FYST420 Advanced electrodynamics Final project Olli Aleksanteri Koskivaara [email protected] he oisson bracket and magnetic T P monopoles Abstract: In this work magnetic monopoles are studied using the Poisson bracket. Using the Hamiltonian formulation of classical mechanics a system can be described by it’s Hamiltonian. The Hamiltonian of a particle in an electromagnetic field is introduced. The Poisson bracket and some of it’s properties are also outlined. Assuming that a magnetic monopole exists, the behaviour of a particle under it’s influence is studied using the machinery presented earlier. It is found that the trajectory of the particle lies on a cone. 1 Introduction Magnetic monopoles are a widely studied subject in physics, even though they remain yet to be found in Nature. Every magnet we have observed so far has both a north and a south pole. The interest in magnetic monopoles arises from the fact that many modern theories in particle physics predict their existence. In this work I study how a charged particle would behave under the influence of a magnetic monopole, main goal being the trajectory of the particle. I start with some basic Hamiltonian mechanics, which I assume the reader is familiar with. I then move on to introduce the Poisson brackets, which will be my main tool in deriving the particle’s trajectory. 2 Hamiltonian mechanics 2.1 Basics In Lagrangian mechanics a system with n degrees of freedom is described by the generalized position and velocity coordinates qi and q˙i, where i = 1, ..., n. The Lagrangian L = L(qi, q˙i, t) is defined as L = T − V, (1) where T is the kinetic energy of the system and V is the potential energy of the system. Once the Lagrangian of the system is known, the equations of motion are directly obtained from the Euler–Lagrange equation ¶L d ¶L − = 0, (2) ¶qi dt ¶q˙i which follows from minimazing the action integral of the system. The transition from Lagrangian mechanics to Hamiltonian mechanics is made by performing a Legendre transformation. This gives rise to the Hamiltonian of the system n H(qi, pi, t) = ∑ q˙j pj − L(qi, q˙i, t), (3) j=1 where pi is the conjugate momentum defined by ¶L pi = . (4) ¶q˙i Now the equations of motion turn out to be the so-called Hamilton’s equations ¶H ¶H q˙i = & p˙i = − . (5) ¶pi ¶qi 2 2.2 A charged particle in an electromagnetic field Let’s take a look at a particle with charge q moving in an electromagnetic field. The electric and magnetic fields E and B can be written in terms of a vector potential A and a scalar potential f as 1 ¶A E = −rf − , B = r × A. (6) c ¶t It turns out, that in this case the Lagrangian solving the Euler–Lagrange equation (2) is of the form [1] 1 q L = m˙r2 − qf + ˙r · A, (7) 2 c where r is the position vector of the particle and c is the speed of light. It is quite straightforward to convince oneself that this is indeed the right Lagrangian by substituting it into equation (2) which eventually yields the familiar Lorentz force law. We can now calculate the conjugate momentum pi using equations (4) and (7): ¶L q pi = = mr˙i + Ai. (8) ¶r˙i c We can immediately see that now the conjugate momentum is not just mass times velocity; instead, there is an extra term given by the vector potential A. Solving equation (8) for r˙i and substituting into equations (3) and (7) gives us the following Hamiltonian: 1 3 q 1 q 2 q q H = ∑ pi − Ai pi − p − A + qf − p − A · A m i=1 c 2m c mc c 1 q 1 q 2 q q2 = p2 − p · A − p − A − p · A + A2 + qf m mc 2m c mc mc2 1 q 2 1 q 2 = p − A − p − A + qf m c 2m c 1 q 2 = p − A + qf. (9) 2m c Substituting this into Hamilton’s equations one can further derive the equations of motion for the particle, again corresponding to the Lorentz force law. I won’t go through the calculations here, because I will only be needing the form of the Hamiltonian. Instead, I move on to introduce my main tool for studying magnetic monopoles. 3 2.3 The Poisson bracket Let f (qi, pi) and g(qi, pi) be two functions of the coordinates and the conjugate momenta on our phase space. The Poisson bracket is defined as n ¶ f ¶g ¶ f ¶g f f , gg = ∑ − . (10) i=1 ¶qi ¶pi ¶pi ¶qi Some properties following straight from the definition are i) f f , f g = 0 ii) f f , gg = −fg, f g iii) fa f + bg, hg = af f , hg + bfg, hg, a, b 2 R iv) f f g, hg = f fg, hg + gf f , hg v) f f , fg, hgg + fh, f f , ggg + fg, fh, f gg = 0. The three last ones are the familiar linearity, Leibniz rule and Jacobi identity. The Poisson brackets seem to behave a lot like matrix commutators. People familiar with differential geometry may also notice the analogy with the Lie bracket or Lie derivative of two vector fields. Maybe the most important application of the Poisson bracket is related to the so-called canonical transformations. A transformation in a phase space is said to be canonical if it leaves the Hamilton’s equations invariant. It turns out, that the Poisson bracket can be used to check whether a coordinate transformation is canonical or not. [2] The Poisson bracket has also an important historical role in physics. When developing quantum mechanics, people related Poisson brackets to commuta- tors and classical quantities to operators in order to quantize the classical sys- tem. Indeed, as noted above, the Poisson bracket has a clear resemblance to commutators. [3] The Poisson bracket has one property that will be especially useful when studying magnetic monopoles. Let f = f (qi, pi, t) be a function such that qi and pi are solutions to the Hamilton’s equations. Then using chain rule we have d f n ¶ f dq n ¶ f dp ¶ f n ¶ f ¶H n ¶ f ¶H ¶ f = ∑ i + ∑ i + = ∑ − ∑ + dt i=1 ¶qi dt i=1 ¶pi dt ¶t i=1 ¶qi ¶pi i=1 ¶pi ¶qi ¶t ¶ f = f f , Hg + , ¶t where on the first line we used the Hamilton’s equations (3). If we further d f assume that dt = 0, i.e., f is a constant of motion, then we obtain the so-called Liouville equation ¶ f f f , Hg + = 0. (11) ¶t Often it is also possible to assume that f = f (qi, pi) does not depend explicitly on time. Then f is a constant of motion if f f , Hg = 0. 4 3 Magnetic monopoles So far we haven’t really done any calculations with magnetic monopoles. Let’s now assume that such a point-like magnetic charge would exist. It would be described by a magnetic field singular at the origin r B = x , (12) jrj3 where x is a constant determining the strength of the field [4]. We see immedi- ately that the field is problematic. One of Maxwell’s equations tells us that the divergence of the magnetic field should vanish, r · B = 0, but in this case the condition cannot be satisfied because of the infinities in the origin. Furthermore, we do not have any vector potential A to play with. However, we can put these problems aside and start to work with the Poisson bracket. When calculating the Hamiltonian for a particle in an electromagnetic field (3) we saw that 1 q r˙ = p − A . i m i c i Using this we can calculate the following Poisson bracket: n q q o fmr˙ , mr˙ g = p − A , p − A i j i c i j c j q q q2 = fp , p g − fp , A g − fA , p g + fA , A g. i j c i j c i j c2 i j Straight from the definition of the Poisson bracket we see that the first term vanishes. The last term vanishes too, since the vector potential A depends only on the position vector r. For the two remaining terms we can use the fact that for any function f (r) depending only on the position vector r we have n ¶pi ¶ f (r) ¶pi ¶ f (r) ¶ f (r) fpi, f (r)g = ∑ − = − a=1 ¶ra ¶pa ¶pa ¶ra ¶ri | {z } |{z} = 0 = dia Using this result our Poisson bracket becomes q ¶Aj ¶Ai q fmr˙i, mr˙jg = − = eijkBk, (13) c ¶ri ¶rj c where in the last step we used the Maxwell’s equation B = r × A and the expression of the cross product in terms of the Levi-Civita symbol eijk. We see that using the Poisson bracket we got rid of the vector potential A, even though it appeared in our derivation. In a similiar fashion we can calculate 5 another Poisson bracket: n q o q fr , mr˙ g = r , p − A = fr , p g − fr , A g i j i j c j i j c i j 3 q ¶ri ¶Aj ¶ri ¶Aj = −fpj, rig − ∑ − c a=1 ¶ra ¶pa ¶pa ¶ra |{z} |{z} = 0 = 0 ¶ri = = dij. (14) ¶rj Now we can use these Poisson brackets (13) and (14) to calculate the Poisson bracket of the angular momentum L = r × m˙r with r and m˙r.