Question

Exercise 3. (12p) (Lorentz boosts) The Maxwell equations (7) are invariant under Lorentz transformations. This implies that given a solution of the Maxwell equa- tions, we obtain another solution by performing a Lorentz transformation to the solution. A particular Lorentz transformation is a Lorentz boost with velocity v in - direction and acts on the electric and magnetic field strength as given in appendix B. (1) Tong) Now consider the electric and magnetic field due to a line along the X-axis with uniform charge density 1 per unit length. Since the electric charge is static, the magnetic field vanishes B = 0. The electric field reads E(t, x) = 27 (2+2(yỹ + zz). (a) (3p) Show, using the Lorentz boost above, that the magnetic fields produced by moving charges with velocity v of an infinitely long wire is given by B(t, x) = - 0 1 260 2612 + 32; (zý- y2). (b) (10) Calculate the inner product E-B before and after the transfor- mation and show that they agree. (©) (10) Calculate the quantity -E2 + eB2 before and after the trans- formation and show that they agree. (2) Consider again the plane wave solutions (2) with k=k, w=w and Emax = cBmax. Perform a Lorentz boost with velocity v in X-direction (this is the direction the light is traveling). Convince yourself that the result will be of the form E' (t', 2') = ý Ex cos (K':' - wt') B'(t',') = BOX cos (K'r' -w't'). This is the electric and magnetic field as seen by an observer which moves with a relative speed u to the original observer in the direction of the light. (a) (2p) Show that the amplitude of E and B change as (6) (b) (2p) Show that the angular frequency changes according to (You just derived the relativistic Doppler effect!) c) (2p) Before the transformation the dispersion relation is given by w(k) = ck. Does the dispersion relation change after the transfor- mation in other words, what is w' ))? (d) (1p) The speed of wave propagation is given by What is the speed before and after the transformation? Since an electromagnetic wave describes light, this is the speed of light.

Answer 1

1)

a) The electric field before the Lorentz boost is

$\vec{E}(t,\vec{x}) = \frac{\eta}{2\pi\epsilon_{0}}\frac{1}{y^{2}+z^{2}}(y\hat{y}+z\hat{z}) \quad (1)$

and

B(t, 7) = 0 (2)

Using the given Lorentz transformations for a boost along $\hat{x}$ direction, we get

y + y (3)

$z\to z \quad (4)$

Ey Ey = 260 V2 + 22

$E_{z}\to \gamma E_{z} = \frac{\eta\gamma}{2\pi\epsilon_{0}}\frac{z}{y^{2}+z^{2}} \quad (6)$

$B_{y}\to \gamma\frac{v}{c^{2}} E_{z} = \frac{\eta}{2\pi\epsilon_{0}}\frac{\gamma v}{c^{2}}\frac{z}{y^{2}+z^{2}} \quad (7)$

$B_{z}\to -\gamma\frac{v}{c^{2}} E_{y} = -\frac{\eta}{2\pi\epsilon_{0}}\frac{\gamma v}{c^{2}}\frac{y}{y^{2}+z^{2}} \quad (8)$

Hence after the boost

$\vec{E}(t,\vec{x}) = \frac{\eta\gamma}{2\pi\epsilon_{0}}\frac{1}{y^{2}+z^{2}}(y\hat{y}+z\hat{z}) \quad (9)$

$\vec{B}(t,\vec{x}) = \frac{\eta}{2\pi\epsilon_{0}}\frac{\gamma v}{c^{2}}\frac{1}{y^{2}+z^{2}}(z\hat{y}-y\hat{z}) \quad (10)$

b) Before the boost

$\vec{E}.\vec{B} = 0 \quad (11)$

because the magnetic field is zero. After the boost

$\vec{E}.\vec{B} = \frac{\eta^{2}}{(2\pi\epsilon_{0})^{2}}\frac{\gamma^{2} v}{c^{2}}\frac{1}{(y^{2}+z^{2})^{2}}(y\hat{y}+z\hat{z}).(z\hat{y}-y\hat{z}) = \frac{\eta^{2}}{(2\pi\epsilon_{0})^{2}}\frac{\gamma^{2} v}{c^{2}}\frac{1}{(y^{2}+z^{2})^{2}}(yz-zy) = 0 \quad (12)$

c) Before the boost

$-\vec{E}^{2}+c^{2}\vec{B}^{2} = -\frac{\eta^{2}}{(2\pi\epsilon_{0})^{2}}\frac{1}{(y^{2}+z^{2})^{2}}(y^{2}+z^{2}) = -\frac{\eta^{2}}{(2\pi\epsilon_{0})^{2}}\frac{1}{y^{2}+z^{2}} \quad (13)$

After the boost

$-\vec{E}^{2}+c^{2}\vec{B}^{2} = -\frac{\eta^{2}\gamma^{2}}{(2\pi\epsilon_{0})^{2}}\frac{1}{(y^{2}+z^{2})^{2}}(y^{2}+z^{2})+c^{2}\frac{\eta^{2}}{(2\pi\epsilon_{0})^{2}}\frac{\gamma^{2} v^{2}}{c^{4}}\frac{1}{(y^{2}+z^{2})^{2}}(y^{2}+z^{2}) = -\frac{\eta^{2}\gamma^{2}}{(2\pi\epsilon_{0})^{2}}\frac{1}{y^{2}+z^{2}}(1-\frac{v^{2}}{c^{2}}) \quad (14)$

Using the definition of $\gamma = \frac{1}{\sqrt{1-v^{2}/c^{2}}}$ we get

$-\vec{E}^{2}+c^{2}\vec{B}^{2} = -\frac{\eta^{2}}{(2\pi\epsilon_{0})^{2}}\frac{1}{y^{2}+z^{2}} \quad (15)$

2) Before the boost the plane wave solution are

$\vec{E}(t,x) = \hat{y}E_{max}\cos(\Tilde{k}x-\Tilde{\omega}t) \quad (16)$

$\vec{B}(t,x) = \hat{z}B_{max}\cos(\Tilde{k}x-\Tilde{\omega}t) \quad (17)$

The plane wave solutions for an observer moving along the $\hat{x}$ direction will also be of the form

Ē(t', 2) = y Emar cos(K x' -wt) (18)

$\vec{B}(t',x') = \hat{z}B'_{max}\cos(k'x'-\omega't') \quad (19)$

because we can see from the given Lorentz transformations that for a boost along x-direction y and z don't mix, but the electric and magnetic fields along y and z direction do change. But since initially

They also remain zero after the boost. The third thing we have to convince oureslves of is that palne waves still remain plane wave, this is because palne waves are translationally invariant along the direction of their propagation, therefore shifting the origin along x-direction does not chnage their form.

a) Using Lorentz transformation the new electric field is

$E_{y} = \gamma\left(E_{max}\cos(\Tilde{k}x-\Tilde{\omega}t)-vB_{max}\cos(\Tilde{k}x-\Tilde{\omega}t) \right ) = \gamma(E_{max}-vB_{max})\cos(\Tilde{k}x-\Tilde{\omega}t) \quad (20)$

Use and write the new field in terms of new coordinates

$E'_{y} = \gamma E_{max}\left(1-\frac{v}{c}\right)\cos\left(\Tilde{k}\gamma(x'+vt')-\Tilde{\omega}\gamma\left(t'+\frac{v}{c^{2}}x' \right )\right) = \gamma E_{max}\left(1-\frac{v}{c}\right)\cos\left(\gamma\left(\Tilde{k}-\Tilde{\omega}\frac{v}{c^{2}}\right)x'-\gamma(\Tilde{\omega}-v\Tilde{k})t'\right) \quad (21)$

Comparing equation (21) with equation (18) we get

$E'_{max} = \gamma\left(1-\frac{v}{c} \right ) E_{max} = \frac{1}{\sqrt{1-v^{2}/c^{2}}}\left(1-\frac{v}{c} \right ) E_{max} = \sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}E_{max} \quad (22)$

b) Again comparing equation (18) and (21) we get

$\omega' = \gamma (\Tilde{\omega}-v\Tilde{k}) \quad (23)$

Using $\Tilde{\omega} = c\Tilde{k}$

$\omega' = \gamma \left(1-\frac{v}{c}\right)\Tilde{\omega} = \sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}\Tilde{\omega} \quad (24)$

c) Comparing equation (18) and (21) it is quite clear that

$k' = \sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}\Tilde{k} \quad (25)$

Now

$\Tilde{\omega} = c\Tilde{k}\Rightarrow \sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}\Tilde{\omega} = c\sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}\Tilde{k}\Rightarrow \omega' = ck' \quad (26)$

d)

From equation (26) we get

$\frac{\mathrm{d} \Tilde{\omega}}{\mathrm{d} \Tilde{k}} = c \quad (27)$

and

$\frac{\mathrm{d} \omega'}{\mathrm{d} k'} = c \quad (28)$

Hence, the speed of light is same in both the frames. This should have been expected as the fact that speed of light is same in every inertial frame is one of the postulates of special relativity.