Modeling High-Energy Gamma-Rays from the Fermi Bubbles Megan Slettstoesser SLAC-TN-15-087 August 015 Abstract In 010, the Fermi Bubbles were discovered at the galactic center of the Milky Way. These giant gamma-ray structures, extending 55 in galactic latitude and 0-30 in galactic longitude, were not redicted. We wish to develo a model for the gamma-ray emission of the Fermi Bubbles. To do so, we assume that second order Fermi acceleration is resonsible for the high-energy emission of the bubbles. Second order Fermi acceleration requires charged articles and irregular magnetic fields both of which are resent in the disk of the Milky Way galaxy. By solving the steady-state case of the transort equation, I comute the roton sectrum due to second order Fermi acceleration. I comare the analytical solutions of the roton sectrum to a numerical solution. I find that the numerical solution to the transort equation converges to the analytical solution in all cases. The gamma-ray sectrum due to roton-roton interaction is comared to Fermi Bubble data from Ackermann et al. 014, and I find that second order Fermi acceleration is a good fit for the gamma-ray sectrum of the Fermi Bubbles at low energies with an injection source term of S = 1.5 10 10 GeV 1 cm 3 yr 1. I find that a non-steady-state solution to the gamma-ray sectrum with an injection source term of S = 10 10 GeV 1 cm 3 yr 1 matches the bubble data at high energies. 1. Introduction The Fermi Large Area Telescoe LAT began to observe the gamma-ray sky at energies over 30 MeV in 008 Atwood et al. 009. Until 010, the gamma-ray sky was known to contain oint sources and diffuse emission. Point sources may be galactic e.g., suernovae, ulsars, binary systems or extragalactic e.g., active galactic nuclei AGN, gamma-ray bursts GRBs, blazars. Diffuse emission is more extended than the angular resolution of the instrument; therefore, it is not considered a oint source. The diffuse emission of the Milky Way galaxy is rimarily roduced by cosmic rays SULI summer intern at Kavli Institute for Particle Astrohysics and Cosmology Home Institution: University of California, Santa Cruz CRs interacting with the interstellar radiation field ISRF or the interstellar medium ISM Strong, Moskalenko, Ptuskin 007. CRs are taken to be redominantly rotons and helium. In 010, however, an addition to the diffusion emission was observed: an unexected gamma-ray structure, termed the Fermi Bubbles Su, Slatyer, Finkbeiner 010. The structure extends 55 in galactic latitude Ackermann et al. 014 and 0-30 in galactic longitude. The bubbles are symmetric with resect to longitude and with resect to the galactic center GC of the Milky Way galaxy. The hard sectrum and high-energy gamma-rays of SLAC National Accelerator Laboratory, 575 Sand Hill Road, Menlo Park, CA 9405 This material is based uon work suorted by the U.S. Deartment of Energy, Office of Science, Office of Workforce Develoment for Teachers and Scientists WDTS under the Science Undergraduate Laboratory Internshi SULI rogram, under Contract No. DE-AC0-76SF00515.
the bubbles are only visible at higher energies. Gamma-ray roduction in the galactic ridge is dominated by Inverse Comton IC scattering, bremsstrahlung emission, and neutral ion decay Strong, Moskalenko, Ptuskin 007; once the gamma-ray emission from the aforementioned rocesses is subtracted from the gammaray sky, the Fermi Bubbles are visible. The ISRF allows for the resence of soft gamma-rays, which creates an environment for IC scattering. A gamma-ray is roduced in IC scattering as a high-energy electron collides with a low-energy hoton γ soft ; in this rocess, the electron transfers some of its energy to the hoton, and as a result, the hoton gains energy. Below is the IC rocess: e CR + γ soft e CR + γ. 1 The cosmic ray electron e CR on the right-hand side RHS of Equation 1 has lost energy and the gamma-ray γ on the RHS has gained energy. Bremsstrahlung emission is caused by the acceleration of an electron in the resence of a stationary nucleus or ion due to the nucleus s or ion s electric field: e CR + N gas e CR + N gas + γ. The cosmic ray electron on the RHS of Equation has lost energy. The gas nucleus N gas on the RHS does not lose energy in the interaction because it is heavy. Neutral ions decay into two gamma-rays: π 0 γ + γ. Neutral ions are roduced by the interaction of cosmic ray rotons CR with a gas nucleus: CR + N gas π 0 + X. 3 In Equation 3, X reresents an arbitrary roduct of the reaction aside from the neutral ion. Examine a more secific case of 3, as shown below: CR + N gas π 0 + +. Take the gas nucleus to be a roton gas. To model the gamma-ray emission of the Fermi Bubbles, we assume that the emission is due to second order Fermi acceleration, an acceleration mechanism in which charged articles interact with moving irregular magnetic fields. In a head-on collision, the article travels with a velocity directionally oosite to the velocity of the magnetic irregularity before collision; the charged article gains energy after this interaction. Conversely, in a following collision, the article and the magnetic irregularity travel in the same direction before collision and the charged article loses energy Fermi 1949. Within the magnetic irregularity, articles diffuse, or undergo a continuous random walk. On average, second order Fermi acceleration causes articles to gain energy because head-on collisions are favored. The energy gain of the articles in our case, rotons, is second order in V in other words, E [ V c c ], where V is the velocity of the magnetic field and c is the seed of the article Longair Vol.. I begin with a solution to the heat equation in Section ; the heat equation is a simlified version of the transort equation. Next, I solve the transort equation for second order Fermi acceleration in Section 3. The roton sectrum given by the analytical solution to the transort equation is comared to the numerical solution of the roton sectrum rovided by Dr. P. Mertsch in Section 4. In Section 5, we comare the numerical solution of the gamma-ray sectrum of the Fermi Bubbles due to roton-roton interaction to data from Ackermann et al. 014.. Analytical Solution to the Heat Equation Examine the most general transort equa-
tion in momentum-sace: f x,, t t = D f x,, t vf x,, t + 1 D f x,, t + 1 ṗ f x,, t v f x,, t. 4 3 τ The first term on the RHS of Equation 4, D f x,, t, describes satial diffusion, the second accounts for satial advection, the third models momentum diffusion, the fourth describes energy losses, the fifth describes adiabatic losses and gains, and the last term models article escae and other losses e.g., decay. Take a secific case of Equation 4 the heat equation given by: fx, t = D fx, t. 5 t x Note that Equation 5 models satial diffusion. The diffusion coefficient D is indeendent of x and t. Assume the following initial condition: fx, 0 = f 0 x. To solve Equation 5 over an infinite interval, emloy Fourier transforms. Denote the Fourier transform of fk, t as fx, t, and define the inverse Fourier transform of fx, t as fk, t, as shown below: fx, t = 1 fk, t = 1 e ikx fk, t dk 6 e ikx fx, t dx. 7 The Fourier transform of the heat equation Equation 5 is given by: d dt fk, t + Dk fk, t = 0. 8 Note that Equation 8 is a searable ordinary differential equation involving differentiation with resect to a single variable. Solving Equation 8 yields: fk, t = f 0 ke Dkt. 9 Substitute the inverse Fourier transform given in Equation 9, into Equation 6, which defines the Fourier transform. The solution to the resulting equation is given by: fx, t = 1 4πDt dx x x f 0 x ex 4Dt 10 The mathematical details leading u to Equation 10 can be found in Aendix A. We see that fx, t is a function of its initial condition. If the initial condition, f 0 x, was given by a Dirac delta function, the solution to fx, t would be as follows: fx, t = 1 4πDt ex x x 4Dt dx 3. Analytical Steady-State Solution to the Transort Equation Second order Fermi acceleration is a urely stochastic rocess; therefore, given the following transort equation in momentum-sace Becker, Le, & Dermer 006 [3] : f, t t = 1 ] Af, t [ f, t D f, t S, t + τ esc 4π, 11 A, which reresents systematic acceleration, is zero. Take f, t to be the hase sace density, D to be the momentum diffusion coefficient this causes energy gains and losses, t esc to be the mean article escae time, and S, t to be a source term Becker, Le, & Dermer 006 [3]. If we take the source term to be a Dirac delta function, the transort equation 11 for second order Fermi acceleration becomes f t = 1 D f f τ esc + S δ 0. 1 4π The first term on the right-hand side RHS reresents momentum diffusion/transort. The coefficient of the source term, S, gives the number 3
of articles injected er GeV er cm 3 er year. Because it will be easier to work with the article sectrum, note the following: Ψ = 4π f f = Ψ 4π. 13 Ψ is the article density and f is the hase sace density. Solving for the steady-state solution, in which f = 0, and using Equation t 13, Equation 1 becomes: 0 = D Ψ Ψ τ esc + Sδ 0. 14 Let Ψ = Ψ. Evaluating the artial derivatives in Equation 14 yields the following steady-state transort equation: 0 = D Ψ + D D Ψ D + D 1 Ψ τ esc + Sδ 0. 15 Now solve for the q = case, which gives the following assumtions: q D = D 0 = 1 q 0 q= == D = 1 τ acc,0 τ esc = τ acc,0 0 16 0 q q= == τ esc = 17 as, Ψ remains infinite. 18 Equation 17 shows that the q = case gives a momentum-indeendent escae time. Noting the momentum-indeendent escae time in the q = case Equation 17 and the relationshi 0 between the diffusion coefficient and momentum Equation 16, and 0 = 1, Equation 15 becomes: 0 = Ψ 1 + τ acc,0 Ψ + δ 0τ acc,0. 19 Examine the homogeneous case of Equation 19, which is linear in Ψ: Ψ 1 + τ acc,0 Ψ = 0. 0 The solution of Ψ a Ψ = 0, where a is a constant, is given by: Ψ = A k 1 + B k, 1 where k 1 and k are the roots of k k a = 0. Solving for the roots: 0 = k k k = 1 ± + τ acc,0 1 + 4 + τ acc,0 k = 1 ± 9. Note that the times τ acc,0 and are always 9 zero or ositive. Therefore, is always ositive. The smallest value 9 can take on, in the case of τ acc,0 = 0, is 3. Thus, by Equation, we have one ositive root and one negative root. However, to satisfy Equation 18, Equation 1 must only contain the negative root. Droing the magnitude of in Equation 1 because the momentum is always ositive, we have: 1 Ψ = A 9, 3 which gives the steady-state solution to the transort equation for Ψ, the article density. Solving the transort equation for f using Equation 13 gives: f 3 9 4 4
A detailed account of this solution is given in Aendix B. We see that for the q = case, the hase sace density is only deendent on the ratio between the acceleration and escae times of the article. Examine the case of τ acc,0 = 4: f 4 Multilying f by 4, as I have done in Figures 1a and 1b, roduces a flat sectrum. 4. Proton Sectra: Analytical and Numerical Solutions The roton sectrum describes the hase sace density f at a fixed satial radius. The numerical solution to the transort equation Equation 1, given by a code by Dr. P. Mertsch which emloys a finitedifference aroximation, can be tested using the analytical solution Equation 4. In Figure 1, the roton sectrum Equation 4 is lotted with the numerical solution. Note that the analytical solution Equation 4 is only valid for > 0. We see that the numerical solution converges to the analytical solution for > 0 in Figure 1b. The analytical solutions to the transort equation given by Becker, Le, & Dermer [3] are valid for all values of. Note that the steadystate solutions to the transort equation given by N G ss q= in Figure and N G ss q< in Figure 3 are Equations [58] and [56], resectively, in Becker, Le, & Dermer [3]. The equations are reroduced below. Note that Ψ and N both reresent article density. Equation [58]: a+1/ λ Nssx G Ṅ 0 x x q= = max. x 0 D λ x 0 x min 5 Equation [56]: N G ssx = Ṅ 0 qx 0 D I β1/ θx q min q a+1/ x xx 0 q/ x 0 θx q max K β1/. q 6 See Becker, Le, & Dermer [3] for details. In Figures b and 3b, we see that the numerical solution converges to the steady-state analytical solution. The numerical solution for the q < case converges to the analytical solution more quickly than the q = case. 5. Gamma-Ray Sectra: Numerical Solution and Fermi Bubble Data The code for the gamma-ray flux due to interaction may be found at htts://github.com/niklask/caramlib. The gamma-ray sectrum is comuted from the gamma-ray flux. The Fermi Bubble data is from Ackermann et al. 014. We wish to find the source term, S, in Equation 1 such that the solution to the transort equation matches the Fermi Bubble data. The numerical solution matches Fermi Bubble data well for lower energies Figure 4a with a source term equal to 1.5 10 10 GeV 1 cm 3 yr 1. Figure 4b is a good fit at high energies for the data in the nonsteady-state case; here, we find a source term of S = 10 10 GeV 1 cm 3 yr 1. 6. Conclusion I solve the transort equation for second order Fermi acceleration and find that the numerical solution converges to the analytical solution. Further, the numerical solution converges to analytical solutions to the transort equation rovided by Becker, Le, & Dermer [3]. The gamma-ray sectra roduced by rotons undergoing second order Fermi acceleration is a good fit for the Fermi Bubbles. The Fermi Bubble 5
a The numerical solution green after a simulated time of 0.3 years. b The numerical solution after a simulated time of 15 years. Figure 1: The roton sectrum given by Equation 4 blue lotted with the numerical solution green in momentum-sace for the steady-state, τ acc,0 = 4 and q = case. a The numerical solution green after a simulated time of 0.6 years. b The numerical solution green after a simulated time of 14.7 years. Figure : The roton sectrum light blue given by Equation 5 lotted with the numerical solution green for steady-state, τ acc,0 = 4 and q = case. Equation 5 is lotted in momentum-sace using the same arameters and assumtions as the numerical solution. data at high energies suggest that the bubbles may not be roduced by a steady-state rocess. Uncertainties exist due to the unknown gas density of the Fermi Bubbles. We find a source term of S = 1.5 10 10 GeV 1 cm 3 yr 1 for lower energies of the Fermi Bubbles and a steady-state gamma-ray sectrum. We find a source term of S = 10 10 GeV 1 cm 3 yr 1 for high energies of the bubbles and non-steady-state solution to the gamma-ray sectrum. 7. Acknowledgements This work was suorted in art by the U.S. Deartment of Energy, Office of Science, Office 6
a The numerical solution after a simulated time of 0.3 years. b The numerical solution after a simulated time of 1.53 years. Figure 3: The roton sectrum dark red lotted with the numerical solution green for τ acc,0 = 4 and q. Equation 6 is lotted with the arameters as the numerical solution. of Workforce Develoment for Teachers and Scientists WDTS under the Science Undergraduate Laboratory Internshis Program SULI and was made ossible thanks to the guidance of Dr. Phili Mertsch. The author would also like to thank Enrique Cuellar for making the SULI interns feel welcome. 8. References [1] Ackermann et al. 014. AJ 793 64 The Sectrum and Morhology of the Fermi Bubbles. [] Atwood, W. B. et al. 009. AJ 697, 1071 [3] Becker, Peter A.; Le, Truong; Dermer, Charles D. 006. AJ 647, 539 Time- Deendent Stochastic Particle Acceleration in Astrohysical Plasmas: Exact Solutions Including Momentum-Deendent Escae. [4] Fermi, E. 1949, Phys. Rev., 75, 1169 [5] Kamae, Tuneyoshi; Karlsson, Niklas; Mizuno, Tsunefumi; Abe, Toshinori; Koi, Tatsumi. 006. AJ 647, 69 Parametrization of γ, e ±, and Neutrino Sectra Produced by Interaction in Astronomical Environments. [6] Karlsson, Niklas. htts://github.com/niklask/caramlib. [7] Longair, Malcolm S. High Energy Astrohysics nd Ed. Vol 1: Particles, hotons, and their detection. Cambridge University Press, New York, New York, 199. [8] Longair, Malcolm S. High Energy Astrohysics. nd Ed. Vol. : Stars, the Galaxy, and the interstellar medium. Cambridge University Press, New York, New York, 1994. [9] Strong, Andrew W.; Moskalenko, Igor V.; Ptuskin, Vladimir S. Annual Review of Nuclear and Particle Science. 007. Vol 57, 85 [10] Su, Meng; Slatyer, Tracy R.; Finkbeiner, Douglas P. 010. AJ 74, 1044 Aendix In this aendix I solve the heat equation Section and transort equation Section 3 in detail. 7
a A source term of S = 1.5 10 10 GeV 1 cm 3 yr 1 after 14.4 years. b A source term of S = 10 10 GeV 1 cm 3 yr 1 after 6.3 years. Figure 4: Fermi Bubble data red Ackermann et al. 014 lotted with numerical solution blue of the gamma-ray sectrum due to roton-roton interaction. The data matches the numerical solution well for lower energies Figure 4a. Figure 4b is a good fit at high energies for the data in the non-steady-state case. Note that the errors bars show statistical errors. A. Analytical Solution to the Heat Equation Examine the heat equation in osition-sace, given by fx, t = D fx, t, 7 t x The coefficient D is indeendent of x and t. Assume the following initial condition: fx, 0 = f 0 x. To solve Equation 7 over an infinite interval, take the Fourier transform of the heat equation. Denote fx, t as the the Fourier transform of fk, t and define fk, t as the inverse Fourier transform of fx, t. fx, t = 1 fk, t = 1 e ikx fk, t dk 8 e ikx fx, t dx. 9 Substitute Equation 8 into the left-hand side LHS of Equation 7 using Equation 8. Note that e ikx is time-indeendent. Therefore, t fx, t = 1 e ikx fk, t dk t = 1 t fk, t e ikx dk 30 Now substitute Equation 8 into the right-hand side RHS of Equation 7. Note that fk, t is 8
osition-indeendent. Thus, fx, t = x x = 1 1 = 1 fk, t fk, t e ikx fk, t dk x eikx k e ikx dk dk 31 Therefore, using the artial derivatives given by Equation 30 and Equation 31, Equation 7 becomes: 0 = fx, t D fx, t t x = 1 t fk, t e ikx dk + D = 1 k fk, t e ikx dk fk, t + Dk fk, t e ikx dk t 0 = t fk, t + Dk fk, t 3 Equation 3 reresents the Fourier transformed heat equation. Note that it is an ordinary differential equation and can be written as: Solving Equation 33 gives: 0 = d dt fk, t + Dk fk, t 33 d fk, t = Dk fk, t dt 1 d fk, fk, t = Dk dt t fk,t d fk, t t f 0 k fk, t = Dk dt 0 fk, t ln = Dk f t 0 k fk, t ex ln = ex Dk f t 0 k fk, t = f 0 ke Dk t 34 9
Using the inverse Fourier transform given by Equation 34 in Equation 8 to solve for fx, t: fx, t = 1 e ikx f 0 ke Dk t dk Using Equation 9 to define fk, 0, = 1 f 0 k = 1 f 0 ke Dtk ixk dk. 35 Substituting Equation 36 in Equation 34, obtain: fx, t = 1 ex Dtk + ix + x k f 0 x e ikx dx 36 f 0 x dk dx. 37 Examine the following Gaussian integral, ex ax + bx dx = π b a ex 4a 38 Integrating Equation 37 with resect to k using Equation 38: fx, t = 1 = 1 = f 0 x dx ex Dtk ix x k dk π Dt ex 1 4πDt f 0 x 39 x x dx 4Dt x x f 0 x ex 4Dt B. Analytical Solution to the Transort Equation dx. 40 Second order Fermi acceleration is a urely stochastic rocess; therefore, given the following transort equation in momentum-sace Becker, Le, & Dermer 006 [3] : f, t = 1 [ ] f, t f, t S, t D t Af, t + τ esc 4π, 41 A, which reresents systematic acceleration, is zero. Take f, t to be the hase sace density, D to be the momentum diffusion coefficient this causes energy gains and losses, t esc to be the mean article escae time, and S, t to be a source term Becker, Le, & Dermer 006; [3]. If we 10
take the source term to be a Dirac delta function, the transort equation Equation 41 for second order Fermi acceleration becomes f = 1 D t f f τ esc + S δ 0 4 4π The first term on the right-hand side RHS reresents momentum diffusion/transort. The coefficient of the source term, S, gives the number of articles injected er GeV er cm 3 er year. Because it will be easier to work with the article sectrum, note the following: Ψ = 4π f f = Ψ 4π 43 Ψ is the article density and f is the hase sace density. Solving for the steady-state solution, in which f = 0, and using Equation 43, Equation 4 becomes: t 0 = 1 D Ψ Ψ 4π 4π τ esc + S δ 0 4π [ = 1 D 4π ] Ψ Ψ τ esc + Sδ 0 = D Ψ Ψ τ esc + Sδ 0 44 Let Ψ = Ψ. Evaluating the inner artial derivative of the first term in Equation 44 using the roduct rule gives: Let Ψ = Ψ Ψ 45 3 Using the artial derivative defined in Equation 45, Equation 44 becomes: 0 = D Ψ D Ψ Ψ τ esc + Sδ 0 46 D = D. Invoking the roduct rule on the first term in Equation 46 roduces: and D Ψ D Ψ = D Ψ + D Ψ 47 D Ψ = + D Ψ D Ψ Using Equation 47 and Equation 48, Equation 46 becomes: 0 = D Ψ + D D Ψ D + D 11 48 1 Ψ + Sδ τ 0 esc 49
Solve for the q = case and assume the following: D = D 0 = Ψ 1 + τ acc,0 0 q= == D = 1 τ acc,0 τ esc = q = 1 τ acc,0 0 q 50 0 q 0 q= == τ esc = 51 as, Ψ remains infinite 5 Equation 51 shows that the q = case gives a momentum-indeendent escae time. Noting Equation 50, Equation 49 becomes: 0 = Ψ τ + Ψ acc,0 τ acc,0 τ + 4 1 Ψ + Sδ acc,0 τ acc,0 τ acc,0 τ 0 esc,0 = Ψ τ + 1 Ψ + Sδ acc,0 τ acc,0 τ 0 esc,0 Ψ + S δ 0τ acc,0 53 Examine the homogeneous case of Equation 53, which is linear in Ψ: Ψ 1 + τ acc,0 The solution of Ψ a Ψ = 0, where a is a constant, is given by: where k 1 and k are the roots of k k a = 0. Solving for the roots: Ψ = 0. 54 Ψ = A k 1 + B k 55 0 = k k k = 1 ± + τ acc,0 1 + 4 + τ acc,0 k = 1 ± 9 56 Note that the times τ acc,0 and are always zero or ositive. Therefore, 9 is always 9 ositive. The smallest value can take on, in the case of τ acc,0 = 0, is 3. Thus, by Equation 56, we have one ositive root and one negative root. However, to satisfy Equation 5, 1
Equation 55 must only contain the negative root. Droing the magnitude of in Equation 55 because the momentum is always ositive, we have: 1 Ψ = A 9, 57 which gives the steady-state solution to the transort equation for Ψ, the article density. Solving for f using Equation 43: f = A 1 9 4π f = A 4π 1 9 f 3 9 58 13