Bremsstrahlung & Synchotron Radiation

Question 1
In an optically thick source the observed flux is dependent on the projected area and the excitation temperature with a given velocity Vobs (Shaw et al., 2013). Consequently, optically thin sources are also dependent on projected area and excitation temperature, but their flux is calculated by obtaining an integration of the emissivity line over the volume.
Therefore, in this case the observed flux will be given by fv=1-e-t2hv4 c31ehv/kTex-14πr2/d2 EQ 1
Where Tex represents the excitation temperature.
In an optically thin source Tv << 1
J(k)T which is the emission coefficient can be given by hvk÷ehvkt-1Therefore, the middle equation in EQ 1 can be represented with Jv assuming it is a free path.
With respect to unit mass Jv can be given by Ev⍴/4πUsing the optical length Ev⍴/4πThe equation can be differentiated to
dEm=Ev4πdvdmdt
dm=⍴dV ………(Equation 2)
Where ⍴ is the density.
Thus, equation one can be written as fv=1-e-t2hv4 c3Ev⍴4π4πr2/d2 ….(Equation 3)
Question 2
The Flux Density Fv=1-e-t2hv4 c31ehv/kTex-14πr2/d2 as in Q1
Whereby,
1ehv/kTex-1 = Ev⍴4πTherefore
Fv=1-(1.602*10-19)-12*1*1(3.0*108)3 Ev⍴4π 4πr2/d2Which is
Fv=1-(1.602*1019)2(3.0*108)3 ⍴dV 4πr2/d2=-1.068×1043 X1011x100x4π10021002=-1.342088x1030JyB)

Question 3
The Lorentz factor ϒ has a velocity of V=ϒv(1+ϒ)In a constant accelerating the relativistic motion is obtained by ω=vr=eBϒMeTherefore, the acceleration ω can be given by ω=ebBϒme Ω with respect to frequency is given by 2πf;
Thus, this equates to 2πf=ebBϒme Where f=1Hz, assuming mass of the electron is 1, the magnetic flux will be given by
B=2πfϒebb=v/c, where V is the velocity vector of a charge which in this case will be assumed to be |1|
B=2πx1x10003.

0x108Hence, B= 1.0472×10-5ii) Frequency is given by f=eBϒmec which equates to f=1.0472×10-51000×3.0x108hence, f=3.4900667×10-17iii) The radius is denoted by r=ϒmec2beBWhich can be written as r=ϒmec1eBvTherefore; r=1000×3.0x1081.0472×10-5Hence; r=2.864782x1016iv) the cone angle or the pitch angle is given by sina=ωϒme/ebB ω=2π radians=2.261x103thus sina=2.261x 103×10001.0472×10-5Therefore, sina=0.2772˚Question 4
Flux Density = 2.0 JyFrequency = 5GHz
Angular Radius 0.15 sec
Linear Radius 3.5 kpc.
Emission Co-efficient?
jv=ℇve4π ℇv=15
FV=2hv4c3 ℇve4π27r4R2 πR2Equipartition magnetic flux density?
Where
Beq=(8πAV)˄(27)
BME = (6πAV) ˄(27)
II.
BME = (6πAV) ˄(27)
=(6πAV) ˄(27)
Volume = 4πr3=583.853
Intensity = UvΩ=IvΩcQuestion 5
In this case, the operator position
xψx=x2Lsin3πxl≠cψxC = constant
In this case, it is not an eigenfunction.
The fourth postulate provides the expectation value. It is
=2lol{sin⁡(3πl)}xsin3πldx=2l0l3x{sin3π}dxUse standard Integral
0l3x(sin3x)2dx=Given the integral
P=2ll/32a/3sin2πxldxP=2l[l6-l4π(sin4π3-sin2π3)]=0.609Sketch of wave function
Recalling from the notes in probability of particles,
Between x=l1and x=l2for ψ3xisPl1l2=l2-l1L+16πsin6πl2L-sin6πl2Lwhere B=2/LSketch

Question 6
The distance of the closest approach given the coulombic fields is;
d=14πℇ0Z1Z2mv2Then calculate the distance by the two protons given;
Energy = 1.5×107 K
Velocity V= 3×10˄5
Mass m = 1.7×10˄-27 Kg.
The total energy expected from the system;
12m1v12+14πℇ0z1z2r+12m2v22+14πℇ0z1z2r=Constant Closest approach = Kinetic Energy = 0 K
mv2+ 12πℇ0z2r = Constant
Constant = 12πℇ0z2r =mv2d=r=12πℇ0z2mv2 In this case,
=[1/(2πx1.5×107)] x [(1.7×10˄-27) x (3×10˄5) / ((1.7×10˄-27) x (300000) ˄2)
194247779.6085.1×10-229x1010D= 6.01252×10-41Height of the Coulomb barrier given radius is 5X10-15r=r0A1/3 Where r= 5X10-15 and r0=1.2 fm and A is mass=1.7×10˄-27.
One major assumption is that the square top of the height has a full with which is unrealistic.
Tunnelling Probability
T=exp⁡(-22mUO-Eah
Reference
Shaw, M. S., Romani, R. W., Cotter, G., Healey, S. E., Michelson, P. F., Readhead, A. C., … & Potter, W. J. (2013). Spectroscopy of the largest ever γ-ray-selected BL Lac sample. The Astrophysical Journal, 764(2), 135.