Skip to main content

Maxwell fourth equation

The Maxwell fourth equation is called "the modified Ampere's circuital law".

Statement:
It states that the line integral of the magnetic field H around any closed part or circuit is equal to the current enclosed by the path.

Differential form (without modification):
That is,
$$\int H.dL = I$$
Let the current be distributed through the current with current density J, then:
$$I = \int J. ds$$
This implies that:
$$\int H.dL = \int J.ds$$ .........(9)
Applying Stokes theorem to the LHS of equ(9) to change line integral to surface integral we have:
$$\int_{s} (\nabla X H).ds = \int_{s} J.ds$$
Since, two surface integrals are equal only if their integrands are equal.
Thus,
$$\nabla X H =J$$ .........(10)
Equ(10) is the differential form of Maxwell fourth equation (without modification)
Take divergence of equ(10)
$$\nabla.(\nabla X H) = \nabla.J$$
Since, the divergence of the curl of a vector is zero.  Therefore,
$$\nabla.(\nabla X H) = 0$$
It means that
$$\nabla.J=0$$
This is the equation of continuity for steady current not for time varying field. The equation for time varying field is given as:
$$\nabla.J = - \frac{d\rho}{dt}$$
This is the reason why Maxwell modify his Ampere's circuital law.


Modification of Ampere's circuital law:
Differential form (with modification):
The Ampere's circuital law was modified by the given concept of displacement current D and also the concept of displacement current density $$J_d$$ for time varying fields.
Equ(10) for time varying field should be rewritten as:
$$\nabla X H = J + J_d$$ ......(11)
By taking divergence of equ(11)
$$\nabla.(\nabla X H) = \nabla J + \nabla J_d$$
As divergence of the curl of a vector is always zero, therefore:
$$\nabla.(\nabla X H) = 0$$
This means that:
$$\nabla.(J + J_d) = 0$$
or
$$\nabla J = - \nabla J_d$$ ..........(α)
But from equation of continuity for time varying fields
$$\nabla.J = \frac{d\rho}{dt}$$ ..........(β)
By comparing equ(α) and equ(β)
$$\nabla.J_d = \frac{d}{dt}(\nabla.D)$$ .....(12)

Because from Maxwell first equation
$$\nabla.D = \rho$$
Since, the divergence of two vectors are equal only if the vectors are equal.
Then:
$$J_d = \frac{dD}{dt}$$
Substitute the above equation into equ(11)
$$\nabla X H = J + \frac{dD}{dt}$$ .....(13)
Here,
$$\frac{dD}{dt}=J_d$$
J = conduction current
D = displacement current.
Equ(13) is the differential form of the Maxwell fourth equation.

Integral form:
Taking surface integral of both sides of equ(13), we have:
$$\int (\nabla X H).ds = \int (J + \frac{dD}{dt}).ds$$
Applying Stokes law to the LHS
$$\int (\nabla X H).ds = \int_{l} H.dL$$
Comparing the above two equation, we get:
$$\int H.dL = \int (J + \frac{dD}{dt}).ds$$ ......(14)
Equ(14) is the integral form of Maxwell fourth equation.


Labels:

Comments

Post a Comment

Popular posts from this blog

Schrodinger equation as a law in physics

The unified theory of wave-particle duality has been used to derive the Schrödinger equations. The Schrodinger equations are generally accepted, by postulate rather than derivation, to be laws of physics. The Schrodinger equations provide a basis for analyzing many kinds of systems (molecular, atomic, and nuclear) in a particular inertial reference frame. The success of the Schrödinger equations constitutes a basis for accepting them, their derivations, and the unified theory of wave-particle duality which makes such derivations possible. This acceptance is completely justified in the favored inertial reference frame. In accord with the principle of relativity, all physical laws must be the same in all inertial reference frames, i.e., all physical laws must be Lorentz invariant. Recall, the relationship: $$\nabla^{2} \psi = \frac{\partial^{2} \psi}{\partial t^{2}}$$ ...........(1) Equ (1) is Lorentz invariant and reduces, by means of the procedure presented in the previou
Post a Comment
Read more

Maxwell first equation

The Maxwell first equation in electrostatics is called the Gauss law in electrostatics. Statement:  It states that the total electric flux \(\psi_E\) passing through a closed hypothetical surface is equal to \(\frac{1}{\epsilon_0}\) enclosed by the surface. Integral Form: $$\phi_E = \int E.ds = \frac{q}{\epsilon_0}$$ $$\int D.ds = q$$ where, $$D = \epsilon_0 E = displacement-vector$$ Let the change be distributed over a volume v and \(\rho\) be the volume charge density. Hence, $$q = \int \rho dv$$ Therefore; $$\int D.ds = \int_{v} \rho dv$$ .........(1) Equ(1) is the integral form of Maxwell first law Differential form: Apply Gauss divergence theorem to the L.H.S of equ(1) from surface integral to volume integral. $$\int D.ds = \int (\nabla.D)dv$$ Substituting this equation to equ(1) $$\int(\nabla.D)dv = \int_{v} \rho dv$$ As two volume integrals are equal only if their integrands are equal. Thus; $$\nabla.D = \rho v$$ ............(2) Equ(2) is the differe
Post a Comment
Read more

Auxiliary time dependent Schrodinger equation

Recall the relationship: $$\phi(t) = C_+ exp[-i(\frac{E}{\hbar})t] + C_- exp[i(\frac{E}{\hbar})t]$$ .............(1) Recall, from the present special case: $$C_+ = 0$$ ........(2) Substitute equ (2) into equ (1), we have that: $$\phi(t) = C_- exp[i(\frac{E}{\hbar})t]$$ .......(3) Also,  recall the relationship: $$\Psi(r,t) = \psi(r) \phi(t)$$ ..........(4) Substitute equ (3) into equ (4): $$\Psi(r,t) = C_- exp \psi (r)  [i(\frac{E}{\hbar})t]$$ ..............(5) Equ (5) can be differentiated with respect to t, and by rearranging we have: $$\Psi(r,t) = -i(\frac{\hbar}{E}) \frac{d \Psi (r, t)}{dt}$$ ..............(6) Recall the relationship: $$-\frac{\hbar^{2}}{2m} \nabla^{2} \Psi(r,t) + V(r) \Psi(r,t) = E \Psi(r,t)$$ ...........(7) Substituting equ (6) into the right hand side of equ (7) we have: $$-\frac{\hbar^{2}}{2m} \nabla^{2} \Psi(r,t) + V(r) \Psi(r,t) = i\hbar \frac{d \Psi(r,t)}{dt}$$ .........(8) Equ (8) is the auxiliary time dependent Schrodinger equation
Post a Comment
Read more