THE INFINITESIMAL TRANSFORMATIONS OF LORENTZ

(in the case of a moving body)

Let us consider two frames S and S_1 in relative motion to each other, which coincide when the times t and t_1 are equal to 0 (i.e. the instants of time in which the frames are coincident constitute the same initial time instant for both) and with the x axis and x_1 axis always superimposed.

In particular, reference is always made to the well-known case in which the origin O_1 of the frame S_1 moves away from the origin O of the frame S. (but with a velocity v which, even if it has only a horizontal component along the abscissa axis x, is not always constant)

Now let’s imagine that in the origin of the frame S_1 a body is positioned which, starting from the origin of the frame S (at the zero initial time t = 0 and at zero initial speed), arrives at a point of the x axis of positive abscissa d with respect to the frame S.

For the frame S, the frame S_1 at time t is in motion at speed v(t) while, for the frame S_1, the frame S at time t_1 is in motion at speed v_1(t_1).
We thus obtain that v(t) = v_1(t_1) (in modulo and for t_1 which corresponds to the travel time measured in S_1 when for the frame S the measured travel time is t) and the two Lorentz factors γ(t) and γ(t_1) coincide.

From the Lorentz transformation x_1 = γ*(x -v*t), for x_1=0 as the body is positioned in the origin of the frame S_1, we obtain x -v*t=0, from which x=v*t. Turning to the differentials we obtain the relation dx = v(t)*dt.
From the other Lorentz transformation x = γ*(x_1+v*t_1), again for x_1=0, we verify that x = γ*v*t_1 and passing to the differentials we obtain the following relation dx = γ(t_1)*v _1(t_1)*dt_1.

The last equation is correct, in fact in the meantime dx_1 = v_1(t_1)*dt_1; for the frame S_1 the infinitesimal distance dx is in motion at speed v_1(t_1) and, in the frame S_1, it is less than dx and measures dx/γ (t_1). From dx_1=v_1(t_1)*dt_1 and from dx_1 = dx/γ(t_1) we get dx = γ(t_1)*v_1 (t_1)*dt_1.
From dx = γ (t_1)*v_1(t_1)*dt_1 we then pass to

v(t)*dt = γ(t_1)*v_1(t_1)*dt_1; finally it is possible to obtain the known result dt_1 = γ^(-1)(t)*dt, and this is true since v(t) = v_1 (t_1) and γ (t)=γ (t_1).

Since dt_1 < dt, proceeding to integrate we also obtain t_1 < t. (the body in motion is “younger” than all the other stationary bodies in the frame S)

dx_1 is less than dx, also x_1(t_1) < x(t) and this is what causes the symmetry to break.

The infinitesimal time dt is the time of the body to travel the infinitesimal distance dx as measured in the frame S, the infinitesimal time dt_1 instead represents the time for the distance dx_1 (dx in motion) “overtakes” the origin of the frame S_1. In the frame S_1, dt_1 therefore corresponds to the infinitesimal travel time of the body which, even if stationary in its frame, waits for the infinitesimal distance dx_1 to meet it with the opposite speed to its own.

For accuracy, if the abscissa d is positive, in the two Lorentz transformations examined the two speeds v(t) and v_1(t_1) are also to be considered as positive values, even the exact expression of the differential dx_1 is

dx_1 = -v_1(t_1)*dt_1. In the demonstration shown dx_1 an infinitesimal distance (and therefore positive) has been considered, also pay attention to the signs of the various quantities in case the body goes back.

It should also be considered that if the body crosses areas where gravitational fields are particularly intense, it will be even “younger” than all the bodies that remain stationary in the frame S, but ideally it is always possible to consider a point-like body of mass negligible in motion (regardless of the gravitational forces acting on massive bodies).

As regards the distance between the two origins of the frames S and S_1 measured at the instant of time t_1 by an observer in the frame S_1, it, indicated with d_1m (t_1), at the instant of time t_1 is equal to x(t)*γ^(-1)(t_1) or d_1m (t_1) = x (t)*γ^(-1) (t).

The last relation is important as it also means that, in the case where the speed of the body is equal to zero, d_1m(t_1)= x(t). When the body is stationary in the frame S all distances have the same measure for all observers of the frame itself. The now stationary body will therefore seem to have traveled a distance equal to d, although this does not correspond to the actual distance traveled from the origin of the frame S with respect to the origin of the frame S_1.

In summary, the useful relationships are:

dx = v(t)*dt

dt_1 = γ^(-1)(t)*dt

dx = γ(t_1)*v_1 (t_1)*dt_1

dx_1 = dx/γ(t_1)= dx/γ(t) = v_1(t_1)*dt_1

d_1m(t_1) = x(t)*γ^(-1) (t) (the only one in which there are no differentials).

This discussion also applies in the event that the body goes back along a further distance d to return to the origin of the frame S with zero final speed. (situation that represents the so-called “twin paradox”)

Although the relation dx = v(t)*dt is trivial at first sight, it is important because it states that there is a body in motion within the frame S. (for example the motion of a body in the Earth’s frame or better in the frame of the fixed stars). If a body is moving within a frame S and travels a distance d, in the body’s frame S_1 it is not only the origin O that moves.(but it is the whole frame S that “travels” a distance less than d)

Remember that the frame S in which the movement takes place is not a privileged frame, it is the frame in which the body at the end of the journey is “younger” than the other bodies that have always remained stationary in the frame S.

And, as has been shown in this article, the Lorentz transformations are sufficient to prove this simply and elegantly.

Massimiliano Dell’Aguzzo

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