Introduction to Mathematical Physics/Relativity/Dynamics

From Wikibooks, open books for an open world
< Introduction to Mathematical Physics‎ | Relativity
Jump to: navigation, search

Fundamental principle of classical mechanics[edit]

Let us state the fundamental principle of classical dynamics for a material point, or particle[1]. A material point is classically described by its mass m, its position r, and its velocity v. It undergoes external actions modelled by forces F_{ext}. The momentum mv of the particle is denoted by p.\index{momentum}

Principle: The fundamental principle of dynamics (or Newton's equation of motion) \index{Newton's equation of motion} states that the time derivative of momentum is equal to the sum of all external forces\index{force}:


\frac{dp}{dt}=\sum F_{ext}

secprinmoindreact

Least action principle[edit]

Principle: Least action principle: \index{least action principle} The function x(t) describing the trajectory of a particle with potential energy U(x) yields a constant action, where the action S is defined by S=\int L dt, where L is the Lagrangian of the particle: L(x,\dot x,t)=-\frac{1}{2}m\dot x ^2+U(x).

This principle can be taken as the basis of material point classical mechanics. But it can also be seen as a consequence of the fundamental dynamics principle presented previously [ma:equad:Arnold83].


m \ddot{x}+\frac{\partial U}{\partial x}=0

Let us multiply by y(t) and integrate over time:


\int (m \ddot{x}y +\frac{\partial U}{\partial x}y)\,dt=0

Using Green's theorem (integration by parts):


\int (-m \dot{x}\dot{y} +\frac{\partial U}{\partial x}y)\,dt=0

\dot{x}\dot{y} is a bilinear form.


-\dot{x}\dot{y}=-\frac{\partial }{\partial x}\frac{1}{2}\dot{x}^2


0= \int m
[-\frac{1}{2}(\dot{x}+\dot{y})^{2}+U(x+y)]
-[-\frac{1}{2}\dot{x}^{2}+U(x)]dt+O(y^{2})

Defining the Lagrangian L by:


L(\dot{x},x,t)=-\frac{1}{2}m\dot x ^2+U(x),

the previous equation can be written


0=-\delta \int L(\dot{x},x,t)dt

meaning that the action S is constant.

Description by energies[edit]

Laws of motion does not tell anything about how to model forces. The force modelization is often physicist's job. Here are two examples of forces expressions:


  1. weight P=mg. g is a vector describing gravitational field around the material point of mass m considered.
  2. electromagnetic force f=qv\wedge B+qE, where q is particle's charge, E is the electric field, B the magnetic field and v the particle's velocity.

This two last forces expressions directly come from physical postulates. However, for other interactions like elastic forces, friction, freedom given to physicist is much greater.

An efficient method to modelize such complex interactions is to use the energy (or power) concept. At chapter chapelectromag, the duality between forces and energy is presented in the case of electromagnetic interaction. At chapters chapapproxconti and chapenermilcon, the concept of energy is developed for the description of continuous media.

Let us recall here some definitions associated to the description of interactions by forces. Elementary work of a force f for an elementary displacement is:

\delta W=f.dr

Instantaneous power emitted by a force f to a material point of velocity v is:

P=f.v

Potential energy gained by the particle during time dt that it needs to move of dr is:

dE=-f.dr=-P.dt

Note that potential energy can be defined only if force field f have conservative circulation[2]. This is the case for weight, for electric force but not for friction. A system that undergoes only conservative forces is hamiltonian. The equations that govern its dynamics are the Hamilton equations: {IMP/label|eqhampa1}}

\frac{dp}{dt}=-\frac{\partial H}{\partial q}

eqhampa2

\frac{dq}{dt}=\frac{\partial H}{\partial p}

where function H(q,p,t) is called hamiltonian of the system. For a particle with a potential energy E_p, the hamiltonian is:

eqformhami

H(q,p,t)=\frac{p^2}{2m}+E_p(q)

where p is particle's momentum and q its position. By extension, every system whose dynamics can be described by equations eqhampa1 and eqhampa2 is called hamiltonian \index{hamiltonian system} even if H is not of the form given by equation eqformhami.

secdynasperel

Dynamics in special relativity[edit]

It has been seen that Lorentz transformations acts on time. Classical dynamics laws have to be modified to take into into account this fact and maintain their invariance under Lorentz transformations as required by relativity postulates. Price to pay is a modification of momentum and energy notions. Let us impose a linear dependence between the impulsion four-vector and velocity four-vector;


P=mU=(m\gamma u,icm\gamma)

where m is the rest mass of the particle, u is the classical speed of the particle u=\frac{dx}{dt}, \gamma=\frac{1}{\sqrt{1-\beta^2}}, with \beta=\frac{u}{c}. Let us call ``relativistic momentum quantity:


p=m \gamma u

and "relativistic energy" quantity:


E=m\gamma c^2

Four-vector P can thus be written:


P=(p,i\frac{E}{c})

Thus, Einstein associates an energy to a mass since at rest:


E=mc^2

This is the matter--energy equivalence . \index{matter--energy equivalence} Fundamental dynamics principle is thus written in the special relativity formalism:


\frac{dP}{dt}=f_\mu

where f_\mu is force four-vector.

Example: Let us give an example of force four-vector. Lorentz force four-vector is defined by:


f^\mu=F^\mu_\nu P^\nu

where F^\mu_\nu is the electromagnetic tensor field (see section seceqmaxcov)

Least action principle in special relativity[edit]

Let us present how the fundamental dynamics can be retrieved from a least action principle. Only the case of a free particle is considered. Action should be written:


S=\int L(u,x,t) dt

where L is invariant by Lorentz transformations, u and x are the classical speed and position of the particle. The simplest solution consists in stating:


L=\gamma k

where k is a constant. Indeed Ldt becomes here:


Ldt=k d\tau

where d\tau=\frac{dt}{\gamma}, (with \gamma=\frac{1}{\sqrt{1-\beta^2}}, and \beta=\frac{u}{c}) is the differential of the eigen time and is thus invariant under any Lorentz transformation. Let us postulate that k=-mc^2. Momentum is then:


p=\frac{\partial L}{\partial u}=\gamma m u

Energy is obtained by a Legendre transformation, E=p.u-L so:


E=\gamma m c^2


Dynamics in general relativity[edit]

The most natural way to introduce the dynamics of a free particle in general relativity is to use the least action principle. Let us define the action S by:


S=\int ds

where ds is the elementary distance in the Rieman space--time space. S is obviously covariant under any frame transformation (because ds does). Consider the fundamental relation eqcovdiff that defines (covariant) differentials. It yields to:


Da^i=da^i+a^k\Gamma^i_{kj}dx^j

where da^i=\frac{\partial a^i}{\partial x^j}dx^j. A deplacement can be represented by dx^i=u^id\lambda where u^i represents the tangent to the trajectory (the velocity). The shortest path corresponds to a movement of the particle such the the tangent is transported parallel to itself, that is

[3]:

Du^i=0.

or

\frac{Du^i}{d\lambda}=0

This yields to

eqdynarelatge

\frac{D u^i}{D\lambda}=\frac{d^2x^i}{d\lambda^2}+\Gamma^{i}_{hk}
\frac{dx^h}{d\lambda} \frac{dx^k}{d\lambda}

where \Gamma^{i}_{hk} is tensor depending on system's metrics (see [ph:relat:Misner73g] for more details). Equation eqdynarelatge is an evolution equation for a free particle. It is the equation of a geodesic. Figure figgeo represents the geodesic between two point A and B on a curved space constituted by a sphere. In this case the geodesic is the arc binding A and B.

figgeo

Geodesic on a sphere.

Note that here, gravitational interaction is contained in the metrics. So the equation for the "free" particle above describes the evolution of a particle undergoing the gravitational interaction. General relativity explain have larges masses can deviate light rays (see figure figlightrayd)

figlightrayd

Curvature of a light.
  1. A formulation adapted to continuous matter will be presented later in the book.
  2. That means:

    \int_C f.dr=0

    for every loop C, or equivalently that

    \mbox{ rot } f=0.

  3. The actual proof that action S is minimal implies that DU^i=0 is given in [ph:relat:Misner73g], [ma:tense:Brillouin64]