Characteristic vibrations of the field. LL2 section 52.

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Presentation transcript:

Characteristic vibrations of the field. LL2 section 52

Suppose electromagnetic fields only, no charges, finite volume V = A x B x C Vector potential A(r,t) = A(t) Exp[i k. r)] (real) Boundary conditions: An integral number of waves fits between walls of box in each direction So that, etc. With n i = positive or negative integer Fields can be expanded in triple Fourier series in x,y, and z

The condition that A is real means that ± k pairs in sum must satisfy

Complex coefficients A k are perpendicular to the wavevectors k. We have freedom to choose gauge: Coulomb Gauge

Substitute the expansion of A for waves in a box into the wave equation A = 0(46.7) Differential equation for coefficients A k (t) Simple harmonic oscillator equation: A k (t) are harmonic functions of time.

For large volumes,is very close to its adjacent value Since the n x differ by unity, the number of possible values of k x within  k x is the number of possible values taken by n x, namely The total number of values for k within the intervals  k x,  k y, and  k z is  n =  n x  n y  n z

oo How many of those values of k in  k are in a specific direction? Spherical coordinates

How many modes with wavevector magnitude k are within  k (all directions)?

Total energy of fields

How to calculate squares of field magnitudes All terms except those with k’ = - k vanish on integration due to the Exp[i(k + k’).r] oscillating factors. All r dependence disappears.

It’s also useful to simplify the cross products but not the time derivatives in 

The expansion can represent standing or running waves, depending on the choice of the harmonic coefficients A k. For running waves: with a k ~ exp[-i  k t],  k = ck. Then each term in the sum is a function of k.r –  k t, and each is a wave propagating in the direction of k.

Since sum is over both signs of k, we can change the sign of k in the second term without changing the sum. But we started with Therefore and (terms a k.a -k and a k *.a -k * appear twice with opposite sign)

Since sum is over both signs of k, we can change the sign of k in the second term without changing the sum. The total energy of the field is the sum of the energies of individual plane waves.

Total momentum of the field = Momentum density, see section 32 Usual relation for momentum of plane wave Unit vector in direction of k Homework

Started with function expressed at every point in space in terms of continuous parameters x, y, and z. Now we have same function expressed in terms of discrete variables k x, k y, and k z. (Simple periodic function known at every point of space)

Put field equations into canonical form by transforming the coefficients a k Define new coefficients (discrete set of generalized coordinates and momenta)

Then the Hamiltonian (energy) of the field is expressed as Hamilton’s equations are the “equations of motion” for the field This coincides with which can therefor be considered a consequence of the equations of motion. (HW)

The other Hamilton’s equation gives This is the same differential equation satisfied by the A k

Since A k was perpendicular to k, a k is too, and so are Q k and P k. Therefore, Q k and P k each has 2 independent components giving the polarization of each traveling wave component k in the sum Same for P k 2 so Each independent term corresponds to one traveling wave of definite wavevector and polarization “Oscillator expansion”

Inverse relationships. (Look like raising and lowering operators!)

Homework

Assignments (10 minute white-board presentations) Present the 3 derivations indicated in the slides. Describe the quantization of EM field and how we get from waves to photons (see LL5 secs 2,3)