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Changing coordinates for the wave equation

The following is based on section 3.3.2 of Electricity and Magnetism for Mathematicians by Thomas A. Garrity (Cambridge UP, 2015). See also blog post Relative Motion and Waves by Conrad Schiff.

The classical wave equation is consistent with the Galilean transformation. Reflected electromagnetic waves are also consistent with classical physics using the dual Galilean transformation, in which linear location is the independent variable. The dual Galilean transformation for motion in one dimension is: x′ = x; t′ = twx = tx/v, where w = 1/v is the relative pace of the moving observer, which is equivalent to their inverse speed.

Suppose we again have two observers, A and B. Let observer B be moving at a constant pace w with respect to A, with A’s and B’s coordinate systems exactly matching up at location x = 0. Think of observer A as at rest, with coordinates x′ for location and t′ for time, and of observer B as moving to the right at pace w, with location coordinate x and time coordinate t. If the two coordinate systems line up at location x = x′ = 0, then the dual Galilean transformations are

x′ = x and t′ = t + wx,

or equivalently,

x = x′ and t = t′wx.

Suppose in the reference frame for B we have a wave y(x, t) satisfying the wave equation

\frac{\partial^2 y}{\partial x^2}-k^{2}\frac{\partial^2 y}{\partial t^2}=0.

In B’s reference frame, the pace of the wave is k = 1/c. From calculus, this pace k must be equal to the rate of change of t with respect to x, in other words, k = dt/dx. This in turn makes

\frac{\partial y}{\partial x}=-k\frac{\partial y}{\partial t}

as will now be seen. Fix a value of y(x, t) = y0. Then there is some t0 such that

y0 = (0, t0).

This means that for all x and t such that t0 = twx, then y0 = y(x, t). The pace of the wave is the rate this point with y-coordinate y0 moves along the t-axis with respect to x. Then

0=\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{\partial y}{\partial t}\frac{\mathrm{d} t}{\mathrm{d} x}+\frac{\partial y}{\partial x}\frac{\mathrm{d} x}{\mathrm{d} x}

=\frac{\partial y}{\partial t}\frac{\mathrm{d} t}{\mathrm{d} x} +\frac{\partial y}{\partial x} =k\frac{\partial y}{\partial t}+\frac{\partial y}{\partial x}

giving us that ∂y/∂x = −ky/∂t.

Observer A is looking at the same wave but measures the wave as having pace k + w. We want to see explicitly that under the appropriate change of coordinates this indeed happens, so this is an exercise with the chain rule.

Our wave y(x, t) can be written as a function of x′ and t′, namely as

y(x, t) = y(x′, t′wx′).

We want to show that this function satisfies

\frac{\partial^2 y}{\partial {x}'^2}-(k+w)^{2}\frac{\partial^2 y}{\partial t^2}=0.

The key to that will be

\frac{\partial }{\partial {x}'} =\frac{\partial x}{\partial {x}'}\frac{\partial }{\partial x}+\frac{\partial }{\partial {x}'}\frac{\partial }{\partial t} =\frac{\partial }{\partial x}-w\frac{\partial }{\partial t}

\frac{\partial }{\partial {t}'} =\frac{\partial t}{\partial {t}'}\frac{\partial }{\partial t} +\frac{\partial }{\partial {t}'}\frac{\partial }{\partial x} =\frac{\partial }{\partial t}

by the chain rule.

Next, start by showing that

\frac{\partial }{\partial {x}'} =\frac{\partial y}{\partial x}-w \frac{\partial y}{\partial t}

and

\frac{\partial y}{\partial {t}'} =\frac{\partial y}{\partial t}.

Turning to second derivatives, we can similarly show that

\frac{\partial^2 y}{\partial {x}'^2} =w^{2} \frac{\partial^2 y}{\partial t^2} -2w \frac{\partial^2 y}{\partial x \partial t} +\frac{\partial^2 y}{\partial x^2}

and

\frac{\partial^2 y}{\partial {t}'^2}=\frac{\partial^2 y}{\partial t^2}.

Knowing that ∂y/∂x = −ky/∂t, we have

\frac{\partial^2 y}{\partial x \partial t} =-k\frac{\partial^2 }{\partial t^2}

This allows us to show that

\frac{\partial^2 y}{\partial {x}'^2} -(k+w)^{2}\frac{\partial^2 y}{\partial {t}'^2}

which is what we desired.

For a wave that is reflected back (two-way wave) a given distance, the forward and reverse pace (±w) cancel, leaving the wave pace k constant. See here.

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