Classical Field Theory

Transcription

1 1 Classical Field Theory Oscar Loaiza-Brito 1 Physics Department División de Ciencias e Ingeniería, Campus León, Universidad de Guanajuato January-June 2014 January-June Series of lectures on classical field theory for undergraduate. 1 address: oloaiza@fisica.ugto.mx

2 Chapter 1 Lagrangian formalism 1.1 Lagrangian for a particle The mechanical state of a particle is specified by a function L depending on some parameters associated to a particle on a given state of motion. The number of parameters are the minimal required to fully describe the trajectory of a particle after given some set of initial conditions. From our courses on Newtonian mechanics, we know that the basic law describing the particle motion is given by second Newton s law: F = m a. (1.1.1) Our goal here is to reproduce such equation from a more basic principle related to that hypothetical function we have called the mechanical state of a particle and represented by the function L. Naively, we know that in order to know where a particle will be in space after some time, it is only necessary to know its position and velocity at a given time, and the force acting on it. Acceleration, for instance, seems to be not necssary since it can be deduced from second Newton s law. This leads us to think that the mechanical state of a particle should be characterized by a function L = L(x, v, F ). However, we know also from newtonian

3 2 1. Lagrangian formalism mechanics, that F = V, (1.1.2) where V = V (x) is the potential energy of the particle. Hence, it seems that L = L(x, v), where x and v are functions on time, and therefore, L depends implicity on time t as well. Let us now establish our main assumption: L is a function such that any (small) variation on the parameters x and v minimize. This means that L(x, v) measures a property of our particle which is minimum through the whole movement of the particle. This is so important we should try to understand this concept in many ways as possible and here is another: consider a particle which moves. At some time t 0, the particle is located with respect to a given refrence system at x 0 and has a velocity given by v 0. If all these parameters do not change their values at any other time, L is a constant and it describes the mechanical state of a particle in rest. If the parameters (x, v) change for different times, the above principle says that L must satisfy that the function S = tf t 0 dt L(x, v; t), (1.1.3) is a minimum with respect to variations on x and v. Assuming that variation on x and v (respectively denoted by δx and δv are transversal to time (i.e, for any given time, we can consider a different trajectory of a particle in the state space), we have that a variation of the above quantity leads us to δs = tf t 0 dt δl = 0, (1.1.4) with δl(x, v; t) = L(x + δx, v + δv; t) L(x, v; t) and leading to L(x + δx, v + δv; t) = L(x, v; t) + L L δx + x v δv + O(δ2, v 2 ), (1.1.5) δl = L L δx + δv = 0, (1.1.6) x v which reduces to ( L x d dt L v ) δx + d ( ) L dt v δx = 0. (1.1.7)

4 1.1. Lagrangian for a particle 3 Since we are assuming that at initial and final time the mechanical state of a particle is known (measured), there are no variation on x or v and δx = 0. Therefore, the function L satisfies the equation of motion given by d L dt ẋ = L x, d L L dt ẋ x = 0. (1.1.8) Comparing this last equation with Newton s second law, we conclude that L(x, v; t) = 1 2 mv2 V (x). (1.1.9) What s the meaning of this? some property related to our particle. We expect L being a constant measurement of A first guess should be that L is the particle s energy, but rather is a difference between kinetic energy and potential energy. Therefore, it seems that the Lagrangian L is a function which measures this difference and gives us the minimal value. In other words, a particle takes a trajectory (solution of the equations of motion) which minimizes the difference between K and V. Hence, it is not only that the total energy is conserved, as we learnt from previous courses, but also that the difference between potential and kinetic energy. Why do we keep just first order corrections? How are the equations of motion modified by considering second order corrections? Compute them Exercices Write explicitly the corresponding Lagrangian, and compute the equations of motion. 1. Free particle (V (x) = 0). 2. A particle in 2d planar motion, V = V ( r). Use polar coordinates. 3. A particle in a constant gravitational force: V = mgx, with m the mass of the particle and g the gravitational constant. 4. Express g in terms of the Newton s gravitational constant G.

5 4 1. Lagrangian formalism 5. A particle in a general gravitational force. 6. An electrical charged particle immersed in an external magnetic field. 7. A particle in a 3d general motion. Write down the corresponding lagrangian and equations of motion using spherical and cylindrical coordinates. 8. Assume L = L(q, q, q; t). Compute the equations of motion Conservation of energy Let us make an assumption. It is quite trivial for us, but it is probably just o reflexion of our ignorance about time. Nevertheless, such an assumption leads us to a very important result in physics: conservation of energy. Since the above function L is used to describe the equations of motion of a particle (or a bunch of n of them for which the generalization is given through the use of the principle of superposition), we do not expect that L changes in time if no interaction is taken into account. In this way: dl dt and since dl dt = 0, (1.1.10) = = n ( L q n i=1 i=1 ) L q + q q ( q L q L ). (1.1.11) Hence, by assuming the last quantity is conserved in the absence of external interactions, we define energy as E = ( n i=1 q ), L L which for the single particle q turns out to be T + V. Compute energy for all the last examples. However, how sure are we that physics laws must keep invariant under time? Classical fields We have seen that a particle under the action of an external force traces out a unique trajectory on the phase space corresponding to the minimum value of S.

6 1.1. Lagrangian for a particle 5 This constraints the possible values of the associated mechanical state of it, or in other words, fixes the particle s dynamics by establishing the corresponding equations of motion, i.e., δs = 0 e.o.m. However, usually we think on forces as the result of variations on potential field values, filling the whole space around some localized source. This field, whose existence is necessary to keep the local interaction between particles and the forces acting on them, is not static. The values are also usually dependent on time establishing a field dynamics. This is the case for the well known electromagnetic field, which satisfy Maxwell s equations. So the question now is if the principle of minimal action can also be used to derive the dynamics of an extended field as the electromagnetic field. Therefore