The Poisson bracket and magnetic monopoles

Transcription

1 FYST420 Advanced electodynamics Olli Aleksante Koskivaaa Final poject The Poisson backet and magnetic monopoles Abstact: In this wok magnetic monopoles ae studied using the Poisson backet. Using the Hamiltonian fomulation of classical mechanics a system can be descbed by it s Hamiltonian. The Hamiltonian of a paticle in an electomagnetic field is intoduced. The Poisson backet and some of it s popeties ae also outlined. Assuming that a magnetic monopole exists, the behaviou of a paticle unde it s influence is studied using the machiney pesented ealie. It is found that the tajectoy of the paticle lies on a cone.

2 1 Intoduction Magnetic monopoles ae a widely studied subject in physics, even though they emain yet to be found in Natue. Evey magnet we have obseved so fa has both a noth and a south pole. The inteest in magnetic monopoles ases fom the fact that many moden theoes in paticle physics pedict thei existence. In this wok I study how a chaged paticle would behave unde the influence of a magnetic monopole, main goal being the tajectoy of the paticle. I stat with some basic Hamiltonian mechanics, which I assume the eade is familia with. I then move on to intoduce the Poisson backets, which will be my main tool in deving the paticle s tajectoy. 2 Hamiltonian mechanics 2.1 Basics In Lagangian mechanics a system with n degees of feedom is descbed by the genealized position and velocity coodinates q i and q i, whee i = 1,..., n. The Lagangian L = Lq i, q i, t is defined as L = T V, 1 whee T is the kinetic enegy of the system and V is the potential enegy of the system. Once the Lagangian of the system is known, the equations of motion ae diectly obtained fom the Eule Lagange equation L q i d dt L q i, 2 which follows fom minimazing the action integal of the system. The tansition fom Lagangian mechanics to Hamiltonian mechanics is made by pefoming a Legende tansfomation. This gives se to the Hamiltonian of the system Hq i, p i, t = n j=1 q j p j Lq i, q i, t, 3 whee p i is the conjugate momentum defined by p i = L. 4 q i Now the equations of motion tun out to be the so-called Hamilton s equations q i = H & ṗ i = H. 5 p i q i 2

3 2.2 A chaged paticle in an electomagnetic field Let s take a look at a paticle with chage q moving in an electomagnetic field. The electc and magnetic fields E and B can be wtten in tems of a vecto potential A and a scala potential φ as E = φ 1 c A, B = A. 6 t It tuns out, that in this case the Lagangian solving the Eule Lagange equation 2 is of the fom [1] L = 1 2 mṙ2 qφ + q ṙ A, 7 c whee is the position vecto of the paticle and c is the speed of light. It is quite staightfowad to convince oneself that this is indeed the ght Lagangian by substituting it into equation 2 which eventually yields the familia Loentz foce law. We can now calculate the conjugate momentum p i using equations 4 and 7: p i = L ṙ i = mṙ i + q c A i. 8 We can immediately see that now the conjugate momentum is not just mass times velocity; instead, thee is an exta tem given by the vecto potential A. Solving equation 8 fo i and substituting into equations 3 and 7 gives us the following Hamiltonian: H = 1 3 m p i q c A i p i 1 2m i=1 = 1 m p2 q mc p A 1 2m = 1 p q 2 m c A 1 2m = 1 2m p q 2 c A q + qφ p q mc c A A p q c A 2 q mc p A + q2 mc 2 A2 + qφ p q c A 2 + qφ p q c A 2 + qφ. 9 Substituting this into Hamilton s equations one can futhe deve the equations of motion fo the paticle, again coesponding to the Loentz foce law. I won t go though the calculations hee, because I will only be needing the fom of the Hamiltonian. Instead, I move on to intoduce my main tool fo studying magnetic monopoles. 3

4 2.3 The Poisson backet Let f q i, p i and gq i, p i be two functions of the coodinates and the conjugate momenta on ou phase space. The Poisson backet is defined as f g f, g = f g. 10 q i p i p i q i n i=1 Some popeties following staight fom the definition ae i f, f ii f, g = g, f iii α f + βg, h = α f, h + βg, h, α, β R iv f g, h = f g, h + g f, h v f, g, h + h, f, g + g, h, f. The thee last ones ae the familia lineaty, Leibniz ule and Jacobi identity. The Poisson backets seem to behave a lot like matx commutatos. People familia with diffeential geomety may also notice the analogy with the Lie backet o Lie devative of two vecto fields. Maybe the most impotant application of the Poisson backet is elated to the so-called canonical tansfomations. A tansfomation in a phase space is said to be canonical if it leaves the Hamilton s equations invaant. It tuns out, that the Poisson backet can be used to check whethe a coodinate tansfomation is canonical o not. [2] The Poisson backet has also an impotant histocal ole in physics. When developing quantum mechanics, people elated Poisson backets to commutatos and classical quantities to opeatos in ode to quantize the classical system. Indeed, as noted above, the Poisson backet has a clea esemblance to commutatos. [3] The Poisson backet has one popety that will be especially useful when studying magnetic monopoles. Let f = f q i, p i, t be a function such that q i and p i ae solutions to the Hamilton s equations. Then using chain ule we have n n d f dt = f dq i q i=1 i dt + i=1 f dp i p i dt + f n t = f H q i=1 i p i n i=1 f H + f p i q i t = f, H + f t, whee on the fist line we used the Hamilton s equations 3. If we futhe assume that d f dt, i.e., f is a constant of motion, then we obtain the so-called Liouville equation f, H + f. 11 t Often it is also possible to assume that f = f q i, p i does not depend explicitly on time. Then f is a constant of motion if f, H. 4

5 3 Magnetic monopoles So fa we haven t eally done any calculations with magnetic monopoles. Let s now assume that such a point-like magnetic chage would exist. It would be descbed by a magnetic field singula at the ogin B = ξ 3, 12 whee ξ is a constant detemining the stength of the field [4]. We see immediately that the field is poblematic. One of Maxwell s equations tells us that the divegence of the magnetic field should vanish, B, but in this case the condition cannot be satisfied because of the infinities in the ogin. Futhemoe, we do not have any vecto potential A to play with. Howeve, we can put these poblems aside and stat to wok with the Poisson backet. When calculating the Hamiltonian fo a paticle in an electomagnetic field 3 we saw that ṙ i = 1 m p i q c A i. Using this we can calculate the following Poisson backet: mṙ i, mṙ j = p i q c A i, p j q c A j = p i, p j q c p i, A j q c A i, p j + q2 c 2 A i, A j. Staight fom the definition of the Poisson backet we see that the fist tem vanishes. The last tem vanishes too, since the vecto potential A depends only on the position vecto. Fo the two emaining tems we can use the fact that fo any function f depending only on the position vecto we have p i, f = n pi a=1 a f p a p i p a = δ ia Using this esult ou Poisson backet becomes mṙ i, mṙ j = q c Aj i f = f a i A i = q j c ɛ ijkb k, 13 whee in the last step we used the Maxwell s equation B = A and the expession of the coss poduct in tems of the Levi-Civita symbol ɛ ijk. We see that using the Poisson backet we got d of the vecto potential A, even though it appeaed in ou devation. In a similia fashion we can calculate 5

6 anothe Poisson backet: i, mṙ j = i, p j q c A j = i, p j q c i, A j = p j, i q 3 A j c a=1 a p a i p a A j a = i j = δ ij. 14 Now we can use these Poisson backets 13 and 14 to calculate the Poisson backet of the angula momentum L = mṙ with and mṙ. Afte some index milling one should end up with the following Poisson backet stuctue: L i, j = ɛ ijk k 15 q L i, mṙ j = ɛ ijk mṙ k + δ ij c B q c B i j. 16 It is easy to see that these indeed wok by explicitely calculating the backets fo a few components of the angula momentum L. If we now substitute the field geneated by the magnetic monopole 12 to the ght hand side of equation 16 we get qξ L i, mṙ j = ɛ ijk mṙ k + δ ij c = ɛ ijk mṙ k + qξ c = ɛ ijk mṙ k + qξ c 3 δ ij i, mṙ j j qξ c i 3 j, 17 whee one can confim the last step by opening the Poisson backet and using the esults and popeties deved ealie. Fom the pevious chapte we emembe that if the Poisson backet of a given function with the Hamiltonian vanishes, then the function is a constant of motion. Using the Poisson backet stuctue obtained above one can see that the Poisson backet of the angula momentum and the Hamiltonian does not necessaly vanish in the case of the magnetic monopole. Howeve, equation 17 suggests that we should intoduce a new quantity of the fom J = L qξ c. 18 This new quantity J is called the geneal o genealized angula momentum [4]. Consisting of the odinay angula momentum and an additional pat it looks a lot like the total angula momentum familia fom quantum mechanics. One 6

7 also sees that if the stength paamete of the magnetic monopole ξ goes to zeo, then J educes back to the odinay angula momentum L. Fo J we get the following Poisson backet stuctue: J i, j = J i, mṙ j = L i qξ c L i qξ c = ɛ ijk mṙ k + qξ c i, j = L i, j qξ c i, mṙ j = L i, mṙ j qξ c, mṙ j qξ c, j = ɛ ijk m k 19, mṙ j, mṙ j = ɛ ijk mṙ k. 20 This looks nice and simple as expected afte the way we defined J. Now we can see what we get if we take the Poisson backet of the geneal angula momentum J with ou Hamiltonian H = 1 2 mṙ2. Fo the fist component J 1 we have H, J 1 = 1 2 mṙ2, J 1 = 1 2 mṙ2 1 + mṙ2 2 + mṙ 2 3, J 1 = 1 2 mṙ2 1, J mṙ2 2, J mṙ2 3, J 1 = 1 2 [ṙ 1mṙ 1, J 1 + ṙ 1 mṙ 1, J 1 ] [ṙ 2mṙ 2, J 1 + ṙ 2 mṙ 2, J 1 ] [ṙ 3mṙ 3, J 1 + ṙ 3 mṙ 3, J 1 ] = ṙ 1 mṙ 1, J 1 + ṙ 2 mṙ 2, J 1 + ṙ 3 mṙ 3, J 1 = ṙ 1 ɛ 11k mṙ k ṙ 2 ɛ 123 mṙ 3 ṙ 3 ɛ 132 = 1 = 1 = mṙ 3 ṙ 2 mṙ 2 ṙ 3. In exactly the same way we can calculate the Poisson backet of the Hamiltonian with J 2 and J 3 to find out that these vanish too. The esult, H, J i, tells us that evey component of J and thus J itself is a constant of motion. This means that the vecto J, descbing the geneal angula momentum of a paticle moving unde the influence of a magnetic monopole, points always in the same diection with a constant magnitude. If we now take the scala poduct between the unit vecto pointing at the position of the paticle and the geneal angula momentum J we find out that J = L qξ = mṙ qξ c c = qξ c., since ṙ = 1 On the othe hand J = J cos θ, 7 ṙ 2

8 whee θ is the angle between J and the position vecto of the paticle. Thus cos θ = qξ c J θ = accos qξ, c J which tells us that the angle between the paticle s position vecto and J is a constant! This is the same esult we obtained on the couse when studying magnetic monopoles in one of the execises, even though the pocess was quite diffeent. We have aved at the main point of this wok, i.e., the tajectoy of the paticle moving nea the magnetic monopole. The fact that θ stays constant means that the tajectoy of the paticle lies on a cone. The cone has it s apex on the magnetic monopole, and it has the vecto J as it s axis. The cone opens to the opposite diection fom J, and it has an angle of accos. Figue 1 below may clafy the situation. qξ c J Figue 1: Paticle unde the influence of a magnetic monopole. 8

9 Refeences [1] P. Goddad and D. I. Olive, Magnetic monopoles in gauge field theoes, Repots on Pogess in Physics , doi: / /41/9/001. [2] H. Goldstein, C. Poole and J. Safko, Classical Mechanics, 3 d Edition, Addison Wesley, San Fancisco, [3] P. A. M. Diac, The Pnciples of Quantum Mechanics, 4 th Edition, Snowball Publishing, [4] Y. M. Shni, Magnetic Monopoles, Spnge-Velag, The Nethelands,