# Why is force a 1-form?

#### bronxman

Hello again,

Yesterday, through your help, I took my understanding to a new level.

I can now phrase my single question and I hope you will allow me to make it a separate thread...

• I understand that the integrand in a line integral is a natural 1-form.
• I understand that to obtain WORK = F.dx that Force is thus a 1-form.
• (In fact, I understand that velocity and acceleration are vectors but force and moments are 1-forms.)

BUT WHY IS FORCE A 1-FORM, PHYSICALLY? I understand the outcome of the MATH, but what is the physical meaning of Force (acceleration weighted with mass) being a 1-form?

Related Linear and Abstract Algebra News on Phys.org

#### bronxman

Is it, perhaps, just this?

Gradient is, mathematically, a 1-form. And in claiming force to be a one-form, we are assuming it is derivable from a potential function?

If so, then can I assume that forces derived from, say, friction, are not 1-forms?

Is that it? Force being a one-form presupposed potential energy?

#### pasmith

Homework Helper
BUT WHY IS FORCE A 1-FORM, PHYSICALLY? I understand the outcome of the MATH, but what is the physical meaning of Force (acceleration weighted with mass) being a 1-form?
There is no physical meaning. Depending on what you want to do, forces can be represented mathematically as either vectors or 1-forms.

If you want to find the acceleration of an object then using a vector is best. If you want to find the work done in moving along a given path then using a 1-form is best.

#### HallsofIvy

Homework Helper
This would be better placed in "Differential Geometry" than "Linear and Abstract Algebra".

• Ssnow

#### ShayanJ

Gold Member
I think its natural. Because when we apply a vector(displacement vector) to the force, we want to get a scalar(work). But something that gets a vector and gives a scalar is a 1-form.

#### micromass

I think its natural. Because when we apply a vector(displacement vector) to the force, we want to get a scalar(work). But something that gets a vector and gives a scalar is a 1-form.
Then we might as well see the displacement vector as the 1-form and the force as the vector, and mathematically that works out fine, but it's (rightfully) not how we do things.
The intuition behind 1-forms is that "anything you want to integrate is a form". Since you want to integrate force, it should be a form.

#### WWGD

Gold Member
A 1-form, formally, is just a linear map, an element of a dual space . Given an f.d vector space V, any linear map defined on v in V into the base field is a 1-form. So, what arguments does a fdx take in order to spit out a scalar? And, yes, we integrate k-forms over k-manifolds.

Last edited:

#### WWGD

Gold Member
Hello again,

Yesterday, through your help, I took my understanding to a new level.

I can now phrase my single question and I hope you will allow me to make it a separate thread...

• I understand that the integrand in a line integral is a natural 1-form.
• I understand that to obtain WORK = F.dx that Force is thus a 1-form.
• (In fact, I understand that velocity and acceleration are vectors but force and moments are 1-forms.)

BUT WHY IS FORCE A 1-FORM, PHYSICALLY? I understand the outcome of the MATH, but what is the physical meaning of Force (acceleration weighted with mass) being a 1-form?
A force, given as a function, is actually a 0-form. fdx is a 1-form, as the wedge of the 0-form f and the 1-form dx.

EDIT: still, as atyy said , if V is a f.d vector space with a distinguished non-degenerate form, then V

and V* are naturally isomorphic, i.e., the isomorphism does not depend on a choice of basis ( you
can make this more precise using language from category theory ) . Still, I don't think the dual of
a 0-form is a 0-vector, but I am not sure.

Last edited:

#### bronxman

A force, given as a function, is actually a 0-form. fdx is a 1-form, as the wedge of the 0-form f and the 1-form dx.
OH! I SEE what I was now misreading. Yes. In Frankel's Geometry of Physics, I see how the integrand in a line integral is a one form. And yes, that means the F function AND the dx TOGETHER. Now I see that with the PULL BACK!!! It seems that when one pulls back the integral of the 1-form F, to a map on the real line, one gets the one-form fdx

#### atyy

Last edited by a moderator:

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving