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Derivation of classical limit in c = 1 units (i.e. without cheating)

  1. Jun 1, 2010 #1
    So, we start with special relativity in c = 1 units. The equations for energy and momentum can be written as:

    [tex]E = E_{0} \gamma[/tex]

    [tex]\vec{P} = E_{0} \gamma \vec{v}[/tex]

    where

    [itex]E_{0}[/itex] is the rest energy,

    [tex]\gamma = \frac{1}{\sqrt{1-v^2}}[/tex]

    and [itex]\vec{v}[/itex] is the (dimensionless) velocity.


    We now want to study the physics of extremely massive objects (i.e. objects with a large rest energy) that move extremely slowly. In particular, we are concerned with describing collisions between such objects. Now, the trivial way of looking at that would be to say that nothing happens, everything stays at rest in the limit that the speeds go to zero. In a collision the speeds change by extremely small amounts that go to zero precisely in the limit we are interested in.

    A more interesing perspective is obtained by zooming in into the low velocity world. As we approach the limit, we fast forward what we're seeing so that everything remains visible. This means that we have to rescale certain variables in an appropriate way. Since we're letting the velocities go to zero we can define a new rescaled velocity:

    [tex]\vec{w}=c\vec{v}[/tex]

    Here c is a dimensionless rescaling variable. We let the velocities go to zero while keeping the rescaled velocity variables finite. Put differently, this amounts to putting v = w/c and letting c go to infinity while keeping w finite. Rewriting the equations for energy and momentum by eliminating [itex]\vec{v}[/tex] in favor of [itex]\vec{w}[/itex] and working to the lowest order in which the velocity dependence appears, gives:


    [tex]E = E_{0} \left(1 + \frac{w^{2}}{2c^{2}}\right)[/tex]

    [tex]\vec{P} = E_{0} \frac{\vec{w}}{c}[/tex]

    So, we're simply omitting higher order terms instead of including cumbersome O((w/c)^3) or O((w/c)^4) terms that will drop out in the end anyway.

    To be able to describe collisions between objects in the scaling limit, the momentum and energy transfers need to be expressed in new variables that stay finite. Now, the rest energies [itex]E_{0}[/itex] have to tend to infinity in the scaling limit, otherwise one cannot guarantee that the rescaled velocities will always stay finite in collisions in the scaling limit. So, let's define a rescaled rest energy:

    [tex]E_{0}= K\tilde{E}_{0}[/tex]

    where K is chosen such that [itex]\tilde{E}_{0}[/itex] stays finite in the scaling limit. It then follows from the momentum equation that we need to rescale the momentum variable according to:

    [tex]\vec{P}= \frac{K}{c}\vec{\tilde{P}}[/tex]

    to get a finite momentum variable in terms of the rescaled velocity and rest energy in the scaling limit. Conservation of momentum equations are linear and thus stay of the same form in terms of the rescaled momentum variables.

    The kinetic energy becomes in terms of the rescaled variables:

    [tex]\frac{K}{2 c^2}w^{2}\tilde{E}_{0}[/tex]

    If this does not stay finite in the scaling limit, we would need to introduce another rescaling constant. But with all the available variables exhausted at this point, you can then no longer describe collisions in terms of the rescaled velocities and rest energies alone anymore. So, to get an interesting scaling limit, we are forced to choose K proportional to c^2. We can call the rescaled rest energy the mass. In the scaling limit, the mass is a new phenomenological parameter, as the relation with the energy becomes singular.

    Also, we see that conservation of energy implies conservation of mass in the scaling limit. Conservation of total kinetic energy is not guaranteed to hold, as energy can be lost in rest energy. In the scaling limit the rest energy change is invisible.
     
  2. jcsd
  3. Jun 1, 2010 #2

    Nabeshin

    User Avatar
    Science Advisor

    Why are you making life so difficult? Why not just taylor expand the lorentz factor like a normal human being?
     
  4. Jun 1, 2010 #3
    But then you have already put the information about the classical limit in the equations and all you are doing is a trivial calculus exercise. It is much more interesting to derive the classical limit ab initio without the factors of c already in the equations telling you exactly how to take the limits.

    Also, by doing it this way, you find that mass is a new parameter of the classical theory that did not exist as an independent quantity in the original theory. This fact is obscured when working in SI units.
     
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