dt Now we will look at the E&M force on moving charges to explore the momentum conservation law in E&M.

Transcription

1 . Momentum Conservation.. Momentum in mechanics In classical mechanics p = m v and nd Newton s law d p F = dt If m is constant with time d v F = m = m a dt Now we will look at the &M force on moving charges to explore the momentum conservation law in &M... Maxwell s Stress Tensor Charge density ρ, current density (or charge flux density) J = ρ v Let s consider the force exerted on a charge element ρdτ within any given volume V with a closed boundary S. The force on the charge in the volume dτ is d F = dq + v B = ρ + v B dτ = ρ + j B dτ / note7

2 Thebodyforcedensityis d F f = dτ = ρ + j B First, let s express ρ and j with fields through Maxwell s equations We obtain from the st and 4th equations = ρ ε B = = B B = µ j + ε So But So ρ = ε j = B ε µ f = ε + B ε B µ = ε + B B ε µ B B = B + B f = ε + B B ε µ = ε B B ε µ B B + ε B + ε = ε h i µ B B ε µ Use product rules (4) near the cover page in the text book Ã! B µ A B = A B + A A + A B + B A / note7

3 Let A = B =,then = + Or = Note also the definition of the Poynting s vector S = µ B Therefor f = ε + µ B B B ε µ S " = ε + # h B + ε i + µ + µ h B B i ε µ S To make the equation more symmetric, we add a term B B (note B =) " f = ε + # h B + ε i + µ + µ h B B + B B i ε µ S It is a very complicated expression. We can write it in a simple form if we define a Maxwell s stress tensor µ T = ε I + B B µ B I where I isaunittensorwithi ij = δ ij, i, j =,, 3or I = In the component form T ij = ε µ i j δ ij + µ µ B i B j B δ ij 3 / note7

4 where i, j = x, y,. For example T xx = ε x x x + y + δ + B x B x B µ x + By + B δ = ε x y + B µ x By B T xy = ε x y δ + µ B x B y B δ = ε x y + µ B x B y We can work out each term and write the tensor explicitly bellow T = ε x y x y x y x y y x x y x y + B x By B Bx B y B x B B µ y B x B y Bx B By B B B x B B y B Bx By The dot product with a vector a T can be carried out as bellow Ã! X X a T = a i T ij e j If we replace a with then = x e + j = (a a a 3 ) x e + T = X j i T T T 3 T T T 3 T 3 T 3 T 33 x 3 e 3, x = x, x = y, x 3 =, Ã X or µ Ã X T = j i i! T ij e j x i! T ij x i 4 / note7

5 It can be shown that the &M force density can now be expressed by f = T ε µ S The total force on the volume is Z µ F = Z T dτ ε µ S dτ I Z T d S = d A ε µ dτ dt Here, I used Gauss law for a tensor Z µ I T T dτ = d A without proof. If S does not change with time, the total force on the volume is the integral of the tensor on the surface: F = I T d A T T T 3 da T d A = T T T 3 da T 3 T 3 T 33 da 3 = (T da + T da + T 3 da 3, T da + T da + T 3 da 3, T 3 da + T 3 da + T 33 da 3 ) The tensor is therefore understood as the force per unit surface area. T ij represent a force per unit surface area in the i-direction on a surface perpendicular to the j-direction. For example, T = T xx isaforceperunitareainthex-direction on a surface parallel to the y-plane (pressure). T = T xy isaforceperunitareainthex-direction on a surface parallel to the x-plane (shear). 5 / note7

6 T x y T x x o y x xample 8.: Find the net force on the northern hemisphere of a solid sphere of radius R and total charge Q (with a uniform charge distribution). o y x Field within the sphere at radius r can be obtained from Gauss law I d A = ε q encl. = ε 4 3 πr3 4 3 πr3 Q = ε r3 R 3 Q 6 / note7

7 or or On the surface at r = R. where 4πr r = r 3 ε R Q 3 = 4πε Q R 3 r = 4πε rq R 3 c r = 4πε Q R c r c r =sinϑ cos φ e x +sinϑ sin φ e y +cosϑ e is a unit radial vector. Due to symmetry, the force should be in the -direction only. I µ T Z µ T Z µ T F = d A = d A + d A On the bowl bowl d A = da c r = R sin ϑdϑdφ [sin ϑ cos φ e x +sinϑ sin φ e y +cosϑ e ] disk µ T d A = T x da x + T y da y + T da = (T x sin ϑ cos φ + T y sin ϑ sin φ + T cos ϑ) R sin ϑdϑdφ Now µ Q T x = ε x = ε 4πε R µ Q T y = ε y = ε 4πε R T = ε x y cos ϑ sin ϑ cos φ cos ϑ sin ϑ sin φ = ε µ Q cos ϑ sin ϑ 4πε R It can be shown µ T d A = ε µ Q R cos ϑ sin ϑdϑdφ 4πε R Z Z µ F bowl = ε Q R cos ϑ sin ϑdϑdφ 4πε R = π µ ε Q Z π/ R cos ϑ sin ϑdϑ 4πε R = Q π/ 4 4πε R sin ϑ = Q 4πε 8R 7 / note7

8 On the disk = 4πε rq R 3 c r = 4πε rq R 3 (cos φ e x +sinφ e y ) d A = rdrdφ ( e ), da = rdrdφ integral T = ε x {} y = ε µ x + y = ε rq 4πε R 3 = µ T d A = T x da x + T y da y + T da = T da Z Z µ F disk = ε rq rdrdφ 4πε R 3 = π µ ε Q Z R r 3 dr 4πε R = π µ ε Q R 4 4πε R 3 4 Q = 4πε 6R F = F bowl + F disk = 4πε 3Q 6R We have shown that a body force can be calculated from a surface integral of a stress tensor Momentum conservation We have shown that f = T ε µ S On the other hand, based on Newton s nd law f = p mech where p mech is the mechanical momentum of the charged particles per unit volume. Let s define p em = ε µ S then the force equation can be written ( p mech + p em )= T 8 / note7

9 integral over the volume P mech + P em = Z µ I T T dτ = d A RHS -the total &M force on the volume. It converts to the change of the mechanical momentum and P em. We interpret that p em = ε µ S is the momentum density stored in the &M fields. The momentum conservation ( p mech + µ p em )= T resembles and Therefor ρ = j, (u mech. + u em )= S. µ T is the momentum flux density. Duality of &M quantities S : Poynting s vector (nergy flux density) ε µ S : Momentum density T : Stress tensor (force per unit area) T : Momentum flux density xample 8.3: momentum in a coaxial cable of inner radius a, outer radius b, and length l,the coaxial cable carries a DC current I and the voltage between the inner and outer conductor is V. Calculate the momentum density. 9 / note7

10 From note7 (example 8. on page 348 of the textbook), we found S = VI πal VI p em = ε µ S = ε µ e πal Z VI P em = ε µ VI e dτ = ε µ πal πal VI = ε µ b a e a h π b a L i e It is NOT ero. In order to make the total momentum to be ero, there must be some other kinds of momentum to balance it. Where is the hidden momentum? We may think that. the cable would recoil. But that did not happen.. Flowing electrons carry momentum. But electrons go one direction and back. Classical mechanics would predicting a perfect cancellation of momentum. 3. The correct answer: Hidden momentum is indeed stored in the moving electrons carrying current. However. this net hidden momentum reveals itself only when the relativity theory is applied (we may discuss that later)..3 Angular Momentum In classical mechanics, the angular momentum is defined by L = r p Phys463,&MIII,C.Xiao /note7

11 m p o y x Similarly, &M fields contain also angular momentum density l em = r p em Note that the angular momentum depends on the choice of the reference point O. We will not further our discussion on the topic. Phys463,&MIII,C.Xiao /note7