4D Curl
Keith Nicewarner
OK, I have another problem for ya'll:
What is the 4th-dimensional equivalent of a curl operator?
By this, I mean a 4D differential operator which will satisfy
such identities as:
curl(curl(A)) = grad(div(A)) - laplacian(A)
The curl is related to rotation and the cross-product.
I know that in 4D, there are 6 rotational degrees of freedom.
For a 4D cross-product to give a 4D vector, *3* vectors are
needed. Does this mean that for a 4D curl to give a 4D vector,
it somehow requires *2* vectors on which to operate?
--
Center for Intelligent Robotic Systems for Space Exploration (CIRSSE)
Rensselaer Polytechnic Institute | Internet: nice...@ral.rpi.edu
CII 8015 | Telephone: (518) 276-2973
Troy, NY 12180-3590 | Fax: (518) 276-8715
Lawrence R. Mead
One generalization of the curl to arbitrary dimension is:
(curl V) (i) = E(ijk) del(j) V(k), where E(ijk) is the Levi-Civita
symbol ( =1 if i=1,j=2,k=3 or cyclic permutation, =0 if any two
indices are equal) and j,k are summed from 1 to 4 .
--
Lawrence R. Mead (lrm...@whale.st.usm.edu) | ESCHEW OBFUSCATION !
Associate Professor of Physics
Lawrence R. Mead
U16...@uicvm.uic.edu
Nicewarner) says:
>
>OK, I have another problem for ya'll:
>
>What is the 4th-dimensional equivalent of a curl operator?
>By this, I mean a 4D differential operator which will satisfy
>such identities as:
> curl(curl(A)) = grad(div(A)) - laplacian(A)
>
>The curl is related to rotation and the cross-product.
>I know that in 4D, there are 6 rotational degrees of freedom.
>For a 4D cross-product to give a 4D vector, *3* vectors are
>needed. Does this mean that for a 4D curl to give a 4D vector,
>it somehow requires *2* vectors on which to operate?
>
Flanders, Differential forms , Accademic Press.
Sen and Nash, title fuzzy
Schultz, An introduction to Differential forms
I also believe
Wald, something General Relativity
--------------------------------------------------
Thaddeus Olczyk, University of Illinois at Chicago
olc...@uicws.phy.uic.edu
John C. Baez
>What is the 4th-dimensional equivalent of a curl operator?
>I know that in 4D, there are 6 rotational degrees of freedom.
>For a 4D cross-product to give a 4D vector, *3* vectors are
>needed. Does this mean that for a 4D curl to give a 4D vector,
>it somehow requires *2* vectors on which to operate?
Good guess! To define things like curls in any number of dimensions is
the job of the exterior differential calculus. This deals with
differential forms; in dimension n you have 0-forms (functions), 1-forms
(which are like vector fields but not the same - see the thread on
sci.math!), 2-forms, up to n-forms. There is an operation called the
exterior derivative d, and d of a p-form is a (p+1)-form. This
operation contains as special cases the divergence, gradient and curl in
3 dimensions. In any number of dimensions, d acts like "curl" on
1-forms. A 1-form in 4 dimensions looks like
a dx + b dy + c dz + e dw
and d of it is
(da/dy) dy dx + (da/dz) dz dx + (da/dw) dw dx +
(db/dx) dx dy + (db/dz) dz dy + (db/dw) dw dy +
2 more lines like that!
Using the rule that dx dy = - dy dx (and so on) we boil this down to 6
terms. The dx dy term, for example, looks like
(db/dx - da/dy) dx dy
so you can see it looks like a curl.
In 3 dimensions d of a 1-form is a 2-form, too, but one can play a
special trick, namely, one can map 2-forms to 1-forms as follows:
dx dy -> dz, dy dz -> dx, dz dx -> dy
(this is an example of the "Hodge star operator"). That's why we can
pretend that d of a 1-form is a 1-form in this case, or, as one said in
the good old days, the curl of a vector field is a vector field. Note
however that this trick is special to 3 dimensions, and is really due to
the fact that in 3 dimensions every infinitesimal rotation is rotation
about some axis. (In higher dimensions all we can say is that its
rotation in some plane!) Also note that this trick is where the
"right-hand rule" comes in - we are making a *choice* at this stage.
One of the charms of differential forms is that one needs no right-hand
rule to define the generalization of the curl!
U16...@uicvm.uic.edu
Baez) says:
>One of the charms of differential forms is that one needs no right-hand
>rule to define the generalization of the curl!
>
Ron Maimon
|>
|> OK, I have another problem for ya'll:
|>
|> What is the 4th-dimensional equivalent of a curl operator?
|> By this, I mean a 4D differential operator which will satisfy
|> such identities as:
|> curl(curl(A)) = grad(div(A)) - laplacian(A)
|>
|> The curl is related to rotation and the cross-product.
|> I know that in 4D, there are 6 rotational degrees of freedom.
|> For a 4D cross-product to give a 4D vector, *3* vectors are
|> needed. Does this mean that for a 4D curl to give a 4D vector,
|> it somehow requires *2* vectors on which to operate?
|>
actually- you have a wonderful four dimensional intuition!
the reason you are confused is because the curl is not a natural notion at all,
and only works in three space. It has many analogs in four space.
The natural thing to do is consider tensor calculus in four dimensions.
a vector is a quadroplet of numbers which transforms in a certain way under
linear transformations.
call the vector Au = (A0,A1,A2,A3)
then, if you have another vector Bu you can form the product:
Puv= AuBv
meaning that you have a quantity Puv with 16 components- one for each value of u
and v between 0 and 3. This is known as a rank two tensor. Any 16 quantities
which have the same transformation laws as a product of vectors is called a rank
two tensor.
Given a vector Au, and the gradient operator vector Dv you can form the curl of a
as the tensor
Fuv= Du Av - Dv Au
this is an antisymmetric n by n component object, and therefore has n(n-1)/2
independent components. In three dimensions, there are three components, and they
are the components of the curl, rearanged in a matrix. Actually, in general, any
cross product is much more naturally written in the language of tensors. I don't
know why people still teach the cross product.
In four dimensions there are six components (as you correctly intuited) they form
a generalized curl. This is actually a very useful thing, since the vector and
scalar potentials in electromagnetism are four quantities (a 4-vector) that
generate the electric and magnetic field (six components). The EM field is the
4-D curl of the vector-scalar potential.
In 4 dimensions you can also define the triple cross product as a 4 by 4 by 4
object- a rank 3 tensor. If you have Au Bv and Cw you can form the tensor
Ruvw = AuBvCw - AvBuCw + AvBwCu - AwBvCu + AwBuCv - AuBwCv
which is the four dimensional `triple cross product' of A B C and has only four
independent components, which can, strangely enough, be arranged as the
components of a vector. This shows that there are many ways of generalizing the
cross product.
I hated the cross product in school because I refused to learn something so
arbitrary as the right hand rule. If you use tensor calculus always and
consistently, you never have to learn it.
- Ron Maimon
Ron Maimon
|>
|> One generalization of the curl to arbitrary dimension is:
|>
|> (curl V) (i) = E(ijk) del(j) V(k), where E(ijk) is the Levi-Civita
|>
|> symbol ( =1 if i=1,j=2,k=3 or cyclic permutation, =0 if any two
|> indices are equal) and j,k are summed from 1 to 4 .
|>
oh dear.
This doesn't work in arbitrary dimensions, because the Levi-Civita symbol has
more than three indices in more than three dimensions. The object you describe as
the Levi Civita symbol is not a tensor.
- Ron Maimon
Jeffrey J. Kosowsky
You only need to remember how to draw the x,y,z axes if you choose a
specific local coordinate system. The beauty of differential forms and
the exterior derivative is that they are coordinate-free and hence
can be defined uniquely on an abstract manifold without reference to
coordinates (or axes). Of course, if one wants to use coordinates,
then one must define their relative orientations.
Jeff
Ron Maimon
|>
|> Good guess! To define things like curls in any number of dimensions is
|> the job of the exterior differential calculus. This deals with
|>
|>
Is it just me, or are there other people who think that the calculus of forms is
the ugliest, most formal, useless, and dryest branches of mathematics?
I am, of course, comparing it to the more general, beautiful, intuitive and
fulfilling tensor calculus (in index notation) that was in fasion about thirty
years ago.
I would never curse anyone to have to read about this theory. I would recommend
books that are out of date way before any of the modern one in this notation.
Is it really just me?
-Ron Maimon
Murali Ramanathan
Subject: Re: Extreme points of convex set
Summary:
Expires:
References: <CBCGz...@ns1.nodak.edu> <23unci$g...@iskut.ucs.ubc.ca>
Sender: rama...@plains.nodak.edu (Murali Ramanathan)
Followup-To: sci.math
Distribution: world
Organization: North Dakota Higher Education Computing Network
Keywords: extreme points, convex set
Thank you all who kindly pointed out the rather egregious error in the
formulation of M(I). Yes, M(I), as stated contains only one element and all
the questions I'd raised regarding it are thus trivial (:---((. In any event,
I think I've got a few leads now to point me in the right direction.
Thanks again.
Murali (rama...@plains.nodak.edu)
P.S. This is to Dr. Robert Israel who answered to my original post: I
forgot to save you email address but, are you the Robert Israel of "Convexity
in the theory of lattice gases" ? Trying to read through the book was what
prompted the original post !!
Daniel E. Platt
Except in 4D, the Levi-Civita tensor is rank 4:
(curl V)(ij) = E(ijkl) del(k) V(l).
If you construct basis vectors u0 = (1 0 0 0), u1 = (0 1 0 0), etc,
then the values of E(ijkl) will be the various determinants you can
constuct out of those basis vectors. If any two of the rows are
the same, the determinant is 0. That leaves 24 non-zero ways
out of 256 possible combinations of basis vectors.
One of the reasons this is interesting is that it is important for
looking at the way volume elements transform (Jacobians and all
that). In this context, it is obvious you need a tensor whose
rank is equal to the dimension of the space.
|> --
|>
|> Lawrence R. Mead (lrm...@whale.st.usm.edu) | ESCHEW OBFUSCATION !
|> Associate Professor of Physics
|>
Dan
--
-------------------------------------------------------------------------------
Daniel E. Platt pl...@watson.ibm.com
The views expressed here do not necessarily reflect those of my employer.
-------------------------------------------------------------------------------
Bradford Holden
Well, the thing is, the two are not incompatible. In the calculus of forms,
you can make tensors, etc. And having used both, I find forms far more
intuitive. But this is just a matter of personal taste. I will tell you
this, the whole concept of div, grad, and curl look really lame to me
compared with the Hodge star and the d operator.
John C. Baez
>|>
>|> Good guess! To define things like curls in any number of dimensions is
>|> the job of the exterior differential calculus. This deals with
>|> differential forms ... [ snip ]
>Is it just me, or are there other people who think that the calculus of
>forms is the ugliest, most formal, useless, and dryest branches of
>mathematics?
>Is it really just me?
Probably not: for any opinion, however erroneous, one can usually find a
few adherents. :-) I have always found differential forms to be one of
the most beautiful, practical and charming branches of mathematics. If
you really prefer good old tensor calculus with lots of indices, you can
translate any formula about differential forms into that language, so it
scarcely seems worth aruging about. For example, if Maxwell's equations
*d*F = j
seem repulsive in differential form notation you are free to write them
out more elegantly using indices. It's really just a matter of taste,
though if you want to read modern mathematics and physics literature
it's best to be comfortable with *both* notations.
Roland Kawakami
>|>
>|> Good guess! To define things like curls in any number of dimensions is
>|> the job of the exterior differential calculus. This deals with
>|> differential forms ... [ snip ]
>|>
>|>
>
>Is it just me, or are there other people who think that the calculus of forms is
>the ugliest, most formal, useless, and dryest branches of mathematics?
>
I'm not sure if there is anything really ground-breaking about differential
forms. Probably anything you can do with forms, you probably could
do without it. (if this is wrong, I would be very very interested
in learning about it)
However, there are some good things about them:
(1) If you are working with tensors (in index notation), and some
of the indices are totally antisymmetric. Then somewhere along
the line, you might run into something like:
nice looking expression ==(do something)==> ugly looking expression
==(amazing cancellation due to anitsymmetry)==> nice expression
I guess some people noticed there was always this "amazing cancellation"
and they wanted to develop a formalism which would automatically
bypass the "ugly looking expression" stage. So differential
forms does this.
(2) A tensor with two lower indices can be viewed as a linear
map from two vectors to a number. A two-form can also be viewed
as a linear map from two vectors to a number.
Now here's the neat part. Since the two-form map is antisymmetric
in it's arguments (i.e. w(V,W) = -w(W,V) W,V are vectors), we can
view the two-form as a map from a parallelogram to a number.
(I'm not so clear on the details of this yet, sorry) Higher
dimensional generalizations are apparent.
So, differential forms can be thought of as "things to
integrate over." I'm not sure if you can have this interpretation
for a general tensor.
I guess I think diff. forms are nice but not necessary.
Roland Kawakami
U16...@uicvm.uic.edu
Maimon) says:
>Is it really just me?
U16...@uicvm.uic.edu
koso...@minerva.harvard.edu (Jeffrey J. Kosowsky) says:
>
>You only need to remember how to draw the x,y,z axes if you choose a
>specific local coordinate system. The beauty of differential forms and
>the exterior derivative is that they are coordinate-free and hence
>can be defined uniquely on an abstract manifold without reference to
>coordinates (or axes). Of course, if one wants to use coordinates,
>then one must define their relative orientations.
>
dx,dy. Which way is *(dx^dy) pointing. A more physical version is
wire carries a current in this direction. You have a point above the wire.
Which direction is the B field. No coordinates , you either need the right
hand rule, or you need a coordinate system to explicitly write your forms.
The specific coordinate system chosen is superfluous.
U16...@uicvm.uic.edu
says:
>
>I'm not sure if there is anything really ground-breaking about differential
>forms. Probably anything you can do with forms, you probably could
>do without it. (if this is wrong, I would be very very interested
>in learning about it)
>
ns, but they probably could be.
2) You can also write out all physics in components. You know;
d_1 E_1 +d_2 E_2 +d_3 E_3 = ....
But do you really want to?
>However, there are some good things about them:
>
>(1) If you are working with tensors (in index notation), and some
>of the indices are totally antisymmetric. Then somewhere along
>the line, you might run into something like:
>
>nice looking expression ==(do something)==> ugly looking expression
> ==(amazing cancellation due to anitsymmetry)==> nice expression
>
>I guess some people noticed there was always this "amazing cancellation"
>and they wanted to develop a formalism which would automatically
>bypass the "ugly looking expression" stage. So differential
>forms does this.
>
Probably not. Differential forms were first developed around the same time
as general relativity.
>(2) A tensor with two lower indices can be viewed as a linear
>map from two vectors to a number. A two-form can also be viewed
>as a linear map from two vectors to a number.
>
Thus they are proportional to determinants ( all forms, not just two forms).
>Now here's the neat part. Since the two-form map is antisymmetric
>in it's arguments (i.e. w(V,W) = -w(W,V) W,V are vectors), we can
>view the two-form as a map from a parallelogram to a number.
>(I'm not so clear on the details of this yet, sorry) Higher
>dimensional generalizations are apparent.
>So, differential forms can be thought of as "things to
>integrate over." I'm not sure if you can have this interpretation
>for a general tensor.
>
Differential forms are volume elements. Tensors have nothing to do with
area.
>
>
You've also forgotten that differential forms form a nice way to treat
connections on principle bundles.
By the way, have we confused the original poster already?
Jeffrey J. Kosowsky
The original thread was talking about finding the exterior derivative
of forms. John Baez said
One of the charms of differential forms is that one needs no right-hand
rule to define the generalization of the curl!
You then said
But you still have to remember how to draw the x,y,z axis.
I then wrote my above post saying that you do not indeed need orientation to
define the exterior derivative.
You then switched topics and started talking about needing orientation
for the Hodge-star operator as in physics applications.
Well, of course you need a notion of orientation, since the Hodge-star
(and the physics analogies) assume a *Riemann* structure on the
manifold -- i.e. you need an inner product. Knowing the inner product
trivially gives you the orientation of the x,y,z axes in any given
coordinate system. But be careful, because knowing the orientation of
the axes does not give you the inner product. In fact knowing the
orientation will only give you the correct sign but not the magnitude
of such quantities as *(dx^dy). So, even in this case the essence is
not the coordinates or the orientation of some arbitrary set of axes
but the Riemann structure.
The point of my posts (and I believe also the point of John's posts)
is that you do *not* need any metrical or inner product structure or
orientation etc. to define the exterior derivative. Of course, as you
pointed out in your last point, there are constructions that do need
an orientation but the exterior derivative is not one of them. In
fact, the exterior derivative and the wedge product are well-defined
also on non-orientable manifolds such as the Mobius band.
Jeff
John C. Baez
>Not just if you need coordinates. Simple example take two orthognal one-forms
>dx,dy. Which way is *(dx^dy) pointing. A more physical version is
>wire carries a current in this direction. You have a point above the wire.
>Which direction is the B field. No coordinates, you either need the right
>hand rule, or you need a coordinate system to explicitly write your forms.
>The specific coordinate system chosen is superfluous.
Okay, now I understand your earlier remark better. You only need a
right-hand rule to define the * operator, not the wedge product.
This is nice because it says that one can define the "cross product" of
two 1-forms to be a 2-form without needing to remember which hand is
which; it's only when you perversely decide to pretend that the 2-form
is a 1-form that you need to pick an arbitrary convention like the
left-hand rule. (I'm left-handed so I prefer that to the right-hand
rule.)
This is not just mathematical sophistry. The point is that the B field
does not "point." It twists. So your question, "which direction is
the B field," is a bit misleading.
For those not in the know about this * operator business, the point is
that in differential form lingo all the "right-hand rule" business is
packed into that operator.
Jeffrey J. Kosowsky
I second John's viewpoint. Paraphrasing the words of ABIAN, I would
say that
The multitudinous PRE-CAMBRIAN SWAMP of Greek and Latin indices
with their associated PREHISTORIC coordinate dependent viewpoint
swamps the stark beauty of differential forms under a ten TON
SLAG HEAP of LAVA and CAMEL DUNG.
In all seriousness, I generally prefer the coordinate free
differential form approach. I do not totally ignore the index point
of view because I will readily admit that some proofs are simplified
by resorting to indices and special coordinate systems. (Of course,
other proofs are simplified by the abstract notation...) So, it is
best to be comfortable with both and to know how to go back and forth.
But I must say that for physical intuition and to help in "guessing"
the correct form of a formulat, I much prefer the abstract approach.
Jeff Kosowsky
Matthew MacIntyre aat the National University of Senegal
: I would never curse anyone to have to read about this theory. I would recommend
: books that are out of date way before any of the modern one in this notation.
: Is it really just me?
Yes, it is.
Ron Maimon
|>
|> I then wrote my above post saying that you do not indeed need orientation to
|> define the exterior derivative.
|>
|> You then switched topics and started talking about needing orientation
|> for the Hodge-star operator as in physics applications.
|>
|> Well, of course you need a notion of orientation, since the Hodge-star
|> (and the physics analogies) assume a *Riemann* structure on the
|> manifold -- i.e. you need an inner product. Knowing the inner product
|> trivially gives you the orientation of the x,y,z axes in any given
|> coordinate system. But be careful, because knowing the orientation of
|> the axes does not give you the inner product. In fact knowing the
|> orientation will only give you the correct sign but not the magnitude
|> of such quantities as *(dx^dy). So, even in this case the essence is
|> not the coordinates or the orientation of some arbitrary set of axes
|> but the Riemann structure.
|>
nuh-uuuh!
the Riemann metric is not enough to specify the orientation since there are
coordinate changes that reverse the orientation but leave the metric unchanged,
e.g. in R^2 flip the roles of x and y. The metric only give you a natural
isomorphism (*sigh*, how I hate that term) between the covariant and contravariant
tensors, which is indeed necessary for defining the Hodge star, which would be
apparent in index notation, since applying the totally antisymmetric totally
covariant pseudotensor to a tensor which is contravariant will give you a
covariant tensor, and you need the metric to raise the indices.
That doesn't change the fact that the totally antisymmetric pseudotensor is just
that- a pseudotensor, and requires an orientation to specify.
On a related topic, tell me, which is more informative
*F
or
ij kl ij ip jq
e F ( where e = g g e )
kl kl pqkl
one is shorter- but it is the kind of shortness derived at the expense of clarity
of thought.
-Ron Maimon
here I am arguing about forms, when my original post lamented about how horrible
they were. (*sigh*)
John C. Baez
(Ron Maimon) writes:
>On a related topic, tell me, which is more informative
>
>*F
>
>or
>
> ij kl ij ip jq
> e F ( where e = g g e )
> kl kl pqkl
>
>one is shorter- but it is the kind of shortness derived at the expense
>of clarity of thought.
I think you're onto something here. The only problem is, you haven't
gone far enough! Don't you think "The sum as k, l, p, and q range
from 1 to 4 of the ip-th coefficient of g times the jq-th coefficient of
g times the pqkl-th coefficient e of times the kl-th coefficient of F"
is much more informative than the heap of symbols you used? That way we
can really see which indices are being summed over, instead of needing
to remember the Einstein summation convention, which after all is a just
mind-numbing trick of notation. In fact, I think the whole idea of
letters for variables is a bit of a sham. Back in the Middle Ages they
would have no truck with this sort of thing. They would have called a
spade a spade, and instead of writing *F they would have simply said
"The sum over four indices as they range from one to four of the entry
of the metric associated to the third index and an index not one of
those four times the entry of the metric associated to the fourth index
and yet another index not one of those four times the entry of the
Levi-Civita symbol associated to the third, fourth, first and second
indices (in that order) times the entry of the field strength associated
to the first and second indices," and there would be no confusion whatsoever!
U16...@uicvm.uic.edu
Baez) says:
>
>This is not just mathematical sophistry. The point is that the B field
>does not "point." It twists. So your question, "which direction is
>the B field," is a bit misleading.
>
one asked in all levels of undergraduate physics classes.
Ron Maimon
ahem,
the reason I used the summation convention is because it is just as coordinate
free as your notation, if you regard the indices as just devices for keeping
track of what is a vector and what is a dual vector.
The Einstein convention is not a sham, nor a trick. It is a way of expressing the
natural inner product between vector and dual vector in a way that keeps most, if
not all, of the symbols of the coordinate dependent method. I don't see how
anyone can regard a properly balanced tensor equation in index notation as
a coordinate dependent construct. It isn't.
I would now challenge you to write the following abstractly
ij kl rp kp
e F F = U
kr
a perfectly reasonable tensor equation that would be insufferably long in
abstract notation. Notice that the index notation allows you to see exactly what
`slots' a tensor has, meaning, what kind of inputs it takes and what output it
gives. Given a rank 3 covariant tensor T and a rank two contravariant tensor R,
how do you distinguish between
ab bc ac ca
T R T R T R T R
abc abc abc abc
In modern notation?
-Ron
Bradford Holden
>
>ahem,
>
>the reason I used the summation convention is because it is just as coordinate
>free as your notation, if you regard the indices as just devices for keeping
>track of what is a vector and what is a dual vector.
>
>The Einstein convention is not a sham, nor a trick. It is a way of expressing the
>natural inner product between vector and dual vector in a way that keeps most, if
>not all, of the symbols of the coordinate dependent method. I don't see how
>anyone can regard a properly balanced tensor equation in index notation as
>a coordinate dependent construct. It isn't.
>
>I would now challenge you to write the following abstractly
>
> ij kl rp kp
>e F F = U
> kr
>
tensor U where the index should be l. I am tempted to make a smart-ass
remark regarding explicitly writing out your components but I am trying
to reform. :)
>a perfectly reasonable tensor equation that would be insufferably long in
>abstract notation. Notice that the index notation allows you to see exactly what
>`slots' a tensor has, meaning, what kind of inputs it takes and what output it
>gives. Given a rank 3 covariant tensor T and a rank two contravariant tensor R,
>how do you distinguish between
>
> ab bc ac ca
>T R T R T R T R
> abc abc abc abc
>
>
>In modern notation?
>
Same way you do, location. The actual indicies on the tensors are not
important, just the location. Of course, what I found in my differential
geometry was that we tended to use a sort of bastardization of the two, ie
used whatever notation illustrated the physics better. There is no law
saying that you have to stay with on or the other.
Can we at least agree that vector calculus should be done away with
completely? I can not stand cross products and curls (well, see its just
like a vector except different.)
>-Ron
John C. Baez
oh-oh...
>the reason I used the summation convention is because it is just as coordinate
>free as your notation, if you regard the indices as just devices for keeping
>track of what is a vector and what is a dual vector.
>The Einstein convention is not a sham, nor a trick. It is a way of
>expressing the natural inner product between vector and dual vector in
>a way that keeps most, if not all, of the symbols of the coordinate
>dependent method.
Wouldn't want to leave out any symbols, would we? :-) Apart from a
mathematical quibble (the pairing between vectors and dual vectors is
not an inner product, the pairing between two vectors using the metric
is an inner product) I agree with all of this. I wasn't seriously
suggesting that the Einstein summation convention was a trick! I use it
a lot. I was just parodying your suggestion that abstract differential
form notation was a trick. I am in favor of whatever notation is
clearest in any given context, and it would be remarkable if *one*
notation were best for everything. It was you, recall, who began this
discussion by claiming that index-free notation was a useless
monstrosity. Naturally this made me want to make fun of that position
by arguing that even the index-ridden notation was too abstract.
>I don't see how anyone can regard a properly balanced
>tensor equation in index notation as a coordinate dependent construct.
>It isn't.
You're right. What Maimon is referring to, by the way, is usually
called the "abstract index notation," and differs from the older
coordinate- or frame-dependent index notation primarily in how one
interprets it, rather than what's written on the page. I think the idea
is due to Penrose. And I agree that it is invaluable in general
relativity and large hunks of differential geometry. I am not opposed
to it! I am opposed to ONLY using it. The vocabulary of differential
forms is more limited than the full-fledged vocabulary of tensors. I
often use the vocabulary of Lie-algebra-valued differential forms, which
is good for a lot of gauge theory. When one is speaking of things of
which one can speak in this more limited vocabulary, it's an excellent
way to go. In part this is because while the abstract index notation is
coordinate-free it is overly cluttered for situations when one is mainly
interested in playing with exterior differential calculus.
>I would now challenge you to write the following abstractly
>
> ij kl rp kp
>e F F = U
> kr
>
>a perfectly reasonable tensor equation that would be insufferably long in
>abstract notation.
Really? If one can write it at all in abstract notation, I doubt it
would be long. I don't see how one can write it at all.
>Notice that the index notation allows you to see exactly what
>`slots' a tensor has, meaning, what kind of inputs it takes and what output it
>gives. Given a rank 3 covariant tensor T and a rank two contravariant
>tensor R, how do you distinguish between
>
> ab bc ac ca
>T R T R T R T R
> abc abc abc abc
>
>in modern notation?
I wouldn't use index-free notation for this sort of thing.
lrud...@vax.clarku.edu
>the "abstract index notation,"... differs from the older
>coordinate- or frame-dependent index notation primarily in how one
>interprets it, rather than what's written on the page. I think the idea
>is due to Penrose.
Maybe so. I first saw it in Edward Nelson's _Tensor Analysis_ (Princeton,
1967), in \S1.12. This excellent little book also has a picture (Fig. 1)
of a tangent vector and a cotangent vector (cf. the recent discussion on
forms), and a proof with Lie algebras that you can get into an arbitrarily
tight parking space. I don't know if it's still in print.
Lee Rudolph
Ron Maimon
> (in some article, I wrote)
|> >how do you distinguish between
|> >
|> > ab bc ac ca
|> >T R T R T R T R
|> > abc abc abc abc
|> >
|> >
|> >In modern notation?
|> >
|> Same way you do, location. The actual indicies on the tensors are not
|> important, just the location. Of course, what I found in my differential
|> geometry was that we tended to use a sort of bastardization of the two, ie
|> used whatever notation illustrated the physics better. There is no law
|> saying that you have to stay with on or the other.
|>
|> Can we at least agree that vector calculus should be done away with
|> completely? I can not stand cross products and curls (well, see its just
|> like a vector except different.)
|>
yup. I actually agree with everything you said. The reason I felt so venomous
about modern notation in the first place is that I learned from an _unreasonable_
source, that never explained the index notation, or how to distinguish between
the three alternatives above. I saw everyone (except me) get confused because
they never got the index notation to fall back on, and they didn't even know that
there was a distinction between the three cases above, since they wrote those
contractions as very high dimensional traces, where they didn't know what to
trace over. I think your approach is extremely reasonable.
-Ron Maimon
Douglas Gillman
Hey Thaddeus, buddy. Big one nine. 2 vectors huh. That a
new result. You from a small college, pardner?
Spotter,
loves them islands,
especially
Truk Island,
South Pacific
--