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.

Covariant Quantities

Covariant Quantity	Components	SI Units
Space-Time Coordinates	$x^{\mu }$	meters
Four-Velocity	$u^{\mu }=dx^{\mu }/d\tau$	meter/second
Four-Momentum	$p^{\mu }$	kilogram•meter/second
Four-Potential	$A^{\mu }$	volts=joules/coulomb
Minkowski Metric	$\eta _{\mu \nu }$	unitless
Electromagnetic Field	$F^{\mu \nu }$	volts/meter=newtons/coulomb

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:

$c^{2}d\tau ^{2}=c^{2}dt^{2}-dx^{2}-dy^{2}-dz^{2}=\eta _{\mu \nu }dx^{\mu }dx^{\nu },$

• 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:

${\dot {x}}^{\mu }\equiv {\frac {dx^{\mu }}{d\vartheta }}$

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

$P_{\mu }=mu_{\mu }+{\frac {q}{c}}A_{\mu }=mp_{\mu }+{\frac {q}{c}}A_{\mu }$

where $u_{\mu }={\frac {dx_{\mu }}{d\tau }}$ 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:

${\dot {p}}^{\mu }=qF^{\mu \nu }u_{\nu }$

$(u_{0},u_{1},u_{2},u_{3})=\gamma (1,-v_{x},-v_{y},-v_{z})$

${\dot {p}}^{1}=\gamma {\frac {d\mathbf {p} _{x}}{dt}}=q\left(F^{10}u_{0}+F^{12}u_{2}+F^{13}u_{3}\right)$

$=\gamma q\left(E_{x}-B_{z}(-\mathbf {v} _{y})+B_{y}(-\mathbf {v} _{z})\right)$

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:

${\mathcal {L}}={\mathcal {L}}_{matter}+A_{\mu }J^{\mu }-{\frac {1}{4\mu _{0}}}F_{\mu \nu }F^{\mu \nu }$

See also

• Relativistic electromagnetism
• Electromagnetic wave equation
• Liénard–Wiechert potential for a charge in arbitrary motion
• Nonhomogeneous electromagnetic wave equation
• Moving magnet and conductor problem
• Electromagnetic tensor
• Proca action
• Stueckelberg action
• Quantum electrodynamics

Notes and references

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

Further reading

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[[:Category:Fundamental physics concepts]]
[[:Category:Electromagnetism]]
[[:Category:Special relativity]]