Is it possible to recover that "energy is a torsor" in the Newtonian limit of GR?

Question

As Baez explains, in the Newtonian framework, energy is a torsor. However, in GR, we have for a particular frame, certain energy conditions, which impose a belief that energy (density) should take a positive value (see also this question), and this also appears to be related to the question here asking why GR is sensitive to absolute energies. As Baez put it, a torsor is like a group that has forgotten its identity. I'm curious in this context, if it's possible to understand how this forgetting happens, both from a physics and group theory perspective, if it does.

Answer 1

A torsor can be better described as an algebra whose operation has been made relative, rather than one in which all members "are equal". In the case of Abelian torsors and Abelian groups you can see how it's connected to the additive identity by rewriting the addition operation $a + c$ as $a - 0 + c$. To make it relative, you replace it by the ternary operator $a - b + c$. Now, it's translation covariant, $$(a + x) - (b + x) + (c + x) = (a - b + c) + x,$$ which shows that the operation is equally-based everywhere and is relative. It's no longer tied down to the $0$.

So, what does an additive torsor look like? Degrees in temperature, before the absolute zero was discovered, is an example. There's no zero, so all you see are relative differences. Entropy in thermodynamics, before the discovery of the Third Law. Pure states in non-relativistic physics have negative relative entropy to mixed states, so that only mixed states are physically relevant. The third law provides an absolute zero for entropy and make pure states physical: it's the place where quantum theory is injected into thermodynamics. Degrees on a circle are also an example of a torsor. In all these places, you have to pick a "conventional" zero or reference point and take differences relative to it, to get actual values.

Energy in Newtonian physics is also such a case ... and (in fact) I would argue that it is still the case in Relativity - ultimately by the Correspondence Principle - but more on that later. In fact, initially Einstein treated energy as residing on a torsor. But more on that later, too.

Actual values in an additive torsor show up, effectively, as difference pairs $a - b$, and in that sense everything is relative. You can add them, formally, by defining $$(a - b) + (c - d) = (a - b + c) - d = a - (d - c + b).$$ In fact, the operation $a - b$ could even be defined formally as the equivalence class of all pairs $(a, b)$ modulo the relation $(a - b + c, d) = (a, d - c + b)$. The result is an Abelian group in which $a - a$ is the identity, independent of $a$ and $b - a$ is the negative of $a - b$.

You can also tack differences onto elements of the torsor, so that $(a - b) + c = a - b + c$. You could also tack differences on the right, after using "commutativity" for both the torsor $a - b + c = c - b + a$ and for the addition operator, thus obtaining $$a + (b - c) = (b - c) + a = b - c + a = a - c + b.$$

To arrive at an axiomatization for Abelian torsors, you need only relativize the operations in the axioms for an Abelian group, e.g. $0 + x = x$ becomes $0 - 0 + x = x$ and $x + 0 = x$ becomes $x - 0 + 0 = x$. So, as axioms, they would read: $$a - a + b = b = b - a + a.$$

The commutativity property $x + y = y + x$ becomes $x - 0 + y = y - 0 + x$, which leads to the Abelian property for torsors: $$a - b + c = c - b + a.$$ Finally, associativity, undergoing similar relativization, becomes $$a - b + (c - d + e) = (a - b + c) - d + e.$$ The additive inverse and its properties come for free, after noting $-x = 0 - x + 0$, for then $$x + -x = x - 0 + (0 - x + 0) = (x - 0 + 0) - x + 0 = x - x + 0 = 0.$$ Similarly $$-x + x = (0 - x + 0) - 0 + x = 0 - x + (0 - 0 + x) = 0 - x + x = 0.$$

The more general case of non-Abelian torsors is discussed here. The axioms are the same as here, except for the commutativity property.

---

As for the energy issue, in an addendum in my reply to Is $E^2 = \left(mc^2\right)^2 + (pc)^2$ or is $E = mc^2$ the correct one?, after recapping the derivation in Einstein's original $E = m c^2$ (which is also linked to from there), I note that some time after this paper, and before c. 1910, energy was put on an absolute basis, with a zero for energy, though it was originally on a torsor scale, when the paper was written.

In contemporary language, this amounts to the reduction of the underlying symmetry group from Poincaré × ℝ to Poincaré. The additional generator makes the zero for energy relative - but also lifts the symmetry group to a form that has a non-relativistic limit to the symmetry group for non-relativistic theory, which is not the Galilean group, but its central extension: the Bargmann group.

To recap the reply itself, a bit, and using Einstein's notation in the $E = m c^2$ paper, $H$ was used for total energy of a body, while $E$ was used for the internal energy for a body at rest, the difference $K = H - E$ being the body's kinetic energy, and $K$ is a difference of torsor quantities $H$ and $E$, so with $K$, the torsor aspect of energy is removed from the picture.

Under Relativity, $H^2 = \left(m c^2\right)^2 + (p c)^2$ for a body of rest mass $m$ moving with a momentum $p$, while $E = m c^2$. Einstein argued that even after emitting light to the left and right in such a way that the body's rest frame remained unchanged, there was still a change in $K$: its kinetic energy decreased. Therefore its rest mass changed. The change in the rest mass matched the energy of the light emitted, divided by $c^2$.

The most general assumption you can make, based on the observations made in the paper is that $E = (m - μ) c^2$, for some "intrinsic mass" $μ$. The paper, itself, actually only drew conclusions concerning $K$, rather than $E$ and never actually stated $E = m c^2$. To draw that conclusion requires the additional assumption $μ = 0$.

However, that's precisely what passes over into the central charge of the Bargmann group which, in non-relativistic theory is the mass! When you set $μ = 0$, what you end up getting in the non-relativistic limit are the zero mass representations. In addition, you lose connection with non-relativistic theory in that the internal energy $E = m c^2$ no longer has a finite non-relativistic limit - except for $m = 0$. Adding $μ$ to the equation fixes that, in that both $m$ and $μ$ have the mass as their non-relativistic limit, so $m - μ → 0$ in the limit in such a way that it balances out $c^2 → ∞$.

So, I challenge the notion that there is an absolute zero for energy.

In fact, the Wightman axioms, in quantum field theory, could even be modified to work with Poincaré × ℝ, instead of with Poincaré. The representations for both groups are the same, except in the former, energy resides in a torsor: there is no absolute zero for energy. This provides additional room for the zero point energy.

Despite the "triviality" of the central extension Poincaré × ℝ, the modification is not trivial: there are subtle differences. With the modification, there is a change in interpretation. The energy-mometum 4-vector $(H,)$ of Relativity becomes, instead, the mass-momentum 4-vector $(M,)$, with the "moving" mass $M = m \sqrt{1 + p^2/(mc)^2}$ taking the place of energy, and $H$ no longer being synonymous with $M c^2$. Instead: $H = (M - μ) c^2$. The most important change is that time translation operator no longer corresponds to $M c^2$, but to $H$.

In field theory, the time translation invariance of the vacuum is used to argue that the vacuum energy is zero. Ultimately, this arises from the boost generators $$ and spatial translation generators $$. Both of these are $$ for the vacuum, and their Lie bracket is $\left[K^i, P_j\right] = i ħ δ^i_j M$, rather than $\left[K^i, P_j\right] = i ħ δ^i_j H/c^2$. The difference is that now $M = μ + H/c^2$, rather than $M = H/c^2$.

With $H$ and $M$ now independent, the Lie brackets for the boost and spatial-translation generators longer yield $H = 0$, but only $M = 0$. As in non-relativistic theory, time translation generator $H$ need no longer be zero. The representations for Poincaré × ℝ now provide room for an additive contribution to energy - even for the homogeneous representations, such as the vacuum.

Answer 2

The physics answer can be seen by looking at the action including a cosmological constant $$ S = \frac{\Lambda c^4}{8\pi G} \int d^4 x \sqrt{-g} + \frac{c^4}{16\pi G}\int d^4 x \sqrt{-g} R + S_{\rm matter} + \cdots $$ where $c$ is the speed of light, $G$ is Newton's constant, $\Lambda$ is the cosmological constant, $S_{\rm matter}$ contains any matter fields (standard model particles), and the $\cdots$ allows for higher order terms.

The cosmological constant term is the first one, $\sim \Lambda \int d^4 x \sqrt{-g}$. You can see that if the metric $g$ is a dynamical field, meaning that we should compute its equation of motion and solve for it (like we do in general relativity), then there will be a term in the equations of motion proportional to $\Lambda$. Therefore, the "constant" term in the energy contributes to the dynamics of the system through the metric. On the other hand, if we fix the metric (which is what we do in special relativity), then $\Lambda$ will not appear in any of the equations of motion for the matter fields, and so the "constant" term in the energy does not contribute to the dynamics of the system.

I am not sure what you are looking for by a "group theory explanation." I think what you want to see is a smooth limit where the energy transitions from being a "1-d vector" to being a "torsor." I don't know if such a limit makes sense. But, at a looser, less precise level, what you can say is that if you make $G$ smaller and smaller (so gravity becomes less and less important), then it becomes harder and harder to see a cosmological constant (or, in other words, you need a larger and larger cosmological constant before you can observe any effect, for a fixed observational precision). As a crude model, say there is some $\Lambda_{\rm min}$ that is the minimum cosmological constant you can observe. If $\Lambda < \Lambda_{\rm min}$, then the constant term effectively doesn't contribute to any observations, and you can treat energy "like a torsor" (meaning, adding constants to the energy won't matter, so long as you stay below $\Lambda_{\rm min}$. If $\Lambda > \Lambda_{\rm min}$, then energy behaves "like a 1D vector", in that the actual value of the energy does matter (not just the value up to an overall constant). Then in the limit that gravity becomes unimportant ($G$ becomes large), $\Lambda_{\rm min}$ becomes larger and larger until you are always in the "torsor like" regime.

Well, that's actually the limit of a fixed metric, where gravity isn't present. You actually asked about Newtonian gravity, where the gravitational field is present but simply small. This case is actually a bit subtle. Without matter, the normal equation for the Newtonian potential $\phi$, without matter, is Laplace's equation $$ \nabla^2 \phi = 0 $$ If we take the Newtonian limit $c\rightarrow \infty$, and assume the background spacetime is Minkowski spacetime (eg, we are working in approximately special relativity with small gravity effects), and ignore the matter action, we can identify $g_{00}\approx -c^2 - 2\phi$ (as well as $g_{0i}\approx 0$ and $g_{ij}\approx \delta_{ij}$), and Laplace's equation get's modified.

What's going on is that the cosmological constant introduces a non-trivial gravitational background, and we strictly speaking can't expand around Minkowski space, because Minkowski isn't a solution of the background equations of motion. One way to think of it is that with a cosmological constant, we shouldn't require the normal asymptotic boundary conditions where $\phi$ falls off to $0$ at large $r$, but rather allow it to grow, to allow for the fact that the gravitational field grows at large distances. The most rigorous way to handle this situation to do is to include $\Lambda$ in the background equations of motion, in which case we will expand around de Sitter space (for a positive cosmological constant, or anti-de Sitter space for a negative cosmological constant) instead of Minkowski space, and then $\phi$ will obey an analogue of Laplace's equation suitable for de Sitter space (assuming a positive cosmological constant; anti-de Sitter space if the CC is negative). However, if we think of the cosmological constant as being "small", we can drop the extra cosmological constant term in Laplace's equation, and recover normal Newtonian gravity where the absolute value of the energy doesn't matter -- essentially "zooming in" to the region well within the horizon in de Sitter space.

The above description is a pretty condensed and informal description of some rigorous calculations that you can find in the literature. The punchline is that for a small cosmological constant, if we're only interested in "local" observations (like the solar system), where the cosmological horizon is irrelevant, then at least as far as the cosmological constant goes, the Newtonian limit behaves a lot like the limit of a fixed metric where the gravitational field is kind of like a matter field that simply decouples from the cosmological constant. This decoupling is what allows energy to move from "vector like" to "torsor like".

Answer 3

The "zero" of energy can be made to become unimportant in general relativity by combining the divergence equation for the energy-momentum tensor with the balance/divergence equation for particle number (or "amount of substance" in a more macroscopic context).

The steps leading to this are carefully explained in

• Eckart: The thermodynamics of irreversible processes. III. Relativistic theory of the simple fluid,

see comments and footnote 3 after equation (27) there. The same sequence of steps appears in many other relativity texts, but Eckart explicitly comments about this point.

Roughly speaking, what happens is this:

Let's say that the absolute amounts of energy $E$ and of matter (particle number) $N$ are important. Call their fluxes $\pmb{\varPhi}$ and $\pmb{J}$. Let's say they satisfy conservation equations $$ \begin{aligned} \partial_t E &= \partial_i \varPhi^i \ , \\ \partial_t N &= \partial_i J^i \ . \end{aligned} $$

Now we choose an arbitrary zero point in the scale for energy, proportional to the amount of matter: $$E = \rho_0 N + U \ , \qquad \pmb{\varPhi} = \rho_0 \pmb{J} + \pmb{Q} \ , $$ where $\rho_0$ is an arbitrary constant of appropriate physical dimensions. The energy value $\rho_0 N$ acts as a zero point, but it is not an absolute zero, because $U$ can be negative. $U$, typically called "internal energy", is the energy measured with respect to the arbitrary zero point $\rho_0 N$. Similarly for the flux. Note that, for the moment, there's nothing physical about the proportionality equations above, because $U$ and $\pmb{Q}$ are free and unrelated to $(N,\pmb{J})$. Misner-Thorne-Wheeler remark on this arbitrary division in § 39.3 p. 1069 of Gravitation:

Here the baryon "mass" density $\rho_0$, despite its name, and despite the fact it is sometimes even more misleadingly called "density of rest mass-energy," is actually a measure of the number density of baryons $n$, and nothing more. It is defined as the product of $n$ with some standard figure for the mass per baryon, $\mu_0$, in some well-defined standard state; thus, $\rho_0 \equiv n \mu_0$.

See also their discussion about "internal energy density" in that chapter and in other parts of their book.

Replacing this proportionality in the conservation for energy we get $$ \rho_0 \partial_t N + \partial_t U = \rho_0 \partial_i J^i + \partial_i Q^i \ ; $$ but taking into account the conservation for matter this can be simplified to $$ \partial_t U = \partial_i Q^i \ . $$ This is an equation of conservation of internal energy, and because of the definition of $U$, we can add any constant to it as we like: this is equivalent to redefining $\rho_0$.

In a manner of speaking, we have divided energy arbitrarily into two parts: one proportional (by a constant) to matter, and which is conserved because matter is conserved; and one that reckons energy with respect to the arbitrary zero set aside in the first part. This second part is also conserved because of conservation of energy, but it appears to be about a "non-absolute" quantity, just because the division between the two parts is arbitrary.

In general relativity the division is such that the first term $c^2$ times larger than the second. So in a first approximation the second can be neglected. This is more or less how "mass" and its conservation come about in Newtonian mechanics; from the point of view of general relativity it takes care of the "absolute part" of energy, as sketched above. Related to this topic is also the theory of "post-Newtonian" approximations; see for instance

• Poisson, Will: Gravity: Newtonian, Post-Newtonian, Relativistic, chapters 6–8.