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Parity and integration in spherical coordinates

  1. Jan 13, 2016 #1
    Hello people!
    I have ended up to this integral ##\int_{φ=0}^{2π} \int_{θ=0}^π \sin θ \ \cos θ~Y_{00}^*~Y_{00}~dθ \, dφ## while I was solving a problem.
    I know that in spherical coordinates when ##\vec r → -\vec r## :
    1) The magnitude of ##\vec r## does not change : ##r' → r##
    2) The angles ##θ## and ##φ## change like ##θ' → π-θ## and ##φ' → π+φ##
    3) So parity of spherical harmonics is ##\hat P## ##Y_{lm}(θ,φ)## ##=## ##Y_{lm}(θ',φ')=(-1)^l## ##Y_{lm}(θ,φ)##
    4) Parity of ##\cos θ## and ##\sin θ## are ##\cos θ'## ##=## ##(-1)## ##\cos θ## and ##\sin θ'## ##=## ##\sin θ## respectively.
    This means that in my case the integrated function has Parity equal to ##(-1)## .

    So my question is:
    Can I say that this integral is zero because of the odd (=parity is equal to (-1)) integrated funtion? Because in the xy plane when we integrate an odd function ##F(x)=-F(-x)## in a symmetric space (e.g. ##\int_{-a}^a F(x) \, dx## with F being odd) we can say that it is zero without to calculate it.
    If yes, can I do this in general? meaning, if i get a random function in spherical coordinates which depends only from angles θ and φ and i want to integrate it with these limits: ##\int_{φ=0}^{2π} \int_{θ=0}^π randomF(θ,φ) \, dθ \, dφ## , can i find its parity and say if it is zero or not?
     
  2. jcsd
  3. Jan 13, 2016 #2

    vanhees71

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    Well, ##\mathrm{Y}_{00}=1/\sqrt{4 \pi}=\text{const}## and your integral thus is
    $$\propto \int_0^{\pi} \mathrm{d} \vartheta \sin \vartheta \cos \vartheta=\int_0^{\pi} \mathrm{d} \vartheta \frac{1}{2} \sin(2 \vartheta)= \left .-\frac{1}{4} \cos(2 \vartheta)\right|_0^{\pi}=0.$$
     
  4. Jan 13, 2016 #3
    Ok that was an easy integral but what if i have something more complicated like ##\int_{φ=0}^{2π} \int_{θ=0}^π \sin θ \ \cos θ~Y_{11}^*~Y_{1-1}~dθ \, dφ## ? If we use parity here we see that the parity of the integrated function is (-1) and the integral must be zero! No calculations ! Just used parity !
    My question is not how to solve the integral making calculations . I want to solve it using parity and i want to tell me if my thoughts are correct on how to use parity in order to solve these integrals!
    Thanks btw for your response :)
     
  5. Jan 14, 2016 #4

    blue_leaf77

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    Yes you can. Just remember that not every function has definite parity.
     
  6. Jan 14, 2016 #5
    Thank you very much ! :)
     
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