Formulation of Galileo’s Transformations

Newtonian physics rests on the principle of relativity modeled through Galileo’s Transformations, where time is established as a universal coordinate for all inertial observers; that is: t=t^\prime. Under this statement, the linear transformation relating observations from two inertial reference frames S and S^\prime reviewed in the class on the principle of special relativity takes the form of a linear transformation:

\begin{array}{rl} t^\prime &= At + Bx,\\ x^\prime &= Dt + Ex,\\ y^\prime &= y, \\ z^\prime &=z, \end{array} [1]

it takes the following form when the inertial frames S and S^\prime are in standard configuration and S^\prime moves with velocity v_{ss^\prime_x}\hat{x} relative to S

\begin{array}{rlr} {}t^\prime &= t \\ x^\prime &= x - v_{ss^\prime_x}t \\ y^\prime &= y \\ z^\prime &= z \end{array} [2]

The Inverse Transformation

From a kind of algebraic symmetry, we can write the inverse transformation:

\begin{array}{rl} t &= t^\prime \\ x &= x^\prime + v_{ss^\prime_x}t \\ y &= y^\prime \\ z &= z^\prime \end{array} [3]

Absolute Time and the Sum of Velocities

From the first equation of Galileo’s transformations (either of the two, [2] or [3]) it is evident that the temporal coordinate of an event does not depend on the frame from which it is observed, while the second allows obtaining what is commonly understood as “common sense” associated with the sum of velocities. If a particle moves with constant velocity v_{ss^\prime_x} along the \hat{x} axis of S, then its velocity in S^\prime is determined by

\displaystyle v^\prime_x = \frac{dx^\prime}{dt^\prime} = \frac{dx^\prime}{dt} = \frac{d}{dt}\left(x - v_{ss^\prime_x} t \right) = v_x - v_{ss^\prime_x}

Deriving in this last expression shows that the acceleration of any particle is the same in S and in S^\prime, that is: dv^\prime_x/dt^\prime = dv_x/dt.

Galilean Geometry of Space and Time

If we consider two events A and B that have coordinates (t_A,x_A,y_A,z_A) and (t_B,x_B,y_B,z_B), respectively. It is easy to see that the quantities \Delta t = t_B - t_A and \Delta r^2 = \Delta x^2 + \Delta y^2 + \Delta z^2 are separately invariant under Galileo’s transformations, leading us to consider space and time as separate entities. On the other hand, \Delta r^2 suggests that this is a geometric property of space itself. We recognize \Delta r^2 as the square of the distance between events in Euclidean space. This defines the geometry of space and time in the context of Newtonian mechanics.

Galileo’s Relativity and Physical Laws

Applied to Newtonian Dynamics

In the previous section, we saw that, in the context of Newtonian physics, any two different inertial frames will always observe the same accelerations. This, coupled with Newton’s second law, implies that all inertial frames will always observe the same dynamics. That is:

\displaystyle F_x = m\frac{dv_x}{dt}= m\frac{dv^\prime_x}{dt^\prime} = F^\prime_x.

This last expression tells us that physics does not change when performing Galileo’s transformations, which is equivalent to saying that: physics is the same for all inertial observers.

Applied to Wave Propagation

While the persistence of physics in the face of changes in inertial observers is expected, first because it is what we observe when moving, and second because it is what has been obtained through previous calculations, it is not always the case. The most notable case of a phenomenon that does not preserve under Galilean transformations is the propagation of waves; generally, the equation that models the propagation of a wave \psi in space and time is in the form

\displaystyle \nabla^2 \psi = \frac{1}{v_0^2}\frac{\partial^2 \psi}{\partial t^2} [4]

where v_0 is the wave propagation speed.

What Effect Do Galilean Transformations Have on Wave Propagation?

For this, there is a short and a long answer. The short answer is that “even observing the same phenomenon, different inertial observers will see ‘different physics'”. The long answer involves examining how the wave propagation equation changes when the Galilean transformation is applied; to do this, we first take the equation [4] and expand it over each of its coordinates obtaining:

\displaystyle \frac{\partial ^2 \psi}{\partial x^2} + \frac{\partial ^2 \psi}{\partial y^2} + \frac{\partial ^2 \psi}{\partial z^2} = \frac{1}{v_0^2} \frac{\partial ^2 \psi}{\partial t^2}. [5]

With this equation at hand, we must now use the equations from [3] to re-express the derivatives in the other inertial frame.

Transformation of the First Derivatives

Following the expressions from [3] and deriving each variable with respect to the prime variables, we obtain:

\displaystyle \frac{\partial x^\prime}{\partial x} = \frac{\partial y^\prime}{\partial y}= \frac{\partial z^\prime}{\partial z}= \frac{\partial t^\prime}{\partial t}= 1

\displaystyle \frac{\partial x^\prime}{\partial t} = - v_{x_0}

While all others are nullified:

\displaystyle \frac{\partial t^\prime}{\partial x} = \frac{\partial t^\prime}{\partial y} = \frac{\partial t^\prime}{\partial z} = \frac{\partial x^\prime}{\partial y} = \frac{\partial x^\prime}{\partial z} = \frac{\partial y^\prime}{\partial x} = \frac{\partial y^\prime}{\partial z} = \frac{\partial y^\prime}{\partial t} = \frac{\partial z^\prime}{\partial x} = \frac{\partial z^\prime}{\partial y} = \frac{\partial z^\prime}{\partial t} = 0

With this at hand, we can now calculate the derivatives of \psi through the chain rule:

\displaystyle\frac{\partial \psi}{\partial x} = \frac{\partial \psi}{\partial x^\prime} \underbrace{\frac{\partial x^\prime}{\partial x}}_{=1} + \frac{\partial \psi}{\partial y^\prime} \underbrace{\frac{\partial y^\prime}{\partial x}}_{=0} + \frac{\partial \psi}{\partial z^\prime} \underbrace{\frac{\partial z^\prime}{\partial x}}_{=0} + \frac{\partial \psi}{\partial t^\prime} \underbrace{\frac{\partial t^\prime}{\partial x}}_{=0} = \frac{\partial \psi}{\partial x^\prime}.

And analogously for the other two spatial variables:

\displaystyle\frac{\partial \psi}{\partial y} = \frac{\partial \psi}{\partial y^\prime}.

\displaystyle\frac{\partial \psi}{\partial z} = \frac{\partial \psi}{\partial z^\prime}.

However, the temporal derivative will show some differences:

\begin{array}{rl} \displaystyle \frac{\partial \psi}{\partial t} &= \displaystyle\frac{\partial \psi}{\partial x^\prime}\underbrace{\frac{\partial x^\prime}{\partial t}}_{=-v_{x_0}} + \frac{\partial \psi}{\partial y^\prime}\underbrace{\frac{\partial y^\prime}{\partial t}}_{=0} + \frac{\partial \psi}{\partial z^\prime}\underbrace{\frac{\partial z^\prime}{\partial t}}_{=0} + \frac{\partial \psi}{\partial t^\prime}\underbrace{\frac{\partial t^\prime}{\partial t}}_{=1}\\ &=\displaystyle -v_{x_0} \frac{\partial \psi}{\partial x^\prime} + \frac{\partial \psi}{\partial t^\prime}, \end{array}

Transformation of the Second Derivatives

For the spatial part, we can continue without major difficulties, the results are:

\displaystyle \frac{\partial^2 \psi}{\partial x^2} = \frac{\partial^2 \psi}{\partial {x^\prime}^2}. [6]

\displaystyle \frac{\partial^2 \psi}{\partial y^2} = \frac{\partial^2 \psi}{\partial {y^\prime}^2} [7]

\displaystyle \frac{\partial^2 \psi}{\partial z^2} = \frac{\partial^2 \psi}{\partial {z^\prime}^2} [8]

But the temporal part, as we could already anticipate from the first derivatives, shows great differences:

\begin{array}{rl} \displaystyle\frac{\partial^2 \psi}{\partial t^2} &=\displaystyle \frac{\partial}{\partial t}\left( -v_{x_0} \frac{\partial \psi}{\partial x^\prime} + \frac{\partial \psi}{\partial t^\prime} \right)\\ & \displaystyle = -v_{x_0} \frac{\partial }{\partial t} \left(\frac{\partial \psi}{\partial x^\prime} \right) + \frac{\partial }{\partial t} \left(\frac{\partial \psi}{\partial t^\prime} \right)\\ &\displaystyle = -v_{x_0} \frac{\partial }{\partial x^\prime} \left(\frac{\partial \psi}{\partial t} \right) + \frac{\partial }{\partial t^\prime} \left(\frac{\partial \psi}{\partial t} \right)\\ &\displaystyle = -v_{x_0} \frac{\partial }{\partial x^\prime} \left(-v_{x_0} \frac{\partial \psi}{\partial x^\prime} + \frac{\partial \psi}{\partial t^\prime} \right) + \frac{\partial }{\partial t^\prime} \left(-v_{x_0} \frac{\partial \psi}{\partial x^\prime} + \frac{\partial \psi}{\partial t^\prime} \right) \end{array}

\displaystyle\frac{\partial^2 \psi}{\partial t^2} = v_{x_0}^2 \frac{\partial^2 \psi}{\partial {x^\prime}^2} - 2v_{x_0}\frac{\partial^2 \psi}{\partial x^\prime \partial t^\prime} + \frac{\partial^2 \psi}{\partial {t^\prime}^2}. [9]

Applying Galileo’s Transformations to Wave Propagation

Thus, it is possible to perform the Galileo transformation on the wave propagation equation by replacing equations [6,7,8] and [9] on [5], resulting in:

\displaystyle \frac{\partial^2 \psi}{\partial {x^\prime}^2} + \frac{\partial^2 \psi}{\partial {y^\prime}^2} + \frac{\partial^2 \psi}{\partial {z^\prime}^2} = \frac{1}{v_0^2} \left(\color{red}{ v_{x_0}^2 \frac{\partial^2 \psi}{\partial {x^\prime}^2} - 2v_{x_0}\frac{\partial^2 \psi}{\partial x^\prime \partial t^\prime}} + \frac{\partial^2 \psi}{\partial {t^\prime}^2} \right). [10]

It is observed that the form of wave propagation does not hold under Galileo transformations due to the appearance of the additional terms marked in red. Although this does not have major consequences for now, in future classes, we will see that this is precisely the point that “breaks”, so to speak, with classical physics, giving way to special relativity.