The Calculus of Variations

[Greg Bernhardt]

Author: Dr. Mark Trodden of Syracuse University

[polyb]

Thanks Greg, that's pretty cool!:biggrin:

Too bad he didn't make any graphs for the visual aid! I gues I'll need to go back to Marion/Thornton for pretty pictures!

I had the chance to meander through both forums and I have to say EXCELLENT WORK Greg! I hope others find it useful as well!

[pi-r8]

What does ds in equation 2 refer to?

[polyb]

If I am reading it correctly, ds is the generalized path. Think of it this way: ds/dt=v. It is a little cryptic and personally I prefer things to always be a little more explicit. Maybe if you were in his class then the representation would be self evident.

Of course I have been wrong before!

[pi-r8]

Ok, I guess that makes sense. I'm just now starting to learn about the calculus of variations, so it's helpful to see another approach to it.

[jcsd]

What does ds in equation 2 refer to?

's' is displacement. If you want to give 'ds' a name it would be infinitessimal displacement, though in this instance it just appears as part of the standard notation for the indefinite integral of 1/v(s) over s.

[Thrice]

Attachment doesn't work for me...

[finchie_88]

Doesn't the ds in the equation refer to distance, not displacement or is that my imagination?
Since length of an arc or path is given by:
s = \int \sqrt{ 1 + \left( \frac{dy}{dx} \right)^2 } dx
I think.

[Perturbation]

The s is the length along the curve, though it can be replaced with any parameter for the curve, which could be length, or time or whatever. The usual parameter for curves in general relativity is proper time, \tau, for example.

[phoenixthoth]

Doesn't the ds in the equation refer to distance, not displacement or is that my imagination?
Since length of an arc or path is given by:
s = \int \sqrt{ 1 + \left( \frac{dy}{dx} \right)^2 } dx
I think.
Seems right. Then it that case, ds would be:
ds=\sqrt{1+\left( \frac{dy}{dx}\right) ^{2}}dx

[adartsesirhc]

Marion-Thornton uses ds as the following:

ds=\sqrt{dx^{2}+dy^{2}}

and from this, they go to

ds=\sqrt{1+(\frac{dy}{dx})^{2}}dx.

Hope this helps.