. Using the "momentum flux" verbiage leads to an incorrect expectation in my opinion. Am I off base on this one?
     
  9. Feb 14, 2017 #8
    I'm not sure I understand that.
    If I have a positive charge flowing through a surface from left to right, it does not matter in what direction I define my surface normal - there is always current to the right and no matter how I orient my surface, the charge to the right of the surface increases whereas that to the left decreases. This is not the case with the momentum flux in the statically stressed case.
    I suspect this is due to the fact that we are dealing with a tensor quantity: current is a vector quantity, so it reverse sign if I flip my x-axis; whereas pressure is a tensor-quantity and [itex]\sigma_{11}[/itex] does not change its sign when I flip the coordinate axis
     
  10. Feb 14, 2017 #9

    Orodruin

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    Yes it is.
     
  11. Feb 14, 2017 #10
    @Orodruin
    I'm sorry, do not understand that.

    How would the momentum on one side of a surface increase and that on the other decrease if the situation is static?

    Consider a cube of material stressed in hydrostatic stress under compression.
    Due to symmetry, the momentum flux through the surface of the cube has to be the same everywhere, if the situation is static, the total momentum flux through the surface (using an outward pointing normal everywhere) is zero.
    If I now consider two opposite faces, and use the same direction of the normal vector (inward on the left, outward on the right, for example), will I then have momentum flowing inward on one of the faces and outward on the other?
     
  12. Feb 14, 2017 #11

    Orodruin

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    So let us go back to Newton 3 (you have exactly the same fallacy). The force on volume 1 across the surface is balanced by an equal and opposite force on volume 2. This force transfers momentum between the volumes. This does not tell you that there cannot be any other forces acting on volume 1 that results in a static situation.

    Consider a cart with two horses attached on opposite sides. Horse 1 pulls to the right and horse 2 pulls to the left. Now, the force from horse 1 on the cart transfers momentum to the cart (force is change in momentum per time). However, horse 2 pulls the cart with an equal but opposite force in the other direction and therefore the cart is in equilibrium because the net change in the momentum of the cart is zero.

    The same works for your situation. In a static situation, any given volume ##V## will experience the force
    $$
    F_i = \int_{\partial V} \sigma_{ij} dS_j = \int_V \partial_j \sigma_{ij} dV
    $$
    due to contact forces at its surface. The total change in the momentum in the volume is therefore given by the volume integral of the divergence of the stress tensor - not by the stress tensor itself.

    Compare this to the flux of an incompressible fluid. For a static incompressible flow, you will have ##\nabla \cdot \vec v = 0##. The mass flow through a surface ##S## is given by
    $$
    \Phi(S) = \rho \int_S \vec v \cdot d\vec S
    $$
    and even though the situation is static - there is a flow (static just means that the system looks the same at all times, not that there are no flows). However, the flux out of any volume is given by
    $$
    \Phi(\partial V) = \rho \oint_{\partial V} \vec v \cdot d\vec S = \rho \int_V \nabla \cdot \vec v \, dV = 0
    $$
    and so the amount of fluid in the volume is conserved - as much fluid is flowing into the volume as out of it.

    In terms of the momentum flows - a static situation just means that the total momentum in a volume is not changing - it does not mean that there are no momentum currents.
     
  13. Feb 14, 2017 #12
    @Orodruin
    Thanks for elaborating. Unfortunately, I still do not fully understand. (Although I do understand your equations.)

    Again, I want to compare to the situation of a charge flux: If I have a wire with two terminals A and B, if a current flows, I can define a direction of the flow of the current (If A is plus and B is minus, technically, the current flows from A to Be, electrons from B to A). At each surface in the wire, I could measure this flux (inserting an amperemeter). Charge accumulates at A and B due to this flow.

    In the case of a wire under compression (or tension) however, A and B are completely symmetric. At each point in the wire, due to Newton III, the forces are balanced. Nor momentum will "accumulate" at A and B and there is no way I can determine the direction of momentum flux.

    The same is the case for your example of a fluid flow. Even if the flow is static, I can still uniquely say for each surface in which direction matter flows through this surface. If I have a ring-shaped pipe, the flow is either clockwise or counterclockwise.

    However, if I have a ring-shaped wire under tension (or compresssion), there is again no way I can ascribe direction to the momentum flux.
     
  14. Feb 14, 2017 #13

    Orodruin

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    Then you are considering one of the terminals. Your situation is more akin to considering a part of the wire in between. The net charge of the wire remains constant even though there is a current flowing through it.

    No they are not, the force on A from the wire is in the opposite direction from the force on B. Therefore, momentum is transferred between A and B through the wire. In order for A and B not to accelerate you need additional forces acting on them which in the charge case would correspond to further passing the charges arriving at A on in a circuit.

    Yes there is. See above.

    Yes there is. The force across a section of the wire is the momentum current density by definition.
     
  15. Feb 14, 2017 #14
    @Orodruin
    Sorry for being dense, but what would be the direction of momentum flow in the wire constrained at A and B?
    From A to B or B to A? How could you decide?
     
  16. Feb 14, 2017 #15

    Orodruin

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    First you would have to decide which is the positive direction. Let us say that it is the direction from A to B and that there is positive tension in the connecting string. This results in a force from the string on B which is in the negative direction and a force on A from the string in the positive direction. Assuming no other forces, B would accelerate in the negative direction (getting less momentum) and A in the positive direction (getting more momentum). The momentum flow would therefore be from B to A.
     
  17. Feb 14, 2017 #16
    But someone who chooses a different coordinate system would decide that the momentum flows in the other direction?
    This is different from the case of a fluid or current flow, where no matter how I choose the coordinate system, the flow direction is unique.
     
  18. Feb 14, 2017 #17

    Orodruin

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    It is actually not any different at all. If you chose positive charge to be negative, you get the same effect. (Note that electrical currents are usually negative charges flowing against the current. We could just as well have defined positive charge as negative.) When you talk about momentum currents, you talk about momentum in a given direction - just as when you talk about charge you talk about what we have defined as positive charge. This is why you have three different momentum components and the current is a tensor (and not a vector) - the stress tensor ##\sigma##, being the momentum current, tells you how much momentum is flowing in a given direction. Its component ##\sigma_{ij}## is the current of momentum in the ##i## direction that flows in the ##j## direction.
     
  19. Feb 14, 2017 #18
    @Orodruin
    Thanks for your patience. But no matter how I define the charge - in the case of charge flow or even clearer in a fluid flow, the direction in which something (electrons or particles) is actually flowing is independent of any coordinate system.
    Mathematically, I think this is all due to the stress being a tensor - if I flip a coordinate direction, a velocity vector (as used in fluid flow or currents) will change sign, but the diagonal terms of the stress tensor will not.
     
  20. Feb 14, 2017 #19
    I think I finally got your point (sorry for being slow):
    The critical difference between momentum and things like charge is this:
    +x-momentum flowing to the right is exactly the same as -x-momentum flowing to the left.

    Therefore, it makes no sense to talk about "momentum" flowing in some direction. I can talk about +x-momentum flowing in some direction. This is invariant to changes in the coordinate system - at each point, the momentum flux is either in the same direction as the component of momentum (+x-momentum flowing in+x-direction or -x-momentum flowing in the -x-direction) or the two directions are opposite. The first case is compressive sress, the second is tensile stress.

    To wrap it up in my example, there are 2 possible ways of loooking at it:
    Surround the wire with a closed surface with outward pointing normal. -x-momentum is flowing out on the left, +x-momentum is flowing out on the right. Alternatively, you can say that +x-momentum is flowing in on the left, -x-momentum is flowing in on the right. Net momentum change: zero.
    Or you use two parallel surface oriented in the same way (e.g., normal to the right). Then +x-momentuum is flowing from the left through both surfaces, so it flows through the volume - or -x-momentum is flowing from the right through both surfaces. Net change is zero again.
     
    Last edited: Feb 14, 2017
  21. Feb 14, 2017 #20

    Dale

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    That sounds like you got it!

    So if you have +x momentum flowing in on the left and -x momentum flowing in on the right then the net x momentum flux is 0.

    So now consider, what happens if you have +x momentum flowing in on the left and no momentum flux on the right. The four-divergence is still 0, so where does the extra momentum flux go?
     
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