# Calculate the exact or approximate length of one period of the wave function

• A
• Ennio
In summary, WolframAlpha and SymboLab calculates a solution involving complicated ellpitical integrals. Could you please help me in finding the right way to calculate the integral? An approximated solution would be also appreciated.
Ennio
TL;DR Summary
exact or approximated calculation of the sinusoidal length in one period [0,2π] or corresponding to [0,λ] including the amplitude of the wave A as well as the wave number k.
The known expression of the wave function is

where A is the amplitude, k the wave number and ω the angular velocity.

The mathematical definition of arc length for a generical function in an interval [a,b] is

where, in our sinusoidal case:

For our purpose (calculation of the length in one period [0,2π] corresponding to [0,λ]), I guess that we can remove the ωt term from the expression, so that

Now, I am really struggling to calculate this integral. There are some instances in internet where mathematicians calculated the length of half period [0,π] of a sinusoidal wave as

which accordingly implies

The fact is that I want to include the amplitude of the wave and the wave number by means the term
as well as the wave number alone in the cos argument
in the main integral, as shown previously.

WolframAlpha and SymboLab calculates a solution involving complicated ellpitical integrals . Could you please help me in finding the right way to calculate the integral? An approximated solution would be also appreciated. The main requirement is the amplitude A and the wave number k are considered in the derivation and in the finalr result. Eventually, l has to be funtion of A and k in the final result.

E.

The period depends on ##k##, should be pretty straightforward to figure out the ##k##-dependence for the curve lenght.

Here is a plot I made for curve length with dependence of the amplitude ##A## keeping ##k=1##.

What is wrong with elliptic integrals btw?

Last edited:
malawi_glenn said:
The period depends on ##k##, should be pretty straightforward to figure out the ##k##-dependence for the curve lenght.

Here is a plot I made for curve length with dependence of the amplitude ##A## keeping ##k=1##.
View attachment 313085

What is wrong with elliptic integrals btw?
Hi malawi_glenn. Thank you for your answer. However, what is the exact mathematical expression for the result? From your graph: if the amplitude is zero then the arc length is 3.2?

Ennio said:
what is the exact mathematical expression for the result
Something with Elliptical functions ;)

Ennio said:
From your graph: if the amplitude is zero then the arc length is 3.2?
Why should it be 3.2?

PeroK
It would be nice if you could share the mathematical procedure to get to the final result. Otherwise, thanks for your partial help.

Ennio said:
It would be nice if you could share the mathematical procedure to get to the final result. Otherwise, thanks for your partial help.
And it would be nice if you could answer my questions. There are tons of information regarding approximations for elliptic integrals. Also it depends on what your application you are going to use this.

btw, is this homework?

The arclength for a single period is $$\begin{split} \int_0^{2\pi/k} \sqrt{1 + (kA)^2\cos^2(kx)}\,dx &= 4\int_0^{\pi/(2k)} \sqrt{1 + (kA)^2\cos^2(kx)}\,dx\qquad\mbox{(by symmetry)} \\ &= 4 \sqrt{1 + (kA)^2} \int_{0}^{\pi/(2k)} \sqrt{1 - \frac{(kA)^2}{1 + (kA)^2}\sin^2(kx)}\,dx \\ &= \frac{4 \sqrt{1 + (kA)^2}}{k} \int_{0}^{\pi/2} \sqrt{1 - \frac{(kA)^2}{1 + (kA)^2}\sin^2(u)}\,du \\ &= \frac{4 \sqrt{1 + (kA)^2}}{k}E\left(\frac{kA}{\sqrt{1 + (kA)^2}}\right)\end{split}$$ where $E(y)$ is the complete elliptic integral of the second kind, which is one of those definite integrals which comes up everywhere in applications but cannot be done analytically, so we give it a name and tabulate its values numerically.

The arclength over a fraction of a quarter period can be reduced to an incomplete elliptic integral of the second kind in a similar manner: $$\int_0^{\alpha\pi/(2k)} \sqrt{1 + (kA)^2\cos^2(kx)}\,dx = \frac{\sqrt{1 + (kA)^2}}{k}E\left(\frac{\alpha\pi}{2}, \frac{kA}{\sqrt{1 + (kA)^2}}\right),\qquad 0 \leq \alpha \leq 1$$

Last edited:
Ennio, dextercioby and malawi_glenn
pasmith said:
The arclength for a single period is $$\begin{split} \int_0^{2\pi/k} \sqrt{1 + (kA)^2\cos^2(kx)}\,dx &= 4\int_0^{\pi/(2k)} \sqrt{1 + (kA)^2\cos^2(kx)}\,dx\qquad\mbox{(by symmetry)} \\ &= 4 \sqrt{1 + (kA)^2} \int_{0}^{\pi/(2k)} \sqrt{1 - \frac{(kA)^2}{1 + (kA)^2}\sin^2(kx)}\,dx \\ &= \frac{4 \sqrt{1 + (kA)^2}}{k} \int_{0}^{\pi/2} \sqrt{1 - \frac{(kA)^2}{1 + (kA)^2}\sin^2(u)}\,du \\ &= \frac{4 \sqrt{1 + (kA)^2}}{k}E\left(\frac{kA}{\sqrt{1 + (kA)^2}}\right)\end{split}$$ where $E(y)$ is the complete elliptic integral of the second kind, which is one of those definite integrals which comes up everywhere in applications but cannot be done analytically, so we give it a name and tabulate its values numerically.

The arclength over a fraction of a quarter period can be reduced to an incomplete elliptic integral of the second kind in a similar manner: $$\int_0^{\alpha\pi/(2k)} \sqrt{1 + (kA)^2\cos^2(kx)}\,dx = \frac{\sqrt{1 + (kA)^2}}{k}E\left(\frac{\alpha\pi}{2}, \frac{kA}{\sqrt{1 + (kA)^2}}\right),\qquad 0 \leq \alpha \leq 1$$

## 1. What is the wave function?

The wave function is a mathematical function that describes the behavior of a quantum system over time. It is used to calculate the probability of finding a particle in a certain state or location.

## 2. How is the length of one period of the wave function calculated?

The length of one period of the wave function can be calculated by determining the wavelength, which is the distance between two consecutive peaks or troughs of the wave. The length of one period is equal to the wavelength.

## 3. Can the length of one period be calculated exactly or only approximately?

The length of one period can be calculated exactly if the wave function is a simple sinusoidal function. However, in more complex systems, it may only be approximated due to the difficulty of solving the mathematical equations.

## 4. What factors can affect the length of one period of the wave function?

The length of one period can be affected by the energy level of the system, the potential energy barrier, and the mass of the particle. It can also be influenced by external factors such as temperature and pressure.

## 5. Why is it important to calculate the length of one period of the wave function?

Calculating the length of one period of the wave function is important in understanding the behavior and properties of quantum systems. It can also help in predicting the behavior of particles and their interactions with other particles in the system.

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