# A The meaning of an integral of a one-form

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1. Jan 25, 2017

### observer1

So I understand that the integral of a differential form ω over the boundary of some orientable manifold Ω is equal to the integral of its exterior derivative dω over the whole of Ω.

And I understand that one can pull back the integral of a 1-form over a line to the line integral between the two endpoints.

I also understand (from my poor calculus training) how to do an integral of some function * dx between two endpoints. I KNOW what that means.

But I do NOT know what it means to integrate a one form. I mean, I get the integral, I get the presence of the integrand, I get the idea of "performing" it over a line. But if there is no "d" symbol on the integral, I cannot figure out what it means.

Note: I am teaching myself differential geometry, forms, calculus on manifolds, all at once. The clouds are clearing and rote operations are becoming more clear. I get that forms make doing these operatoins easier. I can see how the integral of force along a line bewtween two endpoints is work. I get that. But I do not understand what it means to integrate the 1-form force over the line. I cannot attach physical meaning to it.

So... What does it mean to integrate a one form? Where does it come from? (I get how I can pull it back, but I cannot figure out where the integral comes from or what it means when there is no "d")

2. Jan 25, 2017

### Orodruin

Staff Emeritus
But there are "d"s in the differential forms. The coordinate basis for a differential p-form has the structure $dx^{i_1}\wedge\ldots\wedge dx^{i_p}$.

3. Jan 25, 2017

### observer1

AH HA! Yes..... You are right. I see that now... So the "d" is there! I am sorry... I was dense.
So is that then how they quickly decide that force is a one form and it has a potential function such that dV = F

4. Jan 25, 2017

### Orodruin

Staff Emeritus
In order for this to be the case, the force must be conservative (which would mean that dF=0). If this is the case, Poincaré's lemma tells you that a potential exists (at least locally).

In order to conclude that force is a one-form, you would first have to provide details on how "force" has been introduced.

5. Jan 25, 2017

### observer1

Would you mind if I followed up with one more issue?

What motivates me to integrate a potential function alone a line?
Yes, when I do, and use generalized Stokes, and if I realize that the differential of a potential function is a force, and the resulting line integral pulled back to a 1-D mapping, I get the work... sure.

But what motivates me to even want to integate the potential function along the line?

I get the reverse motivation. That is founded in something I "physically" appreciate: the tangent of the force along the line, integrated.
But what motivates me to integrate the potential function along the line?

6. Jan 25, 2017

### Orodruin

Staff Emeritus
Nothing. The potential is a 0-form. It is 1-forms that are integrated along curves.

7. Jan 25, 2017

### observer1

I am sorry... I mis typed. I should have asked: Why integrate the differential of the potential along the line?

Integral of dV=-F
Why integrate the force along a line?

8. Jan 25, 2017

### Orodruin

Staff Emeritus
Are you asking for a mathematical or a physical reason?

9. Jan 25, 2017

### observer1

Was going to say physical, but now that I think.... both? Please?

10. Jan 25, 2017

### Orodruin

Staff Emeritus
Physics wise, because it gives a meaningful and good description of observations for the type of force that we call conservative and lets us relate work done to a potential energy. Mathematics wise, because the natural type of object to integrate a one-form over is a (directed) curve.