Discussion of Planck's units lead me to some thoughts on gravitation (1999/Jan).
indeed, the gravitational field will have to parallel-transport the vicinity of and note that the inner product of their momenta is almost certainly the quantity we should be computing Equally, the momentum of the photon is h.ν/c: and h/c is the product of our mass and length. So consider what we get if, in place of h, we take h/c as fundamental: say h = c.i. We now have mass * length = i and time * c = length: specifically, mass = c.sqrt(i/G), length = sqrt(i.G)/c, time = sqrt(i.G)/c/c.
The first two would be simplified by replacing G with F = G/c/c to sqrt(i/F) and sqrt(i.F) which will make the gravitational force between two bodies F times the product of their masses divided by the square of their light-time separations. Equally, instead of scaling the distance by c, we could have scaled the masses to get a product of momenta divided by the square of a distance.
Equally, consider D = F/c/c giving mass = sqrt(i/D)/c, length = sqrt(i.D).c and time = sqrt(i.D). In this, momentum has unit sqrt(i/D). and Newton's law identifies a force (momentum per time or rate of change of momentum) as D times something bilinear in 4-momenta and divided by the square of a separation-in-time, making momentum and time pivotal. This would appear to describe the gravitational field contributed by a given piece of the world-line of some body as a process propagating outward, on the relevant light-cone, which parallel-transports knowledge of the momentum of the body on that portion of its world-line while, at the same time, attenuating it so as to have constant `total integral' over each space-like section of that light-cone. This is just the formalism of Greens' functions, performed on a smooth manifold while wielding a vector quantity.
The `piece of world-line' is a portion of space-time having extent in all directions, though that extent is small. Its forward light-cone will give us, at any given later time, a volume bounded between two spheres `centered on' the piece in question; the integration discussed above is over this volume. Seen in space-time, this volume is a portion of a space-like sub-manifold, the universe `at that time' and has definite extent in each space-like direction (albeit rather thin in its `radial' direction), so it's a sub-manifold of co-dimension 1: over which integration takes a vector quantity (effectively contracting it with the forward time-like `normal' to the sub-manifold).
So our `graviton' (really, thus far, just a specific contribution to a classical field) propagates itself along geodesics and parallel-transports while attenuating (knowledge of) the momentum of its source: and the attenuation is such as to ensure that the graviton's integral at any one moment doesn't depend on the moment. The integral does depend on the frame whose `simultaneity' is presumed: the observed mass of a `moving' body exceeds its rest mass: it is the forward timelike component of a 4-vector whose true length is the rest mass. This arises naturally from the construction, because the only thing we can integrate up on a spacelike sub-manifold of codimension 1 is indeed a vector quantity: which, having been parallel-transported from a common source along geodesics, is reasonably straightforwardly correlated at each moment (which is as near as we can get to calling the vector field constant throughout the volume of integration).
This classical graviton (-oid) interacts with the time-line of some given other massive entity by imparting some momentum to that entity. If the given entity has extent, we must needs describe it as a momentum density of essentially the same kind as that propagated by the graviton it is encountering, so as to be able to integrate over the entity's extent to find its momentum (or, rather, its mass as a function of choice of frame of reference, which suffices to reconstruct a momentum, albeit a `vector quantity' but not properly based at a point). That fits neatly with Newton's gravitational equation, viewed as an equation p' = Q p with p' the rate of change of p and Q a linear operator, which contracts p with the graviton's attenuated momentum, scales by D = G/c/c/c/c and uses this as a scaling for a vector `back along the geodesic'.
That last notion is interesting. We've got a geodesic running from the entity backwards-lightlike to the source of the graviton, along which the graviton is `propagating' itself. This geodesic has a tangent at our entity, or at least at each point within its extent, and this tangent is the only part of the information available locally for Newton to use in computing the backwards direction. The space-like component of this tangent vector is the direction in which our momentum varies as we follow a time-like trajectory passing through our moment in time. Except that the momentum of the perturbed object may also have its time-like component changed: the 4-vector momentum has constant size, the mass times light's speed, so its time-like component increases or decreases according as the size of the space-like component increases or decreases.
to that entity a momentum equal to an integral over the 4-volume of intersection of the time-line (which must really have space-like extent for an entity with mass) and our `graviton': the integrand contracts our massive entity's 4-momentum with the attenuated vector field, presumably using the metric of space-time (since the things contracted both appear to be vector quantities) to obtain a scalar field. This gets integrated by the measure on our smooth manifold whose nature is `square root of the determinant of the metric'. Actually, since the integrand is a change to the momentum, this has to be multiplied by the `back' direction of the local geodesic-propagation of the graviton.
It knows the momentum of the source it's describing; since it knows this in such terms as to allow us to contract it with the momentum of the target, it must be parallel-transporting this information as it propagates it. The process of propagation should be able to take care of effectively propagating the `direction of propagation' and attenuating its signal in such a manner as to keep the `mass' integral invariant under choice of space-like slice. So I'll discuss the field as if it also remembers (and parallel transports) its direction of propagation and attenuates the momentum to achieve the required scaling.
Given that the field's propagation isn't necessarily in the same direction as the momentum it's describing (thinking about it, the field's going to be decomposable as a superposition of contributions of roughly this form), it now knows two `vector' quantities: and once you're parallel-transporting half the information about space-time, how about the other half ? Given a time-like momentum and a propagation vector which may be light-like, or nearly so, the remaining information could be encoded as a pair of space-like vectors.
Something also has to be propagating the machinery with which to do integration. This is the square root of the determinant of the metric of space-time, and that introduces a problem: but first I must explain what I mean by the square root of a determinant, and why it's relevant.
Consider space-time with tangent bundle T, gradient bundle G = {linear (T| :{scalars})}. For the metric to have a determinant, the dimension of G has to be finite: in which case G has the same dimension as T (they're dual to one another: each is the space of linear maps from the other to scalars). For convenience, I'll take the dimension of space-time to be 4: read 4 as dim(G) = dim(T) if you want a more generic description of determinants..
The metric of space-time is a tensor of kind G&tensor;G or linear (T|:G). The metric is symmetric(2, G) - consequently, it can be expressed as a sum of terms of form, for some g in G and scalar k, (T| t-> g(t).k.g :G). Furthermore: one needs no more terms in the sum than the dimension of G; each non-zero scalar used in place of k can be made either 1 or -1 by scaling the associated g; the gradients used as g can be chosen so that those appearing in the sum are linearly independent. That's just what the metric gets for being symmetric: because it's also invertible, it takes a full basis (4| g :G) to express it as sum(4| i-> k(i).g(i)×g(i) :{linear (T|:G)}) with k = [1,1,1,-1], [-1,-1,-1,1] or some permutation of one of these. Crucially, product(4| k :) is -1.
The measure (the thing that does integration) is properly an antisymmetric(4, G), notionally dx^dy^dz^dt with respect to coordinates x, y, z, t (so these are scalar functions of position) for which dx, dy, dz and dt (which are gradients) are a diagonalising basis for the metric, like the basis g above. If you take four gradients and form their antisymmetric product, you'll get zero if there's a linear dependence among them; shuffling the order in which they're combined always gives either the same answer or -1 times it.
Given a basis of G, not necessarily a diagonaliser of the metric, we can express each term in such a sum in terms of our basis of G, by expressing the gradients it involves as weighted sums of basis gradients and expanding out the products of sums as sums of products. That may give rather more terms than the original sum, but it's of the same form and now each term in the sum is just a scalar times the antisymmetric product of the basis, in some order. By shuffling the order and, where necessary, changing the sign of the scalar, we get a sum of parallel terms, so any member of antisymmetric(4,G) is parallel to the antisymmetric product of any basis of G. This means all members of antisymmetric(4,G) are parallel, so antisymmetric(4,G) is one dimensional; its members `looks like' scalar quantities, indeed (though one cannot multiply its members) there's a well-defined division on antisymmetric(4,G), yielding (truly) scalar ratios - subject to the inevitable ban on zero denominator, though here the zero in question is the zero member of antisymmetric(4,G), not the zero scalar. Take A = antisymmetric(4,G) and, while we're naming things, W = antisymmetric(4,T).
So what I said above is that the measure of space-time is a tensor in A. Just as the metric is invertible, so is the measure (i.e., it is non-zero in a 1-dimensional world, so division by it is feasible). (Formally, the measure is a tensor field on space-time whose rank at each point, p, of the manifold is A(p) = antisymmetric(4,G(p)), which has a zero member, and this isn't the value the measure takes in A(p).) Because we can do division in A (at each point), we can take the ratio of two tensors in A and ask whether their ratio is positive or negative: if positive, we can think of the two A-tensors as having `the same sign', even though there's no clear way to say what the sign of either of them is. Thus {linear (A|:A)} is naturally isomorphic to {scalars}, equating the identity on A with the scalar 1: in particular, it inherits the notion of sign from scalars.
If two A-tensor fields have the same sign at one point and are nowhere zero, we expect them to have the same sign as one another everywhere. However, a tensor field may be defined on a portion of the manifold, not the whole of it; it may even be defined on a disconnected domain - in which case, only expect sign to be `constant' on connected sub-domains. But even then, twisted space-times analogous to the Kleine bottle (of which the Möbius strip is a sort of `meridianal stripe', by analogy with the bits of the Earth's surface within a mile of the equator, except that the strip has a twist in it) don't behave as we expect. A smooth manifold is described as orientable if two nowhere-zero A-tensor fields, each defined on a connected neighbourhood within the manifold, have the same relative sign everywhere that they are both defined. It is usual to suppose that space-time is orientable, and this makes sense to me.
Now, just as G and T are dual to one another, so are A and W: so we can view A as {linear (W|:scalars)}, i.e. dual(W). Equally, A is FAPP {linear (scalars|:A)} and we can compose a member of A, in the first sense, before a member of A, in the second: the composite is, indeed, the tensor product of the two members of A, making it a member of A&tensor;A, aka {linear (W|:A)}. Because A, {scalars} and W are all 1-dimensional, this multiplication behaves very like scalar multiplication: in particular, {linear (W|:A)} is also 1-dimensional. In particular, any two members of A are parallel, so a typical pair of members is a and k.a for some scalar k; their squares are a.a and (k.k).a.a so, since k.k is non-negative, `have the same sign' in {linear (W|:A)}. Since all squares of members of A thus have the same sign, we do get a well-defined natural sign in {linear (W|:A)} by taking the squares of non-zero members of A to be positive. We then find that the product of two members of A has the same sign as their ratio, which is a scalar.
What's a determinant ? Given a linear space V, a basis (dim| q :G) of a linear space and a linear (G| f :V), we can construct A = antisymmetric(dim, G) with ^(q) as a typical member (also written, e.g. when dim is 4 = {3, 2, 1, 0}, as q(3)^q(2)^q(1)^q(0)). With W = antisymmetric(dim, V), we obtain a mapping, det(f) = (A| ^(q)-> f(q(dim-1))^...^f(q(0)) :W), known as the determinant of f. At its most basic, it takes a linear (G| f :G) and gives us back a number, at least roughly-speaking. For any basis (4| q :G), we have ^(q) defined as q(3)^q(2)^q(1)^q(0), with ^ denoting the antisymmetrised tensor product: this is in A. Likewise, (4| f&on;q :G) has its antisymmetrised tensor product, ^(f&on;q) = f(q(3))^f(q(2))^f(q(1))^f(q(0)), which is also in A. Now q is a basis, so ^(q) isn't zero; and is in A, which is one-dimensional, so we can divide ^(f&on;q) by ^(q). The answer you get, det(f) = ^(f&on;q) / ^(q), turns out to be independent of your choice of basis (essentially because we divided by it): it is called the determinant of the linear map f.
Now, just as the determinant of a linear (G|:G) is a linear (A|:A), which is effectively a scalar, so also the determinant of a linear (T|:T) is linear (W|:W) and, likewise, effectively a scalar by virtue of division in W, as in A. Indeed, in both cases, the determinant of the identity corresponds to the scalar 1. However, if we take a linear (G|:T) and apply the same construction, we get a linear (A|:W). For this we don't have a natural equivalence with scalars, except to the extent of having a natural sign, namely that of the determinant of any positive-definite linear (G|:T). Likewise, the determinant of a linear (T|:G), such as our metric, is linear (W|:A).
So take a basis (4| b :T) with dual (4| q :G), so q(i).b(i) = 1 but q(i).b(j) = 0 unless j = i: chose them such as to unit-diagonalise the metric of space-time, with b(3) timelike and b(0), b(1), b(2) spacelike (say - and, by the way, 4 = {3, 2, 1, 0}, 3 = {2, 1, 0}). This gives the metric, give or take an overall sign, as sum(3| i-> q(i)×q(i) :) - q(3)×q(3). Its determinant maps ^(b) -> ^(metric &on; b) = - ^(q) since (metric &on; b) is just q with an odd number of its entries negated. Now ^(q), as a member of dual(W), maps ^(b) to a positive value, and ^(q) × ^(q), as a linear (W|:A), is positive and maps ^(b) to a positive multiple of ^(q). So the determinant of the metric has the opposite sign to ^(q)'s square, hence is negative.
Although {linear (W|:A)} does have a natural notion of sign, it has no natural unit: division by any positive member gives as good an isomorphism with the scalars as any other. So -det(metric) presents itself as a suitable physical unit to use in {linear (W|:A)}: furthermore, if q is some basis of G which unit-diagonalises the metric, -det(metric) = ^(q) × ^(q), providing it with a square-root in A. When we do integration with respect to a chart, (4| i-> r(i) = (manifold: p-> r(p, i) :{scalars}) :), we use (4| i-> dr(i) :G) as a basis: naming the elements as r = [ t, x, y, z ], we get (4| [dt, dx, dy, dz] :G), and the measure takes the form dt^dx^dy^dz if the basis [dt, dx, dy, dz] unit-diagonalises the metric. If it doesn't, we have to scale by a Jacobean
A = ^dim(G), W - ^dim(T) (G|f:T) and (dim|b:G) with dual (dim|p:T) give us (A| ^b -> ^(f&on;b) :W) = `det(f)'.^p duality: double ^b => halve ^p k.^(f&on;b) / (^p / k) = k.k det(f) so this `det' isn't a scalar (but, of course, (A| ^b-> ^(metric&on;b) :W) offers itself as a unit) Crucially, {linear (A|:W)} is iso to W&tensor;W and we expect to be able to build w in W for which ^b -> ^(f&on;b) is k.w×w for some scalar k.
The metric's determinant is a symmetric linear map from antisymmetric(4,T) to antisymmetric(4,G) and equal to -1 times the square of the measure. The measure and metric are perturbed by our graviton, but out graviton doesn't need to carry information about them. They are there for it to exploit.
The measure enables integration over space-time. At any given moment in time, we have a space-like slice over which we can integrate a vector field (e.g. 4-momentum density), to obtain a scalar (mass): contracting the vector field with the measure gives an antisymmetric(3,G) field which we then integrate over the space-like slice (formally, a 3-dimensional sub-manifold of space-time). Our classical `graviton', at any given moment of time, fills a region between two `spheres'; it carries with it a vector field (the attenuated momentum, times G) whose integral over this region doesn't vary with time. The Q-field, which decides how the graviton perturbs the momenta of massive bodies it meets, is the tensor product of this attenuated momentum with a `unit' space-like `inwards'-pointing vector field: the negative of this field is the space-like component of the direction of propagation of the graviton.
We have p' = Q·p and g(p,p') = 0 for the 4-vector momentum, p, of a particle moving through a field we'll be trying to describe by Q, which will depend on the choice of time-like co-ordinate for which p' is rate of change of p; and g is the metric of space-time. Thus g(p,Q·p) = 0 for every vector p, which we can re-write as p·(g&on;Q)·p = 0. We thus obtain p·(g&on;Q)·p = 0 and this must be true for every time-like vector p. Consider time-like b and p; for all sufficiently small k, b-k.p must be time-like also, whence 0 = (b-k.p)·(g&on;Q)·(b-k.p) = -k.(b·(g&on;Q)·p + p·(g&on;Q)·b) = -k.(b-p)·(g&on;Q)·(b-p) whence (b-p)·(g&on;Q)·(b-p) = 0. Any space-like vector may be written as a difference of time-like vectors, hence in fact p·(g&on;Q)·p must be 0 for all vectors p, not just the time-like ones. Thus for any vectors p and b, 0 = -(p-b)·(g&on;Q)·(p-b) = p·(g&on;Q)·b + b·(g&on;Q)·p and g&on;Q is antisymmetric.
Indeed, g&on;Q must be a rotation in the `radial, time' plane (Q annihilates vectors g-orthogonal to this plane), i.e. a multiple of dr^dt; or hence, equally, of the measure contracted with two space-like `oribital' vectors both annihilated by dr and dt.
Written by Eddy.