Spectral Intensity Formula by Using Frequency

benagastov
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Homework Statement
It is known that a hypothetical atom emits light when it transitions from an excited state to a ground state with a frequency w0 = 8 x 10^14 Hz. It is known that the observer receives the frequency of these photons due to the Doppler effect.

Get the intensity of the photon spectrum!
Relevant Equations
\omega \:=\:\omega _0\cdot \left(1\:\pm \frac{v_x}{c}\right)

w = w0 (1 + vx/c)
I found there is kind of solution in Pointon's book: An Introduction to Statistical Physics for Students. But I don't know how to find intensity by using frequency.
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Since no one has replied yet, see if this helps.

benagastov said:
It is known that the observer receives the frequency of these photons due to the Doppler effect.
That's not clear. I guess you mean:
“Since the atoms are moving ramdomly, due to the Doppler effect the observer receives photons with a distribution of frequencies”.

benagastov said:
w = w0 (1 + vx/c)
##\omega## is generally used for angular frequency. 'f' is already used for the distribution function. ##\nu## (Greek letter 'nu') looks too much like 'v' (velocity symbol).

Maybe use 'F' for frequency.

benagastov said:
I found there is kind of solution in Pointon's book: An Introduction to Statistical Physics for Students. But I don't know how to find intensity by using frequency.

View attachment 299395
Since ##\lambda= \frac c F## some simple algebra allows you to rewrite equation 3.23a (from your attachment) in terms of frequency.

Then you can use the same method as shown in your attachment.

Edited to provide only initial guidance.
 
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Steve4Physics said:
Since no one has replied yet, see if this helps.That's not clear. I guess you mean:
“Since the atoms are moving ramdomly, due to the Doppler effect the observer receives photons with a distribution of frequencies”.##\omega## is generally used for angular frequency. 'f' is already used for the distribution function. ##\nu## (Greek letter 'nu') looks too much like 'v' (velocity symbol).

Maybe use 'F' for frequency.Since ##\lambda= \frac c F## some simple algebra allows you to rewrite equation 3.23a (from your attachment) in terms of frequency.

Then you can use the same method as shown in your attachment.

Edited to provide only initial guidance.
Thank you for your reply! Is this how I do it? I use Bose_einstein statistics since it was Photon
 

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benagastov said:
Thank you for your reply! Is this how I do it? I use Bose_einstein statistics since it was Photon
I think you have misunderstood the physics involved.

This is primarily a classical system. The atoms, if stationary, would emit waves of frequency 8x10¹⁴ Hz.

But each atom is a moving source with a velocity component in the observed x-direction of ##v_x##. The Doppler shift is due to having having moving sources (moving atoms). That’s how a distribution of frequencies arises.

The atoms are assumed to have a (classical) Maxwell-Boltzmann velocity distribution. So we know the distribution of the ##v_x##s.

Bose-Einstein statistics are not relevant in the context of this question.

If you don’t understand this, you need to:
- revise the Doppler effect;
- revise the Maxwell-Boltzmann velocity distribution;
- make sure you fully understand the working used in your Post #1 attachment.

You have ignored the instructions I gave in Post #2. So I will repeat them for you:
"Since ##\lambda = \frac c F## some simple algebra allows you to rewrite equation 3.23a (from your attachment) in terms of frequency.

Then you can use the same method as shown in your attachment.”

So your first task is to produce a new version of equation 3.23a with frequency instead of wavelength.

(Also, consider using ‘F’ rather than ##\omega## for frequency, for the reasons I’ve already explained.)
 
Ok, I am sorry took so long. After I discharged from hospital, I took rest time.
So I think I already finish the formula, is this Blue Shift? What I do next? There is no mass in my question.

Problem.PNG
 
benagastov said:
So I think I already finish the formula, is this Blue Shift?
No. The formula (you highlighted in green) gives the Doppler shifted frequency.

If ##v_x## is negative:
- is the source (the atom) moving towards or away from the observer?
- is the observed frequency bigger or smaller than ##f_0##?
- is this a blue shift or a red shift?

If ##v_x## is positive:
- is the source (the atom) moving towards or away from the observer?
- is the observed frequency bigger or smaller than ##f_0##?
- is this a blue shift or a red shift?

You have (correctly) found ##v_x = \frac {c(f-f_0)}{f}##. But you then say ##dv_x = \frac {cf_0}{f} df##. That’s wrong. Can you correct it?

When you have have the correct expression for ##dv_x## you can use it in the distribution function.

benagastov said:
What I do next? There is no mass in my question.
You will have to leave 'm' (mass of an atom of the unspecified element) as a parameter in the equation.

As previously noted, using ##\omega## is confusing as it is usually used for angular frequency.

Maybe you could use ‘g’ for the distribution function rather than ‘f’? Then you could use f for frequency.
 
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