Four-momentum conservation

Required math: calculus

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 3; Problem 3.4.1 & 3.4.2.

A logical extension of the four-velocity {\mathbf{u}} is the four-momentum {\mathbf{p}\equiv m\mathbf{u}}, where {m} is the object’s rest mass. From the definition of {\mathbf{u}} this means that in an inertial frame in which the object’s velocity is {\mathbf{v}}:

\displaystyle  \mathbf{p}=\gamma m\left[1,v_{x},v_{y},v_{z}\right] \ \ \ \ \ (1)

where {\gamma=1/\sqrt{1-v^{2}}}. Experimentally, it is known that the four-momentum of a collection of particles is conserved, and that the old Newtonian law of conservation of momentum is not valid for velocities approaching that of light. As a simple example of this, suppose we have two particles with the same mass travelling towards each other (as viewed in an inertial frame we’ll call the rest frame), so that one particle has a speed of {v_{1}=+v} and the other has {v_{2}=-v}. In Newtonian physics, the total momentum of the system is {mv_{1}-mv_{2}=0}. Now suppose that the particles collide head-on and that the collision is totally inelastic so that the two particles merge to form a single particle of mass {2m} with a speed of {v_{3}=0}. Clearly the momentum after the collision is still zero, so momentum is conserved here.

Now if we observe this collision from an inertial frame moving with a speed {v} then, before the collision one particle is at rest so that {v_{1}^{\prime}=0}. To find the velocity of the other, we need to use the relativistic velocity summation formula to get

\displaystyle   v_{2}^{\prime} \displaystyle  = \displaystyle  -\frac{v_{2}+v}{1+v_{2}v}\ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  -\frac{2v}{1+v^{2}} \ \ \ \ \ (3)

After the collision, the single particle is moving at speed {v_{3}^{\prime}=-v}. If we use the same mass for the particles in this case, the Newtonian law would be violated since

\displaystyle   mv_{1}^{\prime}+mv_{2}^{\prime} \displaystyle  = \displaystyle  0-\frac{2mv}{1+v^{2}}\ \ \ \ \ (4)
\displaystyle  \displaystyle  \ne \displaystyle  2mv_{3}^{\prime} \ \ \ \ \ (5)

What happens if we look at four-momentum instead? In the rest frame the four momentum is

\displaystyle   \mathbf{p} \displaystyle  = \displaystyle  \gamma m\left[1,v,0,0\right]+\gamma m\left[1,-v,0,0\right]\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  \gamma m\left[2,0,0,0\right] \ \ \ \ \ (7)

Since the first component of {\mathbf{p}} is the energy (mass) of the particle, we see that in the rest frame, where the compound particle is at rest, its rest mass is

\displaystyle  M=2\gamma m \ \ \ \ \ (8)

if we’re assuming four-momentum is conserved.

Now in the moving frame, the four-momentum is

\displaystyle  \mathbf{p}^{\prime}=m\left[1,0,0,0\right]+\gamma^{\prime}m[1,v_{2}^{\prime},0,0] \ \ \ \ \ (9)

since the first particle is at rest and the second is now moving with speed {v_{2}^{\prime}}. We can write {\gamma^{\prime}} in terms of {v}:

\displaystyle   \gamma^{\prime} \displaystyle  = \displaystyle  \frac{1}{\sqrt{1-\left(v_{2}^{\prime}\right)^{2}}}\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  \left(1-\frac{4v^{2}}{\left(1+v^{2}\right)^{2}}\right)^{-1/2}\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  \frac{1+v^{2}}{\sqrt{\left(1+v^{2}\right)^{2}-4v^{2}}}\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \frac{1+v^{2}}{\sqrt{1-2v^{2}+v^{4}}}\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  \frac{1+v^{2}}{1-v^{2}} \ \ \ \ \ (14)

Therefore, the total momentum in the moving frame is

\displaystyle   \mathbf{p}^{\prime} \displaystyle  = \displaystyle  m\left[1,0,0,0\right]+\frac{1+v^{2}}{1-v^{2}}m[1,-\frac{2v}{1+v^{2}},0,0]\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  2m\gamma^{2}[1,-v,0,0]\ \ \ \ \ (16)
\displaystyle  \displaystyle  = \displaystyle  \gamma M[1,-v,0,0] \ \ \ \ \ (17)

But this is also the momentum of a particle of rest mass {M} moving with a speed {v}, so in this case, four-momentum is indeed conserved.

16 thoughts on “Four-momentum conservation

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  14. BogdanT


    I was studying Moore book on General Relativity and came across you website.
    I followed you solution to this problem and I remarked the following:
    the rest 4-Momentum is : $P= 2m*\gamma*[1,0,0,0]$
    the moving frame 4-Momentum is: $P’=2m*(\gamma)^2*[1,-v,0,0]$

    But as it looks they are not equal, their magnitude is notequal. And they should.
    The moving $P’$ 4-Momentum has a $\gamma$ coefficient more than the rest $P$ 4-Momentum
    I checked the calculations and did not find any error or miss calculation and cannot understand why they are not equal. Can you please check again the calculations?
    I am curious now where the error could be.

    Thanks for this amazing website.

    1. gwrowe Post author

      The four-momentum itself is not an invariant between inertial frames (four-vectors are not invariants between frames). It is only the magnitude that is invariant and that does actually work out in this case. In the rest frame

      \displaystyle  \left|\mathbf{p}\right|^{2}=p_{\mu}p^{\mu}=-M^{2}

      In the moving frame, we have

      \displaystyle   \left|\mathbf{p}^{\prime}\right|^{2} \displaystyle  = \displaystyle  \gamma^{2}M^{2}\left(-1+v^{2}\right)
      \displaystyle  \displaystyle  = \displaystyle  -\frac{\gamma^{2}}{\gamma^{2}}M^{2}=-M^{2}


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