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Homework Help: Finding a constant of proportionality (Astro)

  1. Aug 1, 2012 #1
    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

    [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] .

    2. Relevant equations

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

    3. 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!
  2. jcsd
  3. Aug 1, 2012 #2
    How did you get -3?
  4. Aug 1, 2012 #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 *@~% ?
  5. Aug 1, 2012 #4
    I really cannot follow this logic. How does this "adding" come about?
  6. Aug 1, 2012 #5

    λ-1 λ-2 = λ-3

    Am I on the right lines?
  7. Aug 1, 2012 #6
    No, you are not. You really need to think how one could go from wavelength to frequency in a formula.
  8. Aug 1, 2012 #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?
  9. Aug 1, 2012 #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?
  10. Aug 1, 2012 #9
    [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.
    Last edited: Aug 1, 2012
  11. Aug 1, 2012 #10
    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?
  12. Aug 1, 2012 #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]


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


    [tex] \beta = 2 [/tex]

    Am I Right?
    Last edited: Aug 1, 2012
  13. Aug 1, 2012 #12
    Or have I messed up still, regarding a sign?
  14. Aug 1, 2012 #13
    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]
  15. Aug 1, 2012 #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]


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

    now i think maybe this should be how i follow:


    [tex] F_v \propto v^{\beta} \propto (\lambda^{-1})\propto(v^{2})^{-1}[/tex]
    Last edited: Aug 1, 2012
  16. Aug 1, 2012 #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?
  17. Aug 1, 2012 #16
    So if [tex] F_v \propto v^{\beta} \propto v^{-2}[/tex]

    [tex] \beta = -2 [/tex]
  18. Aug 1, 2012 #17
    Why is this MINUS 2? Just two messages above it was PLUS 2.
  19. Aug 1, 2012 #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]
  20. Aug 1, 2012 #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].
  21. Aug 1, 2012 #20

    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
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