# Why Is There a Discrepancy in My Potential Function Calculation?

• athrun200
In summary, the conversation discusses different approaches to solving a problem involving finding the potential of a given force field. The first approach involves setting up the equation phi = F_x and F_y and integrating them to find the form of phi. The second approach involves using a method learned from mathematical methods in the physical sciences, in which the potential is only unique up to a scalar additive constant. The conversation also discusses the use of trigonometric identities and finding the constant in the equation. Ultimately, it is determined that the potential is given by phi = (x^2+1)cos^2(y)+c, where c is a constant.
athrun200

## The Attempt at a Solution

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

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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
$$\phi = \int F_x dx= \int 2x cos^2y dx =?$$

lanedance said:
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.

lanedance said:
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?

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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
$$\phi = \int F_y dy= ?$$

lanedance said:
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
$(x^2+1)\frac{2cos^2y-1}{2}+C(x)$

$$\phi = \int F_x dx = \int F_y dy$$

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)

lanedance said:
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)

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athrun200 said:
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

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

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
$$x^2cos^2(y)+ cos^2(y)+h(x) = x^2cos^2(y) + f(y)$$

$$h(x) = 0$$
$$f(y) = cos^2(y)$$

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

where c is any constant

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

## 1. What is potential energy?

Potential energy is a form of energy that an object possesses due to its position or configuration. It is stored energy that has the potential to do work.

## 2. How is potential energy different from kinetic energy?

Potential energy is the energy an object has due to its position or configuration, while kinetic energy is the energy an object has due to its motion. They are two forms of energy that can be converted into each other.

## 3. What are the different types of potential energy?

There are several types of potential energy, including gravitational potential energy, elastic potential energy, chemical potential energy, and electric potential energy. Each type is associated with a specific force or field.

## 4. What is the principle of conservation of energy?

The principle of conservation of energy states that energy cannot be created or destroyed, only transferred or converted from one form to another. In a closed system, the total amount of energy remains constant.

## 5. How does potential energy relate to conservative forces?

Conservative forces are forces that do not dissipate energy and can be expressed as the negative gradient of a potential energy function. This means that the work done by a conservative force is independent of the path taken, only dependent on the initial and final positions. In other words, potential energy is a measure of the work a conservative force can do on an object.

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