Please wait for the animation to completely load.
Momentum, defined in introductory-level physics as mv, needs a revised definition to account for special relativity. To understand why, consider the animation, which shows you a collision between two objects in a laboratory reference frame and in the reference frame of the blue ball. Restart. The switch between reference frames requires a Lorentz transformation from the lab frame to another reference frame (the blue frame). The blue frame is moving at a speed u along the x axis with reference to the lab frame, and the velocity, v, in the lab frame is seen as v' in the blue frame as follows:
vx' = ( vx − u ) / ( 1 − vx u / c2 ), vy' = vy / γ ( 1 − vx u / c2 ) and vz' = vz / γ ( 1 − vx u / c2 ). (3.1)
Since the blue frame is moving to the left at a speed of v relative to the lab frame (u = −v), the transformation equations give the following velocities for the collision of two balls headed toward each other with speed v in the lab frame:
|Blue Reference Frame|
|Before Collision||After Collision|
|Green||vx'green||2v/ (1 + v2/c2)||v|
|vy'green||0||v (1 − v2/c2)1/2|
|vy'blue||0||−v (1 − v2/c2)1/2|
The animation shows the values for the different incident speeds. If you simply use p = mv, momentum is not conserved. Momentum conservation is fundamental, so we redefine momentum (and this is the value that is displayed in the table when you check the "show the relativistic momentum" check box):
p = m0v / ( 1 − v2 / c2 )1/2 (3.2)
where v is the velocity of the particle as measured in a particular reference frame, not the relative velocity of the reference frame (which is why we do not write v2/c2 as β2; β2 is instead u2/c2). Notice that at low enough speeds, the relativistic momentum and non-relativistic momentum give the same values. It is only for speeds near the speed of light that there is a noticeable difference between the two values.
© 2006 by Prentice-Hall, Inc. A Pearson Company