Natural Science Forum / Physics / Research / October 2004

Try this:

A Course in Mathematics for Students of Physics: Volume 1 by Paul
Bamberg, Shlomo Sternberg used at $9.95 at Amazon.

This book re-teaches calculus the way it should have been taught to
you, including distinguishing forms and vectors.

Later, you might want to try the Frankel book "The Geometry of
Physics, an Introduction."

This is not at all true.

1) Forms in themselves are algebraic objects, and not related to
calculus at all. They are generalized from "directed lengths" as the
model for vectors.

2) *Differential* forms allow certain subjects in multivariable
calculus to be better organized and presented than the traditional
operations with Jacobians and change of variables.

3) The objects that appear naturally as differential forms in physics
are just anti-symmetric tensors. Symmetric tensors are just as
important, but unwieldy from the DF viewpoint.

4) There is nothing at all "tied to dimension" about either calculus
or operation with forms (differential or not). All are completely
general concepts.

All that is overrated. In real life, an ability to aptly choose
coordinates often distinguishes the clever (Feynman). There is nothing
wrong with coordinates as such. "Coordinate free" mania is just a sort
of "indicial racism".

/rant

The real problem is that books tend to be either too formalist
(Bourbakian) in their layout, or too triumphalist (Wheelerian) to
enable the student to judge for himself without being swamped by
hyperbole. Unfortunately there is nothing to be done about it, other
than to read older books by proven pedagogues.

"Elementary Mathematics from an Advanced Standpoint", Vols 1 and 2, by
Felix Klein, recently reprinted by Dover. This will provide an
excellent foundation for more advanced study.

-drl

Tom Knight <tk@csail.mit.edu> writes

$US 50 from amazon UK. <gasp>

Is it solidly (3+1)D from the off?

Er, I've seen the 'contents' and I fear its just the sort of book I am
NOT after.

Topics covered include many items previously dealt with at a much more
advanced level, such as algebraic topology (introduced via the analysis
of electrical networks), exterior calculus, Lie derivatives, and star
operators (which are applied to Maxwell's equations and optics).

Danny Ross Lunsford <antimatter33@yahoo.com> writes

Good, that means I am confused....

OK. As usual I have to figure out what you really mean here.
Am I to guess that they are nD equivalents of vectors?
A brief description would be gratefully appreciated.

<ungh> Does that mean I have to put differential (a very long word with
lots of characters) in front every time?
A brief description would be gratefully appreciated (again).

Oh, that's a pity. I rather assumed one tool would do the job neatly and
clearly, at least in (simple) physics.

Indeed, I expressed myself badly.
I rather imagined that if one were to be working an a (3+1)D space that
'conventional' tools and representations might be unwieldy and have a
tendency to obscure rather than clarify. Of course I imagined that those
working in these areas would have developed a notation that conceals the
nuts and bolts and thus showed the physics more clearly. I am getting
the horrible feeling that this might not be so.

Indeed. However indices do clutter things up quite a bit, particularly
when the indices have indices...

Hmm, dunno if I'm up to further anything....
However the book is reasonably priced and a search on the web suggests
it will have a bunch of interesting things, even if it doesn't help me
on this particular quest. So I will follow your advice and get it.

Only seems to be vol 1 about just now but still.

I guess that what you want doesn't exist.
The whole point of introducing vectors to begin with
is that having vectors allows you to be coordinate-free.

And a coordinate-free vector -is- a much simpler concept:
it is just something that transforms in the right way
under a given set of symmetry operations.
(like rotations in space)
A vector is just anything that transforms like a vector.

I hesitate to say it but
(supposing you mastered standard linear algebra in vector form)
you might try the introductory chapters
of the 'Gravitation' phonebook by Misner e.a.

Real mathematicians seem to dislike its pictoral approach thoroughly,
with its 'bongs of the bell', when 'piercing sheets',
and its egg-crates, so perhaps you might like it.

Not cheap, but perhaps you could borrow it somewhere?

Jan

I guess that what you want doesn't exist.
The whole point of introducing vectors to begin with
is that having vectors allows you to be coordinate-free.

And a coordinate-free vector -is- a much simpler concept:
it is just something that transforms in the right way
under a given set of symmetry operations.
(like rotations in space)
A vector is just anything that transforms like a vector.

I hesitate to say it but
(supposing you mastered standard linear algebra in vector form)
you might try the introductory chapters
of the 'Gravitation' phonebook by Misner e.a.

Real mathematicians seem to dislike its pictoral approach thoroughly,
with its 'bongs of the bell', when 'piercing sheets',
and its egg-crates, so perhaps you might like it.

Not cheap, but perhaps you could borrow it somewhere?

Jan

In article <ck0qf2$v11$1@lfa222122.richmond.edu>, Oz
<oz@farmeroz.port995.com> wrote:

[snip]

Here's a brief overview of vectors and forms, from someone who was pretty
much in your shoes a few years ago.

We all have a reasonable intuitive idea of what a vector is:  something,
like a velocity, that has a direction and a magnitude.

Imagine a stone thrown into the air.  If you think about the path the
stone follows in 3-dimensional space, you can mark each point along that
path with the time when the stone is at that point.  So the motion of the
stone gives, not only a path through 3-space, but a path with a
parameter, t, associated with every point.  You can think of drawing the
path with little tick-marks along it:  t=0, t=1, t=2, etc.  And
mathematically, we can think of this path as a function C(t) from the
real numbers to the space the stone moves through.

The velocity of the stone (at any particular time t_0) has a direction
that is tangent to the curve C that it follows through space, at the
particular point C(t_0), and its magnitude will depend on how far along
the path a given increase in t carries the stone.  So if the t=0, t=1,
t=2 tick-marks are *further apart*, the velocity will be bigger, because
the stone will move further in one second.

Exactly what else we can say depends on what space the stone is moving
through, but it's worthwhile just leaving it at that for now, and seeing
that the idea of a vector makes perfect sense when all you have is a path
through some space with a parameter marked along it.  This works whether
the space is flat, curved, 3-dimensional, n-dimensional, whatever.

In 4-dimensional spacetime, you can think of an object's world line, with
the time ticked off by a clock moving with that object as the parameter
along the world line.  The 4-velocity vector defined this way is the
natural velocity to use in special and general relativity.

Now, there's a slightly different concept that also involves direction
and magnitude, which also works in any number of dimensions.  This is
known as a 1-form.

Imagine a map of air pressure, with lines of constant pressure -- isobars
-- marked on it.  It's clear that the pressure gradient has a direction
and a magnitude.  But it's a different concept than the one we get from
marking a parameter along a single curve.  The key to the difference is
that, in a sense, it's opposite and complementary to the notion of a
vector.  A stone's velocity is *greater* if the tickmarks for each time
interval are *further* apart, whereas a pressure gradient becomes *less*
when the isobars are further apart, and greater when they're crowded
closer together.  This leads to opposite rules when we change coordinate
systems (for example, if we simply change units):  a velocity of 1 metre
per second becomes 100 centimetres per second, but a pressure gradient of
1 Pascal per metre becomes 0.01 Pascals per centimetre.

A 1-form is the notion of direction and magnitude you get when you
imagine a stack of curves on a map, or surfaces in 3-space, or, in
general, (n-1)-dimensional hypersurfaces in n-space.

You can combine a vector and a 1-form to get a number:  if you have a
stone moving at 1 metre per second through a pressure gradient of 1
Pascal per metre, then, for the stone, the pressure will change at 1
Pascal per second.  Note that this result is completely independent of
the (spatial) coordinates used, so that if we change units to
centimetres, the 100 centimetres per second x 0.01 Pascals per
centimetre, naturally, gives the same final result.

If this is a 1-form, what is a 2-form?  Well, I won't go into that unless
you really want me to, except to say that you can get an n-form from an
(n-1)-form by taking a kind of derivative.  A scalar function (such as
air pressure) can be thought of as a 0-form, and the process of taking
the gradient to get a 1-form can be generalised.  That generalisation
nicely generalises some operations that might be familiar to you from
3-dimensional calculus/physics:  grad, div, and curl.

When you talk about studying 4-vectors in (3+1)D, I think what you're
after is a good book on special relativity that uses the natural
4-dimensional objects (4-velocity, 4-momentum etc.), rather than making
everything messy by projecting things down to 3 dimensions all the time,
and making SR look like a kind of "distortion" or "correction" to
Newtonian physics.  You might want to look at "Spacetime Physics" by
Taylor and Wheeler.

Dear Oz,

I think what you're looking for is done quite nicely in the language
of Clifford Algebra. Take a look at, for example,

David Hestenes, Space-Time Algebra

The essential idea is to introduce a basis of algebraic objects
\gamma_\mu satisfying,
{\gamma_\mu,\gamma_\nu}=2g_{\mu\nu}
These are what we already know of as "Dirac matrices."

A 4-vector can be expressed in the form A=A^\mu\gamma_\nu, which is
manifestly coordinate invariant.  It is simple to verify that,
AB=A\dot B + A\wedge B, where A\dot B is the Lorentzian dot product
(just like the dot product in 3D) and A\wedge B is the wedge product,
which can be thought of a a generalization of the 3D cross product to
four dimensions.

Unlike in 3D, however, A\wedge B is not another 4-vector---rather it
is something called a bivector, which can be visualized as an oriented
plane in 4D. In four dimensions, there are 6 independent planes so
planes are really different things from vectors, unlike the case in
3D. A good example of a bivector is the electromagnetic field strength
tensor, F=F^{\mu\nu}\gamma_\mu \wedge \gamma_nu.

Clifford Algebra really just a nice way of generalizing the Grassmann
algebra of differential forms to situations---like special
relativity---where we have a metric and some notion of a dot product.
I think
it is the most elegant and essentially unique generalization of the
coordinate invariant 3D vector algebra to 4 dimensions.

I hope you find these comments useful.
Ted

In article <yxFT0qE76TZBFwDx@farmeroz.port995.com>, Oz
<oz@farmeroz.port995.com> wrote:

[snip]

               ^^^^^^
               1-form

Sorry to nitpick, but it's a 1-form in n dimensions, which is *not* the
same thing as an n-form.  In d-dimensional space, you can have 0-forms,
1-forms, 2-forms, ..., d-forms.  Don't confuse the "degree" of the form
with the dimension of the space.  The dimension of the space puts an
upper limit on the degree, but doesn't define it.

Different dimensions are *not* all that different.  I think you're
thinking about converting between a form-based description of EM and the
vector-based description taught in high school, where there are some
details that necessarily depend on the particular dimensions 3 and 4 (and
in a sense also on the history of how the subject was developed) ... but
the generalities of the mathematics really are very, very regular.

But the ability (and sometimes the necessity) to convert between 1-forms
and vector fields is a fact of life in physics, so instead of lamenting
it, you should try to understand it more deeply.

I'll state a few things about the generalities:

(a) If you don't have a metric on your space, then 1-forms and vector
fields cannot be interconverted.  (For example, if there's no metric to
define the notion of orthogonality, you can't define a vector field
orthogonal to the isobars on a weather map, so you can't convert the
pressure gradient from a 1-form into a vector field.)

(b) If you *do* have a metric, then you *can* convert 1-forms into vector
fields and vice versa.

One way to do this is to think of a metric g as something that gives you
a number when you feed it two vectors:  the "dot product" of the two
vectors, v.w=g(v,w).  Also, you can think of a 1-form f as something that
(at a given point in space) gives you a number when you feed it a vector,
w:  this number is usually written <f,w>, and it is found by computing
the rate at which the vector w takes you across the surfaces of the
1-form f (e.g. how fast a velocity takes you across pressure isobars).

Then, given a vector v at each point in space, we can define a
corresponding 1-form f by the equation:

     <f,w> = g(v,w)

for all vectors w at that point.

And given a 1-form f, at each point we can always find a vector v that
satisfies this equation for all vectors w.

What does this interconversion mean geometrically?  In our weather map
example, if a vector field x always points *along* the isobars, i.e.
tangent to them, never crossing them, then we will have <f,x>=0.  And the
vector field v that satisfies our equation for being the equivalent to
the pressure gradient f will satisfy:

   g(v,x) = <f,x> = 0

So the vector field v will have a dot product of zero with the tangents
to the isobars, i.e. it will be orthogonal to the isobars.

(c) On an oriented space with a metric, you can define the Hodge *, which
lets you convert p-forms into (d-p)-forms, where d is the dimension of
the space.  For example, in 4-d spacetime with a metric, you can use the
Hodge * to interconvert 1-forms and 3-forms, and also to convert 2-forms
into other 2-forms.  But I need to explain 2-forms, 3-forms, etc. before
I can say what any of this really means.

There are a number of ways to go from 1-forms to 2-forms (and 3-forms,
etc.), and I'll say a bit about each one.

(a) The geometric picture

You can think of a 1-form as a stack of lines (or planes, or hyperplanes)
in 2-d, 3-d or n-d space.  These are the kind of thing you can get by
taking the contours of a scalar function on the space, though it's
important to understand that not all such "stacks" will be the contour
lines of any scalar function.

Now, imagine superimposing the stacks from two 1-forms.  In two
dimensions, the two stacks of lines will combine to break up the space
into a grid of little quadrilaterals -- so long as the two stacks aren't
the same, or parallel, in which case the result will be degenerate and
the elements of the grid will be empty.

In three dimensions, you have the same general thing, but the two stacks
of planes divide the space up into quadrilateral "tubes" that can (in
general) snake around in complicated ways, disappear, re-appear etc.  In
four dimensions, the two stacks of hyperplanes again divide up the space
into objects with quadrilateral cross-sections, but instead of being
tubes, they now have two free dimensions to snake around in.

The common pattern, though, is that if you have a *two-dimensional*
surface in any of these spaces, it will intersect the pattern formed by
the two stacks to make something that looks like the 2-dimensional case:  
a grid-like pattern of quadrilaterals.

This thing, in any number of dimensions, is known as a 2-form.  If the
1-forms that you combined to make it are f and g, we write this 2-form
(call it b) as:

 b = f ^ g

The quadrilaterals actually have a direction associated with them as well
(a little arrow going clockwise or counterclockwise around the
perimeter), and this changes sign if we combine the two 1-forms in the
opposite order:

 f ^ g = - (g ^ f)

 f ^ f = 0

which also makes sense because if you try to make a grid by combining the
same stack of lines/planes with itself, the result is degenerate.

Now, we can extend this process to include more 1-forms.  If p is any
number from 0 up to d, the dimension of the space, you can combine
1-forms f_1, ... f_p to make a p-form b:

  b = f_1 ^ f_2 ^ f_3 ^ ... ^ f_p

The symbol "^" is called the wedge product.

In general, a p-form created by superimposing p different 1-forms will
break up the d-dimensional space into cells that have p pairs of
hyperplanes as their walls, and (d-p) "free" dimensions to "snake around
in".

In two dimensions:

  1-forms are stacks of lines (for example contour lines), and the gaps
between the lines in these stacks have one free dimension;
  2-forms break up the space into little quadrilaterals, with two pairs
of walls and no free dimension.

In three dimensions:

  1-forms are stacks of planes, and the gaps between those planes have
two free dimensions;
  2-forms are "quadrilateral tubes", with one free dimension;
  3-forms are "cuboids", with 3 pairs of walls and no free dimension.

In four dimensions:

  1-forms are stacks of 3-d hyperplanes, and the gaps between those
hyperplanes have 3 free dimensions;
  2-forms are "quadrilateral regions", with two pairs of 3-d hyperplane
walls, and 2 free dimensions
  3-forms are "cuboidal tubes", with 3 pairs of 3-d hyperplane walls,
and 1 free dimension;
  4-forms are "hypercuboids", with 4 pairs of 3-d hyperplane walls, and
no free dimension.

And in a space of whatever dimension d, if you slice things down to
dimension p by imagining a p-dimensional surface cutting through the
space, the p-form's structures will intersect that p-dimensional surface
by dividing it up into little p-dimensional cuboids, with no free
dimensions left over.

One caveat I need to add is that, as well as the p-forms created as the
wedge products of 1-forms, linear combinations of all these p-forms are
also p-forms, and those linear combinations generally won't be
expressible as wedge products of 1-forms, with the corresponding nice
geometrical picture.

In other words:

a (f ^ g) + b (k ^ l)

where a and b are real numbers and f,g,k, and l are 1-forms, is a 2-form
... but this 2-form generally *won't* be expressible as (q ^ r) where q
and r are 1-forms.  The nice geometrical pictures for f^g and k^l
individually remain, but you need to keep in mind that not all 2-forms
can be described by a *single* geometrical picture.

Also, while this intuitive geometrical picture involves finite numbers of
objects, in reality this is just a way to approximate and visualise
things that really happen in the limit of infinitesimals.  When a stone
is travelling through a pressure gradient, you don't really count a
discrete number of isobars that it crosses, you use calculus to compute
an instantaneous rate at a point.  Similar caveats apply to everything
here:  it's just a discrete approximation, and an aid to understanding
what's going on.

(b) You can think of a 1-form as a linear function that takes a vector at
each point in space and gives you a number.  In the geometrical picture,
that number is "how fast" the vector crosses the stack of planes of the
1-form.

Similarly, you can think of a p-form as a linear function of p different
vectors at each point.  So a p-form b is a function:

  b(v_1, ..., v_p)

that is linear in each of its arguments.  It is also antisymmetric, so
that if you swap any pair of arguments, you change the sign of the result.

If we've built a p-form as the wedge product of p 1-forms f_1, ... f_p,
then:

 (f_1 ^ f_2 ^ ... ^ f_p)(v_1, v_2, ..., v_p)

= sum of signed permutations of the
    arguments of f_1(v_1) ... f_p(v_p).

For example:

 (f_1 ^ f_2)(v_1, v_2) = f_1(v_1)f_2(v_2) - f_1(v_2)f_2(v_1)

In the geometrical picture, p vectors v_1 ... v_p define a little piece
of a p-dimensional surface in the space, and b(v_1, ... v_p) is the
number of p-dimensional hypercuboids of the p-form that intersect this
small surface.  For example, two vectors define a little parallelogram,
and the number of quadrilaterals of a sliced 2-form that lie in that
parallelogram is the value of the 2-form evaluated as a linear function
of those two vectors.

(c) You can get from a p-form to a (p+1)-form by a derivative operation.  
The simplest example of this is when you move from a scalar function on
the space, f, which is a 0-form, to a 1-form which represents the
gradient of f, which is written df.

Exactly what is the 1-form df?  Geometrically, it is the contour lines of
the function f.  Algebraically, at each point df has to define a linear
function of vectors.  Now, if we go back to our notion of vectors as
tangents to parameterised paths through the space, if we have a path C(t)
we can write the vector at some point on that path as @_t (where I'm
using "@" as an ASCII substitute for the partial derivative symbol).  
Using that notation:

   <df,@_t> = df(C(t))/dt, the derivative of the
              function f with respect to t

If we have coordinates x^i, i=1,..d on our space, then these coordinates
are, of course, scalar functions on the space.  We can use these
coordinates to produce both vector fields and 1-forms.

The coordinate 1-forms are just dx^i, the contours of the coordinate
functions.  For example, in R^3 with x, y and z as coordinates, dx is a
stack of planes parallel to the yz plane, i.e. planes which range over
all possible values of y and z, and each have a fixed value of x.

The coordinate vector fields @_x^j at any point are the vectors whose
paths are given by varying one coordinate, x^j, as the path parameter,
and keeping the others fixed.  In R^3, the vector field @_x at any point
(x_0,y_0,z_0) has as its defining path the line C(x)=(x,y_0,z_0).

   <dx^i,@_x^j> = @x^i/@x^j
                = delta^i_j
                = 1 if i=j, otherwise 0

The coordinate 1-forms form a basis of all the 1-forms (at least in the
region of the space where these particular coordinates apply), and
similarly the coordinate vector fields form a basis of all vectors fields
in that region.

Now, getting back to the operator d, which is known as the "exterior
derivative" ... we've defined it for 0-forms, now we need to define it
for 1-forms.  The following properties hold:

  (i)  d (df) = 0
 (ii)  d (h df) = dh ^ df
(iii)  d is linear

where f and h are scalar functions, and df and dh are gradient 1-forms.  
It can be shown that *any* 1-form can be written as a linear combination
of coordinate 1-forms, so this is enough to give the exterior derivative
of any 1-form.

What do these properties mean geometrically?  It turns out that the
exterior derivative can be interpreted as measuring the boundary of the
1-form's geometrical structure.  A gradient 1-form consists of contours,
and contours of a smooth function have no boundary:  they always form
closed curves.  Hence d(df)=0 is consistent with this interpretation.  
Conversely, if we have lines or planes that start and finish, their
boundaries can define a 2-form's structures.

To give an example, consider the 1-form in two dimensions, x dy.  
Geometrically, dy consists of a stack of infinite horizontal lines, with
no endpoints, but x dy vanishes on the y-axis, and the stack of
horizontal lines becomes denser and denser as x increases.  (Give or take
a sign, which we haven't been too careful about tracking in our pictures,
the same thing happens as x decreases.)  So the whole plane is peppered
uniformly with the endpoints of the new lines that are needed.  These
endpoints are really the same kind of structure as a 2-form, which
peppers the plane with quadrilateral cells with no free dimensions:  the
cells are effectively 0-dimensional objects, as they can be shrunk to
infinitesimal size.

And if we compute the exterior derivative, it agrees with this picture:

d(x dy) = dx ^ dy

which is just the superimposition of the coordinate 1-form dx and the
coordinate 1-form dy, i.e. a uniform mesh of quadrilaterals covering the
plane.

And if we contemplate exactly the same example in three dimensions, all
the objects involved just get stretched out along the z-axis.  The 1-form
x dy is a stack of planes, and the boundaries of those planes are lines
which match up with the "quadrilateral tubes" of dx^dy in three
dimensions.

By induction, we can extend the exterior derivative to p-forms.  If a is
a q-form and b is an r-form:

   d(da) = 0
   d(a ^ b) = da ^ b + (-1)^q a ^db

Again, this can be interpreted geometrically in terms of boundaries.

------------------------------------------------------------
OK ... now what is the Hodge *, that converts p-forms into (d-p)-forms?  
I'll just sketch this.  A metric on the space lets you define a d-form
called the volume form, which, geometrically, is a division of the space
into hypercuboids that, if counted over any region, give you the volume
of that region as defined by the metric.  For example, for R^3 with the
ordinary Euclidean metric, dx^dy^dz is the volume form.

The Hodge dual of any p-form b, written as *b, is a (d-p) form that
complements b, in the sense that b ^ *b is a multiple of the volume form.
For example, in R^3:

    *dx = dy ^ dz, so
    dx ^ *dx = dx ^ dy ^ dz

In other words, the Hodge dual of a p-form involves (putting it crudely)
"the other dimensions" of the space, so that when you wedge something
with its dual, all the dimensions get covered by the resulting d-form.

In 3d Euclidean space, if you use the metric to interconvert 1-forms and
vectors, and the Hodge * to interconvert 1-forms and 2-forms, lots of 3d
vector algebra and calculus can be re-expressed in terms of wedge
products and exterior derivatives.  For example, if you convert two
vectors v and w to forms f_v and f_w, then convert the 1-form *(f_v ^
f_w) back to a vector, you get the cross product v X w.  If you compute
the 0-form or number *(f_u ^ f_v ^ f_w) for the forms corresponding to
three vectors u, v, w, it will be the scalar triple product (u X v).w.

On the 3d vector calculus side, the vector correspnding to df, where f is
a scalar function, is just grad f, i.e.

      df = (@f/@x) dx + (@f/@y) dy + (@f/@z) dz

which converts to the grad f vector:

      (@f/@x) @x + (@f/@y) @y + (@f/@z) @z
             

If you convert a 3D vector field v to a 1-form f_v, then take the
exterior derivative d(f_v), convert that 2-form to a 1-form with the
Hodge dual and convert back to a vector ... you get the curl of the
vector field v.

And if you convert a 3D vector field v to a 1-form f_v, convert that to a
2-form with the Hodge dual, take the exterior derivative, then convert
that to a scalar with the Hodge dual, that scalar will be div v, the
divergence of the vector field.
--------------------------------------------------------------

Well, boundaries of sets have one less dimension than the sets
themselves, and forms of increasing degree involve geometrical objects
with fewer "free dimensions", so I guess this will all make sense to you.

Good luck.  If you can, try to get to a good university library or a good
technical bookshop and flick through a few of the possibilities.  
"Gravitation" by Misner, Thorne & Wheeler goes way beyond your (current)
needs, but the sheer bounteousness of it might turn out to be worth the
cost.  In spite of being a GR textbook, it includes quite a bit of SR and
EM, and a few nice pictures of various p-form structures.  Taylor &
Wheeler is also sure to be helpful.

In article <DhcfgARUR2bBFwkA@farmeroz.port995.com>, Oz
<oz@farmeroz.port995.com> wrote:

[snip]

Well, that's a good attitude.  Understanding clearly when you can get by
without doing this is very useful.

It's easy to define mathematical spaces which simply don't have a metric,
and some of those spaces can be useful as abstract spaces in physics.  
But sure, if you're talking about ordinary spacetime in GR, of course
there always really is a metric.  Anyway, we're agreed that it's
instructive sometimes not to use the metric, even where there is one.

Wrong impression.  1-forms consist of stacks of hyperplanes (i.e. objects
of dimension d-1 in d-dimensional space), and if you have a metric you
can always find a vector field orthogonal to those hyperplanes and whose
magnitude is related to the spacing of the hyperplanes.  I gave you the
mathematical details, and you quoted them back to me:

It should be clear that this works in any dimension:  the vector field
corresponding to the form f is the vector field v that satisfies
<f,w>=g(v,w) for any other vector field w.  At any point, the vector will
be orthogonal to any vector that's tangent to the hypersurfaces of the
1-form.

In the language of forms, the 3D electrostatics of a point charge goes
like this:

(a) There is a potential, a scalar function (aka 0-form), U=q/r

(b) The gradient of the potential is dU = -q/r^2 dr, which, viewed
geometrically, consists of the spherical shells that are the contours of
the function U.  (The coordinate 1-form dr consists of evenly spaced
spherical shells, whereas -q/r^2 dr has ever more widely spaced shells as
r increases, and the contours of q/r become further and further apart.)

(c) The electric field E is the vector field corresponding to -dU = q/r^2
dr.  What is that vector field?  Just (q/r^2) @_r, where @_r is the
r-coordinate vector field that points away from the charge.  (Note that
for some metrics, and even some other coordinates in flat-space polar
coordinates, the conversion from 1-form to vector field isn't quite that
simple, but we'll cross that bridge when we come to it.)

So E = (q/r^2) @_r is the electric field, and the force on a test charge
q_2 is given by E q_2.

OK, that's a description of the solution for a point charge, but we
haven't mentioned the equation it's solving.  In vector notation, the
electrostatic equations are:

   div E = rho, where rho is the charge density
  curl E = 0

where E is a vector field.

If instead of being a vector field, though, suppose we think of E as
being the 1-form corresponding to that vector field.  Then these
equations become:

    *d*E = rho
      dE = 0

where * is the Hodge dual.

We've already mentioned that for a point charge, the 1-form version of E
= q/r^2 dr. *E will be a 2-form, which can be thought of geometrically as
a mesh of two intersecting sets of surfaces orthogonal to the spherical
shells of E.  In terms of coordinate forms, *E will be a multiple of
d{theta} ^ d{phi}, the two coordinate 1-forms for the other two polar
coordinates besides r.

The mesh of *E defines quadrilateral tubes that radiate out from the
point charge.  Now, d*E is a 3-form, the "boundary" of *E.  In our
geometrical picture, d*E is a little cuboidal box around every endpoint
of the quadrilateral tubes of *E.  So for the point charge solution, d*E
will be zero everywhere but on the point charge.  And *d*E will be a
0-form, a scalar function on space, equal to the charge density, rho.  
(That this is a delta function for a point particle is something I have
no wish to delve into at all.)

The second equation is simpler.  It just says that the gaps between the
surfaces of the 1-form E (spherical shells, for a point charge) have no
boundary.  The fact that E=-dU actually guarantees this, because ddU is
necessarily zero:  taking the exterior derivative twice always gives zero.

Maxwell's equations in 3+1 spacetime are almost exactly the same as
these, except they apply to a 2-form on spacetime, which is usually
called F:

    *d*F = J
      dF = 0

where J is a 1-form on spacetime which describes the movement and density
of charge.

I won't go into all the details of this; you really, really need to find
a book for this stuff.  I'm just going to give a quick sketch of how the
electrostatic picture extends into 3+1 dimensions.

Basically, the point charge extends into a world line in spacetime.  The
quadrilateral flux tubes poking out from the charge in 3d extend into
planes that terminate on the world line; these are the planes that lie
between the mesh of *F, which being a 2-form in 4 dimensions has two free
dimensions, and two "walled" ones defined by the mesh.

If you cut out one spatial dimension and visualise this in three
dimensions (two of space, one of time) the world line of the charge is
surrounded by planar radial "fins" rather than radial tubes as the gaps
between the mesh of the 2-form *F.  To be a little bit more mathematical,
*F will be a multiple of d{theta} ^ d{phi}.  Planes with fixed values of
theta and phi, and (r,t) taking on all possible values, will radiate out
from the world line, in place of the radial tubes that are the gaps
within the mesh of *E in the 3d version.  That the density and location
of the boundaries of these planes lie on the worldlines of point charges
-- or appear within the world tube of blobs of charge, if you want to
smear away those embarrassing delta functions -- is basically what *d*F =
J is saying.

The 2-form F itself will be a multiple of dr ^ dt.  The hypersurfaces of
dr are three-dimensional, of course, but if you slice through spacetime
and look at just one moment of time, you get spherical shells again.  The
hypersurfaces of dt fill all the spatial dimensions (i.e. all dimensions
orthogonal to the charge's world line).

In a picture where we look at 2 spatial dimensions plus time, we get the
dt hypersurfaces sliced down to horizontal planes, and the dr
hypersurfaces sliced down to concentric cylinders around the world line.

The 2-dimensional gaps between the mesh, though, are just the same kind
of spherical shells as in the 3d case, only here we need the spatial
hypersurfaces of dt to "wall them in" in the time dimension and make them
2-dimensional.  And dF=0 just means, as in the 3d case, that these gaps
have no boundaries:  they form closed spheres.

If you mean special, rather than general, relativity, it's clearer to say
that than "Newtonian".  "Newtonian" suggests pre-relativistic mechanics,
and while that can be described in 3+1 dimensions, I don't think that's
what you want.

Well, you really ought to go and find a suitable textbook at this point
(your whole aim with starting this thread, I know, so I hope you'll
follow through on it).

I'm so used to using units where c=1 that I take no responsibility for
checking the correctness of any factors of c that you use.  I also don't
know why you insist on the added labour of writing the components of
column vectors this way.  m (c,0,0,0)^T says the same thing with less
effort.

Nit-picking out of the way ... a body's 4-velocity is just the vector
defined by taking derivatives with respect to that object's proper time
(the time that elapses for a clock carried along the body's world line).  
If that world line is the t axis in an orthogonal coordinate system
(t,x,y,z) in flat spacetime, then proper time tau will coincide with t,
and the coordinates of the 4-velocity will be
(d/d{tau})(t,x,y,z)=(1,0,0,0).

The 4-momentum, by definition, is just the rest mass times the
4-velocity, so it will have coordinates (m,0,0,0).

Firstly, you're writing matrix multiplication in a non-standard way.  Put
the column vector being multiplied to the right of the matrix, or nobody
else will know what you're talking about.

Secondly, the Lorentz transformation matrix between the frames you
specified is:

   [  w  -vw   0   0]
   [-vw    w   0   0]
   [  0    0   1   0]
   [  0    0   0   1]

You missed out some terms, and you had the wrong sign for the velocity,
at least if I understood you correctly and you want to transform from the
old frame's coordinates to those of a new one whose origin is moving
along the x-axis at a velocity of v.  Those 1's are necessary to keep the
y and z coordinates unchanged (rather than zeroed), and the w you missed
on the diagonal is necessary for the whole thing to approach an identity
matrix as v approaches zero.

In units where c !=1 the v's become (v/c), but please, please just give
up on the c's.  If you want to rotate the time axis towards a spatial
axis, first convert them into the same units, or everything becomes
needlessly tedious.

The result is:

   [  w  -vw   0   0] [1]    [  w]
   [-vw    w   0   0] [0]    [-vw]
   [  0    0   1   0] [0]  = [ 0 ]
   [  0    0   0   1] [0]    [ 0 ]

Note that -w^2 + (-vw)^2 = (w^2)(v^2-1) = -1 by the definition of
gamma(v)=w.  The Lorentzian metric always gives a squared length of minus
1 for a 4-velocity.

I haven't read through your example of a collision carefully, but the
general idea that (a) 4-momentum is conserved, and (b) that it's a pain
trying to solve the resulting non-linear equations, is sound.  Other
people might have useful comments to add on this.  As, of course, will
any good SR textbook ...

In article <vDT0Y$EzihdBFwD9@farmeroz.port995.com>, Oz
<oz@farmeroz.port995.com> wrote:

[snip]

vector,

Usually this is true, but it's not the definition of <,>, so a little
care is need here.  (Unsure why you changed to upper case, but I'll take
that as a typo.)

Vector and 1-form coordinates on their own are always a bit dangerous,
because the reality is that they're totally meaningless numbers unless
they're tied to a particular *basis*.  A vector is not an n-tuple of
numbers, (w^1, ..., w^n), it is the sum w^i e_i, where e_i are basis
vectors.

Now, iff the f_i and w^i are coordinates that refer to "dual bases" for
the vectors and 1-forms is it safe to say <f,w> = f_i w^i.  What are dual
basis?  If {e_i} is the basis for the vector space, and {e^i} is the
basis for the 1-forms, then these two bases are "dual" to each other if:

   <e^i, e_j> = delta^i_j

An example of dual bases is given by the coordinate vector fields @_i and
the coordinate 1-forms dx^i that I defined in an earlier post.  There,
you have:

   <dx^i, @_j> by definition equals dx^i/dx^j = delta^i_j

Anyway, the *reason* that if you are using dual bases you get the f_i w^i
formula is:

   <f, w> = <f_i e^i, w^j e_j>
          = f_i w^j <e^i, e_j>
          = f_i w^j delta^i_j
          = f_i w^i

If the bases aren't dual, you can't expect that formula to work.  
Coordinate bases are automatically dual.

I don't understand your question.  The only forms I've been talking about
all along are also known as differential forms.  So there's no boundary
where we're sometimes talking about differential forms and sometimes not.

No, we agree on the superscript/subscript conventions, and your formula
here is correct (with the caveat about dual bases).  In fact, we can go
one step further and say:

The coordinates of the 1-form f that corresponds to the vector field v
are, if we're working with dual bases, f_n = g_{mn} v^m.  This is known
as "lowering an index" on v to produce f.  Conversely, we can use the
inverse metric g^{mn} to raise an index, and convert a 1-form to a vector
field.

Please don't take my word for anything, think about it.  The
hypersurfaces of a 1-form have *exactly* one less dimension than the
whole space.  Suppose the dimension of the space is d.  Then the
hypersurfaces are (d-1)-dimensional, and the tangent hyperplane to any
point on the hypersurface will be spanned by (d-1) orthogonal vectors.  
The greatest number of vectors in a basis of d-dimensional space is,
obviously, d, so once you pick even one vector orthogonal to all the
(d-1) vectors spanning the hyperplane, you have a complete orthogonal
basis for the whole space, so there's no room left for any ambiguity
except for a scalar multiple.

A scalar is a scalar.  You just write it as a variable, or algebraic
expression, with no enclosing symbols.

I take it "U is a 1-form?" is a typo for "dU is a 1-form?"
If so, then yes, dU is a 1-form:  a scalar multiplied by the 1-form dr

If you mean space is distorted by the mass, what that means for dr all
depends on how r is defined.  We have been talking about standard polar
coordinates in flat space, but if you change that assumption, it's up to
you to specify a new coordinate system for the new physical situation,
and the relationship between metric-defined distances and the coordinate
r need not be the same as in flat-space polar coordinates.

Also, I've been glossing over a small complication:  because the picture
we have of a 1-form involves finite spacing between the planes, and
because the reality involves *infinitesimal* spacing, we always have to
be sure to imagine the planes spaced so closely that neither the 1-form's
value, nor the geometry of space (if there's a metric) are changing much
over the stack of planes we're focusing on.  Real purists would say that
these planes *really* belong in a separate vector space associated with
each point in the manifold:  they are actually planes in the tangent
space.  But in a small enough region, assuming everything varies
smoothly, we can get away with thinking of the stacks of planes forming
nice patterns in real space.

I said dU = -q/r^2 dr, and of course that makes it a 1-form.

Yes we do.

I don't have the patience to give you a precise definition of the Hodge *
in all generality, but I gave you most of it in an earlier post, which
you promised to save and consult but now seem to be ignoring.  Instead of
saying "I forget", why don't you re-read that post?

Sorry, this is gibberish.  "d" is an operator that takes a p-form and
gives you a (p+1)-form, it is not a vector or 1-form, or any other object
with coordinates.  And rho, the charge density, is a scalar, so it has no
coordinates.

This is all random gibberish.  Please re-read the post where I sketched
definitions of these things, or better yet, drop the crap about there
being no suitable textbooks.  I managed to figure out all this stuff from
reading "Gravitation" by Misner, Thorne and Wheeler, and "Gauge Fields,
Knots & Gravity" by Baez and Munian.  Other people have posted other
suggestions.

The idea that the full spacetime version is really implicit in simple
Gaussian electrostatics, plus Lorentz invariance, is spot on, but your
suggestions about the detailed coordinate mechanics of this are wrong.

You really should read the first few chapters of "Gauge Fields, Knots &
Gravity", which is all about converting the old-style Maxwell's equations
to the language of forms.

dr is a 1-form, and hence in 4-dimensional space involves
(4-1)=3-dimensional surfaces of constant r.  Don't say "dr is
one-dimensional", say "dr has degree 1", or "dr is a 1-form".

Try harder.  Everything I'm telling you came out of textbooks.

Have you looked at "Spacetime Physics" by Taylor and Wheeler?

Feynman's lectures on physics vol. 1 chapters 15/17 contain a basic
discussion of 4-vectors.