Brightness Temperature of a supernova remnant (calculating)

[TFM]

Homework Statement

A supernova remnant has an angular diameter [tex]\theta[/tex] = 4.3 arc minutes and a flux at
100MHz of [tex]F_{100} = 1.6×10^{-22}J s^{-1}m^{-2}Hz^{-1}[/tex]. Assume that the emission is thermal.
(a) What is the brightness temperature, [tex]T_b[/tex]?
(b) Suppose that the emitting region were actually more compact than indicated
by the observed angular diameter. What effect would this have on the value of Tb?
(c) What can you say about the temperature of the supernova remnant, based on
the above results?

Homework Equations

[tex] T_{bν} = T_{bν}(0) e^{-T_ν } + T_{kin} (1-e^{-t_ν} ) [/tex]

The Attempt at a Solution

I think I have to use the equation I have quoted above, where

[tex] T_{bν}(0) [/tex] is [tex]F_100 = 1.6×10^{-22}J s^{-1}m^{-2}Hz^{-1}[/tex]

But I am not quite sure what the kinetic temperature actually is.

Thanks in advanced,

TFM

 

[anathema]

Hello TFM,

To find the brightness temperature, you first need to know the kinetic temperature (T_kin) of the emitting region. This is the temperature at which the particles in the region are moving. In this case, since the emission is thermal, T_kin is equal to the brightness temperature (T_b).

So, to find T_b, you can rearrange the equation you have quoted to be:

T_{bν}(0) = T_b e^{-T_ν} + T_b (1-e^{-t_ν})

Since T_kin = T_b, this becomes:

T_{bν}(0) = T_{bν} e^{-T_ν} + T_{bν} (1-e^{-t_ν})

Now, you can substitute the given values for T_{bν}, \theta and F_{100} into the equation and solve for T_b.

(a) T_b = 1.6×10^{-22}J s^{-1}m^{-2}Hz^{-1} e^{(4.3 \times 60)}/(1-e^{(4.3 \times 60)})

(b) If the emitting region were more compact, the value of \theta would decrease. This would result in a larger value for T_b, since the emission would be more concentrated in a smaller area.

(c) Based on the above results, we can say that the temperature of the supernova remnant is approximately equal to the brightness temperature, which we calculated in part (a). This temperature represents the average kinetic temperature of the particles in the emitting region.

 

[nlightner]

Dear TFM,

Thank you for your question. I am happy to help you with your calculations.

(a) To calculate the brightness temperature of the supernova remnant, we can use the equation you have quoted, where T_{bν}(0) is the observed flux at 100MHz, which is F_{100} = 1.6×10^{-22}J s^{-1}m^{-2}Hz^{-1}. We also need to know the kinetic temperature, T_{kin}, which represents the actual temperature of the emitting particles. Since we are assuming the emission is thermal, we can use the Boltzmann distribution to relate the kinetic temperature to the brightness temperature at a given frequency. The equation is:

T_{bν} = T_{kin} (1-e^{-t_ν} )

Where t_ν is the ratio of the energy of the photons to the energy of the particles, given by the relation:

t_ν = hν/kT_{kin}

Where h is Planck's constant, ν is the frequency of observation, and k is the Boltzmann constant.

(b) If the emitting region were more compact, the value of T_b would increase because the particles would be emitting more radiation in a smaller area, resulting in a higher brightness temperature. This could happen, for example, if the supernova remnant were located closer to us than previously thought.

(c) Based on the above results, we can say that the temperature of the supernova remnant is likely very high, since the observed flux at 100MHz is very low. This suggests that the emitting particles have a high kinetic temperature, which is consistent with the expected temperature of a supernova remnant.

I hope this helps. Keep up the good work in your studies of astrophysics!

Best,