Tensors versus differential forms

1. May 1, 2009

RedX

What is the benefit of expressing Maxwell's equation in the language of differential forms? Differential forms seem to be inferior to the language of tensors. Sure you can do fancy things with the exterior derivative and hodge star, but with tensors you can derive those same identities with contraction properties of the permutation tensor.

2. May 2, 2009

Hurkyl

Staff Emeritus
Why does one use the complex numbers, instead of ordered pairs of real numbers?

Why does one use the integers, instead of ordered pairs of natural numbers modulo an equivalence relation?

Why does one study Euclidean geometry, instead of ordered pairs of real numbers?

3. May 2, 2009

Mentz114

Last edited by a moderator: May 4, 2017
4. May 2, 2009

dx

Differential forms are just a special type of tensors, so anything written in the language of differential forms can be written in the language of tensors.

5. May 2, 2009

Mentz114

Flanders thinks not - he states that tensor fields do not behave under transformations. It's in the part of the book available on Amazon.

I don't have a view, but I'm forced to learn diff. geom. to keep up with my subject.
I can't translate the work in Gronwald&Hehl (arXiv:gr-qc/9602013) into tensors only but that may just be because I don't know how.

6. May 2, 2009

dx

Flanders says,
I don't see how this is an argument in favor of differential forms, because the same applies to a 1-form field, which is in fact a contravariant vector field. If you have a differential 1-form on a 2-sphere, and a map from the 2-sphere to R³, there is no natural 1-form induced on R³.

Did I misunderstand what he was saying?

7. May 2, 2009

dx

Also, he goes on to say that for manifolds M and N, if you have a map φ : M → N, and a 0-form ω on N, then there is a natural 0-form induced on M (the pull back φ*ω).

But this is just because 0-forms are contravariant! It wouldn't work the other way, i.e. if ω was a 0-form on M, φ would not induce any natural 0-form on N!

8. May 2, 2009

Mentz114

Thanks for the technical stuff, dx. But it doesn't help me - I still have to learn diff.geom !

Expressing the action of a translation on a Lagrangian seems easy in dg terms but I can't see how to do it with tensors.

This may be off topic so I'll zip up.

9. May 2, 2009

RedX

My knowledge about this subjected is very limited, but can't you just use the inverse mapping of $$\phi$$, and then the pull back of this inverse transformation would be the "natural" form you seek? And doesn't what Flanders say about 0-forms and pull backs apply to k-forms? You always get a natural k-form with a pull back.

Also if 'a' and 'b' are k-forms of the same dimension, then a(*b)=(contraction of a and b) times volume element, which makes physical sense if k=1 since a 1-form is a vector (I guess you should * both sides of the equation to get rid of the volume element to just get the contraction), but what is the contraction of k>1 forms mean?

10. May 2, 2009

dx

No, because φ may not have an inverse.

11. May 2, 2009

MarkParis

A "0-form" is just a function. The label "contravariant" becomes trivial when referring to a function. Said differently, a function can be viewed either as contravariant or covariant.

12. May 2, 2009

Hurkyl

Staff Emeritus
There are pushforward and pullbacks on the individual fibers of the trivial line bundle. But "function on M" is a strictly contravariant idea; if you have a map M-->N, you can turn a "function on N" into a "function on M", but not vice versa. The dual to "real-valued function on M" is "curve in M".

Last edited: May 2, 2009
13. May 3, 2009

dx

No, functions are contravariant.

14. May 3, 2009

MarkParis

You're conflating issues here. The labels contra- and covariant refer to objects defined in the tangent space and its dual of a point in a manifold. This has no bearing on functions defined on the manifold which don't require any reference whatever to the tangent or cotangent space. A function is simply a map from the manifold to the real numbers. While it's true that not all maps phi:M-->N have an inverse, the classical differential geometric definition of contra- and covariance is not at all affected by this fact.

"No, functions are contravariant. "

Keep in mind that we're discussing a definition here. Generally there are no correct or incorrect definitions. If you'd like to say that a definition is consistent with a given published work, you should refer to it. Saying, "No, your definition is wrong," means nothing.

Now, I know that computer scientists refer to contra- and covariant functions. Perhaps that is the definition you're contemplating. ;)

15. May 3, 2009

dx

The definition of covariant and contravariant objects that I'm using is the standard one in mathematics and differential geometry. Let φ be a map φ : M → N. If we have a real-valued function on N, we can get a real valued function on M by composing it with f. This is called the pull-back of f from N to M by φ. This doesn't work in the other direction. Because of the fact that φ goes 'forwards' from M to N, and the pull-back goes backwards from N to M, we say that real valued functions on a manifold are contravariant.

16. May 3, 2009

dx

My posts in this thread were in fact in reference to a published work ("Differential Forms with Applications to the Physical Sciences" by Harley Flanders), as is clear from the discussion. Maybe you should read the thread fully before replying?

17. May 3, 2009

MarkParis

Kindly provide the page reference in Flanders where he says the functions are contravariant.

18. May 4, 2009

MarkParis

The answer to this is: of course it can. A classic treatise on the tensorial description is given by Steven Weinberg in his "Gravitation and Cosmology."

There are, however, interesting issues associated with the description of spin-1/2 fields on differentiable manifolds which do not, to my knowledge, admit a strictly tensorial description.

Last edited: May 4, 2009
19. May 4, 2009

AEM

Suppose you want to do a calculation in order to make a comparison with experiment, for example calculating the equation of a geodesic in the Kerr metric, or even the Schwarzschild metric. Which framework would be more efficient: differential forms or tensorial notation? Suppose you wanted to do something similar in Maxwell's theory, which would be better?

20. May 4, 2009

MarkParis

This is an excellent question and one which can be answered on several levels. For example, we can consider the question from the perspective of geometric content, notation, computation, and others.

The perspective I (perhaps some others, too) find most useful, is the geometric perspective. What I mean by this, in short, is that the objects that we use to describe physical systems are independent of the coordinate systems that we might use to describe them. For example, in 3-dimensional Euclidean space, a vector which points along a particular direction may be described in Cartesian (regular, rectilinear) coordinates, spherical coordinates, etc. The physical content of the vector is unaffected by the selection of a particular system.

So, to address the issue of the electromagnetic field equations -- Maxwell's equations -- we see in the differential-form version of these equations a geometric description of the field quantities which is coordinate independent. And this, I think, isn't really just a "benefit." From the geometric perspective it's a prerequisite to have a meaningful physical theory.

The issue of notation -- this is a two-edged sword. For simplifying notation, which differential forms usually afford, can have the effect of making complicated ideas and/or their proofs more apparent. The other edge is that they can serve to obscure complexity and lead to confusion. Usually though, I think the risk of the latter is outweighed by the former.

Computation is another interesting and important point. Let's get something formal out of the way first: both tensors and differential forms are geometric constructions. Tensors are defined in a coordinate independent manner as multilinear maps from the products of vector spaces and its dual to (typically) the real numbers. "Forms" are simply the multilinear, alternating variety of these from products of the vector space to the reals and obey the exterior algebra. Differential forms are the generalization of forms to differentiable manifolds where the vector space (and its dual) is replaced by the tangent (and the cotangent) space. And in flat (Minkowski) space, the differential forms are just a particular type of tensor field defined on the four-dimensional spacetime.

Computations of a given physical observable always require a specific coordinate system. Whether one wishes to use the formalism of differential forms or not is really just a matter of preference. While Maxwell's equations are compact and elegant when expressed in terms of differential forms, I'm not aware of any specific calculations that are carried out very far in that formalism. Usually, one computes in terms of the electromagnetic field tensor $$F_{\mu\nu}$$ in a given frame. In general relativity, however, the computation of the curvature tensor, in particular, is greatly simplified by "sticking with" differential forms for a long way into the calculation.