Module II: Relativity and Electrodynamics

Transcription

1 Module II: Relativity and Electrodynamics Lecture 2: Lorentz transformations of observables Amol Dighe TIFR, Mumbai

2 Outline Length, time, velocity, acceleration Transformations of electric and magnetic fields EM wave: Aberration, Doppler effect, intensity

3 Coming up... Length, time, velocity, acceleration Transformations of electric and magnetic fields EM wave: Aberration, Doppler effect, intensity

4 Length contraction The question: If the length of an object measured in frame S is L, what is the length measured in S? Formal statement: An object is stationary in frame S. The coordinates of two ends of the object in this frame are x 1 and x 2, independent of t 1 or t 2. The measurement of length L corresponds to x 2 x 1 = L. In frame S, the coordinates of these ends are (x 1, t 1 ) and (x 2, t 2 ). The measurement of length in this frame corresponds to L = x 2 x 1, when t 1 = t 2. We have This gives x 1 = γx 1 + γβct 1, x 2 = γx 2 + γβct 2. x 2 x 1 = γ(x 2 x 1) + γβc(t 2 t 1). When t 1 = t 2, one then gets L = L/γ, (1) which is length contraction.

5 Time dilation Formal statement: The time interval between two events at the same location in frame S (at x 1 = x 2 ) is T = t 2 t 1, what is t 2 t 1 in frame S? We have Since x 1 = x 2, this gives which is time dilation. ct 1 = γct 1 γβx 1, ct 2 = γct 2 γβx 2. T = (t 2 t 1) = γ(t 2 t 1 ) = γt, (2)

6 Velocity measured from a moving frame Let the velocity of a particle in the frame S be (u x, u y, u z ). What will be the velocity of this particle as measured in a frame S moving with a speed v in the x-direction (in the S frame)? Formalizing: Given dx/dt = u x, dy/dt = u y, dz/dt = u z, determine u x = dx /dt, u y = dy /dt, u z = dz /dt. Using the Lorentz transformations c dt γ γβ 0 0 c dt dx dy = γβ γ 0 0 dx dy. dz dz One gets u x = u x v 1 u x v/c 2, u y = u y γ(1 u x v/c 2 ), u z = u z γ(1 u x v/c 2 ). (3)

7 Velocity addition The results about velocities measured in different frames lead directly to the velocity addition formula: If two particles A and B are moving towards each other with speeds u A and u B, then the speed of approach of A as measured by B is obtained simply by using v = u B in the earlier results. This gives u A u B = u A + u B 1 + u A u B /c 2. (4) Note that this can never exceed c. Also, if either u A or u B equals c, then u A u B = c.

8 Acceleration In frame S, a particle has instantaneous velocity (u x, u y, u z ) and acceleration (a x, a y, a z ). What is the velocity and acceleration in frame S (moving with speed v along the x-direction)? The velocity in the new frame is (u x, u y, u z) as calculated earlier. The acceleration components are obtained simply through a x = du x dt, a y = du y dt, a z = du z dt, where one has to use du x /dt = a x, du y /dt = a y, du z /dt = a z. Problem Calculate the components of acceleration in frame S.

9 Coming up... Length, time, velocity, acceleration Transformations of electric and magnetic fields EM wave: Aberration, Doppler effect, intensity

10 Invariance of Maxwell s equations We would like Maxwell s equations to be invariant under Lorentz transformations. That is, in vacuum, we would like to have E = 0, E = B / t, B = 0, B = µ 0 ɛ 0 E / t. (5) This condition allows us to determine the transformation properties of E and B. (See Einstein s original paper.) From the Lorentz transformations and the chain rule for derivatives, x (ct = ) (ct ) x + (ct) (ct ) (ct) = γβ x + γ (ct) x = x x x + (ct) x (ct) = γ x + γβ (ct) y = y, z = z (6)

11 Transformations of E and B fields Problem The transformations for components of and / t were obtained earlier in this lecture, in eq. 6. Assume the components of E and B to be some linear combinations of the components of E and B, with coefficients that are functions of v. Show that E x = E x, E y = γ(e y vb z ), E z = γ(e z + vb y ), B x = B x, B y = γ(b y + v c 2 E z), B z = γ(b z v c 2 E y).

12 Lorentz force from relativistic transformations We thus know how E and B fields should transform under a change of frame. The transformations can be written in the form: E = E, E T = γ( E T + v B T ), B = B, B T = γ( B T v c 2 E T ), where the subscript T denotes components transverse to the relative motion of frames, and denotes components along the relative motion of frames. Note that in the limit of small velocities (γ 1), the above set of equations give E = E + E T E + v B, (7) the Lorentz force law. Thus, the principle of invariance of Maxwell s equations under Lorentz transformations has led us to the Lorentz force law directly.

13 More about the Lorentz force Relativity tells us that Lorentz force law is not an arbitrary addition to Maxwell s equations, but just a consequence of the additional principle of invariance of Maxwell s equations under frame changes. It shows that the E and B fields are connected. A pure B field in a frame give rise to an E field in another. On the other hand, magnetic fields B may be produced by purely electric fields E in another inertial frame. The standard Lorentz force law is valid only at low velocities. At large velocities, the law E = E + v B has to be modified to take care of the factors of γ in some components.

14 Coming up... Length, time, velocity, acceleration Transformations of electric and magnetic fields EM wave: Aberration, Doppler effect, intensity

15 Invariance of the wave equation From the Lorentz transformations and the chain rule for derivatives, (ct) x This gives = x (ct) x + (ct ) (ct) (ct ) = γβ x + γ (ct ) = x x x + (ct ) x (ct ) = γ x γβ (ct (8) ) 2 (ct) 2 2 = 2 (ct ) 2 2, (9) so that the electromagnetic fields in free space, which are solutions of ( ) 2 (ct) 2 2 V( x, t) = 0 are the solutions of ( ) 2 (ct ) 2 2 V( x, t ) = 0 in any other inertial frame.

16 Invariance of the plane wave solution Since the wave equation does not change, the plane wave solution in one frame stays a plane wave solution in another frame. That is, if a plane wave has the form V 0 exp[i( k x ωt)], with ω = c k in one frame S, it takes the form in the other frame S. V 0 exp[i( k x ω t )], with ω = c k The relationship between the primed and unprimed values of k and ω can be obtained by equating the phases at (x, y, z, t) and the corresponding (x, y, z, t ). If the frame S is moxing with speed v along the x direction, then for a wave with k = (k x, k y, k z ) and k = (k x, k y, k z), the phase-equality k x ωt = k x ω t gives k x x + k y y + k z z k ct = (γk x + γβ k )x + k yy + k zz (γβk x + γ k )ct. (10)

17 Aberration Let a source be stationary in the frame S, i.e. it is moving with a speed v in the x-direction in the S frame. Let the wavevector in the S frame be in the x y plane, making an angle θ with the x -axis. That is, tan θ = k y/k x. As seen from frame S, the wavevector makes an angle θ with the x-axis, with tan θ = k y k y = k x γk x + γβ k = sin θ γ(cos θ + β) (11) Thus the direction of emission of an EM wave from a source changes if the source is moving. This is the phenomenon of aberration. Note that θ = 0 θ = 0, θ = π θ = π, however θ = π/2 θ = 1/γβ. Thus, aberration tends to focus the directions of emitted waves towards the direction of motion of the source.

18 Doppler effect Equating the coefficients of (ct) on the two sides of the phase-equality, one gets k = γβk x + γ k = γ(1 + β cos θ ) k (12) Thus, ω = ω γ(1 + β cos θ ). This change of frequency due to the motion of the source is Doppler effect. Note the three special cases: θ = 0 ω = ω (1 + β)/(1 β), θ = π ω = ω (1 β)/(1 + β), θ = π/2 ω = ω γ. The first two cases are the blue-shift and red-shift associated with approaching and receding sources, respectively, present even without relativity (though the magnitudes are different). The third is the transverse Doppler shift, which is absent if relativistic effects are ignored.

19 Intensity of the wave Intensity power transmitted Poynting vector The magnitude of the Poynting vector N is We know that N = E H = 1 µ 0 c E 2 = 1 µ 0 c ( E 2 + E T 2 ) (13) E = E, ET = γ( E T v B T ) (14) Let us define k to be in the x-y plane. separate the E into two components: Component in the x-y plane Component normal to the x-y plane

20 k and E in the x-y plane The components of k, E, B are k = k cos θ ˆx + k sin θ ŷ E = E sin θ ˆx + E cos θ ŷ B = ( E /c) ẑ (15) The Poynting vector becomes N = = N 1 µ 0 c E 2 [ ( sin 2 θ + γ 2 cos θ + v ) 2 ] c ) 2 ] [ sin 2 θ + γ 2 ( cos θ + v c (16) Enhancement along the direction of motion of the source Problem Repeat the calculation for E along the z axis.

21 Take-home message from this lecture The transformations of velocity and acceleration while going from one inertial frame to another Transformations of E and B fields, and derivation of Lorentz force law from Maxwell s equations Relativistic aberration, Doppler shift, and change in intensity of an EM wave See simulations at