Surely fault about relativistic momentum

[BarbaraDav]

Dear Friends

Two identical particles are running, in opposite directions, along the x-axis of a reference frame having the origin coinciding with their barycentre; they hits frontally and glue together.

Their speeds are ("b" for before collision, "a" for after) :

V1b = u
V2b = -u
Va = 0

Let's watch the event from a reference frame moving toward right at speed u.

Relativistic speed composition gives:

V1b' = 0
V2b' = - 2u / ( 1 + u^2/c^2)
Va' = -u

This way the relativistic momentum is not conserved:

m ( 0 )+ m ( - 2u / ( 1 + u^2/c^2) ) != 2m ( -u )

Obviously, I'm wrong but really can't see what is going the bad way.
Please, can you give any hints ?

Best regards!

Barbara Da Vinci
Rome

 

[JesseM]

Relativistic momentum is not equal to mv, it's equal to $$mv\gamma$$, i.e. $$\frac{mv}{\sqrt{1 - v^2/c^2}}$$

 

[Entropia]

, Italy

Dear Barbara,

Thank you for sharing your thoughts on this topic. It is understandable that you may be confused about the conservation of relativistic momentum in this scenario. However, there are a few key concepts to keep in mind that can help clear up any misunderstandings.

Firstly, it is important to note that the concept of relativistic momentum is based on the principle of relativity, which states that the laws of physics should be the same in all inertial reference frames. This means that the laws of physics, including the conservation of momentum, should hold true regardless of the observer's frame of reference. In your example, you have correctly applied the relativistic speed composition formula to calculate the velocities in the moving reference frame. However, you may have overlooked the fact that the masses of the particles also change in the moving frame due to the relativistic effects of time dilation and length contraction. This means that the momentum equation should also take into account the changes in mass, as shown below:

m0 ( 0 )+ m0 ( - 2u / ( 1 + u^2/c^2) ) = 2m ( -u )

Where m0 is the rest mass and m is the relativistic mass of the particles. This equation shows that the total relativistic momentum is conserved in the moving frame, as it should be according to the principle of relativity.

Furthermore, it is worth noting that the scenario you described is not a physically possible situation. Gluing two particles together after a collision would result in a larger, more massive particle with a different rest mass and therefore a different relativistic mass. This would affect the calculation of the relativistic momentum in the moving frame.

I hope this helps to clarify the concept of relativistic momentum and its conservation in different reference frames. Keep exploring and questioning, and you will continue to deepen your understanding of this fascinating topic.

Best regards,