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Local Lorentz invariance or local Poincaré invariance?

+ 7 like - 0 dislike

Fast question.

I know that the group of all the isometries in Minkowski's space-time is the Poincaré group that is just Lorentz's group (rotations and boosts) and translations in space-time.

Now, in general relativity texts I often read local Lorentz invariance. I figure that in GR since we have that locally SR is fulfilled what is really meant here is local Poincaré invariance.

So, is this terminology abuse or is it really local Lorentz? And if it is Local Lorentz, is it the full Lorentz group, just the proper orthocronous or any other thing?

This post imported from StackExchange Physics at 2014-03-31 16:06 (UCT), posted by SE-user silvrfuck

asked Mar 30, 2014 in Theoretical Physics by Dmitry hand me the Kalashnikov (685 points) [ revision history ]
edited Jul 21, 2015 by Dilaton
The basic variables you can talk about local Lorentz invariance in GR are tetrad $e^a_\mu$ and spin-connection $\omega^{a,b}_\mu$ (sort of Christoffel symbols for the tetrad). If the Torsion constaint $D e^a_\mu=0$ is imposed then local translations can be identified with diffeomorphisms. So if you consider tetrad independently of anything else, then there is only local Lorentz invariance. But if you are going to define the covariant derivative and relate tetrad to a spin-connection then local translations are there but they are not genuine symmetries, they are identical to diffeos.

This post imported from StackExchange Physics at 2014-03-31 16:06 (UCT), posted by SE-user John

4 Answers

+ 5 like - 0 dislike

There is a deep difference between Lorentz invariance and Poincaré invariance in GR. The former is exact,  whereas  the latter is only approximated. Mathematically speaking, the reason is that Lorentz invariance can be completely described referring to the tangent space only which is unaffected from curvatures. Conversely Poincarè invariance concerns movements of points in the manifold. Alternatively it can be discussed in the tangent space, where it is exact, but the tangent space is just an approximate model of the underlying manifold (using the exponential map for instance).  

Local Poincaré invariance pratically means that in a neighbourhood of a timelike geodesics $\gamma$ you can build up a coordinate system (a so-called geodesically normal coordinate system, whose existence is guarnteed form differential geometry) where the metric takes the Minkowskian expression and the first derivative of the metric vanish, both  exactly on $\gamma$ (consequently the connection coefficients $\Gamma_{\mu\nu}^\alpha(\gamma)$ also vanish on $\gamma$). So, translations along the four coordinates, starting from $\gamma$, are approximated symmetries: the components of the metrics do not change, keeping the Minkowskian form, at the first order of Taylor expansion around $\gamma$.  In this way, around $\gamma$ there is an approximeted Poincaré symmtery in addition to the exact Lorentz symmetry already present separately at each point (each tangent space) of the manifold.

Physically speaking this geometric result also implies a, quite precise, mathematical formulation of Einstein's equivalence principle.

To see this fact consider a second timelike geodesic $\alpha$ crossing $\gamma$ exactly at the event $p$. This second geodesic describes the story of a point of matter freely falling with the observer $\gamma$, with some non vanishing initial velocity with respect to the observer.

Let us describe $\alpha$ in the said normal coordinates system around $\gamma$, as $x^\mu= x^\mu(\lambda)$, assuming to fix the origin of both the normal coordinates and of the affine parameter $\lambda$ just at the crossing point $p$. One has the Taylor expansion which makes sense as soon as $\lambda \to 0$, namely when the point of matter is close to the observer:

$$x^\mu(t) = 0 + \lambda \frac{dx^\mu}{d\lambda}|_{\lambda=0} + \frac{\lambda^2}{2!}\frac{d^2x^\mu}{d\lambda^2}|_{\lambda=0}+ O^\alpha(\lambda^3)\:.\quad (1)$$

However, since $\alpha$ satisfies the geodesic equation:

$$\frac{d^2x^\mu}{d\lambda^2} = - \Gamma^\mu_{\nu\sigma}\frac{dx^\nu}{d\lambda}\frac{dx^\sigma}{d\lambda}$$

and  since $\Gamma^\mu_{\nu\sigma}(\rho)=0$ as said above, we also have:

$$\frac{d^2x^\mu}{d\lambda^2}|_{\lambda=0} = - \Gamma^\mu_{\nu\sigma}(\rho)\frac{dx^\nu}{d\lambda}|_{\lambda=0}\frac{dx^\sigma}{d\lambda}|_{\lambda=0} = 0\:.\qquad (2)$$

We conclude that, in normal coordinates around the free falling observer $\gamma$, the motion of the free falling point of matter $\alpha$ is described by (1), taking (2) into account:

$$x^\mu(\lambda) =  \lambda \frac{dx^\mu}{d\lambda}|_{\lambda=0} + O^\alpha(\lambda^3)\:. \qquad (3)$$

Since both geodesics are timelike, it must hold $\frac{dx^0}{d\lambda}|_{\lambda=0}\neq 0$. Thus we can use the time coordinate of the normal coordinate system, $t=x^0$, as a new, more natural parameter along $\alpha$. If $v^i$, $i=1,2,3$, denote  the components of the spatial  $3$-velocity of the geodesic $\alpha$ at $t=0$ (when it crosses $\gamma$), computed with respect to $\gamma$ its-self

$$v^i = \frac{\frac{dx^i}{d\lambda}|_{\lambda=0}}{\frac{dx^0}{d\lambda}|_{\lambda=0}}\,$$

(3) can be re-arranged to, for $i=1,2,3$:

$$x^i(t) =  t v^i  + O^i(t^3)\:. \qquad (3)$$

The fact that no terms of order $t^2$ show up in the right-hand side is the mathematical expression of Einstein's equivalence principle: Around a free falling observer, the gravitational acceleration disappears exactly at the origin of coordinates and, for short intervals of time and small regions around it, the gravitational field is suppressed and  free falling bodies are seen to be in inertial motion.

The existence of this natural approximated Poincaré simmetry around each timelike geodesic, permits to export some physical laws from special relativity to general relativity. It can be done for laws which  (1) are local and (2)  do not involve spacetime derivatives of order $>1$. For instance $\partial_\mu F^{\mu\nu} = J^\nu$ is an equation valid in special relativity for the electromagnetic field.  Exactly on $\gamma$, since the connection coefficients vanish, the expression of first derivative of tensors in coordinate  and the expression in coordinate of the first covariant derivative coincide.  So, assuming that, for free falling observers, physics is (locally) described by the same laws as in special relativity, at least for law involving first derivatives at most, we can conclude that $\partial_\mu F^{\mu\nu} = J^\nu$ holds even in general relativity, in normal coordinates around $\gamma$. Passing to generic coordinates, the same law can be written:

$$\nabla_\mu F^{\mu\nu} = J^\nu$$

where we have introduced  the covariant derivative $\nabla$.
Since we are free to suppose that every event of the spacetime can be crossed by a free falling observer, we can eventually conclude that $\nabla_\mu F^{\mu\nu} = J^\nu$ is valid everywhere in general relativity.

This is another stronger mathematical version of the equivalence principle, again relying upon the appearance of this approximated Poincaré symmetry due to the existence of normal coordinates.

Regarding the choice between full Lorentz group/ orthochronous one, the point is that, mathematically you can describe and use  both. Physics decides which is the appropriate symmetry for a given physical system.

answered Apr 2, 2014 by Valter Moretti (2,005 points) [ revision history ]
edited Apr 13, 2014 by Valter Moretti
Most voted comments show all comments

I like these very clear and interesting clarifications, +1!

@V. Moretti Could you please explain with more detail how  this is related to the  equivalence principle? that's the kind of answer I wanted

@silvrfuck  Please have a look at the details I have now inserted in my previous answer.

@V. Moretti a couple of questions.

1)There is soething I don't see. You have showed how around a freely falling observer you get rid of the second derivative when considering freely falling particles. I get that in a neighborhood of $\gamma$ we get inertial motion of the test particle as the $O(t^3)$ terms will be small, but why does the second derivative have such a meaning? I mean even if we had that second derivative, if we moved infinitesimally the terms $O(t^2)$ would still be small and to small enough regions of space-time we would have inertial motion, and thus the equivalen principle again. Why do you care of getting rid of the t^2 term and don't bother that we have a t^3?

2)when you say $\partial_{mu}F^{\mu\nu}=J^{\nu}$ holds around $\gamma$ you mean holds to first order, right?

Hi. Concerning (1), notice that in the presence of $O(t^2)$ you could not conclude that the acceleration of $\alpha$ for $t=0$ is suppressed! Instead we know (Einstein's equivalence principle)  that it is suppressed and we have here an "infinitesimal" estimation of that.

Regarding (2), sorry I wrote $\partial_\mu$ in placeof $\nabla_\mu$, now I have corrected the text.  However   $\nabla_\mu F^{\mu \nu} = J^\nu$ is equivalent to  $\partial_\mu F^{\mu \nu} = J^\nu$ exactly on the considered geodesic using normal coordinates. Passing to GR there are two possibilities, noticing that the connection coefficients vanishe exactly on $\gamma$.

(a) We can conclude that, in GR,  $\nabla_\mu F^{\mu \nu} = J^\nu$ is exact at the considered point on $\gamma$ (but every point of spacetime is crossed from a timelike geodesic and so it means that the law is exact everywhere).

(b) We can differently conclude that law in GR may be different from its form in SR, because it involves some added terms related to curvatures (think of something like $J^\mu R_{\mu\nu}$) which vanish in SR but cannot be suppressed (as $\Gamma^\mu_{\alpha\beta}$) suitably fixing the coordinates in GR.

Most recent comments show all comments

yeah, thanks

Good! bye, Valter

+ 2 like - 1 dislike

All the other answers are very thorough, I just wanted to add some physical intuition:

When restricting to "local" you can think about it as studying the physics at a single point in spacetime. We assume different observers are all at that point and this means they can only be boosted with respect to each other. Hence the importance of local Lorentz invariance. To discuss invariance under translations you need two different points and how this works is described in the other answers...

This post imported from StackExchange Physics at 2014-03-31 16:06 (UCT), posted by SE-user Heterotic
answered Mar 31, 2014 by Heterotic (515 points) [ no revision ]
+ 0 like - 0 dislike

Local Lorentz invariance is usually invoked when talking about some aspects of the equivalence principle: that different observers correspond to different inertial reference frames. Inertial frames are related to one another by local Lorentz transformations.

In addition, the metric is invariant under a local Lorentz transformation. In formulations of gravity using tetrads, the rotation coefficients obey a gauge transformation under local Lorentz transformations, but the Riemann tensor is still Lorentz covariant (hence why it's called a gauge transformation, after all).

These notions don't really make sense for the full Poincare group, as translations don't relate to the notion of observers in different reference frames. So I don't think wikipedia is correct on this point.

wikipedia says that local Lorentz invariance can be generalized to the Poincare group. From what I know of gravity using tetrads, I think this corresponds to the transformation law of the tetrad: that under a translation (or a general coordinate transformation), the tetrad must transform in a certain manner so that the inner prouducts of vectors and covectors remain invariant. This too can be considered a gauge transformation. Tetrads transform under local Lorentz transformations as well, so this aspect of coordinate transformation in addition to LLTs may correspond to full "local Poincare covariance," but I would investigate further to be sure.

This post imported from StackExchange Physics at 2014-03-31 16:06 (UCT), posted by SE-user Muphrid
answered Mar 30, 2014 by Muphrid (20 points) [ no revision ]
my knowledge on tetrads is quite limited so I don't really follow your last paragraph, but thanks for the answer. I'll try to check what you say

This post imported from StackExchange Physics at 2014-03-31 16:06 (UCT), posted by SE-user silvrfuck
+ 0 like - 0 dislike

There is a risk of confusion here because in some sense Minkowski space is too nice. What I mean by this is that in the setting of Minkowski space, because of its simple structure, there are identifications and globalizations possible that do not make sense in a general spacetime.

In a general spacetime at every point you can define the tangent vector space. It roughly has one direction for every direction you can move in. This does not mean that the space is an affine space (an affine space is like a vector space without a preferred origin), it could be a sphere for example. But for Minkowski space, the spacetime is indeed an affine space and this leads you confusion of the whole space with the tangent space.

Let us talk about relativity. An observer in spacetime can find three spacelike curves and one timelike curve through his or her time and position. The principle of Lorentz invariance is that any choice is fine! For the space part this is just that you can rotate your laboratory and get the same results. That you are allowed to mix time and space comes from that the speed of light should be the same for observers in relative motion.

So really, what local Lorentz invariance means is that you can rotate your laboratory without changed results, and observers moving relative to it see the same physics. This is an expression of symmetry in the tangent space.

Now in Minkowski spacetime pick an arbitrary origin. Then Minkowski spacetime has the same structure as the (1+3) tangent spaces of general relativity, so the local Lorentz invariance can be made global. Since the origin was arbitrary you have also four translation symmetries. This is the Poincare symmetry.

Local Lorentz invariance is a statement about how your local choice of time and space axis is unimportant. Global Lorentz and Poincare invariance is a much stronger statement about the symmetries of spacetime itself. In particular, a spacetime need not have any symmetries at all (and there are many known examples of solutions to Einstein's equations that don't).

This post imported from StackExchange Physics at 2014-03-31 16:06 (UCT), posted by SE-user Robin Ekman
answered Mar 30, 2014 by Robin Ekman (210 points) [ no revision ]
so, if i understood right, you say that it is just Lorentz (with parity and everything) but not Poincaré, right?

This post imported from StackExchange Physics at 2014-03-31 16:06 (UCT), posted by SE-user silvrfuck
Not necessarily with parity and time inversion. The standard model has violations of both (But CPT is always a symmetry.)

This post imported from StackExchange Physics at 2014-03-31 16:06 (UCT), posted by SE-user Robin Ekman

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