*Leibniz Equivalence*, stating that spacetime models, appropriately related, "represent the same physical world". In Robert Wald's monograph (

*General Relativity*, 1984), the principle is explained like this:

... if $\phi : M \rightarrow N$ is a diffeomorphism, then $M$ and $N$ have identical manifold structure. If a theory describes nature in terms of a spacetime manifold $M$ and tensor fields $T^{(i)}$ defined on the manifold, then if $\phi : M \rightarrow N$ is a diffeomorphism, the solutions $(M, T^{(i)})$ and $(N, \phi^{\ast}T^{(i)})$ have physically identical properties. Any physically meaningful statement about $(M, T^{(i)})$ will hold with equal validity for $(N, \phi^{\ast}T^{(i)})$. On the other hand, if $(N, T^{\prime (i)})$ is not related to $(M, T^{(i)})$ by a diffeomorphism and if the tensor fields $T^{(i)}$ represent measurable quantities, then $(N, T^{\prime (i)})$ will be physically distinguishable from $(M, T^{(i)})$. Thus, the diffeomorphisms comprise thePerhaps surprisingly, Wald's formulation is equivalent to:gauge freedomof any theory formulated in terms of tensor fields on a spacetime manifold. (Wald 1984: 438)

If spacetime models $S_1$ and $S_2$ areIf this is the correct expression of gauge freedom in $GR$, then:isomorphic, then they represent the same world.

a gauge choice in $GR$ = tIf $S$ is a spacetime model with underlying point set $X$, one can define its image $S^{\pi}$ under an arbitrary permutation $\pi : X \to X$: here one transforms everything, topology included. So, suppose $S$ has the form $S = (X, \mathcal{T}, g, \dots)$, where we have made the topology $\mathcal{T} \subseteq \mathcal{P}(X)$ explicit. (The elements of $\mathcal{T}$ are open sets.) Then define the transformed topology, and fields:he selection of a representing spacetime model from an isomorphism class.

$\mathcal{T}^{\pi} \hspace{4mm} := \{\pi[O] \mid O \in \mathcal{T}\}$.And the permuted model is then:

$g^{\pi}(p) := g(\pi^{-1}(p))$, for each $p \in X$.

and so on.

$S^{\pi} := (X, \mathcal{T}^{\pi}, g^{\pi}, \dots)$Then it's almost trivial that:

$S^{\pi} \cong S$So, $S^{\pi}$ and $S$ represent the same world: this means that an

*arbitrary permutation*of the point set of a spacetime model generates a physically equivalent model. And, if so, then there is no reason for this permutation to be a

*diffeomorphism*of the underlying

*manifold*(i.e., an automorphism---symmetry---of $M$).

If this is right, the gauge freedom in GR is not given by $\text{Diff}(M)$.

Although I am not a specialist, I think that since an arbitrary permutation is not a topological isomorphism, you cannot claim that X and permutated-X are essential the same. Generally speaking, if the permutation is "very bad" and you consider the induced topology on permutated-X (ie U is open in permutated-X iff its backward image is open) then it is not the case that X can admit a metric tensor at all.

ReplyDeleteThanks, anon - this is the question I always get!

ReplyDeleteI updated a bit to explain how the permutation group $Sym(X)$ acts on the model $S = (X, \mathcal{T}, g, \dots)$.

You're right *if* we try and keep the topology $\mathcal{T}$ on the underlying manifold fixed, and leave it invariant. But, in this case, given the permutation $\pi : X \to X$, we transform the topology of $S$ as well, along with all other structure defined the base set $X$ (metric, matter fields, ...). By construction, the resulting model, $S^{\pi}$, is isomorphic to the original, no matter how crazy $\pi$ is.

In a sense, in addition to the metric tensor $g$ and other fields, we also treat the topology as physical.

Cheers,

Jeff

I think that the formulation of Leibnitz Invariance as isomorphism invariance is a quite nice way to think about it. But you seem to miss covariance, that is that physicists demand that the form of physical laws does not change under symmetry transformations. In the case of your example, the Einstein Equations of $S^\pi$ would be $\mathcal{G}^\pi(g^\pi) =1/m_p^{2} T^\pi$, where covariance demands the form $\mathcal{G}(g^\pi)=1/m_p^2 T^\pi$, with the Einstein Tensor $\mathcal{G}$. ( I am not sure if there is a less hand waving definition, than "I do not want to learn my formulas more than once." though.)

ReplyDeleteThanks, Yoshi,

ReplyDeleteWe have covariance, because we formulate laws in terms of tensors $T, \dots$, over a manifold $M$. This guarantees co-ordinate independence. Suppose our law is:

$T(p) = 0$, for all $p \in M$

saying that the tensor quantity $T$ vanishes everywhere.

Let $T^{\phi_1} and T^{\phi_2}$ be the co-ordinate representations of $T$ under co-ordinate systems $\phi_1, \phi_2$. We get covariance because:

[$T^{\phi_1}(p) = 0$, for all $p$] iff, [$T^{\phi_2}(p) = 0$, for all $p$].

Cheers,

Jeff

(For a co-ordinate system $\phi : U \to V \subseteq \mathbb{R}^4$, the co-ordinate representation $T^{\phi}$ is defined (on $V$) to be: $T \circ \phi^{-1}$.)

ReplyDelete