- #1
Sonderval
- 234
- 11
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)?
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)?