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.(adsbygoogle = window.adsbygoogle || []).push({});

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 momentumin 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.

Please help me see what I'm doing wrong.

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

In frame S, particle A moves at u_{1}= 0.5c (in the positive x direction), and has a mass of 10 kg. Particle B moves at u_{2}= -.05c (in the negative x direction), and has a mass of 10 kg.

So, prior to collision, here is my data:

γ_{1}= (1 - (0.5)^{2})^{-1/2}= 1.1547

γ_{2}= (1 - (-0.5)^{2})^{-1/2}= 1.1547

So,

P_{1}= γ_{1}(10 kg) (0.5c)

P_{2}= γ_{2}(10 kg) (-.05c)

So P_{1}+ P_{2}= 0

After collision:

P_{F}= (γ_{1}*m + γ_{2}*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 u_{1}' = (0.5c - 0.9c)/(1 - (0.5c)(0.9c)) = -0.7273c

u_{2}' = (-0.5c - 0.9c)/(1 - (-0.5c)(0.9c)) = -0.9655c

γ_{1}' = (1 - (-0.7273c)^{2})^{-1/2}= 1.457

γ_{2}' = (1 - (-0.9655c)^{2})^{-1/2}= 3.840

So the initial momentum in S' is:

γ_{1}'(10 kg)(-0.7273c) + γ_{2}'(10kg)(-.9655c) =

(1.457)(10 kg)(-0.7273c) + (3.840)(10 kg)(-0.9655c) =-47.67 kg c

But the final momentum is:

(γ_{1}'*m + γ_{2}'*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!

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# Relativistic Momentum: Please help me to see that it is conserved.

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