( θ ) appear in the angular part:

Transcription

1 lectroagnetic Theory (MT) Prof Ruiz, UNC Asheville, doctorphys on YouTue Chapter S Notes Lorentz Force Law S1 MT Other Physics Courses We have seen the figure elow with its triad of courses: Optics, lectroagnetic Theory, Special Relativity (Modern Physics) ħ + V We have also seen a connection to quantu echanics through our deep analysis of Laplace's equation Poisson's equation: V ρ V ε The Laplacian appears in the Schrödinger equation: In spherical coordinates, we found for our Laplacian solutions of the for V ( r, θ ) Ar + P (cos ) l l l 1 l θ + l r In quantu echanics things are ore coplicated For spherical proles, the associated Legendre polynoials Pl ( θ ) appear in the angular part: Y (, ) ( ) i l θ φ NlPl θ e φ, where N l are noralization constants The functions Y ( θ, φ ) are called spherical haronics The general solution for spherical l proles is a wave function of the following for ψ ( r, θ, φ) R ( r) Y ( θ, φ) nl nl l Michael J Ruiz, Creative Coons Attriution-NonCoercial-ShareAlike 3 Unported License

2 Today we add a fifth course to the list: Classical Mechanics Mechanics especially connects to our course via the Lorentz Force Law lectroagnetic Theory (The Maxwell quations) Optics (lectroagnetic Waves as a Solution to the Free-Space Maxwell quations) Special Relativity (Invariance of the Maxwell quations) Quantu Mechanics (The Matheatics of the Laplacian) Classical Mechanics (The Lorentz Force Law) Other connections with echanics include the following: The Concept of the Potential Kinetic Potential nergy Mechanical nergy converted to lectrical nergy The Lorentz force law is perhaps the est since we can coine Newton's Law with it Mechanics F a Newton's nd Law lectroagnetic Theory F q( + v ) Lorentz Force Law We have already seen the following two siple cases F q F qv x( t) v t y( t) q t v r qv v r q In the next section we consider the siplest prole where there is oth a constant electric field a constant agnetic field Michael J Ruiz, Creative Coons Attriution-NonCoercial-ShareAlike 3 Unported License

3 S Lorentz Force Place a charge q at the origin so that it is at rest at t What happens? GIVN: iˆ Charge q with r () j ˆ v () SOLUTION: We start the solution y coining Newton's Second Law with the Lorentz Force Law F a F q( + v ) We will use the notation where a dot eans a derivative with respect to tie For the x- variale we have vx ɺ x ax ɺɺ x iˆ ˆj kˆ F q( + v ) qiˆ + q xɺ yɺ zɺ qiˆ + qiˆ ( zɺ ) + qkˆ ( x ɺ ) Coining with Newton's equations we arrive at three differential equations x ɺɺ q qzɺ y ɺɺ z ɺɺ qxɺ The y-equation is the easiest The solution to ɺɺ y gives yɺ ( t) c1, where c 1 is the constant of integration Then y( t) c + c t 1 ut c y() c1 y() Therefore, y( t ) ɺ There is no otion along the y-axis Michael J Ruiz, Creative Coons Attriution-NonCoercial-ShareAlike 3 Unported License

4 This leaves xɺɺ q qzɺ z qx ɺ ɺɺ Divide y the ass Let q q q q xɺɺ zɺ zɺɺ xɺ x ɺɺ z ɺ z ɺɺ x ɺ The trick now is to differential x with respect to tie to otain a ter with zɺɺ which we then get rid of using our second equation, ie, zɺɺ xɺ ɺɺɺ x ɺɺ z ɺɺɺ ɺ x x When you take two derivatives get the sae thing ack with a inus sign out solutions are the cosine sine xɺ ( t) a cost + sint, where a are constants The initial condition for the speed is xɺ () a, which takes care of one constant Integrating leads to xɺ ( t) sint x( t) cost + c, where c is a constant Fro the initial condition x() + c, we find c x( t) (1 cos t) Then, Michael J Ruiz, Creative Coons Attriution-NonCoercial-ShareAlike 3 Unported License

5 Suary: x ɺɺ z ɺ z x ɺɺ ɺ x( t) (1 cos t) xɺ ( t) sint zɺɺ xɺ sin t zɺ cost + d Since zɺ () + d fro the initial condition, d ecoes Integrating again leads to ɺ zɺ cost + or z (1 cos t) z t t The initial condition sin + +, where z() eans is a constant is gone Then we have our equation z sint + t z ( sin t + t) ut where did this constant creep in We have to get rid of that All final answers ust e in ters of the given paraeters We need to work with one of these x ɺɺ z ɺ z ɺɺ x ɺ PS1 (Practice Prole) Why does the second equation not work for us? We will work with xɺɺ zɺ, z ( t sin t), x( t) (1 cos t) Michael J Ruiz, Creative Coons Attriution-NonCoercial-ShareAlike 3 Unported License

6 Sustitute zɺ (1 cos t) in xɺɺ zɺ to find xɺɺ z (1 cos t) ɺ Then set this equal to The result is d x( t) d xɺɺ (1 cos t) cost dt dt cos t (1 cos t) Note that the The constant ters on the right side ust cancel: cosine ters on each side of the equation alance we learn nothing fro that The constant coes fro the constant ters: The solutions x( t) (1 cos t) z( t) ( t sin t) ecoe Let R watch x( t) (1 cos t) z( t) ( t sin t) Then x( t) R(1 cos t) z( t) R( t sin t) Now x R R cost z Rt Rsint R cos t R x R sint Rt z R cos t + R sin t ( R x) + ( Rt z) Michael J Ruiz, Creative Coons Attriution-NonCoercial-ShareAlike 3 Unported License

7 R (cos t + sin t) ( x R) + ( z Rt) R ( x R) + ( z Rt) We can use our shift rule to figure out this is a circle whose center is at ( x, y, x) ( R,, Rt) Note that the center oves along the z-axis The charge q follows along the perieter of the oving circle according to x( t) R(1 cos t) z( t) R( t sin t) The path is a cycloid Courtesy Wikipedia Michael J Ruiz, Creative Coons Attriution-NonCoercial-ShareAlike 3 Unported License

8 S3 The Vector Potential We are in search for a potential for the agnetic field After all, the electric field has one In electrostatics, we have ρ ε In agnetostatics, we have Note that Here is why,, V µ J eans that the electric field is a gradient of a scalar: f f f ε ˆ ˆ ijk ek εijk ek x x x x i j i j f You get zero ecause the ε ijk is antisyetric in i j, ut you have syetry in i j with the partial derivatives which can e taken in any order The result is zip! PS (Practice) Work this out the long way without instein's suation convention A ( A) A A ( A) ( ) x x x x x Note that eans that the agnetic field is a curl of a vector: Here is why i k k ε jki εijk i i j i j I used ε jki ε ijk since you can always ake a cyclic change Then it is easier to see that you get zero ecause the ε ijk is antisyetric in i j while the derivative are syetric in i j This is essentially the sae arguent as efore PS3 (Practice) Work this out the long way without instein's suation convention Our suary is ρ ε µ J,,, V,, A Michael J Ruiz, Creative Coons Attriution-NonCoercial-ShareAlike 3 Unported License