Subject: Einstein's Equation in Pictures: Back in murky waters?
Date: Sat, 03 Apr 2004 19:15:59 +0300
From: Dimi Chakalov <>
To: Matthew Frank <>

Hi Matt,

Two years ago, on Fri, 29 Mar 2002 05:37:05 -0600 (CST), responding to my question Q2 (please see your email printed below), you said that "there's no way of isolating something called "the gravitational field stress-energy" at all in general relativity".

I'm still trying to find an answer to my Q2:

Q2. Specifically, what *contradiction* will be reached if it were a

The story is from the crucial month of November 1917,

I tried to elaborate at the front web page of my web site; see also
ReadMeFirst from the left navigation menu. Let me try to be more

My guess about Q2 above is the following: the reason why the
gravitational field stress-energy cannot be a tensor comes from the
requirement that the "thickness" of spacetime hypersurface needs to be zero. The 4-D spacetime is not embedded into some "outer spacetime", as we know from standard diff geometry; see the well-known balloon metaphor at

Thus, in the first picture from your "Einstein's Equation in Pictures",

you started with a *unit timelike vector*  v  at a point  p  , which is
a measure of the thickness of the *3-D space* hypersurface. Then you wrote, after the last picture, "Einstein's equation says that energy is the curvature of space. What does this mean?"

Are you talking about the dark energy that is casing the *accelerated* expansion of our good old spacetime along the cosmological time arrow? If yes, I think you may have to provide more pictures.

If the unit timelike vector  v  at a point  p  points to the "direction"
of the cosmological time arrow, the first question we should ask is how to define the elementary step, or elementary increment, of that
cosmological time. Surely we're measuring this time with our physical clocks confined "inside" 4-D spacetime, but these clocks cannot, do not, and should not be able to read the elementary increment of *that same cosmological time* along the unit timelike vector  v . The putative global mode of spacetime is suggested to provide some room for this unit timelike vector  v , as you might have noticed.

But how can we identify the increment of time along the unit timelike vector  v , if the latter should be set to approach *asymptotically* zero?

Look at Axiom IV in "Axioms for Geodesics", by Herbert Busemann,

"IV. Every point p has a neighborhood S(p, rp), rp > 0, such that for any two distinct points x, y in the neighborhood a point z with (xyz) exists."

Bingo! The point  z  is the so-called 'third point' in the solution to the paradox of continuum, after Robin Le Poidevin,

This point  z  belongs to the global mode of spacetime. You can try it with your brain,

However, since you're firmly confined inside the local mode of spacetime, which pertains to physical clocks and inertial frames, you cannot at all isolate something called "the gravitational field stress-energy" in general relativity, and are back in murky waters, since November 1917. And you are not alone,

More at

If all this exercise is a bit dense for you, I'll be happy to elaborate.

Take care,


On Fri, 29 Mar 2002 05:37:05 -0600 (CST), Matthew Frank wrote:
> Hi there.
> > It is a real pleasure to read your "Einstein's Equation in Pictures", > > gr-qc/0203100 [Ref. 1]. Thank you!
> I'm glad you liked it.
> > Q1. Why is it that the quantity proposed by Einstein for the
> > gravitational field stress-energy turned out not to be a tensor
> > [Refs. 2 and 3]?
> >
> > Q2. Specifically, what *contradiction* will be reached if it were
> > a tensor?
> The problem is not that "the gravitational field stress-energy" is not
> a tensor, but that there's no way of isolating something called "the
> gravitational field stress-energy" at all in general relativity.  All of
> the scalar or tensor quantities that one might try to use to capture
> this notion fail for one reason or other, and there are some general
> proofs to this effect.
> > Q3. Can you cast your answer to Q2 in Herbert Busemann SDG
> > and/or in Lawvere-Kock SDG?
> I'm not sure that either form of synthetic differential geometry is
> relevant to the question, though I am fond of both (and intrigued
> that, though they go by the same name, they diverge from
> standard differential geometry in opposite ways).
> --Matt

Note: There are many more subtle points in Matthew Frank's paper of March 15, 2002. I will highlight them in red.

"The metric also determines distances and lengths of curves. A geodesic is a locally straightest path, a path y0 such that for any continuous variation of curves yu, the derivative (d/du)(length of yu) vanishes at u=0. Note that any vector can be followed into a geodesic path, as indicated in the third picture for Einstein's equation."

A vector can be literally instructed to follow a geodesic path 'by hand', but it is not at all clear how this is being done by Mother Nature: the requirement  u=0  might not be actually achieved. See the affine structure (G. Nerlich) here; more on similar oxymorons from classical mechanics here.

"With the notion of distance comes a notion of volume (...). However, these r4 inequalities (cf. Appendix on Riemannian volumes - D.C.) are strict enough to give a procedure for estimating volumes to within an arbitrarily small factor of  1±e : simply divide the region into countably many balls of radius at most  e  and sum their Euclidean volumes."

No way, Matt! Have you tried to cover your bathroom wall with a countable set of finite tiles with dimensions at most  e ? It's an old task, Roger Penrose tried it many years ago. There are always gaps on the wall (they perhaps can be covered entirely by some fractals, but you never know). This is the old story of the dynamical, always running infinitesimal. See also the Thompson's lamp paradox here.

"It is the energy tensor which makes the left-hand side of the equation meaningful. This concept of "energy density in the direction of v" or "local energy as measured by the observer v" is best illustrated by examples. In a vacuum, the energy in any direction is always 0. (...)

"Having this at hand makes it possible to go through an example of Einstein's equation in detail; this may in particular help clarify the dimensionality of the vector spaces and manifolds. Consider Minkowski space (the space-time of special relativity), coordinatized as (x,y,z,t). Also consider an observer at the origin moving in the t-direction. For this observer, the spacelike vectors are all the vectors which have no t-component. The geodesics form the 3-d hypersurface t=0. The ball of radius r is all points (x,y,z,0) with x2 + y2 + z2 < r2, and it has volume exactly (4/3)[pi]r3. Hence the curvature of the hypersurface is 0, as it should be since Minkowski space is a vacuum."

We know that Minkowski spacetime is a crude approximation to the real picture in General Relativity, so the first thing we should do is to unravel the approximation which produces the physically wrong notion of zero curvature: you can enjoy zero curvature iff  you can count those tiny little Euclidean volumes which build up the volume that is "exactly (4/3)[pi]r3".

Only you can't. See the story of Archimedes and [pi] here. Again, I'll be happy to elaborate. Only don't ask me what is 'curvature of spacetime', because I really don't know. Ask John Baez or Jerzy Lewandowski instead.

John Baez, for example, posed the following question: "Is gravity really curvature, or what else -- and why does it then look like curvature?"

Why does it look like curvature, really? It's like asking why the quantum waves look like ordinary waves in the double-split experiment. Or why is it that we observe some 'cosmic equator'. These are quantum phenomena, but they have to look in some way in the local mode of spacetime. We know that there are no empty waves nor some 'absolute axis' for the rotation of the universe. Likewise, we talk about 'spacetime curvature' but we know very well that this animal is being cast in the local mode of spacetime. We define it by referring to the global mode of spacetime, like 'the context of a sentence', but I don't think that we would see it like 'spacetime curvature' in the global mode of spacetime. Generally speaking, the effects of the Holon are always cast in the local mode of spacetime, and always with the inevitable distortion.

One way to demonstrate this conjecture would be to eliminate the inertial mass of my body and levitate over the head of John Baez and his colleagues at GR17 in Dublin. Then perhaps they would be interested in the nature of gravity.

I'm joking, of course. They wouldn't. Nobody has shown any interest in the elementary increment of the cosmological time arrow. Do you wonder why? The answer is here.

D. Chakalov
April 3, 2004
Last update: May 15, 2004