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.

The following statements are all logically equivalent (i.e. they are interlinked by a $\Leftrightarrow$ relation)
4) F is the gradient of some scalar function V: $\vec{F}(x,y,z) =-\nabla V(x,y,z)$