# Poincare transformations- parametrization independent

1. Oct 17, 2014

### ChrisVer

Well if I have a worldline given by $x^{\mu}(\tau)$
And I want to make a Poincare transformation: $x^{\mu} (\tau) \rightarrow \Lambda^{\mu}_{\nu} x^{\nu}(\tau) + a^{\mu}$.

I have one question,why can't $a, \Lambda$ explicitly depend on $\tau$? that is to have:

$x^{\mu}(\tau) \rightarrow \Lambda^{\mu}_{\nu}(\tau) x^{\nu} (\tau)+ a^{\mu}(\tau)$.

Somehow I have a feeling this would make the Poincare transformations local....

2. Oct 17, 2014

### ShayanJ

Is it consistent with $\Lambda^\alpha _{\ \mu} \Lambda^\beta_{\ \nu} \ \eta_{\alpha \beta}=\eta_{\mu \nu}$? I doubt it!
But even if it is, There will be some restrictions on the $\tau$ dependence of the transformations!

3. Oct 18, 2014

### ChrisVer

maybe you can find someway to make this work.
But for example if that was the case, then the free particle relativistic action $\int d^4x \sqrt{ \dot{x} \cdot \dot{x}}$ will not be invariant as it is, because of the $\dot{a}, \dot{\Lambda}$ terms... For simplicity I can put the time dependence only on the Lambdas.... this would introduce a connection term as it does for gauge symmetries.

4. Oct 19, 2014

### ShayanJ

You should note that $\eta_{\mu \nu}$ is spacetime independent and so $\Lambda^\alpha_{\ \mu} \Lambda^\beta_{\ \nu} \ \eta_{\alpha \beta}$ should be. So if the Lorentz transformation depends on spacetime coordinates, it should be somehow that is eliminated in the product $\Lambda^\alpha_{\ \mu}\Lambda^\beta_{\ \nu} \ \eta_{\alpha \beta}$, but it seems impossible to me!
Remember both $\Lambda$s are indicating the same transformation and so the condition means a restriction on the form of transformation. But I fail to see any physical interpretation of restrictions coming from such considerations and so I suspect its not possible.
I should confess I really wish I had enough mathematical experience so that I could do the calculations but I don't know what to do with it.
Where are all the real physicists of the PF?

5. Oct 19, 2014

### dextercioby

Are you looking for Poincare gauge theory? It starts with gauging the Poincare transformations (i.e. the 10 group parameters are not constant wrt $x^{\mu}$) and then goes on to explore the consequences on field theory.

6. Oct 19, 2014

### ShayanJ

Last edited by a moderator: May 7, 2017
7. Oct 19, 2014

### ChrisVer

The Poincare transformations are not so simple I think... why? because, the relation for $\eta= \Lambda^T \eta \Lambda$ is a result of the isometry, that is you asked for: $(x',y')= (x,y)$ and this for Lorentz transformations gave you the above result.

However that stops being the case for the Poincare transformations.
$(x',y')= g_{\mu \nu} x'^{\mu} x'^{\nu}= g_{\mu \nu} (\Lambda^{\mu}_{\rho} x^{\rho} + a^{\mu})(\Lambda^{\nu}_{\sigma} x^{\rho} + a^{\nu}) =(\?) g_{\rho \sigma} x^{\rho} x^{\sigma}$
... So because of that I can't see why the $\eta= \Lambda^T \eta \Lambda$ should hold.

Partially I just did not see the reason why not assigning a parameter dependence on your transformations. Eg in U(1) let's say global symmetry you had a spacetime independent transformation parameter $a$: $e^{ia}$... which then you turned into spacetime dependent to get the local U(1).
What stops from doing that in Poincare case? I didn't even try to make them spacetime dependent, but "parameterization" dependent.

8. Oct 20, 2014

### ShayanJ

$\Lambda^\alpha _{\ \mu} \Lambda^\beta_{\ \nu} \ \eta_{\alpha \beta}=\eta_{\mu \nu}$ is the definition of a Lorentz transformation regardless of the fact that its going to be followed by a translation or not!
In the U(1) case, the only condition on the transformation is unitarity which is untouched by making the transformation local. But here it doesn't seem trivial that a spacetime dependent $\Lambda$ satisfies the definition of a Lorentz transformation or not.

9. Oct 20, 2014

### ChrisVer

This relation is just a result of the isometry you impose on the Minkowski space, that is for lorentz invariant quantities to remain invariant under Lorentz transformations. That is not true for the Poincare transformations and that's what I wanted to show above (you can't impose the same isometry for Poincare transfs). In other words minkowski space by itself is not invariant under poincare transformations but only under a subgroup of the Poincare group [that are the lorentz transformations of boosts and rotations], or not?!
I could as well say the SU(N).
For example if I impose the Poincare transformation $\Lambda x +a(\tau)$ to keep the particle invariant, I would have to change the derivatives to covariant derivatives.

10. Oct 21, 2014

### ShayanJ

I know what you mean, but here we aren't imposing the condition on the Poincare transformation.
I'm just telling, in order for $x'^{\mu}=\Lambda^\mu_\rho x^\rho+a^\mu$ to be a Poincare transformation, $\Lambda^\mu_\rho$ should be a Lorentz transformation, i.e. should satisfy $\Lambda^\mu_{\ \rho}\Lambda^\nu_{\ \sigma} \eta_{\mu\nu}=\eta_{\rho \sigma}$.
I'm OK with making the translation part spacetime dependent, I'm just telling that its not trivial that a spacetime dependent matrix can be a Lorentz transformation. So either its impossible or there will be a restriction on the kind of dependence.
So I think this issue is treated in a paper before all those I mentioned and all physicists working in this field know it well enough that they don't bother mentioning it in every paper.
Also I should say that, it seems to me you can have a spacetime dependent transformation that is not parametrization dependent but you can't have a parametrization dependent transformation that is not spacetime dependent.

Last edited: Oct 21, 2014
11. Oct 21, 2014

### ShayanJ

Also, note that the condition $\Lambda^\mu_{\ \rho}\Lambda^\nu_{\ \sigma} \eta_{\mu\nu}=\eta_{\rho \sigma}$ doesn't come from $\eta_{\mu \nu} x^\mu x^\nu=\eta_{\rho \sigma} x'^\rho x'^\sigma$ but from $\eta_{\mu \nu} dx^\mu dx^\nu=\eta_{\rho \sigma} dx'^\rho dx'^\sigma$, so (global) Poincare transformations are isometries too.

12. Oct 21, 2014

### ChrisVer

I think I understand what you are trying to say.

Yes you are right. That was a bad mistake.
However if I made the Poincare transformations local (at least their translation parts), that would break, because $da$ would not be trivially zero... Also I'd try to stand on the local translation parts, since local lorentz transformations indeed sound weird, and I think it's impossible to have them. You would need a local metric tensor as well, which would bring GR into game, and GR is not Poincare invariant (it gets replaced by coordinate transformation invariance)

How can you say that?

13. Oct 21, 2014

### ShayanJ

You're mixing up things!

1) A Lorentz transformation is a coordinate transformation $\Lambda^\mu_{\ \rho}$ satisfying $\Lambda^\mu_{\ \rho}\Lambda^\nu_{\ \sigma} \eta_{\mu\nu}=\eta_{\rho \sigma}$.

2) A Poincare transformation is a Lorentz transformation followed by a translation.

3) Global Lorentz and Poincare transformations are isometries of the Minkowski spacetime.

4) Making the translation part of a Poincare transformation to depend on parametrization, has nothing to do with the Lorentz transformation part of the Poincare transformation. The non-vanishing of $da$ only means a Poincare transformation with a local translation part, is not an isometry of Minkowski spacetime. Its still a Poincare transformation, its just not an isometry of the Minkowski spacetime. But its special case, the global Poincare transformation, is!

5) 1 means, a Lorentz transformation is defined as the most general homogeneous isometry of the Minkowski spacetime. So here, if anything causes a homogeneous coordinate transformation to change the length of spacetime intervals, it means that transformation is not a Lorentz transformation. So if you want to have a local Lorentz transformation, you should look for local homogeneous coordinate transformations that are also isometries of the Minkowski spacetime. The existence of such a thing is not trivial.(That's the reason I'm confused why every one is that much OK with Poincare gauge theory without asking such questions! It can't be because "we're going to talk about curved spacetime, so we should forget Minkowskian things" because spacetime is still locally Minkowskian.)

Its not clear to me too but It seems to me it should be the case. I try to give an argument.
Let's assume there is a parametrization dependent transformation that is spacetime independent. But it means this transformation is global i.e. changes things the same amount everywhere. Oh...that's the flaw! Different parametrizations can see different global constants! But I think that's the only way the transformation can be parametrization dependent without being spacetime dependent. It should be a global constant depending on the parametrization. I should stress it doesn't depend on the coordinates but only on which parametrization you're using.

Last edited: Oct 21, 2014
14. Oct 21, 2014

### haushofer

For the non-rel. particle this is how you can couple the particle to Newtonian gravity, see

http://arxiv.org/abs/1206.5176

Making the translations local, you can introduce a potential which transforms inhomogeneously, such that the action is invariant under local translations, i.e. accelerations. The usual gauging procedure however would consist of introducing D gauge fields which have a Stuckelberg transformation, i.e. a shift symmetry.

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