User:Jambaugh/Sandbox/EM

Introduction
In Einstein's Special Relativity space and time are unified into a four dimensional space-time in which observations as seen by observers moving at different relative velocities transform under the Lorentz group, SO(3,1). The three dimensional vectors (3-vectors) and 3x3 tensors of Euclidean space are extended to four dimensional vectors (4-vectors) and 4x4 tensors of Minkowski_space-time. Including translations in space and time yields the larger Poincaré group, ISO(3,1).

A manifestly covariant formulation of Electromagnetism means a formulation expressing the physical quantities in terms of their transformation representation under the Lorentz and Poincaré groups. When this is done...
 * the usual scalar and 3-vector potentials are recognized as components of a single 4-vector,
 * the usual 3-vector electric and 3-vector magnetic fields become components of a single rank 2 anti-symmetric electro-magnetic field tensor,
 * The Coulomb and Lorentz forces are unified into a single, 4-dimensional covariant Lorentz force equation.
 * All eight of Maxwell's Equations (two scalar and two 3-vector) may then be expressed as two 4-vector equations.

Conventions

 * Einstein's summation convention is used throughout and space-time indices run from 0 to 3. Repeated indices are summed over their values.
 * e.g. $$a^\mu b_\mu = a^0 b_0 + a^1 b_1 + a^2 b_2 + a^3 b_3$$


 * The standard coordinates (and their differentials) will be expressed using raised greek indice:
 * $$dx^\mu\sim(dx^0,dx^1,dx^2,dx^3)=(cdt,dx,dy,dz)$$


 * The covector of partial derivatives will be expressed using subscripted partial symbol:
 * $$ \partial_\mu \sim (\partial_0,\partial_1,\partial_2,\partial_3)

= \left(\frac{\partial}{\partial ct},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)$$
 * The Minkowski metric used in this article will be the proper-time metric:
 * $$c^2d\tau^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 = \eta_{\mu\nu}dx^\mu dx^\nu,$$


 * Indices are lowered using this metric, $$dx_\mu = \eta_{\mu\nu}dx^\nu$$ and raised using the reciprocal metric.

The 4-Vector Potential
Electromagnetism is a U(1) gauge theory which is to say it is derivable by assuming a complex phase degree of freedom for particles moving through space-time. The way this complex phase, $$\theta$$ changes as a particle with charge $$q$$ is translated defines the 4-vector potential as the gauge connection:
 * $$d\theta = \frac{q}{c}A_\mu(x) dx^\mu$$

NOTE: Neither the phase nor the components of the phase connection are physically observable although differences in phase connection may be observed via interference experiments. (Ref: Aharanov-Bhom effect.) TODO: discuss gauge transformations and canonical momentum's role as generator of translations.


 * $$(A^0,A^1,A^2,A^3) \equiv (\phi,c\mathbf{A}_x,c\mathbf{A}_y,c\mathbf{A}_z)$$( volts in SI units)

NOTE: Gauge transformations: $$\tilde{A}_\mu = A_\mu + \partial_\mu \chi(x)$$

Test Particle Lagrangian and The Covariant Lorentz Force
TODO: Make note of the fact that we cannot use proper-time parametrization as proper-time is not defined until after we impose the dynamic constraints a la Euler-Lagrange equations.

We chose an arbitrary (time-like) parametrization of a test particle's path and define an action:


 * $$ S = \int_{\vartheta_1}^{\vartheta_2} L(x^\mu(\vartheta),\dot{x}^\mu(\vartheta))d\vartheta$$

where the dotted coordinates correspond to parameter derivatives:
 * $$ \dot{x}^\mu \equiv \frac{d x^\mu}{d\vartheta}$$

The Lagrangian is:
 * $$ L = m\dot{\tau}+\dot{\theta} = m\sqrt{\dot{x}^\mu \dot{x}_\mu} + \frac{q}{c}A_\mu(x) \dot{x}^\mu $$

Using the standard variational methods we obtain the Euler-Lagrange equations:
 * $$ \frac{d}{d\theta}\left(\frac{\partial L}{\partial \dot{x}^\mu}\right) = \frac{\partial L}{\partial x^\mu}$$

The canonical momentum is:
 * $$ P_\mu \equiv \frac{\partial L}{\partial \dot{x}^\mu} = m\frac{\dot{x}_\mu}{\sqrt{\dot{x}^\nu \dot{x}_\nu}}+\frac{q}{c}A_\mu=

m\frac{\dot{x}_\mu}{\dot{\tau}} + qA_\mu\dot{x}$$ or
 * $$ P_\mu = m u_\mu + \frac{q}{c}A_\mu = mp_\mu + \frac{q}{c}A_\mu$$

where $$u_\mu = \frac{d x_\mu}{d\tau}$$ is the proper-time 4-velocity of the particle and $$p_\mu$$ is then the kinetic 4-momentum.

The E-L equations then take the form:
 * $$\dot{P}_\mu = \frac{q}{c}\partial_\mu A_\nu \dot{x}^\nu$$

expanding
 * $$\dot{P}_\mu = \dot{p}_\mu + \frac{q}{c} \dot{A}_\mu = \dot{p}_\mu + \frac{q}{c}(\partial_\nu A_\mu) \dot{x}^\nu$$

yields the E-L equations in the form of the covariant Lorentz force:
 * $$ \dot{p}_\mu = \frac{q}{c}\left(\partial_\mu A_\nu - \partial_\nu A_\mu\right)\dot{x}^\nu \equiv \frac{q}{c} F_{\mu\nu}\dot{x}^\nu$$

This defines the electro-magnetic field tensor.


 * $$ F_{\mu\nu} \equiv \partial_\mu A_\nu - \partial_\nu A_\mu$$

Note that the electro-magnetic field tensor is anti-symmetric, $$F_{\mu\nu} = - F_{\nu\mu}$$

The Covariant Lorentz Force
The electromagnetic field tensor is defined as:


 * $$ F_{\mu\nu}\equiv \partial_\mu A_\nu- \partial_\nu A_\mu$$ (in units of volts/meter)

Evaluating typical components in terms of the conventional scalar and vector potentials gives us these components in terms of the E and B fields:
 * $$F_{01} = \partial_0 A_1 - \partial_1 A_0 = \partial_{ct}(-c\mathbf{A}_x)-\partial_x\phi =  E_x$$


 * $$F_{23} = \partial_2 A_3 - \partial_3 A_2 = \partial_y(-c\mathbf{A}_z) - \partial_z(-c\mathbf{A}_y) =c[\nabla\times\mathbf{A}]_x=cB_x$$

Likewise expanding each term gives :


 * $$F_{\mu\nu} \sim \left( \begin{array}{cccc}

0&E_x& E_y&E_z\\ -E_x&0&cB_z&-cB_y\\ -E_y&-cB_z&0&cB_x\\ -E_z&cB_y&-cB_x&0 \end{array}\right)$$ and
 * $$F^{\mu\nu} \sim \left( \begin{array}{cccc}

0&-E_x&-E_y&-E_z\\ E_x&0&-cB_z&cB_y\\ E_y&cB_z&0&-cB_x\\ E_z&-cB_y&cB_x&0 \end{array}\right)$$


 * Edit Line: Changing sign convention.

The covariant Lorentz force becomes:


 * $$\dot{p}^\mu = q F^{\mu\nu}u_\nu$$


 * $$ (u_0,u_1,u_2,u_3) = \gamma(1,-v_x,-v_y,-v_z)$$


 * $$\dot{p}^1 = \gamma \frac{d \mathbf{p}_x}{dt} = q\left(F^{10}u_0 + F^{12}u_2 + F^{13}u_3 \right)$$
 * $$=\gamma q\left(E_x - B_z(- \mathbf{v}_y) + B_y(-\mathbf{v}_z)\right) $$

hence


 * $$ \frac{d \mathbf{p}_x}{dt} = \hat{\imath}\cdot q\left( \mathbf{E} + \mathbf{v}\times \mathbf{B}\right)$$

The other components are similarly calculated and we have the combined Lorentz and Coulomb forces:


 * $$ \frac{d \mathbf{p}}{dt} = q\mathbf{E} + q\mathbf{v}\times \mathbf{B}$$

We also have the energy component:
 * $$\dot{p}^0 = \gamma \frac{d \mathcal{E}}{dct} = q\left(F^{01}u_1 + F^{02}u_2 + F^{03}u_3 \right)=\frac{\gamma q}{c}\left(E_x\mathbf{v}_x + E_y \mathbf{v}_y + E_z\mathbf{v}_z\right) $$

Thus
 * $$\frac{d \mathcal{E}}{dt} = q\mathbf{E}\cdot \mathbf{v}$$

This is the work done on the particle by the Coulomb force.

The Canonical Hamiltonian
The generally covariant Hamiltonian is...


 * $$H = \dot{x}^\mu P_\mu - L$$
 * $$ = \dot{x}^\mu(p_\mu -qA^\mu) - m\sqrt{\dot{x}^\alpha\dot{x}_\alpha} + qA_\mu\dot{x}^\mu$$
 * $$ = m\frac{\dot{x}^\mu\dot{x}_\mu}{\sqrt{\dot{x}^\alpha\dot{x}_\alpha}}- m\sqrt{\dot{x}^\alpha\dot{x}_\alpha}$$
 * $$ = m\sqrt{\dot{x}^\alpha\dot{x}_\alpha}- m\sqrt{\dot{x}^\alpha\dot{x}_\alpha}=0$$

This zero Hamiltonian is typical of generally covariant dynamics. The zero value however must be understood as resulting from the dynamic constraints and so we seek the constraint equation $$H(p,x)\dot{=}0$$ which will define the Hamiltonian.

(Badly worded and reasoned... fix)


 * $$L = m\sqrt{\dot{x}^\mu \dot{x}_\mu} - qA_\mu\dot{x}^\mu= \dot{\tau}\left(m-\frac{q}{m}A_\mu (P^\mu-A^\mu)\right)$$

$$m\dot{\tau}= \sqrt{p^\mu p_\mu} = \sqrt{(P^\mu + qA^\mu)(P_\mu+qA_\mu)}$$
 * $$ p^\mu = (P^\mu + qA^\mu)$$


 * $$ H = \dot{x}^\mu P_\mu - L =

\frac{\dot{\tau}}{m}(P^\mu + qA^\mu)P_\mu - \dot{\tau}\left(m-\frac{q}{m}A_\mu (P^\mu+qA^\mu)\right)$$


 * $$H = \frac{\dot{\tau}}{m}\left[(P^\mu+qA^\mu)(P_\mu+qA_\mu) - m^2\right]$$
 * $$= \frac{1}{m^2}\left[(P^\mu+qA^\mu)(P_\mu+qA_\mu)\right]^{3/2} - \left[(P^\mu+qA^\mu)(P_\mu+qA_\mu)\right]^{1/2}$$

Note that on the mass shell $$p^\mu p_\mu = m^2$$ and thus
 * $$H= \frac{1}{m^2}[ m^3] - m = 0$$

'''!!!! This is Horrible!!!! Try again...'''

Lagrangian Density of Electro-Magnetic Field in the presence of Currents
Lagrangian Density:
 * $$ \mathcal{L} = \mathcal{L}_{matter} + A_\mu J^\mu -\frac{1}{4\mu_0} F_{\mu\nu}F^{\mu\nu}$$

Notes and references

 * W. Rindler, Introduction to Special Relativity, 2nd edition, Oxford Science Publications, 1991, ISBN 0-19-853952-5.