Proper time and four-velocity
I remember how proper time and relativistic velocity had utterly confused me until 2020. Again, the difficulty lies in the correct physical interpretation and understanding of these definitions as opposed to imagination of spurious relationships, which is perhaps common in many who fail to understand special relativity.
First of all, why is proper time qualified with the word “proper”? To clarify this, we begin by defining a worldline of an object as the path that objects traces in 4-dimensional spacetime. If the object exists in reality, then its worldline must be timelike, i.e. , since no object with mass can reach or exceed the speed of light. Proper time is then naturally defined as the time measured by a clock following that line. Since proper time is restricted to information solely in the frame of reference of some world line, it must be a Lorentz scalar or Lorentz covariant/invariant as defined in .
As of , we derived the Lorentz transformation for two dimensions:
Inversion of it gives
Now we ask how to define based on this the relativistic velocity of an arbitrary object. At some point on the worldline of that object in the reference frame wherein that the object has instantaneous speed zero, i.e. , if perceived in another reference frame such that moves at an arbitrary velocity with respect to , the ordinary velocity of course per definition of the two frames. It does not give any additional information. However, we would interested in the rate of change in spacetime coordinates in the observer frame with respect to proper time, in this case, which is given by, noting how
Generalizing to four coordinates and using Einstein notation gives us for the four-velocity or relativistic velocity
thereby the magnitude of four-velocity is also a Lorentz invariant.
Lorentz covariance in Einstein notation
Take any Lorentz transformation and arbitrary four-vectors along with transformed values . In Einstein notation, the Lorentz invariance of the spacetime interval is expressed as
Note that in matrix form, it would be
wherein we multiply a row vector with a matrix with a column vector. However, in the tensor form of , we are multiplying a rank 2 covariant tensor (which is the collection of sums of tensor products of pairs of rank 1 covariant tensors) with the tensor product of two rank 1 contravariant tensors. Then, how are they equivalent? The answer is simply that in , the rank 2 covariant tensor corresponding to the matrix was defined with values such that upon its application to contravariant rank-1 tensors in Einstein notation form,
where are the values of expressed in column vector for. In the case that , the result is the entry of in its th row and th column. Thus, the values of are the values of the th row, th entry of its corresponding matrix. Under the definition of lowering of an index in the specification of Einstein notation, i.e. , in order for to hold in generality, .
Now going back to applying the tensor product identity
to it, we get
which tells us
Similarly, the invariance generalizes to the inner product as follows
Using this property, we can take inner products of four-vectors for different physical quantities which transform via equivalent Lorentz representations to derive more Lorentz invariants. I might write a subsequent article in which I take the inner product of the velocity four-vector with the yet to be defined four-momentum, four-force, and four-acceleration to derive more Lorentz invariants.