Does pure math have anything to say about the Lorentz transformation?

Question

The Lorentz factor, $${1 \over {\sqrt {1 - {{{v^2}} \over {{c^2}}}} }}$$ appears intuitively correct from a mathematical viewpoint. According to special relativity, the ratio of the velocity of a body to the speed of light cannot ever equal one or be greater than one. Furthermore, whichever direction a body is travelling with uniform velocity, its measurement of the speed of light is the same as all other inertial frames. In order to do this, we must consider imposing mathematical restrictions upon the ratio of velocity to the speed of light.

There are three mathematical principles at work here, the first and only categorical imperative in mathematics is 'division by zero is undefined'; secondary, the square root of a negative number is not a real number and the square of any real number has a positive value.

Let's begin with the ratio of velocity of a body to the speed of light. $${v \over c}$$ In order to prevent things from travelling at the speed of light, consider the following,$${1 \over {1 - {v \over c}}}$$ If $v$ = $c$ then ${v \over c}$ equals one, the denominator here would be zero which is undefined. Furthermore, according to special relativity, things cannot go faster than the speed of light. This can be constrained on this expression by taking its square root as taking the square root of a negative number is not a real number.$$\sqrt {{1 \over {1 - {v \over c}}}} $$

Physicists always square the ratio of velocity to the speed of light. My only guess is because the speed of light is constant regardless of which way a body is travelling; this means it must be squared in order to ensure its value is positive, otherwise, depending upon which way a body is moving, we would end up with different values of the factor.

So the Lorentz factor takes the shape it is. $${1 \over {\sqrt {1 - {{{v^2}} \over {{c^2}}}} }}$$ Is my reasoning way off beam?

Answer 1

Special relativity "prevents" objects from attaining speeds greater gthan $c$ in two senses. First, the formula for finding the relative velocity of two objects, which in some third frame are travelling toward each other with velocities $v_1$ and $v_2$, is $$ \frac{v}{c} = \frac{ \frac{v_1}{c}+\frac{v_2}{c}}{1+\frac{v_1}{c}\frac{v_2}{c}} $$ If we work in a system of units where $c = 1$ (for example, if we use 1 light-second instead of a meter to denominate distance) this looks like $$ v = \frac{v_1 + v_2}{1+v_1 v_2} $$ and this should be familliar as the formula for hyperbolic tangent of the sum of two numbers.

Since the net velocity $v$ is $\tanh(x)$ for some value of $x$, and the hyperbolic tangent of a real number can never exceed 1, the velocity of two objects relative to one another cannot exceed 1 (the speed of light).

The other sense in which special relativity prevents objects from exceeding the speed of light is that if you add some finite impulse (change in momentum) to an object, you change its velocity via a similar formula, which approaches $c$ asymptotically. Thus to achieve the speed of light you would have to inject infinite impulse or infinite energy.

Answer 2

Your derivation is an argument as to why such a quantity as the Lorentz factor would probably show up in formulas of special relativity, as it constrains $v$ in the precise manner as does special relativity. But it says nothing about how or where this quantity would show up, or why it would be significant in the theory. Remember that the Lorentz factor is a measure of the discrepancy between the coordinates two inertial observers with relative velocity $v$ will assign to events in spacetime. As such, the Lorentz factor has no meaning of its own with respect to a single observer traveling in spacetime.