Four-Vector Kinematics

Back to Contents

Imagine occupying an inertial frame of reference, a frame that neither accelerates nor rotates. Two events occur at two separate places and two separate times. Because we can establish a coordinate grid anywhere in that frame, we choose, for convenience, to place it such that the first of the events occurs at the origin. In that case the four-dimensional vector describing the distance and duration between the two events consists simply of the coordinates of the second event. Measuring from one event to the other parallel to the axes of the Cartesian coordinate grid gives us

(Eq’n 1)

If some particle or body can come arbitrarily close to both events, then there exists an inertial frame in which both events occur at the same point in space. In that frame, which we call the proper frame, the vector pointing from one event to the other takes the form

(Eq’n 2)

Minkowski’s theorem relates those two equations to each other through their squares;

(Eq’n 3)

That equality expresses a theorem analogous, in four dimensions, to the Pythagorean theorem. That fact means that we can convert one of the vectors of Equations 1 and 2 into the other by way of the Lorentz Transformation and that fact, in turn, marks the 4-plexes of Equations 1 and 2 as proper 4-vectors.

If we specify that the proper frame moves through the improper frame at a speed vx in the two frames’ mutual positive x-direction, then we can rewrite Equation 3 as

(Eq’n 4)

(Eq’n 5)

in which we have

(Eq’n 6)

as the Lorentz factor between the two frames.

If a particle touches both of the above defined events without accelerating as it travels between them, then we can assign to it a uniform velocity,

(Eq’n 7)

That expression represents the classical vector velocity of Newtonian mechanics, but it doesn’t have the form of a proper relativistic 4-vector, even if we add a temporal component. But if distance and duration can combine into a 4-vector, then surely their ratio can do so as well. In order to stand as a proper 4-vector, whatever we devise as the 4-velocity must have an invariant square in accordance with Minkowski’s theorem.

If we differentiate something that remains invariant under the Lorentz Transformation with respect to something else that remains invariant under the Lorentz Transformation, then we get something that also remains invariant under the Lorentz Transformation. So if we differentiate the square of the distance-duration 4-vector with respect to the proper time, then we get the inner (dot) product of the distance-duration 4-vector with a 4-component velocity vector;

(Eq’n 8)

But we have a theorem that tells us that the inner product of any two 4-vectors remains invariant under the Lorentz Transformation, so we must infer that the velocity vector in Equation 8 represents a proper 4-vector. To test that proposition we need only square the presumed 4-velocity to see whether it yields an invariant. In that calculation we get

(Eq’n 9)

In that calculation I had to make the velocity in the Lorentz factor equal to the magnitude of the three spatial components of the measured velocity as calculated through the Pythagorean theorem.

That analysis applies to the velocity of one inertial frame with respect to one that we designate as the proper frame. But we assign velocities primarily to particles and bodies, so now we want to ask whether those give us proper 4-vectors. Certainly an unaccelerated body occupies and marks an inertial frame of reference, so it does have a proper 4-velocity. But we don’t want to bring a third inertial frame into our analysis, so we consider the body moving in the two frames that we analyzed above. In the proper frame we measure the velocity of the body as u’=(u’x,u’y,u’z) and in the improper frame we have u=(ux,uy,uz).

As measured from the proper frame, the 4-velocity of the body takes the same form as does the inertial frame it occupies,

(Eq’n 10)

in which we have the Lorentz factor between the proper frame and the body’s frame as

(Eq’n 11)

Now how would an observer occupying the improper frame write that 4-velocity?

We should start by applying the Lorentz Transformation to the differentials that go into our calculations of the measured velocities;

(Eq’ns 12)

Taking the appropriate ratios gives us then,

(Eq’ns 13)

Those components should fit into Equation 10 with an appropriate Lorentz factor multiplying them, in this instance the Lorentz factor between the improper frame and the body’s frame. We have

(Eq’n 14)

We that Lorentz factor we can write the 4-velocity of the body as seen from the improper frame,

(Eq’n 15)

Squaring that 4-vector yields -c2, as we require, whether we express that 4-velocity in terms of u or in terms of u’ and vx.

Thus we obtain our description of measured velocity and 4-velocity and of the relationship between them. Now we want to do the same for acceleration.

Inertial frames don’t accelerate, so we can’t apply a simple geometric analysis to determining whether we can find an acceleration 4-vector (though in General Relativity we do deal with accelerating frames of reference, but that give us an entirely different problem). We obtain one clue as to what we need by differentiating Equation 9 with respect to the elapse of proper time,

(Eq’n 16)

That equation tells us that in four dimensions the presumed 4-acceleration stands perpendicular to the associated 4-velocity.

Imagine that a body under acceleration exists, temporarily, in the proper frame. Observers in that frame obtain from their measurements of the body’s locations at various times an acceleration of a’=(ax’,ay’,az’). Observers in the improper frame will measure a different acceleration, which we can calculate by differentiating the body’s velocity with respect to time. Those observers have dt=γdt’, so they calculate

(Eq’n 17)

Note that at the instant in question ux=vx. If the observers multiply that equation by the relativistic mass of the body, they will calculate the relativistic force acting on the body.

We can’t transform that acceleration into a proper 4-vector by adding a temporal component to it, as we did with velocity. Instead, in order to get a proper invariant, we must differentiate the 4-velocity measured in the improper frame (which I am now using as if it were a proper inertial frame, a ploy that underscores the meaning of relativity) with respect to the time that has the same value for all observers, the time measured at the accelerating body. Because we will have to differentiate the Lorentz factor associated with the body, we start with

(Eq’n 18)

Now we differentiate the 4-vector and get

(Eq’n 19)

That has the appropriate form for a 4-vector. To see whether it transforms properly under the Lorentz Transformation we square it to see whether we get an invariant. We get that square as

(Eq’n 20)

We have an invariant in that result if and only if

(Eq’n 21)

That criterion doesn’t conform to Equation 17. If we multiply the right side of that equation by the Lorentz factor and square the result, we don’t get a number independent of the Lorentz factor, so we don’t get an invariant. That means that {A} does not represent a proper 4-vector. That means, in turn, that we cannot legitimately apply the Lorentz Transformation to the measured acceleration. Equation 17 alone gives us the relativistic transformation of acceleration.

Thus we have the basic relativistic kinematics. On this foundation we can now build a description of relativistic dynamics. That construction provides the subject of the next essay in this series.

eabf

Back to Contents