Lagrangian (field theory)

Lagrangian field theory is a formalism in classical field theory. It is the field theoretic analogue of Lagrangian mechanics. Lagrangian mechanics is used for discrete particles each with a finite number of degrees of freedom. Lagrangian field theory applies to continua and fields.

Lagrangians and Lagrangian densities in field theory

The time integral of the Lagrangian is called the action denoted by S. In field theory, a distinction is occasionally made between the Lagrangian L, of which the action is the time integral:

${\mathcal {S}}=\int {L\,\mathrm {d} t}$

and the Lagrangian density $\scriptstyle {\mathcal {L}}$, which one integrates over all spacetime to get the action:

${\mathcal {S}}[\varphi _{i}]=\int {{\mathcal {L}}(\varphi _{i}(x))\,\mathrm {d} ^{4}x}.$

• General form of Lagrangian density: ${\mathcal {L}}={\mathcal {L}}(\varphi _{i},\varphi _{i,\mu })$ where $\varphi _{i,\mu }\equiv {\frac {\partial \varphi _{i}}{\partial x^{\mu }}}\equiv \partial _{\mu }\varphi _{i}$ (see 4-gradient)
• The relationship between ${\mathcal {L}}$ and $L$: $L=\int {{\mathcal {L}}\,d^{n-1}x}$, where $n$ is the space-time dimension similar to $q=\int \rho \,dV$.
• In field theory, the independent variable t was replaced by an event in spacetime (x, y, z, t) or still more generally by a point s on a manifold.

The Lagrangian is then the spatial integral of the Lagrangian density. However, $\scriptstyle {\mathcal {L}}$ is also frequently simply called the Lagrangian, especially in modern use; it is far more useful in relativistic theories since it is a locally defined, Lorentz scalar field. Both definitions of the Lagrangian can be seen as special cases of the general form, depending on whether the spatial variable $\scriptstyle \mathbf {x}$ is incorporated into the index i or the parameters s in φ_i(s). Quantum field theories in particle physics, such as quantum electrodynamics, are usually described in terms of $\scriptstyle {\mathcal {L}}$, and the terms in this form of the Lagrangian translate quickly to the rules used in evaluating Feynman diagrams.

Notice that, in the presence of gravity or when using general curvilinear coordinates, the Lagrangian density ${\mathcal {L}}$ will include a factor of ${\sqrt {\vert g\vert }}$ or its equivalent to ensure that it is a scalar density so that the integral will be invariant.

Selected fields

To go with the section on test particles above, here are the equations for the fields in which they move. The equations below pertain to the fields in which the test particles described above move and allow the calculation of those fields. The equations below will not give you the equations of motion of a test particle in the field but will instead give you the potential (field) induced by quantities such as mass or charge density at any point $\scriptstyle (\mathbf {x} ,t)$. For example, in the case of Newtonian gravity, the Lagrangian density integrated over spacetime gives you an equation which, if solved, would yield $\scriptstyle \Phi (\mathbf {x} ,t)$. This $\scriptstyle \Phi (\mathbf {x} ,t)$, when substituted back in equation (1), the Lagrangian equation for the test particle in a Newtonian gravitational field, provides the information needed to calculate the acceleration of the particle.

Newtonian gravity

The Lagrangian (density) is $\scriptstyle {\mathcal {L}}$ in J·m⁻³. The interaction term mΦ is replaced by a term involving a continuous mass density ρ in kg·m⁻³. This is necessary because using a point source for a field would result in mathematical difficulties. The resulting Lagrangian for the classical gravitational field is:

${\mathcal {L}}(\mathbf {x} ,t)=-\rho (\mathbf {x} ,t)\Phi (\mathbf {x} ,t)-{1 \over 8\pi G}(\nabla \Phi (\mathbf {x} ,t))^{2}$

where G in m³·kg⁻¹·s⁻² is the gravitational constant. Variation of the integral with respect to Φ gives:

$\delta {\mathcal {L}}(\mathbf {x} ,t)=-\rho (\mathbf {x} ,t)\delta \Phi (\mathbf {x} ,t)-{2 \over 8\pi G}(\nabla \Phi (\mathbf {x} ,t))\cdot (\nabla \delta \Phi (\mathbf {x} ,t)).$

Integrate by parts and discard the total integral. Then divide out by δΦ to get:

$0=-\rho (\mathbf {x} ,t)+{1 \over 4\pi G}\nabla \cdot \nabla \Phi (\mathbf {x} ,t)$

and thus

$4\pi G\rho (\mathbf {x} ,t)=\nabla ^{2}\Phi (\mathbf {x} ,t)$

which yields Gauss's law for gravity.

Einstein Gravity

Main article: Einstein–Hilbert action

The Lagrange density for general relativity in the presence of matter fields is

${\mathcal {L}}_{\text{GR}}={\mathcal {L}}_{\text{EH}}+{\mathcal {L}}_{\text{matter}}={\frac {c^{4}}{16\pi G}}\left(R-2\Lambda \right)+{\mathcal {L}}_{\text{matter}}$

$\scriptstyle R$ is the curvature scalar, which is the Ricci tensor contracted with the metric tensor, and the Ricci tensor is the Riemann tensor contracted with a Kronecker delta. The integral of ${\mathcal {L}}_{\text{EH}}$ is known as the Einstein-Hilbert action. The Riemann tensor is the tidal force tensor, and is constructed out of Christoffel symbols and derivatives of Christoffel symbols, which are the gravitational force field. Plugging this Lagrangian into the Euler-Lagrange equation and taking the metric tensor $g_{\mu \nu }$ as the field, we obtain the Einstein field equations

$R_{\mu \nu }-{\frac {1}{2}}Rg_{\mu \nu }+g_{\mu \nu }\Lambda ={\frac {8\pi G}{c^{4}}}T_{\mu \nu }$

The last tensor is the energy momentum tensor and is defined by

$T_{\mu \nu }\equiv {\frac {-2}{\sqrt {-g}}}{\frac {\delta ({\mathcal {L}}_{\mathrm {matter} }{\sqrt {-g}})}{\delta g^{\mu \nu }}}=-2{\frac {\delta {\mathcal {L}}_{\mathrm {matter} }}{\delta g^{\mu \nu }}}+g_{\mu \nu }{\mathcal {L}}_{\mathrm {matter} }.$

$\scriptstyle g$ is the determinant of the metric tensor when regarded as a matrix. $\Lambda$ is the Cosmological constant. Generally, in general relativity, the integration measure of the action of Lagrange density is ${\sqrt {-g}}d^{4}x$. This makes the integral coordinate independent, as the root of the metric determinant is equivalent to the Jacobian determinant. The minus sign is a consequence of the metric signature (the determinant by itself is negative).

Electromagnetism in special relativity

The interaction terms

$-q\phi (\mathbf {x} (t),t)+q{\dot {\mathbf {x} }}(t)\cdot \mathbf {A} (\mathbf {x} (t),t)$

are replaced by terms involving a continuous charge density ρ in A·s·m⁻³ and current density $\scriptstyle \mathbf {j}$ in A·m⁻². The resulting Lagrangian for the electromagnetic field is:

${\mathcal {L}}(\mathbf {x} ,t)=-\rho (\mathbf {x} ,t)\phi (\mathbf {x} ,t)+\mathbf {j} (\mathbf {x} ,t)\cdot \mathbf {A} (\mathbf {x} ,t)+{\epsilon _{0} \over 2}{E}^{2}(\mathbf {x} ,t)-{1 \over {2\mu _{0}}}{B}^{2}(\mathbf {x} ,t).$

Varying this with respect to ϕ, we get

$0=-\rho (\mathbf {x} ,t)+\epsilon _{0}\nabla \cdot \mathbf {E} (\mathbf {x} ,t)$

which yields Gauss' law.

Varying instead with respect to $\scriptstyle \mathbf {A}$, we get

$0=\mathbf {j} (\mathbf {x} ,t)+\epsilon _{0}{\dot {\mathbf {E} }}(\mathbf {x} ,t)-{1 \over \mu _{0}}\nabla \times \mathbf {B} (\mathbf {x} ,t)$

which yields Ampère's law.

Using tensor notation, we can write all this more compactly. The term $-\rho \phi (\mathbf {x} ,t)+\mathbf {j} \cdot \mathbf {A}$ is actually the inner product of two four-vectors. We package the charge density into the current 4-vector and the potential into the potential 4-vector. These two new vectors are

$j^{\mu }=(\rho ,\mathbf {j} )\quad {\text{and}}\quad A_{\mu }=(-\phi ,\mathbf {A} )$

We can then write the interaction term as

$-\rho \phi +\mathbf {j} \cdot \mathbf {A} =j^{\mu }A_{\mu }$

Additionally, we can package the E and B fields into what is known as the electromagnetic tensor $F_{\mu \nu }$. We define this tensor as

$F_{\mu \nu }=\partial _{\mu }A_{\nu }-\partial _{\nu }A_{\mu }$

The term we are looking out for turns out to be

${\epsilon _{0} \over 2}{E}^{2}-{1 \over {2\mu _{0}}}{B}^{2}=-{\frac {1}{4\mu _{0}}}F_{\mu \nu }F^{\mu \nu }=-{\frac {1}{4\mu _{0}}}F_{\mu \nu }F_{\rho \sigma }\eta ^{\mu \rho }\eta ^{\nu \sigma }$

We have made use of the Minkowski metric to raise the indices on the EMF tensor. In this notation, Maxwell's equations are

$\partial _{\mu }F^{\mu \nu }=-\mu _{0}j^{\nu }\quad {\text{and}}\quad \epsilon ^{\mu \nu \lambda \sigma }\partial _{\nu }F_{\lambda \sigma }=0$

where ε is the Levi-Civita tensor. So the Lagrange density for electromagnetism in special relativity written in terms of Lorentz vectors and tensors is

${\mathcal {L}}(x)=j^{\mu }(x)A_{\mu }(x)-{\frac {1}{4\mu _{0}}}F_{\mu \nu }(x)F^{\mu \nu }(x)$

In this notation it is apparent that classical electromagnetism is a Lorentz-invariant theory. By the equivalence principle, it becomes simple to extend the notion of electromagnetism to curved spacetime.

Electromagnetism in general relativity

The Lagrange density of electromagnetism in general relativity also contains the Einstein-Hilbert action from above. The pure electromagnetic Lagrangian is precisely a matter Lagrangian ${\mathcal {L}}_{\text{matter}}$. The Lagrangian is

${\begin{aligned}{\mathcal {L}}(x)&=j^{\mu }(x)A_{\mu }(x)-{1 \over 4\mu _{0}}F_{\mu \nu }(x)F_{\rho \sigma }(x)g^{\mu \rho }(x)g^{\nu \sigma }(x)+{\frac {c^{4}}{16\pi G}}R(x)\\&={\mathcal {L}}_{\text{Maxwell}}+{\mathcal {L}}_{\text{Einstein-Hilbert}}.\end{aligned}}$

This Lagrangian is obtained by simply replacing the Minkowski metric in the above flat Lagrangian with a more general (possibly curved) metric $g_{\mu \nu }(x)$. We can generate the Einstein Field Equations in the presence of an EM field using this lagrangian. The energy-momentum tensor is

$T^{\mu \nu }(x)={\frac {2}{\sqrt {-g(x)}}}{\frac {\delta }{\delta g_{\mu \nu }(x)}}{\mathcal {S}}_{\text{Maxwell}}={\frac {1}{\mu _{0}}}\left(F_{{\text{ }}\lambda }^{\mu }(x)F^{\nu \lambda }(x)-{\frac {1}{4}}g^{\mu \nu }(x)F_{\rho \sigma }(x)F^{\rho \sigma }(x)\right)$

It can be shown that this energy momentum tensor is traceless, i.e. that

$T=g_{\mu \nu }T^{\mu \nu }=0$

If we take the trace of both sides of the Einstein Field Equations, we obtain

$R=-{\frac {8\pi G}{c^{4}}}T$

So the tracelessness of the energy momentum tensor implies that the curvature scalar in an electromagnetic field vanishes. The Einstein equations are then

$R^{\mu \nu }={\frac {8\pi G}{c^{4}}}{\frac {1}{\mu _{0}}}\left(F_{{\text{ }}\lambda }^{\mu }(x)F^{\nu \lambda }(x)-{\frac {1}{4}}g^{\mu \nu }(x)F_{\rho \sigma }(x)F^{\rho \sigma }(x)\right)$

Additionally, Maxwell's equations are

$D_{\mu }F^{\mu \nu }=-\mu _{0}j^{\nu }$

where $D_{\mu }$ is the covariant derivative. For free space, we can set the current tensor equal to zero, $j^{\mu }=0$. Solving both Einstein and Maxwell's equations around a spherically symmetric mass distribution in free space leads to the Reissner-Nordstrom charged black hole, with the defining line element (written in natural units and with charge Q):

$ds^{2}=\left(1-{\frac {2M}{r}}+{\frac {Q^{2}}{r^{2}}}\right)dt^{2}-\left(1-{\frac {2M}{r}}+{\frac {Q^{2}}{r^{2}}}\right)^{-1}dr^{2}-r^{2}d\Omega ^{2}$

Electromagnetism using differential forms

Using differential forms, the electromagnetic action S in vacuum on a (pseudo-) Riemannian manifold $\scriptstyle {\mathcal {M}}$ can be written (using natural units, c = ε₀ = 1) as

${\mathcal {S}}[\mathbf {A} ]=\int _{\mathcal {M}}\left(-{\frac {1}{2}}\,\mathbf {F} \wedge \star \mathbf {F} +\mathbf {A} \wedge \star \mathbf {J} \right).$

Here, A stands for the electromagnetic potential 1-form, J is the current 1-form, F is the field strength 2-form and the star denotes the Hodge star operator. This is exactly the same Lagrangian as in the section above, except that the treatment here is coordinate-free; expanding the integrand into a basis yields the identical, lengthy expression. Note that with forms, an additional integration measure is not necessary because forms have coordinate differentials built in. Variation of the action leads to

$\mathrm {d} {\star }\mathbf {F} =\mathbf {J} .$

These are Maxwell's equations for the electromagnetic potential. Substituting F = dA immediately yields the equation for the fields,

$\mathrm {d} \mathbf {F} =0$

because F is an exact form.

Dirac Lagrangian

The Lagrangian density for a Dirac field is:

${\mathcal {L}}=i\hbar c{\bar {\psi }}{\partial }\!\!/\ \psi -mc^{2}{\bar {\psi }}\psi$

where ψ is a Dirac spinor (annihilation operator), $\scriptstyle {\bar {\psi }}=\psi ^{\dagger }\gamma ^{0}$ is its Dirac adjoint (creation operator), and ${\partial }\!\!/$ is Feynman slash notation for $\scriptstyle \gamma ^{\sigma }\partial _{\sigma }\!$.

Quantum electrodynamic Lagrangian

The Lagrangian density for QED is:

${\mathcal {L}}_{\mathrm {QED} }=i\hbar c{\bar {\psi }}{D}\!\!\!\!/\ \psi -mc^{2}{\bar {\psi }}\psi -{1 \over 4\mu _{0}}F_{\mu \nu }F^{\mu \nu }$

where $\scriptstyle F^{\mu \nu }\!$ is the electromagnetic tensor, D is the gauge covariant derivative, and ${D}\!\!\!\!/$ is Feynman notation for $\scriptstyle \gamma ^{\sigma }D_{\sigma }\!$.

Quantum chromodynamic Lagrangian

The Lagrangian density for quantum chromodynamics is:

${\mathcal {L}}_{\mathrm {QCD} }=\sum _{n}\left(i\hbar c{\bar {\psi }}_{n}{D}\!\!\!\!/\ \psi _{n}-m_{n}c^{2}{\bar {\psi }}_{n}\psi _{n}\right)-{1 \over 4}G^{\alpha }{}_{\mu \nu }G_{\alpha }{}^{\mu \nu }$

where D is the QCD gauge covariant derivative, n = 1, 2, ...6 counts the quark types, and $\scriptstyle G^{\alpha }{}_{\mu \nu }\!$ is the gluon field strength tensor.

See also

• Hamiltonian field theory