# Poynting theorem and derivation

Poynting Theorem

Statement. This theorem states that the cross product of electric field vector, E and magnetic field vector, H at any point is a measure of the rate of flow of electromagnetic energy per unit area at that point, that is

P = E x H

Here P → Poynting vector and it is named after its discoverer, J.H. Poynting. The direction of P is perpendicular to E and H and in the direction of vector E x H

Proof. Consider Maxwell’s fourth equation (Modified Ampere’s Circuital Law), that is

del x H = J + ε dE/dt

or                                 J = (del x H)  –  ε dE/dt

The above equation has the dimensions of current density. Now, to convert the dimensions into rate of energy flow per unit volume, take dot product of both sides of above equation by E, that is

E. J = E. (del x H) – εE.  dE/dt         (1)

Use vector Indentity

del. (E x H) = H. (del x E) – E. (del x H)

or         E. (del x H) = H. (del x E) – del (  E x H )

By substituting value of E. (del x H) in equation (1) , we get

E. J. =H . (del x E) – del . (E x H) – εE  dE/dt                     (2)

also from Maxwell’s  third equation (Faraday’s law of electromagnetic induction).

del x E =  μdH/dt

By substituting value of del x E in equation (2) we get

E. J. =μH . (dH)/dt  – εE.  dE/dt –  del . (E x H)                   (3)

We can write

H. dH/dt  = 1/2   dH2/dt                                 (4a)

E. dE/dt  = 1/2  dE2/dt                                               (4b)

By substituting equations 4a and 4b in equation 3 , we get

E. J. =  -μ/2  dH 2/dt  –   ε/2    dE 2/dt  – del . (E x H)

E. J. =  -d/dt  [ dH 2/2  +   εE2 /2] – del . (E x H)

By taking volume integral on both sides, we get

∫E. J. dV =  -d/dt  ∫ [μ H 2 /2 +   εE2/2 ]  dV – ∫del . (E x H) dV      (5)

apply Gauss’s Divergence theorem to second term of R.H.S., to change volume integral into surface integral, that is

∫del . (E x H) dV = ∫ (E x H) . dS

Substitute above equation in equation 5

Thus

∫E. J. dV = -d/dt  ∫[ ε E2/2 +  μ H2/2]  dV – ∫(E x H) . dS     (6)

or        ∫ (E x H) . dS = ∫-d/dt   [ ε E 2 /2 +  μ H 2 /2]  dV –∫ E. J. dV

Interpretation of above equation :

L.H.S. Term

∫ (E x H) . dS  → It represents the rate of outward flow of energy from a volume S

V and the integral is over the closed surface surrounding the volume. This rate of outward flow of power from a volume V is represented by

∫ P . dS = ∫ (E x H) . dS

where Poynting vector, P = E x H

Inward flow of power is represented by

– ∫ P . dS = – φ∫ (E x H) . ds

R.H.S. First Term

-d /dt  [μ H 2/2   +  ε E 2 /2]  dV → If the energy is flowing out of the region, there must be  a corresponding decrease of electromagnetic energy. So here negative sign indicates decrease. Electromagnetic energy is the sum of magnetic energy, μ H 2/2 and electric

energy, ε E 2/2 . So first term of R.H.S. represents rate of decrease of stored electromagnetic  2  energy.

R.H.S. Second Term

∫ (E. J) dV →Total ohmic power dissipated within the volume.

So from the law of conservation of energy, equation (6) can be written in words as

Rate of energy disripation in volume V = Rate at which stored electromagnetic energy is decreasing in V + Inward rate of flow of energy through the surface of the volume.

## 5 Replies to “Poynting theorem and derivation”

1. Avirup Mullick says:

How these two identities used?

H. dH/dt = 1/2 dH2/dt

E. dE/dt = 1/2 dE2/dt

1. amsh says:

2. Lone Shehzan says:

As we k/t dH^2/dt=2HdH/dt
= HdH/dt =1/2 dH^2/dt

3. Roshan verma says:

because d/dt [H^2] = 2H.dH/dt

2. Mudit Srivastava says:

d/dt(H^2) =d/dt(H.H)= H.d/dt(H) +H.d/dt(H) => 2H.d/dt(H) = d/dt(H^2)

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