Why elementary work is not an exact differential?

In summary, elementary work is defined as δW=Fdr, but it cannot be written as dW=Fdr in general. This is because work, like heat, is not a function of state and depends on the path. In the case of a conservative force, work is an exact differential and can be written as δW=-dU, where U is the potential energy function. However, for non-conservative forces, work is not a derivative of any function. This is why the use of δW is more appropriate in general, as it allows for the consideration of non-conservative forces.
  • #1
phydev
20
0
Why elementary work is defined as δW=Fdr?
My ques. is not on the definition; it is on why it cannot be dW=Fdr?
 
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  • #2
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.
 
  • #3
tommyli said:
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!
 
  • #4
Work is a line integral:

[tex]W=\oint \mathbf F \cdot d\mathbf r[/tex]

It is path independent in the case of a conservative force, but in general the integral depends on the path. In other words, [itex]\mathbf F \cdot d\mathbf r[/itex] is not an exact differential by definition (an exact differential is path independent).
 
  • #5
D H said:
Work is a line integral:

[tex]W=\oint \mathbf F \cdot d\mathbf r[/tex]

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

Thanks, it was helpful!
 
  • #6
phydev said:
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.
 
  • #7
tommyli said:
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!
 
  • #8
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.
 
  • #9
phydev said:
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.
 
  • #10
tommyli said:
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!
 

1. Why is elementary work not an exact differential?

Elementary work is not an exact differential because it depends on the path taken between two points in a thermodynamic system. This means that the work done on or by the system will be different depending on the specific path taken, even if the starting and ending points are the same. In other words, the value of work is not solely determined by the state of the system but also by the process used to change the system's state.

2. What does "exact differential" mean in the context of thermodynamics?

In thermodynamics, an exact differential means that the value of a property only depends on the initial and final states of the system, and not on the path taken between these states. This is in contrast to non-exact differentials, such as elementary work, which depend on the specific path taken in changing the system's state.

3. How does this concept relate to the first law of thermodynamics?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted. This means that the total change in energy of a system is independent of the path taken, and therefore energy is an exact differential. However, work, which is a form of energy transfer, is not an exact differential as it depends on the path taken between two states of the system.

4. Can you give an example of a process where elementary work is not an exact differential?

An example of a process where elementary work is not an exact differential is the expansion of a gas against a constant external pressure. If the gas expands slowly and reversibly, the work done is equal to the external pressure multiplied by the change in volume. However, if the gas expands rapidly and irreversibly, the work done will be greater than the external pressure multiplied by the change in volume. This is because the rapid expansion process involves dissipative forces, such as friction, that contribute to the work done.

5. How is the concept of exact differentials used in thermodynamic calculations?

In thermodynamic calculations, exact differentials are used to determine the total change in a property of a system, such as internal energy or entropy, by integrating over the entire path between the initial and final states. This allows for the calculation of thermodynamic quantities that are independent of the specific path taken, such as the change in internal energy of a system. In contrast, non-exact differentials, like elementary work, must be calculated separately for each specific path and cannot be integrated to determine a total change in a property.

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