How Does Relativistic Two-Way Momentum Appear to Different Observers?

[Dale]

You could only measure momentum by its true relative velocity to the object it is traveling relative to in the collision.

No, you can measure momentum relative to any frame you like. That is the whole point of the principle of relativity (whether Galilean relativity in Newtonian mechanics or special relativity in Einstein's mechanics).

In each frame the momentum will be different, but in each frame using the momentum for that frame will give you the correct answers in that frame for the collision.

 

[John232]

No, you can measure momentum relative to any frame you like. That is the whole point of the principle of relativity (whether Galilean relativity in Newtonian mechanics or special relativity in Einstein's mechanics).

In each frame the momentum will be different, but in each frame using the momentum for that frame will give you the correct answers in that frame for the collision.

Really? So there is a parallel universe that pops up that shows different results for a collision? I still don't see how you could use a different frame to calculate momentum correctly, the whole reason for this discussion. Could you show an example where the collision is the same in two separate frames of references?

 

[Dale]

Really? So there is a parallel universe that pops up that shows different results for a collision?

Nonsense.

I still don't see how you could use a different frame to calculate momentum correctly, the whole reason for this discussion. Could you show an example where the collision is the same in two separate frames of references?

Sure, work any collision problem you like in any frame you like and post the results, and I will transform it to another frame and post the results.

 

[John232]

Sure, work any collision problem you like in any frame you like and post the results, and I will transform it to another frame and post the results.

Newton hits a bowling ball that weights 1kg with another bowling ball that weights 1kg. One is at rest on a table, and the other is moveing at 1 m/s. The ball at rest then gets hit and then moves 1m/s in the opposite direction.

 

[lugita15]

Newton hits a bowling ball that weights 1kg with another bowling ball that weights 1kg. One is at rest on a table, and the other is moveing at 1 m/s. The ball at rest then gets hit and then moves 1m/s in the opposite direction.

I think you mean the same direction; in frame S1 initially ball A is moving to the right at 1m/s, and it collides elastically with the stationary ball B, so then ball A comes to a stop and ball B moves to the right at 1m/s. Now consider frame S2, which is moving at 3m/s to the left with respect to frame S1. In frame S2 initially ball A is moving at 4m/s to the right and ball B is moving at 3m/s to the right, and they collide elastically, so then ball A is moving at 3m/s to the right and ball B is moving at 4m/s to the right.

 

[John232]

I think you mean the same direction; in frame S1 initially ball A is moving to the right at 1m/s, and it collides elastically with the stationary ball B, so then ball A comes to a stop and ball B moves to the right at 1m/s. Now consider frame S2, which is moving at 3m/s to the left with respect to frame S1. In frame S2 initially ball A is moving at 4m/s to the right and ball B is moving at 3m/s to the right, and they collide elastically, so then ball A is moving at 3m/s to the right and ball B is moving at 4m/s to the right.

I meant the opposite side of the ball. So then how did you find that an object traveling at 4m/s would only transfer 1 Newton to the other object that is traveling at 3m/s?

 

[Dale]

Newton hits a bowling ball that weights 1kg with another bowling ball that weights 1kg. One is at rest on a table, and the other is moveing at 1 m/s. The ball at rest then gets hit and then moves 1m/s in the opposite direction.

Since you didn't work the problem out you made a mistake. The other ball will move in the same direction as the original ball. But anyway, consider the frame where the other ball is at initially at rest.

One ball is moving at -1 m/s and is on a table which is also moving at -1 m/s, the other ball is at rest. The ball at rest gets hit and then moves at -1 m/s.

EDIT: lugita15's analysis also works (and is more interesting), the point is that you can use any frame you feel like using.

 

[lugita15]

I meant the opposite side of the ball. So then how did you find that an object traveling at 4m/s would only transfer 1 Newton to the other object that is traveling at 3m/s?

They're transferring momentum, measured in kg m/s, not force, measured in kg m/s^2 or Newtons (although they presumably will exert forces on each other while they are colliding). The momentum transferred between the two balls is (1 kg)(4m/s)-(1 kg)(3m/s) = 1 kg m/s.

 

[John232]

They're transferring momentum, measured in kg m/s, not force, measured in kg m/s^2 or Newtons (although they presumably will exert forces on each other while they are colliding). The momentum transferred between the two balls is (1 kg)(4m/s)-(1 kg)(3m/s) = 1 kg m/s.

Sounds a lot like what I was saying before. You have to find the difference in the two objects themselves to find the true transfer of momentum. That is what you just did by doing that calculation.

 

[lugita15]

Sounds a lot like what I was saying before. You have to find the difference in the two objects themselves to find the true transfer of momentum. That is what you just did by doing that calculation.

Didn't I just cover this in post #28?

 

[John232]

Didn't I just cover this in post #28?

I thought I did in my quote on post #28. I didn't see anything in the posted link that would show it would be used in a tranfer of momentum. It simply doesn't have some key variables in it to determine that. I think if it was used to determine a transfer of momentum by adding the velocities in this way it wouldn't give the correct value that an observer would get being at rest relative to one of the two objects. The addition of velocity formula in SR wouldn't be used to determine the transfer of momentum because the added distance between two objects is arbritray to the calculation. The two objects would be hitting each other not moveing away from each other closer to the speed of light.

 

[Dale]

I think if it was used to determine a transfer of momentum by adding the velocities in this way it wouldn't give the correct value that an observer would get being at rest relative to one of the two objects. The addition of velocity formula in SR wouldn't be used to determine the transfer of momentum because the added distance between two objects is arbritray to the calculation. The two objects would be hitting each other not moveing away from each other closer to the speed of light.

In SR it is actually more convenient to work with momentum than with velocity. Momentum adds like normal in SR, and it is only the relationship between momentum and velocity which is non-linear. The conservation of the 4-momentum encapsulates the conservation of energy, conservation of mass, and conservation of momentum into one nice convenient mathematical package.

 

[Dale]

I meant the opposite side of the ball. So then how did you find that an object traveling at 4m/s would only transfer 1 Newton to the other object that is traveling at 3m/s?

The same way you found it in the original. Conservation of KE for an elastic collision.

 

[John232]

It doesn't seem like it would be that convenient to work to me.

I got Ʃ F = m1(v1'-v1°)/(t√(1-(v1'+v1°)^2/4c^2) + m2(v2'-v2°)/(t√(1-(v2'+v2°)^2/4c^2)

 

[Dale]

Use the four-momentum instead. That is what I was referring to above.

 

[John232]

Use the four-momentum instead. That is what I was referring to above.

I would but I wouldn't even know what to put into that thing. I figured if you found one dimension you would have them all. What is a U^μ? The collision i gave was two vectors coming at each other, different velocities would have different vectors.