# Finding a constant of proportionality (Astro)

1. Aug 1, 2012

### lquinnl

1. The problem statement, all variables and given/known data

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

$$F_\lambda \propto \lambda^{-2}$$ .

If the spectrum in frequency units is described by

$$F_v \propto v^\beta$$,

calculate $$\beta$$ .

2. Relevant equations

Obviously we have $$v = \frac{c}{\lambda}$$

3. The attempt at a solution

so if $$v = \frac{c}{\lambda}$$

then does that mean that $$\beta = -3$$

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

2. Aug 1, 2012

### voko

How did you get -3?

3. Aug 1, 2012

### lquinnl

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. Aug 1, 2012

5. Aug 1, 2012

### lquinnl

Because

λ-1 λ-2 = λ-3

Am I on the right lines?

6. Aug 1, 2012

### voko

No, you are not. You really need to think how one could go from wavelength to frequency in a formula.

7. Aug 1, 2012

### lquinnl

Looking at this again I am not convinced of my reasoning. I have this:

$$F_\lambda \propto \lambda^{-2}$$ .

$$F_v \propto v^{-\beta}$$ .

$$F_v \propto v^{-\beta} \propto {\frac{c}{\lambda}}^{-\beta}$$ .

Any chance of a poke in the right direction, if you know?

8. Aug 1, 2012

### voko

You have a formula that contains $\lambda^{-2}$.

You need to express this formula via $\nu$. How would you do it?

9. Aug 1, 2012

### lquinnl

$$F_\lambda \propto \lambda^{-2} \propto (v^{-1})^{-2} \propto v^{-3}$$

$$v \propto \lambda^{-1}$$

$$F_v \propto v^{\beta} \propto \lambda^{-1} \propto (v^{-3})^{-1} \propto v^{-4}$$

so $$\beta = 4$$

????

P.s. Thanks for all this help.

Last edited: Aug 1, 2012
10. Aug 1, 2012

### voko

This part is correct.

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

11. Aug 1, 2012

### lquinnl

But $(\nu^{-1})^{-2}$ is NOT $\nu^{-3}$. 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!

$$(v^{-1})^{-2} = v^{2}$$

so

$$F_v \propto v^{\beta} \propto \lambda^{-1} \propto (v^{2})^{-1} \propto v^{-2}$$

and

$$\beta = 2$$

Am I Right?

Last edited: Aug 1, 2012
12. Aug 1, 2012

### lquinnl

Or have I messed up still, regarding a sign?

13. Aug 1, 2012

### voko

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

$\lambda^{-2} \propto \nu^{2}$

we have

$F \propto \lambda^{-2} \propto \nu ^{2}$

14. Aug 1, 2012

### lquinnl

What I'm saying is that

$$F_\lambda \propto \lambda^{-2} \propto v^{2}$$

from $$v = \frac{c}{\lambda}$$ we have $$v \propto \lambda^{-1}$$

so

$$F_v \propto \lambda^{-1}$$

now i think maybe this should be how i follow:

so

$$F_v \propto v^{\beta} \propto (\lambda^{-1})\propto(v^{2})^{-1}$$

Last edited: Aug 1, 2012
15. Aug 1, 2012

### voko

Again, you just got the expression in terms of $v^{2}$. While you know it should be in terms of $v^{\beta}$. What does this mean to you?

16. Aug 1, 2012

### lquinnl

So if $$F_v \propto v^{\beta} \propto v^{-2}$$

$$\beta = -2$$

17. Aug 1, 2012

### voko

Why is this MINUS 2? Just two messages above it was PLUS 2.

18. Aug 1, 2012

### lquinnl

I thought it should be minus 2 from this statement:

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

19. Aug 1, 2012

### voko

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 $\lambda^{-2}$, which means $F = k\lambda^{-2}$

You are asked to find out to what degree of $\nu$ that is proportional. This is done by expressing $\lambda$ via $\nu$, which is $\lambda = c\nu^{-1}$, so you end up with $F = k(c\nu^{-1})^{-2} = kc^{-2}\nu^{2}$, which simply means that $F \propto \nu^{2}$ and $\beta = 2$.

20. Aug 1, 2012

### lquinnl

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