# Shortest Path?

1. Jul 3, 2007

### Winzer

Ok, so I was eating an apple and I asked myself: If I had two points on the surface and I wanted to find the shortest path between those two points, how would I go about that mathimatically. How about if I had an ellipsoid?
Thanks

2. Jul 4, 2007

### matt grime

The standard method is to use calculus of variations, or something like it.

3. Jul 4, 2007

### HallsofIvy

Staff Emeritus
On a sphere, which an apple approximates, The shortest distance between two points is measured along a great circle. More generally matt grime is correct: a path is a function and "calculus of variations" deals with functions having minimum and maximum properties.

4. Jul 4, 2007

### Kummer

The two posters before me already said this sort of problem is called "Calculus of Variations". I just wanted to add that given some "object" (I am not going to be formal to be easier to follow). Given an object (a sphere, a torus, a cube ... ) and two points on that object, the shortest path between them is called a geodesic. It is of interest in higher Geometry problems.

5. Jul 4, 2007

### Winzer

Geodesic, yes I have heard that word before; it involves the shortest path on a curved surface right?

Level wise, where is Calc of variations relative to Calc II and III?

6. Jul 4, 2007

### Kummer

Caculus of Variations is more advanced than Calculus III. It usually involves solving a differential equation (sometimes even a system of differential equations).

Most students learn Calculus of Variations in a "Mathematical Physics"/"Mathematical Methods for Engineers" class. But they only learn a very small part of this theory. Though there are books entirely written on this subject only a few pages are ever necessary for Engineering students. Those are referred to as the Lagrange-Euler Equations. Named in honor of the great Swiss mathematician Leonard Euler and after Joseph Louis Lagrange who created the fundamental ideas of the Calculus of Variations at the age of 19 when he sent Euler his solution to the Isoperimetric Problem.

7. Jul 4, 2007

### Winzer

Well I guess this question will have to wait till I cover Mathematical Physics my junoir year.

8. Jul 4, 2007

### Kummer

Here is the classic Calculus of Variations problem.
-----

In Calculus you study functions (real), basically a function is something that takes a number and turns it into another number. For instance, $$f(x)=x^2$$ means for each number you give in $$x$$ the output number is $$x^2$$.

In Integral Trasforms you study transforms, for example the Laplace Transform, these are traditionally covered in a Differential Equations course. I am assuming you know them. Another example, which you might later encounter in a Partial Differential Equations course are called (or perhaps in Math/Physics class) [url="http://en.wikipedia.org/wiki/Fourier_transforms]Fourier Transforms[/url]. Basically what the Laplace/Fourier transforms are is something that transforms a function into another function. So for example, the Laplace transform on $$f(x)=x$$ is $$F(s)=s^{-2}$$. Which means the function gets turned into another funtion.

In Calculus of Variations you study a similar situation called a functional. And basically a functional is something that turns a function into a number. I will give you an example below and hopefully you will understand.

Here is a classical application of Calulcus of Varations I promised in the begining of this thread.

Given two distinct points $$(x_1,y_1)$$ and $$(x_2,y_2)$$. What is the shortest path between them? You certainly know the answer is a straight line. But look at how this problem is approach. First we set us an equation involving a functional. Here is what we do. Let $$y=f(x)$$ be any curve between those two points. And the length of that curve is $$\int_{x_1}^{x_2} \sqrt{1+[f'(x)]^2} \ dx$$, this is simply the arc length formula from Calculus II. So we write,
$$I[f(x)] = \int_{x_1}^{x_2}\sqrt{1+[f'(x)]^2} \ dx$$.
This means if $$f(x)$$ is any curve passing through those two points then $$I[f(x)]$$ is the value of this functional. Note, the input is a function but the output is a number.

So the way the problem is stated is:
Given $$I[f(x)]=\int_{x_1}^{x_2}\sqrt{1+[f'(x)]^2} \ dx \mbox{ with }f(x_1)=y_1 \mbox{ and }f(x_2)=y_2$$ minimize the functional.

The conditions $$f(x_1)=y_1 \ f(x_2)=y_2$$ are referred to as the boundary condition, because that assures us the curve passes through those two points.

But the above problem is a typical application of Calculus of Variations which is easily solved with Euler-Lagrange Equation if you read the Wikipedia article.

9. Jul 6, 2007

### Winzer

MMmmm that is a pretty cool way to go about it. Yah functional is new to my vocab since I am in only Calc II . But I can't wait to learn more about those, they sound really interesting.
Thanks

10. Jul 7, 2007

### Gib Z

Halls: When you say "great circle", i'm thinking of a sphere being made up of a revolved circle. We find the circle that goes through both the points we want to find the distance between, they will be points on the circle and we go from there. Am I thinking correctly?

And the geodesic word reminded me of an extremely impractical method, since I hear that word so much of Special Relativity. One could beam a photon from the first point to the next, and it will always follow the geodesic path. We measure how long it took to reach there, and knowing c = 299, 792, 458 m/s, we can work out the distance :) However the will be inaccuracies, as measurement can not be perfect even theoretically (HUP).

11. Jul 7, 2007

### HallsofIvy

Staff Emeritus
As long as the center of a circle, lying on the sphere, is the same as the center of the sphere, then it is a great circle. On the surface of the earth, lines of longitude are great circles, lines of latitude are not.

As far as you "photon measure" is concerned, how are you constraining the photon to follow the curvature of the sphere?

12. Jul 7, 2007

### Gib Z

Yes that was what I was trying to explain :)

Quiet you!

13. Jul 7, 2007

### matt grime

Not to mention the idea that we actually do know the speed of the photon is very dodgy too - what you wrote would be the speed in a vacuum, perchance....

14. Jul 7, 2007

### Gib Z

O well that, it shouldn't be too hard to find the refractive index of an apple.

15. Jul 7, 2007

### Kummer

It is interesting! In fact most of the advanced mathematical techniques you learn in "Mathematical Physics/Mathematical Methods" are all interesting. But the sad thing is only a little bit is covered on each type of Method. It is understandable why that is done, to save time, but is it much nicer to be exposed to all the techinques from a mathematical point of view.

16. Jul 7, 2007

### Winzer

SO in Math Physics, do they pretty much jump around material wise?

17. Jul 7, 2007

### Kummer

Yes. For example, Complex Analysis which is a beautiful subject and an entire semester is devoted to it for math majors is broken down into a few theorems and laws.

18. Jul 7, 2007

### Winzer

Krummer you have been more then helpful but I have one more question:
What course do you get to study about the Riemman Zeta Function?
Just curoius, it looks most interesting, even though I can barly understand it.
That is probably math major stuff right?
I am an engineering major but I have been taking such a keen interest in pure mathematics.

19. Jul 7, 2007

### ice109

probably number theory since it deals with primes

20. Jul 8, 2007

### Kummer

It is a course in Number Theory. A good understanding Riemann's Zeta Function requires good knowledge of Complex Analysis. It is a type of number theory called Analytic Number Theory, it gets its named because we apply results from analysis (mostly complex) to the study of numbers. Such a course is an advanced Graduate course. And yes, it is a math course, it is never studied in Engineering.

Chapter 2 of this free book does explain it a little.