Finding a constant of proportionality (Astro)

In summary, the conversation discusses the process of finding the value of β in a formula for the spectrum of a central galaxy in the Perseus cluster. The formula is expressed in wavelength units as F_λ ∝ λ^−2 and in frequency units as F_v ∝ v^β. Through a series of calculations and mistakes, it is determined that the value of β is 2. The conversation also clarifies the relationship between wavelength and frequency and the concept of proportionality in equations.
  • #1
lquinnl
26
0

Homework Statement



OK so I'm doing a past exam paper as some revision:

The central galaxy in the Perseus cluster has an X-ray spectrum in wavelength units
which is well described by the power law

[tex] F_\lambda \propto \lambda^{-2} [/tex] .

If the spectrum in frequency units is described by

[tex] F_v \propto v^\beta [/tex],

calculate [tex]\beta[/tex] .

Homework Equations



Obviously we have [tex] v = \frac{c}{\lambda} [/tex]


The Attempt at a Solution



so if [tex] v = \frac{c}{\lambda} [/tex]

then does that mean that [tex] \beta = -3 [/tex]

I feel like I am really missing the point here. Any help would be greatly appreciated.

Thanks in advance!
 
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  • #2
How did you get -3?
 
  • #3
Because lambda is to the power of -2 and lambda is related to frequency as v = c lambda^{-1} therefore adding an extra -1

does that make sense or is it total *@~% ?
 
  • #4
I really cannot follow this logic. How does this "adding" come about?
 
  • #5
Because

λ-1 λ-2 = λ-3

Am I on the right lines?
 
  • #6
No, you are not. You really need to think how one could go from wavelength to frequency in a formula.
 
  • #7
Looking at this again I am not convinced of my reasoning. I have this:

[tex] F_\lambda \propto \lambda^{-2} [/tex] .



[tex] F_v \propto v^{-\beta} [/tex] .



[tex] F_v \propto v^{-\beta} \propto {\frac{c}{\lambda}}^{-\beta} [/tex] .

Any chance of a poke in the right direction, if you know?
 
  • #8
You have a formula that contains [itex]\lambda^{-2}[/itex].

You need to express this formula via [itex]\nu[/itex]. How would you do it?
 
  • #9
voko said:
You have a formula that contains [itex]\lambda^{-2}[/itex].

You need to express this formula via [itex]\nu[/itex]. How would you do it?

[tex]F_\lambda \propto \lambda^{-2} \propto (v^{-1})^{-2} \propto v^{-3}[/tex]

[tex]v \propto \lambda^{-1}[/tex]

[tex]F_v \propto v^{\beta} \propto \lambda^{-1} \propto (v^{-3})^{-1} \propto v^{-4}[/tex]

so [tex] \beta = 4[/tex]

?

P.s. Thanks for all this help.
 
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  • #10
lquinnl said:
[tex]F_\lambda \propto \lambda^{-2} \propto (v^{-1})^{-2}[/tex]

This part is correct.

But [itex](\nu^{-1})^{-2}[/itex] is NOT [itex]\nu^{-3}[/itex]. I note that you made a similar mistake in another derivation. What is the proper result of such an operation?
 
  • #11
But [itex](\nu^{-1})^{-2}[/itex] is NOT [itex]\nu^{-3}[/itex]. I note that you made a similar mistake in another derivation. What is the proper result of such an operation?

OF COURSE! I feel so stupid today, really need to wake up!

[tex] (v^{-1})^{-2} = v^{2} [/tex]

so

[tex] F_v \propto v^{\beta} \propto \lambda^{-1} \propto (v^{2})^{-1} \propto v^{-2}[/tex]

and

[tex] \beta = 2 [/tex]

Am I Right?
 
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  • #12
Or have I messed up still, regarding a sign?
 
  • #13
lquinnl said:
But [itex](\nu^{-1})^{-2}[/itex] is NOT [itex]\nu^{-3}[/itex]. I note that you made a similar mistake in another derivation. What is the proper result of such an operation?

OF COURSE! I feel so stupid today, really need to wake up!

[tex] (v^{-1})^{-2} = v^{2} [/tex]

This is correct. But what you do then puzzles me. You have just proved that since

[itex]\lambda^{-2} \propto \nu^{2}[/itex]

we have

[itex]F \propto \lambda^{-2} \propto \nu ^{2}[/itex]
 
  • #14
What I'm saying is that

[tex]F_\lambda \propto \lambda^{-2} \propto v^{2} [/tex]

from [tex] v = \frac{c}{\lambda}[/tex] we have [tex] v \propto \lambda^{-1}[/tex]

so

[tex] F_v \propto \lambda^{-1}[/tex]

now i think maybe this should be how i follow:so

[tex] F_v \propto v^{\beta} \propto (\lambda^{-1})\propto(v^{2})^{-1}[/tex]
 
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  • #15
Again, you just got the expression in terms of [itex]v^{2}[/itex]. While you know it should be in terms of [itex]v^{\beta}[/itex]. What does this mean to you?
 
  • #16
voko said:
Again, you just got the expression in terms of [itex]v^{2}[/itex]. While you know it should be in terms of [itex]v^{\beta}[/itex]. What does this mean to you?

So if [tex] F_v \propto v^{\beta} \propto v^{-2}[/tex]

[tex] \beta = -2 [/tex]
 
  • #17
Why is this MINUS 2? Just two messages above it was PLUS 2.
 
  • #18
I thought it should be minus 2 from this statement:

[tex] F_v \propto v^{\beta} \propto (\lambda^{-1})\propto(v^{2})^{-1}[/tex][/QUOTE]
 
  • #19
I think you are very confused. Let me explain the whole thing to you.

When something (A) is proportional something else (B), then A = CB, where C does not contain B in any way.

In your case, you have F proportional to [itex]\lambda^{-2}[/itex], which means [itex]F = k\lambda^{-2}[/itex]

You are asked to find out to what degree of [itex]\nu[/itex] that is proportional. This is done by expressing [itex]\lambda[/itex] via [itex]\nu[/itex], which is [itex]\lambda = c\nu^{-1}[/itex], so you end up with [itex]F = k(c\nu^{-1})^{-2} = kc^{-2}\nu^{2}[/itex], which simply means that [itex]F \propto \nu^{2}[/itex] and [itex]\beta = 2[/itex].
 
  • #20
THANK YOU VERY MUCH!

I feel like I have done an incredible job of over-complicating this problem!

I think I was getting confused by the fact we have F_v and F_lambda.

So it was just a case of expressing wavelength as frequency and amalgamating the powers.

Seems incredibly straightforward now!

Thanks again
 

What is a constant of proportionality?

A constant of proportionality is a value that relates two quantities that are directly proportional to each other. It is represented by the letter "k" and remains the same regardless of the values of the two quantities.

Why is finding a constant of proportionality important in astronomy?

In astronomy, finding a constant of proportionality allows us to accurately calculate and predict relationships between different astronomical quantities, such as the distance between objects, their masses, and their velocities. It is essential for understanding and studying the universe.

How do you find a constant of proportionality?

To find a constant of proportionality, you need to have two quantities that are directly proportional to each other. Then, you can use the formula k = y/x, where k is the constant of proportionality, y is one of the quantities, and x is the other quantity. By plugging in the values and solving for k, you can find the constant of proportionality.

What are some examples of constant of proportionality in astronomy?

One example of a constant of proportionality in astronomy is the Hubble constant, which relates the distance between galaxies and their recession velocity. Another example is the gravitational constant, which relates the masses and distances of celestial bodies in Newton's law of gravitation.

Can a constant of proportionality ever change?

No, a constant of proportionality remains the same as long as the relationship between the two quantities remains directly proportional. However, different values of a constant of proportionality may exist for different relationships between the same two quantities.

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