Why Is There a Discrepancy in My Potential Function Calculation?

[athrun200]

Homework Statement

Homework Equations

The Attempt at a Solution

The answer should be -(x^{2}-1)cos^{2}\theta
But I get a constant -1.
Why?

 

[lanedance]

can you explain what you are attempting to do? I would approach as follows

say phi exists then you know
F_x = \frac{\partial\phi}{\partial x}
F_y = \frac{\partial\phi}{\partial y}

so integrating the first gives
\phi = \int F_x dx= \int 2x cos^2y dx =?

 

[athrun200]

can you explain what you are attempting to do? I would approach as follows

Sorry that I skip some steps and forget to give explanation.

This is my approach.

say phi exists then you know
F_x = \frac{\partial\phi}{\partial x}
F_y = \frac{\partial\phi}{\partial y}

so integrating the first gives
\phi = \int F_x dx= \int 2x cos^2y dx =?

This yields x^2 cos^2y+C

But how to find the constant?

 

[lanedance]

not quite as you are only integrating w.r.t. to x, so c could be a function of y
\phi = \int F_x dx= \int 2x cos^2y dx = x^2cos^2y + c(y)

now also do it from the other direction
\phi = \int F_y dy= ?

 

[athrun200]

not quite as you are only integrating w.r.t. to x, so c could be a function of y
\phi = \int F_x dx= \int 2x cos^2y dx = x^2cos^2y + c(y)

now also do it from the other direction
\phi = \int F_y dy= ?

(x^2+1)\frac{cos2y}{2}+C(x)

it can then simplfied as
(x^2+1)\frac{2cos^2y-1}{2}+C(x)

 

[lanedance]

i can't quite follow your simplification, you should equate
\phi = \int F_x dx = \int F_y dy

also I would use a different constant function for the 2nd integral (say c(y) for first and d(x) for the second)

and from there you should be able to deduce withther it is possible to to solve c(x) & d(y)

 

[athrun200]

i can't quite follow your simplification, you should equate
\phi = \int F_x dx = \int F_y dy

This is just a identity.
http://en.wikipedia.org/wiki/List_o...ouble-.2C_triple-.2C_and_half-angle_formulae"

 

[athrun200]

and from there you should be able to deduce withther it is possible to to solve c(x) & d(y)

It's my first time to solve the problem like this.
So I stuck in here

I would like to ask, what's wrong with my method above?
I have solved a lot of problems by using that method.(I learn it from mathematical methods in the physical sciences Chapter 6 sec8)

 

[lanedance]

This is just a identity.
http://en.wikipedia.org/wiki/List_o...ouble-.2C_triple-.2C_and_half-angle_formulae"

i don't think angles apply here...

 

[lanedance]

It's my first time to solve the problem like this.
So I stuck in here

I would like to ask, what's wrong with my method above?
I have solved a lot of problems by using that method.(I learn it from mathematical methods in the physical sciences Chapter 6 sec8)

stick to whatever you're comfortable with... took me awhile to figure out what you're doing as there's a few jumps (also the big pics are difficult to read on a laptop screen), but i think you set phi=0 at the origin and do a path interval over F to find the form of phi if it exists, in essence i think they're the same thing though yours has the advantage of setting one value to zero in the integral

 

[lanedance]

also note the potential is only unique upto a scalar additive constant ie (phi) or (phi +1) give the same force field

 

[lanedance]

this is how i would approach it using my method (i re-named c & d as f & g to be clear they are functions)
\phi(x,y) = \int F_x dx = \int 2x cos^2(y) = x^2cos^2(y) + f(y)
\phi(x,y) = \int F_y dx = \int -(x^2+1) sin(2y) dy = (x^2+1) \frac{1}{2}cos(2y) +g(x) = (x^2+1) (cos^2(y)-\frac{1}{2})+g(x)

in the last line we may as well group all the x only terms in a function, so let
h(x) = +\frac{1}{2}(x^2-1)+g(x)

equating and re-arranging a little gives
x^2cos^2(y)+ cos^2(y)+h(x) = x^2cos^2(y) + f(y)

then we can read off (up to an additive constant)
h(x) = 0
f(y) = cos^2(y)

so we get
\phi(x,y) = (x^2+1)cos^2(y)+ c

where c is any constant

 

[lanedance]

after all that... and re-reading the posts i think it was only post 11 you needed!

you can only ever determine phi upto a additive scalar constant, as the constant will always be sent to zero in the differentiation to get the field