The relativistic invariance of Maxwell’s field equations is to be established now. The field equations for free space are
∇
→
⋅
E
=
0
,
…
(
a
)
∇
→
⋅
B
=
0
,
…
(
b
)
∇
→
×
E
=
−
∂
B
∂
t
…
(
c
)
∇
→
×
B
=
1
c
2
∂
E
∂
t
…
(
d
)
∇
→
⋅
E
=
0
,
…
a
∇
→
⋅
B
=
0
,
…
b
∇
→
×
E
=
−
∂
B
∂
t
…
c
∇
→
×
B
=
1
c
2
∂
E
∂
t
…
d
{:[ vec(grad)*E=0″,” dots(a)],[ vec(grad)*B=0″,” dots(b)],[ vec(grad)xxE=-(delB)/(del t) dots(c)],[ vec(grad)xxB=(1)/(c^(2))(delE)/(del t) dots(d)]:} \begin{aligned}
\vec{\nabla} \cdot \mathbf{E} & =0, & \ldots\left(\mathrm{a}\right) \\
\vec{\nabla} \cdot \mathbf{B} & =0, & \ldots\left(\mathrm{b}\right) \\
\vec{\nabla} \times \mathbf{E} & =-\frac{\partial \mathbf{B}}{\partial t} & \ldots\left(\mathrm{c}\right) \\
\vec{\nabla} \times \mathbf{B} & =\frac {1} {c^{2}}\frac{\partial \mathbf{E}}{\partial t} & \ldots\left(\mathrm{d}\right)
\end{aligned} ∇ → ⋅ E = 0 , … ( a ) ∇ → ⋅ B = 0 , … ( b ) ∇ → × E = − ∂ B ∂ t … ( c ) ∇ → × B = 1 c 2 ∂ E ∂ t … ( d ) because
ρ
=
0
,
J
=
0
,
ρ
=
0
,
J
=
0
,
rho=0,J=0, \rho=0, J=0, ρ = 0 , J = 0 , for free space.
Let a frame of reference
F
′
(
x
′
,
y
′
,
z
′
,
t
′
)
F
′
x
′
,
y
′
,
z
′
,
t
′
F^(‘)(x^(‘),y^(‘),z^(‘),t^(‘)) F^{\prime}\left(x^{\prime}, y^{\prime}, z^{\prime}, t^{\prime}\right) F ′ ( x ′ , y ′ , z ′ , t ′ ) move with uniform velocity
v
v
v v v in positive
x
x
x x x -direction with respect to frame
F
(
x
,
y
,
z
,
t
)
F
(
x
,
y
,
z
,
t
)
F(x,y,z,t) F(x, y, z, t) F ( x , y , z , t ) . Then the eqs. (a-d) in
F
F
F F F should retain the same form in
F
′
F
′
F^(‘) F^{\prime} F ′ as given below by eqs. (a’-d’) to establish their invariance,
(
∇
→
⋅
E
)
′
=
0
,
…
(
a
′
)
(
∇
→
⋅
H
)
′
=
0
,
…
(
b
′
)
(
∇
→
×
E
)
′
=
−
∂
B
′
∂
t
′
,
…
(
c
′
)
(
∇
→
×
B
)
′
=
1
c
2
∂
E
′
∂
t
′
,
…
(
d
′
)
(
∇
→
⋅
E
)
′
=
0
,
…
a
′
(
∇
→
⋅
H
)
′
=
0
,
…
b
′
(
∇
→
×
E
)
′
=
−
∂
B
′
∂
t
′
,
…
c
′
(
∇
→
×
B
)
′
=
1
c
2
∂
E
′
∂
t
′
,
…
d
′
{:[( vec(grad)*E)^(‘)=0″,” dots(a^(‘))],[( vec(grad)*H)^(‘)=0″,” dots(b^(‘))],[( vec(grad)xxE)^(‘)=-(delB^(‘))/(delt^(‘))”,” dots(c^(‘))],[( vec(grad)xxB)^(‘)=(1)/(c^(2))(delE^(‘))/(delt^(‘))”,” dots(d^(‘))]:} \begin{aligned}
(\vec{\nabla} \cdot \mathbf{E})^{\prime} & =0, & \ldots\left(\mathrm{a}^{\prime}\right) \\
(\vec{\nabla} \cdot \mathbf{H})^{\prime} & =0, & \ldots\left(\mathrm{b}^{\prime}\right) \\
(\vec{\nabla} \times \mathbf{E})^{\prime} & =-\frac{\partial \mathbf{B}^{\prime}}{\partial t^{\prime}}, & \ldots\left(\mathrm{c}^{\prime}\right) \\
(\vec{\nabla} \times \mathbf{B})^{\prime} & =\frac {1} {c^{2}}\frac{\partial \mathbf{E}^{\prime}}{\partial t^{\prime}}, & \ldots\left(\mathrm{d}^{\prime}\right)
\end{aligned} ( ∇ → ⋅ E ) ′ = 0 , … ( a ′ ) ( ∇ → ⋅ H ) ′ = 0 , … ( b ′ ) ( ∇ → × E ) ′ = − ∂ B ′ ∂ t ′ , … ( c ′ ) ( ∇ → × B ) ′ = 1 c 2 ∂ E ′ ∂ t ′ , … ( d ′ ) In order to show the relativistic invariance of eqs. (a-d), we write them in terms of components of
E
E
E \mathbf{E} E and
B
B
B \mathbf{B} B as
∂
E
x
∂
x
+
∂
E
y
∂
y
+
∂
E
z
∂
z
=
0
…
(
a
1
)
∂
B
x
∂
x
+
∂
B
y
∂
y
+
∂
B
z
∂
z
=
0
…
(
b
1
)
∂
E
x
∂
x
+
∂
E
y
∂
y
+
∂
E
z
∂
z
=
0
…
a
1
∂
B
x
∂
x
+
∂
B
y
∂
y
+
∂
B
z
∂
z
=
0
…
b
1
{:[(delE_(x))/(del x)+(delE_(y))/(del y)+(delE_(z))/(del z)=0 dots(a1)],[(delB_(x))/(del x)+(delB_(y))/(del y)+(delB_(z))/(del z)=0 dots(b1)]:} \begin{aligned}
& \frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z}=0 & \ldots\left(\mathrm{a1}\right) \\
& \frac{\partial B_x}{\partial x}+\frac{\partial B_y}{\partial y}+\frac{\partial B_z}{\partial z}=0 & \ldots\left(\mathrm{b1}\right) \\
\end{aligned} ∂ E x ∂ x + ∂ E y ∂ y + ∂ E z ∂ z = 0 … ( a 1 ) ∂ B x ∂ x + ∂ B y ∂ y + ∂ B z ∂ z = 0 … ( b 1 )
(
∂
E
z
∂
y
−
∂
E
y
∂
z
)
=
−
∂
B
x
∂
t
,
…
(
c
1
)
(
∂
E
x
∂
z
−
∂
E
z
∂
x
)
=
−
∂
B
y
∂
t
,
…
(
c
2
)
(
∂
E
y
∂
x
−
∂
E
x
∂
y
)
=
−
∂
B
z
∂
t
,
…
(
c
3
)
∂
E
z
∂
y
−
∂
E
y
∂
z
=
−
∂
B
x
∂
t
,
…
c
1
∂
E
x
∂
z
−
∂
E
z
∂
x
=
−
∂
B
y
∂
t
,
…
c
2
∂
E
y
∂
x
−
∂
E
x
∂
y
=
−
∂
B
z
∂
t
,
…
c
3
{:[((delE_(z))/(del y)-(delE_(y))/(del z))=-(delB_(x))/(del t)”,” dots(c1)],[((delE_(x))/(del z)-(delE_(z))/(del x))=-(delB_(y))/(del t)”,” dots(c2)],[((delE_(y))/(del x)-(delE_(x))/(del y))=-(delB_(z))/(del t)”,” dots(c3)]:} \begin{aligned}
\left(\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}\right) & =-\frac{\partial B_x}{\partial t}, & \ldots\left(\mathrm{c1}\right) \\
\left(\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x}\right) & =-\frac{\partial B_y}{\partial t}, & \ldots\left(\mathrm{c2}\right) \\
\left(\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}\right) & =-\frac{\partial B_z}{\partial t}, & \ldots\left(\mathrm{c3}\right) \\
\end{aligned} ( ∂ E z ∂ y − ∂ E y ∂ z ) = − ∂ B x ∂ t , … ( c 1 ) ( ∂ E x ∂ z − ∂ E z ∂ x ) = − ∂ B y ∂ t , … ( c 2 ) ( ∂ E y ∂ x − ∂ E x ∂ y ) = − ∂ B z ∂ t , … ( c 3 ) and,
(
∂
B
z
∂
y
−
∂
B
y
∂
z
)
=
1
c
2
∂
E
x
∂
t
,
…
(
d
1
)
(
∂
B
x
∂
z
−
∂
B
z
∂
x
)
=
1
c
2
∂
E
y
∂
t
,
…
(
d
2
)
(
∂
B
y
∂
x
−
∂
B
x
∂
y
)
=
1
c
2
∂
E
z
∂
t
,
…
(
d
3
)
∂
B
z
∂
y
−
∂
B
y
∂
z
=
1
c
2
∂
E
x
∂
t
,
…
d
1
∂
B
x
∂
z
−
∂
B
z
∂
x
=
1
c
2
∂
E
y
∂
t
,
…
d
2
∂
B
y
∂
x
−
∂
B
x
∂
y
=
1
c
2
∂
E
z
∂
t
,
…
d
3
{:[((delB_(z))/(del y)-(delB_(y))/(del z))=(1)/(c^(2))(delE_(x))/(del t)”,” dots(d1)],[((delB_(x))/(del z)-(delB_(z))/(del x))=(1)/(c^(2))(delE_(y))/(del t)”,” dots(d2)],[((delB_(y))/(del x)-(delB_(x))/(del y))=(1)/(c^(2))(delE_(z))/(del t)”,” dots(d3)]:} \begin{aligned}
\left(\frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z}\right)=\frac {1} {c^{2}} \frac{\partial E_x}{\partial t}, & \ldots\left(\mathrm{d1}\right) \\
\left(\frac{\partial B_x}{\partial z}-\frac{\partial B_z}{\partial x}\right)=\frac {1} {c^{2}} \frac{\partial E_y}{\partial t}, & \ldots\left(\mathrm{d2}\right) \\
\left(\frac{\partial B_y}{\partial x}-\frac{\partial B_x}{\partial y}\right)=\frac {1} {c^{2}} \frac{\partial E_z}{\partial t}, & \ldots\left(\mathrm{d3}\right) \\
\end{aligned} ( ∂ B z ∂ y − ∂ B y ∂ z ) = 1 c 2 ∂ E x ∂ t , … ( d 1 ) ( ∂ B x ∂ z − ∂ B z ∂ x ) = 1 c 2 ∂ E y ∂ t , … ( d 2 ) ( ∂ B y ∂ x − ∂ B x ∂ y ) = 1 c 2 ∂ E z ∂ t , … ( d 3 ) Physically these equations represent fields which are space-time dependent.
The Lorentz Transformation equations are given as:
x
′
=
γ
(
x
−
v
t
)
y
′
=
y
z
′
=
z
t
′
=
γ
(
t
−
v
x
c
2
)
x
′
=
γ
(
x
−
v
t
)
y
′
=
y
z
′
=
z
t
′
=
γ
(
t
−
v
x
c
2
)
{:[x’=gamma(x-vt)],[y’=y],[z’=z],[t’=gamma(t-(vx)/(c^(2)))]:} \begin{aligned}
x{\prime} & = \gamma(x-vt) \\
y{\prime} & = y \\
z{\prime} & = z \\
t{\prime} & = \gamma(t-\frac {vx} {c^{2}}) \\
\end{aligned} x ′ = γ ( x − v t ) y ′ = y z ′ = z t ′ = γ ( t − v x c 2 ) These equations give the following results:
∂
∂
x
=
γ
∂
∂
x
′
−
γ
v
c
2
∂
∂
t
′
,
…
(
e
)
∂
∂
y
=
∂
∂
y
′
,
…
(
f
)
∂
∂
z
=
∂
∂
z
′
,
…
(
g
)
∂
∂
t
=
γ
∂
∂
t
′
−
γ
v
∂
∂
x
′
,
…
(
h
)
∂
∂
x
=
γ
∂
∂
x
′
−
γ
v
c
2
∂
∂
t
′
,
…
e
∂
∂
y
=
∂
∂
y
′
,
…
f
∂
∂
z
=
∂
∂
z
′
,
…
g
∂
∂
t
=
γ
∂
∂
t
′
−
γ
v
∂
∂
x
′
,
…
h
{:[(del)/(del x)=gamma(del)/(delx^(‘))-gamma(v)/(c^(2))(del)/(delt^(‘))”,” dots(e)],[quad(del)/(del y)=(del)/(dely^(‘))”,” dots(f)],[quad(del)/(del z)=(del)/(delz^(‘))”,” dots(g)],[(del)/(del t)=gamma(del)/(delt^(‘))-gamma v(del)/(delx^(‘))”,” dots(h)]:} \begin{aligned}
\frac{\partial}{\partial x} & =\gamma \frac{\partial}{\partial x^{\prime}}-\gamma \frac{v}{c^2} \frac{\partial}{\partial t^{\prime}}, & \ldots\left(\mathrm{e}\right)\\
\quad \frac{\partial}{\partial y} & =\frac{\partial}{\partial y^{\prime}}, & \ldots\left(\mathrm{f}\right)\\
\quad \frac{\partial}{\partial z} & =\frac{\partial}{\partial z^{\prime}}, & \ldots\left(\mathrm{g}\right)\\
\frac{\partial}{\partial t} & =\gamma \frac{\partial}{\partial t^{\prime}}-\gamma v \frac{\partial}{\partial x^{\prime}}, & \ldots\left(\mathrm{h}\right)\\
\end{aligned} ∂ ∂ x = γ ∂ ∂ x ′ − γ v c 2 ∂ ∂ t ′ , … ( e ) ∂ ∂ y = ∂ ∂ y ′ , … ( f ) ∂ ∂ z = ∂ ∂ z ′ , … ( g ) ∂ ∂ t = γ ∂ ∂ t ′ − γ v ∂ ∂ x ′ , … ( h ) Now substituting the results of transformation given by eq. (e), (f), (g) and (h) in eq. (b1) and (c1), we get
∂
E
z
∂
y
′
−
∂
E
y
∂
z
′
=
−
γ
∂
B
x
∂
t
′
+
γ
v
∂
B
x
∂
x
′
,
…
(
i
)
∂
E
z
∂
y
′
−
∂
E
y
∂
z
′
=
−
γ
∂
B
x
∂
t
′
+
γ
v
∂
B
x
∂
x
′
,
…
i
{:(delE_(z))/(dely^(‘))-(delE_(y))/(delz^(‘))=-gamma(delB_(x))/(delt^(‘))+gammav(delB_(x))/(delx^(‘))”,” dots(i):} \begin{aligned}
\frac{\partial E_z}{\partial y^{\prime}}-\frac{\partial E_y}{\partial z^{\prime}}=-\gamma \frac{\partial B_x}{\partial t^{\prime}}+{\gamma}{v} \frac{\partial B_x}{\partial x^{\prime}}, & \ldots\left(\mathrm{i}\right)\\
\end{aligned} ∂ E z ∂ y ′ − ∂ E y ∂ z ′ = − γ ∂ B x ∂ t ′ + γ v ∂ B x ∂ x ′ , … ( i ) and
γ
∂
B
x
∂
x
′
−
γ
v
c
2
∂
B
x
∂
t
′
+
∂
B
y
∂
y
′
+
∂
B
z
∂
z
′
=
0
,
…
(
j
)
γ
∂
B
x
∂
x
′
−
γ
v
c
2
∂
B
x
∂
t
′
+
∂
B
y
∂
y
′
+
∂
B
z
∂
z
′
=
0
,
…
j
{:gamma(delB_(x))/(delx^(‘))-gamma(v)/(c^(2))(delB_(x))/(delt^(‘))+(delB_(y))/(dely^(‘))+(delB_(z))/(delz^(‘))=0″,” dots(j):} \begin{aligned}
\gamma \frac{\partial B_x}{\partial x^{\prime}}-\gamma \frac{v}{c^2} \frac{\partial B_x}{\partial t^{\prime}}+\frac{\partial B_y}{\partial y^{\prime}}+\frac{\partial B_z}{\partial z^{\prime}}=0, & \ldots\left(\mathrm{j}\right)
\end{aligned} γ ∂ B x ∂ x ′ − γ v c 2 ∂ B x ∂ t ′ + ∂ B y ∂ y ′ + ∂ B z ∂ z ′ = 0 , … ( j ) Eliminating
(
∂
B
x
/
∂
x
′
)
∂
B
x
/
∂
x
′
(delB_(x)//delx^(‘)) \left(\partial B_x / \partial x^{\prime}\right) ( ∂ B x / ∂ x ′ ) from these equations, we get
∂
∂
y
′
[
γ
(
E
z
+
v
B
y
)
]
−
∂
∂
z
′
[
γ
(
E
y
−
v
B
z
)
]
=
−
∂
B
x
∂
t
′
,
…
(
k
)
∂
∂
y
′
γ
E
z
+
v
B
y
−
∂
∂
z
′
γ
E
y
−
v
B
z
=
−
∂
B
x
∂
t
′
,
…
k
{:(del)/(dely^(‘))[gamma(E_(z)+vB_(y))]-(del)/(delz^(‘))[gamma(E_(y)-vB_(z))]=-(delB_(x))/(delt^(‘))”,” dots(k):} \begin{aligned}
\frac{\partial}{\partial y^{\prime}}\left[\gamma\left(E_z+v B_y\right)\right]-\frac{\partial}{\partial z^{\prime}}\left[\gamma\left(E_y-v B_z\right)\right]=- \frac{\partial B_x}{\partial t^{\prime}}, & \ldots\left(\mathrm{k}\right)\\
\end{aligned} ∂ ∂ y ′ [ γ ( E z + v B y ) ] − ∂ ∂ z ′ [ γ ( E y − v B z ) ] = − ∂ B x ∂ t ′ , … ( k ) As derived previously, we can write the components of the field in frame
F
′
F
′
F^(‘) F^{\prime} F ′ in the following way:
γ
(
E
z
+
v
B
y
)
=
E
z
′
;
γ
(
E
y
−
v
B
z
)
=
E
y
′
;
E
x
=
E
x
′
γ
(
B
z
−
v
c
2
E
y
)
=
B
z
′
;
γ
(
B
y
+
v
c
2
E
z
)
=
B
y
′
;
B
x
=
B
x
′
,
…
(
l
)
γ
E
z
+
v
B
y
=
E
z
′
;
γ
E
y
−
v
B
z
=
E
y
′
;
E
x
=
E
x
′
γ
B
z
−
v
c
2
E
y
=
B
z
′
;
γ
B
y
+
v
c
2
E
z
=
B
y
′
;
B
x
=
B
x
′
,
…
l
{:[gamma(E_(z)+vB_(y))=E_(z)^(‘);quad gamma(E_(y)-vB_(z))=E_(y)^(‘);quadE_(x)=E_(x)^(‘)],[gamma(B_(z)-(v)/(c^(2))E_(y))=B_(z)^(‘);quad gamma(B_(y)+(v)/(c^(2))E_(z))=B_(y)^(‘);quadB_(x)=B_(x)^(‘)”,” dots(l)]:} \begin{aligned}
\gamma\left(E_z+v B_y\right)=E_z{ }^{\prime} ; \quad \gamma\left(E_y-v B_z\right)=E_y{ }^{\prime}; \quad E_x=E_x^{\prime}\\
\gamma\left(B_z-\frac {v}{c^2} E_y\right)=B_z{ }^{\prime} ; \quad \gamma\left(B_y+\frac {v}{c^2} E_z\right)=B_y{ }^{\prime}; \quad B_x=B_x^{\prime}, & \ldots\left(\mathrm{l}\right)\\
\end{aligned} γ ( E z + v B y ) = E z ′ ; γ ( E y − v B z ) = E y ′ ; E x = E x ′ γ ( B z − v c 2 E y ) = B z ′ ; γ ( B y + v c 2 E z ) = B y ′ ; B x = B x ′ , … ( l ) Then eq. (k) assumes the form :
[
∂
E
z
′
∂
y
′
−
∂
E
y
′
∂
z
′
]
=
−
∂
B
x
′
∂
t
′
,
∂
E
z
′
∂
y
′
−
∂
E
y
′
∂
z
′
=
−
∂
B
x
′
∂
t
′
,
[(delE_(z)^(‘))/(dely^(‘))-(delE_(y)^(‘))/(delz^(‘))]=-(delB_(x)^(‘))/(delt^(‘)), \left[\frac{\partial E_z{ }^{\prime}}{\partial y^{\prime}}-\frac{\partial E_y{ }^{\prime}}{\partial z^{\prime}}\right]=-\frac{\partial B_x^{\prime}}{\partial t^{\prime}}, [ ∂ E z ′ ∂ y ′ − ∂ E y ′ ∂ z ′ ] = − ∂ B x ′ ∂ t ′ , which is similar to eq. (c1). Similarly, the other two sets assume the form;
[
∂
E
x
′
∂
z
′
−
∂
E
z
′
∂
x
′
]
=
−
∂
B
y
′
∂
t
′
[
∂
E
y
′
′
∂
x
′
−
∂
E
x
′
∂
y
′
]
=
−
∂
B
z
′
∂
t
′
∂
E
x
′
∂
z
′
−
∂
E
z
′
∂
x
′
=
−
∂
B
y
′
∂
t
′
∂
E
y
′
′
∂
x
′
−
∂
E
x
′
∂
y
′
=
−
∂
B
z
′
∂
t
′
{:[[(delE_(x)^(‘))/(delz^(‘))-(delE_(z)^(‘))/(delx^(‘))]=-(delB_(y)^(‘))/(delt^(‘))],[[(delE_(y^(‘))^(‘))/(delx^(‘))-(delE_(x)^(‘))/(dely^(‘))]=-(delB_(z)^(‘))/(delt^(‘))]:} \begin{aligned}
& \left[\frac{\partial E_x^{\prime}}{\partial z^{\prime}}-\frac{\partial E_z^{\prime}}{\partial x^{\prime}}\right]=-\frac{\partial B_y^{\prime}}{\partial t^{\prime}} \\
& \left[\frac{\partial E_{y^{\prime}}^{\prime}}{\partial x^{\prime}}-\frac{\partial E_x^{\prime}}{\partial y^{\prime}}\right]=-\frac{\partial B_z^{\prime}}{\partial t^{\prime}}
\end{aligned} [ ∂ E x ′ ∂ z ′ − ∂ E z ′ ∂ x ′ ] = − ∂ B y ′ ∂ t ′ [ ∂ E y ′ ′ ∂ x ′ − ∂ E x ′ ∂ y ′ ] = − ∂ B z ′ ∂ t ′ or, in short;
(
∇
→
×
E
)
′
=
∂
B
′
∂
t
′
(
∇
→
×
E
)
′
=
∂
B
′
∂
t
′
( vec(grad)xxE)^(‘)=(delB^(‘))/(delt^(‘)) (\vec{\nabla} \times \mathbf{E})^{\prime}=\frac{\partial B^{\prime}}{\partial t^{\prime}} ( ∇ → × E ) ′ = ∂ B ′ ∂ t ′ which is eq. (
c
′
c
′
c^(‘) c^{\prime} c ′ ) in frame
F
′
F
′
F^(‘) F^{\prime} F ′ and bears the same form.
Similarly, substituting the results of transformation given by eq. (e), (f), (g) and (h) in eq. (d1) and (a1), we get
∂
B
z
∂
y
′
−
∂
B
y
∂
z
′
=
γ
c
2
∂
E
x
∂
t
′
−
γ
v
c
2
∂
E
x
∂
x
′
,
…
(
m
)
∂
B
z
∂
y
′
−
∂
B
y
∂
z
′
=
γ
c
2
∂
E
x
∂
t
′
−
γ
v
c
2
∂
E
x
∂
x
′
,
…
m
{:(delB_(z))/(dely^(‘))-(delB_(y))/(delz^(‘))=(gamma)/(c^(2))(delE_(x))/(delt^(‘))-(gamma v)/(c^(2))(delE_(x))/(delx^(‘))”,” dots(m):} \begin{aligned}
\frac{\partial B_z}{\partial y^{\prime}}-\frac{\partial B_y}{\partial z^{\prime}}=\frac{\gamma}{c^2} \frac{\partial E_x}{\partial t^{\prime}}-\frac{\gamma v}{c^2} \frac{\partial E_x}{\partial x^{\prime}}, & \ldots\left(\mathrm{m}\right) \\
\end{aligned} ∂ B z ∂ y ′ − ∂ B y ∂ z ′ = γ c 2 ∂ E x ∂ t ′ − γ v c 2 ∂ E x ∂ x ′ , … ( m ) and
γ
∂
E
x
∂
x
′
−
γ
v
c
2
∂
E
x
∂
t
′
+
∂
E
y
∂
y
′
+
∂
E
z
∂
z
′
=
0
,
…
(
n
)
γ
∂
E
x
∂
x
′
−
γ
v
c
2
∂
E
x
∂
t
′
+
∂
E
y
∂
y
′
+
∂
E
z
∂
z
′
=
0
,
…
n
{:gamma(delE_(x))/(delx^(‘))-gamma(v)/(c^(2))(delE_(x))/(delt^(‘))+(delE_(y))/(dely^(‘))+(delE_(z))/(delz^(‘))=0″,” dots(n):} \begin{aligned}
\gamma \frac{\partial E_x}{\partial x^{\prime}}-\gamma \frac{v}{c^2} \frac{\partial E_x}{\partial t^{\prime}}+\frac{\partial E_y}{\partial y^{\prime}}+\frac{\partial E_z}{\partial z^{\prime}}=0, & \ldots\left(\mathrm{n}\right)
\end{aligned} γ ∂ E x ∂ x ′ − γ v c 2 ∂ E x ∂ t ′ + ∂ E y ∂ y ′ + ∂ E z ∂ z ′ = 0 , … ( n ) Eliminating
(
∂
E
x
/
∂
x
′
)
∂
E
x
/
∂
x
′
(delE_(x)//delx^(‘)) \left(\partial E_x / \partial x^{\prime}\right) ( ∂ E x / ∂ x ′ ) from these equations, we get
∂
∂
y
′
[
γ
(
B
z
−
v
c
2
E
y
)
]
−
∂
∂
z
′
[
γ
(
B
y
+
v
c
2
E
z
)
]
=
1
c
2
∂
E
x
∂
t
′
,
…
(
o
)
∂
∂
y
′
γ
B
z
−
v
c
2
E
y
−
∂
∂
z
′
γ
B
y
+
v
c
2
E
z
=
1
c
2
∂
E
x
∂
t
′
,
…
o
{:(del)/(dely^(‘))[gamma(B_(z)-(v)/(c^(2))E_(y))]-(del)/(delz^(‘))[gamma(B_(y)+(v)/(c^(2))E_(z))]=(1)/(c^(2))(delE_(x))/(delt^(‘))”,” dots(o):} \begin{aligned}
\frac{\partial}{\partial y^{\prime}}\left[\gamma\left(B_z-\frac {v}{c^2} E_y\right)\right]-\frac{\partial}{\partial z^{\prime}}\left[\gamma\left(B_y+\frac {v}{c^2} E_z\right)\right]=\frac{1}{c^2}\frac{\partial E_x}{\partial t^{\prime}}, & \ldots\left(\mathrm{o}\right)\\
\end{aligned} ∂ ∂ y ′ [ γ ( B z − v c 2 E y ) ] − ∂ ∂ z ′ [ γ ( B y + v c 2 E z ) ] = 1 c 2 ∂ E x ∂ t ′ , … ( o ) As derived previously, we can write the components of the field in frame
F
′
F
′
F^(‘) F^{\prime} F ′ in the following way:
γ
(
E
z
+
v
B
y
)
=
E
z
′
;
γ
(
E
y
−
v
B
z
)
=
E
y
′
;
E
x
=
E
x
′
γ
(
B
z
−
v
c
2
E
y
)
=
B
z
′
;
γ
(
B
y
+
v
c
2
E
z
)
=
B
y
′
;
B
x
=
B
x
′
,
…
(
l
)
γ
E
z
+
v
B
y
=
E
z
′
;
γ
E
y
−
v
B
z
=
E
y
′
;
E
x
=
E
x
′
γ
B
z
−
v
c
2
E
y
=
B
z
′
;
γ
B
y
+
v
c
2
E
z
=
B
y
′
;
B
x
=
B
x
′
,
…
l
{:[gamma(E_(z)+vB_(y))=E_(z)^(‘);quad gamma(E_(y)-vB_(z))=E_(y)^(‘);quadE_(x)=E_(x)^(‘)],[gamma(B_(z)-(v)/(c^(2))E_(y))=B_(z)^(‘);quad gamma(B_(y)+(v)/(c^(2))E_(z))=B_(y)^(‘);quadB_(x)=B_(x)^(‘)”,” dots(l)]:} \begin{aligned}
\gamma\left(E_z+v B_y\right)=E_z{ }^{\prime} ; \quad \gamma\left(E_y-v B_z\right)=E_y{ }^{\prime}; \quad E_x=E_x^{\prime}\\
\gamma\left(B_z-\frac {v}{c^2} E_y\right)=B_z{ }^{\prime} ; \quad \gamma\left(B_y+\frac {v}{c^2} E_z\right)=B_y{ }^{\prime}; \quad B_x=B_x^{\prime}, & \ldots\left(\mathrm{l}\right)\\
\end{aligned} γ ( E z + v B y ) = E z ′ ; γ ( E y − v B z ) = E y ′ ; E x = E x ′ γ ( B z − v c 2 E y ) = B z ′ ; γ ( B y + v c 2 E z ) = B y ′ ; B x = B x ′ , … ( l ) Then eq. (o) assumes the form :
[
∂
B
z
′
∂
y
′
−
∂
B
y
′
∂
z
′
]
=
1
c
2
∂
E
x
′
∂
t
′
,
∂
B
z
′
∂
y
′
−
∂
B
y
′
∂
z
′
=
1
c
2
∂
E
x
′
∂
t
′
,
[(delB_(z)^(‘))/(dely^(‘))-(delB_(y)^(‘))/(delz^(‘))]=(1)/(c^(2))(delE_(x)^(‘))/(delt^(‘)), \left[\frac{\partial B_z{ }^{\prime}}{\partial y^{\prime}}-\frac{\partial B_y{ }^{\prime}}{\partial z^{\prime}}\right]=\frac {1}{c^2}\frac{\partial E_x^{\prime}}{\partial t^{\prime}}, [ ∂ B z ′ ∂ y ′ − ∂ B y ′ ∂ z ′ ] = 1 c 2 ∂ E x ′ ∂ t ′ , which is similar to eq. (d1). Similarly, the other two sets assume the form;
[
∂
B
x
′
∂
z
′
−
∂
B
z
′
∂
x
′
]
=
1
c
2
∂
E
y
′
∂
t
′
[
∂
B
y
′
∂
x
′
−
∂
B
x
′
∂
y
′
]
=
1
c
2
∂
E
z
′
∂
t
′
∂
B
x
′
∂
z
′
−
∂
B
z
′
∂
x
′
=
1
c
2
∂
E
y
′
∂
t
′
∂
B
y
′
∂
x
′
−
∂
B
x
′
∂
y
′
=
1
c
2
∂
E
z
′
∂
t
′
{:[[(delB_(x)^(‘))/(delz^(‘))-(delB_(z)^(‘))/(delx^(‘))]=(1)/(c^(2))(delE_(y)^(‘))/(delt^(‘))],[[(delB_(y)^(‘))/(delx^(‘))-(delB_(x)^(‘))/(dely^(‘))]=(1)/(c^(2))(delE_(z)^(‘))/(delt^(‘))]:} \begin{aligned}
& \left[\frac{\partial B_x^{\prime}}{\partial z^{\prime}}-\frac{\partial B_z^{\prime}}{\partial x^{\prime}}\right]=\frac{1}{c^2}\frac{\partial E_y^{\prime}}{\partial t^{\prime}} \\
& \left[\frac{\partial B_y^{\prime}}{\partial x^{\prime}}-\frac{\partial B_x^{\prime}}{\partial y^{\prime}}\right]=\frac{1}{c^2}\frac{\partial E_z^{\prime}}{\partial t^{\prime}}
\end{aligned} [ ∂ B x ′ ∂ z ′ − ∂ B z ′ ∂ x ′ ] = 1 c 2 ∂ E y ′ ∂ t ′ [ ∂ B y ′ ∂ x ′ − ∂ B x ′ ∂ y ′ ] = 1 c 2 ∂ E z ′ ∂ t ′ or, in short;
(
∇
→
×
B
)
′
=
1
c
2
∂
E
′
∂
t
′
(
∇
→
×
B
)
′
=
1
c
2
∂
E
′
∂
t
′
( vec(grad)xxB)^(‘)=(1)/(c^(2))(delE^(‘))/(delt^(‘)) (\vec{\nabla} \times \mathbf{B})^{\prime}=\frac{1}{c^2}\frac{\partial E^{\prime}}{\partial t^{\prime}} ( ∇ → × B ) ′ = 1 c 2 ∂ E ′ ∂ t ′ which is eq. (
d
′
d
′
d^(‘) d^{\prime} d ′ ) in frame
F
′
F
′
F^(‘) F^{\prime} F ′ and bears the same form.
Using eq. (l) and the transformation equations (e-h) in eq. (a1), we get;
γ
∂
E
x
′
∂
x
′
+
1
γ
∂
E
y
′
∂
y
′
+
1
γ
∂
E
z
′
∂
z
′
+
v
(
∂
B
z
∂
y
′
−
∂
B
y
∂
z
′
)
−
γ
v
c
2
∂
E
x
′
∂
t
′
=
0
γ
∂
E
x
′
∂
x
′
+
1
γ
∂
E
y
′
∂
y
′
+
1
γ
∂
E
z
′
∂
z
′
+
v
∂
B
z
∂
y
′
−
∂
B
y
∂
z
′
−
γ
v
c
2
∂
E
x
′
∂
t
′
=
0
{:gamma(delE_(x)^(‘))/(delx^(‘))+(1)/(gamma)(delE_(y)^(‘))/(dely^(‘))+(1)/(gamma)(delE_(z)^(‘))/(delz^(‘))+v((delB_(z))/(dely^(‘))-(delB_(y))/(delz^(‘)))-gamma(v)/(c^(2))(delE_(x^(‘)))/(delt^(‘))=0:} \begin{aligned}
{\gamma} \frac{\partial E_x^{\prime}}{\partial x^{\prime}}+\frac{1}{\gamma}\frac{\partial E_y^{\prime}}{\partial y^{\prime}}+\frac{1}{\gamma}\frac{\partial E_z^{\prime}}{\partial z^{\prime}}+v\left(\frac{\partial B_z}{\partial y^{\prime}}-\frac{\partial B_y}{\partial z^{\prime}}\right)-\gamma \frac{v}{c^2} \frac{\partial E_{x^{\prime}}}{\partial t^{\prime}}=0\\
\end{aligned} γ ∂ E x ′ ∂ x ′ + 1 γ ∂ E y ′ ∂ y ′ + 1 γ ∂ E z ′ ∂ z ′ + v ( ∂ B z ∂ y ′ − ∂ B y ∂ z ′ ) − γ v c 2 ∂ E x ′ ∂ t ′ = 0 Substituting eq. (m) in above eq. and using eq. (l), we get
1
γ
(
∂
E
x
′
∂
x
′
+
∂
E
y
′
∂
y
′
+
∂
E
z
′
∂
z
′
)
=
0
1
γ
∂
E
x
′
∂
x
′
+
∂
E
y
′
∂
y
′
+
∂
E
z
′
∂
z
′
=
0
(1)/(gamma)((delE_(x)^(‘))/(delx^(‘))+(delE_(y)^(‘))/(dely^(‘))+(delE_(z)^(‘))/(delz^(‘)))=0 \frac{1}{\gamma}\left(\frac{\partial E_x^{\prime}}{\partial x^{\prime}}+\frac{\partial E_y^{\prime}}{\partial y^{\prime}}+\frac{\partial E_z^{\prime}}{\partial z^{\prime}}\right)=0 1 γ ( ∂ E x ′ ∂ x ′ + ∂ E y ′ ∂ y ′ + ∂ E z ′ ∂ z ′ ) = 0 or
(
∇
→
.
E
)
′
=
0
(
∇
→
.
E
)
′
=
0
( vec(grad).E)^(‘)=0 (\vec{\nabla}. \mathbf{E})^{\prime}=0 ( ∇ → . E ) ′ = 0 This shows that the first Maxwell equation bears the same form in frame
F
′
F
′
F^(‘) F^{\prime} F ′ .
Similarly, using eq. (l) and the transformation equations (e-h) in eq. (b1), we get;
γ
∂
B
x
′
∂
x
′
+
1
γ
∂
B
y
′
∂
y
′
+
1
γ
∂
B
z
′
∂
z
′
−
v
c
2
(
∂
E
z
∂
y
′
−
∂
E
y
∂
z
′
)
−
γ
v
c
2
∂
B
x
′
∂
t
′
=
0
γ
∂
B
x
′
∂
x
′
+
1
γ
∂
B
y
′
∂
y
′
+
1
γ
∂
B
z
′
∂
z
′
−
v
c
2
∂
E
z
∂
y
′
−
∂
E
y
∂
z
′
−
γ
v
c
2
∂
B
x
′
∂
t
′
=
0
{:gamma(delB_(x)^(‘))/(delx^(‘))+(1)/(gamma)(delB_(y)^(‘))/(dely^(‘))+(1)/(gamma)(delB_(z)^(‘))/(delz^(‘))-(v)/(c^(2))((delE_(z))/(dely^(‘))-(delE_(y))/(delz^(‘)))-gamma(v)/(c^(2))(delB_(x^(‘)))/(delt^(‘))=0:} \begin{aligned}
{\gamma} \frac{\partial B_x^{\prime}}{\partial x^{\prime}}+\frac{1}{\gamma}\frac{\partial B_y^{\prime}}{\partial y^{\prime}}+\frac{1}{\gamma}\frac{\partial B_z^{\prime}}{\partial z^{\prime}}-\frac{v}{c^2}\left(\frac{\partial E_z}{\partial y^{\prime}}-\frac{\partial E_y}{\partial z^{\prime}}\right)-\gamma \frac{v}{c^2} \frac{\partial B_{x^{\prime}}}{\partial t^{\prime}}=0\\
\end{aligned} γ ∂ B x ′ ∂ x ′ + 1 γ ∂ B y ′ ∂ y ′ + 1 γ ∂ B z ′ ∂ z ′ − v c 2 ( ∂ E z ∂ y ′ − ∂ E y ∂ z ′ ) − γ v c 2 ∂ B x ′ ∂ t ′ = 0 Substituting eq. (i) in above eq. and using eq. (l), we get
1
γ
(
∂
B
x
′
∂
x
′
+
∂
B
y
′
∂
y
′
+
∂
B
z
′
∂
z
′
)
=
0
1
γ
∂
B
x
′
∂
x
′
+
∂
B
y
′
∂
y
′
+
∂
B
z
′
∂
z
′
=
0
(1)/(gamma)((delB_(x)^(‘))/(delx^(‘))+(delB_(y)^(‘))/(dely^(‘))+(delB_(z)^(‘))/(delz^(‘)))=0 \frac{1}{\gamma}\left(\frac{\partial B_x^{\prime}}{\partial x^{\prime}}+\frac{\partial B_y^{\prime}}{\partial y^{\prime}}+\frac{\partial B_z^{\prime}}{\partial z^{\prime}}\right)=0 1 γ ( ∂ B x ′ ∂ x ′ + ∂ B y ′ ∂ y ′ + ∂ B z ′ ∂ z ′ ) = 0 or
(
∇
→
.
B
)
′
=
0
(
∇
→
.
B
)
′
=
0
( vec(grad).B)^(‘)=0 (\vec{\nabla}. \mathbf{B})^{\prime}=0 ( ∇ → . B ) ′ = 0 This shows that the second Maxwell equation bears the same form in frame
F
′
F
′
F^(‘) F^{\prime} F ′ .
This derivation, to prove the invariance of Maxwell’s equations under Lorentz transformation, shows that field components inter-transform in such a way that the form of field equations is retained.
