2024-03-18 Conservation laws

Topics today:

• Storage of energy in magnetic and electric fields
• Linear and Angular momentum
• Energy and momentum of electromagnetic fields within the framework of maxwell’s equations
  • Energy density of EM field
  • Poynting Vector
  • Maxwell stress/strain tensors
  • Force $\vec{F}=q\vec{E}+q\vec{v}\times \vec{B}$, known as the Lorentz Force.

Phys 2 review :

• Electric charge is quantized in electrons
  • All electric charges is stored as integer multiples of the electron
• It is globally conserved

Charge conservation in EM theory

• Continuity equation states there the change in charge is equal to the flux through a surface around the change in charge

$$\frac{d\rho }{dt}+\vec{\nabla }\cdot \vec{j}=0$$ $$⟹\frac{d}{dt}\int \rho dV=-\int \vec{\nabla }\cdot \vec{j}dV$$ $$\frac{d{Q}_{\text{in Vol}}}{dt}=-\int \vec{j}\cdot d\vec{a}$$

Via Divergence Theorem

The Coulomb and Lorentz forces work over distance in a vacuum. This is interesting and unique.

The energy density of electric and magnetic fields can be easily seen in capacitors. Ex: Parallel plate capacitor

$$C={ϵ}_0\frac{A}{d}$$ $$U_E=\frac{1}{2}{ϵ}_0\frac{A}{d}{E}^2{d}^2$$

Inductors as well!

Conservation of electromagnetic field energy:

$$\frac{d}{dt}{E}_{1mech}+\frac{d}{dt}\int \frac{1}{2}\vec{E}\cdot \vec{D}+\vec{B}+\vec{h}dV+\underset{a(v)}{\int }\vec{s}\cdot d\vec{a}$$

The last term of this is also known as the Poynting vector In laymans terms, the above equation states that the energy lost or gained electromagnetically in the system has to be in the form of a poynting vector

Flashback to Phys 1:

Mechanical power is from the change in mechanical energy

Using the lorentz force equation

$$\vec{F}=q\vec{E}+\vec{q}\times \vec{B}$$

The mechanical power is :

$$F\cdot \vec{v}=q\vec{v}\cdot \vec{E}+q\vec{v}\text{cancelto}0\cdot (\vec{v}\times \vec{B})K$$

Therefore the electrical power is given by

$$\vec{F}\cdot \vec{v}=q\vec{v}\cdot \vec{E}$$

Electrical Power density is given by

$$\frac{q\vec{v}\cdot \vec{E}}{\text{volume}}\to \vec{j}\cdot E$$

And current density is given by

$$\vec{j}=\frac{q\vec{v}}{\text{ volume}}$$

Electrical power is given by

$${\int }_V\vec{j}\cdot \vec{E}dV=\frac{d{E}_{mech}}{dE}$$

Example of the utility of these fields

$$\vec{j}=\vec{\nabla }\times \vec{H}-\dot{\vec{D}}$$ $$\vec{\nabla }\cdot (\vec{E}\times \vec{H})=\vec{H}\cdot (\vec{\nabla }\times \vec{E})-\vec{E}\cdot (\vec{\nabla }\times \vec{H})$$ $$\vec{\nabla }\times \vec{E}=-\dot{\vec{B}}$$ $$\vec{E}(\vec{\nabla }\times \vec{H})=-\vec{H}\cdot \vec{B}-\nabla \cdot (\vec{E}\times \vec{H})$$ $$\int \vec{j}\cdot \vec{E}dV={\int }_V\vec{E}\cdot (\vec{\nabla }\times \vec{H})-\vec{\dot{D}}dV$$ $$={\int }_V(-\vec{H}\cdot \overrightarrow{\dot{B}-\stackrel{⃗}{\nabla }\cdot (\stackrel{⃗}{E}\times \stackrel{⃗}{H}-\stackrel{⃗}{E}\cdot \dot{D}\stackrel{⃗}{})})dV$$ $$={\int }_V(-\mu \vec{H}\dot{\vec{H}}-ϵ\vec{E}\dot{\vec{E}}-\vec{\nabla }\cdot (\vec{E}\times \vec{H}))dV$$

Change in time of $E_{mech}$ is

$$\frac{d{E}_{\text{mech}}}{dt}=\int \vec{j}\cdot \vec{E}dV=-\frac{d}{dt}{\int }_V\frac{1}{2}ϵ{\vec{E}}^2\frac{1}{2}\mu {\vec{H}}^{2}dV-\int \vec{E}\times \vec{H}da$$

Energy Flux Density term is given by the poynting vector $\int \vec{S}\cdot d\vec{a}$

Momentum

Momentum of the electromagnetic field Lorentz force

$$\vec{F}=q(\vec{E}+\vec{v}\times \vec{B})=\frac{d\vec{p}m}{dt}\text{\hspace{1em}(1)}$$

When we introduce rho and cancel $\vec{j}$, equation 1 is conserved and converted to

$$\frac{d\vec{p}m}{dt}={\int }_v(\rho \vec{E}+\vec{j}\times \vec{B})dV$$

From maxwells eqs.

$$\rho =\vec{\nabla }\cdot \vec{D},\rho =\vec{\nabla }\cdot \vec{D},\vec{j}=\vec{\nabla }\times \vec{H}-\dot{\vec{D}}$$

The converted eq 1 becomes

$${\int }_V(\vec{E}(\vec{\nabla }\cdot \vec{D})+(\vec{\nabla }\times \vec{H}-\dot{\vec{D}})\times \vec{B})dV$$ $$\frac{d}{dt}(\vec{D}\times \vec{B})=\vec{D}\times \vec{B}+\vec{D}\times \dot{\vec{B}}$$ $$=\frac{d}{dt}(\vec{D}\times \vec{B})-\vec{D}\times (-\vec{\nabla }\times \vec{E})$$ $$\frac{d\vec{p}m}{dt}={\int }_v\vec{E}(\vec{\nabla }\cdot \vec{D})-\vec{D}\times (\vec{\nabla }\times \vec{E})+(\vec{\nabla }\times \vec{H})\times \vec{B}-\frac{d}{dt}(\vec{D}\times \vec{B})dV$$

There is a lot of components in this integral that correspond to real quantities. Not the first term though Adding “0”

$$0=\vec{H}\cdot (\vec{\nabla }\cdot \vec{B})$$ $$\frac{d\vec{p}m}{dt}={\int }_v\vec{E}(\vec{\nabla }\cdot \vec{D})-\vec{D}\times (\vec{\nabla }\times \vec{E})+(\vec{\nabla }\times \vec{H})\times \vec{B}-\frac{d}{dt}(\vec{D}\times \vec{B})+\vec{H}\cdot (\vec{\nabla }\cdot \vec{B})dV$$

Which yields the following:

$$\frac{d\vec{p}m}{dt}={\int }_v\vec{E}(\vec{\nabla }\cdot \vec{D})-\vec{D}\times (\vec{\nabla }\times \vec{E})+(\vec{\nabla }\times \vec{H})\times \vec{B}-\frac{d}{dt}(\vec{D}\times \vec{B})+\vec{H}\cdot (\vec{\nabla }\cdot \vec{B})dV$$ $$\frac{d\vec{p}m}{dt}={\int }_v(\vec{E}(\vec{\nabla }\cdot \vec{D})-\vec{D}\times \vec{\nabla }\times \vec{E}+(\vec{\nabla }\times \vec{H})\times \vec{B}-\frac{d}{dt}(\vec{d}\times \vec{B})+\vec{H}(\vec{\nabla }\cdot \vec{B}))dV$$

This yields us the final result:

The change of momentum with time is

$$\frac{d\vec{p}m}{dt}+\frac{d}{dt}{\int }_V\vec{D}\times \vec{B}dV$$ $$={\int }_V[\vec{E}(\vec{\nabla }\cdot \vec{D}-\vec{D}\times (\vec{\nabla }\times \vec{E})+\vec{H}(\vec{\nabla }\cdot \vec{B})-\vec{B}\times (\vec{\nabla }\times \vec{H}))]dV$$

Where $\vec{D}\times \vec{B}$ is the electromagnetic momentum density

and

$$(\vec{E}(\vec{\nabla }\cdot \vec{D})-\vec{D}\times (\vec{\nabla }\times \vec{E})+\vec{H}(\vec{\nabla }\cdot \vec{B})-\vec{B}\times (\vec{\nabla }\times \vec{H}))$$

This is known as the Maxwell stress-strain tensor:

$$T_{ij}=D_i{E}_j-\frac{1}{2}{\delta }_{ij}(\vec{D}\cdot \vec{E})+B_i{H}_j-\frac{1}{2}{\delta }_{ij}(\vec{B}\cdot \vec{H})$$

Angular Momentum

Defined in mechanics as

$$\vec{l}=\vec{t}\times \vec{p}$$

For the angular momentum density of the electromagnetic field we can use

$$\vec{l}=\vec{t}\times (\vec{D}\times \vec{B})$$

This is analogous to classical/mechanical angular momentum.

Electromagnetic waves have $\vec{E}$ and $\vec{B}$ fields, current topic of research is generating EM waves with a specific angular momenta.

In class problems 8.1 and 8.7