Lagrangian, Lagrangian Function, Λαγκρανζιανή


Λαγρασιανή Καθιερωμένου Σωματιδιακού Προτύπου


Η Λαγρασιανή του Καθιερωμένου Σωματιδιακού Προτύπου.


Η Λαγρασιανή του Καθιερωμένου Σωματιδιακού Προτύπου.


Λαγρασιανή Καθιερωμένο Σωματιδιακό Πρότυπο


Ενέργεια (φυσική οντότητα) Ακτινοβολία Ενέργεια (φυσικό μέγεθος)
Μηχανική Ενέργεια Κινητική Ενέργεια Δυναμική Ενέργεια Έργο Ελαστική Ενέργεια Δυναμικό Βαρυτική Ενέργεια Λαγρασιανή (Lagrangian) Χαμιλτονιανή (Hamiltonian)
Θερμική Ενέργεια Θερμότητα Εσωτερική Ενέργεια Ενθαλπία Ελεύθερη Ενέργεια Helmholtz Ελεύθερη Ενέργεια Gibbs Χημική Ενέργεια
Ηλεκτρομαγνητική Ενέργεια Ηλεκτρική Ενέργεια Μαγνητική Ενέργεια Ηλεκτρικό Δυναμικό Μαγνητικό Δυναμικό Ηλεκτρική Τάση Επαγωγική Τάση
Ιονίζουσα Ενέργεια Πυρηνική Ενέργεια Ενέργεια Μηδενικού Σημείου
Σκοτεινή Ενέργεια Big Bang
Αρχή Διατήρησης Ενέργειας Ισοδυναμία Μάζας - Ενέργειας 1ος θερμοδυναμικός Νόμος 2ος Θερμοδυναμικός Νόμος
Ενέργειες Ενεργειακές ΠηγέςΠεδία
Οικονομική Ενέργεια Χρήμα Φυσικός ΠόροςΕνεργειακή Παραγωγή Υδροηλεκτρικό Εργοστάσιο Ατμοηλεκτρικό Εργοστάσιο
Υπερβατικές Ενέργειες


Ανανεώσιμη Ενέργεια Μη-Ανανεώσιμη Ενέργεια


Joseph-Louis Lagrange
Αναλυτική Μηχανική Λαγρασιανή


Λαγρασιανή Καθιερωμένου Σωματιδιακού Προτύπου

- Ένα Φυσικό Μέγεθος.


Έλαβε το όνομά της από τον διάσημο φυσικό Lagrange.


Είναι μια μαθηματική συνάρτηση που παριστάνει μαθηματικά τον τρόπο με τον οποίο αλληλεπιδρούν μεταξύ τους, αλλά και με τα πεδία, τα διάφορα σωματίδια.


Στην Κλασσική Φυσική, η Λαγρασιανή L είναι η διαφορά της κινητικής ενέργειας Κ μείον την δυναμική ενέργεια V που κατέχει το εξεταζόμενο φυσικό σύστημα .

Η εξίσωση που την καθορίζει είναι η εξής:

L = K - V


The Lagrangian, $ \mathcal{L} $, of a dynamical system is a function that summarizes the dynamics of the system. It is named after Joseph Louis Lagrange. The concept of a Lagrangian was originally introduced in a reformulation of Κλασσική Μηχανική known as Αναλυτική Μηχανική.

In classical mechanics, the Lagrangian is defined as the Κινητική Ενέργεια (kinetic energy), $ T $, του φυσικού συστήματος μείον την Δυναμική Ενέργεια (potential energy) του, $ V $.

In symbols,

$ \mathcal{L} = T - V $.

Κάτω από συγκεκριμένες προϋποθέσεις, οι οποίες δίδονται από Lagrangian mechanics, εάν η Λαγρασιανή του συστήματος είναι γνωστή, τότε οι Εξισώσεις Κίνησης (equations of motion) του συστήματος μπορούν να ληφθούν με άμεση αντικατάσταση της συναρτησιακής έκφρασης της Λαγρασιανής στις Εξισώσεις Euler - Lagrange, οι οποίες αποτελούν ειδική οικογένεια Μερικών Διαφορικών Εξισώσεων (differential equations).

Λαγρασιανός ΦορμαμαλισμόςEdit

Σημασία Edit

The Lagrange formulation of mechanics is important not just for its broad applications, but also for its role in advancing βαθεία κατανόηση της Φυσικής. Although Lagrange only sought to describe Κλασσική Μηχανική, η "Αρχή της Δράσης (action principle)" that is used to derive the Lagrange equation is now recognized to be applicable to Κβαντική Φυσική.

Physical action and quantum-mechanical phase (waves) are related via Σταθερά Planck, and the principle of stationary action can be understood in terms of constructive interference of wave functions.

The same principle, and the Lagrange formalism, are tied closely to Θεώρημα Noether, which relates physical conserved quantities to continuous συμμετρίες of a physical system.

Lagrangian mechanics and Θεώρημα Noether together yield a φυσικό φορμαλισμό for Πρώτη Κβάντωση (first quantization) by including Μεταθέτες (commutators) between certain terms of the Lagrangian equations of motion for a physical system.


  • The formulation is not tied to any one coordinate system -- rather, any convenient variables $ \varphi_i(s) $ may be used to describe the system; these variables are called "generalized coordinates" and may be any independent variable of the system (for example, strength of the magnetic field at a particular location; angle of a pulley; position of a particle in space; or degree of excitation of a particular eigenmode in a complex system). This makes it easy to incorporate constraints into a theory by defining coordinates which only describe states of the system which satisfy the constraints.
  • If the Lagrangian is invariant under a symmetry, then the resulting equations of motion are also invariant under that symmetry. This is very helpful in showing that theories are consistent with either special relativity or general relativity.
  • Equations derived from a Lagrangian will almost automatically be unambiguous and consistent, unlike equations just thrown together from various sources.


The Εξισώσεις Κίνησης (equations of motion) are obtained by means of an Φυσική Δράση (action) principle, written as:

$ \frac{\delta \mathcal{S}}{\delta \varphi_i} = 0 $

where the action, S, είναι ένα Μαθηματικό Συναρτησιοειδές (functional)

$ \mathcal{S}[\varphi_i] = \int{\mathcal{L}[\varphi_i(s)]{}\,\mathrm{d}^ns}, $

and where $ {}{}{}{}\ s_\alpha $ denotes the set of parameters of the system.

The equations of motion obtained by means of the [[Συναρτησιακή Παράγωγος (functional derivative) are identical to the usual Εξισώσεις Euler - Lagrange.

Dynamical systems whose equations of motion are obtainable by means of an action principle on a suitably chosen Lagrangian are known as Lagrangian dynamical systems.

Examples of Lagrangian dynamical systems range from the classical version of the Σωματιδιακό Πρότυπο, to Newton's equations, to purely mathematical problems such as geodesic equations and Πρόβλημα Plateau.

Παραδείγμα Κλασσικής ΜηχανικήςEdit

Ορθογώνιο Σύστημα ΣυντεταγμένωνEdit

Suppose we have a Τρισδιάστατος Χώρος and the Λαγρασιανή (Lagrangian)

$ L(\vec{x}, \dot{\vec{x}}) \ = \ \frac{1}{2} \ m \ \dot{\vec{x}}^2 \ - \ V(\vec{x}) $.

Τότε, η Εξίσωση Euler-Lagrange είναι:

$ \frac{d~}{dt} \ \left( \, \frac{\partial L}{\partial \dot{x}_i} \, \right) \ - \ \frac{\partial L}{\partial x_i} \ = \ 0 $

where $ i = 1, 2, 3 $.

Η παραγώγιση δίδει:

$ \frac{\partial L}{\partial x_i} \ = \ - \ \frac{\partial V}{\partial x_i} $
$ \frac{\partial L}{\partial \dot{x}_i} \ = \ \frac{\partial ~}{\partial \dot{x}_i} \, \left( \, \frac{1}{2} \ m \ \dot{\vec{x}}^2 \, \right) \ = \ \frac{1}{2} \ m \ \frac{\partial ~}{\partial \dot{x}_i} \, \left( \, \dot{x}_i \, \dot{x}_i \, \right) = \ m \, \dot{x}_i $
$ \frac{d~}{dt} \ \left( \, \frac{\partial L}{\partial \dot{x}_i} \, \right) \ = \ m \, \ddot{x}_i $

The Euler-Lagrange equations can therefore be written as:

$ m\ddot{\vec{x}}+\nabla V=0 $

where the time derivative is written conventionally as a dot above the quantity being differentiated, and $ \nabla $ is the del operator.

Using this result, it can easily be shown that the Lagrangian approach is equivalent to the Newtonian one.

If the force is written in terms of the potential $ \vec{F}=- \nabla V(x) $; the resulting equation is $ \vec{F}=m\ddot{\vec{x}} $, which is exactly the same equation as in a Newtonian approach for a constant mass object.

A very similar deduction gives us the expression $ \vec{F}=\mathrm{d}\vec{p}/\mathrm{d}t $, which is Newton's Second Law in its general form.

Σφαιρικό Σύστημα ΣυντεταγμένωνEdit

Suppose we have a three-dimensional space using spherical coordinates $ r, \theta, \phi $ with the Lagrangian

$ \frac{m}{2}(\dot{r}^2+r^2\dot{\theta}^2 +r^2\sin^2\theta\dot{\varphi}^2)-V(r). $

Τότε οι Εξισώσεις Euler-Lagrange είναι:

$ m\ddot{r}-mr(\dot{\theta}^2+\sin^2\theta\dot{\varphi}^2)+V' =0, $
$ \frac{\mathrm{d}}{\mathrm{d}t}(mr^2\dot{\theta}) -mr^2\sin\theta\cos\theta\dot{\varphi}^2=0, $
$ \frac{\mathrm{d}}{\mathrm{d}t}(mr^2\sin^2\theta\dot{\varphi})=0. $

Here the set of parameters $ s_i $ is just the time $ t $, and the dynamical variables $ \phi_i(s) $ are the trajectories $ \vec x(t) $ of the particle.

Despite the use of standard variables such as $ x $, the Lagrangian allows the use of any coordinates, which do not need to be orthogonal. These are "generalized coordinates".

Μηχανική Υλικού ΣωματιδίουEdit

A Υλικό Σωματίδιο (test particle) is a hypothetical simplified point particle with no properties other than mass and charge. Real particles like electrons and up-quarks are more complex and have additional terms in their Lagrangians.

Κλασσική περίπτωση με Νευτώνεια ΒαρύτηταEdit

The Lagrangian is $ L \! $ joules. Given a particle with mass $ m \! $ kilograms, and position $ \vec{x} $ meters in a Newtonian gravitation field with potential $ \zeta \! $ joules per kilogram. The particle's world line is parameterized by time $ t\! $ seconds. The particle's kinetic energy is:

$ T[t] = {1 \over 2} m \dot{\vec{x}}[t] \cdot \dot{\vec{x}}[t] $

and the particle's gravitational potential energy is:

$ V[t] = m \zeta [\vec{x} [t],t] . $

Thus the Lagrangian is:

$ L[t] = T[t] - V[t] = {1 \over 2} m \dot{\vec{x}}[t] \cdot \dot{\vec{x}}[t] - m \zeta [\vec{x} [t],t] . $

Varying $ \vec{x}\! $ in the integral (equivalent to the Euler Lagrange differential equation), we get

$ 0 = \delta\int{L[t] \, \mathrm{d}t} = \int{\delta L[t] \, \mathrm{d}t} $
$ = \int{m \dot{\vec{x}}[t] \cdot \dot{\delta \vec{x}}[t] - m \nabla \zeta [\vec{x} [t],t] \cdot \delta \vec{x}[t] \, \mathrm{d}t}. $

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

$ 0 = - m \ddot{\vec{x}}[t] - m \nabla \zeta [\vec{x} [t],t] $

and thus

$ m \ddot{\vec{x}}[t] = - m \nabla \zeta [\vec{x} [t],t] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1) $

is the equation of motion — two different expressions for the force.

Ειδική Σχετιστικιστική περίπτωση με ηλεκτρομαγνητισμόEdit

In Ειδική Σχετικότητα (special relativity), the form of the term which gives rise to the derivative of the momentum must be changed; it is no longer the kinetic energy.

It becomes:

$ - m c^2 \frac{d \tau[t]}{d t} = - m c^2 \sqrt {1 - \frac{v^2 [t]}{c^2}} $
$ = -m c^2 + {1 \over 2} m v^2 [t] + {1 \over 8} m \frac{v^4 [t]}{c^2} + \dots $

(In special relativity, the energy of a free test particle is $ m c^2 \frac{dt}{d \tau [t]} = \frac{m c^2}{\sqrt {1 - \frac{v^2 [t]}{c^2}}} = +m c^2 + {1 \over 2} m v^2 [t] + {3 \over 8} m \frac{v^4 [t]}{c^2} + \dots $)

where $ c \! $ meters per second is the speed of light in vacuum, $ \tau \! $ seconds is the proper time (i.e. time measured by a clock moving with the particle) and $ v^2 [t] = \dot{\vec{x}}[t] \cdot \dot{\vec{x}}[t]. $ Notice that the second term in the series is just the classical kinetic energy. Suppose the particle has electrical charge $ q\! $ coulombs and is in an electromagnetic field with scalar potential $ \phi \! $ volts (a volt is a joule per coulomb) and vector potential $ \vec{A} $ volt seconds per meter. The Lagrangian of a special relativistic test particle in an electromagnetic field is:

$ L[t] = - m c^2 \sqrt {1 - \frac{v^2 [t]}{c^2}} - q \phi [\vec{x}[t],t] + q \dot{\vec{x}}[t] \cdot \vec{A} [\vec{x}[t],t] $

Varying this with respect to $ \vec{x} $, we get

$ 0 = - \frac{d}{d t}\left(\frac{m \dot{\vec{x}}[t]} {\sqrt {1 - \frac{v^2 [t]}{c^2}}}\right) - q \nabla\phi [\vec{x}[t],t] - q \partial_t{\vec{A}} [\vec{x}[t],t] - q \dot{\vec{x}}[t] \cdot \nabla\vec{A} [\vec{x}[t],t] + q \nabla{\vec{A}} [\vec{x}[t],t] \cdot \dot{\vec{x}}[t] $

which is:

$ \frac{d}{d t}\left(\frac{m \dot{\vec{x}}[t]} {\sqrt {1 - \frac{v^2 [t]}{c^2}}}\right) = q \vec{E}[\vec{x}[t],t] + q \dot{\vec{x}}[t] \times \vec{B} [\vec{x}[t],t] $

which is the equation for the Lorentz force where

$ \vec{E}[\vec{x},t] = - \nabla\phi [\vec{x},t] - \partial_t{\vec{A}} [\vec{x},t] $
$ \vec{B}[\vec{x},t] = \nabla \times \vec{A} [\vec{x},t] $

Γενική Σχετικιστική περίπτωσηEdit

In Γενική Σχετικότητα (general relativity), the first term generalizes (includes) both the classical kinetic energy and interaction with the Newtonian gravitational potential.

It becomes:

$ - m c^2 \frac{d \tau[t]}{d t} $
$ = - m c \sqrt {- g_{\alpha\beta}[x[t]] \frac{d x^{\alpha}[t]}{d t} \frac{d x^{\beta}[t]}{d t}} . $

The Lagrangian of a general relativistic test particle in an electromagnetic field is:

$ L[t] = - m c \sqrt {- g_{\alpha\beta}[x[t]] \frac{d x^{\alpha}[t]}{d t} \frac{d x^{\beta}[t]}{d t}} + q \frac{d x^{\gamma}[t]}{d t} A_{\gamma}[x[t]] . $

If the four space-time coordinates $ x^{\alpha}\! $ are given in arbitrary units (i.e. unit-less), then $ g_{\alpha\beta}\! $ meters squared is the rank 2 symmetric metric tensor which is also the gravitational potential. Also, $ A_{\gamma}\! $ volt seconds is the electromagnetic 4-vector potential. Notice that a factor of c has been absorbed into the square root because it is the equivalent of

$ c\, \sqrt {1 - \frac{v^2 [t]}{c^2}} = \sqrt {- ( - c^2 + v^2 [t])} . $

Note that this notion has been directly generalized from special relativity

Κλασσική Πεδιακή ΘεωρίαEdit

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 $ \mathcal{L} $, which one integrates over all space-time to get the action:

$ \mathcal{S} [\varphi_i] = \int{\mathcal{L} [\varphi_i (x)]\, \mathrm{d}^4x} $

The Lagrangian is then the spatial integral of the Lagrangian density. However, $ \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 $ \vec x $ is incorporated into the index $ i $ or the parameters $ s $ in $ \varphi_i(s) $. Quantum field theories in particle physics, such as quantum electrodynamics, are usually described in terms of $ \mathcal{L} $, and the terms in this form of the Lagrangian translate quickly to the rules used in evaluating Διαγράμματα Feynman.

Επιλεγμένα ΠεδίαEdit

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 $ [\vec{x},t] $.

For example, in the case of Newtonian gravity, the Lagrangian density integrated over space-time gives you an equation which, if solved, would yield $ \zeta [\vec{x},t] $.

This $ \zeta [\vec{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.

Νευτώνιο Βαρυτικό ΠεδίοEdit

The Lagrangian (density) is $ \mathcal{L} $ joules per cubic meter. The interaction term $ m \zeta \! $ is replaced by a term involving a continuous mass density $ \mu \! $ kilograms per cubic meter. 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}[\vec{x},t] = - \mu [\vec{x},t] \zeta [\vec{x},t] - {1 \over 8 \pi G} (\nabla \zeta [\vec{x},t])^2 $

where $ G \! $ meters cubed per kilogram second squared is the gravitational constant. Variation of the integral with respect to $ \zeta \! $ gives:

$ 0 = - \mu [\vec{x},t] \delta\zeta [\vec{x},t] - {2 \over 8 \pi G} (\nabla \zeta [\vec{x},t]) \cdot (\nabla \delta\zeta [\vec{x},t]) . $

Integrate by parts and discard the total integral. Then divide out by $ \delta\zeta \! $ to get:

$ 0 = - \mu [\vec{x},t] + {1 \over 4 \pi G} \nabla \cdot \nabla \zeta [\vec{x},t] $

and thus

$ 4 \pi G \mu [\vec{x},t] = \nabla^2 \zeta [\vec{x},t] . $

(Ειδικό Σχετικιστικό) Ηλεκτρομαγνητικό ΠεδίοEdit

The interaction terms $ - q \phi [\vec{x}[t],t] + q \dot{\vec{x}}[t] \cdot \vec{A} [\vec{x}[t],t] $ are replaced by terms involving a continuous charge density $ \rho \! $ coulombs per cubic meter and current density $ \vec{j} \! $ amperes per square meter. The resulting Lagrangian for the electromagnetic field is:

$ \mathcal{L}[\vec{x},t] = - \rho [\vec{x},t] \phi [\vec{x},t] + \vec{j} [\vec{x},t] \cdot \vec{A} [\vec{x},t] + {\epsilon_0 \over 2} {E}^2 [\vec{x},t] - {1 \over {2 \mu_0}} {B}^2 [\vec{x},t] . $

Varying this with respect to $ \phi \! $, we get

$ 0 = - \rho [\vec{x},t] + \epsilon_0 \nabla \cdot \vec{E} [\vec{x},t] $

which yields Gauss' law.

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

$ 0 = \vec{j} [\vec{x},t] + \epsilon_0 \partial_t \vec{E} [\vec{x},t] - {1 \over \mu_0} \nabla \times \vec{B} [\vec{x},t] $

which yields Ampère's law.

(Γενικό Σχετικιστικό) Ηλεκτρομαγνητικό ΠεδίοEdit

For the Lagrangian of gravity in general relativity, see Einstein-Hilbert action. The Lagrangian of the electromagnetic field is:

$ \mathcal{L}[x] = + J^{\gamma}[x] A_{\gamma}[x] - {1 \over 4\mu_0} F_{\mu \nu}[x] F_{\alpha \beta}[x] g^{\mu\alpha}[x] g^{\nu\beta}[x] \sqrt{\frac{-1}{c^2} \mathrm{det} [g[x]]} $

If the four space-time coordinates $ x^{\alpha}\! $ are given in arbitrary units, then: $ \mathcal{L} $ joule seconds is the Lagrangian, a scalar density; $ J^{\gamma}\! $ coulombs is the current, a vector density; and $ F_{\mu \nu}\! $ volt seconds is the electromagnetic tensor, a covariant antisymmetric tensor of rank two. Notice that the determinant under the square root sign is applied to the matrix of components of the covariant metric tensor $ g_{\alpha\beta}\! $, and $ g^{\alpha\beta}\! $ is its inverse. Notice that the units of the Lagrangian changed because we are integrating over $ x^0, x^1, x^2, x^3\! $ which are unit-less rather than over $ t, x, y, z \! $ which have units of seconds meters cubed. The electromagnetic field tensor is formed by anti-symmetrizing the partial derivative of the electromagnetic vector potential; so it is not an independent variable. The square root is needed to convert that term into a scalar density instead of just a scalar, and also to compensate for the change in the units of the variables of integration. The factor of $ \frac{-1}{c^2} $ inside the square root is needed to normalize it so that the square root will reduce to one in special relativity (since the determinant is $ - c^2 \! $ in special relativity).

Κβαντική Πεδιακή ΘεωρίαEdit

Dirac LagrangianEdit

The Lagrangian density for a Dirac field is:

$ \mathcal{L} = \bar \psi (i \hbar c \not\!D - mc^2) \psi $

where $ \psi\! $ is a spinor, $ \bar \psi = \psi^\dagger \gamma^0 $ is its Dirac adjoint, $ D\! $ is the gauge covariant derivative, and $ \not\!D $ is Feynman notation for $ \gamma^\sigma D_\sigma\! $.

Κβαντική ΗλεκτροδυναμικήEdit

The Lagrangian density for Κβαντική Ηλεκτροδυναμική (QED) is:

$ \mathcal{L}_{\mathrm{QED}} = \bar \psi (i \hbar c\not\!D - mc^2) \psi - {1 \over 4\mu_0} F_{\mu \nu} F^{\mu \nu} $

where $ F^{\mu \nu}\! $ is the electromagnetic tensor

Κβαντική ΧρωμοδυναμικήEdit

The Lagrangian density for Κβαντική Χρωμοδυναμική (QCD) is [1] [2] [3]:

$ \mathcal{L}_{\mathrm{QCD}} = \sum_n \bar \psi_n (i \hbar c\not\!D - m_n c^2) \psi_n - {1\over 4} G^\alpha {}_{\mu\nu} G_\alpha {}^{\mu\nu} $

where $ D\! $ is the QCD gauge covariant derivative, and $ G^\alpha {}_{\mu\nu}\! $ is the gluon field strength tensor.

Μαθηματικός ΦορμαλισμόςEdit

Suppose we have an n-dimensional manifold, $ M $, and a target manifold, $ T $. Let $ \mathcal{C} $ be the configuration space of smooth functions from $ M $ to $ T $.

Παραδείγματα Edit

Μαθηματική ΑνάπτυξηEdit

Consider a functional, $ \mathcal{S}:\mathcal{C}\rightarrow \mathbb{R} $, called the action. Physical reasons determine that it is a mapping to $ \mathbb{R} $, not $ \mathbb{C} $.

In order for the action to be local, we need additional restrictions on the action. If $ \varphi\in\mathcal{C} $, we assume $ \mathcal{S}[\varphi] $ is the integral over $ M $ of a function of $ \phi $, its derivatives and the position called the Lagrangian, $ \mathcal{L}(\varphi,\partial\varphi,\partial\partial\varphi, ...,x) $. In other words,

$ \forall\varphi\in\mathcal{C}, \ \ \mathcal{S}[\varphi]\equiv\int_M \mathrm{d}^nx \mathcal{L} \big( \varphi(x),\partial\varphi(x),\partial\partial\varphi(x), ...,x \big). $

It is assumed below, in addition, that the Lagrangian depends on only the field value and its first derivative but not the higher derivatives.

Given boundary conditions, basically a specification of the value of $ \phi $ at the boundary if $ M $ is compact or some limit on $ \phi $ as x approaches $ \infty $ (this will help in doing integration by parts), the subspace of $ \mathcal{C} $ consisting of functions, $ \phi $ such that all functional derivatives of $ S $ at $ \phi $ are zero and $ \phi $ satisfies the given boundary conditions is the subspace of on shell solutions.

The solution is given by the Euler-Lagrange equations (thanks to the boundary conditions),

$ \frac{\delta\mathcal{S}}{\delta\varphi}=-\partial_\mu \left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\varphi)}\right)+ \frac{\partial\mathcal{L}}{\partial\varphi}=0. $

The left hand side is the Συναρτησιακή Παράγωγος (functional derivative) of the Φυσική Δράση (action) with respect to $ \phi $.

Electromagnetism without charges and currentsEdit

When there are no electric charges (ρ=0) and no electric currents (j=0), Classical electromagnetism and Maxwell's equations can be derived from the action defined:

$ \mathcal{S} = \int \left( -\begin{matrix} \frac{1}{4 \mu_0} \end{matrix} F_{\mu\nu} F^{\mu\nu} \right) \mathrm{d}^4 x \, $


$ \mathrm{d}^4 x \; $   is over space and time.

This means the Lagrangian density is

$ \mathcal{L} \, $ $ = -\begin{matrix} \frac{1}{4\mu_0} \end{matrix} F_{\mu\nu} F^{\mu\nu} \, $
$ = -\begin{matrix} \frac{1}{4\mu_0} \end{matrix} \left( \partial_\mu A_\nu - \partial_\nu A_\mu \right) \left( \partial^\mu A^\nu - \partial^\nu A^\mu \right). \, $
$ = -\begin{matrix} \frac{1}{4\mu_0} \end{matrix} \left( \partial_\mu A_\nu \partial^\mu A^\nu - \partial_\nu A_\mu \partial^\mu A^\nu - \partial_\mu A_\nu \partial^\nu A^\mu + \partial_\nu A_\mu \partial^\nu A^\mu \right). $

The far left and far right terms are the same because $ \mu $ and $ \nu $ are just dummy indices after all. The two middle terms are also the same, so the Lagrangian density is

$ \mathcal{L} \, $ $ = -\begin{matrix} \frac{1}{2\mu_0} \end{matrix} \left( \partial_\mu A_\nu \partial^\mu A^\nu - \partial_\nu A_\mu \partial^\mu A^\nu \right). $

We can then plug this into the Euler-Lagrange equation of motion for a field:

$ \partial_\nu \left( \frac{\partial \mathcal{L}}{\partial ( \partial_\nu A_\mu )} \right) - \frac{\partial \mathcal{L}}{\partial A_\mu} = 0 . \, $

The second term is zero because the Lagrangian in this case only contains derivatives. So the Euler-Lagrange equation becomes:

$ \partial_\nu \left( \partial^\mu A^\nu - \partial^\nu A^\mu \right) = 0. \, $

The quantity in parentheses above is just the field tensor, so this finally simplifies to

$ \partial_\nu F^{\mu \nu} = 0. \, $

That equation is just another way of writing the two inhomogeneous Maxwell's equations as long as you make the substitutions:

$ ~E^i /c \ \ = -F^{0 i} \, $
$ \epsilon^{ijk} B^k = -F^{ij} \, $

where $ i \, $ and $ j \, $ take on the values of 1, 2, and 3.

When there are charges or currents, the Lagrangian needs an extra term to account for the coupling between them and the electromagnetic field. In that case $ \partial_\nu F^{\mu \nu} $ is equal to the 4-current instead of zero.

Role in quantum electrodynamics and field theoryEdit

The Lagrangian of quantum electrodynamics extends beyond the classical Lagrangian established in relativity, from $ \mathcal{L}=\bar\psi(i\hbar c \, \gamma^\alpha D_\alpha - mc^2)\psi -\frac{1}{4 \mu_0}F_{\alpha\beta}F^{\alpha\beta}, $  to incorporate the creation and annihilation of photons (and electrons).

In quantum field theory it is used as the template for the gauge field strength tensor. By being employed in addition to the local interaction Lagrangian it reprises its usual role in QED.

Electromagnetism in MatterEdit

Separating the free currents from the bound currents, another way to write the Lagrangian density is as follows:

$ \mathcal{L} \, = \, - \frac{1}{4 \mu_0} F^{\alpha \beta} F_{\alpha \beta} - A_{\alpha} J^{\alpha}_{\text{free}} + \frac12 F_{\alpha \beta} \mathcal{M}^{\alpha \beta} \,. $

Using Euler–Lagrange equation, the equations of motion for $ \mathcal{D}^{\mu\nu} $ can be derived.

The equivalent expression in non-relativistic vector notation is

$ \mathcal{L} \, = \, \frac12 \left(\epsilon_{0} E^2 - \frac{1}{\mu_{0}} B^2\right) - \phi \, \rho_{\text{free}} + \bold{A} \cdot \bold{J}_{\text{free}} + \bold{E} \cdot \bold{P} + \bold{B} \cdot \bold{M} \,. $


Εσωτερική ΑρθρογραφίαEdit



Ikl Κίνδυνοι ΧρήσηςIkl

Αν και θα βρείτε εξακριβωμένες πληροφορίες
σε αυτήν την εγκυκλοπαίδεια
ωστόσο, παρακαλούμε να λάβετε σοβαρά υπ' όψη ότι
η "Sciencepedia" δεν μπορεί να εγγυηθεί, από καμιά άποψη,
την εγκυρότητα των πληροφοριών που περιλαμβάνει.

"Οι πληροφορίες αυτές μπορεί πρόσφατα
να έχουν αλλοιωθεί, βανδαλισθεί ή μεταβληθεί από κάποιο άτομο,
η άποψη του οποίου δεν συνάδει με το "επίπεδο γνώσης"
του ιδιαίτερου γνωστικού τομέα που σας ενδιαφέρει."

Πρέπει να λάβετε υπ' όψη ότι
όλα τα άρθρα μπορεί να είναι ακριβή, γενικώς,
και για μακρά χρονική περίοδο,
αλλά να υποστούν κάποιο βανδαλισμό ή ακατάλληλη επεξεργασία,
ελάχιστο χρονικό διάστημα, πριν τα δείτε.

Οι διάφοροι "Εξωτερικοί Σύνδεσμοι (Links)"
(όχι μόνον, της Sciencepedia
αλλά και κάθε διαδικτυακού ιστότοπου (ή αλλιώς site)),
αν και άκρως απαραίτητοι,
είναι αδύνατον να ελεγχθούν
(λόγω της ρευστής φύσης του Web),
και επομένως είναι ενδεχόμενο να οδηγήσουν
σε παραπλανητικό, κακόβουλο ή άσεμνο περιεχόμενο.
Ο αναγνώστης πρέπει να είναι
εξαιρετικά προσεκτικός όταν τους χρησιμοποιεί.

- Μην κάνετε χρήση του περιεχομένου της παρούσας εγκυκλοπαίδειας
αν διαφωνείτε με όσα αναγράφονται σε αυτήν


>>Διαμαρτυρία προς την wikia<<

- Όχι, στις διαφημίσεις που περιέχουν απαράδεκτο περιεχόμενο (άσεμνες εικόνες, ροζ αγγελίες κλπ.)