Between QM and QFT

In his textbook Quantum Field Theory: A Modern Approach, Michio Kaku first describes quantum field theory as

\textrm{Quantum field theory} = \left\{\begin{array}{l}\textrm{Group theory} \\ \textrm{Quantum mechanics}\end{array}\right.

Later on he says,

\lim_{N\to\infty}\textrm{Quantum Mechanics} = \textrm{Quantum Field Theory}

where N is the number of degrees of freedom of the system. How compatible are these descriptions?

Consider the group-theoretic description of n-sided polygons given by dihedral groups. A regular polygon with n sides has 2n symmetries, half rotational and half reflective. As n\to\infty one intuitively expects the n-gon to approach a circle and thus would expect that the infinite dihedral group would thus describe the group of symmetries of a circle. But, if I understand this correctly, it doesn’t. In fact the infinite dihedral group is actually an isometry group of \mathbb{Z} which is a line!

Given Kaku’s two definitions, then, I am impelled to ask,

  1. is QFT <em>really</em> the limit of QM for as N \to\infty; and
  2. if it is, then what might we learn by instead considering the limit of large N?

Does anyone know if this has been done?

Spacetime curvature redux

Apparently the argument I’ve been having isn’t about whether E&M fields can produce curvature or not. Rather the argument is about whether curvature is gravity (which, admittedly, is the standard interpretation but is not the interpretation I’ve used for the past decade). If so, there is no gravitational field, per se – it flat out doesn’t exist. There’s only curvature which can be interpreted as being gravity. But then the word “gravity” simply becomes a label and we end up with interpretational contradictions related to mass. In fact it verily begs for a redefinition of mass. Either way, it’s not a simple problem to be swept under the rug.

Rainich unification

In the responses to my previous post, Cristi Stoica provided some very useful references that, for now, I am going to classify as “Rainich unification” since I needed something for the title of this post.  What caught my eye was the following from the abstract of a paper by Misner and Wheeler from 1957:

Maxwell’s equations then reduce, as shown thirty years ago by Rainich, to a simple statement connecting the Ricci curvature and its rate of change. In contrast to unified field theories, one then secures from the standard theory of Maxwell and Einstein an already unified field theory. (Misner, C.W. and J.A. Wheeler (1957). “Classical physics as geometry”. Ann. Phy. 2: 525-603.)

Some related work was done by Penrose and Rindler in the 1980s but not much seems to have come of it. Whatever happened to this approach? Why was it dropped? What are its inherent flaws?

Spacetime: a more convincing argument

Following up on my previous post, here’s an even more convincing argument that it is possible to construct a toy universe in which the curvature of spacetime is due to electromagnetism and not gravity and that I was unjustly vilified over at nForum (remember while reading the rest of my post below that it was asserted that I lacked a rudimentary knowledge of general relativity).

Consider again a charged massless universe, that is a universe in which there are electromagnetic fields but no gravitational fields. In such a case the complete stress-energy tensor only includes the electromagnetic portion. Einstein’s field equations are

G^{\alpha\beta}=8\pi T^{\alpha\beta}.

In order to prove my point I need to show that G is solely determined by T (which I always thought was the standard interpretation, but after the tongue-lashing I received I’ll prove it just to be on the safe side).

In my toy universe I will assume that Riemannian geometry still exists (since, as a mathematical tool, it is independent of anything physical anyway). The definition of G is

G \equiv R^{\alpha\beta}-\frac{1}{2}g^{\alpha\beta}R

where the Rs may be written in terms of g. Thus it boils down to determining whether g, which is the metric, can be independently determined (i.e. from conditions not present in Einstein’s field equations).

As a simple case, let’s take the weak field approximation,

g_{\alpha\beta} = \eta_{\alpha\beta} + h_{\alpha\beta}

where |h_{\alpha\beta}|<<1 and \eta_{\alpha\beta} is the Minkowski metric for flat spacetime. If we further restrict ourselves to the linearized theory, the linearized, weak field Einstein equations are

\Box\bar{h}^{\mu\nu} = -16\pi T^{\mu\nu}

where \Box is the d'Alembertian. It is easier to explain the next step by showing what is done in normal GR, i.e. not in my toy universe. In such a case, for example,


and this is compared to the Newtonian


where \phi is a scalar potential identified with Newtonian gravity. Thus, we choose

\bar{h}^{00} = -4\phi

in order to force this to match the Newtonian gravity!

Let’s switch back to my toy universe now. The energy density, \rho, is given by T^{00} which, in normal GR is the density of the gravitational field. But in my toy universe,

T^{00} = \frac{1}{2}(E^{2} + B^{2})

where we have employed units such that \mu_{0} = \epsilon_{0} = c = 1. Further, in the electrostatic case, we note that,

\bigtriangledown^{2}\phi = -\rho

where \rho is the charge density and where we again are employing units with \mu_{0} = \epsilon_{0} = c = 1. Thus in the electrostatic case of my toy universe we may choose

\bar{h}^{00} = -16\pi\phi

where \phi is the charge density! The metric and thus the curvature of spacetime in my toy universe has absolutely nothing to do with gravity!

Addendum: If the full stress-energy tensor is employed in general relativity, i.e. with both gravitational and electromagnetic portions, one could presumably do a similar weak field construction in which \bar{h} depends on both gravitational and electromagnetic fields which means I’m even right in this universe: the metric encodes the curvature of spacetime due to field sources and is not necessarily due solely to gravity.

A little thought experiment

I got into a long and protracted argument with someone on an online forum recently over a seemingly minor point and got accused of not knowing basic physics.  It would be funny if the guy weren’t so rude and dismissive about it.  At any rate, part of the argument came out of the following thought experiment I concocted (actually the version I present is a much better version than the one that sparked the argument).  In it, I devise a situation (albeit highly contrived) in which it is impossible to tell whether a certain object is attracted to a certain type of box due to gravitation or electrostatics.  This highly contrived story takes place in a universe in which we have massless, charged particles (something we don’t presently see in nature) and assumes that the given materials are all that you have in your possession (actually, in theory, you wouldn’t even need massless, charged particles as long as it was plausible that you could have a small black hole inside the box).


You are in possession of a neutral object of some mass like, for instance, a piece of Scotch tape or a balloon or something of that nature.  You are also in possession of  special type of box that happens to be both massless and neutral.  In the box is either a strong gravitational field source (a neutral, massive object) or a strong electromagnetic field source (a massless, charged object).  You do not know which.  You begin to bring the object (i.e. the balloon or tape or whatever) closer to the box.  At some point it becomes attracted to the box.  It is your job to determine whether the attraction is due to gravity or electrostatics.

If it is an electrostatic attraction that occurs, this is due to polarization which means the sign of the charge producing it doesn’t matter (this is a simple experiment/demo I do with Scotch tape in one of my classes).  Since you can’t see inside the box you have no idea how much charge or mass is present (so, for example, the box might even contain a small black hole) which means you can’t determine what the source is based on the rate at which the object was attracted to the box.  Therefore, you have absolutely no way to tell whether it was attracted due to gravity or electromagnetism!

Are there holes in the argument?  Remember that this is all you have access to (i.e. you can’t go running out to get an electrically charged object to see if it is repelled).

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