Published March 16, 2023 | Version 1

Special Relativity, Legendre Transformations and Fluctuations

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  Momentum and velocity are two well-known physical observables in classical physics (and quantum physics as well because quantum particles have impulse and move through space in a certain amount of time based on an external clock and ruler). They are, however, measured in different ways. Velocity describes motion through x in some time t, while p is measured through impulses. Thus one does not need a clock to measure momentum. It may be measured in terms of the damage done to a foil, compression of a spring etc. Thus p=m1v1=m2v2 etc., i.e. a numerical number associated with momentum does not automatically yield the velocity unless one knows rest mass. In many situations rest mass is known (i.e. one deals with an electron etc) and so there seems to be a tendency to consider momentum and velocity to be almost the same thing when they are distinctly different.

    In fact, we argue that the notion of wavefunction W(x) and density P(x)=W*(x)W(x) are associated with momentum and velocity respectively. In fact, for a free particle these two measurables are linked with completely different spatial probabilities. Constant velocity impies P(x)=constant while exp(ipx) describes the two-dimensional x-probability associated with p which is based on a constant mo and v.

    In special relativity for a free particle (i.e. p constant, v constant), there are two well known Lorentz invariants:  EE = pp + momo  (c=1) and  A = -Et + px. One may note that v does not appear explicitly even though one may write E and p in terms of v i.e. E = mo/sqrt(1-vv) and p=Ev. The first invariant describes E in terms of rest mass and p (measured through impulse). The second involves p,E and x,t with x and t appearing as independent even though ultimately x(t) and v=dx/dt=constant = x/t if x and t both start at 0. This independence is crucial because it means one may consider fluctuations which keep A constant separately. For example, if t→ t+ hbar/E and x→ x+hbar/p, then A remains constant. This leads to the idea of the quantum wavelength and a spatial probability independent of t, namely exp(ipx). One average, however, x=vt, but how does this appear in Lorentz invariants? There is no explicit v. 

   We argue that the Lorentz invariant A = -Et + px must convert E from a function of p to one in v and that this occurs through a Legendre transformation. Thus the form of the invariant is itself a Legendre transformation with E being transformed into L, the Lagrangian. There is, however, a second important feature because p=dL/dv (partial) and given that p is constant, dp/dt=0 so d/dt dL/dv partial =0 which is in the form of an extremization. We argue that this is not an accident. dL/dv=p follows from E being constant, so, in other words, it follows from the px term. This same term leads to exp(ipx) which means that one does not have smooth x=vt motion within a wavelength. Thus, we argue that when one transforms to the second measurable v, the notion of this fluctuation behaviour must also be manifested. In other words, v=constant or dv/dt is not enough. The extremization behaviour of the Lagrangian, we argue, is thus not a mathematical convenience, but gives an indication that even though the extreme path x=vt is followed on average, other paths are sampled. We argue that this idea is incorporated into special relativity, in fact within a single Lorentz invariant  A = -Et + px.


 

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