Galilean invariance of the wave equation

Question

Given the wave equation for a material wave: $$\frac{\partial^2 \phi}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2 \phi}{\partial t^2},$$ we can apply the Galilean transformation $x'=x-Vt$ and $t'= t$ which results in $$\left(1-\frac{V^2}{c^2}\right)\frac{\partial^2 \phi}{\partial x'^2} = \frac{1}{c^2} \frac{\partial^2 \phi}{\partial t'^2} - \frac{2V}{c^2} \frac{\partial^2 \phi}{\partial x' \partial t'}$$ from which, we can conclude is clearly not invariant under a Galilean transformation.

I wonder how this is possible when looking at the derivation of the wave equation. In the derivation of the wave equation, we use Newton's second law $F = ma$. It is often claimed that Newton's 2nd law is Galilean invariant by proofing that the acceleration $a$ is invariant under a Galilean transformation.

Now, how is it possible that the wave equation is not invariant under a Galilean transformation but the basic ingredient in deriving the same wave equation, Newton's 2nd law, is Galilean invariant?

Answer 1

You derived the wave equation assuming the material was stationary, and that is the only case for which that equation is valid. When you transform reference frames you no longer have a stationary material, so of course you will need a generalized form of the wave equation which can account for the motion of the material.

The generalized form which you arrived at however will be invariant under Galilean transformations since it describes waves in a moving material, and after a Galilean transformation you still have a moving material, the new situation is still one which can be described by the equation.

In one reference frame you have a material moving with velocity u, with waves described by,

$$\left(1-\frac{u^2}{c^2}\right)\frac{\partial^2 \phi}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2 \phi}{\partial t^2} + \frac{2u}{c^2} \frac{\partial^2 \phi}{\partial x \partial t}$$

You apply a Galilean transformation with velocity v,

$$t' = t$$

$$x' = x - vt$$

Resulting in,

$$u' = u - v$$

$$\left(1-\frac{u'^2}{c^2}\right)\frac{\partial^2 \phi}{\partial x'^2} = \frac{1}{c^2} \frac{\partial^2 \phi}{\partial t'^2} + \frac{2u'}{c^2} \frac{\partial^2 \phi}{\partial x' \partial t'}$$

Answer 2

A simple derivation of the wave equation from Newton’s second law is to consider harmonic oscillators in series: $$ \ddot u_n = (u_{n+1}+u_{n-1}-2u_n) $$ with $u$ the displacement and normalized mass and spring constant. The Galilean invariance is indeed preserved: $$ u_n\to u_n-Vt $$ From this perspective, it is still present in the continuum limit. Your transformation was just the wrong one: $$ \phi\to\phi-Vt $$

Another way to get the wave equation in the classical setting is to consider the Navier-Stokes equation (neglecting the nonlinear advection): $$ \partial_tv=-\nabla p $$ with the compressibility: $$ p=-\nabla\cdot u $$ with $u$ the displacement so that: $$ \partial_t^2v=\nabla(\nabla\cdot v) $$ Again Galilean invariance is preserved as long as you implement it with the correct transformation: $$ v(x,t)\to v(x-Vt,t)-V $$

In short, the implementation of Galilean invariance depends on the context of the system.

Answer 3

As we know, a continuous wave extends both in space and in time. That is, in the case of a continuous wave spatial properties and time properties are in a relation to each other.

Also, there is something about taking a second derivative and squaring; they tend to go hand in hand.

Kevin Brown, the author of the website mathpages.com, features on that website a series of articles under the umbrella name, 'Reflections on Relativity'.

In an article labeled '1.4 The Relativity of Light' the ordinary wave equation is discussed.

The german physicist Woldemar Voigt (1850-1919) had an interest in ramifications of the Doppler effect.

The following, starting below the first set of two horizontal lines, to the second set of horizontal lines, is a transcript of the discussion by Kevin Brown:

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[...] Galilean transformations are not the most general possible linear transformations. Voigt considered the question of whether there is any linear transformation that leaves the wave equation unchanged.

A one dimensional wave-equation, with $u$ the propagation speed of the wave.

$$\frac{\partial^2 \phi}{\partial x^2} = \frac{1}{u^2} \frac{\partial^2 \phi}{\partial t^2} $$

The general linear transformation between two coordinate systems X,T and x,t moving relative to each other with a given speed v is of the form

$$ x = AX + BT \qquad t = CX + DT $$

for constants A,B,C,D (which may be functions of v). If we choose units of space and time so that the characteristic speed u equals 1, the wave equation in terms of x,t coordinates is simply $\partial^2\phi/\partial x^2= \partial^2\phi/\partial t^2$. We seek constants A,B,C,D (for a given relative velocity $v$ between the coordinate systems) such that if $\phi$ satisfies the wave equation in terms of x,t then it also satisfies the wave equation $\partial^2\phi/\partial X^2= \partial^2\phi/\partial T^2$ in terms of X,T. To express the latter equation in terms of the x,t coordinates, recall that the total differential of $\phi$ can be written in the form

$$ d\phi = \frac{\partial \phi}{\partial x}dx + \frac{\partial \phi}{\partial t}dt $$

Also, at any constant T, the value of $\phi$ is purely a function of X, so we can divide through the above equation by dX to give

$$ \frac{\partial \phi}{\partial X} = \left( \frac{d \phi}{dX} \right)_T = \ \frac{\partial \phi}{\partial x} \left( \frac{dx}{dX} \right)_T + \frac{\partial \phi}{\partial t} \left( \frac{dt}{dX} \right)_T \ = \ A\frac{\partial \phi}{\partial x} + C\frac{\partial \phi}{\partial t} $$

Taking the partial derivative of this with respect to X then gives

$$ \frac{\partial^2 \phi}{\partial X^2} = A\frac{\partial^2 \phi}{\partial X \partial x} + C \frac{\partial^2 \phi}{\partial X \partial t} $$

Since partial differentiation is commutative, this can be written as

$$ \frac{\partial^2 \phi}{\partial X^2} = A \frac{\partial}{\partial x} \left( \frac{\partial \phi}{\partial X} \right) + C \frac{\partial}{\partial t} \left( \frac{\partial \phi}{\partial X} \right) $$

Substituting the prior expression for $\partial \phi/\partial X$ and carrying out the partial differentiations gives an expression for $\partial^2 / \partial X^2$ in terms of partials of $\phi$ with respect to x and t. Likewise we can derive a similar expression for $\partial^2 / \partial T^2$. Making use of these expressions, the wave equation $\partial^2 / \partial X^2 = \partial^2 / \partial T^2$ can be written as

$$ A^2 \frac{\partial^2 \phi}{\partial x^2} + 2AC\frac{\partial^2 \phi}{\partial x \partial t} + C^2 \frac{\partial^2 \phi}{\partial t^2} = B^2 \frac{\partial^2 \phi}{\partial x^2} + 2BD\frac{\partial^2 \phi}{\partial x \partial t} + D^2 \frac{\partial^2 \phi}{\partial t^2} $$

Since $\phi$ satisfies the equation $\partial^2 / \partial x^2 = \partial^2 / \partial t^2$ , the above equation reduces to

$$ \left( A^2 - B^2 + C^2 = D^2 \right) \frac{\partial^2 \phi}{\partial x^2} + 2(AC - BD)\frac{\partial^2 \phi}{\partial x \partial t} = 0 $$

The mixed partial derivative is not proportional to the second derivative with respect to x, so the coefficients of each individual terms must vanish. Thus the necessary and sufficient condition for $\phi$ to satisfy the wave equation in terms of the X,T coordinates (given that it satisfies the wave equation in terms of the x,t coordinates) is that the constants A,B,C,D of our linear transformation satisfy

$$ A^2 + C^2 = B^2 + D^2 \qquad AC = BD $$

Furthermore, the differential of the space transformation is $dx = AdX + BdT$, so an increment with $dx = 0$ satisfies $dX/dT = −B/A$. This represents the velocity v at which the spatial origin of the x,t coordinates is moving relative to the X,T coordinates. We also have the inverse transformation from (X,T) to (x,t):

$$ X = \frac{D}{AD - BC} x \ + \ \frac{-B}{AD - BC} t $$

$$ T = \frac{-C}{AD - BC} x \ + \ \frac{A}{AD - BC} t $$

Proceeding as before, the differential of this space transformation gives $dx/dt = B/D$ for the velocity of the spatial origin of the X,T coordinates with respect to the x,t coordinates, and this must equal −v. Therefore we have $B = −Av = −Dv$, and so $A = D$. It follows from the condition imposed by the wave equation that $B = C$, so both of these equal $−Av$. Our transformation can then be written in the form

$$ x = A(X - vT) \qquad t = A(T - vX) $$

The same analysis shows that the perpendicular coordinates y and z of the transformed system must be given by

$$ y = A \sqrt{1 - v^2} \ Y \qquad z = A \sqrt{1 - v^2} \ Z $$

In order to make the transformation formula for x agree with the Galilean transformation, Voigt chose $A = 1$, so he did not actually arrive at the Lorentz transformation, but nevertheless he had shown roughly how the wave equation could actually be relativistic – just like the dynamic behavior of inertial particles – provided we are willing to consider a transformation of the space and time coordinates that differs from the Galilean transformation. Had he considered the inverse transformation

$$ X = \frac{1}{A(1-v^2)}(x+vt) \qquad T = \frac{1}{A(1-v^2)}(t+vx) $$

he might have noticed that the determinant is $A^2(1−v^2)$, so to make this equal to 1 we must have $A = 1/\sqrt{1−v2}$, which not only implies y = Y and z = Z, but also makes the transformation formally identical to its inverse. In other words, he would have arrived at a completely relativistic framework for the wave equation. However, this was not Voigt’s objective, and he evidently regarded the transformed coordinates x, y, z and t as merely a convenient parameterization for purposes of calculation, without attaching any greater significance to them.

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Overall:
We see that the wave equation and the Lorentz transformations slot together seamlessly.

Here's another consideration:
As we know, Maxwell's equations allow a solution that describes a propagating electromagnetic $wave$.

Historically, Hendrik Antoon Lorentz arrived at the Lorentz transformations in the process of exploring how solutions to Maxwell's equation transfrom from one inertial coordinate system to another inertial coordinate system.

A recurring question on physics forums is: how does it come about that in exploring solutions to Maxwell's equations Lorentz anticipated special relativity? How do Maxwell's equations support that?

Well, it would appear that Maxwell's equation anticipate special relativity by virtue of supporting a solution that describes a propagating electromagnetic wave.