Invariance of Maxwell Field Equations under Lorentz Transformation

untitled-document-2-dd0fe481-d771-46c0-8a55-cbae0e52e3fc
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=Bt(c)×B=1c2Et(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 vvv in positive x x xxx-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 FFF 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)=Bt,(c)(×B)=1c2Et,(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}Exx+Eyy+Ezz=0(a1)Bxx+Byy+Bzz=0(b1)
( 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}(EzyEyz)=Bxt,(c1)(ExzEzx)=Byt,(c2)(EyxExy)=Bzt,(c3)
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}(BzyByz)=1c2Ext,(d1)(BxzBzx)=1c2Eyt,(d2)(ByxBxy)=1c2Ezt,(d3)
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(xvt)y=yz=zt=γ(tvxc2)
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=γxγvc2t,(e)y=y,(f)z=z,(g)t=γtγvx,(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}EzyEyz=γBxt+γvBxx,(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}γBxxγvc2Bxt+Byy+Bzz=0,(j)
Eliminating ( B x / x ) B x / x (delB_(x)//delx^(‘))\left(\partial B_x / \partial x^{\prime}\right)(Bx/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(Ez+vBy)]z[γ(EyvBz)]=Bxt,(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(Ez+vBy)=Ez;γ(EyvBz)=Ey;Ex=Exγ(Bzvc2Ey)=Bz;γ(By+vc2Ez)=By;Bx=Bx,(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}},[EzyEyz]=Bxt,
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}[ExzEzx]=Byt[EyxExy]=Bzt
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)=Bt
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}BzyByz=γc2Extγvc2Exx,(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}γExxγvc2Ext+Eyy+Ezz=0,(n)
Eliminating ( E x / x ) E x / x (delE_(x)//delx^(‘))\left(\partial E_x / \partial x^{\prime}\right)(Ex/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(Bzvc2Ey)]z[γ(By+vc2Ez)]=1c2Ext,(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(Ez+vBy)=Ez;γ(EyvBz)=Ey;Ex=Exγ(Bzvc2Ey)=Bz;γ(By+vc2Ez)=By;Bx=Bx,(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}},[BzyByz]=1c2Ext,
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}[BxzBzx]=1c2Eyt[ByxBxy]=1c2Ezt
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)=1c2Et
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}