Gradient one-form: normal or tangent

[joneall]

Working through Schutz "First course in general relativity" + Carroll, Hartle and Collier, with some help from Wikipedia and older posts on this forum. I am confused about the gradient one-form and whether or not it is normal to a surface.

In the words of Wikipedia (gradient):

If f is differentiable, then the dot product (∇f )x ⋅ v of the gradient at a point x with a vector v gives the directional derivative of f at x in the direction v. It follows that in this case the gradient of f is orthogonal to the level sets of f.

The first sentence is almost clear. What do they mean by "dot product"? Is this a scalar product, which I think requires a metric, or a contraction, which does not? When written out, the formula for the directional derivative contains no metric and is clearly the contraction of a vector and the gradient.

It's the second sentence that gets me. Just how it "follows" that the gradient is orthogonal to the level sets does not jump out at me. (Sorry, I don't know how to write formulas on the forum.)

In good ol'-fashioned 2d Euclidan geometry, if y = f(x), then the gradient dy/dx is the slope of the tangent to the curve at a particular value of x. So how do we get from tangents to orthogonals?

To continue, Schutz (64) talks about the model of one-forms as surfaces crossed by vectors. He says: "... the one-form's surfaces are straight and parallel.This is because we deal with one-forms at a point, not over an extended region: 'tangent' one-forms in the same sense as tangent vectors."

So now one-forms are tangents, even tho the gradient is normal. As Schutz (66) says 2 pages later: "A one-form is said to be normal to a surface if its value is zero on every vector tangent to the surface." By "on", I presume he means "associated with". That, at least, seems clear. A normal to a surface has no components on a tangent to the surface.

Schutz again (72): "A vector is said to be normal to a surface if its associated one-form is a normal one-form."

I find all this very confusing and difficult to visualize. Yet visualize I must.

 

[Orodruin]

I would strongly suggest against using Wikipedia as a reliable source.

You might debate whether or not to consider the differential df as a "gradient". It is a one-form with components $\partial_\mu f$ in all coordinate systems. However, it is a dual vector and not a tangent vector and as such not in the tangent vector space. What people would normally refer to as the gradient is defined when you have a metric. It is then the unique tangent vector $X$ such that $g(X,Y) = df(Y)$ for all tangent vectors $Y$.

Part of your confusion also seems to stem from the meaning of "tangent vector". A tangent vector is any vector in the tangent space of a manifold. For any given tangent vector and a surface, the tangent vector may be normal to the surface or not.

In good ol'-fashioned 2d Euclidan geometry, if y = f(x), then the gradient dy/dx is the slope of the tangent to the curve at a particular value of x. So how do we get from tangents to orthogonals?

You are drawing the wrong inference here. Your "2d case" is actually a function on a one-dimensional manifold. Your "gradient" is a vector in a one-dimensional vector space. In the one-dimensional case, level surfaces are points and your gradient is indeed orthogonal to all displacements in the surface (there are no displacements in the surface!)

 

[joneall]

I would strongly suggest against using Wikipedia as a reliable source.

Sure. I just did because it was simple and they agreed with what else I had read.

You might debate whether or not to consider the differential df as a "gradient". It is a one-form with components $\partial_\mu f$ in all coordinate systems. However, it is a dual vector and not a tangent vector and as such not in the tangent vector space. What people would normally refer to as the gradient is defined when you have a metric. It is then the unique tangent vector $X$ such that $g(X,Y) = df(Y)$ for all tangent vectors $Y$.

Ok, that is clearer. Of course, it's not I who call the one-form a gradient, it's Schutz, Hartle and others.

Part of your confusion also seems to stem from the meaning of "tangent vector". A tangent vector is any vector in the tangent space of a manifold. For any given tangent vector and a surface, the tangent vector may be normal to the surface or not.

Aha, that is indeed much clearer. It is hard to get away from imagining the tangent space as a 2d plane tangent to the surface at the point. I'll have to get used to that.

You are drawing the wrong inference here. Your "2d case" is actually a function on a one-dimensional manifold. Your "gradient" is a vector in a one-dimensional vector space. In the one-dimensional case, level surfaces are points and your gradient is indeed orthogonal to all displacements in the surface (there are no displacements in the surface!)

Sorry, I don't get that. What is the use of a gradient in a one-dimensional vector space. Or any vector? They would all point either up or down, with varying lengths?

Thanks for your reply!

 

[stevendaryl]

In vector calculus for Euclidean space, people often use the word "gradient" and the symbol $\nabla f$ to mean the vector whose components are $\partial_\mu f$. Then the directional derivative of $f$ along a vector $v$ is given by:

$(\nabla f) \cdot v$

This notion of "gradient" involves two uses of the metric, which cancel each other out, in a sense.

1. To construct a vector from the components $\partial_\mu f$, you have to use the metric: $(\nabla f)^\nu = g^{\mu \nu} \partial_\mu f$
2. To construct the dot-product, you use the metric again: $(\nabla f) \cdot v = g_{\mu \nu} (\nabla f)^\mu v^\nu$

But the metric drops out of the final result: $(\nabla f) \cdot v = g_{\mu \nu} (\nabla f)^\mu v^\nu = g_{\mu \nu} (g^{\mu \lambda} \partial_\lambda f) v^\nu = (g_{\mu \nu} g^{\mu \lambda}) \partial_\lambda f v^\nu = \delta^\lambda_\nu \partial_\lambda v^\nu = \partial_\nu v^\nu$

As a vector, $\nabla f$ is perpendicular to the surfaces of constant $f$, the "level curves" of f.

 

[robphy]

Enlightening reading...
https://people.ucsc.edu/~rmont/papers/Burke_DivGradCurl.pdf
unfinished draft by Bill Burke
http://www.ucolick.org/~burke/home.html

It might be good to have a physical model of these objects. Electromagnetism provides good examples. However, one has to take the view of what is called pre-metric electromagnetism... where the familiar Maxwell Equations for the electric and magnetic vector fields have been--in some sense--had the metric and volume element factored out. One ends up with Gauss' Laws being associated with two-forms [describing flux] and Faraday and Ampere's Laws being associated with one-forms [describing circulation] and time-derivatives of two-forms [describing changing flux].

In this picture, the electric-field one-form can be visualized like a family of equipotential surfaces... with units Volts/meter.
In doing a line-integral, one considers tangent vectors to curves. The non-metrical contraction yields the the change in voltage along the path. This piercing of surfaces is associated with Misner-Thorne-Wheeler's "bongs of a bell".

I believe the origin of this idea comes from van Dantzig (1934) (https://www.google.com/search?q=van+Dantzig+electromagnetism ) and Schouten (https://www.amazon.com/dp/0486655822/?tag=pfamazon01-20 ).

I have created some visualizations in VRML.. but they need to be updated into a format more folks can view.

 

[vanhees71]

Well, the question can be answered also for the general case of a general differentiable manifold (you don't even need a metric or pseudo-metric). The gradient of a scalar field is, by definition, the one form
$$\mathrm{d} f=\mathrm{d} q^{\mu} \partial_{\mu} f.$$
Now, if you consider a surface, implicitly defined by
$$f=c=\text{const}, \qquad (*)$$
then of course
$$\mathrm{d} f=0$$
along the surface. If now you have a (pseudo-)metric, you can write
$$\mathrm{d} f=\mathrm{d} q^{\mu} \partial_{\mu} f= g_{\mu \nu} \mathrm{d} q^{\mu} \partial^{\nu} f,$$
in this sense
$$\partial^{\nu} f=g^{\nu \rho} \partial_{\rho} f$$
is a vector field perpendicular to the surface defined by (*).

 

[Orodruin]

Well, the question can be answered also for the general case of a general differentiable manifold (you don't even need a metric or pseudo-metric). The gradient of a scalar field is, by definition, the one form
$$\mathrm{d} f=\mathrm{d} q^{\mu} \partial_{\mu} f.$$

I know some mathematicians that would not agree with this. They would tell you that the gradient is a tangent vector defined only when there is a metric and that the corresponding one-form is just to be called "the differential of f".

Of course, the bottom line is that df is a one-form that maps all displacements in a level surface to zero and has components $\partial_\mu f$. Also, the language is not helped by the fact that the set of one-forms is also a vector space and consequently one-forms are also vectors in that sense - although "vector" is sometimes taken to refer exclusively to tangent vectors...

 

[vanhees71]

Ok, but that's semantics. For me the "natural" view is that the gradient of a scalar field is a one form. If you have a (pseudo-)metric you can map it basis-independently to a vector (1st-rank tensor). Maybe, some mathematicians call this the gradient.

All this is, of course, not systematically taught from the very beginning, when one considers vector calculus on the Euclidean $\mathbb{R}^3$ using Cartesian coordinates or orthonormal curved coordinates (usually cylinder and spherical coordinates). Then usually one doesn't even distinguish between vectors and one-forms at all! Of course the components are the same, because then $g_{\mu \nu}=\delta_{\mu \nu}$, and you consider objects whose components are only tensors in the restricted sense of rotations and not general changes between bases.

 

[Orodruin]

Ok, but that's semantics. For me the "natural" view is that the gradient of a scalar field is a one form. If you have a (pseudo-)metric you can map it basis-independently to a vector (1st-rank tensor). Maybe, some mathematicians call this the gradient.

I agree that it is semantics, but semantics can sometimes be important in order to make yourself understood, which is why I pointed it out. In the end, the object with components $\partial_\mu f$ in a coordinate basis is a one-form and it is a natural extension of the gradient.