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8 Answers 8

up vote 10 down vote accepted


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and the relativistic dispersion relation:



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@user12345, B.Elliot Huh? This (as in the first sentence of this answer) is wrong. If you set the reference frame to be moving with the light beam, for example, then the observed velocity of the light beam would be: $\frac{c_0+c_0}{1+\frac{c_0^2}{c_0^2}}=c_0\neq0 $... But I won't -1, just in case you didn't mean that... –  Dimensio1n0 Jul 6 '13 at 9:59

We don't actually delete the units. They are still there.


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Despite the indoctrination by high school physics teachers, physics is dimensionless. So, all units and dimensionfull constants can indeed be deleted. Now, this looks to be impossible as it then looks like you are throwing away information from equations that you can then no longer get back from using only those equations. This is not true, but recovering the original equations does require one to study the appropriate scaling limits of the theory in question, in this case special relativity in a way that is not usually done in textbooks.

Properly restoring c in the equations of special relativity should proceed as follows. We need to study what happens in the limit of zero velocities when one imposes conservation of energy and momentum. Of course, one can say that then everything stops moving. But the more interesting question is what happens if we make a movie of processes happening at slower and slower velocities and we simultaneously speed up the playback speed of the movie so that we keep on seeing the motion of the involved objects. We can then take the limit of scaling all the velocities to zero while the movie will still appear to show the objects moving at more or less the same speeds on the screen. The way the object appear to move and interact with each other on the screen will then be described by classical laws of physics.

The equations for energy and momentum of a free particle are:

$$E = \gamma(v) E_{0}$$

$$\vec{P} = \gamma(v) E_{0} \vec{v}$$


$$\gamma(v) = \frac{1}{\sqrt{1-v^{2}}}$$

We put $\vec{v} = \vec{v'}/c$ and assume that v' is kept finite while c is sent to infinity. Note that here c is just a dimensionless scaling parameter. This amounts to sending the velocity to zero while zooming in into the low velocity world so that it remains visible. To make the velocity dependence of the energy visible we need to expand it to second order in $\vec{v'}/c$, while the velocity dependence of momentum appears at zeroth order. Replacing the expressions for the energy and momentum by these expansions, gives:

$$E = E_{0}\left(1 + \frac{v'^2}{2c^2}\right)$$

$$\vec{P} = E_{0} \frac{\vec{v'}}{c}$$

Now, consider an elastic collision for which the above expressions for the energy and momentum are applicable, i.e. v' is finite. Then, demanding that we have conservation of momentum in any arbitrary frame, yields that both the sum of the rest rest energies and the sum of the momenta are separately conserved. Energy conservation then implies that that the the sum of the kinetic energies

$$T = E_{0}\frac{v'^2}{2c^2}$$

is conserved. The two terms in the energy equation scale in different ways, so we have to introduce a new variable to make sure that both terms remain visible in the c to infinity limit. We can e.g. make the kinetic energy finite by putting

$$E_{0} = m c^2$$

where m is assumed to stay finite in the c to infinity limit. We then have to rescale the momentum:

$$\vec{P'}= \frac{\vec{P}}{c}= m \vec{v'}$$

The total rescaled momentum, the kinetic energy and the "mass" m are then finite quantities that are conserved in the c to infinity limit. Note that we end up with this result irrespective of how we make the quantities finite. We could just as well have kept $E_{0}$ finite. If we then call this the mass m:

$$E_{0} = m $$

then we need to rescale the momentum according to:

$$\vec{P'}= \vec{P} c = m \vec{v'}$$

and the kinetic energy needs to be rescaled according to:

$$T' = T c^2 = \frac{1}{2} m v'^2$$

What then happens is that the rest energy $E_{0}$ does not scale in the same way as the kinetic energy. We can define a rescaled rest energy $E_{0}'$ that does scale in the same way as the kinetic energy:

$$E_{0}' = E_{0} c^{2} = m c^{2}$$

So, we see that in the scaling limit we end up with three finite independent conserved quantities: mass, kinetic energy and momentum and that if we formally keep the rescaling parameter c, the relation between the rest energy expressed in the finite kinetic energy units and the mass is $E_{0} = m c^2$.

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protected by Qmechanic Jul 6 '13 at 12:48


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