Is a radial force always conservative?
Sam Lachterman
= f(r)R
where R = (x,y) and r = magnitude(R). Show that such a force field is
conservative.
I have not gotten very far with this at all. I have been trying to
show that F is a gradient of some function, but I do not understand
how to do this for *any* f which depends on the magnitude of R.
All help is appreciated.
Thanks in advance,
Sam
Ron Verrall
Sam Lachterman wrote:
I assume you mean that the x- and y-components of the force F is given
by:
F_x = xf(r)
F_y = yf(r)
Consider what happens when you take the gradient of a function G(r)
Grad G(r) = [G'(r)x/r, G'(r)y/r]
since the partial derivative of r wrt x is x/r. Sim for y.
Now you set this equal to your force F and get
xf(r) = G'(r)x/r
yf(r) = G'(r)y/r
Both equations are:
f(r) = G'(r)/r or
G'(r) = rf(r)
An integration gives you G(r). This is true whether or not you can write
down the integral in terms of algebraic functions. Since your force is
the gradient of a function, it is conservative.
Another way to show it is to take the curl of the force.
Curl F = (d/dx)F_y - (d/dy)F_x
Where the derivatives are partials.
Curl F = yf'(r)x/r - xf'(r)y/r = 0
Where again I've used the partial of r wrt x is x/r, etc.
Since the curl of the vector field is zero, the field is conservative.
Cheers,
Ron Verrall
--
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Casey Hawthorne
Then how does that apply to your force?
Regards,
Casey
lena
Isn't any vector field conservative in the plane. There are no holes, so
the cohomology is trivial, isn't it?
lena
You need something like curl to be zero to guarantee that the vector
field is the gradient of some function, or something like that.
David C. Ullrich
wrote:
>A radial force field F in the plane can be written in the form F(x,y)
>= f(r)R
>where R = (x,y) and r = magnitude(R). Show that such a force field is
>conservative.
>
>I have not gotten very far with this at all. I have been trying to
>show that F is a gradient of some function, but I do not understand
>how to do this for *any* f which depends on the magnitude of R.
Suppose g is a function of one variable and g' = f. Define
p(x,y) = g(||(x,y)||). What is the gradient of p?
>All help is appreciated.
>
>Thanks in advance,
>Sam
******************
David C. Ullrich
David C. Ullrich
<ull...@math.okstate.edu> wrote:
>On 10 Mar 2003 15:19:33 -0800, slach...@fuse.net (Sam Lachterman)
>wrote:
>
>>A radial force field F in the plane can be written in the form F(x,y)
>>= f(r)R
>>where R = (x,y) and r = magnitude(R). Show that such a force field is
>>conservative.
>>
>>I have not gotten very far with this at all. I have been trying to
>>show that F is a gradient of some function, but I do not understand
>>how to do this for *any* f which depends on the magnitude of R.
>
>Suppose g is a function of one variable and g' = f. Define
>p(x,y) = g(||(x,y)||). What is the gradient of p?
Come to think of it maybe I should add a hint: Don't try to
find the gradient of p by calculating partial derivatives.
Instead use the fact that the gradient of p points in the
direction in which p is increasing fastest (and has
magnitude equal to the directional derivative of p in
that direction.)
******************
David C. Ullrich
Sam Lachterman
you my progress on this, and perhaps you could clear things up for me.
Let g(x,y) = ||(x,y)||
Then F(x,y) = xf(g(x,y))i + yf(g(x,y))j
If F is conservative, there exists some scalar field p(g(x,y)) such
that grad(p(g(x,y))) = F.
let h = p(g(x,y))
then grad(h) = (p'g_x, p'g_y) where g_x denotes the partial derivative
of g w.r.t to x, etc.
Now g_x = x/g and g_y = y/g
So if grad(h) = F, p' = f(g(x,y))g(x,y)
From the 1st FTC (in one dimension), p = Int[f(t)t,t,0,||(x,y)||]
Are we done?
Thanks,
Sam
David C. Ullrich <ull...@math.okstate.edu> wrote in message news:<5fqr6v480qfjicgg9...@4ax.com>...
David C. Ullrich
wrote:
>I do not understand the hint you are trying to give me. Let me show
>you my progress on this, and perhaps you could clear things up for me.
>
>Let g(x,y) = ||(x,y)||
>Then F(x,y) = xf(g(x,y))i + yf(g(x,y))j
>If F is conservative, there exists some scalar field p(g(x,y)) such
>that grad(p(g(x,y))) = F.
>
>let h = p(g(x,y))
>then grad(h) = (p'g_x, p'g_y) where g_x denotes the partial derivative
>of g w.r.t to x, etc.
>
>Now g_x = x/g and g_y = y/g
>So if grad(h) = F, p' = f(g(x,y))g(x,y)
>From the 1st FTC (in one dimension), p = Int[f(t)t,t,0,||(x,y)||]
>
>Are we done?
Good for you to show your progress, but it looks like you're
ignoring a few things I said (as well as totally changing the
notation - your g and p are totally different things from what
I called g and p. Changing the meaning of the letters for
no reason makes it harder to carry on a conversation - if
you didn't realize you were changing what the letters meant
then read what I wrote again, more slowly this time. Also
don't miss the "hint".)
>Thanks,
>Sam
>
>
>David C. Ullrich <ull...@math.okstate.edu> wrote in message news:<5fqr6v480qfjicgg9...@4ax.com>...
>> On Tue, 11 Mar 2003 06:32:44 -0600, David C. Ullrich
>> <ull...@math.okstate.edu> wrote:
>>
>> >On 10 Mar 2003 15:19:33 -0800, slach...@fuse.net (Sam Lachterman)
>> >wrote:
>> >
>> >>A radial force field F in the plane can be written in the form F(x,y)
>> >>= f(r)R
>> >>where R = (x,y) and r = magnitude(R). Show that such a force field is
>> >>conservative.
>> >>
>>>I have not gotten very far with this at all. I have been trying to
>> >>show that F is a gradient of some function, but I do not understand
>> >>how to do this for *any* f which depends on the magnitude of R.
>> >
>> >Suppose g is a function of one variable and g' = f. Define
>> >p(x,y) = g(||(x,y)||). What is the gradient of p?
>>
>> Come to think of it maybe I should add a hint: Don't try to
>> find the gradient of p by calculating partial derivatives.
>> Instead use the fact that the gradient of p points in the
>> direction in which p is increasing fastest (and has
>> magnitude equal to the directional derivative of p in
>> that direction.)
>>
>>
>>
>> ******************
>>
>> David C. Ullrich
******************
David C. Ullrich
Rob Johnson
slach...@fuse.net (Sam Lachterman) wrote:
>A radial force field F in the plane can be written in the form F(x,y)
>= f(r)R
>where R = (x,y) and r = magnitude(R). Show that such a force field is
>conservative.
Yes. Define the energy, E, by
|\
E(r) = | r f(r) dr
\|
Then F(x,y) = grad(E(|(x,y)|).
Since F is a gradient, it is conservative. It is also true that
a body moving under the sole influence of F will sweep out equal
areas in equal times.
Rob Johnson
r...@whim.org
David C. Ullrich
wrote:
>In article <40dd692a.03031...@posting.google.com>,
>slach...@fuse.net (Sam Lachterman) wrote:
>>A radial force field F in the plane can be written in the form F(x,y)
>>= f(r)R
>>where R = (x,y) and r = magnitude(R). Show that such a force field is
>>conservative.
>
>Yes. Define the energy, E, by
>
> |\
> E(r) = | r f(r) dr
> \|
>
>Then F(x,y) = grad(E(|(x,y)|).
Looked at first like you meant the integral of f(r), not r f(r).
But of course you're right - I was inadvertently taking f to
be defined by F = f(r) R/||R||.
>Since F is a gradient, it is conservative. It is also true that
>a body moving under the sole influence of F will sweep out equal
>areas in equal times.
>
>Rob Johnson
>r...@whim.org
******************
David C. Ullrich
Bob Pease
"Casey Hawthorne" <cas...@istar.ca> wrote in message
news:eqhq6vk021anphkav...@4ax.com...
The other posters are using theorems which follow from the definition..
A force is "conservative" if the work done in moving a particle from one
place to another is independent of the path taken.
Not all force fields are conservative, but the form presented here seems to
be.
Bob Pease