# Total linear momentum does not conserved as viewed from other frame?

1. Apr 4, 2013

### m2sg

Relativitic momentum p =γ m v

Consider explosion happens with two particles of equal mass m moving in opposite directions with the same speed. So, the total momentum before and after explosion is zero and conserved.

However, if I try to calculated the total momentum before and after explosion as veiwed from another intial frame of reference moving with speed v, they are not conserved!

Before explosion, p = (m+m)v
After explosion, p = p1 + p2 = γ1 m v1 + γ2 m v2

v1 is the velocity of m1 as viewed from another frame
v2 is the velocity of m2 as viewed from another frame
γ1 is 1/(1-v1^2/c^2)^0.5
γ2 is 1/(1-v2^2/c^2)^0.5

My calculation shows the total linear momentum does NOT conserved! Why?
Thank you.

2. Apr 4, 2013

### Mentz114

You are mixing quantities in different frames. Choose one frame and analyse the problem from that frame.

http://www.phy.duke.edu/courses/100/lectures/Rel_2/Rel2.html [Broken]

Last edited by a moderator: May 6, 2017
3. Apr 4, 2013

### m2sg

I did choose one frame. v1 and v2 are the quantities as viewed from the chosen frame.

4. Apr 4, 2013

### Mentz114

OK. But ( as the article explains ) 3-momentum is not conserved. The conserved quantity is energy-momentum.

5. Apr 4, 2013

### Staff: Mentor

Shouldn't this be p = γ(m+m)v?

How did you find v1 and v2?

6. Apr 4, 2013

### Staff: Mentor

Your calculation has several problems. First, you missed the gamma before the explosion. Second, the mass of each of the post-explosion particles should be less than half of the mass of the pre-explosion particle. Third, your calculation doesn't actually show any non-conservation since you never calculated v1 and v2.

The best way to do this calculation is using four-vectors. That guarantees conservation.

7. Apr 4, 2013

### m2sg

Thank you very much for your replies.

DaleSpam and jtbell,

You are right. It was typo.
Before explosion I actually calculated p with γ, that is p= γ(m+m)v

I found the v1 and v2 using the usual Lorentz velocity transformation equations.

The mass before explosion is m + m
The mass after explosion is m1=m and m2=m.

But, the total p does NOT conserved.

8. Apr 5, 2013

### m2sg

Let the frame S' be the total p = 0 before and after explosion. Let v = 0.4 c be the speed the frame S' moved relative to frame S.
Now, I would to check is the total p conserved viewed in frame S.
Let v1' = -0.5 c be velocity of m1=m in frame S'; v2' = +0.5 c be velocity of m2=m.
Using v1 = (v1' + v)/(1+v1'v/c^2), I got the velocity of m1 relative to frame S, v1 = -0.125 c.
Using this v1, the momemtum p1 = γ m1 v1 = 1/(1-v1^2/c^2)^0.5 m1 v1 = -0.126 m c.
Simliarly, v2 = (v2' + v)/(1+v2'v/c^2) = 0.75 c; p2 = γ m2 v2 = 1/(1-v2^2/c^2)^0.5 m2 v2 = 1.134 mc
So, after explosion total p = -0.126 mc + 1.134 mc = 1.008 mc
However, before explosion total p = γ(m+m)v = 1/(1-v^2/c^2)^0.5 (m+m) v = 0.873 mc !

9. Apr 5, 2013

### Staff: Mentor

This is wrong. Total energy (kinetic + mass) is not conserved in the center of momentum frame when you do this. Because of how energy and momentum transform (components of the same four vector) this mistaken non conservstion of energy in one frame leads to a failure of momentum conservation in other frames.

10. Apr 5, 2013

### Staff: Mentor

In other words: in the reference frame in which the two masses are initially stationary, there is no kinetic energy before the "explosion". Afterwards, the masses have kinetic energy. This must come from some of the original rest-energy E = mc2 of the two masses. Therefore their rest-energies afterwards must be smaller, along with their masses.

For a similar example, consider the decay of a lambda baryon to a proton and a pion:

$$\Lambda^0 \rightarrow p + \pi^-$$

Look up the masses of these particles, and compare the mass of the lambda to the sum of the masses of the proton and pion.

Last edited: Apr 5, 2013
11. Apr 6, 2013

### m2sg

Thank you, DaleSpam and jtbell for pointing out my misconception of assuming the masses before and after explosion.
So, I can use the conservation of total energy in S' frame to find the rest mass M before explosion.
total energy before explosion = after explosion
γ M c^2 = γ1 m1 c^2 + γ2 m2 c^2
With v = 0.4 c, I found M = 2.3094 m.
Since the rest M does not changed with speed, M is constant independent of frame.
Now, before explosion total p = γ(M)v = 1/(1-v^2/c^2)^0.5 (M) v = 1.008 mc. Yes, the total p is now conserved as viewed from other frame.
Is the way I reached the conservation of momentum correct? Thank you.

12. Apr 6, 2013

### Staff: Mentor

Yes, it is correct in principle, but I think there is a small numerical mistake. I get M = 2.18 m for v = 0.4 c.

There is another way to reach conservation of momentum which is also correct but is quite a bit easier. This is by using the four-momentum:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html
http://en.wikipedia.org/wiki/Four-momentum

If you take this then a particle of mass m at rest has a four-momentum of (m,0,0,0) in units where c=1. That same particle, travelling at .4 in the x direction has a four-momentum of (1.09 m, 0.44 m, 0,0) and travelling at .4 in the -x direction it has a four-momentum of (1.09 m, -0.44 m, 0,0). Adding those two gives (2.18 m, 0,0,0), which is the four-momentum of a particle of mass M = 2.18 m at rest.

13. Apr 8, 2013

### m2sg

Thank you for introducing me the 4-vector of energy-momentum and the illustrative example.
The v in the first link is correct, which v = 0.4 c. However, when calculating the 4-vector for the 2 particles after explosion the v was actually v1 = 0.5c and v2 = -0.5c. Hence, the 2 4-vectors are (1.155, 0.5774, 0, 0) and (1.155, -0.5774, 0, 0). Adding these two gives (2.31, 0, 0, 0), which is of mass M = 2.31 m.
Yes, using the given transformation in the link it also shows total p is conserved as viewed from other frame. Thanks.

14. Apr 8, 2013

### Staff: Mentor

Ahh, my misunderstanding, sorry. My numbers agree with yours now.