Special relativity and light RL 4.1, 4.9, 5.4, (6.7)

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1 Special relativity and light RL 4.1, 4.9, 5.4, (6.7)

2 First: Bremsstrahlung recap Braking radiation, free-free emission Important in hot plasma (e.g. coronae) Most relevant: thermal Bremsstrahlung What does its spectrum look like?

3 First: Bremsstrahlung recap Answer: blackbody when optically thick (because it s thermal), ~flat otherwise (with exp. cut-off), Note: not *completely* flat, due to ḡ ff (T, )

4 Two postulates: Special relativity and light F 0 Say frame moves with speed along x w.r.t. frame F (frames coincide at ) Lorentz transformations: v t=0 F F 0 (=comoving) x 0 = (x vt) y 0 = y z 0 = z where t 0 v = t c 2 x Lorentz factor ( 1) 1 q 1 v 1 2 c 2 2 1/2

5 Special relativity and light Most basic consequences: Length contraction: L = 1 L 0 (objects in F seem shorter from ) 0 F Time dilation: T = T 0 (clocks in F 0 seem to tick slower from F ) Note: relation is symmetric, because F 0 considers F to be moving at and temporal simultaneity is not Lorentz invariant (relativity of simultaneity) ) But velocities also transform: 9 dx = (dx 0 + vdt 0 ) = dt = dt 0 + v c 2 dx0 ; ) u x = dx dt = u y = u0 x + v 1+vu 0 x/c 2 u 0 y (1 + vu 0 x/c 2 ) v ( v) = (v) Velocity addition equation u z = u 0 z (1 + vu 0 x/c 2 )

6 v We chose along arbitrarily, so more generally: u k = u0 k + v 1+vu 0, u u 0? k /c2? = (1 + vu 0 k /c2 ) (where k and? are defined w.r.t. v) Since and transform differently, Therefore aberration: (where u 0 = u 0 ) Aberration of light ( u 0 = c ): Special relativity and light x u k u? 6= 0 tan = cos = tan = u? = u k sin 0 (cos 0 + v/c) cos 0 + v/c 1+(v/c) cos 0 u 0 sin 0 (u 0 cos 0 + v) y 0 u 0 v 0 x 0

7 Special relativity and light v 0 = /2 Say light is emitted perpendicularly to ( ): sin 0 tan = (cos 0 + v/c) = c 9 v >= cos = >; cos 0 + v/c 1+(v/c) cos 0 = v c ) sin = 1 Then for large, and so F 0 sin 1/ F Light from a relativistic emitter is beamed in the forward direction!

8 Isotropic emitter moving to the right

9 { Relativistic doppler effect Signal from moving source: Doppler shift In emitter s frame, wavefronts are but in observer s frame that distance is: obs = c t d = c t v t cos = c t(1 cos ) em = c t 0 apart, Since t = t 0, we find: obs = em (1 cos ) ) obs em t 2 t 1 = t µ = em obs = (1 cos ) { Doppler factor: Can be both <1 and >1! 1 (1 µ) = (1 + µ0 ) cos 0 () obs = em )

10 Isotropic emitter moving to the right

exclusively from the front and is blue-shifted Monochromatic flux is

11 Isotropic emitter moving to the right Same goes for radiation the moving particle sees coming in: appears to be coming almost exclusively from the front and is blue-shifted Monochromatic flux is spread over frequencies (angle-dependently) Total intensity boosted by factor (!!!) 4

12 Redshift Say we observe a photon from an object moving at line of sight away from us v along the Redshift: 1+z obs em = [1 + cos(0)] = p s 1+ = 1 Non-relativistic limit: 1 ) z

13 Lorentz invariant quantities How do other RT quantities transform? Solid angle: 9 d =sin d d d 0 =sin 0 d 0 d 0 >= sin d = d 0 = d dcos >; ) d d 0 = d cos d cos 0 = 2 Using Lorentz invariance of either phase-space volumes or photon counts, one can show that is also Lorentz invariant I / 3 So: I = S = 3 I 0 3 S 0 I = 4 I 0 (since = 0 ) S I (since transforms like ) (since we integrate I over = 0 )

14 The fraction of photons passing through a material with optical depth is Counts are Lorentz invariant, so But we also know that l = l 0 Lorentz invariant quantities exp( ) = l sin v Since (perpendicular to ): 9 sin = 0 sin 0 >= sin = tan cos ) 0 = 1 sin 0 >; exp( ) =exp( 0 ) ) = 0 = 1 0 Since j = S, we then also know that j = 2 j 0

15 Lorentz invariant quantities If the emitter is a blackbody, then I = B (T )= 2h 3 /c 2 exp(h /kt ) 1 I / 3 exp(h /kt ) But since is Lorentz invariant, must be too! e h /kt = e h 0 /kt 0 ) T = 0 T 0 ) T T 0 = Therefore, a relativistic blackbody still has a blackbody spectrum in the observed frame, but with a temperature T 0

16 Cosmic microwave dipole Part of the sky moves towards us, part away from us (due to our own movement) therefore radiation from the CMB (blackbody) looks hotter in one direction and colder in the other (dipole effect)!

17 Emitted vs received Let s consider the Lorentz transformation of de I = dt d d da h dn = dt d d da = I h( 0 ) dn 0 (dt 0 / )( d 0 )(d 0 / 2 ) da 0 = 2 I 0 again: But we should have I = 3 I 0 ; what gives? We have to distinguish emitted radiation from received radiation: the source is moving in between emitting photons, so the proper time interval to use here is not dt = dt 0 /, but dt = dt 0 /

18 Relativistic Bremsstrahlung Consider Bremsstrahlung from the electron s rest frame Ion accelerating toward it, hence sending out dipole emission (see lecture 4) When speeds are relativistic, this radiation appears beamed (roughly) towards the electron As the electron sees it, it is not sending out Bremsstrahlung itself instead, it Thomson scatters the virtual quanta of the ion! Relativistic correction to earlier Bremsstrahlung results

19 Cyclotron and synchrotron emission RL (4.8), (6.1), 6.6, (6.7)

20 Cyclotron radiation Recall: an accelerated charge will radiate like a dipole Larmor s formula: P = 2q2 u 2 3c 3 Radiation pattern: dw dtd = q2 u 2 4 c 3 sin2 u Angle is relative to u Radiation pattern True regardless of source of acceleration

21 Cyclotron radiation Remember the right-hand rule: charge moving in a magnetic field will be accelerated and therefore radiate Lorentz force: F = q c v B (cgs) When perpendicular, F = q c vb Equate to centripetal force: F = mv2 r L = q c vb ) r L = mcv qb Larmor radius The angular gyration frequency is then:! B = v r L = qb mc (SI:! B!! g = qb ) m B-field out of the page

22 Gyration period: T = 2 = 2 mc! B qb Frequency of radiation: B =! B 2 = 1 T = Power spectrum just has a single peak, at Cyclotron radiation qb 2 mc! =! B a v P (!) E(t) T t 1!/! B

23 Cyclotron radiation More generally, a particle will also have a velocity component parallel to the field replace with ) Resulting motion is a helix around the magnetic field lines v v? We find for the acceleration: u = a = v?! B Total radiated power is then: P = 2q2 v? 2!2 B 3c 3 = 2q4 2? B 2 3m 2 c 3

24 If the charge is an electron, and ; the classical electron radius is: We can therefore write the power as: For isotropic velocities, with the pitch angle (angle between Z field and velocity): Thus: P = 4 9 r2 ec 2 B 2 Cyclotron radiation e2 q = e m = m e r e = m e c 2 P = 2 3 r2 ec?b 2 2 2? 2 = sin 2 d = Finally, using U B = B 2 /8 (Thomson cross section) and (magnetic energy density): T =8 r 2 e/3 P = 4 3 T c 2 U B

25 Cyclotron radiation First detected cyclotron: Hercules X-1 (1977, Trumper) B 55 kev Hot electrons spiraling around neutron star s magnetic field

26 From cyclotron to synchrotron So far we ve been assuming that the particles are non-relativistic We know what should happen relativistically: radiation is beamed in the direction of motion, and spread out over frequency!

27 From cyclotron to synchrotron Before we saw what happens for an isotropic emitter For a dipole: Cyclotron Synchrotron

28 From cyclotron to synchrotron Non-relativistic 1 Slightly relativistic Second harmonic Highly relativistic

29 From cyclotron to synchrotron Remember: in relativity, we need to distinguish emitted from received radiation The observed fundamental harmonic is:! B =! 0 B/ sin 2 Same factor for higher-order harmonics Next week: synchrotron emission

Radiative Processes in Astrophysics

Radiative Processes in Astrophysics 9. Synchrotron Radiation Eline Tolstoy http://www.astro.rug.nl/~etolstoy/astroa07/ Useful reminders relativistic terms, and simplifications for very high velocities