Galilean non-invariance of classical electromagnetism

Had Galilean transformation holds for not only mechanics but also electromagnetism, Newtonian relativity would hold for the whole of the physics. However, we know from Maxwell's equation that ${\displaystyle c={\frac {1}{\sqrt {\nu _{0}\epsilon _{0}}}}=2.997925\times 10^{8}m/sec}$, which is the velocity of the propagation of electromagnetic waves in vacuum.^[1] Hence, it is important to check if Maxwell's equation is invariant under Galilean relativity. For this, we have to find the difference (if any), in the observed force of charge when it is moving at a certain velocity and observed by two reference frames ${\displaystyle S^{\prime }}$ and ${\displaystyle S}$ in such a way that the velocity of ${\displaystyle S^{\prime }}$ is ${\displaystyle {\vec {v_{0}}}}$ more than ${\displaystyle S}$ (which is at absolute rest).^[2]

Electric and magnetic field under Galilean relativityEdit

In order to check whether Maxwell's equation is invariant under Galilean transformation, we have to check how the electric and magnetic field transforms under Galilean transformation.Let a charged particle/s or body is moving at a velocity ${\displaystyle {\vec {v}}}$ with respect to S frame.
So, we know that ${\displaystyle {\vec {F}}=q({\vec {E}}+{\vec {v}}\times {\vec {B}})}$ in ${\displaystyle S}$ frame and ${\displaystyle {\vec {F}}\prime =q\prime ({\vec {E}}\prime +({\vec {v}}-{\vec {v_{0}}})\times {\vec {B}}\prime )}$ in${\displaystyle S\prime }$ frame from Lorentz Force.
Now, we assume that Galilean invariance holds. That is, ${\displaystyle {\vec {F}}={\vec {F}}\prime }$ and ${\displaystyle q=q\prime }$(from observation).
${\displaystyle \implies q({\vec {E}}+{\vec {v}}\times {\vec {B}})=q\prime ({\vec {E}}\prime +({\vec {v}}-{\vec {v_{0}}})\times {\vec {B}}\prime )}$

${\displaystyle \implies {\vec {E}}+{\vec {v}}\times {\vec {B}}={\vec {E}}\prime +({\vec {v}}-{\vec {v_{0}}})\times {\vec {B}}\prime }$

(1)

${\displaystyle \implies {\vec {E}}\prime ={\vec {E}}+({\vec {v}}\times {\vec {B}})+({\vec {v_{0}}}\times {\vec {B}}\prime )-({\vec {v}}\times {\vec {B}}\prime )}$
This equation is valid for all ${\displaystyle {\vec {v}}}$.
Let, ${\displaystyle {\vec {v}}=0}$

${\displaystyle \therefore {\vec {E}}\prime ={\vec {E}}+({\vec {v_{0}}}\times {\vec {B}}\prime )}$

(a)

By using equation (a) in (1), we get

${\displaystyle {\vec {B}}\prime ={\vec {B}}}$

(b)

Transformation of ${\displaystyle \rho }$ and ${\displaystyle {\vec {J}}}$Edit

Now, we have to find the transformation(if any) of charge and current densities under Galilean transformation.
Let, ${\displaystyle \rho }$ and ${\displaystyle {\vec {J}}}$ be charge and current densities with respective to S frame respectively. Then, ${\displaystyle \rho \prime }$ and ${\displaystyle {\vec {J}}\prime }$ be the charge and current densities in ${\displaystyle S\prime }$ frame respectively.
We know, ${\displaystyle {\vec {J}}=\rho {\vec {v}}}$
Again, we know that ${\displaystyle \rho =\rho \prime }$
Thus, ${\displaystyle {\vec {J}}\prime =\rho \prime {\vec {v}}\prime =\rho ({\vec {v}}-{\vec {v_{0}}})}$
${\displaystyle \implies {\vec {J}}\prime ={\vec {J}}-{\vec {J_{0}}}here,{\vec {J_{0}}}=\rho {\vec {v_{0}}}}$
Thus, we have

${\displaystyle \rho =\rho \prime }$

(c)

and

${\displaystyle {\vec {J}}\prime ={\vec {J}}-{\vec {J_{0}}}}$

(d)

Transformation of ${\displaystyle \nabla }$, ${\displaystyle \mu _{0}}$ and ${\displaystyle {\frac {\partial }{\partial t}}}$Edit

We know that ${\displaystyle \mu _{0}=N/A^{2}}$. Here, ${\displaystyle A={\frac {q}{t}}}$. Since q'=q, ${\displaystyle {\vec {F}}\prime ={\vec {F}}}$ and t'=t(Galilean principle), we get

${\displaystyle \mu _{0}\prime =\mu _{0}}$

(e)

Now, Let ${\displaystyle f(x,y,z,t),\therefore f(x',y',z',t')}$
t'=t
${\displaystyle {\vec {r}}\prime ={\vec {r}}-{\vec {v_{0}}}t}$
${\displaystyle \therefore {\frac {\partial f}{\partial x}}={\frac {\partial f}{\partial x'}}}$
As, ${\displaystyle {\frac {\partial x'}{\partial x}}=1,{\frac {\partial y'}{\partial x}}=0,{\frac {\partial z'}{\partial x}}=0,{\frac {\partial t'}{\partial x}}=0}$
Similarly, ${\displaystyle {\frac {\partial f}{\partial y}}={\frac {\partial f}{\partial y'}},{\frac {\partial f}{\partial z}}={\frac {\partial f}{\partial z'}}and{\frac {\partial }{\partial t'}}={\frac {\partial }{\partial t}}+{\vec {v}}.{\vec {\nabla }}}$
Thus, we get

${\displaystyle \nabla \prime =\nabla }$

(f)

${\displaystyle {\frac {\partial }{\partial t'}}={\frac {\partial }{\partial t}}+{\vec {v}}.{\vec {\nabla }}}$

(g)

Transformation of Maxwell's equationEdit

Now by using equations (a) to (g) we can easily see that Gauss's law and Ampère's circuital law doesn't preserve its form. That is, it non-invariant under Galilean transformation. Whereas, Gauss's law for magnetism and Faraday's law preserve its form under Galilean transformation. Thus, we can see that Maxwell's equation does not preserve its form under Galilean transformation, i.e., it is not invariant under Galilean transformation. 

This article uses material from the Wikipedia article  Metasyntactic variable, which is released under the  Creative Commons Attribution-ShareAlike 3.0 Unported License.