New Transformation Equations and the Electric Field Four-vector

9. Transformation of Velocity

A body is moving with uniform four-velocity, , as measured by an observer at rest in the primed frame. The primed frame, in turn, is moving with uniform four-velocity, , relative to the unprimed frame. We wish to find the four-velocity, , of the body as measured by an observer at rest in the unprimed frame. Since the components of all four-vectors must transform like the coordinates of an event in spacetime, we can use (6.4) to find the components of . We replace the coordinates and , in (6.4), with the components of the velocities, and , respectively. We then find the components of the four-velocity of the body, as measured by the unprimed observer, to be

If the four-velocity of the primed frame is , and the four-velocity of the body relative to the primed frame is, , the components of the four-velocity of the body relative to the unprimed frame, using (6.4), are and

At spatial velocities much less than c, we have , and we get the Galilean result . The signs in the previous sentence are "approximately equal to", not "equal to". We see that, for low velocities, most of the four-velocity is still in the time direction in both primed and unprimed frames of reference. Incidentally, any time the magnitude of the spatial part of a four-velocity is less than c, we will also have a time component, and vice versa.

For velocities in transverse directions, for example and , we have

For combined velocities along the x-axis at the speed of light, and , we get , but

In the unprimed frame, there are no spatial components in the transformed four-velocity, , but there is a component directed along the negative t-axis. The magnitude of , however, is still

Its direction is rotated 180 degrees relative to the direction of the four-velocity of the unprimed frame, which we will call , where , since it is "at rest" in space, but not in time.

For spatial velocities at the speed of light in transverse directions, for example and , we have and

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