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RedX

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RedX

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Hurkyl

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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

Mentz114

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Read what Harley Flanders has to say on this subject here -

https://www.amazon.com/dp/0486661695/?tag=pfamazon01-20

I've never seen gauge gravity theory theories written in anything but diff. geom. terms and I wonder if it can be done with tensors.

https://www.amazon.com/dp/0486661695/?tag=pfamazon01-20

I've never seen gauge gravity theory theories written in anything but diff. geom. terms and I wonder if it can be done with tensors.

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dx

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I've never seen gauge gravity theory theories written in anything but diff. geom. terms and I wonder if it can be done with tensors.

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

Mentz114

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

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

dx

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Flanders said:Tensor fields do not behave themselves under mappings. For example, given a contravariant vector field on x-space, and a mapping φ from x-space to y-space, there is no naturally induced field on the y-space

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?

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dx

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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

Mentz114

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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

RedX

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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!

My knowledge about this subjected is very limited, but can't you just use the inverse mapping of [tex]\phi[/tex], 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?

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dx

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My knowledge about this subjected is very limited, but can't you just use the inverse mapping of [tex]\phi[/tex], and then the pull back of this inverse transformation would be the "natural" form you seek?

No, because φ may not have an inverse.

- #11

MarkParis

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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!

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.

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Hurkyl

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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".

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dx

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Said differently, a function can be viewed either as contravariant or covariant.

No, functions are contravariant.

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MarkParis

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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".

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. ;)

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dx

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dx

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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.

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

MarkParis

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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?

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

- #18

MarkParis

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...

I've never seen gauge gravity theory theories written in anything but diff. geom. terms and I wonder if it can be done with tensors.

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.

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AEM

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- #20

MarkParis

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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

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 [tex]F_{\mu\nu}[/tex] 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.

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MarkParis

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This is a good but difficult question to answer in all generality. I think it depends on the calculation. For example, when Einstein computed the advance of the perihelion of Mercury, he selected a coordinate system and computed the relevant tensors to obtain the physical observable, [tex]\dot\phi_p[/tex]. But he may have saved himself some time if he'd calculated the curvature tensor in the formalism of differential forms. Differential forms make the calculation of curvature much easier. See, for example, Meissner, Thorne and Wheeler, section 14.5, p. 348. Contrast the curvature calculations in MTW with those in Weinberg's "Gravitation and Cosmology" sec. 6.1 (and others).

In Maxwell's theory, I don't think most people would keep the form language for very long into the calculation. Most scenarios aren't complicated enough to keep the equations in that language and usually one just ends up going to the electromagnetic tensor (or its components). Basically, the larger the rank of the tensor, the easier differential forms make your job.

- #22

Mentz114

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Diff geom does help to understand frames in GR and the tangent and cotangent spaces on a manifold are easily written in DG terms ( as MarkParis has said ).

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AEM

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- #24

dx

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