Definite and indefinite integration in the definition of work

[etotheipi]

This is going to sound like a silly question, but here we go anyway! I've always thought about a definite integral being used for modelling a change in some quantity whilst an indefinite integral is employed to find the defining function of that quantity.

For example, consider the force-momentum relationship $$F dt = dP$$ If we integrate definitely from t₁ to t₂, we will get a change in momentum or impulse:$$\int_{t_{1}}^{t_{2}} F dt = \int_{t_{1}}^{t_{2}} dP = P_{2} - P_{1} = \Delta P = I$$whilst a definite integral will instead allow us to pick a constant of integration to find the function which outputs the momentum at any given time:$$\int F dt = P + C$$It is a commonly used definition that work is the integral of force with respect to displacement, but it is usually written as an indefinite integral. This is confusing to me, since the indefinite integral of a force with respect to displacement appears to define the kinetic energy of the body:$$\int F dx = \int mv dv = E_{k} + C$$

On the other hand, a definite integral from position x₁ to x₂ defines a change in kinetic energy, and the net work is also commonly defined as the change in kinetic energy of a body, like the following

$$W = \int_{x_{1}}^{x_{2}} F dx = E_{k2} - E_{k1} = \Delta E_{k}$$

Is the quantity work more precisely defined as a definite integral as opposed to the indefinite definition which seems to be everywhere?

 

[PeroK]

This is going to sound like a silly question, but here we go anyway! I've always thought about a definite integral being used for modelling a change in some quantity whilst an indefinite integral is employed to find the defining function of that quantity.

For example, consider the force-momentum relationship $$F dt = dP$$ If we integrate definitely from t₁ to t₂, we will get a change in momentum or impulse:$$\int_{t_{1}}^{t_{2}} F dt = \int_{t_{1}}^{t_{2}} dP = P_{2} - P_{1} = \Delta P = I$$whilst a definite integral will instead allow us to pick a constant of integration to find the function which outputs the momentum at any given time:$$\int F dt = P + C$$It is a commonly used definition that work is the integral of force with respect to displacement, but it is usually written as an indefinite integral. This is confusing to me, since the indefinite integral of a force with respect to displacement appears to define the kinetic energy of the body:$$\int F dx = \int mv dv = E_{k} + C$$

On the other hand, a definite integral from position x₁ to x₂ defines a change in kinetic energy, and the net work is also commonly defined as the change in kinetic energy of a body, like the following

$$W = \int_{x_{1}}^{x_{2}} F dx = E_{k2} - E_{k1} = \Delta E_{k}$$

Is the quantity work more precisely defined as a definite integral as opposed to the indefinite definition which seems to be everywhere?

In general, these things are always definite integrals.

You can, however, have a definite integral with variable bounds - usually the end point. This creates the variable KE or momentum as a force acts over a variable time interval or distance. So, for example:

$p(t) = p(0) + \int_0^t F(t')dt'$

Writing:

$p(t) = \int F(t)dt$

Is a bit sloppy, I would say.

 

[etotheipi]

In general, these things are always definite integrals.

You can, however, have a definite integral with variable bounds - usually the end point. This creates the variable KE or momentum as a force acts over a variable time interval or distance. So, for example:

$p(t) = p(0) + \int_0^t F(t')dt'$

Writing:

$p(t) = \int F(t)dt$

Is a bit sloppy, I would say.

Thank you, that clears up a lot of confusion!

 

[sophiecentaur]

Is the quantity work more precisely defined as a definite integral as opposed to the indefinite definition which seems to be everywhere?

It's the value of C that seems to be the difficulty. How are you going to find it without working out, effectively, a definite integral somewhere? The indefinite integral is only half the story.

 

[PeroK]

Thank you, that clears up a lot of confusion!

Actually, a further thought. In classical mechanics we are normally using an inertial reference frame, as defined by Newton's first law. If we pick a reference frame then the momentum has a definite value, as a function of time. That's represented by the definite integral above.

But, you might ask whether the momentum can be represented more generally, independent of any particular inertial reference frame. For that you could use the indefinite integral. Note that mathematically an indefinite integral is not a function, but an equivalence class of functions. The momentum, therefore, is also an equivalence class of functions, one for each inertial reference frame.

The quantity common to all inertial reference frames is the change in momentum. The actual momentum itself depends on choice of frame.

In this case, however, you would need to make clear that's what is meant. Once you choose a reference frame, then the integral becomes definite; and the momentum is resolved to a specific function.

In fact, not to get too far ahead, being able to think of a quantity like momentum in this way eventually becomes essential in General Relativity, where there are no global inertial reference frames. But that's a story for another day.