Re: [Phys-L] momentum conservation, particle annihilation, different reference frames

[John Denker <jsd@av8n.com>]

On 11/26/2014 07:01 PM, Philip Keller wrote:
>  
> But wasn't the momentum of the two anti-particles 2mv before the annihilation?

My bad.  I skipped too many steps.  Let's start over:

You can make it work in any frame if you use the full
four-dimensional momentum.

For the incident particles, you can write the momentum as
   p  =  m u

where the 4-velocity is
   u  =  dR / dτ

where τ is the proper time.  That is not to be confused
with the 3-velocity in some frame
   v  =  dR / dt

where t is the wall-clock time, i.e. the projection of
time onto that frame.

Let's assume |v| is small compared to c, so we can neglect
the KE of the incident particles.  Then to a sufficient 
approximation, the 4-velocity of each fuel particle is
    u[f]  =  [ 1  v_1  0  0 ]

The 4-velocity is almost entirely in the timelike direction.
It is proceeding toward the future at a rate of 60 minutes
per hour.

The 4-momentum is m times that, i.e.
    p[f]  =  m [ 1  v_1  0  0 ]

The 4-momentum of the two fuel particles together is
    p[2f]  =  2m [ 1  v_1  0  0 ]

After the annihilation, each photon has 4-momentum
    p[γ]  =  m [ 1  v_1  0  ±w ]

where w is on the order of 1, huge compared to v_1.
Each photon is massless.  It's momentum is not in the
timelike direction, but rather in the lightlike (null)
direction.

We could evaluate w exactly, but we don't care, because 
it is going to drop out of next equation.

If you add the two photons together, you get
   p[2γ]  =  2m [ 1  v_1  0  0 ]

which is exactly the same as before, the same as p[2f].
Duh.  Momentum is conserved.  Each of the four components
of the momentum vector is separately conserved.  You
can see by inspection that the mass of the /pair/ of
photons is the same as the mass of the pair of fuel
particles.  Remember that p•p = -m^2.  The pair has
mass even though either photon separately is massless.

Conservation of momentum guarantees that you cannot
possibly change the mass of the spaceship unless you
throw something overboard.  Annihilations are no
exception.

To say the same thing another way:  You learned in 
HS chemistry that the mass of a mixture is equal to
the mass of all the ingredients ... but that's not
true in special relativity, even when the ingredients
are non-interacting, as we see in the two photons here.
For details, see
  https://www.av8n.com/physics/spacetime-welcome.htm
especially
  https://www.av8n.com/physics/spacetime-welcome.htm#sec-invariance-conservation



ps FWIW:  The spatial part of p[2γ] is "almost" zero
i.e. negligible compared to the spatial part of p[γ].
That is mostly what I was saying in my previous note.