How are spacetime symmetries different from simple general coordinate invariance? Physical laws should be coordinate independent. Are Poincare invariances not simply changing coordinates?

  • 4
    $\begingroup$ No, spacetime symmetries in the sense you're describing are physical symmetries (i.e. moving points of the manifold around, otherwise known as isometries), whereas coordinate invariance (or covariance) leaves the points unchanged. Writing a theory a generally covariant way says nothing about the isometries of the spacetime. $\endgroup$
    – Eletie
    Dec 12 '20 at 13:39
  • 1
    $\begingroup$ See also: active/passive transformations. $\endgroup$
    – Charlie
    Dec 12 '20 at 14:08

Perhaps it would be useful to consider a system which does not exhibit Poincare invariance. A typical relativistic action for a worldline $\gamma$ in Minkowski space $\mathcal M$ might look like this:

$$S[\gamma] = \int d\sigma \left[mc^2\sqrt{\eta_{\gamma}(\dot \gamma,\dot \gamma)} - U(\gamma) \right]$$

where $\gamma=\gamma(\sigma)$ and $\eta_\gamma$ is the flat spacetime metric at the point $\gamma(\sigma)$. This makes no reference to any coordinates, so it is manifestly coordinate-invariant. It is not Poincare invariant, however.

A Poincare transformation is a diffeomorphism $\phi:\mathcal M \rightarrow \mathcal M$ which preserves the flat spacetime metric, i.e.

$$ \eta_{\phi(\gamma)}(\phi_*\dot \gamma,\phi_*\dot \gamma) = \eta(\dot \gamma,\dot \gamma)$$

where $\phi_*$ is the pushforward along $\phi$. If you aren't familiar with that terminology, it simply means that when you map spacetime points to other spacetime points via $\phi$, you also map tangent vectors to other tangent vectors via $\phi_*$. Concretely, $\phi$ might implement a spatial rotation, a boost, or a spacetime translation on spacetime points, while $\phi_*$ implements the corresponding transformation on the tangent vectors.

The new action obtained after this transformation is

$$S[\phi\circ\gamma] = \int d\sigma\left[mc^2\sqrt{\eta_{\phi(\gamma)}(\phi_*\dot \gamma,\phi_*\dot \gamma)}-U\big(\phi(\gamma)\big)\right] $$ $$=\int d\sigma\left[mc^2\sqrt{\eta_\gamma(\dot \gamma,\dot \gamma)}-U\big(\phi(\gamma)\big)\right]$$

where we've used the fact that $\phi$ is an isometry of $\eta$. Clearly this is not the same as the original action, because the second term has become $U\big(\phi(\gamma)\big)$. As a result, Poincare invariance for a single particle in an external potential requires that the external potential be a constant function on spacetime - but that's the same as no external potential at all.

In more down-to-Earth terms, coordinate invariance is trivial because when we change coordinates $x^\mu\mapsto x'^\mu$, we also change the form of e.g. the potential function $U_{(x)}(\mathbf x)\mapsto U_{(x')}(\mathbf x')$, where the latter transformation is defined in such a way as to preserve the physics. In Physics 101 we say the gravitational potential is $U(x,y)=mgy$; if we were to implement a 90$^\circ$ clockwise rotation, then this function would become $U'(x',y')=-mgx'$. This is a different function, which is specifically chosen so that solutions of the old equations of motion are mapped to solutions of the new equations of motion under our 90$^\circ$ rotation.

A physical symmetry is very different. It contains physical content because it asserts that when we change coordinates $x^\mu \mapsto x'^\mu$, solutions to the equations of motion are preserved even when we don't change $U$. That is, $U$ is insensitive to the transformation we performed on the coordinates. This requires very specific conditions on $U$ (which depend on the transformation in question, of course), and so the presence of a physical symmetry is a powerful constraint on what $U$ could be.


There are two Poincare transformation that we can think about. The first is spacetime isometries and the other is change of observer. All of them is coordinate independent.

  1. Space time isometries. I will consider $\textbf{spacetime}$ as $(M,\eta)$ where $M$ is a four dimensional $\textbf{manifold}$ and $g$ the metric which in this coordinates $$ \begin{align*} x \colon M &\longrightarrow \mathbb{R}^4\\ p &\mapsto x(p)=(x_0,x_1,x_2,x_3).\tag 1 \end{align*} $$ is given by $$g^{(x)}=dx^0\otimes dx^0-dx^1\otimes dx^1-dx^2\otimes dx^2-dx^3\otimes dx^3 \tag2$$ Suppose we have the equation of motion for particle given by $$\nabla_{v_\gamma}v_\gamma=F \tag3 $$ where $v_\gamma$ is the tangent vector associated with the curve $\gamma$ and $F$ is a force field. Let $\phi:M\rightarrow M$ be a diffeomorphism, we define the induced metric by $$\widetilde{g}(X,Y)=g(\phi_*X,\phi_*Y) \tag 4$$.Associated with this metric we have the connection $\widetilde{\nabla}$ wich we can show that $\varphi_{*} \nabla_{X} Y=\widetilde{\nabla}_{\phi_{*} X}\left(\phi_{*} Y\right)$ and so we have $$ \widetilde{\nabla}_{\phi_*v_\gamma}\phi_*v_\gamma=\phi_*F \tag 5 $$ So if $\gamma$ is a a solution of eq $(3)$ ,$\phi(\gamma$) is a solution of eq $(5)$ this is the meaning of the expression diffeomorphism map solution into solution. Now in special relativity the metric $g$ is not a dynamic variable that is ,it is fixed. So in $(4)$ we need that $\phi$ satisfy $$\widetilde{g}(X,Y)=g(\phi_*X,\phi_*Y)=g(X,Y) \tag 6$$ Diffeomorphism satisfying condition $(6)$ are called Poincare transformation. I hope the examples bellow illustrate the difference between coordinate transformation and diffeomorphism.

Example 1 Suppose we change from coordinates $(1)$ to coordinates $$y_0=x_0, \quad y_1=x_1+x_2, \quad y_2=x_2, \quad y_3=x_3 \tag 7 $$ than $(2)$ becomes $$g^{(y)}=dx^0\otimes dy^0- \frac{1}{2}dy^1\otimes dy^1-\frac{1}{2}dy^2\otimes dy^2-dy^3\otimes dy^3 \tag8 $$ From $(1)$ we have that $g^{(x)}(\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_1})=-1$. In coordinates $(7)$ we have $\frac{\partial}{\partial x}=\frac{\partial}{\partial y_1}+\frac{\partial}{\partial y_2}$ and so using $(8)$ we have that $g^{(y)}(\frac{\partial}{\partial y_1}+\frac{\partial}{\partial y_2},\frac{\partial}{\partial y_1}+\frac{\partial}{\partial y_2})=-1$

Now suppose that we have a diffeomorphism given in coordinates $(1)$ by $$x_0 \rightarrow x_0, \quad x_1\rightarrow x_1+x_2, \quad x_2 \rightarrow x_2, \quad x_3\rightarrow x_3 \tag 9 $$ Under this diffeomorphism we have that $ \frac{\partial}{\partial x}\rightarrow\frac{\partial}{\partial x_1}+\frac{\partial}{\partial x_2}$ and since the metric is fixed we have from $(2)$ that $g^{(x)}(\frac{\partial}{\partial x_1}+\frac{\partial}{\partial x_2},\frac{\partial}{\partial x_1}+\frac{\partial}{\partial x_2})=-2$

As we see in this example for coordinates transformation the result remain invariant while for diffeomorphism it changed.

Example 2. Suppose we use cylindrical coordinates $(t,r,\theta,z)$ whose coordinates transformation from $2$ are given by $$x_0=t,\quad x_1= r\cos \theta ,\quad x_2= r\sin \theta, \quad x_3=z $$Now suppose we have the diffeomorphism $$t \rightarrow \gamma(t-vr \cos \theta), \quad r \rightarrow \sqrt{\gamma^2(r \cos \theta -vt)^2 +r^2 \sin^2 \theta},\\ \theta \rightarrow \tan ^{-1}(\frac{r \sin \theta}{\gamma(r \cos \theta -vt)}), \quad z\rightarrow z \\ \tag 9$$ In this coordinates we can show that $$ \frac{\partial}{\partial x}\rightarrow \gamma \cos \theta \frac{\partial}{\partial r}+\gamma\frac{\sin \theta }{ r}\frac{\partial}{\partial \theta} + \gamma v \frac{\partial}{\partial r}\tag 10$$

In cylindrical coordinates we can show that the metric $g$ is given by $$g^{(c)}=dt\otimes dt-dr\otimes dr-r^2d\theta\otimes d\theta-dz\otimes dz \tag11 $$ From $(10)$ and $(11)$ we can se that the $g(\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_1})$ is invariant under diffeomorphism $(9)$ Diffeomorphism $(9)$ is called Lorentz transformation

  1. Change of observers. An $\textbf{observer}$ is a worldline $\gamma$ with together with a choice of basis $ O=v_{\gamma,\gamma(\lambda)} \equiv e_0(\lambda) , e_1(\lambda), e_2(\lambda), e_3(\lambda) $ of each $T_{\gamma(\lambda)}M$ where the observer worldline passes, if $$ g(e_a(\lambda), e_b(\lambda)) = \eta_{ab} = \left[ \begin{matrix} 1 & & & \\ & -1 & & \\ & & -1 & \\ & & & -1 \end{matrix} \right]_{ab} \tag2 $$

$v_{\gamma,\gamma(\lambda)}$ is the tangent vector of the the curve $\gamma$ at the point $\gamma(\lambda)$. Suppose we have two inertial observer $\gamma^A(\lambda)$ and $\gamma^B(\lambda)$.In coordinates $(1)$ we have $x^\mu(\gamma^A)=(\lambda,0,0,0)$ and $x^\mu(\gamma^B)=(\Lambda^0_0\lambda+q^0,\Lambda^1_0\lambda+q^1,\Lambda^2_0\lambda+q^2,\Lambda^3_0\lambda+q^3)$.Suppose also that $e^A_0=\frac{\partial}{\partial x^0},\quad e^A_1=\frac{\partial}{\partial x^1},\quad e^A_2=\frac{\partial}{\partial x^2},\quad e^A_3=\frac{\partial}{\partial x^3}$.We can show that $e^B_{\mu}=\Lambda^\nu_\mu e_{\nu}$ with $$\eta_{ab}=\Lambda^\nu_a\Lambda^\mu_b\eta_{\nu \mu}$$. Let $\phi$ be a map that take us from observer $A$ to observer $B$ that is $\phi(\gamma^A)=\gamma^B$ . Than we have $$\phi_*\Bigg(\frac{\partial}{\partial x^0}\Bigg)=\frac{\partial \phi^\nu}{\partial x^0}\frac{\partial}{\partial x^\nu}=\Lambda^\nu_0\frac{\partial}{\partial x^\nu}$$ if we impose additionally that $\frac{\partial \phi^\nu}{\partial x^\mu}=\Lambda^\nu_\mu$ we would have the solution $$\phi^\nu=\Lambda^\nu_\mu x^\mu+q^\nu$$ The last expression has this format in the coordinates $(1)$ .Of course if we choose another coordinate it would have another format.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.