Last month’s column dealt with the transformation of the wave equation under a Galilean transformation. Under this transformation, a wave moving with speed $$c$$ in a medium (e.g. a string or slinky or whatever) itself moving with speed $$v$$ with respect to some observer will be seen by that observer to be moving with speed $$c \pm v$$ depending on direction. To get this result, some clever manipulations had to be made that related a set of mixed partial derivatives in one frame with respect to the other.
It is natural enough to ask if the wave equation admits other transformations and, if so, do any of these represent anything physical. The idea lurking around is special relativity and the famous null result of the Michelson-Morley interferometer. But for the sake of this post, we will assume only simple physics, dictated by symmetry, and will discover that the Lorentz transformation falls out of the mathematical constraints that arise from the wave equation.
The situation is the same as in the previous post, with an observer $${\mathcal O}’$$ seeing the wave generated in the frame of $${\mathcal O}$$.
This time the transformation used will be symmetric in space and time, with the form
\[ x’ = \gamma( x + vt ) \]
and
\[ t’ = \gamma( t + \alpha x) \; .\]
The term $$\gamma$$ is a dimensionless quantity (whose value, at this point, may be unity) while $$\alpha$$ has units of an inverse velocity. Both of them will be determined from the transformation and some simple physical principles.
Assume, as before, that the wave equation as described by $${\mathcal O}$$ is given by
\[ c^2 \frac{\partial^2 f}{\partial x^2} – \frac{\partial^2 f}{\partial t^2} = 0 \; .\]
The transformation of the partial derivative with respect to the spatial variable in the unprimed to the primed frame is obtained through the chain rule
\[ \frac{\partial f}{\partial x} = \frac{\partial f}{\partial x’} \frac{\partial x’}{\partial x} + \frac{\partial f}{\partial t’} \frac{\partial t’}{\partial x} \; , \]
with an analogous relation for the partial derivatives with respect to time.
Using the transformation above gives
\[ \frac{\partial f}{\partial x} = \gamma \frac{\partial f}{\partial x’} + \alpha \frac{\partial f}{\partial t’} \]
and
\[ \frac{\partial f}{\partial t} = \gamma v \frac{\partial f}{\partial x’} + \gamma \frac{\partial f}{\partial t’} \; .\]
The second partial derivatives are obtained in the same way, but because of the transformation, four terms result from the expansion of
\[ \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} \left( \gamma \frac{\partial f}{\partial x’} + \alpha \frac{\partial f}{\partial t’} \right) \]
rather than 2 or 3 in the Galilean case considered before.
Assuming that the mixed partials commute, the middle two terms combine to give
\[ \frac{\partial^2 f}{\partial x^2} = \gamma^2 \frac{\partial^2 f}{\partial x’^2} + 2 \gamma^2 \alpha \frac{\partial^2 f}{\partial x’ \partial t’} + \gamma^2 \alpha^2 \frac{\partial^2 f}{\partial t’^2} \; .\]
Likewise, the second partial derivative with respect to time
\[ \frac{\partial^2 f}{\partial t^2} = \frac{\partial}{\partial t} \left( \gamma v \frac{\partial f}{\partial x’} + \gamma \frac{\partial f}{\partial t’} \right) \]
becomes
\[ \frac{\partial^2 f}{\partial t^2} = \gamma^2 v^2 \frac{\partial^2 f}{\partial x’^2} + 2 \gamma^2 v \frac{\partial^2 f}{\partial t’ \partial x’} + \gamma^2 \alpha^2 \frac{\partial^2 f}{\partial t’^2} \; .\]
Substituting these results into the original wave equation leads to a wave equation expressed in $${\mathcal O}’$$’s frame of
\[ c^2 \frac{\partial^2 f}{\partial x^2} – \frac{\partial^2 f}{\partial t^2} = \gamma^2(c^2 – v^2) \frac{\partial^2 f}{\partial x’^2} – \gamma^2 \left(1 – \frac{v^2}{c^2} \right) \frac{\partial^2 f}{\partial t’^2} \\ + 2 \gamma^2(\alpha c^2 – v) \frac{\partial^2 f}{\partial t’ \partial x’} = 0 \; .\]
There is no need to argue the mixed partial derivative into a different form (as was the case in the Galelean case) since a simple selection of
\[ \alpha c^2 – v = 0 \]
eliminates the term entirely. Solving this equation determines the unknown term $$\alpha$$ as
\[ \alpha = \frac{v^2}{c^2} \; ,\]
leaving the wave equation to take the invariant form of
\[ c^2 \frac{\partial^2 f}{\partial x’^2} – \frac{\partial^2 f}{\partial t’^2} = 0 \; .\]
In other words, the speed of the wave is constant regardless of how either the source or the observer move with respect to each other.
The only unknown term, $$\gamma$$, can be determined by the requirement that the transformation be invertible and that the determinant of the matrix be unity. The reason for the first requirement rests with the fact that there is no preferred frame – that the physics described on one frame can be related to the other no matter which frame is picked as the starting point. Failure to impose the second requirement would mean that the physics in one frame could be changed simply by transforming to another frame and then transforming back.
Since the transformation can be written matrix form as
\[ \left[ \begin{array}{c} x’ \\ t’ \end{array} \right] = \left[ \begin{array}{cc} \gamma & \gamma v \\ \gamma \frac{v}{c^2} & \gamma \end{array} \right] \left[ \begin{array}{c} x \\ t \end{array} \right] \]
then the requirement of a unit determinant means
\[ \gamma^2 – \gamma^2 \frac{v^2}{c^2} = 1 \; .\]
Solving for $$\gamma$$ give the famous Lorentz-Fitzgerald contraction
\[ \gamma = \frac{1}{\sqrt{1-v^2/c^2}} \; .\]
Note that in the limit as $$v \rightarrow 0$$, $$\gamma \rightarrow 1$$ and $$\alpha \rightarrow 0$$, thus restoring the Galilean transformation.
Thus only by starting with a symmetry principle that treats space and time on equal footing and some simple physical requirements on the wave equation, we can find a transformation that predicts that the speed of the wave is constant with respect to all observers. This is exactly what is seen in experiments and the resulting transformation is the gateway into the full machinery of special relativity.