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Energy, momentum, & mass of a box of moving particles

  1. Apr 17, 2006 #1


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    This is one of those problems I've been meaning to take a closer look at for a long time, and I have finally done it.

    Note that by "mass" of the box, I mean the invariant mass of the system of particles, which, as we will see, is not an invariant in this case because the system is not isolated.

    Suppose we have a pair of flat plates, with a bunch of particles bouncing around in between the plates


    \unitlength 1mm

















    In the rest frame of the box, we have n particles, of rest mass m, all with a velocity v. Half the particles are moving to the left - the other half, to the right.

    The quesiton is, how do we describe this box in a moving frame of reference, and what is its energy, momentum, and mass in a moving frame of reference? The walls of the box are assumed to be of negligible weight.

    We start by writing down the 3 and 4 velocities of the particles in the rest frame of the box, using geometric units where c=1.

    3-velocity= v
    4=velocity = [itex][1/\sqrt{1-v^2},v/\sqrt{1-v^2},0,0][/itex]
    3-velocity = -v
    4-velocity= [itex][1/\sqrt{1-v^2},-v/\sqrt{1-v^2},0,0][/itex]

    Now we do a boost. Let us have an observer moving to the left with a velocity u, so that the box appears to move to his right with a velocity u.

    The transformed velocities are
    3-velocity = (u+v)/(1+uv)
    4-velocity = [itex][(1+uv)/\sqrt{(1-u^2)(1-v^2)},(u+v)/\sqrt{(1-u^2)(1-v^2)},0,0][/itex]

    3-velocity = (u-v)/(1-uv)
    4-velocity = [itex][(1-uv)/\sqrt{(1-u^2)(1-v^2)},(u-v)/\sqrt{(1-u^2)(1-v^2)},0,0][/itex]

    In the rest frame of the box, the number of left-moving particles is the same as the number of right-moving particles. This is not true, however, in the moving frame of the box. Because our definition of simultaneity has changed when we perform the boost, we have to compute how many particles are moving left, and how many are moving right.

    To do this, we note that in the rest frame of the box, the rate at which particles impact the right side of the box in particles/second is equal to the rate at which particles impact the left side of the box.

    In the moving frame, this must still be true, though the actual rate will be different due to time dilation. While a clock on the left side of the box will no longer be synchronized with a clock on the right side of the box after the boost, it will still tick at the same rate, and hence the rate at which particles impact must be the same on the left side as it is on the right side.

    The rate at which particles impact will be given by

    \frac{\textrm{particles}} {\textrm{meter}} \, \frac{\textrm{meters}} {\textrm{second}} = \frac{\textrm{particles}} {\textrm{second}}

    This leads to the equation

    \frac{n1 \left( \frac{u+v}{1+uv} - u \right) }{L} = \frac{n2 \left( u - \frac{u-v}{1-uv} \right) }{L}

    where n1 is the number of right-moving particles and n2 is the number of left moving particles. L is the length of the box, which cancels out.

    Solving this equation, we get
    [tex]n1 = n \frac{1+uv}{2}[/tex]
    [tex]n2 = n \frac{1-uv}{2} [/tex]

    We can now compute the total energy of the particles from the 4-velocities

    E_{\textrm{tot}} = n1 \, m \frac{1+uv}{\sqrt{(1-u^2)(1-v^2)}} + n2 \, m \frac{1-uv}{\sqrt{(1-u^2)(1-v^2)}} = \frac{n m \left(1+u^2v^2 \right)}{\sqrt{(1-u^2)(1-v^2)}}
    P_{\textrm{tot}} = n1 \frac{u + v}{\sqrt{(1-u^2)(1-v^2)}} + n2 \frac{u-v}{\sqrt{(1-u^2)(1-v^2)}} = \frac{n m u \left(1+v^2 \right)}{\sqrt{(1-u^2)(1-v^2)}}

    Note that [itex]m^2 = E_{\textrm{tot}}^2 - P_{\textrm{tot}}^2[/itex] is not constant!
    Because this system is not an isolated system, the invariant mass depends on the frame of reference and is not a true invariant. Thus the energy and mass of this non-isolated system does not transform as a 4-vector.

    This result can also be obtained (in the continuum limit, for a large number of particles) with a stress-energy tensor approach, which I will only sketch.

    The energy component [itex]T^{00}[/itex] of the stress energy tensor in the rest frame of the box is given by the total energy in the box divided by the volume of the box, i.e.

    [tex]T^{00} = \frac{n m}{V \sqrt{1-v^2}}[/tex]

    where V is the volume of the box. We also need to compute the pressure in the box, which is anisotropic - the only pressure is in the 'x' direction. We compute the pressure by force / area, which is momentum / area * time.

    We know that a total momentum of

    [tex]2 \frac{n}{2} m v / \sqrt{1-v^2} [/tex]

    will be transferred to each wall of the box in a time L/v.

    This leads to the relationship
    [tex]T^{11} = v^2 T^{00}[/tex]

    We can then boost the stress-energy tensor in the usual manner, and confirm that
    [tex]E = \int T^{00} dV[/tex]
    [tex]P = \int T^{01} dV[/tex]

    gives the total system energy and the x-component of the system momentum, and that these match the results computed earlier.
    Last edited: Apr 17, 2006
  2. jcsd
  3. Apr 17, 2006 #2
    I did this years ago and placed it on my website and posted a link to it here if I recall corrrectly. Its at


    The reason the system is not invariant is that you're never supposed to add tensors which are not located at the same event. The sum of two 4-vectors located at two different events is not a 4-vector. Although it can still be a very useful thing, i.e. when the 4-vectors move in straight lines inbetween collisions. Instead of a box I used a magnetic field as the external force acting on the system. There you can see that there is no unique way to add the 4-vectors. in different frames you want to add all energies and momentums "at the same time." But when you use external forces that can't yield a unique quantity.

  4. Apr 17, 2006 #3


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    The unbalanced stresses in the problem (there is no tension to balance the outward pressure on the walls) are what prevents the system from being isolated, and from having a conserved invariant mass.

    Did you also get the result that the ratio of momentum / velocity depended on the direction of the velocity, because of the anisotropic pressure?

    \frac{P}{u} = \frac{n m \left( 1 + v^2 \right)}{\sqrt{(1-u^2)(1-v^2)}}

    but if we perform the boost in the y or z direction

    \frac{P}{u} = \frac{n m }{\sqrt{(1-u^2)(1-v^2)}}

    [add]I did look at your web-page, BTW, but while the problem you worked there was related to the issue, it wasn't exactly the same one I did above.
    Last edited: Apr 17, 2006
  5. Apr 17, 2006 #4
    If you include the stresses on the wall then you'll get an invariant mass for the "gas + box" system.

    Did you also get the result that the ratio of momentum / velocity depended on the direction of the velocity, because of the anisotropic pressure?

    Yes but instead of doing a system of particles I did a magnetic field since it was a homework problem in Ohanian ... who was very happy that I sent him my work on mass density. See the mass density of a magnetic field paradox.


    The point I was hoping to make was that the expected invariance is absent because the system is not closed.

    Did you know that Tolman mentions this problem in his text and that Griffiths wrote a paper for AJP where he mentions that momentum is not parallel to velocity in a non-closed system? I can post refs if you'd like.

  6. Apr 17, 2006 #5


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    Would that be Tolman text be "Relativity, Thermodynamics and Cosmology"


    I've been thinking about picking it up - it would be $2 to borrow it via interlibrary loan, and it only costs about $14 to buy.
  7. Apr 17, 2006 #6
  8. Apr 19, 2006 #7
    I'm attaching a part of Rinder's 1982 intro to SR book. Its a good two pages concerning this issuse and the actual physics of why rather that relying only on the math.

  9. Apr 19, 2006 #8
    Last edited: Apr 19, 2006
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