Undergrad Find potential integrating on segments parallel to axes

Click For Summary
To find the potential of a conservative vector field using segments parallel to the coordinate axes, the domain D must allow any point to connect to a chosen point (x0, y0, z0) via such segments. While D must be connected, this alone is insufficient; it must also permit connections that strictly follow the axes. An example provided illustrates that a connected domain can still fail this requirement if points cannot be reached through axis-aligned paths. The discussion suggests that compactness may play a role in determining the necessary conditions for D. Overall, a deeper exploration of the geometric properties of D is needed to establish precise conditions for this method.
crick
Messages
39
Reaction score
4
A simple method to find the potential of a conservative vector field defined on a domain ##D## is to calculate the integral
$$U(x,y,z)=\int_{\gamma} F \cdot ds$$

On a curve ##\gamma## that is made of segments parallel to the coordinate axes, that start from a chosen point ##(x_0,y_0,z_0)##.

I would like to know what are precisely the conditions that ##D## must satisfy to use this method. ##D## should be made in such way that "any point can be connected to ##(x_0,y_0,z_0)## with, indeed, a segment parallel to the coordinate axes".

But what are the sufficient mathematical conditions for $D$ in order to have this property?**

I would say that it surely has to be connected, but that seems not to be enough. For example taking
$$D= \{ (x,y) : y>x-1\} \,\,\,\, \,\,\,\,\,(x_0,y_0)=(0,0)$$
##D## is connected but I do not think that any point can be connected to ##(0,0)## via a segment parallel to the coordinate axes.
 
Physics news on Phys.org
crick said:
I do not think that any point can be connected to (0,0) via a segment parallel to the coordinate axes.
It does not need to be. It only has to be connected by some path consisting of segments parallel to the axes. Of course, it could get quite messy.
A simple example which does not work is two regions connected only by a diagonal line.
 
Last edited:
In my mind it seems like a compactness argument should be useful, but I do not presently have the energy to follow up.
 

Similar threads

Replies
2
Views
2K
  • · Replies 24 ·
Replies
24
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K