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Tensors without coordinates

by dEdt
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dEdt
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Nov16-12, 04:05 PM
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My class is starting to cover E&M in Lorentz covariant form, and obviously the subject of tensors came up. The problem is that my prof defines tensors in terms of coordinates, which is ugly and against the spirit of relativity. Is there a way of doing tensors coordinate-free in a physics friendly fashion?
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DaleSpam
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Nov16-12, 04:11 PM
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What do you mean he defines tensors in terms of coordinates? Does he use expressions like [itex]g_{\mu\nu}x^{\mu}x^{\nu}[/itex], or do you mean that he would write [itex](t,x,y,z).(t,x,y,z)[/itex] for the same expression?
bcrowell
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Nov16-12, 04:34 PM
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Relativity can certainly be done in coordinate-free notation. You might want to look at Bertel Laurent, Introduction to spacetime: a first course on relativity.

Matterwave
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Nov16-12, 04:34 PM
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Tensors without coordinates

The advantage of coordinates is that they make calculations far less abstract and easier to grasp. They give you the power to change an abstract geometrical problem into a much more concrete algebraic problem with known, simple roads to the solution.

For a topic such as E&M which requires only special relativity, I would suggest that doing it the coordinate-dependent method is enough. It's clear and does not require any discourse into differential geometry. If you really want to learn tensors in a coordinate-independent form, you might try any good book on differential geometry. I suggest Bernard Schutz's Geometrical methods of mathematical physics.
robphy
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Nov16-12, 05:15 PM
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Note that there is a middle ground between
coordinate-based index-notation... like [itex]g_{\mu\nu}U^\mu V^\nu[/itex], where there is an implied summation,
and coordinate-free non-indexed-notation... like [itex]g(U,V)[/itex] or [itex] \widetilde{U} \cdot \widetilde{V}[/itex]
...and that is the coordinate-free abstract-index notation: [itex]g_{ab}U^a V^b[/itex], where there is no summation (but a mapping of vector spaces).

I second the suggestion of Laurent's book
http://www.worldcat.org/title/introd.../oclc/65217217

These lecture notes might be helpful:
http://www.pma.caltech.edu/Courses/p...1/1101.2.K.pdf
http://www.pma.caltech.edu/Courses/p...1/1102.3.K.pdf
from Blandford and Thorne http://www.pma.caltech.edu/Courses/ph136/yr2011/
Chestermiller
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Nov16-12, 05:18 PM
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Quote Quote by dEdt View Post
My class is starting to cover E&M in Lorentz covariant form, and obviously the subject of tensors came up. The problem is that my prof defines tensors in terms of coordinates, which is ugly and against the spirit of relativity. Is there a way of doing tensors coordinate-free in a physics friendly fashion?
Use of dyadic tensor notation works well for tensors up to second rank. There was a nice tutorial on tensor analysis a short while ago in the Educational Materials section of PF, and it was mostly done in dyadic notation. You should be able to find it rather easily.

Actually, here it is: http://www.grc.nasa.gov/WWW/K-12/Num...2002211716.pdf
haushofer
#7
Nov17-12, 12:42 PM
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I've never understood this fetish of people to write down everything in a coordinate-free way. As long as you are aware of the fact that tensors don't care about coordinate choices while their components and the chosen basis do, everything is fine. And in doing explicit calculations you have to choose coordinates anyway.
Muphrid
#8
Nov17-12, 02:28 PM
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Quote Quote by dEdt View Post
My class is starting to cover E&M in Lorentz covariant form, and obviously the subject of tensors came up. The problem is that my prof defines tensors in terms of coordinates, which is ugly and against the spirit of relativity. Is there a way of doing tensors coordinate-free in a physics friendly fashion?
One approach is to use multivectors and wedge products.

What are bivectors? Oriented planes, the same way vectors can be taken to represent oriented lines with magnitudes. If you have two basis vectors ##e_x, e_y##, then the bivector ##e_x \wedge e_y## represents the unit bivector of the xy-plane. In 3+1 spacetime, there are six such planes: tx, ty, tz, xy, yz, and zx.

6 components...sound familiar? It should. These are exactly the components of the electromagnetic field tensor. What we call ##F_{\mu \nu}## is a bivector (field). We can extract the components of ##F## the way you would with vectors:

$$A_\mu = A \cdot e_\mu \\
F_{\mu \nu} = F \cdot (e_\nu \wedge e_\mu)$$

(We've slightly extended the dot product's definition, but this is the general idea.)

The use of the wedge makes a lot of expressions simpler that otherwise, in index notation, would require explicit use of antisymmetric summations that are harder to geometrically interpret. We know, for instance, that the EM field obeys

$$\partial_\alpha F_{\beta \gamma} + \partial_\beta F_{\gamma \alpha} + \partial_\gamma F_{\alpha \beta} = 0$$

But using the wedge product and the vector derivative $\nabla = e^\mu \partial_\mu$, we get a more easily condensed result,

$$\nabla \wedge F = 0$$

This is a 4D analogue to curl. The other equation is ##\nabla \cdot F = -\mu_0 j## (constants and sign depending on exact conventions). This capture's Maxwell's equations outside of material media.

The use of bivectors and wedges makes the process of converting from traditional EM fields to expressions that are covariant straightforward. Consider, for example, the EM bivector for a stationary point charge at the origin.

$$F = -Q \mu_0 c \frac{e_t \wedge r}{4\pi|r|^3}$$

We identify ##e_t## as the four-velocity of the point-charge, ##u##. ##r##, the position vector in three-space, must be entirely orthogonal to ##e_0##, so if four-position is ##s##, then ##r = -e_0 \cdot (e_0 \wedge s)##. Again, we replace ##e_0## by ##u## to get

$$F = -\mu_0 Qc\frac{u \wedge s}{4\pi |u \cdot (u \wedge s)|^3}$$

In this way, one can compute the EM bivector for any uniformly moving point charge given the four-velocity $u$, and in whatever coordinate system we choose.
Bill_K
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Nov17-12, 02:48 PM
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Yes indeed, index free notation is available in special cases. Dyadics cover symmetric rank two tensors, and bivectors cover the antisymmetric ones. Now, without using indices, let's see you write a Christoffel symbol.
Muphrid
#10
Nov17-12, 03:05 PM
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I personally have no need of dyadics--most linear operators can be directly expressed as functions of the vector they act on. For instance, a reflection operator is

$$\underline N(a) = a - 2(a \cdot n)n$$

for any unit vector ##n##. Ultimately, that's what linear operators are--functions of vectors (or bivectors, etc.) that are linear in their arguments. Matrix representation, dyadics--they can be used, but many linear operators have additional structure to them that lends to easy, concise definitions.

This is one thing that I dislike about traditional formulations of tensor analysis--there is no distinction between linear operators and multivectors. They can all be expressed with indices or in abstract index notation, but they are very, very different.


Christoffel symbols are not tensors, of course. But as long as we're in flat spacetime, they are unnecessary. A linear operator ##\overline h## captures the Jacobian of the coordinate transformation and acts on the vector derivative ##\nabla## to form the covariant derivative. ##a \cdot D = a \cdot \overline h(\nabla)##. This result is the same as when using ##D_\mu = \partial_\mu + \Gamma_{\mu \nu}^\lambda##, but involves no non-tensor objects.
Bill_K
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Nov17-12, 04:13 PM
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What do you consider the stress-energy tensor to be? Isn't it just a symmetric tensor? Or do you feel there is an essential distinction to be made between Tμν, Tμν and Tμν.

How about the Riemann tensor Rμνστ - a linear operator on bivectors? In that case I should have written it Rμνστ. But if I do that, then how do I write the cyclic identity Rμ[νστ] = 0.

What about the angular momentum density, Mμνσ.

Doing away with index notation is a noble aim if it could be done simply and uniformly. But IMO all the contortions and special cases needed wind up making things more, rather than less, complicated.
DrGreg
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Nov17-12, 06:01 PM
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Penrose graphical notation is one proposal to manipulate tensors without indices, but it's had little success, is hard to learn and even harder to draw online.
bcrowell
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Nov17-12, 06:10 PM
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Re the practicability doing 100% of GR using coordinate-free methods, folks may want to look at this free book by Winitzki: https://sites.google.com/site/winitz...ral-relativity He presents an awful lot of GR (way more techniques and ideas than I have under my own belt) using coordinate-free notation. In a few places, IIRC, he does say that a certain calculation is too cumbersome to carry out without reverting to index notation, but I think they were few and far between. Section 1.1.1 explains his philosophy about this.

AFAICT mathematicians today use nothing but coordinate-free notation for differential geometry. (This is my impression from looking at WP articles and posts on mathoverflow and math.stackexchange.) I bet a lot of them wouldn't even understand index-gymnastics notation if they saw it.

So I really don't believe that this is any more than a matter of taste and convenience. Pick your favorite flavor, chocolate or vanilla. Enjoy.

And in any case abstract index notation *is* coordinate-free notation, and personally, I don't find Winitzki's arguments, for example, to be persuasive when applied to abstract index notation. To me it seems like he's attacking a straw man, which is index notation from the era before abstract index notation.
Muphrid
#14
Nov17-12, 06:50 PM
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Quote Quote by Bill_K View Post
What do you consider the stress-energy tensor to be? Isn't it just a symmetric tensor? Or do you feel there is an essential distinction to be made between Tμν, Tμν and Tμν.

How about the Riemann tensor Rμνστ - a linear operator on bivectors? In that case I should have written it Rμνστ. But if I do that, then how do I write the cyclic identity Rμ[νστ] = 0.

What about the angular momentum density, Mμνσ.

Doing away with index notation is a noble aim if it could be done simply and uniformly. But IMO all the contortions and special cases needed wind up making things more, rather than less, complicated.
The stress energy tensor is a linear operator on a vector. Feeding it different vectors (or evaluating the resultant vector by its covariant or contravariant components) doesn't change that.

For example, the stress-energy tensor of an ideal fluid is

$$\underline T(a) = (\rho + p)(a \cdot u)u - pa$$

If I recall, the interpretation of this is the flux of energy-momentum across a hypersurface orthogonal to ##a##.

Thus, with the Riemann tensor, you can put the indices wherever you like; you're just evaluating the object differently (with respect to the covariant basis, the contravariant, or some combination).

The cyclic identity (along with several others) is captured in ##\partial_a \wedge \underline R(a \wedge b)##--I admit, this is a pretty significant and non-trivial result that is really not well-covered in Lasenby, Doran, and Gull, but in may be covered better in the former pair's book (which I've left at work, unfortunately). But if your question is more "what is the Reimann tensor?" then yes, it is a linear operator from bivectors to bivectors. Again, raising and lowering indices isn't really a problem--that's just done via the metric, which is itself a linear operator.

I admit, I'm not too familiar with the angular momentum density and so can't comment on its interpretation as a linear operator vs. a multivector.


Overall, the reason I prefer this stuff with wedges and multivectors is because it's a smaller jump to get into it from traditional vector calculus and analysis. Just replacing cross products with wedge products gives a great idea of how to work traditional EM problems (prior to relativity) with it. I feel differential forms tries to shoehorn everything into forms instead of just building off of the usual notions of scalar and vector fields (and often, one resorts to duality instead of just using the inner product directly, which I find clumsy and dumb; the exterior derivative has many great properties, but so does the coderivative, and phrasing everything in terms of one over the other makes certain things needlessly complicated).

(Abstract) index notation feels very divorced from geometric interpretations, and while symmetric and antisymmetric notations help make up for the lack of ease in translating the wedge, it's a small benefit.

Honestly, though, I would be happy from a pedagogical standpoint if the mathematical framework from the time of starting undergrad EM to GR were more uniform, instead of going from vector calculus to tensors in abstract index notation or whatever else.
dextercioby
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Nov18-12, 04:18 AM
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Quote Quote by dEdt View Post
My class is starting to cover E&M in Lorentz covariant form, and obviously the subject of tensors came up. The problem is that my prof defines tensors in terms of coordinates, which is ugly and against the spirit of relativity. Is there a way of doing tensors coordinate-free in a physics friendly fashion?
Obviously the <ugly> and <against the spirit of relativity> is subjective, as the 'spirit of relativity' has definitely changed in the last 100 years. Einstein, Pauli and Eddington wrote relativity with index notation, and to me they and their works represent the true 'spirit of relativity'.

As the mathematics along the 20th century advanced by developing DiffGeo in a coordinate independent way, there was the need to adjust SR and GR to the modern notions and notations, it's true. But, (under)graduate courses in electrodynamics will always use the standard textbooks and the predecessors' notes to teach students the theory in agreement with their mathematical knowledge.

From this perspective, I'd say that specially relativistic electrodynamics would always be taught using tensor components. GR OTOH, different enchilada. I would argue for modern diffgeo as a prerequisite and lectures in agreement with modern mathematics.
Bill_K
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Nov18-12, 06:29 AM
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folks may want to look at this free book by Winitzki:
Ben, thanks for this very interesting reference. Index-free issue aside, he covers some interesting material.

First of all, I note in the preface and introduction he expects his readers to have already been exposed to a General Relativity course using the standard approach. And, at least partly for that reason, the book strives to be "bilingual". That is, he often presents the same argument twice, with and without indices, making it a good test for the relative utility of the two methods. It's interesting to see where the index-free method struggles.

One place it struggles, apparently, is the trace operation. Using indices, of course, the trace is trivial and worth only a moment's thought, but Winitzki spends almost three pages explaining what a trace "really" is, and finally comes up with:
The tensor T (a, b, c, ...) is first written as a linear function of a ⊗ b, i.e. T (a ⊗ b, c, ...), and then the trace is found by substituting the inverse metric g−1, which is a (2,0)-tensor, instead of the argument a ⊗ b: Tr(a,b)T (a, b, c, ...) = Tr(a,b)T (a ⊗ b, c, ...) ≡ T (g−1, c, ...). In this way, the trace operation can be understood as a simple substitution a ⊗ b = g−1.
And later, for example, in discussing the variation of the Hilbert action, he states,
The index-free computations above are cumbersome, mainly because different traces of a high-rank tensor are required. The index-free notation is best for low-rank tensor computations that do not involve complicated traces. After arriving at Eq. (5.11), it is definitely easier to proceed using the index notation.
Feynman, who was of course very practical-minded, once characterized the difference between physicists and mathematicians: when a physicist builds something, he's interested in whether it is sturdy enough to remain standing and gets to the right place. Whereas a mathematician will spend most of his time sanding and polishing the boards.


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