In his 1905 paper, Einstein applies Lorentz transformation on Maxwell's equations, and relates the electric force $(X,Y,Z)$ and magnetic force $(L,M,N)$ in an inertial frame $K$ with spectime coordinates $(x,y,z,t)$ at rest, and $(X',Y',Z')$ and $(L',M',N')$ in $k$ with coordinates $(\xi,\eta,\zeta,\tau)$ that is moving with $v$ with respect to $K$.

So first we have Maxwell's equations in $K$ :

$$\begin{array}{lcllcl} \displaystyle \frac1c\frac{\partial\rm X}{\partial t}&\displaystyle=&\displaystyle\frac{\partial\rm N}{\partial y}-\frac{\partial\rm M}{\partial z},&\displaystyle\frac1c\frac{\partial\rm L}{\partial t}&\displaystyle=&\displaystyle\frac{\partial\rm Y}{\partial z}-\frac{\partial\rm Z}{\partial y},\\\displaystyle \frac1c\frac{\partial\rm Y}{\partial t}&\displaystyle=&\displaystyle\frac{\partial\rm L}{\partial z}-\frac{\partial\rm N}{\partial x},&\displaystyle\frac1c\frac{\partial\rm M}{\partial t}&\displaystyle=&\displaystyle\frac{\partial\rm Z}{\partial x}-\frac{\partial\rm X}{\partial z},\\\displaystyle \frac1c\frac{\partial\rm Z}{\partial t}&\displaystyle=&\displaystyle\frac{\partial\rm M}{\partial x}-\frac{\partial\rm L}{\partial y},&\displaystyle\frac1c\frac{\partial\rm N}{\partial t}&\displaystyle=&\displaystyle\frac{\partial\rm X}{\partial y}-\frac{\partial\rm Y}{\partial x}, \end{array}$$ where $(\rm X,Y,Z)$ denotes the vector of the electric force, and $(\rm L,M,N)$ that of the magnetic force.

Applying the LT we get:

$\quad$ If we apply to these equations the transformation developed in $\S \text{3}$, by referring to the electromagnetic processes to the system of co-ordinates there introduced, moving with the velocity $v$, we obtain the equations $$\begin{array}{rll} \frac1c\frac{\partial\rm X}{\partial\tau} & = \frac{\partial}{\partial\eta}\Big\{\beta\Big(\mathrm N-\frac vc\rm Y\Big)\Big\} & -\frac{\partial}{\partial\zeta}\Big\{\beta\Big(\mathrm M+\frac vc\rm Z\Big)\Big\},\\ \frac1c\frac{\partial}{\partial\tau}\Big\{\beta\Big(\mathrm Y-\frac vc\rm N\Big)\Big\}&=\frac{\partial\rm L}{\partial\xi}&-\frac{\partial}{\partial\zeta}\Big\{\beta\Big(\mathrm N-\frac vc\rm Y\Big)\Big\},\\ \frac1c\frac{\partial}{\partial\tau}\Big\{\beta\Big(\mathrm Z+\frac vc\rm M\Big)\Big\}&=\frac{\partial}{\partial\xi}\Big\{\beta\Big(\mathrm M+\frac vc\rm Z\Big)\Big\}&-\frac{\partial\rm L}{\partial\eta},\\ \frac1c\frac{\partial\rm L}{\partial\tau}&=\frac{\partial}{\partial\zeta}\Big\{\beta\Big(\mathrm Y-\frac vc\rm N\Big)\Big\}&-\frac{\partial}{\partial\eta}\Big\{\beta\Big(\mathrm Z+\frac vc\rm M\Big)\Big\},\\ \frac1c\frac{\partial}{\partial\tau}\Big\{\beta\Big(\mathrm M+\frac vc\rm Z\Big)\Big\}&=\frac{\partial}{\partial\xi}\Big\{\beta\Big(\mathrm Z+\frac vc\rm M\Big)\Big\}&-\frac{\partial\rm X}{\partial\zeta},\\ \frac1c\frac{\partial}{\partial\tau}\Big\{\beta\Big(\mathrm N-\frac vc\rm Y\Big)\Big\}&=\frac{\partial\rm X}{\partial\eta}&-\frac\partial{\partial\zeta}\Big\{\beta\Big(\mathrm Y-\frac vc\rm N\Big)\Big\}, \end{array}$$ where $$\beta=1/\sqrt{1-v^2/c^2}.$$

Since the principle of relativity requires Maxwell's equation to have the same form in $k$ we have:

$$\begin{array}{lcllcl} \displaystyle \frac1c\frac{\partial\rm X'}{\partial\tau}&\displaystyle=&\displaystyle\frac{\partial\rm N'}{\partial \eta}-\frac{\partial\rm M'}{\partial \zeta},&\displaystyle\frac1c\frac{\partial\rm L'}{\partial \tau}&\displaystyle=&\displaystyle\frac{\partial\rm Y'}{\partial \zeta}-\frac{\partial\rm Z'}{\partial \eta},\\\displaystyle \frac1c\frac{\partial\rm Y'}{\partial \tau}&\displaystyle=&\displaystyle\frac{\partial\rm L'}{\partial \zeta}-\frac{\partial\rm N'}{\partial \xi},&\displaystyle\frac1c\frac{\partial\rm M'}{\partial \tau}&\displaystyle=&\displaystyle\frac{\partial\rm Z'}{\partial \xi}-\frac{\partial\rm X'}{\partial \zeta},\\\displaystyle \frac1c\frac{\partial\rm Z'}{\partial \tau}&\displaystyle=&\displaystyle\frac{\partial\rm M'}{\partial \xi}-\frac{\partial\rm L'}{\partial \eta},&\displaystyle\frac1c\frac{\partial\rm N'}{\partial \tau}&\displaystyle=&\displaystyle\frac{\partial\rm X'}{\partial \eta}-\frac{\partial\rm Y'}{\partial \xi}, \end{array}$$

Then Einstein goes on to say:

$\quad$ Evidently the two systems of equations found for system $k$ must express exactly the same thing, since both systems of equations are equivalent to the Maxwell-Hertz equations for system $\rm K$. Since, further, the equations of the two systems agree, with the exception of the symbols for the vectors, it follows that the functions occurring in the systems of equations at corresponding places must agree, with the exception of a factor $\psi(v)$, which is common for all functions of the one system of equations, and is independent of $\xi,\eta,\zeta$ and $\tau$ but depends upon $v$. Thus we have the relations $$\begin{array}{lcllcl} \mathrm X' & = & \psi(v)\mathrm X, & \mathrm L' & = & \psi(v)\mathrm L,\\ \mathrm Y' & = & \psi(v)\beta(\mathrm Y-\frac vc\mathrm N), & \mathrm M' & = & \psi(v)\beta(\mathrm M+\frac vc\mathrm Z),\\ \mathrm Z' & = & \psi(v)\beta(\mathrm Z+\frac vc\mathrm M), & \mathrm N' & = & \psi(v)\beta(\mathrm N-\frac vc\mathrm Y).\\ \end{array}$$

My question: If we compared the two system of equations of the system $k$ we should arrive at

$$\begin{array}{lcllcl} \mathrm X' & = & \mathrm X, & \mathrm L' & = & \mathrm L,\\ \mathrm Y' & = & \beta(\mathrm Y-\frac vc\mathrm N), & \mathrm M' & = & \beta(\mathrm M+\frac vc\mathrm Z),\\ \mathrm Z' & = & \beta(\mathrm Z+\frac vc\mathrm M), & \mathrm N' & = & \beta(\mathrm N-\frac vc\mathrm Y).\\ \end{array}$$

But instead we have a factor of $\psi(v)$ in every equation on the right hand side.

I would like to know why it's the case that

it follows that the functions occurring in the systems of equations at corresponding places must agree, with the exception of a factor $\psi(v)$

Where does $\psi(v)$ come from? and what does it represent? And why it's always positive?

Side note: He later proves that $\psi(v)=1$.


1 Answer 1


He's just trying to discuss the most general case, as happens during all previous paragraphs (the third one also discusses a $\phi(v)$ that turns out must be 1 as well):

In this paragraph, he assumes the non-prime and prime fields as already given both satisfying Maxwell equations (boxes 1 & 3). He then tries to connect them, similar to his reasoning in paragraph 3. When you compare equations from the second box to Maxwell's eqs for the prime fields (third box), you can multiply the first set by a factor $\psi(v)$ and its equations still hold, as it depends on $v$ only. Thus the most general identification of the prime fields in terms of the non-prime is including such $\psi(v)$. This factor doesn't just "appear"; he adds it ad-hoc for the sake of generality.

Up to this point, no assumption of positiveness for $\psi$ is required, just that it is a function of $v$ only.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.