Light Sphere Equation Invariance
Maxwell’s Equations Transformation
Transformation of the Wave Amplitude
Maxwell-Hertz 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}
We begin with the description of a plane electromagnetic wave in a stationary reference frame K. The electric and magnetic field components are represented as sinusoidal functions that depend on a phase factor, \Phi:
\begin{array}{l|l} X = X_0\sin(\Phi) & L = L_0\sin(\Phi)\\ Y = Y_0\sin(\Phi) & M = M_0\sin(\Phi)\\ Z = Z_0\sin(\Phi) & N = N_0\sin(\Phi)\\ \end{array}
With:
\Phi = \omega \left\{t - \frac{1}{c} (lx + my + nz) \right\}
and X_0,\dots are the vector defining the amplitude, and L_0,\dots the direction-cosine of the wave-normal.
We can apply the transformation equation we found previously for the electric and magnetic field components:
\begin{array}{l|l} X^\prime = X_0\sin(\Phi^\prime) & L^\prime = L_0\sin(\Phi^\prime) \\ Y^\prime = \beta \left(Y_0 - \frac{v}{c} N_0\right)\sin(\Phi^\prime) & M^\prime = \beta \left(M_0 + \frac{v}{c} Z_0\right)\sin(\Phi^\prime)\\ Z^\prime = \beta \left(Z_0 + \frac{v}{c} M_0\right)\sin(\Phi^\prime) & N^\prime = \beta \left(N_0 - \frac{v}{c} Y_0\right)\sin(\Phi^\prime) \end{array}
The phase in the new frame, \Phi^\prime, is a function of the coordinates (x^\prime, y^\prime, z^\prime, t^\prime):
\Phi^\prime = \omega^\prime \left(t^\prime - \frac{1}{c} (l^\prime x^\prime + m^\prime y^\prime + n^\prime z^\prime) \right)
To determine the transformed frequency \omega^\prime, we reformulate the original phase \Phi using the coordinates of the moving frame. This is achieved by substituting the Lorentz transformation equations for t, x, y, and z into the expression for \Phi^\prime:
\begin{aligned} \Phi^\prime &= \omega \left\{ \beta \left(t^\prime + \frac{v}{c^2} x^\prime\right) - \frac{1}{c} \left[ l \beta \left(x^\prime + v t^\prime\right) + m y^\prime + n z^\prime \right] \right\} \\ &= \omega \left\{ \beta t^\prime \left(1 - vl/c\right) - \frac{1}{c} \left[ x^\prime \left(-\beta\frac{v}{c} + l\beta\right) + m y^\prime + n z^\prime \right] \right\} \\ &= \omega \beta \left(1 - vl/c\right) \left\{ t^\prime - \frac{1}{c\left(1 - vl/c\right)} \left[ x^\prime \left(l - \frac{v}{c}\right) + \frac{m}{\beta} y^\prime + \frac{n}{\beta} z^\prime \right] \right\} \end{aligned}
By comparing the functional form of this transformed phase with the definition of \Phi^\prime, we can directly identify the expression for the coefficients:
\begin{aligned} & \omega^\prime = \omega \beta \left(1 - lv/c\right) \\ & l^\prime = \frac{l - v/c}{1 - lv/c} \\ & m^\prime = \frac{m}{\beta(1 - lv/c)} \\ & n^\prime = \frac{n}{\beta(1 - lv/c)} \\ \end{aligned}
The angle \phi represents the angle between the wave-normal (the direction of the wave’s propagation) and the x-axis, which is the direction of motion of the reference frame K^\prime. The direction-cosine l is, by definition, the cosine of the angle between the direction vector and the x-axis. Therefore, we have the direct relationship:
l = \cos(\phi)
Similarly, in the moving frame K^\prime, the observed angle of propagation \phi^\prime is related to the direction-cosine l^\prime by:
l^\prime = \cos(\phi^\prime)
The relationship between angular frequency \omega and frequency \nu (number of oscillations per unit time) is given by:
\omega = 2\pi\nu
We can use this to express the Doppler effect in terms of frequency, which is often more common in physical discussions. Starting with the transformation for the angular frequency:
\omega^\prime = \omega \beta \left(1 - lv/c\right)
We substitute \omega = 2\pi\nu, \omega^\prime = 2\pi\nu^\prime, and l = \cos(\phi):
2\pi\nu^\prime = 2\pi\nu \beta \left(1 - \frac{v}{c}\cos(\phi)\right)
This formula is the transformation for the frequency \nu:
\nu^\prime = \nu \beta \left(1 - \cos(\phi) \cdot v/c\right) = \nu \frac{1 - \cos(\phi) \cdot v/c}{\sqrt{1 - v^2/c^2}}
We can now take the previously derived expression for l^\prime and substitute the cosine of the angles. Starting with the transformation for the direction-cosine l^\prime:
l^\prime = \frac{l - v/c}{1 - lv/c}
We replace l with \cos(\phi) and l^\prime with \cos(\phi^\prime) to obtain the law of aberration in its angular form:
\cos(\phi^\prime) = \frac{\cos(\phi) - v/c}{1 - \cos(\phi) \cdot v/c}
This equation shows how the observed direction of a light ray changes when viewed from a moving reference frame.
The final piece of the transformation concerns the amplitude of the wave. The intensity of the light is proportional to the square of its amplitude, and it is this squared amplitude, A^{\prime 2}, which we will determine. We define the squared amplitude in each frame as the sum of the squares of the electric field component amplitudes:
A^2 = X_0^2 + Y_0^2 + Z_0^2
In the moving frame K^\prime, the squared amplitude is correspondingly:
A^{\prime 2} = X_0^{\prime 2} + Y_0^{\prime 2} + Z_0^{\prime 2}
To find the relationship between A^2 and A^{\prime 2}, we begin with the transformation equations for the electric field component amplitudes:
\begin{aligned} X_0^\prime &= X_0 \\ Y_0^\prime &= \beta \left(Y_0 - \frac{v}{c} N_0\right) \\ Z_0^\prime &= \beta \left(Z_0 + \frac{v}{c} M_0\right) \end{aligned}
For a plane wave, the magnetic field components (L_0, M_0, N_0) are related to the electric field components and the direction of propagation (l, m, n):
\begin{aligned} L_0 &= mZ_0 - nY_0 \\ M_0 &= nX_0 - lZ_0 \\ N_0 &= lY_0 - mX_0 \end{aligned}
Substituting these relations into the transformation equations yields expressions for Y_0^\prime and Z_0^\prime purely in terms of the electric field components in the stationary frame.
\begin{aligned} Y_0^\prime &= \beta \left(Y_0 - \frac{v}{c} (lY_0 - mX_0)\right) = \beta \left[Y_0\left(1 - \frac{vl}{c}\right) + X_0\frac{vm}{c}\right] \\ Z_0^\prime &= \beta \left(Z_0 + \frac{v}{c} (nX_0 - lZ_0)\right) = \beta \left[Z_0\left(1 - \frac{vl}{c}\right) + X_0\frac{vn}{c}\right] \end{aligned}
Now we can compute the squared amplitude A^{\prime 2} by squaring and summing the transformed components:
\begin{aligned} A^{\prime 2} = & X_0^2 + \beta^2 \left[Y_0\left(1 - vl/c\right) + X_0\frac{vm}{c}\right]^2 \\ & + \beta^2 \left[Z_0\left(1 - vl/c\right) + X_0\frac{vn}{c}\right]^2 \\ = & X_0^2 + \beta^2\left(1 - vl/c\right)^2 \left(Y_0^2+Z_0^2\right) \\ & + \beta^2\left(\frac{v}{c}\right)^2 X_0^2 \left(m^2+n^2\right) \\ & + 2\beta^2 \left(1 - vl/c\right)\frac{v}{c} X_0(mY_0+nZ_0) \end{aligned}
This expression can be simplified by using two fundamental properties of plane waves. First, the electric field vector is perpendicular to the direction of propagation, which implies:
\begin{aligned} & lX_0 + mY_0 + nZ_0 = 0 \\ & mY_0+nZ_0 = -lX_0 \end{aligned}
Second, the sum of the squares of the direction-cosines is one:
\begin{aligned} & l^2+m^2+n^2=1 \\ & m^2+n^2=1-l^2 \end{aligned}
Substituting these into the equation for A^{\prime 2}:
\begin{aligned} A^{\prime 2} = & X_0^2 + \beta^2\left(1 - vl/c\right)^2 \left(Y_0^2+Z_0^2\right) \\ & + \beta^2\left(v/c\right)^2 X_0^2 \left(1-l^2\right) \\ & + 2\beta^2 \left(1 - vl/c\right)v/c X_0(lX_0) \\ = & X_0^2 \left[ 1 + \beta^2\left(v/c\right)^2\left(1-l^2\right) - 2\beta^2 l v/c\left(1 - vl/c\right) \right] \\ & + \beta^2\left(1 - vl/c\right)^2 \left(Y_0^2+Z_0^2\right) \\ = & C_{X0} X_0^2 + \beta^2\left(1 - vl/c\right)^2 \left(Y_0^2+Z_0^2\right) \\ \end{aligned}
Let’s focus on the coefficient of the X_0^2 term. Using the relation \beta^2(1-(v/c)^2) =1:
\begin{aligned} C_{X0} = & 1 + \beta^2(v/c)^2 - \beta^2(v/c)^2l^2 - 2\beta^2vl/c + 2\beta^2(v/c)^2l^2 \\ = & 1 + \beta^2(v/c)^2 - 2\beta^2vl/c + \left[2\beta^2(v/c)^2l^2 - \beta^2(v/c)^2l^2\right] \\ = & 1 + \beta^2(v/c)^2 - 2\beta^2vl/c + \beta^2(v/c)^2l^2 \\ = & \beta^2(1 - (v/c)^2) + \beta^2(v/c)^2 - 2\beta^2vl/c + \beta^2(v/c)^2l^2 \\ = & \beta^2 - 2\beta^2vl/c + \beta^2(v/c)^2l^2 \\ = & \beta^2\left[1 - 2vl/c + (vl/c)^2\right] = \beta^2\left(1 - vl/c\right)^2 \end{aligned}
The coefficient of X_0^2 is identical to the coefficient of (Y_0^2+Z_0^2):
A^{\prime 2} = (X_0^2 + Y_0^2 + Z_0^2) \beta^2\left(1 - vl/c\right)^2
Replacing X_0^2 + Y_0^2 + Z_0^2 with A^2 and the direction-cosine l with \cos(\phi), we arrive at the final transformation for the squared amplitude:
A^{\prime 2} = A^2 \frac{(1 - \cos(\phi) \cdot v/c)^2}{1 - v^2/c^2}
This result shows that the transformation factor for the wave’s amplitude A^\prime is exactly the same as the transformation factor for its frequency \nu^\prime.
We begin with the sphere equation:
(x-lct)^2+(y-mct)^2+(z-nct)^2=R^2
To obtain the shape of that surface at a single instant in the moving frame set \tau=0, we use the transformation:
\begin{aligned} t &= \beta \left(\frac{v}{c^2}\xi\right) \\ x &= \beta \xi \\ y &= \eta \\ z &= \zeta \end{aligned}
We substitute those into the sphere equation:
\left(\beta\xi - l\beta v\xi/c\right)^2 +(\eta - m\beta v\xi/c)^2 +(\zeta - n\beta v\xi/c)^2 =R^2
To compute the volume of the ellipsoid at \tau=0 we a the linear map from (\xi,\eta,\zeta) to (x-lct,\;y-mct,\;z-nct):
M=\begin{bmatrix} \beta(1 - vl/c) & 0 & 0\\ - \beta m v/c & 1 & 0\\ - \beta n v/c & 0 & 1 \end{bmatrix}
The symmetric object is A=M^TM, and the surface is:
\|Mv\|^2=R^2
so the quadratic form matrix is A = M^TM, which is symmetric positive-definite. Volumes scale by |\det( M)|, equivalently by \sqrt{\det (A)}.
We can compute A:
\begin{aligned} A & = M^T M \\ & = \begin{bmatrix} \beta(1 - vl/c) & -\beta mv/c & - \beta nv/c \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \beta(1 - vl/c) & 0 & 0\\ - \beta mv/c & 1 & 0\\ - \beta nv/c & 0 & 1 \end{bmatrix} \\ & = \begin{bmatrix} \beta^{2}(1-2lv/c+v^2/c^2) & -\beta m v/c & -\beta n v/c \\ -\beta m v/c & 1 & 0\\ -\beta n v/c & 0 & 1 \end{bmatrix} \end{aligned}
for A_{11} we used the relationship m^2 + n^2 = 1 -l^2 to simplify the formula.
We then compute the determinant:
\begin{aligned} \det (A) & = \beta^{2}(1-2lv/c+v^2/c^2) - \beta^2 m^2 v^2/c^2 - \beta^2 n^2 v^2/c^2 \\ & = \beta^{2}(1-2lv/c+v^2/c^2) - \beta^2 v^2/c^2 (1 - l^2) \\ & = \beta^{2}(1-2lv/c+l^2v^2/c^2) \\ & = \beta^{2}(1-lv/c)^2 \end{aligned}
The volume of the ellipsoid is:
S^\prime=\frac{\frac{4}{3}\pi R^3}{\sqrt{\det A}} =\frac{\frac{4}{3}\pi R^3}{\beta(1-lv/c)}
Since the volume of the sphere is:
S = \frac{4}{3}\pi R^3
The ratio of the volumes is:
\frac{S^\prime}{S} = \frac{1}{\beta(1-lv/c)} = \frac{\sqrt{1 - v^2/c^2}}{1 - \cos(\phi) \cdot v/c}
as l = \cos(\phi).
The ratio of the energy enclose by the surface is then:
\begin{aligned} \frac{E^\prime}{E} & = \frac{A^{\prime 2}S^\prime}{A S} = A^2 \frac{(1 - \cos(\phi) \cdot v/c)^2}{1 - v^2/c^2} \frac{1}{A^2} \frac{\sqrt{1 - v^2/c^2}}{1 - \cos(\phi) \cdot v/c} \\ & = \frac{1 - \cos(\phi) \cdot v/c}{1 - v^2/c^2}\sqrt{1 - v^2/c^2} \\ & = \frac{1 - \cos(\phi) \cdot v/c}{\sqrt{1 - v^2/c^2}} \end{aligned}
Having established the transformation of energy, we can now derive the characteristics of the incident and reflected light waves as measured in the rest frame of the mirror. This will provide the necessary formulas to calculate the light pressure.
We will define three states for the light wave:
We begin with the properties of the incident plane wave in the stationary reference frame K. These are the initial values before any transformation:
Next, we describe the wave as it is observed from the reference frame K^\prime, which is moving with velocity v along the x-axis. These are the quantities an observer on the moving surface would measure for the incoming light. The transformation formulas are derived directly from the principles of special relativity.
The transformed amplitude, A^\prime, is given by:
A^\prime = A \frac{1 - (v/c)\cos(\phi)}{\sqrt{1-v^2/c^2}} = A \beta(1 - (v/c)\cos(\phi))
The transformed frequency, \nu^\prime, is a result of the relativistic Doppler effect:
\nu^\prime = \nu \frac{1 - (v/c)\cos(\phi)}{\sqrt{1-v^2/c^2}} = \nu \beta(1 - (v/c)\cos(\phi))
The transformed angle, \phi^\prime, is determined by the relativistic aberration of light:
\cos(\phi^\prime) = \frac{\cos(\phi) - v/c}{1 - (v/c)\cos(\phi)}
Now, we consider the wave after it reflects off the surface. This reflection event is analyzed entirely within the moving frame K^\prime. For a perfect mirror, the reflection is elastic. The frequency and amplitude of the wave do not change upon reflection within this frame:
\begin{aligned} A^{\prime\prime} &= A^\prime \\ \nu^{\prime\prime} &= \nu^\prime \end{aligned}
The law of reflection holds in the rest frame of the mirror. This means the angle of reflection equals the angle of incidence, but the direction of the x-component of propagation is reversed:
\cos(\phi^{\prime\prime}) = -\cos(\phi^\prime)
These quantities describe the reflected wave as measured by an observer in the moving frame K^\prime.
Now we perform the final step, which is to transform the properties of the reflected wave from the moving frame K^\prime back to the original stationary frame K. This will give us the quantities A^{\prime\prime\prime}, \cos(\phi^{\prime\prime\prime}), and \nu^{\prime\prime\prime}, which describe the reflected wave as seen by an observer in the stationary frame.
To do this, we apply the inverse Lorentz transformation formulas. The inverse transformation is found by taking the original formulas and replacing v with -v.
The inverse transformation for frequency is:
\nu^{\prime\prime\prime} = \nu^{\prime\prime} \beta(1 + (v/c)\cos(\phi^{\prime\prime}))
We proceed by substituting the known relationships for the double-primed quantities, which are the laws of reflection in the moving frame K^\prime.
First, substitute \nu^{\prime\prime} = \nu^\prime and \cos(\phi^{\prime\prime}) = -\cos(\phi^\prime):
\nu^{\prime\prime\prime} = \nu^\prime \beta(1 - (v/c)\cos(\phi^\prime))
Now, we substitute the expressions for \nu^\prime and \cos(\phi^\prime) in terms of the original incident wave properties (\nu, \phi):
\begin{aligned} \nu^{\prime\prime\prime} &= \left( \nu \beta(1 - (v/c)\cos(\phi)) \right) \cdot \beta\left(1 - (v/c)\frac{\cos(\phi) - v/c}{1 - (v/c)\cos(\phi)}\right) \\ &= \nu\beta^2 (1 - (v/c)\cos(\phi)) \left( \frac{(1 - (v/c)\cos(\phi)) - (v/c)(\cos(\phi) - v/c)}{1 - (v/c)\cos(\phi)} \right) \end{aligned}
The term (1 - (v/c)\cos(\phi)) cancels out. We are left with simplifying the numerator:
\begin{aligned} \nu^{\prime\prime\prime} &= \nu\beta^2 \left( 1 - (v/c)\cos(\phi) - (v/c)\cos(\phi) + v^2/c^2 \right) \\ &= \nu\beta^2 \left( 1 - 2(v/c)\cos(\phi) + v^2/c^2 \right) \end{aligned}
Finally, substituting \beta^2 = (1 - v^2/c^2)^{-1}:
\nu^{\prime\prime\prime} = \nu \frac{1 - 2(v/c)\cos(\phi) + v^2/c^2}{1 - v^2/c^2}
The inverse transformation for the angle is:
\cos(\phi^{\prime\prime\prime}) = \frac{\cos(\phi^{\prime\prime}) + v/c}{1 + (v/c)\cos(\phi^{\prime\prime})}
Substituting \cos(\phi^{\prime\prime}) = -\cos(\phi^\prime):
\cos(\phi^{\prime\prime\prime}) = \frac{-\cos(\phi^\prime) + v/c}{1 - (v/c)\cos(\phi^\prime)}
Now, Substituting the expression for \cos(\phi^\prime):
\cos(\phi^{\prime\prime\prime}) = \frac{-\left(\frac{\cos(\phi) - v/c}{1 - (v/c)\cos(\phi)}\right) + v/c}{1 - (v/c)\left(\frac{\cos(\phi) - v/c}{1 - (v/c)\cos(\phi)}\right)}
To simplify, multiply the numerator and denominator by (1 - (v/c)\cos(\phi)):
\begin{aligned} \cos(\phi^{\prime\prime\prime}) &= \frac{-(\cos(\phi) - v/c) + (v/c)(1 - (v/c)\cos(\phi))}{(1 - (v/c)\cos(\phi)) - (v/c)(\cos(\phi) - v/c)} \\ &= \frac{-\cos(\phi) + v/c + v/c - (v^2/c^2)\cos(\phi)}{1 - (v/c)\cos(\phi) - (v/c)\cos(\phi) + v^2/c^2} \\ &= \frac{-(1+v^2/c^2)\cos(\phi) + 2v/c}{1 - 2(v/c)\cos(\phi) + v^2/c^2} \end{aligned}
The inverse transformation for the amplitude has the same form as the frequency transformation:
A^{\prime\prime\prime} = A^{\prime\prime} \beta(1 + (v/c)\cos(\phi^{\prime\prime}))
The algebraic steps are identical to the derivation for the frequency. We substitute A^{\prime\prime} = A^\prime and follow the same simplifications:
\begin{aligned} A^{\prime\prime\prime} &= A^\prime \beta(1 - (v/c)\cos(\phi^\prime)) \\ &= A \beta^2 (1 - 2(v/c)\cos(\phi) + v^2/c^2) \end{aligned}
This gives the final expression for the amplitude of the reflected wave as measured in the stationary frame K:
A^{\prime\prime\prime} = A \frac{1 - 2(v/c)\cos(\phi) + v^2/c^2}{1 - v^2/c^2}
The pressure exerted by light on a reflecting surface can be determined from the principle of conservation of energy. The work done on the mirror per unit area, per unit time, must equal the net energy lost by the electromagnetic field per unit area, per unit time. This net energy change is the difference between the incident energy flux and the emergent (reflected) energy flux.
We define the pressure P through the relation:
P v = \Phi_{\text{in}} - \Phi_{\text{out}}
Here, \Phi_{\text{in}} is the energy arriving at the mirror, and \Phi_{\text{out}} is the energy leaving the mirror, both per unit area and time as measured in the stationary frame K.
The incident energy flux is given by:
\Phi_{\text{in}} = \frac{A^2}{8\pi}(c\cos(\phi) - v)
The emergent energy flux depends on the properties of the reflected wave after being transformed back into the stationary frame K:
\Phi_{\text{out}} = \frac{A^{\prime\prime\prime2}}{8\pi}(-c\cos(\phi)''' + v)
Our first task is to construct the expression for \Phi_{\text{out}} using the transformation formulas for the reflected wave’s amplitude A^{\prime\prime\prime} and angle \phi^{\prime\prime\prime}.
We start by evaluating the term (-c\cos(\phi)^{\prime\prime\prime} + v). Substituting the expression for \cos(\phi)^{\prime\prime\prime}:
\begin{aligned} -c\cos(\phi)^{\prime\prime\prime} + v &= -c\left(-\frac{\left(1+v^2/c^2\right)\cos(\phi)-2v/c}{1-2v/c\cos(\phi)+v^2/c^2}\right) + v \\ &= c\left(\frac{\left(1+v^2/c^2\right)\cos(\phi)-2v/c}{1-2v/c\cos(\phi)+v^2/c^2}\right) + v \end{aligned}
To simplify, we bring both terms to a common denominator:
\begin{aligned} -c\cos(\phi)^{\prime\prime\prime} + v &= c\left[\frac{\left(1+v^2/c^2\right)\cos(\phi)-2v/c + v/c\left(1-2v/c\cos(\phi)+v^2/c^2\right)}{1-2v/c\cos(\phi)+v^2/c^2}\right] \\ &= c\left[\frac{\cos(\phi) + v^2/c^2\cos(\phi) - 2v/c + v/c - 2v^2/c^2\cos(\phi) + v^3/c^3}{1-2v/c\cos(\phi)+v^2/c^2}\right] \end{aligned}
Collecting terms in the numerator gives:
\begin{aligned} \cos(\phi) - v^2/c^2\cos(\phi) - v/c + v^3/c^3 & = \cos(\phi)\left(1-v^2/c^2\right) - v/c\left(1-v^2/c^2\right) \\ &= \left(1-v^2/c^2\right)\left(\cos(\phi) - v/c\right) \end{aligned}
This provides an expression for the velocity-dependent part of the flux:
-c\cos(\phi)^{\prime\prime\prime} + v = c\,\frac{\left(1-v^2/c^2\right)\left(\cos(\phi)-v/c\right)}{1-2v/c\cos(\phi)+v^2/c^2}
Now we can assemble the full expression for \Phi_{\text{out}} by substituting this result along with the formula for A^{\prime\prime\prime2}:
\begin{aligned} \Phi_{\text{out}} &= \frac{1}{8\pi}\left[A\,\frac{1-2v/c\cos(\phi)+v^2/c^2}{1-v^2/c^2}\right]^2 \left[c\,\frac{\left(1-v^2/c^2\right)\left(\cos(\phi)-v/c\right)}{1-2v/c\cos(\phi)+v^2/c^2}\right] \\ &= \frac{A^2}{8\pi} \frac{\left(1-2v/c\cos(\phi)+v^2/c^2\right)^2}{\left(1-v^2/c^2\right)^2} c\,\frac{\left(1-v^2/c^2\right)\left(\cos(\phi)-v/c\right)}{1-2v/c\cos(\phi)+v^2/c^2} \end{aligned}
Canceling common factors yields the final form for the emergent flux:
\Phi_{\text{out}} = \frac{A^2}{8\pi}c\left(\cos(\phi)-v/c\right) \frac{1-2v/c\cos(\phi)+v^2/c^2}{1-v^2/c^2}
The net energy flux \Delta\Phi is the difference \Phi_{\text{in}} - \Phi_{\text{out}}. We can factor out the common term \frac{A^2}{8\pi}c\left(\cos(\phi) - v/c\right):
\begin{aligned} \Delta\Phi &= \Phi_{\text{in}} - \Phi_{\text{out}} \\ &= \frac{A^2}{8\pi}c\left(\cos(\phi) - v/c\right) \left[1 - \frac{1-2v/c\cos(\phi)+v^2/c^2}{1-v^2/c^2}\right] \end{aligned}
We simplify the term inside the square brackets:
\begin{aligned} \left[\dots\right] &= \frac{\left(1-v^2/c^2\right) - \left(1-2v/c\cos(\phi)+v^2/c^2\right)}{1-v^2/c^2} \\ &= \frac{1-v^2/c^2 - 1+2v/c\cos(\phi)-v^2/c^2}{1-v^2/c^2} \\ &= \frac{2v/c\cos(\phi)-2v^2/c^2}{1-v^2/c^2} = \frac{2v/c\left(\cos(\phi)-v/c\right)}{1-v^2/c^2} \end{aligned}
Substituting this back into the expression for \Delta\Phi:
\begin{aligned} \Delta\Phi &= \frac{A^2}{8\pi}c\left(\cos(\phi) - v/c\right) \frac{2v/c\left(\cos(\phi)-v/c\right)}{1-v^2/c^2} \\ &= \frac{2A^2}{8\pi} v \frac{\left(\cos(\phi)-v/c\right)^2}{1-v^2/c^2} \end{aligned}
Finally, we find the pressure P by dividing the work rate \Delta\Phi by the velocity v:
P = \frac{\Delta\Phi}{v} = 2\cdot\frac{A^2}{8\pi} \frac{\left(\cos(\phi)-v/c\right)^2}{1-v^2/c^2}
For velocities that are small compared to the speed of light, we can examine the first-order approximation. In this case, terms of order v/c are kept, while higher-order terms like (v/c)^2 are neglected. The denominator approaches 1 and the numerator approaches \cos^2(\phi).
P \approx 2\cdot\frac{A^2}{8\pi}\cos^2(\phi)
The transformation of the right-hand side of the equation, which contains the spatial derivatives of the fields, is identical to the procedure we have already detailed for the source-free case here. We can therefore reference that result and focus only on the new algebraic steps introduced by the charge density \rho and the convection current \rho \mathbf u.
We begin with the first Maxwell-Hertz equation in the stationary frame K:
\frac{1}{c}\left(\frac{\partial X}{\partial t} + \rho u_x \right) = \frac{\partial N}{\partial y} - \frac{\partial M}{\partial z}
Here \rho is the volume charge density and u_x is the x-component of the charge velocity. Applying the Lorentz transformation for the partial derivatives to the left-hand side gives:
\frac{1}{c}\left[\beta\left(\frac{\partial X}{\partial \tau} - v \frac{\partial X}{\partial \xi}\right) + \rho u_x \right] = \frac{\partial N}{\partial \eta} - \frac{\partial M}{\partial \zeta}
The difference from the source-free case is in the divergence equation (Gauss’s Law), which now includes the charge density:
\frac{\partial X}{\partial x} + \frac{\partial Y}{\partial y} + \frac{\partial Z}{\partial z} = \rho
Transforming this equation into the moving frame coordinates provides our substitution term:
\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} = \rho
Solving for \partial X/\partial \xi yields a new expression that incorporates \rho:
\frac{\partial X}{\partial \xi} = \frac{v}{c^2}\frac{\partial X}{\partial \tau} + \frac{1}{\beta}\left(\rho - \frac{\partial Y}{\partial \eta} - \frac{\partial Z}{\partial \zeta}\right)
We now substitute this back into the main transformed equation:
\frac{1}{c}\left[\beta\left(\frac{\partial X}{\partial \tau} - v\left\{\frac{v}{c^2}\frac{\partial X}{\partial \tau} + \frac{1}{\beta}\left(\rho - \frac{\partial Y}{\partial \eta} - \frac{\partial Z}{\partial \zeta}\right)\right\}\right) + \rho u_x \right] = \frac{\partial N}{\partial \eta} - \frac{\partial M}{\partial \zeta}
Distributing the terms and using the relation 1-v^2/c^2 = 1/\beta^2:
\frac{1}{c}\left[\frac{1}{\beta}\frac{\partial X}{\partial \tau} - v\rho + v\left(\frac{\partial Y}{\partial \eta} + \frac{\partial Z}{\partial \zeta}\right) + \rho u_x \right] = \frac{\partial N}{\partial \eta} - \frac{\partial M}{\partial \zeta}
Isolating the time derivative \partial X/\partial \tau:
\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) - \frac{\beta}{c}\rho(u_x-v)
The first part of the right-hand side, involving the derivatives of the fields, transforms as demonstrated here. This combination is equal to:
\frac{\partial N^\prime}{\partial \eta} - \frac{\partial M^\prime}{\partial \zeta}
The new task is to transform the remaining term, -\frac{\beta}{c}\rho(u_x-v), which contains the charge and current. To do this, we must express it using quantities measured in the moving frame: the charge density \rho^\prime and the charge velocity u_\xi. The transformation for charge density is:
\rho^\prime = \beta\rho\left(1 -u_x v / c^2\right)
The velocity component u_\xi is given by the relativistic velocity addition formula:
u_\xi = \frac{u_x - v}{1 -u_x v / c^2}
By combining these, we find the transformation for the \xi-component of the current density, \rho^\prime u_\xi:
\begin{aligned} \rho^\prime u_\xi &= \beta\rho\left(1 -u_x v / c^2\right) \left(\frac{u_x - v}{1 -u_x v / c^2}\right) \\ &= \beta \rho (u_x - v) \end{aligned}
Substituting these known transformations back into our equation gives:
\frac{1}{c}\frac{\partial X}{\partial \tau} = \left(\frac{\partial N^\prime}{\partial \eta} - \frac{\partial M^\prime}{\partial \zeta}\right) - \frac{1}{c}\rho^\prime u_\xi
Finally, since the electric field component parallel to the direction of motion is invariant, X^\prime = X, we can rewrite the equation in its final form, which is identical in structure to the original:
\frac{1}{c}\left(\frac{\partial X^\prime}{\partial \tau} + \rho^\prime u_\xi \right) = \frac{\partial N^\prime}{\partial \eta} - \frac{\partial M^\prime}{\partial \zeta}
We start with the Maxwell-Hertz equation for the Y-component in the stationary frame K:
\frac{1}{c}\left(\frac{\partial Y}{\partial t} + \rho u_y \right) = \frac{\partial L}{\partial z} - \frac{\partial N}{\partial x}
Transforming the partial derivatives and rearranging the terms to group the combinations that form the transformed fields yields:
\frac{\beta}{c}\frac{\partial}{\partial \tau}\left(Y - \frac{v}{c}N\right) = \frac{\partial L}{\partial \zeta} - \beta\frac{\partial}{\partial \xi}\left(N - \frac{v}{c}Y\right) - \frac{1}{c}\rho u_y
The combination of the spatial derivatives on the right-hand side transforms as demonstrated here. This combination is equal to:
\frac{\partial L^\prime}{\partial \zeta} - \frac{\partial N^\prime}{\partial \xi}
The left-hand side of the equation contains the time derivative of the transformed field component Y^\prime = \beta(Y - \frac{v}{c}N). The remaining term is the current density, which transforms as \rho^\prime u_\eta = \rho u_y. Substituting these known transformations, the equation becomes:
\frac{1}{c}\frac{\partial Y^\prime}{\partial \tau} = \frac{\partial L^\prime}{\partial \zeta} - \frac{\partial N^\prime}{\partial \xi} - \frac{1}{c}\rho^\prime u_\eta
Rearranging this expression gives the final equation in a form identical to the original:
\frac{1}{c}\left(\frac{\partial Y^\prime}{\partial \tau} + \rho^\prime u_\eta\right) = \frac{\partial L^\prime}{\partial \zeta} - \frac{\partial N^\prime}{\partial \xi}
The derivation for the Z-component follows the same logic. We begin with the equation in frame K:
\frac{1}{c}\left(\frac{\partial Z}{\partial t} + \rho u_z \right) = \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}
After transforming the partial derivatives and regrouping the terms, we have:
\frac{\beta}{c}\frac{\partial}{\partial \tau}\left(Z + \frac{v}{c}M\right) = \beta\frac{\partial}{\partial \xi}\left(M + \frac{v}{c}Z\right) - \frac{\partial L}{\partial \eta} - \frac{1}{c}\rho u_z
The combination of the spatial derivatives on the right-hand side transforms as demonstrated here. This combination is equal to:
\frac{\partial M^\prime}{\partial \xi} - \frac{\partial L^\prime}{\partial \eta}
The left-hand side is the time derivative of the transformed field Z^\prime = \beta(Z + \frac{v}{c}M), and the current density transforms as \rho^\prime u_\zeta = \rho u_z. By substituting these transformed parts into the equation, we obtain:
\frac{1}{c}\frac{\partial Z^\prime}{\partial \tau} = \frac{\partial M^\prime}{\partial \xi} - \frac{\partial L^\prime}{\partial \eta} - \frac{1}{c}\rho^\prime u_\zeta
This is rearranged into the final covariant form:
\frac{1}{c}\left(\frac{\partial Z^\prime}{\partial \tau} + \rho^\prime u_\zeta\right) = \frac{\partial M^\prime}{\partial \xi} - \frac{\partial L^\prime}{\partial \eta}
Let us consider the motion of the electron. In the system K fixed to the field, the equations of motion are written in the usual Newtonian form with the electric field components (X, Y, Z). In the system k, moving with the electron, the same equations must hold but expressed in terms of the new coordinates \xi,\eta,\zeta,\tau. The task is then to determine how the derivatives with respect to \tau in k transform into derivatives with respect to t in K.
Let us begin with the equation of motion in the X-direction, valid in the system K:
m \frac{\mathrm d^2 \xi}{\mathrm d\tau^2} = \varepsilon X^\prime
Transforming to the system k, moving with velocity v along the x-axis, we use the Lorentz transformations. Differentiating, the velocity in the new coordinates is:
\begin{aligned} u_\xi = \frac{\mathrm d\xi}{\mathrm d\tau} & = \frac{\mathrm d\xi}{\mathrm dt}\frac{\mathrm dt}{\mathrm d\tau} = \frac{\mathrm d\xi /\mathrm dt}{\mathrm d\tau /\mathrm d t} = \frac{\beta(u_x - v)}{\beta (1-vu_x/c^2)} = \frac{u_x - v}{1-vu_x/c^2} \end{aligned}
where we used:
\frac{\mathrm d\tau}{\mathrm dt} = \frac{\mathrm d}{\mathrm dt}\left[\beta (1-vx/c^2)\right] = \beta (1-vu_x/c^2)
and differentiating again using the quotient rule:
\frac{\mathrm d}{\mathrm dt}\left(\frac{u}{w}\right) = \frac{u^\prime w - u w^\prime}{w^2}
Defining w \equiv 1-v u_x/c^2 we have w^\prime = -va_x/c^2:
\begin{aligned} a_\xi & = \frac{\mathrm d}{\mathrm d\tau}\left(\frac{\mathrm d\xi}{\mathrm d\tau}\right) = \frac{\mathrm d}{\mathrm d\tau}\left(u^\prime_\xi\right) = \frac{1}{\beta w}\frac{\mathrm d}{\mathrm dt}\left( \frac{u_x - v}{1-vu_x/c^2}\right) \\ & = \frac{1}{\beta w}\frac{\mathrm d}{\mathrm dt} \left( \frac{u_x - v}{w}\right) = \frac{1}{\beta w} \frac{a_x w + (u_x -v) v a_x /c^2}{w^2} = \frac{a_x}{\beta w^3} (w + (u_x -v)v/c^2) \\ & \frac{a_x}{\beta w^3} (1-v u_x/c^2 + u_xv/c^2 - v^2/c^2) = \frac{a_x}{\beta w^3} (1 - v^2/c^2) = \frac{a_x}{\beta^3 w^3} \end{aligned}
We then consider that, at t=0, u_x = v, and w=1-v^2/c^2 = \beta^{-2} and the above becomes:
\frac{\mathrm d^2\xi}{\mathrm d\tau^2} = \frac{a_x}{\beta^3 w^3} = a_x \beta^3
Using the equation X^\prime = X previously derived, we have:
m \frac{\mathrm d^2 \xi}{\mathrm d\tau^2} = m \beta^3 \frac{\mathrm d^2x}{\mathrm dt^2} = \varepsilon X^\prime = \varepsilon X
Solving for the acceleration gives:
\frac{\mathrm d^2x}{\mathrm dt^2} = \frac{\varepsilon}{m\beta^3}X
For the y component we can apply similar steps and find the velocity transformation for the transverse component, differentiate it with respect to proper time \tau using the chain rule, and then apply the instantaneous rest frame condition:
u_\eta = \frac{\mathrm d\eta}{\mathrm d\tau} = \frac{\mathrm d\eta/\mathrm dt}{\mathrm d\tau/\mathrm dt} = \frac{u_y}{\beta(1-vu_x/c^2)}= \frac{u_y}{\beta w}
Now, we find the acceleration in frame k:
a_\eta = \frac{\mathrm d^2\eta}{\mathrm d\tau^2} =\frac{1}{\beta w} \frac{\mathrm d}{\mathrm dt}\left(\frac{u_y}{\beta w}\right) = \frac{1}{\beta w}\left(\frac{u_y}{\beta w}\right) = \frac{a_y \beta w - v_yv v_x/c^2}{\beta^3 w^3}
We then apply the instantaneous rest frame condition. At the moment of observation, the velocity of the electron is purely along the x-axis and is equal to the velocity of the frame k:
\begin{aligned} & u_x = v \\ & u_y = 0 \end{aligned}
Substituting these conditions into the general transformation:
\frac{\mathrm d^2\eta}{\mathrm d\tau^2} = \frac{a_y \beta w}{\beta^3 w^3} = \frac{a_y}{\beta^2 w^2} = a_y \beta^2 = \beta^2 \frac{\mathrm d^2y}{\mathrm dt^2}
We use the equation of motion in the rest frame, m \frac{\mathrm d^2\eta}{\mathrm d\tau^2} = \varepsilon Y^\prime, and the field transformation Y^\prime = \beta(Y - \frac{v}{c}N_z):
m\left(\beta^2 \frac{\mathrm d^2y}{\mathrm dt^2}\right) = \varepsilon \beta\left(Y - \frac{v}{c}N_z\right)
We obtain the equation of motion for the transverse component:
\frac{\mathrm d^2y}{\mathrm dt^2} = \frac{\varepsilon}{m\beta}\left(Y - \frac{v}{c}N_z\right)
For the z component we can apply the same step as y:
u_\zeta = \frac{\mathrm d\zeta}{\mathrm d\tau} = \frac{\mathrm d\zeta/\mathrm dt}{\mathrm d\tau/\mathrm dt} = \frac{u_z}{\beta(1-vu_x/c^2)}= \frac{u_z}{\beta w}
Now, we find the acceleration in frame k, \frac{\mathrm d^2\zeta}{\mathrm d\tau^2}, by differentiating u^\prime_\zeta with respect to \tau:
a_\zeta = \frac{\mathrm d^2\zeta}{\mathrm d\tau^2} = \frac{1}{\beta w} \frac{\mathrm d}{\mathrm dt}\left(\frac{u_z}{\beta w}\right) = \frac{a_z \beta w + u_z \beta v a_x/c^2}{\beta^3 w^3}
Now we apply the instantaneous rest frame condition. At the moment of observation, the velocity of the electron is purely along the x-axis and is equal to the velocity of the frame k:
\begin{aligned} & u_x = v \\ & u_z = 0 \end{aligned}
Substituting these conditions into the general transformation, the second term in the numerator vanishes:
\frac{\mathrm d^2\zeta}{\mathrm d\tau^2} = \frac{a_z \beta w}{\beta^3 w^3} = \frac{a_z}{\beta^2 w^2}
Since u_x=v, we have w = 1 - v^2/c^2 = 1/\beta^2. Substituting this in:
\frac{a_z}{\beta^2 (1/\beta^2)^2} = a_z \beta^2 = \beta^2 \frac{\mathrm d^2z}{\mathrm dt^2}
Finally, we use the equation of motion in the rest frame, m \frac{\mathrm d^2\zeta}{\mathrm d\tau^2} = \varepsilon Z^\prime, and the field transformation for the z-component, Z^\prime = \beta(Z + \frac{v}{c}N_y):
m\left(\beta^2 \frac{\mathrm d^2z}{\mathrm dt^2}\right) = \varepsilon \beta\left(Z + \frac{v}{c}N_y\right)
We obtain the equation of motion for the transverse component:
\frac{\mathrm d^2z}{\mathrm dt^2} = \frac{\varepsilon}{m\beta}\left(Z + \frac{v}{c}N_y\right)
Let us consider the work, W, done by the electric force component eX on the electron as it moves along the x-axis from an initial position to a final position. The work done is given by the integral of the force over the displacement:
W = \int F_x \, \mathrm dx = \int eX \, \mathrm dx
From the longitudinal equation of motion derived previously, we have the expression for the force:
eX = m \beta^3 \frac{\mathrm d^2x}{\mathrm dt^2}
The acceleration term, \frac{\mathrm d^2x}{\mathrm dt^2}, can be rewritten in a way that facilitates integration with respect to position. Using the chain rule, we express acceleration in terms of velocity v and position x:
\frac{\mathrm d^2x}{\mathrm dt^2} = \frac{\mathrm dv}{\mathrm dt} = \frac{\mathrm dv}{\mathrm dx}\frac{\mathrm dx}{\mathrm dt} = v \frac{\mathrm dv}{\mathrm dx}
Substituting this into the force equation gives:
eX = m \beta^3 v \frac{\mathrm dv}{\mathrm dx}
We can now substitute this expression for the force into the work integral. The differential element \mathrm dx cancels, changing the variable of integration from position x to velocity v. We integrate from an initial velocity of 0 to a final velocity of v:
W = \int_0^v \left(m \beta^3 v \frac{\mathrm dv}{\mathrm dx}\right) \mathrm dx = m \int_0^v \beta^3 v \, \mathrm dv
The kinetic energy, T, is equal to this work W. To compute the integral, we substitute the definition of the Lorentz factor, \beta = (1 - v^2/c^2)^{-1/2}:
T = m \int_0^v \left(1 - \frac{v^2}{c^2}\right)^{-3/2} v \, \mathrm dv
This integral can be solved using a change of variables. Let s = 1 - v^2/c^2. The differential is then \mathrm ds = -2v/c^2 \, \mathrm dv, which implies v \, \mathrm dv = - \frac{c^2}{2} \, \mathrm ds. We must also transform the limits of integration: v=0, s = 1 - 0 = 1 and when v=v, s = 1 - v^2/c^2.
Substituting these into the integral:
\begin{aligned} T &= m \int_1^{1-v^2/c^2} s^{-3/2} \left(-\frac{c^2}{2}\right) \mathrm ds \\ &= -\frac{mc^2}{2} \int_1^{1-v^2/c^2} s^{-3/2} \, \mathrm ds\\ & = -\frac{mc^2}{2} \left. \frac{s^{-1/2}}{-1/2} \right|_1^{1-v^2/c^2} \\ &= mc^2 \left. s^{-1/2} \right|_1^{1-v^2/c^2} \\ &= mc^2 \left( \frac{1}{\sqrt{1-v^2/c^2}} - 1 \right) \end{aligned}
This result expresses the kinetic energy of a relativistic particle, and demonstrates that the kinetic energy is the difference between the total energy of the particle, E = m c^2 \beta, and its rest energy, E_0 = mc^2.
In the low-velocity limit (v \ll c), a binomial expansion of \beta \approx 1 + \frac{1}{2}\frac{v^2}{c^2} shows that this expression correctly reduces to the classical kinetic energy, T \approx \frac{1}{2}mv^2.
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.