Why elementary work is not an exact differential?

[phydev]

Why elementary work is defined as δW=Fdr?
My ques. is not on the definition; it is on why it cannot be dW=Fdr?

 

[tommyli]

It is only an exact differential int he case of a conservative force. Precisely, if there is a potential we write

dU = -Fdr

which implies that F = -dU/dr.

 

[phydev]

It is only an exact differential int he case of a conservative force. Precisely, if there is a potential we write

dU = -Fdr

which implies that F = -dU/dr.

I'm not talking about U(potential energy function), I'm asking about W.
I know that in case of conservative/potential field δW=-dU.

Reference: Fundamental Laws of Mechanics, IE irodov
from equation 3.1 to 3.49
wherever needed he used δA for elementary work, in general!

 

[D H]

Work is a line integral:

$$W=\oint \mathbf F \cdot d\mathbf r$$

It is path independent in the case of a conservative force, but in general the integral depends on the path. In other words, $\mathbf F \cdot d\mathbf r$ is not an exact differential by definition (an exact differential is path independent).

 

[phydev]

Work is a line integral:

$$W=\oint \mathbf F \cdot d\mathbf r$$

It is path independent in the case of a conservative force, but in general the integral depends on the path. In other words, $\mathbf F \cdot d\mathbf r$ is not an exact differential by definition (an exact differential is path independent).

Thanks, it was helpful!

 

[tommyli]

I'm not talking about U(potential energy function), I'm asking about W.
I know that in case of conservative/potential field δW=-dU.

Reference: Fundamental Laws of Mechanics, IE irodov
from equation 3.1 to 3.49
wherever needed he used δA for elementary work, in general!

Whenever you write df it implies that the differential operator is applied to the function f, so if you write work as an exact differentail dW implicitly you are saying force can be written as the derivative of some function of spatial coordinates.

 

[phydev]

Whenever you write df it implies that the differential operator is applied to the function f, so if you write work as an exact differentail dW implicitly you are saying force can be written as the derivative of some function of spatial coordinates.

Yeah! right!
Now, what does it further imply?
Cannot force be derivative of a function of spatial coordinates?

I think I have got it, but request you to elaborate so that I may confirm.

Thanks!

 

[Hobold]

If work was an exact differential, for any two points a and b, you could write that the work to go from one point is F(b) - F(a), where F' is work. But this is most certainly not true, as this is saying work is a function of state, i.e. if you have a point, you'd have a work associated to it. This is false, as work is something you use to go from one state to another. It's pretty much like heat. Heat is also not a function of state and depends on the path.

 

[tommyli]

Yeah! right!
Now, what does it further imply?
Cannot force be derivative of a function of spatial coordinates?

I think I have got it, but request you to elaborate so that I may confirm.

Thanks!

If force is a derivative of some function of spatial coordinates, this function is called the potential energy, and the force is conservative. Non-conservative forces are certainly not derivatives of any function.

 

[phydev]

If force is a derivative of some function of spatial coordinates, this function is called the potential energy, and the force is conservative. Non-conservative forces are certainly not derivatives of any function.

well... thanks,
I concluded the same!