Associated Legendre Function with Angles

[bryanso]

Homework Statement

In Associated Legendre Function with Angles ... why is the following argument used?

Relevant Equations

$\sqrt{1 - x^2} = sin\,\theta$

In Wikipedia https://en.m.wikipedia.org/wiki/Associated_Legendre_polynomials, Section Reparameterization in terms of angles, I see this argument:

Let $x = cos\,\theta$

$\sqrt{1 - x^2} = sin\,\theta$

This is also in Griffiths' Introduction to Quantum Mechanics.

Why is this a valid argument?

The LHS is always positive. The RHS is sometimes positive and sometimes negative. $\theta = -1$ makes it wrong.

 

[fresh_42]

Yes, one has to manage the range of the angels. For $\theta > \pi$ we need the negative root. I assume that most applications have angels from $[0,\pi]$ and in the other case: switch to the opposite orientation of the coordinate system. The problem only occurs if angels occur in both ranges.

 

[bryanso]

Thanks!

 

[Mark44]

Homework Statement:: In Associated Legendre Function with Angles ... why is the following argument used?
Relevant Equations:: $\sqrt{(1 - x)^2} = sin\,\theta$

In Wikipedia https://en.m.wikipedia.org/wiki/Associated_Legendre_polynomials, Section Reparameterization in terms of angles, I see this argument:

Let $x = cos\,\theta$

$\sqrt{(1 - x)^2} = sin\,\theta$

No, that's not what it says. Instead, it's $\sqrt{1 - x^2}$. This is a standard substitution used in trig substitution in integrals.

You also have the same incorrect equation in your Relevant Equations.

This is also in Griffiths' Introduction to Quantum Mechanics.

Why is this a valid argument?

The LHS is always positive. The RHS is sometimes positive and sometimes negative. $\theta = -1$ makes it wrong.

 

[Mark44]

Yes, one has to manage the range of the angels.

It's better to manage the range of the angles. I don't think we'd have much luck trying to manage angels.

 

[bryanso]

Yes I had a typo. I will try to edit it. The question remains.

 

[bryanso]

No, that's not what it says. Instead, it's $\sqrt{1 - x^2}$. This is a standard substitution used in trig substitution in integrals.

You also have the same incorrect equation in your Relevant Equations.

Thanks. I have edited the equations.

 

[Orodruin]

The range of $\theta$ in spherical coordinates (which is where the associated Legendre functions are used) is 0 to $\pi$. Therefore $\sin(\theta)$ is always positive.

 

[Orodruin]

The range of $\theta$ in spherical coordinates (which is where the associated Legendre functions are used) is 0 to $\pi$. Therefore $\sin(\theta)$ is always positive.

Put differently, the domain of the associated Legendre functions (just as that of the Legendre polynomials) is [-1,1]. This is why $x = \cos\theta$ works so well with $\theta \in [0,\pi]$ (and, consequently, $\sin\theta \geq 0$).