# Diffeomorphism invariance in special relativity

Suppose space time is the manifold $$M$$ isomorphic $$\mathbb{R^4}$$ whit the metric $$-\eta_{00}=\eta_{11}=\eta_{22}=\eta_{33}=1$$ in the Cartesian coordinates $$\Psi(p)=(x^0,x^1,x^2,x^3)$$ for $$p \in M$$ .

So we have model $$A=\bigg \langle {M,\eta,\phi,\frac{\partial}{\partial x^a }} \bigg \rangle$$ for general relativity and a model $$B=\bigg \langle {M,\eta,\phi,\frac{\partial}{\partial x^a }} \bigg \rangle$$

for special relativity where $$\phi$$ is a scalar field and $$\frac{\partial}{\partial x^a }$$ vector field.

It is often said that general relativity is diffeomorphism invariant and special relativity not.
For example, suppose we have a diffeomorphism $$F:M\rightarrow M$$ and the wave equation $$\eta^{ab}\frac{\partial \phi}{\partial x^a }\frac{\partial \phi}{\partial x^b }=0$$

We have by pullback $$F^*\left(\frac{\partial }{\partial x^a }\right)=\frac{\partial G^{b}}{\partial x^a }\frac{\partial}{\partial x^b },$$ $$F^*\phi=\phi \circ F$$ and $$F^*(\eta_{ab})=\frac{\partial F^{k}}{\partial x^a }\frac{\partial F^{j}}{\partial x^b }\eta_{kj}$$ where $$G=F^{-1}$$

So $$F^*\left(\eta^{ab}\frac{\partial \phi }{\partial x^a }\frac{\partial \phi}{\partial x^b }\right)=\eta^{ab}\frac{\partial (\phi \circ F)}{\partial x^a }\frac{\partial (\phi \circ F)}{\partial x^b}$$

that is $$\phi \circ F$$ is a solution if is a solution $$\phi$$.

Is the reason that this does not work in special relativity is that in special relativity the metric $$\eta$$ has to be fix. That is $$F^*B=\bigg \langle {M,F^*\eta,F^*\phi,F^*\phi\frac{\partial}{\partial x^a }} \bigg \rangle$$ is not a model of special relativity because $$F^*\eta \neq \eta$$, and in general relativity if $$A$$ is model than $$C=\bigg \langle {M,g,\phi,\frac{\partial}{\partial x^a }} \bigg \rangle$$ is a different model than $$A$$, but a model of general relativity.

What i mean is,though $$F^*B$$ and $$B$$ have equivalent solution they are not equivalent models because $$F^*B$$ is not a model of special relativity.

Is this the reason why special relativity is not diffeomorphism invariant?

• Possible duplicate of Special relativity and diffeomorphism invariance – Umaxo Oct 18 '19 at 10:20
• Relativistic equations are only required to be invariant under some representation of the Poincaré group. This is true in special relativity, where spacetime is flat; but in GR, spacetime is curved, and this restriction doesn't apply. Basically, this is because in flat spacetime, the Poincare group is always the isometry group of the spacetime; but a general curved spacetime may not have an isometry group at all. So the only way to write laws of physics that are invariant in a general curved spacetime is to make them invariant under general diffeomorphisms. – Lagrangian Oct 24 '19 at 9:56
• If I understand what you mean by diffeomorphism invariance, then endowing special relativity with that property is easy: just replace the condition $$\text{metric}=\sum_\mu \eta_{\mu\nu}dx^\mu\otimes dx^\nu$$ with the condition $$\text{Riemann curvature tensor}=0.$$ The resulting theory is "diffeomorphism invariant" in the same sense that general relativity is, even though spacetime is constrained to be flat. Am I interpreting the question correctly? – Chiral Anomaly Oct 24 '19 at 13:13
• if we treat special relativity as general relativity , than special relativity diffeomorphism invariant since it is contained in general relativity and general relativity is diffeomorphism invariant . my question is when we consider special relativity not as special case of general relativity, if the reason to special relativity not be diffeomorphism invariant is because $F^*\eta \neq \eta$, since.... – amilton moreira Oct 24 '19 at 15:22
• when special relativity is not as special case of general relativity it is considered a theory with a fixed metric $\eta$ so a manifold with metric $A$ $F^*\eta \neq \eta$ is not as special relativity although $F^*B$ and $B$ have equivalent solution – amilton moreira Oct 24 '19 at 15:38

The only possible sense in which SR is any "less" diffeomorphism invariant than GR is in that Minkowski space happens to admit a natural family of global coordinate systems (Lorentz coordinates) in which the metric happens to take a particularly simple form, which is identically equal to $$\pm \mathrm{diag}(-1, 1, 1, 1)$$ - so simple that you can completely hide it into the raised and lowered indices, which have a trivial mathematical relation (just flippig the sign of the timelike component). In these coordinates, all of the covariant derivatives reduce to partial derivatives, so for elementary applications it's often easiest to always work in those coordinates and not worry about coordinate transformations. But it's still nothing more than just a particularly convenient choice of coordinates, and you're always free to switch to other coordinates. (Of course, the Riemann curvature tensor will vanish in any system coordinates.) In GR, there's no global choice of coordinates in general, and certainly not one as simple as Lorentz coordinates, so we're forced to deal with all the subtleties of changes of coordinates that we can get away with ignoring in SR.
• @Dwagg One approach to diffeomorphism invariance in special relativity is writing everything in terms of a metric tensor. The metric tensors in this theory all satisfy $R_{κλμν}=0$. If you drop that condition, the extra freedom you get is exactly the gravitational field. So the physical content is that the mathematical framework that gives you (unphysical) diffeomorphism invariance also gives you a physical force of gravity; it's "unreasonably effective", I suppose. Tparker, could you add something about this to the answer? – benrg Oct 27 '19 at 2:42