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Integration of A exp( (− 3 R^2)/(2Na^2)) (whilst following forum rules; I think)

  1. Sep 22, 2014 #1
    1. The problem statement, all variables and given/known data
    A exp( (− 3 R^2)/(2Na^2))
    I wish to integrate ^
    The expression is equal to P(N,R)
    The limits are infinity to -infinity.
    The intention is to find the normalisation constant / A in terms of N and a (then to cube it).


    3. The attempt at a solution

    This is what I've managed.

    A S e^(-3R^2) dR + S e^(0.5Na^2) dN

    (S being integral)

    Am I on the right tracks?
    Should I separate e^0.5 and integrate with respect to N, or should I not integrate it?

    Thank you everyone who's helped me along so far.
     
  2. jcsd
  3. Sep 22, 2014 #2

    Mark44

    Staff: Mentor

    Much better! Thank you!
     
  4. Sep 22, 2014 #3

    Mark44

    Staff: Mentor

    Is the following what you're trying to integrate?
    $$A\int_{-\infty}^{\infty} e^{-3R^2}dR + \int_{-\infty}^{\infty} e^{0.5Na^2}dN$$

    Can you show how you got to this from the integral at the top of your post?
     
  5. Sep 22, 2014 #4

    Almost. I stuffed up again.

    A S e^(-3R^2) dR + S e^(1/(2Na^2)) dN

    ^is what I intended.

    I separated A and e^(-3R^2) and e^(1/(2Na^2)) from each other.

    Next, I intend to integrate e^(-3R^2) with respect to R; infinity to - infinity
    and e^(1/(2Na^2)) with respect to N; presumably indefinitely.

    I am unsure whether I should remove e^0.5 from e^(1/(2Na^2)) or integrate it all with dN.

    I am also not completely sure about how much maths I may have just made up.
    This may be an extra big stuff up.
    Thanks for taking the time!
     
  6. Sep 22, 2014 #5

    Mark44

    Staff: Mentor

    You are not using the rules for exponents correctly. e^(a + b) = e^a * e^b and e^(a - b) = (e^a)/(e^b), but what you're doing for e^(a/b) is incorrect.

    Also, in the first post you say you want to integrate exp( (− 3 R^2)/(2Na^2)). With respect to what? What is the variable here - R? If so, are N and a constants as far as the integration goes? I'm confused as to what you're trying to do.
     
  7. Sep 22, 2014 #6

    'Ideal chains form a random coil conformation in solution. The polymer's end-to-end length (see figure) is the distance between the chain ends: this can be used as a measure of the coil size. The end-to-end length has an average value, but there are fluctuations observed about that average value: how can we characterize these fluctuations? You will address this question in this problem (which believe it or not is closely analogous to the velocity distribution of an ideal gas!)

    Consider a chain made up of N linkers of length a.
    The end to end vector distribution R = (Rx,Ry,Rz), is given by

    P(N,R)= exp( (− 3 R^2)/(2Na^2))

    where R^2 = IRI^2 = Rx^2 + Ry^2 + Rz^2

    What is A (express in terms of N and a)?'

    xyz are subscripts

    Does this make things more easily decipherable?

    I guess N and a are constant as far as the integration goes.

    I think what I'm trying to do is to find the normalisation constant for, say P(N,Rx),
    then to cube that to find the normalisation constant for R.

    I didn't want to post all of that but I hope it makes things clearer than I may have.
     
  8. Sep 22, 2014 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You should have posted all of that to begin with; it describes your problem and we don't need to guess anymore. So, to summarize: for constants ##A, a, N## (##a \neq 0, N > 0##), you want to find the triple integral
    [tex] \int_{R^3} A e^{-\left(\frac{R^2}{2N a^2}\right)} dx dy dz,\\
    \text{where } \; R^2 = x^2 + y^2 + z^2 [/tex]
    Is that your problem?

    So, what is the next thing you would do?
     
  9. Sep 22, 2014 #8
    Sub in x2+y2+z2!

    ∫Ae^−((x2+y2+z2)/2Na^2)dxdydz,

    A ∫ (e^-(x^2)/2Na^2))dx (e^-(y^2)/2Na^2))dy (e^-(z^2)/2Na^2))dz

    Right?
     
  10. Sep 22, 2014 #9

    Ray Vickson

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    Keep going.
     
  11. Sep 22, 2014 #10
    A(√π2Na^2)(√π2Na^2)(√π2Na^2)
    A(√π2Na^2)^3

    Still right?
     
  12. Sep 22, 2014 #11

    Ray Vickson

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    Science Advisor
    Homework Helper

    That depends on whether √π2Na^2 means ##\sqrt{\pi} 2Na^2##, ##\sqrt{2 \pi} N a^2## or ##\sqrt{2 \pi N a^2}##; I suspect you mean the latter. If so, you are right. If you use parentheses like this: sqrt(2 * pi * N a^2), it would be 100% clear. Alternatively, use LaTeX, as I did.
     
  13. Sep 22, 2014 #12
    Hey! Thanks so much for helping!

    I seem to have sorted it out.

    However, am I correct in believing that we left out the 3 from A e^( (− 3 R^2)/(2Na^2)) somewhere along the way?

    A= (1/(2PiNa^2))^3/2 if not

    (3/(2PiNa^2))^3/2 if so.
     
  14. Sep 22, 2014 #13

    Ray Vickson

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    Science Advisor
    Homework Helper

     
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