1. May 17, 2010

### Battlemage!

This is NOT homework (school is out), but it has the form of a problem because making up problems is how I "test" these ideas to convince myself they work. I tried the search function and couldn't find what I was looking for.

EDIT- I am not trying to "show my own theory," etc. When I have been learning special relativity I've tried to derive everything from scratch. I've got the Lorentz transformation and the energy equation, but I needed to START with the relativistic momentum equation in order to derive it, and I'm trying to learn WHY we use it, so that's what this thread is about.

So, here goes:

I want to show (to myself) that in special relativity, relativistic momentum is conserved. I am clearly misunderstanding something about how this works, because in one frame I am getting that it is conserved, but in another frame it is not(in S' the initial momentum ≠ the final momentum in my made up problem). I am getting that I have an additional absolute value of 5 kg c on my final momentum for S' than I had in my initial momentum.

There are two particles moving toward each other. They collied perfectly inelastically and form a third mass.

In frame S, particle A moves at u1 = 0.5c (in the positive x direction), and has a mass of 10 kg. Particle B moves at u2 = -.05c (in the negative x direction), and has a mass of 10 kg.

So, prior to collision, here is my data:

γ1 = (1 - (0.5)2)-1/2 = 1.1547
γ2 = (1 - (-0.5)2)-1/2 = 1.1547

So,

P1 = γ1(10 kg) (0.5c)

P2 = γ2(10 kg) (-.05c)

So P1 + P2 = 0

After collision:

PF = (γ1*m + γ2*m)( (0.5c - 0.5c)/(1 - (0.5)(-0.5)) = 0

So momentum is conserved in frame S.

Now, frame S' moves in the positive x direction at speed v = 0.9c

So u1' = (0.5c - 0.9c)/(1 - (0.5c)(0.9c)) = -0.7273c

u2' = (-0.5c - 0.9c)/(1 - (-0.5c)(0.9c)) = -0.9655c

γ1' = (1 - (-0.7273c)2)-1/2 = 1.457

γ2' = (1 - (-0.9655c)2)-1/2 = 3.840

So the initial momentum in S' is:

γ1'(10 kg)(-0.7273c) + γ2'(10kg)(-.9655c) =
(1.457)(10 kg)(-0.7273c) + (3.840)(10 kg)(-0.9655c) = -47.67 kg c

But the final momentum is:

1'*m + γ2'*m) ( (-0.7273c - -0.9655c)/(1 - (-0.7273)(-0.9655)) =

(1.457*10 kg + 3.840*10 kg) (-0.9945c) = -52.68 kg c

Now obviously -47.67 kg c ≠ -52.68 kg c. It's off by about 5 kg c.

Clearly I am not understanding how to calculate momentum in special relativity. If anyone wants to take the time to examine what I've done, please show me where I am messing up.

Again, this is NOT homework. This is just me trying to "see" that relativistic momentum is conserved for myself. But if it needs to be moved I don't mind, but I do apologize if it is in the wrong section.

Thank you for whoever tries to help me!

Last edited: May 17, 2010
2. May 17, 2010

### starthaus

What you are asking is not a trivial subject. It is covered in an excellent way by C.Moller (The Theory of Relativity) in no less than 4 chapters! (28-31). You will need to get the book (about 20\$ on Abebooks.com).

OK.

OK

You sure about this? The two masses now move together. What is their speed in the primed frame?

Last edited: May 17, 2010
3. May 17, 2010

### phyzguy

This is an interesting problem. The discrepancy is because you didn't calculate the final momentum in the primed frame correctly. After the collision in the primed frame, the two particles are moving together with a velocity of 0.9c. So their total momentum is m (0.9)/sqrt(1-0.9^2). But what value of m is correct to use? Since m is invariant, we can evaluate it in any frame, and it is easiest to evaluate it in the CM frame. Since energy is conserved, the total energy of the (now stationary) particles after the collision is 2*(10kg)*(1/sqrt(1-0.5^2), the same as it is before the collision. Basically their kinetic energy of motion has been converted into rest mass. So the final momentum in the primed frame is given by
p(final) = 2(10kg)(1/sqrt(1-0.5^2))(0.9)(1/sqrt(1-0.9^2)) = 47.67, the same as before the collision.

Last edited: May 17, 2010
4. May 17, 2010

### Battlemage!

Ha! I think I got it! Thanks a lot you guys (course I still could be off...)

I know that I made at least two mistakes.

(1) I should have just took the third velocity (which was 0) from the S frame and just used Einstein's velocity transformation to get that velocity in the S' frame, which is -0.9c.

(2) The masses did not add up to 20 kg in the S' frame after collision, because as was pointed out by phyzguy, some energy was converted into mass.

Instead what I did this time was I took the Energy of each particle in the S frame, added them together and divided by c2 to get the new third mass, which was 23.1 kg (11.55 kg + 11.55 kg).

So, here is the final result:

with u3' = -0.9c, then γ3' = 2.294

So the final momentum is:

(2.294) (23.1 kg) (-0.9c) = -47.68 kg c

which is pretty close to what I had before the collision with the two 10 kg masses (I'm assuming some rounding error).

... well, either that is right or I got really lucky.

Again, thank you all for your hints. If I am still missing it, feel free to correct me.

5. May 17, 2010

### starthaus

Yes, now it is correct. You can produce a fully symbolic solution if you keep in mind that:

$$\gamma(u_1)m_1u_1+\gamma(u_2)m_2u_2=\gamma(u)(m_1+m_2)u$$ (conservation of momentum)

$$\gamma(u_1)m_1c^2+\gamma(u_2)m_2c^2=\gamma(u)(m_1+m_2)c^2$$ (conservation of energy).

To the above, add that, in a primed frame, moving at V:

$$u_i=\frac{u'_i+V}{1+u'_i V/c^2}$$

with the immediate consequence:

$$\gamma(u_i)=\gamma(u'_i) \gamma(V) (1+u'_i V/c^2)$$

Substitute in the above, do a little algebra and you should be getting:

$$\gamma(u'_1)m_1u'_1+\gamma(u'_2)m_2u'_2=\gamma(u')(m_1+m_2)u'$$

$$\gamma(u'_1)m_1c^2+\gamma(u'_2)m_2c^2=\gamma(u')(m_1+m_2)c^2$$

Last edited: May 17, 2010
6. May 20, 2010

### yuiop

Yep, it is interesting. Presumably the additional rest mass (about 3.1kgs) would be observable by weighing the (very hot) combined mass after the collision.