6. Maxwell Equations and Conservation Laws 6A. Maxwell’s Equations Ampere’s Law says Take the divergence: But conservation of charge says: So Ampere’s.

6. Maxwell Equations and Conservation Laws 6A. Maxwell’s Equations Ampere’s Law says Take the divergence: But conservation of charge says: So Ampere’s Law (as written) only applies to static situations Substitute Coulomb’s Law We now have So it is not J, but this combination that has no divergence And this combination is therefore writeable as a curl This suggests Ampere’s Law becomes Inconsistency of Ampere’s Law

Maxwell’s Equations We now have all four equations known as Maxwell’s Equations We also need constitutive equations relating D to E and B to H Most commonly, we’ll just assume linear relations In general, these will also be functions of frequency, so In vacuum these relations are simply In vacuum, generally rewrite equations in terms of only B and E The constant  0  0 has units of an inverse velocity squared c is, of course, the speed of light

The two homogenous Maxwell equations: Anything with a vanishing divergence is a curl Substitute this into Faraday’s law Anything with a vanishing curl is a divergence In summary, we have Vector and Scalar Potentials

Gauge Choice and Gauge Transformations B and E are physical; they can be directly measured A and  cannot be measured, so they may be non-unique Suppose (A,  ) and (A',  ') both produce the same electric and magnetic fields, so that This implies that Anything with a vanishing curl is a divergence, so –  is an arbitrary function of space and time Substitute into our other equation This implies These are the equations for a gauge transformation

Coulomb and Lorentz Gauges We previously defined Coulomb gauge by the demand that Suppose this is not satisfied, so We want to find a gauge transformation such that We therefore want: Can we solve: We already know the solution to this equation: Another useful gauge is Lorentz gauge, defined by Suppose this is not satisfied, so We want to find a gauge transformation such that We therefore want We will be solving this equation soon, and it has a solution (actually several)

Non-Uniqueness of Solutions We want to solve Maxwell’s Equation in vacuum Given  and J, can we find E and B? There are solutions that have no sources We can always add these homogenous solutions to non- homogenous solutions We will first try to get solutions assuming there are no fields without the sources We assume the fields can only appear after their sources We will try to find the vector and scalar potential This form automatically satisfies homogenous equations Because we have gauge freedom, A and  are not unique We will have to pick a gauge –Coulomb or Lorentz gauge 6B. Solving Maxwell in Vacuum

Equations for the Vector and Scalar Potentials We still have two Maxwell equations to satisfy Substitute into each of them These equations are hard because they are coupled An appropriate choice of gauge can help us solve them

Solving in Coulomb Gauge Let’s try solving these in Coulomb gauge,  A = 0 Then these formulas simplify to The first of these we know how to solve: Treating this as completed, we rewrite the second one as This is the second time we’ve encountered the equation We still need to solve this We will do so shortly

Solving in Lorentz Gauge Let’s try solving these in Lorentz gauge, in which Get rid of  A in each equation Cancel and rewrite slightly: This is the third and fourth time we encountered this equation If we solve this, then we are done with both cases So, it’s time to solve it

Green’s Function for d’Alambert Operator (1) We need to solve this equation Strategy: Solve it for a point source –I’m not including 4  like the book We will call the solution the Green’s function Pretty clearly, the Green’s function will depend only on the difference between x and x', and between t and t', so we write This simpler Green’s function is a solution of It is certainly going to be spherically symmetric, so Write it in spherical coordinates:

Green’s Function for d’Alambert Operator (2) Recall, we are interested in solutions where the source causes the fields This implies that for t < 0, we must have For t > 0, right side is zero, multiply by – c 2 r and “factor” the derivatives One of the two derivatives must vanish The solution to this equation is of the form For t > 0, right side must be well-behaved at r = 0, so Function and time derivative must also be continuous at t = 0 if r > 0, so The only solution to these equations is f(r) = constant, which then cancels out So effectively, f = 0 except at zero It must be a delta-function

Green’s Function for d’Alambert Operator (3) We note that since t > 0 and r > 0, the second term actually never contributes And for t 0, first term vanishes everywhere, so To find f 0, integrate the equation over time from –  to +  First term and right side can be done via delta-function Second term can be done by fundamental theorem of calculus Second term vanishes at both limits But we already know So f 0 = 1/4 

Green’s Function for d’Alambert Operator (4) Now change the problem back to a different point source: We can translate the solution in space and time: To solve the more general problem We now use superposition Do the t' integral using the delta function

Application to Coulomb and Lorentz Gauge In Coulomb gauge, we have We now have the solution to the second equation: In Lorentz gauge, we have We now have the solution to these equations

Sample Problem 6.1 (1) A point charge q moves along the z-axis at velocity v. Find the fields everywhere. The equation of motion for the particle is The charge density is The current density is Work in Lorentz gauge Scalar and vector potential are related It remains to find the scalar potential

Sample Problem 6.1 (2) Recall the rule for doing integrals with delta functions: We therefore have: Let’s switch to cylindrical coordinates A point charge q moves along the z-axis at velocity v. Find the fields everywhere.

Sample Problem 6.1 (3) In this equation, z' is the root of the argument of the  -function Do a LOT of algebra: Recall our expression relating vector and scalar potentials: Now we find the electric field: A point charge q moves along the z-axis at velocity v. Find the fields everywhere.

Sample Problem 6.1 (4) Now for the magnetic fields: Interestingly, electric fields always point directly away from the charge’s current position A point charge q moves along the z-axis at velocity v. Find the fields everywhere.

Comments on Fields from a Point Charge It is interesting that E always points away from the current position of the charge According to relativity, information must travel no faster than light The formulas we used do not contradict this: Suppose at t = 0, the charge suddenly stopped at the origin At this point the electric fields all point back towards the origin For a while, they must continue to point as if the charge kept moving Until the information reaches them There is a smarter/easier way to do this whole computation –Using relativity

Relativity and Gauge Choice According to relativity, information cannot travel faster than light In Lorentz gauge, the scalar and vector potentials respond to the charge distribution and current at the retarded time Lorentz gauge is, in fact, Lorentz invariant In contrast, in Coulomb gauge, the potential reacts instantaneously to the charge distribution –Apparent causality violation! Nonetheless, E and B are invariant, and hence the same in all gauges Causality violation in  cancels causality violation in A in Coulomb gauge Coulomb gauge is manifestly not Lorentz invariant

Energy Density For linear media, we know the energy density: Even when this is not true, we were able to demonstrate that the rate of change of the total electromagnetic energy is So we would guess that the local energy density satisfies For electric charge conservation we had a local version: Can we find a similar formula for conservation of electromagnetic energy? 6C. Cons. of Energy & the Poynting Vector

The Poynting Vector Faraday’s Law and Ampere’s Law: Rewrite using these Pull out one of those fancy product rules: We therefore have We define the Poynting vector: We therefore have:

Where Does the Energy Go? This looks like conservation of energy, except for the right-hand side For example, if we integrate it over a volume V, it yields Second term on the right is work being done on particles being pushed by electric fields Mechanical energy may appear as macroscopic motion, heating, etc. Right hand side is now flow of energy out of the region –Poynting vector shows the direction energy is flowing

Sample Problem 6.2 (1) A point charge q moves at velocity v. Find the energy flow out of a cylinder of radius a centered on the path of the charge (a) ahead of the charge, and (b) behind the charge We can arbitrarily have the charge move along z-axis, then we know the fields: The Poynting vector is then The flow out of the cylinder is

Sample Problem 6.2 (2) A point charge q moves at velocity v. Find the energy flow out of a cylinder of radius a centered on the path of the charge (a) ahead of the charge, and (b) behind the charge Let radius be a, and change variables to We now substitute limits –From 0 to  ahead of –From –  to 0 behind

Rate of Change of Mechanical Momentum Is there a similar formula for conservation of momentum? We expect three formulas, one for each component of p The forces on particles is given by If we have a continuum of different types of particles with number density n i, this becomes This expression can be simplified using So we have Force is the time derivative of momentum, so 6D. Conservation of Momentum in E and M

Go Nuts with Maxwell’s Equations Use Ampere’s Law and Coulomb’s Law So we have: Use the product rule So we have Use Faraday’s Law, and Gauss’s Law for magnetism

When the Going Gets Tough: We’d like to interpret second term on the left as momentum Then we would have momentum density: To make this identification, we need to show right side is a total divergence It is tricky because there may be momentum shared with the medium Let’s assume we are in vacuum: There is clearly a similarity between the E and the H terms

Can We Write It as a Divergence? Consider just the z-component of the E-terms Very similar formula works for the H-terms Define the z vector components of the stress tensor Then we have

The Maxwell Stress Tensor Use the divergence theorem: Interpretation: g z is the electromagnetic momentum density in the z-direction T z is (minus) the flow of z-momentum Generalize this to get the Maxwell stress tensor: Then we find: So g is momentum density And T ij is (minus) the flow of i-momentum in the j-direction

Comments on Momentum Density Recall, we have really only worked this out for vacuum –Jackson says it gets complicated otherwise So we have:

Rotations It will be helpful, for the moment, to denote components of a vector by subscripts: We believe the universe is invariant under translations and rotations –Laws of physics do not depend on choice of origin or choice of directions for axes What makes rotations and translations the correct transformations? Consider the distance formula between two points x and y: If you translate both x and y by the same amount, the distance between them won’t change If you rotate them with a rotation matrix R, they will also not change –But only if we pick the matrix R judiciously! 6E. Symmetries of Fields

Proper and Improper Rotations We want to keep, for example, distance to origin fixed We therefore have: This will work if (and only if) Written as a 3  3 matrix, this implies Take the determinant of this equation, using the fact that det(R T ) = det(R) Those with det(R) = +1 are called proper rotations –They can be achieved by a combination of small rotations Those with det(R) = –1 are called improper rotations Simple example of improper is the parity rotation All other improper are combinations of parity plus proper

Scalars, Vectors, and Tensors So, we say that (for example) energy U is a scalar and momentum p is a vector –What does this mean? A scalar quantity is a quantity that is unchanged when you rotate it A vector quantity is a quantity that rotates like the coordinates We can also have (rank 2) tensors with two indices, both of which must be rotated You can make them out of lower rank tensors, for example This can be generalized to tensors of arbitrary order –A scalar is a rank-0 tensor –A vector is a rank-1 tensor –Rank-2 tensors are sometimes simply called tensors

Scalars, Vector, and Tensor Fields We also talk about  as a scalar function and A as a vector function –What does this mean? A scalar function, under a rotation, has its coordinates change A vector function also has its components get mixed up You can make tensor functions as well

Invariant Tensors There are a some tensors that are invariant under rotations For example, if we rotate the Kronecker delta-function, we get Consider the Levi-Civita tensor, defined by It is completely anti-symmetric, that is if you change any pair of indices you get same thing times –1 Under a rotation, this becomes Easy to show that  ' is completely antisymmetric –Which means it is proportional to  The proportionality constant is the determinant –This is definition of determinant So it is invariant under proper rotations but not improper rotations

Dot and Cross Products, Pseudovectors Suppose we wanted to multiply two vectors a and b We could simply make a rank-2 tensor out of them: Or we could combine them using Kronecker delta: Or we could combine them using Levi-Civita: That’s why these are the particular ways we normally combine them Not hard to show that S transforms as a scalar under all rotations In contrast, if a and b are two vectors, because Levi-Civita has a minus sign A vector that behaves like this is called a pseudovector –Those that don’t behave this way are called true vectors or vectors We can similarly make scalar quantities with odd behavior –For example, consider the triple scalar product of three true vectors Under rotation, pretty easy to see that

Examples True vectors and true scalars behave the way you expect when you perform an improper rotation Consider, for example, a rotating moving object Reflect it in a mirror The velocity will reflect like you expect But the angular momentum is reversed! This implies angular momentum is a pseudovector Not surprising, since v L vRvR LRLR

Symmetry of Gradient and Divergence We can take derivatives to change fields of one type into another We note that Consider first the gradient Suppose  is a true scalar field Then  transforms as a true vector: Now consider the divergence The divergence converts a true vector into a true scalar

The Curl The curl is a bit more complicated Let it act on a true vector field, then when we perform a rotation, we have Use the fact that RR T = 1 to rewrite this as We already know how the Levi-Civita symbol rotates So it turns vectors into pseudovectors

Electricity and Magnetism Under Rotations Are the laws of electricity and magnetism invariant under rotations? We have to make sensible guesses of how all electromagnetic fields quantities transform under rotations If there is any sense to it, the number density of particles is a true scalar field And charge is a true scalar Therefore, charge density will be a scalar: Velocity is true vector quantity Therefore, current is a vector field If you look at the Lorentz force formula: Since F = ma, and a is a true vector, E must be a true vector. If B were a true vector, then v  B would be a pseudovector, which is wrong So B must be a pseudovector

Maxwell’s Equations and Time Reversal So far we have the following guesses: Let’s look at Maxwell’s Equations in vacuum Coulomb’s Law: both sides are true scalar Faraday’s Law: all terms are pseudovector Ampere’s Law: all terms are true vector Gauss’s Law for B-fields: left side is pseudoscalar Summary: All equations are invariant under both proper and improper rotations What if we are in a medium? For example, look at relationship between D, E and P This implies that D and P must also be true vectors Similarly, H and M must be pseudovectors Everything still works, though some materials may lack certain symmetries  True scalar J True vector E True vector B Pseudovector  True scalar J True vector E, D, P True vector B, H, M Pseudovector

Some Other Quantities What about scalar and vector potential? These depend on gauge choice Gauge choice could violate rotational invariance –Axial gauge is defined by A 3 = 0, for example But the two choices we made are rotationally invariant We recall that This implies that  is a true scalar and A a true vector The latter is consistent with Energy density is: So it’s a true scalar Poynting vector and momentum density True vectors Maxwell-stress tensor: True (rank-2) tensor  True scalar J True vector E, D, P True vector B, H, M Pseudovector  True scalar A True vector S, g True vector T ij True tensor u True scalar

Using Symmetries to Solve Problems Whenever faced with a new problem, try to identify symmetries of the problem Try to choose coordinates that maximize your ability to take advantage of symmetries First, use any translations or proper rotations to argue which coordinates must be irrelevant You may also be able to use these to completely eliminate certain components of the potential answer Then consider one or more improper rotations to restrict it further –Being careful to remember which fields are pseudovectors! Then you can often use simple formulas (like integral versions of Gauss’s Law or Ampere’s Law) to finish the problem. –Try to orient your Gaussian surface or Ampere loop so it picks up any unknown field

Sample Problem 6.3 (1) A solid infinite cylinder of charge has radius a and uniform charge density  throughout. Find the electric field everywhere. Pick coordinates –Cylindrical, with the z-axis along the axis of the cylinder The problem is translation independent in z-direction –It must have same answer if z  z + k The problem is rotation independent about the z-axis –It must have same answer if    + k We now know Now let’s try reflection across the z = 0 plane This is improper, but E is a true vector This reverses z-direction, but leaves  and  unchanged So the electric field will be And therefore Similarly, you can reflect across y = 0 plane, reversing 

Sample Problem 6.3 (2) A solid infinite cylinder of charge has radius a and uniform charge density  throughout. Find the electric field everywhere. We now have Pick an appropriate Gaussian surface Gauss’s Law tells us Repeat for a Gaussian surface outside the cylinder Combine to a final formula

Sample Problem 6.4 (1) A tokamak is in the shape of a rectangular cross-section donut centered on the z-axis, as sketched in the cutaway view. A total current I is then sent around the rectangular direction of the tokamak, so it goes up through the hole in the center, across the top, down on the outside, and then back to the center, equally at all angles. Find B at all points inside or outside the tokamak. Use cylindrical coordinates passing through the middle No translation symmetry, but rotation around the z-axis All quantities must be independent of  Current is also symmetric under reflection across the xz-plane Under this reflection, z and  are unchanged, but  is reversed If B were a true vector, this would imply But B is a pseudovector, so instead, This implies that

Sample Problem 6.4 (2) A tokamak is in the shape of a rectangular cross-section donut centered on the z-axis, as sketched in the cutaway view. A total current I is then sent around the rectangular direction of the tokamak, so it goes up through the hole in the center, across the top, down on the outside, and then back to the center, equally at all angles. Find B at all points inside or outside the tokamak. Our field is now Pick an Ampere loop in the  direction For example, to get it in the “cake” of the tokamak, how about this loop? The full current passes through this loop, so A similar loop that is not in the “cake” of the donut can also be drawn

Electricity and Magnetism and Time Reversal Are the laws of electricity and magnetism invariant under time reversal? We have to make sensible guesses of how all electromagnetic fields quantities transform under time reversal If there is any sense to it, the number density of particles is unchanged under time reversal And charge remains unchanged Therefore, charge density will be invariant: Velocity though, will change sign Therefore, current changes sign If you look at the Lorentz force formula: Since F = ma, and a is the second derivative, it will be unchanged Therefore E must be unchanged But since v changes sign, B must change sign

Maxwell’s Equations and Time Reversal So far we have the following guesses: Let’s look at Maxwell’s Equations in vacuum Coulomb’s Law: both sides are positive Faraday’s Law: all terms are positive Ampere’s Law: all terms are negative Gauss’s Law for B-fields: left side is negative Summary: All equations are invariant under time reversal What if we are in a medium? For example, look at relationship between D, E and P This implies that D and P must also be positive Similarly, H and M must also be negative  + J – E + B –  + J – E, D, P + B, H, M –

Some Other Quantities This is getting boring Sometimes, time reversal can be useful to solve the problem But often, boundary conditions are not time symmetric Consider the fields from an arbitrary source in Lorentz gauge: We explicitly assumed the fields were “caused” by the charges This solution is not time symmetric, even if the problem is The laws of E and M are time symmetric It is a deep question why the universe around us is not  + J – E, D, P + B, H, M –  + A – S, g – T ij + u +

6. Maxwell Equations and Conservation Laws 6A. Maxwell’s Equations Ampere’s Law says Take the divergence: But conservation of charge says: So Ampere’s.

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6. Maxwell Equations and Conservation Laws 6A. Maxwell’s Equations Ampere’s Law says Take the divergence: But conservation of charge says: So Ampere’s.

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Presentation on theme: "6. Maxwell Equations and Conservation Laws 6A. Maxwell’s Equations Ampere’s Law says Take the divergence: But conservation of charge says: So Ampere’s."— Presentation transcript:

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