Kinetic theory of gases

  • #1
Hello. There is something I don't understand about the momentum tensor [itex]t_{ik}=nm\int f(|v|) v_iv_k[/itex] of
an ideal gas with an isotropic velocities distribution where n is the number of molecules per unit volume and m is the
mass of a molecule. Since the the velocity distrubution is isotropic clearly t_ik=0 for i<>k, the problem is that when
i=k we obtein [itex]t_{ii}=1/3nm\overline{v^2}[/itex] which means that if we fix a point in space inside the gas and take some
direction given by a versor n, then an element of surface ds perpendicular to n will experience a momentum flux in same
direction given by n.
The reason why I don't understand this result is because being the velocities distribution isotropic I should find the
same amount of momentun traversing the element of suface in opposite directions thus canceling each other out meaning I should
have t_ik identically zero for all indices. How am I wrong?
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Answers and Replies

  • #2
Which textbook did you get this from.
Would like to take a look at this ... :)
  • #3
Pauli Lectures on Physics, volume 3 - Thermodynamics and the Kinecti Theory of Gases,
and could be volum4 too - Statistical Mechanics
  • #4
Oh , I have bought that book. It is Vol.3 Pg 96.
You notice the integral( top of the page) is only for v1>0? So it is only for 1 direction.
  • #5
yes, in that part he is calculating pressure and is ok, however at the bottom of that page he defines the tensor and I'm trying to interpret the momentum tensor, by the way I do have problems relating the tensor with pressure in the general case, but that is another issue. Notice that in page 98 he asserts what I first posted about the value of the components of the tensor
  • #6
I think the surface divides the molecules into 2 populations. Those leaving the surface and approaching the surface. Those leaving are excluded because they will not create any pressure, and so are not included in the tensor. The reason why they say is isotropic has to do with the fact that the orientation of the surface does not affect the integral because the molecules are completely random in directions.
  • #7
The problem is the evaluation of tensor. Your interpretation makes sense but how are you going to calculate the tensor in one point of space inside the gas?What is the region of integration. For intance for t_ik to be zero when i<>k you need to integrate over the whole region of the gas in that way the simetry of the distribution function annihilates the integral.
  • #8
The region of integration is over all of velocity space. However since v1^2 is an even function, the integral is equal to that of 2 times the integral over half of the velocity space for v1>0.
  • #9
Yes, I was mistaken when I said the whole region of space I should have said velocity space, anyway this lead us to problem of interpertation of the momentum tensor. This tensor gives the momentum flux of all the particles in the gas. How can the total momentum flux diferent from zero?
  • #10
How can the integral of (V1^2)*f over all all v1 be zero if the gas molecules are moving (so (V1^2)>0) and f>0?
  • #11
Yes, that is the probem it isn't, that is why I don't understand.
  • #12
You mean you believe that the integral can be 0?
  • #13
I believe the momentum flux tensor must be zero because the distribution of velocities is isotropic. How can you explain the net momentum flux through an element of surface when there should be same amount of momentum passing in equal quantities in opposite directions?
  • #14
You are right, the net momentum flux is zero. But they are interested in only one direction - the momentum flux outward, the direction outward being that of the container to which the surface belongs.
  • #15
Now we agree the net momentum flux is zero. We also agree that they are interested in only one direccton. The problem remains in that the calculation gives the total momentun and this result is not zero and this is what I don't understand.
Do yo agree that they are calculating the total tensor momentum?
  • #16
Let say we are dealing with a cube with 6 faces. If we are only dealing with momentum flowing outward at each surface (which could be negative if the momentum just so is flowing inward) then the total momentum flowing outward of the cube is not necessarilty zero , and can be postive or negative , correct?
  • #17
yes, I think that's correct
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  • #18
Suddenly it occurs to me that it is possible for momentum flux across a surface not to be zero, by interpreting it this way-

Forget about walls. Let say a molecule A on the left is colliding with another molecule B on the right. Molecule A has to lose momentum and molecule B has to gain momentum in the x direction , if we set the sign convention as the right = positive. Then we draw a vertical surface to separate these 2 molecules.

So there is a net flow of momentum across the surface and it is not zero. The velocities are still isotropic at the points of collisions.

Any flaw in this argument?
  • #19
I'm not sure because isotorpy demands the same thing must happen in the opposite direction,that is, from right to left canceling once more the first positive flux

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