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Black Hole Astrophysics Chapters 9.2.2&9.4 All figures extracted from online sources of from the textbook. Laws of Conservation of Charge and Current (Ch9.2.2) Conservation of Charge Very much mass density, the charge density is the sum of all species. ππ β‘ π΄ ππ ππ π However, in this case, it can be zero everywhere in the plasma. From the continuity equation in classical physics easily translate that to π»πΌ π½πΌ = 0. πππ ππ‘ + π» · π½ = 0, we can very This equation enables us to determine only ππ , therefore we will later introduce the Generalized Ohmβs law to find the π½ part. *NoteοΌ I sort of find it confusing to have all these 3-vectors and 4-vectos together in 1 slide, so to avoid having to label which is a 3-vector and which is a 4-vector, I chose to put vector/tensor symbols only for 3-vectors and all 4-vectors will be written in component form. Conservation of Charge Without loss of generality, we can separate the 4-current into π½πΌ = ππ π πΌ + ππΌ Containing a component parallel to U and another perpendicular to it. Therefore, ππΌ β‘ πΞ±Ξ² π½π½ And it automatically satisfies π πΌ ππΌ = 0 as we derived last week for the projection tensor. Recap for Plasma Astrophysics Recall that we had defined the velocity of a fluid element as π£ = ππ ππ π£ π +ππ ππ π£ π ππ β π£ π, this is due to the fact that momentum is dominated by protons and therefore the center of momentum velocity can be approximated by the proton velocity. Next, the current is defined as π½ = ππ ππ π£ π + ππ ππ π£ π β ππ ππ π£ π β π£ π . The interesting thing to observe here is that given not too much of a density difference between the protons and electrons, the current becomes proportional to the difference of velocities π£ π β π£ π . A useful (at least I find it usefulβ¦) way of thinking of this is that current, π½ is caused motion of electrons with respect to the c.m. velocity (which is π£ π ) of the fluid elements. Thus, fluid element motion is driven by protons current is mainly caused by electrons. Using these two simple concepts we can understand the Generalized Ohmβs law quite simply. The generalized Ohmβs law β Review The generalized Ohmβs law reads as : π π½ ππ ππ ππ 2 ππ ππ ππ 2 =β π»· ππ+ πΈ+π£×π΅ + π½ ×π΅ β π· π½ ππ‘ ππ ππ ππ ππ In simple language, it simply says: What are the different ways we can cause the current to change? ππ½ ππ‘ =β― π 1. The Pressure gradient term β ππ π» · π π π As we just explained, current is caused by velocity difference between π β &π+ , however, if the pressure on π β &π+ are similar, π+ , being heavier, is harder to push. Thus, mostly itβs only the electrons being accelerated. π£ π (t) π£ π (t+dt) π½ (t) π£ π (t)~ π£ π (t+dt) π½ (t+dt) π£ π (t)~ π£ π (t+dt) ππ ππ omitted for all figures The generalized Ohmβs law β Review The generalized Ohmβs law reads as : π π½ ππ ππ ππ 2 ππ ππ ππ 2 =β π»· ππ+ πΈ+π£×π΅ + π½ ×π΅ β π· π½ ππ‘ ππ ππ ππ ππ In simple language, it simply says: What are the different ways we can cause the current to change? 2. The collision term ππ ππ 2 β π π π ππ½ ππ‘ =β― · π½ Again, similar to the pressure gradient term, forces mainly affect the velocity of electrons. π£ π (t) π£ π (t+dt) π½ (t) π£ π (t)~ π£ π (t+dt) π½ (t+dt) π£ π (t)~ π£ π (t+dt) The generalized Ohmβs law β Review The generalized Ohmβs law reads as : π π½ ππ ππ ππ 2 ππ ππ ππ 2 =β π»· ππ+ πΈ+π£×π΅ + π½ ×π΅ β π· π½ ππ‘ ππ ππ ππ ππ In simple language, it simply says: What are the different ways we can cause the current to change? 3. The Electric field term ππ ππ 2 ππ ππ½ ππ‘ =β― πΈ This again is because that a given electric field will mainly accelerate electrons. π£ π (t) π£ π (t+dt) π½ (t) π£ π (t)~ π£ π (t+dt) π½ (t+dt) π£ π (t)~ π£ π (t+dt) The generalized Ohmβs law β Review The generalized Ohmβs law reads as : π π½ ππ ππ ππ 2 ππ ππ ππ 2 =β π»· ππ+ πΈ+π£×π΅ + π½ ×π΅ β π· π½ ππ‘ ππ ππ ππ ππ In simple language, it simply says: What are the different ways we can cause the current to change? 4. The plasma motion term ππ ππ 2 ππ ππ½ ππ‘ =β― π£×π΅ As we mentioned earlier, the plasma bulk velocity mainly follows that of protons, thus this term tells us the change in current due to protons deflected by B field. π΅ π£ π (t) π΅ π½ (t) π£ π (t) π£ π (t) π£ π (t+dt) × π΅ π£ π (t) × π΅ π£ π (t) π£ π (t+dt) π½ (t) π½ (t+dt) The generalized Ohmβs law β Review The generalized Ohmβs law reads as : π π½ ππ ππ ππ 2 ππ ππ ππ 2 =β π»· ππ+ πΈ+π£×π΅ + π½ ×π΅ β π· π½ ππ‘ ππ ππ ππ ππ In simple language, it simply says: What are the different ways we can cause the current to change? 5. The Hall effect term ππ ππ ππ½ ππ‘ =β― π½ ×π΅ This term is slightly more complicated, it contains π½ , which means that both rotation of electron velocity and ion velocity contribute. (It is that Hall effect which is commonly used to determine semiconductor type) π΅ π£ π (t) β π£ π (t) × π΅ π½ (t+dt) π½ (t) π£ π (t) π£ π (t) × π΅ π΅ π£ π (t) π£ π (t+dt) π½ (t) π£ π (t) π£ π (t+dt) The generalized Ohmβs law Slightly modified, the generalized Ohmβs law reads as : π π½ ππ ππ ππ 2 ππ ππ ππ 2 + π»· ππ= πΈ+π£ ×π΅ + π½ ×π΅ β π· π½ ππ‘ ππ ππ ππ ππ Now, the fully General Relativistic version: ππ2 1 π»πΌ πΆ = ππΌ + βπ ππΌ πΉ Ξ±Ξ² β ππ ππ ππ½ + ππ½ 4π π ππ + ππ πΌ π½ πΆ Ξ±Ξ² β‘ ππ + π π + π πΌ πβ² π½ + πβ² πΌ ππ½ + ππ πΞ±Ξ² 2 π Ξ±Ξ² In the rest frame of the fluid, it reads as ππ ππ + 2 π πβ² π₯ πΆ Ξ±Ξ² = πβ² π¦ πβ² π§ πβ² ππ 0 0 π₯ πβ² 0 ππ 0 π¦ πβ² π§ 0 0 ππ OKβ¦ I have to admit, my first response to this was β¦ WTH The generalized Ohmβs law π»πΌ πΆ πΆ Ξ±Ξ² Ξ±Ξ² ππ2 1 = ππΌ + βπ ππΌ πΉ Ξ±Ξ² β ππ ππ ππ½ + ππ½ 4π π ππ + ππ πΌ π½ πΌ β‘ ππ + π π + π πβ² π2 ππ¦ 1 ππ₯ ππ§ ππΌ + βπ ππΌ πΉ Ξ±Ξ² = 1, ββπ , ββπ , ββπ π π π π βπ π·πΈ π βπ πΈ+ π×π΅ π = βπ πΈ+ π×π΅ π βπ πΈ+ π×π΅ π 0 βπΈπ₯ βπΈπ¦ βπΈπ§ πΈπ₯ 0 βπ΅π§ π΅π¦ π½ + πβ² πΌ ππ½ + ππ πΞ±Ξ² πΈπ¦ π΅π§ 0 βπ΅π₯ πΈπ§ βπ΅π¦ π΅π₯ 0 = π₯ π¦ ππ ππ ππ½ + ππ½ = ππ π§ ππ π ππ₯ ππ¦ ππ§ βπ π·πΈ π βπ πΈπ₯ + ππ¦ π΅π§ β ππ§ π΅π¦ π βπ πΈπ¦ + ππ§ π΅π₯ β ππ₯ π΅π§ π βπ πΈπ§ + ππ₯ π΅π¦ β ππ¦ π΅π₯ π The generalized Ohmβs law π»πΌ πΆ πΆ Ξ±Ξ² Ξ±Ξ² ππ2 1 = ππΌ + βπ ππΌ πΉ Ξ±Ξ² β ππ ππ ππ½ + ππ½ 4π π ππ + ππ πΌ π½ πΌ β‘ ππ + π π + π πβ² π2 ππ + π»πΌ πΆ Ξ±Ξ² = = π π π π , , , ππ‘ ππ₯ ππ¦ ππ§ ππ π ππ + 2 + π» · π β² ππ‘ π π πβ² π₯ πππ + ππ‘ ππ₯ π¦ πππ π πβ² + ππ‘ ππ¦ π§ πππ π πβ² + ππ‘ ππ§ πβ² πβ² πβ² π½ ππ π2 πβ² π¦ ππ 0 0 π₯ π§ + πβ² πΌ ππ½ + ππ πΞ±Ξ² π₯ πβ² π¦ πβ² 0 ππ 0 0 0 ππ ππ π ππ + 2 + π» · π β² ππ‘ π π₯ ππβ² + π»ππ ππ‘ = ππβ² + π»ππ ππ‘ ππβ² + π»ππ ππ‘ π¦ π§ π§ The generalized Ohmβs lawβ putting it together ππ2 1 π»πΌ πΆ = ππΌ + βπ ππΌ πΉ Ξ±Ξ² β ππ ππ ππ½ + ππ½ 4π π ππ + ππ πΌ π½ πΆ Ξ±Ξ² β‘ ππ + π π + π πΌ πβ² π½ + πβ² πΌ ππ½ + ππ πΞ±Ξ² 2 π βπ I have no idea what ππ π π · πΈ β ππ ππ π ππ + 2 + π» · π β² these two terms meanβ¦ π ππ‘ π Ξ±Ξ² The GR version of the generalized Ohmβs law in the βLocal frame of the fluidβ ππβ² + π»ππ ππ‘ ππβ² + π»ππ ππ‘ ππβ² + π»ππ ππ‘ The generalized Ohmβs law in classical theory. π₯ π¦ π§ ππ2 = 4π βπ πΈ+ π×π΅ π βπ πΈ+ π×π΅ π βπ πΈ+ π×π΅ π π₯ β ππ π π₯ π¦ β ππ π π¦ π§ β ππ π π§ In the local frame, π£ = 0 π π½ ππ ππ ππ 2 ππ ππ ππ 2 + π»· ππ= πΈ+π£ ×π΅ + π½ ×π΅ β π· π½ ππ‘ ππ ππ ππ ππ Pressure Gradient Plasma motion Hall Resistive Comparison with the 3D version π π½ ππ ππ ππ 2 ππ ππ ππ 2 + π»· ππ= πΈ+π£ ×π΅ + π½ ×π΅ β π· π½ ππ‘ ππ ππ ππ ππ Convective terms that were ignored in Plasma class. In a general frame, the π£ × π΅ term comes back. The terms in the Generalized Ohmβs Law The GR version of the generalized Ohmβs law in the βLocal frame of the fluidβ ππ π ππ + 2 + π» · π β² ππ‘ π π₯ ππβ² + π»ππ ππ‘ ππβ² + π»ππ ππ‘ π¦ ππβ² + π»ππ ππ‘ π§ Spatial current Lorentz enhanced version of the spatial current ππ2 = 4π βπ π · πΈ β ππ ππ π π π₯ βπ πΈ+ π×π΅ β ππ ππ₯ π βπ πΈ+ π×π΅ π βπ πΈ+ π×π΅ π π β‘ π΄ ππ β² π π¦ β ππ ππ¦ π§ β ππ ππ§ π£ π π£ π3 π£ π β‘ π΄ ππ πΎ π£ π π£ π 3 π£ π The terms in the Generalized Ohmβs Law The GR version of the generalized Ohmβs law in the βLocal frame of the fluidβ ππ π ππ + 2 + π» · π β² ππ‘ π π₯ ππβ² + π»ππ ππ‘ ππβ² + π»ππ ππ‘ π¦ ππβ² + π»ππ ππ‘ π§ ππ2 = 4π βπ π · πΈ β ππ ππ π π π₯ βπ πΈ+ π×π΅ β ππ ππ₯ π βπ πΈ+ π×π΅ π βπ πΈ+ π×π΅ π π¦ β ππ ππ¦ π§ β ππ ππ§ ππ 2 ππ ππ π 2 β‘ 4ππ΄ β 4π ππ π ππ 4π π 1 βπ β‘ 2 π΄ ππ | π π | β π π ππ | π| π π π πcoll ππ β‘ 4π π2 π ππ2 Plasma frequency Hall coefficient Resistivity These terms can be determined through the equation of states: ππ = ππ π, π ππ = ππ π, π ππ = ππ π, π ππ = ππ π, π βπ = βπ π, π Full stress-energy tensor for gas π Ξ±Ξ² gas = π Ξ±Ξ² fluid + π Ξ±Ξ² Conduction + π Ξ±Ξ² Viscosity π 1 π½ = π + p + 2 π πΌ ππ½ + πΞ±Ξ² π + 2 πππΌ ππ½ + π πΌ ππ π π π Ξ±Ξ² π Ξ±Ξ² fluid π+π 0 0 0 = Viscosity = Οc 2 + ππ π Ξ±Ξ² gas = πππ₯ π¦ ππ πππ§ 0 0 0 0 0 π 0 0 0 0 π 0 0 0 0 π 0 xx β2ππ£,π π΄ β ππ£,π π© β2ππ£,π π΄ yx β2ππ£,π π΄ zx π Ξ±Ξ² Conduction + β2ππ£,π π΄ Ξ±Ξ² β ππ£,π π©πΞ±Ξ² = 0 0 πππ₯ π¦ ππ πππ§ π¦ πππ₯ 0 ππ 0 πππ§ 0 0 0 0 0 0 0 0 xy β2ππ£,π π΄ β2ππ£,π π΄ yy β ππ£,π π© β2ππ£,π π΄ zy π¦ β2ππ£,π π΄ xz β2ππ£,π π΄ yz β2ππ£,π π΄ zz β ππ£,π π© πππ₯ β2ππ£,π π΄ xx β ππ£,π π© + ππ ππ β2ππ£,π π΄ xy πππ§ β2ππ£,π π΄ xz β2ππ£,π π΄ yx β2ππ£,π π΄ zx β2ππ£,π π΄ yy β ππ£,π π© + ππ β2ππ£,π π΄ zy β2ππ£,π π΄ yz β2ππ£,π π΄ zz β ππ£,π π© + ππ Structure similarity with the energymomentum tensor 2 π 1 π Ξ±Ξ² π»πΌ πΆ = ππΌ + βπ ππΌ πΉ Ξ±Ξ² β ππ ππ ππ½ + ππ½ 4π π ππ + ππ πΌ π½ Ξ±Ξ² πΌ π½ πΌ π½ Ξ±Ξ² πΆ β‘ ππ + π π + π πβ² + πβ² π + π π π π2 ππ ππ + π 2 πβ² π₯ πβ² π¦ πβ² π§ In the rest frame of the fluid, πΆ Οc 2 + ππ π Ξ±Ξ² gas = πππ₯ π¦ ππ πππ§ Ξ±Ξ² = πβ² πβ² πβ² π₯ π¦ π§ ππ 0 0 0 ππ 0 π¦ 0 0 ππ πππ₯ β2ππ£,π π΄ xx β ππ£,π π© + ππ ππ β2ππ£,π π΄ xy πππ§ β2ππ£,π π΄ xz β2ππ£,π π΄ yx β2ππ£,π π΄ zx β2ππ£,π π΄ yy β ππ£,π π© + ππ β2ππ£,π π΄ zy β2ππ£,π π΄ yz β2ππ£,π π΄ zz β ππ£,π π© + ππ How do we solve them? For simplicity, letβs look at the local frame of the fluid. The generalized Ohmβs law then looks like: β ππ π ππ + 2 + π» · π β² ππ‘ π π₯ ππβ² + π»ππ ππ‘ ππβ² + π»ππ ππ‘ ππβ² + π»ππ ππ‘ π¦ π§ π ππ2 = 4π π · πΈ β ππ ππ π π π₯ βπ πΈ+ π×π΅ β ππ π π₯ π βπ πΈ+ π×π΅ π βπ πΈ+ π×π΅ π π¦ β ππ π π¦ π§ β ππ π π§ These three equations allow us to take a time step for πβ². This equation allows us to determine πΎπ . (How?) π β² = πΎπ π helps us to find π and π πΌ ππΌ = 0 helps to find ππΌ Then from charge continuity, π»πΌ π½πΌ = 0 and π½πΌ = ππ π πΌ + ππΌ we can update ππ . With the new 4-current, we can update the fields using Maxwellβs equations. Relativistic Hall MHD The generalized Ohmβs law reads as: 2 π 1 π π»πΌ πΆ Ξ±Ξ² = ππΌ + βπ ππΌ πΉ Ξ±Ξ² β ππ ππ ππ½ + ππ½ 4π π 2 ππ However, in most cases, 4π is very large that we can forget about the π»πΌ πΆ Ξ±Ξ² term. This reduces the equation to 1 ππΌ + βπ ππΌ πΉ Ξ±Ξ² = ππ ππ ππ½ + ππ½ π In which we find that the beamed current (πβ²)π½ is gone (it is in the πΆ Ξ±Ξ² term) Therefore, we no longer need πΎπ . This in turn means that the βtβ component of the Ohmβs law which was used to determine πΎπ now is useless. To simplify, we can project out the spatial current ππ½ : ππΌ Ξ±Ξ² ππΎ Ξ±Ξ² πΉ + βπ πΉ πΞ³Ξ± = ππ ππ½ π π In non-relativistic case, itβs simply πΈ + βπ π½ π × π΅ = ππ π½ Relativistic Resistive MHD Another simplification that is often used it to drop the Hall term in the following equation we got just now: ππΌ Ξ±Ξ² ππΎ Ξ±Ξ² πΉ + βπ πΉ πΞ³Ξ± = ππ ππ½ π π Hall term By comparing the Hall and resistive(collision) terms, βπ |ππΌ || π΅ |/π eB 1 ππΏ β = << 1 ππ |ππΌ | ππ π πcoll πcoll we can argue that since many collisions can occur within a Larmor orbit, the term will not contribute muchβ¦???????? π½ π ×π΅ Plugging in the numbers givesβ¦ Ξ½L = 1.76 × 107 B for B~1mG, this is of order 10000β¦. 3.41 Ξ½coll = 3 2 n for T~108 , n~1015 this is of order 1000β¦ T What the hell happened with the ratio!? Relativistic Ideal MHD The last simplification to make is to consider dropping off the resistive term. ππΌ Ξ±Ξ² πΉ = ππ ππ½ π We estimate the conductivity as follows πth β ππ β‘ ππ2 2 1 ππ π = β ππ 4ΟΞ½coll ππ kT ;π ππ π = π2 kT : 3 2 1 2kT β ~2.1 × 108 π β1 π 1.5 2 2 ππ Οrπ πth Οe ππ For π = 106 ~1010 πΎ, this gives ππ = 1017 ~1023 π β1 >>5 × 1016 for copper! Given the extreme lack of electrical resistance in astrophysical plasmas, one can safely ignore the resistivity term, leading to the simple βidealβ Ohmβs law expression ππΌ Ξ±Ξ² πΉ =0 π Relativistic Ideal MHD In the non-relativistic limit and evaluating in the lab frame, UΞ± Ξ±Ξ² F c = 0 becomes π£ π πΈ = β × π΅ which is the familiar ideal Ohmβs law expression. I personally think that it should not only be Ideal MHD that deserves this comment, ππ½ starting from the step when we dropped off the ππ‘ term to get the Hall-MHD, it becomes basically impossible to directly update the currentβ¦ Flux freezing in actual simulations Optically thin Radiative Emission (Ch9.4) Optically thin radiation heat transfer βan Introduction Scattering matter matter ε Ahhhhβ¦ Absroption Ow! Previously, we discuss the heat transfer by radiation by considering the opacity, which means that weβre considering radiation transferring heat within the system β being eaten or scattered within the system. Now we consider photons that are allowed to escape the system and thereby carrying energy with them. And the spectrum remains approximately that of the emission process. . 1 π π»ππ ππ = π» · π π = β π»· β π 3π π π ,bf We discuss the following processes: Inverse Compton Bremsstrahlung Synchrotron Emission The BIG assumption We consider the different types of emission radiated from a thermal distribution of particles! (For example, Maxwellian for a classical gas) π2 β π2 e 2π0kT Bremsstrahlung (Free-Free Emission) π β βπππ; π + β ion πβπβ; π+π+ πβπ+ How do we understand these trends in a simpler manner? Bremsstrahlung (Free-Free Emission) π£π π~ b Assuming that the v magnitude doesnβt change much, π then the time that the charge takes to fly past is π~ π£. The emitted frequency (being the Fourier transform π£ of time) therefore would have a cut off at πcut ~ π. π π£ + π£π π£π Ξv Constructive! For the low frequency waves, the interaction time is much shorter than the inverse of its frequency. Thus, these waves will be coherent with those emitted a small time ago of the same frequency. This leads to dE dΟ β Ξv 2 πcut ~ π£ π Bremsstrahlung (Free-Free Emission) dE β Ξv 2 dΟ π£π b Now, we can estimate the change in velocity Ξv by integrating over the acceleration history of the particle β + Ξv = π£π π£π Ze2 πdt π ββ Ξv This gives dE dΟ π2 + vt 2 3 = 2ππ 2 πππ£ 1 β πππ£ 1 β π2 π2 π£ 2 Since in astrophysics, there are usually a large number of particles, we now consider a flux of particles shooting at stationary targets. ππ ππ‘ The emitted power per unit volume per unit frequency is then πmax dE 1 β π π π£ 2ππdb 2 π2 π£ 2 π‘ π dΟdVdt π πmin Number of incident particles per unit time at a particular impact parameter Thermal Bremsstrahlung dE β dΟdVdt πmax πmin 1 [π π π£ 2ππdb π2 π2 π£ 2 π‘ π β ππ‘ ππ πmax βn π2 π£ πmin β Gaunt factor (π), in general some complicated function of order unity. For a thermal population of particles, for a quick estimate, we can characterized the velocity by the thermal velocity π£th β π π which gives dE ππ‘ ππ β β π dΟdVdt π1.5 π π2 β π2 e 2π0kT Thermal Bremsstrahlungβtotal power dE = dVdt dE dΟ β dΟdVdt ππ‘ ππ β β ππcut 1.5 π π ππ‘ ππ π1.5 β π πdΟ Using the fact that the spectra can be approximated by a box function, π£ πcut ~ π For πcut , the βbβ that gives the highest frequency is obviously the smallest b, i.e. πmin . Here, we can have two cases: 1. The emitting particles have less energy and a large portion is radiated. β π£ mv 2 π 1 πmin = β β β mv π 1 β β min This idea works but seems slightly incompatible to the picture given in Rybicki & Lightman p158β¦ 1.5 1 π£ π 2. The emitting particles are very energetic 2 πmin β β 2 β mv 3 β 2 and only give away a small portion of their mv π πmin kinetic energy to radiation. Thermal Bremsstrahlungβtotal power Power per unit volume 1. The emitting particles have less energy and a large portion is radiated. 1 πmin β π£ mv 2 π = β β β mv π 1 β β min Power per unit volume: dE ππ‘ ππ β β ππ dVdt βπ1.5 Power per unit mass: dE 1 ππ‘ ππ β β ππ dt dmtot ππ ππ βπ1.5 dE dVdt β ππ‘ ππ β ππcut π1.5 π 2. The emitting particles are very energetic and only give away a small portion of their kinetic energy to radiation. 2 πmin 1 π£ π 1.5 3 β β β mv β mv 2 π 2 π min Power per unit volume: dE ππ‘ ππ β β 2 ππ dVdt π Power per unit mass: dE 1 ππ‘ ππ β β Tπ dt dmtot ππ ππ π2 Thermal Bremsstrahlung Case1: Low energy dE 1 ππ‘ ππ β dt dmtot ππ ππ βπ1.5 π β βπππ; π + β ion Case2: High energy dE 1 ππ‘ ππ β β T π dt dmtot ππ ππ π2 Whatβs caused the difference in this case? β π π πβπβ; π+π+ πβπ+ Iβm not quite sure whether the π1.5 causes the difference between 147 and 337 for the π β βπππ; π + β ion v.s. π β π + case. Using the reduced mass gives an overestimate of 20%... Thermal Synchrotron The local synchrotron cooling rate is approximated as a sum of optically thick and thin emission: . ππ β 2ΟkTπ 3 = π + 3Hc 2 π β ππ π ππ ππ Turnover frequency Break frequency *Note that there. is some confusion in the units here, ππ β should be erg/g/s but for the synchrotron part, all units are erg/s/ππ3 I think just dividing by the mean density of the plasma should fix the small bug: β 1 2πkTπ 3 .β ππ = π + ππ π dΞ½ π 3π»π 2 π ππ Optically thick emission Optically thin power law Synchrotron emission Therefore, for the Synchrotron section, we will follow the book but keep in mind the small difference. Thermal Synchrotron βOptically thick part Since we are considering a Thermal distribution of emitting particles, the optically thick emission can be approximated by Rayleigh-Jeans Law Then, calculating the power per unit volume gives: dE π 2πkTπ 3 β π΅π π dΞ½ = π dVdt π» 3Hc 2 π As given in the textbook. 2kT 2 π΅π π = 2 π π . ππ β 2ΟkTπ 3 = π + 3Hc 2 π β ππ π ππ ππ *Note that in most texts, one will see that the intensity of optically thick synchrotron is πΌπ,thick π β π 2.5 . That is derived using a Power Law population of emitters, not thermal (as is assumed here)! Thermal Synchrotron βOptically thin part Normalizing the integration Integrate over relativistic Maxwellian??? Radio astrophysics. Nonthermal processes in galactic and extragalactic sources by Pacholczyk, A. G 1996ApJ...465..327M 1996ApJ...465..312E Thermal Synchrotron βOptically thin part Comptonization β Why we are interested From what we introduced last week about the opacities of electron scattering and absorption, we find from the figure below that electron scattering is more important. Electron Scattering Dominates Comptonization β Inverse Compton In each of these photon scatterings, if the electron and photon energies are very different, an inelastic scattering (the Compton process) will occur, either raising or lowering the photon energy in the process. In most cases we will encounter, the photons will be cooler than the electrons, resulting in raising the energy of the photons and, therefore, cooling the hot, electron-scattering plasma in the process. Comptonization Since Compton scattering itself doesnβt create new photons, what is does is to use the photons emitted by synchrotron and Bremstrahlung to steal more energy from the system. Therefore, one way to write the cooling rate is to multiply some Comptonenhancement factor ππΆ π . .β . π = πbr,πΆ πbr + ππ ,πΆ ππ β Comptonization π = 10β10 π · cmβ3 ; π΅ = 8380πΊ ; π» = 2.7 × 107 cm 109 πΎ β² ππ Compton Enhanced Synchrotron dominated 109 πΎ β² ππ Synchrotron dominated 104 πΎ β² ππ β² 109 πΎ Bremsstrahlung dominated Bremsstrahlung quenched due to recombination of free electrons with ions. 2009ApJ...693..771F Repeated Scattering After the initial emission of Bremsstrahlung or synchrotron photons, if the optical depth πes is high, these photons do not immediately leave the plasma. They remain within the hot medium, gaining energy from most of the inelastic scatterings they encounter. Depending on how large the Compton y parameter is, this can drastically affect the ο¬nal spectrum that emerges from the plasma. Comptonized spectra for different values of Sunyaev and Titarchukβs parameter πΎ β π¦ β1 Wien peak is produced as plasma is Compton thick. Single freq. photon injection π0 = 10β3 kTπ β Compton thin but still produces a power law spectrum up to the characteristic thermal frequency kTπ β As long as the injected photons are much cooler than the hot electron scattering plasma, the nature of the output spectrum from Comptonization depends mainly on y, not on the nature of the input spectrum.
