Light Sphere Equation Invariance
Maxwell’s Equations Transformation
In his 1905 paper, “On the Electrodynamics of Moving Bodies”, Albert Einstein resolved the apparent conflict between the laws of mechanics, as described by Newton, and the laws of electromagnetism, as formulated by Maxwell.
He based his theory on two postulates:
From these two postulates, Einstein first derived the new rules for transforming coordinates and times between reference frames moving relative to one another. These are the Lorentz transformations. The first part of his paper establishes this new kinematics.
The second, and arguably more dense, part of the paper addresses the question if Maxwell’s equations for electromagnetism are compatible with this new relativistic framework.
The following sections will provide the calculations that the original paper assumes the reader can perform.
We begin with the synchronization equation:
\frac{1}{2} \left[ \tau(0,0,0,t) + \tau\left(0,0,0,t + \frac{x^\prime}{c-v} + \frac{x^\prime}{c+v}\right) \right] = \tau\left(x^\prime,0,0,t + \frac{x^\prime}{c-v}\right)
Our objective is to understand how the time \tau changes for an infinitesimally small displacement x^\prime. We achieve this by expanding the function \tau(x^\prime, t) around the point (0, t), considering only terms of the first order. For a function of two variables f(x,y), the first-order Taylor expansion around a point (a,b) is given by:
f(x,y) \approx f(a,b) + \frac{\partial f}{\partial x}\bigg|_{(a,b)}(x-a) + \frac{\partial f}{\partial y}\bigg|_{(a,b)}(y-b)
We will apply this principle to the terms in the synchronization equation. For simplicity, we can write \tau(x^\prime,0,0,t) as \tau(x^\prime, t).
First, let’s address the terms on the left-hand side of the equation. The second term involves an evaluation of \tau at x^\prime=0 but at a later time. Let us first simplify the time argument:
t + \frac{x^\prime}{c-v} + \frac{x^\prime}{c+v} = t + \frac{x^\prime(c+v) + x^\prime(c-v)}{(c-v)(c+v)} = t + \frac{2cx^\prime}{c^2-v^2}
The term becomes \tau\left(0, t + \frac{2cx^\prime}{c^2-v^2}\right). We expand this as a function of time around t:
\tau\left(0, t + \frac{2cx^\prime}{c^2-v^2}\right) \approx \tau(0,t) + \frac{\partial \tau}{\partial t}\bigg|_{(0,t)} \left(\frac{2cx^\prime}{c^2-v^2}\right)
Now we expand the right-hand side of the original equation. This is an expansion in both variables, x^\prime and t, around the point (0,t):
\tau\left(x^\prime, t + \frac{x^\prime}{c-v}\right) \approx \tau(0,t) + \frac{\partial \tau}{\partial x^\prime}\bigg|_{(0,t)} (x^\prime - 0) + \frac{\partial \tau}{\partial t}\bigg|_{(0,t)} \left(t + \frac{x^\prime}{c-v} - t\right)
This simplifies to:
\tau\left(x^\prime, t + \frac{x^\prime}{c-v}\right) \approx \tau(0,t) + \frac{\partial \tau}{\partial x^\prime}x^\prime + \frac{\partial \tau}{\partial t}\frac{x^\prime}{c-v}
We can now substitute these expanded forms back into the initial equation. All partial derivatives are evaluated at (0,t):
\frac{1}{2} \left[ \tau(0,t) + \left( \tau(0,t) + \frac{\partial \tau}{\partial t} \frac{2cx^\prime}{c^2-v^2} \right) \right] = \tau(0,t) + \frac{\partial \tau}{\partial x^\prime}x^\prime + \frac{\partial \tau}{\partial t}\frac{x^\prime}{c-v}
Simplifying the left-hand side gives:
\tau(0,t) + \frac{1}{2}\frac{\partial \tau}{\partial t} \frac{2cx^\prime}{c^2-v^2} = \tau(0,t) + \frac{\partial \tau}{\partial x^\prime}x^\prime + \frac{\partial \tau}{\partial t}\frac{x^\prime}{c-v}
The \tau(0,t) terms on both sides cancel out:
\frac{\partial \tau}{\partial t} \frac{cx^\prime}{c^2-v^2} = \frac{\partial \tau}{\partial x^\prime}x^\prime + \frac{\partial \tau}{\partial t}\frac{x^\prime}{c-v}
Since we are considering an infinitesimal but non-zero displacement x^\prime, we can divide the entire equation by x^\prime:
\frac{\partial \tau}{\partial t} \frac{c}{c^2-v^2} = \frac{\partial \tau}{\partial x^\prime} + \frac{\partial \tau}{\partial t}\frac{1}{c-v}
To find a relation for the spatial derivative \frac{\partial \tau}{\partial x^\prime}, we rearrange the terms:
\begin{aligned} \frac{\partial \tau}{\partial x^\prime} &= \frac{\partial \tau}{\partial t} \frac{c}{c^2-v^2} - \frac{\partial \tau}{\partial t}\frac{1}{c-v} \\ &= \frac{\partial \tau}{\partial t} \left( \frac{c}{c^2-v^2} - \frac{c+v}{c^2-v^2} \right) \\ &= \frac{\partial \tau}{\partial t} \left( \frac{c - (c+v)}{c^2-v^2} \right) \\ &= \frac{\partial \tau}{\partial t} \left( \frac{-v}{c^2-v^2} \right) \end{aligned}
This manipulation leads to the final partial differential equation:
\frac{\partial \tau}{\partial x^\prime} + \frac{v}{c^2 - v^2} \frac{\partial \tau}{\partial t} = 0
Einstein initially includes a function \phi(v) in the Lorentz transformations, which is independent of the coordinates and whose value is determined later in the analysis. Let us re-examine the transformations including this function and derive their inverse, as well as prove the invariance of the form of the equation for a light wave.
The transformations from the stationary system (x, y, z, t) to the moving system (\xi, \eta, \zeta, \tau) are given by:
\begin{aligned} \tau &= \phi(v) \beta \left(t - \frac{v}{c^2}x\right) \\ \xi &= \phi(v) \beta (x - vt) \\ \eta &= \phi(v) y \\ \zeta &= \phi(v) z \end{aligned}
The term \beta is the familiar Lorentz factor:
\beta = \frac{1}{\sqrt{1 - v^2/c^2}}
Our goal is to express the stationary coordinates x, y, z, t in terms of the moving coordinates \xi, \eta, \zeta, \tau. The transformations for the transverse coordinates y and z are straightforward to invert:
\begin{aligned} & y = \frac{1}{\phi(v)} \eta = \phi^{-1}(v) \eta \\ & z = \frac{1}{\phi(v)} \zeta = \phi^{-1}(v) \zeta \end{aligned}
To find x and t, we must solve the first two equations as a system. Let’s start by isolating x from the equation for \xi:
x = \frac{\xi}{\phi(v)\beta} + vt
Now, we substitute this expression for x into the equation for \tau:
\begin{aligned} & \tau = \phi(v) \beta \left( t - \frac{v}{c^2} \left( \frac{\xi}{\phi(v)\beta} + vt \right) \right) \\ & \frac{\tau}{\phi(v)\beta} = t - \frac{v\xi}{\phi(v)\beta c^2} - \frac{v^2}{c^2}t \\ & \frac{\tau}{\phi(v)\beta} + \frac{v\xi}{\phi(v)\beta c^2} = t\left(1 - \frac{v^2}{c^2}\right) \end{aligned}
We recognize that the term (1 - v^2/c^2) is equal to 1/\beta^2:
\begin{aligned} & \frac{1}{\phi(v)\beta}\left(\tau + \frac{v\xi}{c^2}\right) = t\frac{1}{\beta^2} \\ & t = \frac{\beta}{\phi(v)}\left(\tau + \frac{v\xi}{c^2}\right) \end{aligned}
With the expression for t found, we substitute it back into our equation for x.
\begin{aligned} x &= \frac{\xi}{\phi(v)\beta} + v \left[ \frac{\beta}{\phi(v)}\left(\tau + \frac{v\xi}{c^2}\right) \right] \\ &= \frac{1}{\phi(v)}\left( \frac{\xi}{\beta} + v\beta\tau + \frac{v^2\beta\xi}{c^2} \right) \\ &= \frac{\beta}{\phi(v)}\left( \frac{\xi}{\beta^2} + v\tau + \frac{v^2\xi}{c^2} \right) \\ &= \frac{\beta}{\phi(v)}\left( \xi\left(1 - \frac{v^2}{c^2}\right) + v\tau + \frac{v^2\xi}{c^2} \right) \\ &= \frac{\beta}{\phi(v)}\left( \xi - \frac{v^2\xi}{c^2} + v\tau + \frac{v^2\xi}{c^2} \right) \\ &= \frac{\beta}{\phi(v)}(\xi + v\tau) \end{aligned}
The complete inverse transformations are therefore:
\begin{aligned} t &= \phi^{-1}(v) \beta \left(\tau + \frac{v}{c^2}\xi\right) \\ x &= \phi^{-1}(v) \beta (\xi + v\tau) \\ y &= \phi^{-1}(v) \eta \\ z &= \phi^{-1}(v) \zeta \end{aligned}
Since the speed of light is constant for all inertial observers, this means that an expanding sphere of light described by the equation x^2 + y^2 + z^2 = c^2t^2 in a stationary frame must be described by a similar equation, \xi^2 + \eta^2 + \zeta^2 = c^2\tau^2, in a moving frame. We can demonstrate this property by directly substituting the inverse Lorentz transformation equations into the first equation and showing that it yields the second.
The square of the x-coordinate is:
\begin{aligned} x^2 &= \left[ \phi^{-1}(v) \beta (\xi + v\tau) \right]^2 \\ &= \phi^{-2}(v) \beta^2 (\xi^2 + 2v\xi\tau + v^2\tau^2) \end{aligned}
The squares of the transverse coordinates are:
\begin{aligned} y^2 &= \left[ \phi^{-1}(v) \eta \right]^2 = \phi^{-2}(v) \eta^2 \\ z^2 &= \left[ \phi^{-1}(v) \zeta \right]^2 = \phi^{-2}(v) \zeta^2 \end{aligned}
Finally, we compute the time component:
\begin{aligned} c^2t^2 &= c^2 \left[ \phi^{-1}(v) \beta \left(\tau + \frac{v}{c^2}\xi\right) \right]^2 \\ &= c^2 \phi^{-2}(v) \beta^2 \left(\tau^2 + \frac{2v}{c^2}\xi\tau + \frac{v^2}{c^4}\xi^2\right) \\ &= \phi^{-2}(v) \beta^2 \left(c^2\tau^2 + 2v\xi\tau + \frac{v^2}{c^2}\xi^2\right) \end{aligned}
Now we combine these parts into the expression for the spacetime interval:
\begin{aligned} x^2 + y^2 + z^2 - c^2t^2 = & \phi^{-2}(v) \beta^2 (\xi^2 + 2v\xi\tau + v^2\tau^2) + \phi^{-2}(v) \eta^2 \\ & + \phi^{-2}(v) \zeta^2 - \phi^{-2}(v) \beta^2 \left(c^2\tau^2 + 2v\xi\tau + \frac{v^2}{c^2}\xi^2\right) \\ = & \phi^{-2}(v) \left[ \beta^2 (\xi^2 + 2v\xi\tau + v^2\tau^2) + \eta^2 + \zeta^2 \right. \\ & \left.- \beta^2 \left(c^2\tau^2 + 2v\xi\tau + \frac{v^2}{c^2}\xi^2\right) \right] \\ = & \phi^{-2}(v) \left[ \beta^2\xi^2 + 2\beta^2v\xi\tau + \beta^2v^2\tau^2 + \eta^2 + \zeta^2 \right. \\ & \left. - \beta^2c^2\tau^2 - 2\beta^2v\xi\tau - \beta^2\frac{v^2}{c^2}\xi^2 \right] \end{aligned}
The term 2\beta^2v\xi\tau cancels out. We can now group the remaining terms by the moving frame coordinates \xi and \tau:
\begin{aligned} x^2 + y^2 + z^2 - c^2t^2 & = \phi^{-2}(v) \left[ \left(\beta^2\xi^2 - \beta^2\frac{v^2}{c^2}\xi^2\right) + \left(\beta^2v^2\tau^2 - \beta^2c^2\tau^2\right) + \eta^2 + \zeta^2 \right] \\ & = \phi^{-2}(v) \left[ \xi^2\beta^2\left(1 - \frac{v^2}{c^2}\right) - \tau^2\beta^2(c^2 - v^2) + \eta^2 + \zeta^2 \right] \end{aligned}
We use the definition of the Lorentz factor, \beta^2 = 1/(1 - v^2/c^2), which means \beta^2(1 - v^2/c^2) = 1. Let’s apply this and also factor c^2 from the \tau^2 group:
\begin{aligned} x^2 + y^2 + z^2 - c^2t^2 &= \phi^{-2}(v) \left[ \xi^2(1) - \tau^2\beta^2c^2\left(1 - \frac{v^2}{c^2}\right) + \eta^2 + \zeta^2 \right] \\ &= \phi^{-2}(v) \left[ \xi^2 - c^2\tau^2(1) + \eta^2 + \zeta^2 \right] \end{aligned}
Rearranging the terms inside the bracket gives us the final relationship:
x^2 + y^2 + z^2 - c^2t^2 = \phi^{-2}(v) \left( \xi^2 + \eta^2 + \zeta^2 - c^2\tau^2 \right)
This result shows how the spacetime interval transforms between frames. If the coordinates in the stationary frame describe a light ray such that x^2 + y^2 + z^2 - c^2t^2 = 0, then it must follow that:
0 = \phi^{-2}(v) \left( \xi^2 + \eta^2 + \zeta^2 - c^2\tau^2 \right)
As \phi(v) cannot be infinite, this implies:
\xi^2 + \eta^2 + \zeta^2 - c^2\tau^2 = 0
Which is the equation for a light sphere in the moving frame, \xi^2 + \eta^2 + \zeta^2 = c^2\tau^2. This confirms that the form of the equation is invariant under the Lorentz transformations.
Let a point move in accordance with the equations:
\begin{aligned} & \xi = w_\xi \tau \\ & \eta = w_\eta \tau \\ & \zeta = 0 \end{aligned}
where w_\xi and w_\eta are constant.
In the paper Einstein previously derived that \phi(v) \equiv 1 so the inverse transformations from the moving system to the stationary system are given by:
\begin{aligned} t &= \beta \left(\tau + \frac{v}{c^2}\xi\right) \\ x &= \beta (\xi + v\tau) \\ y &= \eta \\ z &= \zeta \end{aligned}
Our first step is to substitute the motion equations into the transformation equations. This replaces the spatial coordinates \xi, \eta, and \zeta with expressions involving the time \tau.
For the time t:
t = \beta \left(\tau + \frac{v}{c^2}(w_\xi \tau)\right) = \beta \tau (1 + v w_\xi / c^2)
For the coordinate x:
x = \beta ((w_\xi \tau) + v\tau) = \beta \tau (w_\xi + v)
For the coordinate y:
y = \eta = w_\eta \tau
The coordinate z is simply:
z = \zeta = 0
We now have a system of equations where t, x, and y are all expressed in terms of the single parameter \tau. To find the velocity components in the stationary frame, we need to express x and y as functions of t. We can achieve this by first solving the equation for t to find an expression for \tau.
From the equation for t:
\tau = \frac{t}{\beta (1 + v w_\xi / c^2)}
Now we substitute this expression for \tau into the equations for x and y.
For x:
\begin{aligned} x &= \beta (w_\xi + v) \tau \\ &= \beta (w_\xi + v) \left( \frac{t}{\beta (1 + v w_\xi / c^2)} \right) \\ &= \frac{v + w_\xi}{1 + v w_\xi/c^2} t \end{aligned}
For y:
\begin{aligned} y &= w_\eta \tau \\ &= w_\eta \left( \frac{t}{\beta (1 + v w_\xi / c^2)} \right) \\ &= \frac{w_\eta}{\beta (1 + v w_\xi / c^2)} t \\ &= \frac{\sqrt{1 - v^2/c^2}}{ 1 + v w_\xi / c^2}w_\eta t \\ \end{aligned}
Then, we define:
\begin{aligned} & V^2 = \left(\frac{\mathrm dx}{\mathrm dt}\right)^2 + \left(\frac{\mathrm dy}{\mathrm dt}\right)^2 \\ & w^2 = w^2_\xi + w^2_\eta \\ & a = \tan^{-1} \left(w_\eta / w_\xi \right) \end{aligned}
where a is the angle between the velocities v and w so that:
\begin{aligned} & w_\xi = w \cos(a) \\ & w_\eta = w \sin(a) \end{aligned}
We determine the velocity components in the stationary frame:
\begin{aligned} \frac{\mathrm dx}{\mathrm dt} &= \frac{v + w_\xi}{1 + v w_\xi/c^2} \\ \frac{\mathrm dy}{\mathrm dt} &= \frac{w_\eta \sqrt{1 - v^2/c^2}}{ 1 + v w_\xi / c^2} \end{aligned}
We then square them:
\begin{aligned} & \left(\frac{\mathrm dx}{\mathrm dt}\right)^2 = \frac{(v + w_\xi)^2}{\left(1 + v w_\xi/c^2\right)^2}\\ & \left(\frac{\mathrm dy}{\mathrm dt}\right)^2 = \frac{\left(w_\eta \sqrt{1 - v^2/c^2}\right)^2}{\left(1 + \frac{v w_\xi}{c^2}\right)^2} = \frac{w_\eta^2 (1 - v^2/c^2)}{\left(1 + v w_\xi/c^2\right)^2} \end{aligned}
Finally we add them:
\begin{aligned} V^2 & = \left(\frac{\mathrm dx}{\mathrm dt}\right)^2 + \left(\frac{\mathrm dy}{\mathrm dt}\right)^2 \\ & = \frac{(v + w_\xi)^2 + w_\eta^2 (1 - v^2/c^2)}{\left(1 + v w_\xi/c^2\right)^2} \\ & = \frac{\left(v^2 + 2vw_\xi + w_\xi^2\right) + \left(w_\eta^2 - v^2 w_\eta^2/c^2\right)}{\left(1 + v w \cos(a)/c^2\right)^2} \\ & = \frac{v^2 + 2vw_\xi + (w_\xi^2 + w_\eta^2) - v^2 w_\eta^2/c^2}{\left(1 + v w \cos(a)/c^2\right)^2} \\ & = \frac{v^2 + w^2 + 2vw \cos(a) - (v w \sin(a)/c)^2}{\left(1 + v w \cos(a)/c^2\right)^2} \end{aligned}
So the velocity is:
V = \frac{\sqrt{v^2 + w^2 + 2vw \cos(a) - (v w \sin(a)/c)^2)}}{1 + v w \cos(a)/c^2}
If w has the same direction as the axis of X, then \cos(a)=1, \sin(a) =0 and the expression simplifies as:
V = \frac{v + w}{1 + vw/c^2}
An analysis of Maxwell’s equations in a moving reference frame requires a direct application of the chain rule to the partial derivatives. This process reveals the interdependence of electric and magnetic fields.
To express the laws of electromagnetism in the moving frame, we must first find how the partial derivative operators transform. This is achieved using the chain rule.
For any function f, the derivative with respect to a stationary coordinate is a sum of derivatives with respect to the moving coordinates.
For the time derivative, we have:
\frac{\partial f}{\partial t} = \frac{\partial f}{\partial \xi}\frac{\partial \xi}{\partial t} + \frac{\partial f}{\partial \eta}\frac{\partial \eta}{\partial t} + \frac{\partial f}{\partial \zeta}\frac{\partial \zeta}{\partial t} + \frac{\partial f}{\partial \tau}\frac{\partial \tau}{\partial t}
The required coefficients are found by differentiating the Lorentz transformation equations with respect to t:
\begin{aligned} & \frac{\partial \xi}{\partial t} = -\beta v \\ & \frac{\partial \eta}{\partial t} = 0\\ &\frac{\partial \zeta}{\partial t} = 0 \\ &\frac{\partial \tau}{\partial t} = \beta \end{aligned}
Substituting these yields the transformation for the time derivative operator:
\frac{\partial}{\partial t} = \beta \left(\frac{\partial}{\partial \tau} - v \frac{\partial}{\partial \xi}\right)
Similarly, for the spatial derivatives, we find:
\begin{aligned} \frac{\partial}{\partial x} &= \beta \left(\frac{\partial}{\partial \xi} - \frac{v}{c^2} \frac{\partial}{\partial \tau}\right) \\ \frac{\partial}{\partial y} &= \frac{\partial}{\partial \eta} \\ \frac{\partial}{\partial z} &= \frac{\partial}{\partial \zeta} \end{aligned}
Let the electric field components in the stationary frame be (X, Y, Z) and the magnetic field components be (L, M, N). We consider the first of Maxwell’s equations as presented in the paper:
\frac{1}{c} \frac{\partial X}{\partial t} = \frac{\partial N}{\partial y} - \frac{\partial M}{\partial z}
Following the initial application of the Lorentz transformation to the partial differential operators, we arrive at the following expression for the first of Maxwell’s equations:
\frac{1}{c} \beta \left(\frac{\partial X}{\partial \tau} - v \frac{\partial X}{\partial \xi}\right) = \frac{\partial N}{\partial \eta} - \frac{\partial M}{\partial \zeta}
The presence of the term \partial X/\partial \xi on the left-hand side complicates a direct interpretation of this equation.
To resolve this, we employ another of Maxwell’s equations, Gauss’s Law for electricity, which states that the divergence of the electric field is zero. In the stationary frame, this is:
\frac{\partial X}{\partial x} + \frac{\partial Y}{\partial y} + \frac{\partial Z}{\partial z} = 0
We transform this law into the moving coordinate system by applying the same operator transformations:
\beta \left(\frac{\partial X}{\partial \xi} - \frac{v}{c^2} \frac{\partial X}{\partial \tau}\right) + \frac{\partial Y}{\partial \eta} + \frac{\partial Z}{\partial \zeta} = 0
This equation provides a relationship that can be used to eliminate \partial X/\partial \xi. We solve it for \partial X/\partial \xi:
\frac{\partial X}{\partial \xi} = \frac{v}{c^2} \frac{\partial X}{\partial \tau} - \frac{1}{\beta}\left(\frac{\partial Y}{\partial \eta} + \frac{\partial Z}{\partial \zeta}\right)
Now, we substitute this expression back into our primary transformed equation:
\begin{aligned} & \frac{1}{c} \beta \left[ \frac{\partial X}{\partial \tau} - v \left( \frac{v}{c^2} \frac{\partial X}{\partial \tau} - \frac{1}{\beta}\left(\frac{\partial Y}{\partial \eta} + \frac{\partial Z}{\partial \zeta}\right) \right) \right] = \frac{\partial N}{\partial \eta} - \frac{\partial M}{\partial \zeta} \\ & \frac{1}{c} \beta \left[ \frac{\partial X}{\partial \tau} - \frac{v^2}{c^2} \frac{\partial X}{\partial \tau} + \frac{v}{\beta}\left(\frac{\partial Y}{\partial \eta} + \frac{\partial Z}{\partial \zeta}\right) \right] = \frac{\partial N}{\partial \eta} - \frac{\partial M}{\partial \zeta} \end{aligned}
We can combine the terms containing \partial X/\partial \tau by recalling the definition of the Lorentz factor, \beta = 1/\sqrt{1 - v^2/c^2}, which implies that 1 - v^2/c^2 = 1/\beta^2:
\begin{aligned} & \frac{1}{c} \beta \left[ \frac{1}{\beta^2} \frac{\partial X}{\partial \tau} + \frac{v}{\beta}\left(\frac{\partial Y}{\partial \eta} + \frac{\partial Z}{\partial \zeta}\right) \right] = \frac{\partial N}{\partial \eta} - \frac{\partial M}{\partial \zeta} \\ & \frac{1}{c} \left[ \frac{1}{\beta} \frac{\partial X}{\partial \tau} + v\left(\frac{\partial Y}{\partial \eta} + \frac{\partial Z}{\partial \zeta}\right) \right] = \frac{\partial N}{\partial \eta} - \frac{\partial M}{\partial \zeta} \\ & \frac{1}{c\beta} \frac{\partial X}{\partial \tau} = \frac{\partial N}{\partial \eta} - \frac{\partial M}{\partial \zeta} - \frac{v}{c}\left(\frac{\partial Y}{\partial \eta} + \frac{\partial Z}{\partial \zeta}\right) \\ & \frac{1}{c} \frac{\partial X}{\partial \tau} = \beta \left( \frac{\partial N}{\partial \eta} - \frac{\partial M}{\partial \zeta} \right) - \frac{\beta v}{c} \left( \frac{\partial Y}{\partial \eta} + \frac{\partial Z}{\partial \zeta} \right) \end{aligned}
Which gives us the final form:
\frac{1}{c} \frac{\partial X}{\partial \tau} = \dfrac{\partial}{\partial \eta} \left[ \beta \left( N - \dfrac{v}{c} Y \right) \right] - \dfrac{\partial}{\partial \zeta} \left[ \beta \left( M + \dfrac{v}{c} Z \right) \right]
We begin with the Maxwell’s equation relating the time derivative of Y to the curl of the magnetic field:
\frac{1}{c}\frac{\partial Y}{\partial t} = \frac{\partial L}{\partial z} - \frac{\partial N}{\partial x}
We substitute the operator transformations into this equation:
\frac{1}{c} \beta \left(\frac{\partial Y}{\partial \tau} - v \frac{\partial Y}{\partial \xi}\right) = \frac{\partial L}{\partial \zeta} - \beta \left(\frac{\partial N}{\partial \xi} - \frac{v}{c^2} \frac{\partial N}{\partial \tau}\right)
Expanding the terms on both sides gives:
\frac{\beta}{c} \frac{\partial Y}{\partial \tau} - \frac{\beta v}{c} \frac{\partial Y}{\partial \xi} = \frac{\partial L}{\partial \zeta} - \beta \frac{\partial N}{\partial \xi} + \frac{\beta v}{c^2} \frac{\partial N}{\partial \tau}
We rearrange this expression by gathering all terms involving the time derivative \partial/\partial \tau on the left-hand side and all terms involving spatial derivatives on the right-hand side:
\frac{\beta}{c} \frac{\partial Y}{\partial \tau} - \frac{\beta v}{c^2} \frac{\partial N}{\partial \tau} = \frac{\partial L}{\partial \zeta} - \beta \frac{\partial N}{\partial \xi} + \frac{\beta v}{c} \frac{\partial Y}{\partial \xi}
Factoring out the derivative operators gives us the final form:
\frac{1}{c} \frac{\partial}{\partial \tau} \left[ \beta \left( Y - \frac{v}{c} N \right) \right] = \frac{\partial L}{\partial \zeta} - \frac{\partial}{\partial \xi} \left[ \beta \left( N - \frac{v}{c} Y \right) \right]
The procedure for the Z-component is analogous. We start with the corresponding Maxwell’s equation:
\frac{1}{c}\frac{\partial Z}{\partial t} = \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}
Again, we apply the operator transformations:
\frac{1}{c} \beta \left(\frac{\partial Z}{\partial \tau} - v \frac{\partial Z}{\partial \xi}\right) = \beta \left(\frac{\partial M}{\partial \xi} - \frac{v}{c^2} \frac{\partial M}{\partial \tau}\right) - \frac{\partial L}{\partial \eta}
Expanding the terms results in:
\frac{\beta}{c} \frac{\partial Z}{\partial \tau} - \frac{\beta v}{c} \frac{\partial Z}{\partial \xi} = \beta \frac{\partial M}{\partial \xi} - \frac{\beta v}{c^2} \frac{\partial M}{\partial \tau} - \frac{\partial L}{\partial \eta}
We rearrange the equation by collecting the \partial/\partial \tau terms on the left and the spatial derivative terms on the right:
\frac{\beta}{c} \frac{\partial Z}{\partial \tau} + \frac{\beta v}{c^2} \frac{\partial M}{\partial \tau} = \beta \frac{\partial M}{\partial \xi} + \frac{\beta v}{c} \frac{\partial Z}{\partial \xi} - \frac{\partial L}{\partial \eta}
Factoring out the derivative operators gives us the final form:
\frac{1}{c} \frac{\partial}{\partial \tau} \left[ \beta \left( Z + \frac{v}{c} M \right) \right] = \frac{\partial}{\partial \xi} \left[ \beta \left( M + \frac{v}{c} Z \right) \right] - \frac{\partial L}{\partial \eta}
We begin with the equation for the time derivative of the longitudinal magnetic field component, L:
\frac{1}{c} \frac{\partial L}{\partial t} = \frac{\partial Y}{\partial z} - \frac{\partial Z}{\partial y}
As with the X-component equation, a direct transformation is complicated by the presence of a derivative with respect to the direction of motion. We first apply the operator transformations:
\frac{1}{c} \beta \left(\frac{\partial L}{\partial \tau} - v \frac{\partial L}{\partial \xi}\right) = \frac{\partial Y}{\partial \zeta} - \frac{\partial Z}{\partial \eta}
To proceed, we must find an expression for the term \partial L/\partial \xi. We use Gauss’s law for magnetism, which states that the divergence of the magnetic field is zero:
\frac{\partial L}{\partial x} + \frac{\partial M}{\partial y} + \frac{\partial N}{\partial z} = 0
Transforming this law into the moving coordinates yields:
\beta \left(\frac{\partial L}{\partial \xi} - \frac{v}{c^2} \frac{\partial L}{\partial \tau}\right) + \frac{\partial M}{\partial \eta} + \frac{\partial N}{\partial \zeta} = 0
We solve this for the term we wish to eliminate, \partial L/\partial \xi:
\frac{\partial L}{\partial \xi} = \frac{v}{c^2} \frac{\partial L}{\partial \tau} - \frac{1}{\beta}\left(\frac{\partial M}{\partial \eta} + \frac{\partial N}{\partial \zeta}\right)
This expression is substituted back into our primary transformed equation:
\frac{1}{c} \beta \left[ \frac{\partial L}{\partial \tau} - v \left( \frac{v}{c^2} \frac{\partial L}{\partial \tau} - \frac{1}{\beta}\left(\frac{\partial M}{\partial \eta} + \frac{\partial N}{\partial \zeta}\right) \right) \right] = \frac{\partial Y}{\partial \zeta} - \frac{\partial Z}{\partial \eta}
Simplifying the left-hand side using the relation 1 - v^2/c^2 = 1/\beta^2 leads to:
\begin{aligned} & \frac{1}{c} \beta \left[ \frac{1}{\beta^2} \frac{\partial L}{\partial \tau} + \frac{v}{\beta}\left(\frac{\partial M}{\partial \eta} + \frac{\partial N}{\partial \zeta}\right) \right] = \frac{\partial Y}{\partial \zeta} - \frac{\partial Z}{\partial \eta}\\ & \frac{1}{c} \left[ \frac{1}{\beta} \frac{\partial L}{\partial \tau} + v\left(\frac{\partial M}{\partial \eta} + \frac{\partial N}{\partial \zeta}\right) \right] = \frac{\partial Y}{\partial \zeta} - \frac{\partial Z}{\partial \eta}\\ & \frac{1}{c} \frac{\partial L}{\partial \tau} = \beta \left( \frac{\partial Y}{\partial \zeta} - \frac{\partial Z}{\partial \eta} \right) - \frac{\beta v}{c} \left( \frac{\partial M}{\partial \eta} + \frac{\partial N}{\partial \zeta} \right) \end{aligned}
Finally, we group the terms on the right-hand side by their partial derivative operators to achieve the final transformed equation:
\frac{1}{c} \frac{\partial L}{\partial \tau} = \frac{\partial}{\partial \zeta} \left[ \beta \left( Y - \frac{v}{c} N \right) \right] - \frac{\partial}{\partial \eta} \left[ \beta \left( Z + \frac{v}{c} M \right) \right]
The derivation for the transverse M-component is more direct. We start with the equation:
\frac{1}{c}\frac{\partial M}{\partial t} = \frac{\partial Z}{\partial x} - \frac{\partial X}{\partial z}
Applying the operator transformations gives:
\frac{1}{c} \beta \left(\frac{\partial M}{\partial \tau} - v \frac{\partial M}{\partial \xi}\right) = \beta \left(\frac{\partial Z}{\partial \xi} - \frac{v}{c^2} \frac{\partial Z}{\partial \tau}\right) - \frac{\partial X}{\partial \zeta}
We rearrange the terms to collect all time derivatives on the left and all spatial derivatives on the right:
\frac{\beta}{c} \frac{\partial M}{\partial \tau} + \frac{\beta v}{c^2} \frac{\partial Z}{\partial \tau} = \beta \frac{\partial Z}{\partial \xi} + \frac{\beta v}{c} \frac{\partial M}{\partial \xi} - \frac{\partial X}{\partial \zeta}
Factoring out the derivative operators reveals the underlying structure:
\frac{1}{c} \frac{\partial}{\partial \tau} \left[ \beta \left( M + \frac{v}{c} Z \right) \right] = \frac{\partial}{\partial \xi} \left[ \beta \left( Z + \frac{v}{c} M \right) \right] - \frac{\partial X}{\partial \zeta}
The final equation of this set follows the same straightforward pattern. We begin with:
\frac{1}{c}\frac{\partial N}{\partial t} = \frac{\partial X}{\partial y} - \frac{\partial Y}{\partial x}
Substituting the operator transformations yields:
\frac{1}{c} \beta \left(\frac{\partial N}{\partial \tau} - v \frac{\partial N}{\partial \xi}\right) = \frac{\partial X}{\partial \eta} - \beta \left(\frac{\partial Y}{\partial \xi} - \frac{v}{c^2} \frac{\partial Y}{\partial \tau}\right)
We again gather the time derivatives on the left and the spatial derivatives on the right:
\frac{\beta}{c} \frac{\partial N}{\partial \tau} - \frac{\beta v}{c^2} \frac{\partial Y}{\partial \tau} = \frac{\partial X}{\partial \eta} - \beta \frac{\partial Y}{\partial \xi} + \frac{\beta v}{c} \frac{\partial N}{\partial \xi}
Factoring out the derivative operators provides the final transformed equation for the N-component:
\frac{1}{c} \frac{\partial}{\partial \tau} \left[ \beta \left( N - \frac{v}{c} Y \right) \right] = \frac{\partial X}{\partial \eta} - \frac{\partial}{\partial \xi} \left[ \beta \left( Y - \frac{v}{c} N \right) \right]
The six transformed equations are:
\begin{array}{lcll} \dfrac{1}{c} \dfrac{\partial X}{\partial \tau} & = & \dfrac{\partial}{\partial \eta} \left[ \beta \left( N - \dfrac{v}{c} Y \right) \right] & - \dfrac{\partial}{\partial \zeta} \left[ \beta \left( M + \dfrac{v}{c} Z \right) \right] \\[10px] \dfrac{1}{c} \dfrac{\partial}{\partial \tau} \left[ \beta \left( Y - \dfrac{v}{c} N \right) \right] & = & \dfrac{\partial L}{\partial \zeta} & - \dfrac{\partial}{\partial \xi} \left[ \beta \left( N - \dfrac{v}{c} Y \right) \right] \\[10px] \dfrac{1}{c} \dfrac{\partial}{\partial \tau} \left[ \beta \left( Z + \dfrac{v}{c} M \right) \right] & = & \dfrac{\partial}{\partial \xi} \left[ \beta \left( M + \dfrac{v}{c} Z \right) \right] & - \dfrac{\partial L}{\partial \eta} \\[15px] \dfrac{1}{c} \dfrac{\partial L}{\partial \tau} & = & \dfrac{\partial}{\partial \zeta} \left[ \beta \left( Y - \dfrac{v}{c} N \right) \right] & - \dfrac{\partial}{\partial \eta} \left[ \beta \left( Z + \dfrac{v}{c} M \right) \right] \\[10px] \dfrac{1}{c} \dfrac{\partial}{\partial \tau} \left[ \beta \left( M + \dfrac{v}{c} Z \right) \right] & = & \dfrac{\partial}{\partial \xi} \left[ \beta \left( Z + \dfrac{v}{c} M \right) \right] & - \dfrac{\partial X}{\partial \zeta} \\[10px] \dfrac{1}{c} \dfrac{\partial}{\partial \tau} \left[ \beta \left( N - \dfrac{v}{c} Y \right) \right] & = & \dfrac{\partial X}{\partial \eta} & - \dfrac{\partial}{\partial \xi} \left[ \beta \left( Y - \dfrac{v}{c} N \right) \right] \end{array}
Now, we introduce the physical postulate known as the Principle of Relativity. This principle asserts that the laws of physics must retain the same mathematical form in all inertial frames of reference. Consequently, in the moving frame (\xi, \eta, \zeta, \tau), Maxwell’s equations must be written with new field components (X^\prime, Y^\prime, Z^\prime) and (L^\prime, M^\prime, N^\prime), but the structure of the equations must be identical to the original set.
The required form of Maxwell’s equations in the primed frame is therefore:
\begin{array}{lcll} \dfrac{1}{c} \dfrac{\partial X^\prime}{\partial \tau} & = & \dfrac{\partial N^\prime}{\partial \eta} & - \dfrac{\partial M^\prime}{\partial \zeta} \\[10px] \dfrac{1}{c} \dfrac{\partial Y^\prime}{\partial \tau} & = & \dfrac{\partial L^\prime}{\partial \zeta} & - \dfrac{\partial N^\prime}{\partial \xi} \\[10px] \dfrac{1}{c} \dfrac{\partial Z^\prime}{\partial \tau} & = & \dfrac{\partial M^\prime}{\partial \xi} & - \dfrac{\partial L^\prime}{\partial \eta} \\[15px] \dfrac{1}{c} \dfrac{\partial L^\prime}{\partial \tau} & = & \dfrac{\partial Y^\prime}{\partial \zeta} & - \dfrac{\partial Z^\prime}{\partial \eta} \\[10px] \dfrac{1}{c} \dfrac{\partial M^\prime}{\partial \tau} & = & \dfrac{\partial Z^\prime}{\partial \xi} & - \dfrac{\partial X^\prime}{\partial \zeta} \\[10px] \dfrac{1}{c} \dfrac{\partial N^\prime}{\partial \tau} & = & \dfrac{\partial X^\prime}{\partial \eta} & - \dfrac{\partial Y^\prime}{\partial \xi} \end{array}
The transformation for the fields is found by demanding that the first set of equations be formally identical to the second set. By comparing the two sets of equations term by term, we can make a unique identification for each primed field component.
For the equations to be identical, the quantities being differentiated must be equal. This comparison yields the following set of transformations for the electromagnetic field components:
\begin{aligned} X^\prime &= X \\ Y^\prime &= \beta \left( Y - \frac{v}{c} N \right) \\ Z^\prime &= \beta \left( Z + \frac{v}{c} M \right) \\[15px] L^\prime &= L \\ M^\prime &= \beta \left( M + \frac{v}{c} Z \right) \\ N^\prime &= \beta \left( N - \frac{v}{c} Y \right) \end{aligned}
EINSTEIN, Albert, 1905. On the Electrodynamics of Moving Bodies. English translation from "Zur elektrodynamik bewegter Korper", prepared by John Walker (1999). Annalen der Physik, 17, 891-921.