# Bunch o' Questions

1. Sep 28, 2005

### Noone1982

This thread is kind of an extension of my last, so pardon any overlap.

1) dW = F • dL

My teacher says "careful the path you integrate this on." But isnt there only one possible let of limits for something? I mean, if the particle is traveling on say y = x^2 from 0 to 5, what other path could there be? How do conservative and non-conservative vector fields play into these limits?

I noticed some problems you can just integrate dx with the x limits, dy with the y limits and dz with the z limits and get the right answer. However, some I notice you have to put everything in terms of say x and just integrate over x to get the right answer. Integrating over x,y,z limits gives me the right answers for some but not others. Why?

I would think the answer would be the same for an integral of dx integrated over x limits + dy integrated over y limits + dz integrated over z limits compared to an all x or all y or all z integral. Why does it matter?

2. Sep 28, 2005

### quasar987

For conservative fields, you can integrate along ANY path that has those limits and the result will be the same! This allows for easier computation. For non-conservative field, you MUST integrate along the path taken by the particle and thus you must parametrize the path, which is often times a lot of pain.

It is not entirely clear to me what you mean, but if it's what I think it means the field is conservative and what you'Re doing when you integraate "dx with the x limit, then dy with the y limits and then dz with the z limits" is you integrate along a straight line on the x axis (for which path y and z are constant), then along a straight line along the y axis (for with x and z are constant), and finally along the z axis. This is the tactic I was refering to in paragraph one when talking about how with conservative field, you can integrate along any path, which eases computation.

3. Sep 29, 2005

### Noone1982

Thank you, it is becoming clearer now. Now it is a conservative field if the curl is equal to zero?

4. Sep 29, 2005

### quasar987

It is a necessary and sufficient condition, yes.

Here's the answer to all your conservative fields needs and demands.

Theorem (Helmholtz):
The following statements are all logically equivalent (i.e. they are interlinked by a $\Leftrightarrow$ relation)

1) curl of F is 0
2) integral of F.dl is independant of path for any given end points
3) integral of F.dl = 0 for any closed loop.
4) F is the gradient of some scalar function V: $\vec{F}(x,y,z) =-\nabla V(x,y,z)$
5) F is said to be a conservative vector field.

Last edited: Sep 29, 2005