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Summary of Lagrangian/Hamiltonian/Action derivatives

$\mathbf {\dot {q}} =+{\frac {\partial H}{\partial \mathbf {p} }}$

$\mathbf {p} ={\frac {\partial {\mathcal {S}}}{\partial \mathbf {q} }}={\frac {\partial L}{\partial \mathbf {\dot {q}} }}$

$\mathbf {\dot {p}} =-{\frac {\partial H}{\partial \mathbf {q} }}={\frac {\partial L}{\partial \mathbf {q} }}={\boldsymbol {\mathcal {Q}}}$

$L={\frac {d{\mathcal {S}}}{dt}}\,,\quad H=-{\frac {\partial {\mathcal {S}}}{\partial t}}$

${\frac {dH}{dt}}={\frac {\partial L}{\partial t}}$

scattering diff. cross-section (to merge)

Differential cross-section

See also: spherical polar coordinates

For a two-body collision, the angular distribution of one of the scattered particles is described using the differential solid angle:

${\rm {d}}\Omega ={\rm {d}}\cos \theta {\rm {d}}\phi$

The scattered path of the particle is within a cone of solid angle Ω, and its axis in the direction (θ, φ).

The differential cross-section dσ/dΩ for N identical scattered particles is given by:

${\rm {d}}W=JN{\frac {{\rm {d}}\sigma (\theta ,\phi )}{{\rm {d}}\Omega }}{\rm {d}}\Omega \,,$

where

$W=JN\sigma \,,$

is measured rate of the paricles entering the cone described above, N is the number of particles

$J=nv\,,$

is the particle flux (number of particles passing through unit area per unit time) corresponding to the number density n (number of particles per unit volume) and velocity v (in the rest frame of the collision point). Integrating the equation gives the scattering cross-section:

$\sigma =\int _{0}^{2\pi }{\rm {d}}\phi \int _{-1}^{+1}{\rm {d}}\cos \theta {\frac {{\rm {d}}\sigma (\theta ,\phi )}{{\rm {d}}\Omega }}\,.$

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Application to one spin particle in three spatial dimensions
For a particle, with spin, in all three spatial dimensions, the wavefunction is

Ket Ψ, ket bases, and orthonormality

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Noether's theorem

A simple way to formulate Noether's theorem using Lagrangian mechanics is as follows.

If there is quantity that is a constant of the motion, it means the Lagrangian can be parameterized by a continuous variable and still describe the same motion. Let the continuous parameter be s (such as length, angle of rotation, etc.). Then for L independent of s it immediately follows that:

$L[Q(s,t),{\dot {Q}}(s,t)]=L[q(t),{\dot {q}}(t)]$

and

${\frac {d}{ds}}L[Q(s,t),{\dot {Q}}(s,t)]=0$

where q is the solution of the Euler-Lagrange equation for s = 0, and Q for any s, that is; q(t) = Q(s = 0, t). Since the same L describes the same motion for all s. By the chain rule:

${\frac {dL}{ds}}={\frac {\partial L}{\partial Q}}{\frac {dQ}{ds}}+{\frac {\partial L}{\partial {\dot {Q}}}}{\frac {d{\dot {Q}}}{ds}}=0$

Using the Euler-Lagrange equation:

${\frac {\partial L}{\partial Q}}={\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {Q}}}}$

gives

${\frac {dL}{ds}}={\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {Q}}}}{\frac {dQ}{ds}}+{\frac {\partial L}{\partial {\dot {Q}}}}{\frac {\partial {\dot {Q}}}{\partial s}}={\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {Q}}}}{\frac {dQ}{ds}}\right)=0$

and the definition of generalized momentum conjugate to the generalized coordinate Q:

$p={\frac {\partial L}{\partial {\dot {Q}}}}$

we have

$A(q,{\dot {q}})=\left.\left(p{\frac {dQ}{ds}}\right)\right|_{s=0}=C$

where C is a constant, and A is a quantity that is time independent, and the equation is evaluated at s = 0 for convenience. It follows that A is a conserved quantity, with respect to the parameter s.

In general, for n lots of s parameters, and N generalized coordinates, writing them as the vectors: s = (s₁, s₂, ... s_n), q = (q₁, q₂, ... q_N), Q = (Q₁, Q₂, ... Q_N) and also q(t) = Q(s = 0, t), the generalized version is:

$A_{j}(\mathbf {q} ,{\dot {\mathbf {q} }})=\left.\left(\mathbf {p} \cdot {\frac {d\mathbf {Q} }{ds_{j}}}\right)\right|_{s_{j}=0}=C_{j}$

where · denotes the dot product, written explicitly:

$A_{j}(q_{1},q_{2}\cdots q_{N},{\dot {q}}_{1}{\dot {,q}}_{2}\cdots {\dot {q}}_{N})=\sum _{i=1}^{N}\left.\left(p_{i}{\frac {dQ_{i}}{ds_{j}}}\right)\right|_{s_{j}=0}=C_{j}$

i.e. for every continuous parameter that leaves the Lagrangian invariant, there is a component of the conserved quantity. For translational lengths s_j = x_j, it follows A_j is a component of the system's total linear momentum p_j. For angles s_j = θ_j, then A_j is a component of the system's total angular momentum L_j. If s = t, then A is the total energy E of the system.

The Levi-Civita tensor and Hodge dual

Levi-Civita tensor

The covariant Levi-Civita tensor in an n-D metric space may be defined as the unique (up to a sign) n-form (completely antisymmetric order-n covariant tensor) that obeys the relation

$\left|\epsilon _{\alpha _{1}\dots \alpha _{n}}g^{\alpha _{1}\beta _{1}}\dots g^{\alpha _{n}\beta _{n}}\epsilon _{\beta _{1}\dots \beta _{n}}\right|=n!$

The choice of sign defines an orientation in the space.

The contravariant Levi-Civita tensor is an n-vector that may be defined by raising each of the indices of the corresponding covariant tensor in the normal fashion:

$\epsilon ^{\alpha _{1}\dots \alpha _{n}}=g^{\alpha _{1}\beta _{1}}\dots g^{\alpha _{n}\beta _{n}}\epsilon _{\beta _{1}\dots \beta _{n}}$

Levi-Civita symbols

The Levi-Civita symbols $\varepsilon _{\alpha _{1}\dots \alpha _{n}}=\varepsilon ^{\alpha _{1}\dots \alpha _{n}}$ are defined for an n-dimensional space as +1, −1 or 0 respectively when the indices are an even permutation of the indices of the ordered basis vectors, an odd permutation thereof, or neither.

These symbols do not form the components of a tensor, but are related to the components of the Levi-Civita tensor by a scalar:

$\pm {\sqrt {|\det(g_{\mu \nu })|}}\varepsilon _{\alpha _{1}\dots \alpha _{n}}=\epsilon _{\alpha _{1}\dots \alpha _{n}}$

In this article, variants of the epsilon glyph are used to distinguish the tensor from the symbol. This notation is not standard.

Hodge dual

Not to be confused with covector, dual vector or bra, which are related abstractions of functionals in dual spaces.

Fully contracting a p-form S with the contravariant Levi-Civita tensor produces its Hodge dual T = ∗S, which is a q-vector, where p + q = n:

$T^{\beta _{1}\cdots \beta _{q}}={\frac {1}{p!}}S_{\alpha _{1}\dots \alpha _{p}}\epsilon ^{\alpha _{1}\dots \alpha _{p}\beta _{1}\dots \beta _{q}}$

Similarly, fully contracting a q-vector T with the covariant Levi-Civita tensor produces its Hodge dual U = ∗T, which is a p-form, where p + q = n:

$U_{\gamma _{1}\dots \gamma _{p}}={\frac {1}{q!}}T^{\beta _{1}\dots \beta _{q}}\epsilon _{\beta _{1}\dots \beta _{q}\gamma _{1}\cdots \gamma _{p}}$

where the star denotes the operation of taking the dual of T. Taking the dual again returns the same tensor, up to a sign:

${*{*S}}=\pm S$

Maxwell's equations (yet more...)

Formulation	Homogeneous equations		Nonhomogeneous equations
Vacuum
Vector calculus (fields)	$\nabla \cdot \mathbf {B} =0$	$\nabla \times \mathbf {E} +{\frac {\partial \mathbf {B} }{\partial t}}=0$	$\nabla \cdot \mathbf {E} ={\frac {\rho }{\varepsilon _{0}}}$	$\nabla \times \mathbf {B} -{\frac {1}{c^{2}}}{\frac {\partial \mathbf {E} }{\partial t}}=\mu _{0}\mathbf {J}$
Vector calculus (potentials, any gauge)	identities		$\nabla ^{2}\varphi +{\frac {\partial }{\partial t}}\left(\mathbf {\nabla } \cdot \mathbf {A} \right)=-{\frac {\rho }{\varepsilon _{0}}}$	$\Box \mathbf {A} +\mathbf {\nabla } \left(\mathbf {\nabla } \cdot \mathbf {A} +{\frac {1}{c^{2}}}{\frac {\partial \varphi }{\partial t}}\right)=\mu _{0}\mathbf {J}$
QED, vector calculus (potentials, Lorenz gauge)	identities		$\Box \varphi =-{\frac {1}{\varepsilon _{0}}}e\psi ^{\dagger }\psi$	$\Box \mathbf {A} =-\mu _{0}e\psi ^{\dagger }{\boldsymbol {\alpha }}\psi$
Tensor calculus (potentials, Lorenz gauge)	identities		$\Box A^{\mu }=\mu _{0}J^{\mu }$
Tensor calculus (fields)	${\dfrac {\partial F_{\alpha \beta }}{\partial x^{\gamma }}}+{\dfrac {\partial F_{\gamma \alpha }}{\partial x^{\beta }}}+{\dfrac {\partial F_{\beta \gamma }}{\partial x^{\alpha }}}=0$		${\dfrac {\partial F^{\beta \alpha }}{\partial x^{\alpha }}}=\mu _{0}J^{\beta }$
Differential forms (fields)	$\mathrm {d} \mathbf {F} =0$		$\mathrm {d} \star \mathbf {F} =\mathbf {J}$
Geometric algebra (fields)	$\nabla F=\mu _{0}cJ$
Algebra of physical space (fields)	$\left({\frac {1}{c}}{\dfrac {\partial }{\partial t}}+{\boldsymbol {\nabla }}\right)F=\mu _{0}cJ$
Matter
Vector calculus (fields)	$\nabla \cdot \mathbf {B} =0$	$\nabla \times \mathbf {E} +{\frac {\partial \mathbf {B} }{\partial t}}=0$	$\nabla \cdot \mathbf {D} =\rho _{f}$	$\nabla \times \mathbf {H} -{\frac {\partial \mathbf {D} }{\partial t}}=\mathbf {J} _{f}$
Tensor calculus (fields)	${\dfrac {\partial F_{\alpha \beta }}{\partial x^{\gamma }}}+{\dfrac {\partial F_{\gamma \alpha }}{\partial x^{\beta }}}+{\dfrac {\partial F_{\beta \gamma }}{\partial x^{\alpha }}}=0$		${\frac {\partial {\mathcal {D}}^{\mu \nu }}{\partial x^{\mu }}}={J^{\nu }}_{\mathrm {free} }$
Differential forms (fields)	$\mathrm {d} \mathbf {F} =0$		$\mathrm {d} \star {\boldsymbol {\mathcal {D}}}=\mathbf {J} _{\mathrm {free} }$

Special relativity

4-vectors

Velocity and acceleration in 4D

Recognising other physical quantities as tensors also simplifies their transformation laws. First note that the velocity four-vector U^μ is given by

$U^{\mu }={\frac {dx^{\mu }}{d\tau }}={\begin{pmatrix}\gamma c\\\gamma v_{x}\\\gamma v_{y}\\\gamma v_{z}\end{pmatrix}}$

Recognising this, we can turn the awkward looking law about composition of velocities into a simple statement about transforming the velocity four-vector of one particle from one frame to another. U^μ also has an invariant form:

${\mathbf {U} }^{2}=\eta _{\nu \mu }U^{\nu }U^{\mu }=-c^{2}.$

So all velocity four-vectors have a magnitude of c. This is an expression of the fact that there is no such thing as being at coordinate rest in relativity: at the least, you are always moving forward through time. The acceleration 4-vector is given by $A^{\mu }=d{\mathbf {U} ^{\mu }}/d\tau$. Given this, differentiating the above equation by τ produces

$2\eta _{\mu \nu }A^{\mu }U^{\nu }=0.\!$

So in relativity, the acceleration four-vector and the velocity four-vector are orthogonal.

Momentum in 4D

The momentum and energy combine into a covariant 4-vector:

$p_{\nu }=m\,\,\eta _{\nu \mu }U^{\mu }={\begin{pmatrix}-E/c\\p_{x}\\p_{y}\\p_{z}\end{pmatrix}}.$

where m is the invariant mass.

The invariant magnitude of the momentum 4-vector is:

$\mathbf {p} ^{2}=\eta ^{\mu \nu }p_{\mu }p_{\nu }=-(E/c)^{2}+p^{2}.$

We can work out what this invariant is by first arguing that, since it is a scalar, it doesn't matter which reference frame we calculate it, and then by transforming to a frame where the total momentum is zero.

$\mathbf {p} ^{2}=-(E_{rest}/c)^{2}=-(m\cdot c)^{2}.$

We see that the rest energy is an independent invariant. A rest energy can be calculated even for particles and systems in motion, by translating to a frame in which momentum is zero.

The rest energy is related to the mass according to the celebrated equation discussed above:

$E_{rest}=mc^{2}\,$

Note that the mass of systems measured in their center of momentum frame (where total momentum is zero) is given by the total energy of the system in this frame. It may not be equal to the sum of individual system masses measured in other frames.

Force in 4D

To use Newton's third law of motion, both forces must be defined as the rate of change of momentum with respect to the same time coordinate. That is, it requires the 3D force defined above. Unfortunately, there is no tensor in 4D which contains the components of the 3D force vector among its components.

If a particle is not traveling at c, one can transform the 3D force from the particle's co-moving reference frame into the observer's reference frame. This yields a 4-vector called the four-force. It is the rate of change of the above energy momentum four-vector with respect to proper time. The covariant version of the four-force is:

$F_{\nu }={\frac {dp_{\nu }}{d\tau }}={\begin{pmatrix}-{d(E/c)}/{d\tau }\\{dp_{x}}/{d\tau }\\{dp_{y}}/{d\tau }\\{dp_{z}}/{d\tau }\end{pmatrix}}$

where $\tau \,$ is the proper time.

In the rest frame of the object, the time component of the four force is zero unless the "invariant mass" of the object is changing (this requires a non-closed system in which energy/mass is being directly added or removed from the object) in which case it is the negative of that rate of change of mass, times c. In general, though, the components of the four force are not equal to the components of the three-force, because the three force is defined by the rate of change of momentum with respect to coordinate time, i.e. ${\frac {dp}{dt}}$ while the four force is defined by the rate of change of momentum with respect to proper time, i.e. ${\frac {dp}{d\tau }}$.

In a continuous medium, the 3D density of force combines with the density of power to form a covariant 4-vector. The spatial part is the result of dividing the force on a small cell (in 3-space) by the volume of that cell. The time component is −1/c times the power transferred to that cell divided by the volume of the cell. This will be used below in the section on electromagnetism.

EM in 4d spacetime

Maxwell's equations in the 3D form are already consistent with the physical content of special relativity. But we must rewrite them to make them manifestly invariant.

The charge density $\rho \!$ and current density $[J_{x},J_{y},J_{z}]\!$ are unified into the current-charge 4-vector:

$J^{\mu }={\begin{pmatrix}\rho c\\J_{x}\\J_{y}\\J_{z}\end{pmatrix}}.$

The law of charge conservation, ${\frac {\partial \rho }{\partial t}}+\nabla \cdot \mathbf {J} =0$, becomes:

$\partial _{\mu }J^{\mu }=0.\!$

The electric field $[E_{x},E_{y},E_{z}]\!$ and the magnetic induction $[B_{x},B_{y},B_{z}]\!$ are now unified into the (rank 2 antisymmetric covariant) electromagnetic field tensor:

$F_{\mu \nu }={\begin{pmatrix}0&-E_{x}/c&-E_{y}/c&-E_{z}/c\\E_{x}/c&0&B_{z}&-B_{y}\\E_{y}/c&-B_{z}&0&B_{x}\\E_{z}/c&B_{y}&-B_{x}&0\end{pmatrix}}.$

The density, $f_{\mu }\!$, of the Lorentz force, $\mathbf {f} =\rho \mathbf {E} +\mathbf {J} \times \mathbf {B}$, exerted on matter by the electromagnetic field becomes:

$f_{\mu }=F_{\mu \nu }J^{\nu }.\!$

Faraday's law of induction, $\nabla \times \mathbf {E} =-{\frac {\partial \mathbf {B} }{\partial t}}$, and Gauss's law for magnetism, $\nabla \cdot \mathbf {B} =0$, combine to form:

$\partial _{\lambda }F_{\mu \nu }+\partial _{\mu }F_{\nu \lambda }+\partial _{\nu }F_{\lambda \mu }=0.\!$

Although there appear to be 64 equations here, it actually reduces to just four independent equations. Using the antisymmetry of the electromagnetic field one can either reduce to an identity (0=0) or render redundant all the equations except for those with λ,μ,ν = either 1,2,3 or 2,3,0 or 3,0,1 or 0,1,2.

The electric displacement $[D_{x},D_{y},D_{z}]\!$ and the magnetic field $[H_{x},H_{y},H_{z}]\!$ are now unified into the (rank 2 antisymmetric contravariant) electromagnetic displacement tensor:

${\mathcal {D}}^{\mu \nu }={\begin{pmatrix}0&D_{x}c&D_{y}c&D_{z}c\\-D_{x}c&0&H_{z}&-H_{y}\\-D_{y}c&-H_{z}&0&H_{x}\\-D_{z}c&H_{y}&-H_{x}&0\end{pmatrix}}.$

Ampère's law, $\nabla \times \mathbf {H} =\mathbf {J} +{\frac {\partial \mathbf {D} }{\partial t}}$, and Gauss's law, $\nabla \cdot \mathbf {D} =\rho$, combine to form:

$\partial _{\nu }{\mathcal {D}}^{\mu \nu }=J^{\mu }.\!$

In a vacuum, the constitutive equations are:

$\mu _{0}{\mathcal {D}}^{\mu \nu }=\eta ^{\mu \alpha }F_{\alpha \beta }\eta ^{\beta \nu }\,.$

Antisymmetry reduces these 16 equations to just six independent equations. Because it is usual to define $F^{\mu \nu }\,$ by

$F^{\mu \nu }=\eta ^{\mu \alpha }F_{\alpha \beta }\eta ^{\beta \nu }\,$

the constitutive equations may, in a vacuum, be combined with Ampère's law etc. to get:

$\partial _{\beta }F^{\alpha \beta }=\mu _{0}J^{\alpha }.\!$

The energy density of the electromagnetic field combines with Poynting vector and the Maxwell stress tensor to form the 4D electromagnetic stress-energy tensor. It is the flux (density) of the momentum 4-vector and as a rank 2 mixed tensor it is:

$T_{\alpha }^{\pi }=F_{\alpha \beta }{\mathcal {D}}^{\pi \beta }-{\frac {1}{4}}\delta _{\alpha }^{\pi }F_{\mu \nu }{\mathcal {D}}^{\mu \nu }$

where $\delta _{\alpha }^{\pi }$ is the Kronecker delta. When upper index is lowered with η, it becomes symmetric and is part of the source of the gravitational field.

The conservation of linear momentum and energy by the electromagnetic field is expressed by:

$f_{\mu }+\partial _{\nu }T_{\mu }^{\nu }=0\!$

where $f_{\mu }\!$ is again the density of the Lorentz force. This equation can be deduced from the equations above (with considerable effort).