Work of a non conservative field

• Granger
In summary: Basically, the integral is complicated, and that's life. Sometimes problems are not easy, and we just have to deal with them anyway.
Granger

Homework Statement

Compute the work of the vector field $$H: \mathbb{R^2} \setminus{(0,0}) \to \mathbb{R}$$

$$H(x,y)=\bigg(y^2-\frac{y}{x^2+y^2},1+2xy+\frac{x}{x^2+y^2}\bigg)$$

in the path $$g(t) = (1-t^2, t^2+t-1) with t\in[-1,1]$$

Homework Equations

3. The Attempt at a Solution [/B]

So first I considered my vector field as a sum of 2 vector fields: $$H = F + G$$

$$F(x,y)=\bigg(y^2,1+2xy\bigg)$$

$$G(x,y)=\bigg(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\bigg)$$

The vector field $$F$$ is conservative with one of many potentials $$A(x,y) = y^2x+y$$
Then I worked out the work using the definition and fundamental theorem of calculus obtaining the value 2.

So no problems at this point.

But $$G$$ is not a conservative vector field (it doesn't have a potential, even though it's a closed field). How should I proceed? I tried the definition but we get to a very complicated integral... The path isn't closed so we can't apply Green's theorem... What should I do?

Granger said:

Homework Statement

Compute the work of the vector field $$H: \mathbb{R^2} \setminus{(0,0}) \to \mathbb{R}$$

$$H(x,y)=\bigg(y^2-\frac{y}{x^2+y^2},1+2xy+\frac{x}{x^2+y^2}\bigg)$$

in the path $$g(t) = (1-t^2, t^2+t-1) with t\in[-1,1]$$

Homework Equations

3. The Attempt at a Solution [/B]

So first I considered my vector field as a sum of 2 vector fields: $$H = F + G$$

$$F(x,y)=\bigg(y^2,1+2xy\bigg)$$

$$G(x,y)=\bigg(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\bigg)$$

The vector field $$F$$ is conservative with one of many potentials $$A(x,y) = y^2x+y$$
Then I worked out the work using the definition and fundamental theorem of calculus obtaining the value 2.

So no problems at this point.

But $$G$$ is not a conservative vector field (it doesn't have a potential, even though it's a closed field). How should I proceed? I tried the definition but we get to a very complicated integral... The path isn't closed so we can't apply Green's theorem... What should I do?

What is preventing you from just going ahead and computing ##\int_{-1}^1 \vec{H}(t) \cdot \nabla g(t) \, dt?##

Ray Vickson said:
What is preventing you from just going ahead and computing ##\int_{-1}^1 \vec{H}(t) \cdot \nabla g(t) \, dt?##

You mean H(g(t)) (with the vector sign). Because the integral gets complicated.

Granger said:
You mean H(g(t)) (with the vector sign). Because the integral gets complicated.

Yes, I know, but if you simplify the integrand first (to a single ratio of two polynomials) you DO get a tractable integral that is not too bad. (The integral is horrible if you do not simplify the integrand first.) For example, you could submit the integration to Wolfram Alpha and get a usable answer. Alternatively, you could use a numerical integration method to get a numerical answer to as much accuracy as you want.

Basically, the integral is complicated, and that's life. Sometimes problems are not easy, and we just have to deal with them anyway.

What is a non conservative field?

A non conservative field is a type of vector field in which the work done by the field on an object depends on the path taken, rather than just the starting and ending points. This means that the work done is not independent of the path, unlike in a conservative field.

How is work calculated in a non conservative field?

In a non conservative field, work is calculated by taking the line integral of the vector field along the path of an object. This involves breaking up the path into small segments and summing the work done by the field along each segment. The total work is then the sum of all these small segments.

What is the significance of a non conservative field in physics?

Non conservative fields are important in physics because they represent situations where the energy of a system is not conserved. This can occur in real-world scenarios such as friction or air resistance, where the work done on an object is not completely converted into kinetic or potential energy.

How can a non conservative field be identified?

A non conservative field can be identified by looking at the path independence of work. In a conservative field, the work done is the same regardless of the path taken, while in a non conservative field, the work done is path dependent. Additionally, a non conservative field can be identified by the presence of external forces or sources that are not included in the vector field.

What are some real-life examples of a non conservative field?

Some common examples of non conservative fields include friction, air resistance, and magnetic fields. In these situations, the work done on an object is not completely converted into kinetic or potential energy, and the work done depends on the path taken. Other examples include electric fields in the presence of non-conservative sources, such as a battery or generator.

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