Solving Problem of Electrodynamic - Assignment Example

Institute Question 2.2 a) The potential Inside Sphere As implied by definition of conducting on the surface V = 0. We must place animage charge outside the sphere on the axis defined by the real charge q and the sphere’s center. Use a Cartesian coordinate system and set the x-axis to be the axis defined by the charge, its image, and its center. a) The induced surface charge density The surface charge density will simply be the same as calculated by Jackson for the invers problem. For a charge outside a sphere which is conducting, the surface charge density is such. b) The scale and direction of the force acting on q. The force acting on q can be obtained by Coulomb’s law. c) Is there any change in the solution if the sphere is kept at fixed potential Φ? Is the sphere has a total charge Q on its inner and outer surface? If the sphere is kept at a fixed potential _, we must add an image charge at the origin so that the potential at R is _. If the sphere has a total charge Q on its inner and outermost surfaces, we figure out what image charge would create a surface charge equal to Q and place this image at the origin. Question 3.3 a) Find the potential for r > R. The same integration can be carried out for r > R. b) What is the capacitance of the disc? Question 3.22 As before, the procedure for determining the Greens function is to split. The region of interest in to two parts (one on each side of the observation point), and separate solutions of the Laplace equation that satisfy the boundary conditions of every region, and then join the two solutions at the source point such that their values match up but the first derivative (in whichever dimension we chose sides) has a finite discontinuity. Suppose the observation point is , let’s break the region into two sub-regions, defined by. The general solution of the Laplace equation in two-dimensional polar coordinates is; The solution in the first region must be admissible down to which excludes the in term and the negative powers of p. However, these terms may be included in the solutions for region in second place. In individually regions, the solution must vanish at which excludes the cos terms ( ) .The solution must also vanish at which requires that with these considerations we may write down the solutions for G in the two regions. The final step to choose the constant in (7) such as to make Referring to 7 we have, Subtracting (12) from (11) we obtain, Then from (10) we read off; And plugging this into (9) gives; Plugging this into (7) we finally obtain; Question 6.2 a) Let’s first assume that the charge is traveling along the z axis , so that its position is given by Henceforth when the electromagnetic disturbance has reached the origin, the particle has traveled as far as the electromagnetic disturbance did, but in the opposite direction, so it is now twice as far from the origin as it was when the disturbance we are just now feeling was generated. Question 4.5 a) The force in general is Let us keep the track of components Enlarge the external electric field in a Taylor series and only keep the focus on the first few terms because it varies slowly over space. Note we have labeled the gradients as being done with respect to primed variables to tell them apart from the variables to be intergrated. Substitute the expansion into the force equation; The electric fields do not depend on the unprimed variables and come out of the integrals, which was the point of the Taylor sequences expansion. After a little manipulation, we recognize the integrals that are left as the dipole moment and quadrupole moments Next note that the external field is created by charges outside our volume of interest, so This is the exact factor found in the final term shown, dropping that entire term out. Replacing the index notation with notation of vector, we finally have; Here we have switched primed variables to unprimed to match Jackson, and because the originally unprimed variables (the integration variables) are neatly tucked away now in the multipole moments. When we write the first term in terms of the potential, we can factor out the gradient operator. Let us compare this to the expansion (Jackson 4.24) of the energy W We see that using the two equations, we recover the familiar expression b) Electrostatic torque on a charge distribution ρ as a result of the external field E(0) is Let us look at one component, say component And inserting these into the torque equation to find We end up with; Question 9.10 a) Find the effective transitional “Magnetization”, calculate, and evaluate all the non-vanishing radiation multi-poles in the long-wavelength limit. The magnetization is b) In the electric dipole approximation calculate the total time averaged power radiation .Express your answer in units of c) Interpreting the classically calculated power as the photon energy times the transition probability, evaluate numerically the transition probability in units of reciprocal seconds. d) If, instead of the semi-classical charge density used above, the electron in the 2p state was described by a circular Bohr orbit of radius , rotating with the transition frequency w0, what would the predicted power be? Express your answer in the same units as in part b and evaluate the ratio of the powers numerically. For Bohr transition, a dipole transition, Question 9.6 a) Start with the expressions for the potentials in the Lorentz gauge This can be written in terms of the dipole moment as follows: b) We now want to calculate the fields from the potentials. This is mostly straight forward, albeit tedious, algebra. We need to remember that the dipole moment is evaluated at the retarded time, so that there is an implicit time dependence in p (t0) Therefore Then for the magnetic field we have For the electric field we have We have more terms to calculate now. The time derivative of A is easy: The gradient of Φ is a bit further involved. As shown below Combining all of these terms, we The last term (which is the “radiation term”) can be rewritten using the vector triple product, c) These results clearly reproduce the results for the harmonic fields with the sub-situations Question 12.14 a) Deriving the Euler-Lagrange equation of motion? Under what assumption? (Where’s the verb in the last sentence?) The Euler-Lagrange theorem says; b) Show explicitly, and with what assumptions, that this Lagrangian density differs from (12.85) by a 4 – divergence. Does this added 4-divergence affect the action or the equation of motion? The other Lagrangian is Question 11.6 a) Taking into account the possibility of convection current as well as conduction current, show that the covariant generalization of Ohm’s law is That will allow us to realize the electric field as a contraction of the Maxwell field strength tensor with the 4-velocity Therefore the right hand side of Ohm’s law can be written as The left hand side of Ohm’s law ought to be the 3-vector current density. It is natural to take However, this contains more than the 3-current, since it includes the charge density as well. Because is not involved in Ohm’s law, we need to subtract it out. This can be done by noting that Hence Combining (3) and (4) gives the covariant form of Ohm’s law Where we have now let go all primes, since the equation is in covariant form (i.e. frame self-determining). Note that this can equivalently be written as b) Show that if the medium has a velocity with respect to some inertial frame that the 3-vector current in that frame is Where ρ is the charge density observed in that frame. Working in the lab’s frame, we take Substituting this into (5) and noting that We obtain the time component equation Together with the space components equation Solving (6) for and substituting this into (7) gives That will be giving an explicit realization of the conduction and convection currents (the latter being the term in the above) c) If the medium is uncharged in its rest frame what is the charge density and the expression for in the frame of part b)? It is therefore the relativistic generalization of the equation Question 12.16 a) Starting with the Proca Lagrangian density (12.91) and following the same procedure as for the electro magnetized fields, show that the symmetric stress-energy momentum tensor for the Proca fields is The Proca Lagrangian density is b) For these fields in interaction with the external source as in (12.91), show that the differential conservation laws take the same form as for the electro magnetized fields, namely; Using a 4-divergence of the symmetric stress tensor (9) give Note that in the second line we have used the fact that which is automatic for the Proca’s equation. To obtain the last line, we used the Bianchi identity as well as the Proca equation of motion c) Given the explicit form of the Maxwell’s tensor, it is straightforward to show that Thus The time-time component of this is Work Cited Armero, F., and J. C. Simo. "Long-term dissipativity of time-stepping algorithms for an abstract evolution equation with applications to the incompressible MHD and Navier-Stokes equations." Computer Methods in Applied Mechanics and Engineering 131.1 (1996): 41-90. Read More