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In summary: I'm not sure how to say this without coming off as condescending, but it just doesn't seem like a very well-defined term to me.It's an easy check to make; the inhomogenous term of the transformation disappears under such transformations (although these Poincare transformations are of course not the most general transformations to do this). So it is completely justified to say that "the connection coefficients transform as a tensor under the group of Poincare transformations". Just as e.g. Newton's second law is a tensor equation under the group of Galilei-transformations but not under accelerations or, more generally, general coordinate transformations (this is the whole idea of inertial observers being equivalent, but accelerating

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That said, the author has probably specified that the origin of coordinates in the inertial frame is at the event of interest, whether it's necessary or not.

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Yes, you are certainly right about that.Ibix said:But is it necessary that the origin of the inertial frame is at the event of interest? I mean, I can't see why you'd want to choose a different location (I think it just adds terms to everything), but could you?

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It's not necessary, but often it's convenient. You can always perform a constant translation as general coordinate transformation which doesn't change your connection coefficients; just look at their transformation. So if they're zero at the origin of your local inertial frame, they also will be zero in your translated local inertial frame.GR191511 said:

Actually, that a constant translation doesn't change the connection coefficients is just part of the fact that connection coefficients are tensors under the Poincaré group.

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No, they aren't. If they were, it would be impossible to make them all vanish at a point by a coordinate transformation. You can't make a non-vanishing tensor vanish by a coordinate transformation.haushofer said:connection coefficients are tensors

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Yes they are. They transform as tensors under the group of Poincaré transformations. That's an easy check to make; the inhomogenous term of the transformation disappears under such transformations (although these Poincare transformations are of course not the most general transformations to do this). So it is completely justified to say that "the connection coefficients transform as a tensor under the group of Poincare transformations". Just as e.g. Newton's second law is a tensor equation under the group of Galilei-transformations but not under accelerations or, more generally, general coordinate transformations (this is the whole idea of inertial observers being equivalent, but accelerating observers are not).PeterDonis said:No, they aren't. If they were, it would be impossible to make them all vanish at a point by a coordinate transformation. You can't make a non-vanishing tensor vanish by a coordinate transformation.

So whether something is a tensor depends on the group of coordinate transformations you specify. By "coordinate transformation" you (in your post I quote here above) implicitly mean "GENERAL coordinate transformation", but as far as I know, that implicit "GENERAL" is not part of the definition of a tensor. At least not how I used it during my graduate or PhD.

To clarify: I'd call e.g. the Klein-Gordon equation if one uses just partial derivatives instead of covariant ones (as you encounter it in everey QFT-book) also a tensor equation, because it transforms as a tensor under the Poincare group. I'd say it's pretty pedantic to point at the Klein-Gordon equation in a QFT-textbook and claim it's not a tensor equation "because it doesn't contain covariant derivatives". Everyone who is not autistic understands that in a QFT book the group of Poincare transformations (and not GCT's) is implied when you mention the word "tensor". That's why I think it's a bit silly you only partially quote my post, removing the addendum "under the Poincaré group".

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It is in GR, since in GR the laws of physics must be invariant under general coordinate transformations.haushofer said:as far as I know, that implicit "GENERAL" is not part of the definition of a tensor

Even in the context of SR, where the laws only have to be invariant under Poincare transformations, it seems to me that your statement cannot be correct as it stands, since, for example, I can take a nonzero Christoffel symbol at a particular point in spherical polar coordinates and make it vanish by transforming to Cartesian coordinates. So your definition of "Poincare transformations" would have to only include Cartesian coordinates, which makes your claim trivial since in flat spacetime all Christoffel symbols vanish in such coordinates.

Can you give a reference that describes and justifies the definition of tensors you are using? I ask because it is not one I have seen in any textbook on relativity that I have read.haushofer said:At least not how I used it during my graduate or PhD.

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The gradient of a scalar field, which is what appears in the Klein-Gordon equation,haushofer said:I'd say it's pretty pedantic to point at the Klein-Gordon equation in a QFT-textbook and claim it's not a tensor equation "because it doesn't contain covariant derivatives".

In any case, your claim was not that the partial derivative of a scalar field is a tensor. It was that the Christoffel symbol is a tensor "under the Poincare group". As I pointed out in post #8, it seems to me that that claim can only be true if we restrict to Cartesian coordinates, which makes the claim true but trivial.

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Hmm. But does the Jacobian transformation matrix "##M##" for Cartesian ##\leftrightarrow## Spherical Polar satisfy ##M\,M^T = 1##? (Doesn't look like it to me. -- I could type it out here if you don't have it at hand.)PeterDonis said:I can take a nonzero Christoffel symbol at a particular point in spherical polar coordinates and make it vanish by transforming to Cartesian coordinates. So your definition of "Poincare transformations" would have to only include Cartesian coordinates, [...]

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What would it mean if it didn't?strangerep said:does the Jacobian transformation matrix "##M##" for Cartesian ##\leftrightarrow## Spherical Polar satisfy ##M\,M^T = 1##? (Doesn't look like it to me. -- I could type it out here if you don't have it at hand.)

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I was trying to see how going between Cartesian and SphPol could possibly be a Lorentz transformation.PeterDonis said:What would it mean if it didn't?

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I wasn't claiming that it was. Indeed, my point was, as I said in posts #8 and #9, that, as far as I can tell, "a tensor under the Poincare group" requires restricting to Cartesian coordinates, which makes the claim that the Christoffel symbols are a "tensor" under such transformations trivial, since they vanish. And if we reject that and allow anythingstrangerep said:I was trying to see how going between Cartesian and SphPol could possibly be a Lorentz transformation.

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You're missing the point.strangerep said:Unless I'm missing something, this means transforms like a tensor under 3D rotations if we're working in spherical polar coords.

Whether or not the transformation from spherical polar to Cartesian coordinates on 4D flat spacetime is a "Poincare transformation" or not, it is a

The claim by @haushofer that I originally responded to was that Christoffel symbols are "tensors under the Poincare group". But what does this mean? The fact that non-vanishing Christoffel symbols in spherical polar coordinates can be made to vanish when transforming to Cartesian coordinates means that there are only two possibilities:

(1) The transformation from spherical polar to Cartesian coordinates on flat spacetime

But let's suppose your argument, suitably extended and completed, is valid. That means that Christoffel symbols are

(2) The transformation from spherical polar to Cartesian coordinates on flat spacetime is

That would mean that the

I see no other possibilities, and @haushofer has not responded to this argument. Can you?

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I think you might have misunderstood my posts above, and built a strawman as a result.PeterDonis said:You're missing the point.

I'll make an attempt to clarify...

Yes, they are not tensors under the full group of general coordinate transformations.PeterDonis said:Whether or not the transformation from spherical polar to Cartesian coordinates on 4D flat spacetime is a "Poincare transformation" or not, it is avalid general coordinate transformation, which makes formerly non-vanishing Christoffel symbols vanish.And in every GR textbook I have read, this is the sort of argument that is given to show why Christoffel symbols arenottensors.

I'm surprised you need to ask that. Was it a rhetorical question?PeterDonis said:The claim by @haushofer that I originally responded to was that Christoffel symbols are "tensors under the Poincare group". But what does this mean?

No, that is not my argument. There has been a misunderstanding.PeterDonis said:The fact that non-vanishing Christoffel symbols in spherical polar coordinates can be made to vanish when transforming to Cartesian coordinates means that there are only two possibilities:

(1) The transformation from spherical polar to Cartesian coordinates on flat spacetimeisa valid Poincare transformation. Your post makes an argument that it should be, [...]

That is my understanding. Am I wrong about that?PeterDonis said:(2) The transformation from spherical polar to Cartesian coordinates on flat spacetime isnota valid Poincare transformation.

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I think you have somehow missed the fact that this whole subthread was started by a response by me to a claim by @haushofer which I will quote again:strangerep said:I think you might have misunderstood my posts above, and built a strawman as a result.

I am trying to make sense of that claim.haushofer said:connection coefficients are tensors under the Poincaré group.

Everyone in this discussion agrees with that. That's not the point at issue.strangerep said:they are not tensors under the full group of general coordinate transformations.

No, it was a genuine question. I'm surprised you didn't realize that since all of my previous posts in this thread should have made it obvious.strangerep said:I'm surprised you need to ask that. Was it a rhetorical question?

I can only make sense of the claim in two ways, which I described in post #15. One way makes the claim false; the other makes it trivial. Since I don't think @haushofer meant his claim to be either false or trivial, I am trying to find another way to make sense of it, but I have not found one.

If your understanding, which you have now clarified, is that the transformation from spherical polar to Cartesian coordinates is

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Since apparently what that claim meansstrangerep said:I'm surprised you need to ask that. Was it a rhetorical question?

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No, IPeterDonis said:I think you have somehow missed the fact that this whole subthread was started by a response by me to a claim by @haushofer which I will quote again:

[...]

I'll respond to your question about what I "think it means" tomorrow (my time), if haushofer or someone else doesn't clarify it first.

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I think he understands haushofer's calim, he just disagrees with the use of terminology.strangerep said:No, Ididrealize that the whole time. I was just having an extremely hard time understanding how or why or what you didn't understand about haushofer's claim. It happens.

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I believe haushofer refers here to the following argument:haushofer said:Actually, that a constant translation doesn't change the connection coefficients is just part of the fact that connection coefficients are tensors under the Poincaré group.

In any spacetime, flat or curved, the general change of coordinates ##x^{\sigma}\rightarrow x^{\prime\sigma}(x)## transforms the Christoffel connection ##\Gamma_{\alpha\mu}^{\sigma}## according to: $$\Gamma_{\alpha\mu}^{\prime\sigma}=\frac{\partial x^{\prime\sigma}}{\partial x^{\beta}}\frac{\partial x^{\rho}}{\partial x^{\prime\mu}}\frac{\partial x^{\tau}}{\partial x^{\prime\alpha}}\Gamma_{\tau\rho}^{\beta}+\frac{\partial x^{\prime\sigma}}{\partial x^{\beta}}\frac{\partial^{2}x^{\beta}}{\partial x^{\prime\alpha}\partial x^{\prime\mu}}\\=\frac{\partial x^{\prime\sigma}}{\partial x^{\beta}}\frac{\partial x^{\rho}}{\partial x^{\prime\mu}}\frac{\partial x^{\tau}}{\partial x^{\prime\alpha}}\Gamma_{\tau\rho}^{\beta}-\frac{\partial x^{\tau}}{\partial x^{\prime\alpha}}\frac{\partial^{2}x^{\prime\sigma}}{\partial x^{\tau}\partial x^{\beta}}\frac{\partial x^{\beta}}{\partial x^{\prime\mu}}$$ It's clear from the last two terms that ##\Gamma## transforms

But is there any utility in these Poincaré transformations, especially in a curved spacetime? They might prove useful in rectangular-coordinates ##(t,x,y,z)## for something like gravitational plane waves, where ##M## and ##\xi## could be chosen to boost the momentum vector, rotate polarization directions and move the origin, thereby inducing a tensor-transformation of the connection coefficients. But what about ##(t,r,\theta,\phi)## in Schwarzschild spacetime? The same choice of ##M## and ##\xi## would scramble those coordinates in a completely nonsensical way. And first converting that spacetime to rectangular coordinates, Poincaré transforming, and then converting back would be an algebraic nightmare. So I remain skeptical.

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PeterDonis said:It is in GR, since in GR the laws of physics must be invariant under general coordinate transformations.

Even in the context of SR, where the laws only have to be invariant under Poincare transformations, it seems to me that your statement cannot be correct as it stands, since, for example, I can take a nonzero Christoffel symbol at a particular point in spherical polar coordinates and make it vanish by transforming to Cartesian coordinates. So your definition of "Poincare transformations" would have to only include Cartesian coordinates, which makes your claim trivial since in flat spacetime all Christoffel symbols vanish in such coordinates.

[1] Well, yes, but in GR one often gauge-fixes the coordinates and restrict oneself to a subgroup of gct's. One can then perfectly ask the question whether a field transforms as a tensor under this subgroup. My own research was a lot about using the Newton-Cartan formalism, where you constantly gauge fix coordinates and see how the resulting transformation laws change. We used this terminology all the time.

[2] Well, yes, that's a good point, but I wouldn't call a transformation from Cartesian to polar coordinates a Poincare transformation. Of course, the Christoffel symbols don't transform tensorially under such a transformation. A Poincare transformation is a Lorentz transformation and/or a spacetime translation. So if you've picked Cartesian coordinates, the christoffel symbol is zero; every Poincare transformation keeps it zero. If you've picked spherical coordinates, the christoffel symbol is non-zero, and stays that way under Poincare transformation.

[3] I have to check, but I just use the philosophy "if it quacks like a duck and walks like a duck, let's call it a duck". So if something transforms tensorially (i.e. multilinearly) under a subgroup of gct's, let's call it a tensor under this subgroup.

Renormalize indeed shows what I mean.

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I'm not sure. But even then, someone like Wald calls the Christoffel symbol "a tensor field associated with the covariant derivative" after eqn. 3.1.15. Just to say that "what textbooks say" is not always a good argument; I think many of us would say Wald's statement there can be confusing.martinbn said:

As I said, I'm perfectly fine with people who think"my terminology" is unusal; it's just how we used it during my PhD in my particular field of research. I think it's a helpful way to think about it, because it naturally arises whenever one restricts himself to certain subgroups of GCT's and checks how tensors or other fields transform under this restriction. E.g., for me the expression "all inertial observers are equivalent" is similar to "Newton's second law only transforms tensorially under the restricted group of Galilei-transformations." But if other people find that confusing, be my guest :)

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haushofer said:"if it quacks like a duck and walks like a duck, let's call it a duck"

What would you say about pseudovectors, then? They are not geometrical objects, but still transform like vectors under certain transformations. Would you call these vectors, too?

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I'm not sure what you mean exactly by "geometrical object". If it can be considered to be a multilinear mapping from a dual space to the reals and if the components transform multilinearly according to the very definition under a group of transformations, I'd call it a vector (tensor) under these transformations.Cepatiti said:I was taught in GR that a tensor ought to be a "geometrical object". Hence in particular, it cannot be annihilated by a mere coordinate transformation.

What would you say about pseudovectors, then? They are not geometrical objects, but still transform like vectors under certain transformations. Would you call these vectors, too?

Look, I'm not saying something silly here, as far as I know. I just say that if one talks about tensors, one should specify the group of transformations one considers. Often this happens implicitly, as in GR or quantum field theory.

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Given what you mean by this, yes, one can. But if that subgroup is the subgroup of Poincare transformations between Cartesian coordinate systems, then the claim that the Christoffel symbols are a tensor under this subgroup is trivial, since you're "transforming" a tensor that vanishes. It appears that that is indeed what you meant.haushofer said:in GR one often gauge-fixes the coordinates and restrict oneself to a subgroup of gct's. One can then perfectly ask the question whether a field transforms as a tensor under this subgroup.

Yes, and that just reinforces what I said above, since the "Poincare transformations" you speak of are restricted to transformations between Cartesian coordinate charts.haushofer said:I wouldn't call a transformation from Cartesian to polar coordinates a Poincare transformation.

Even if it vanishes, apparently.haushofer said:if something transforms tensorially (i.e. multilinearly) under a subgroup of gct's, let's call it a tensor under this subgroup.

But actually, now that you have confirmed what you meant, there is more. I'll address that in a follow-up post.

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Your conclusion does not follow. How I see it is that Poincaré transformations connect different inertial coordinate systems (you could call such a coordinate system "an observer"). Within such a coordinate system (or: "observer", if you like), you can choose Cartesian or cylindrical or spherical or whatever spatial coordinates you want. But such a transformation is not a Poincare transformation. If I interpret such a transformation with a mental picture in my mind, I just see the same observer pondering how to label her events in space. If that observer picture confuses you, just leave it out; that's not the point.PeterDonis said:(2) The transformation from spherical polar to Cartesian coordinates on flat spacetime isnota valid Poincare transformation. Your argument, if correct, would contradict this, but let's suppose that, for whatever reason, your argument cannot be extended to translations or boosts.

That would mean that theonlykind of coordinates that would be included in "Poincare transformations" would be Cartesian coordinates. Andthatwould meant that the claim that Christoffel symbols are "tensors under Poincare transformations" would be true only because it was trivial--because in Cartesian coordinates in flat spacetime all Christoffel symbols vanish.

So once you've chosen your spatial coordinates (Cartesian, spherical,...), you can calculate your christoffel symbol. If you then apply a Poincaré transformation (Lorentz transformation, rotation or spacetime translation), this symbol you've calculated transforms tensorially (because the inhomogeneous term in the transformation drops out). If you want the symbol to become zero again, you have to step out of the group of Poincaré transformations. One way is to choose spherical coordinates (a mere spatial coordinate transformation); another way is to start accelerating (in which you involve time in your coordinate transformations).

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[1] I never said it's not trivial. Within the choice of cartesian coordinates you effectively transform a "zero tensor" if you apply poincare transformations on the christoffel symbol, just like a rotation of the zero vector doesn't magically make your tensor become non-zero.PeterDonis said:Given what you mean by this, yes, one can. But if that subgroup is the subgroup of Poincare transformations between Cartesian coordinate systems, then the claim that the Christoffel symbols are a tensor under this subgroup is trivial, since you're "transforming" a tensor that vanishes. It appears that that is indeed what you meant.

Yes, and that just reinforces what I said above, since the "Poincare transformations" you speak of are restricted to transformations between Cartesian coordinate charts.

Even if it vanishes, apparently.

But actually, now that you have confirmed what you meant, there is more. I'll address that in a follow-up post.

[2] Or spherical coordinates. Or whatever fancy spatial coordinate system you want. Maybe I missed something, but since when do we call the transformation from cartesian to spherical coordinates possibly a poincare transformation?

[3] Ok. :)

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Careful. The posts in this thread have been conflating twoIbix said:is it necessary that the origin of the inertial frame is at the event of interest? I mean, I can't see why you'd want to choose a different location (I think it just adds terms to everything), but could you?

Strictly speaking, a "local inertial frame" is a frame on the tangent space at a point, and it describes, not displacements, but tangent vectors at that point. So it doesn't even make sense to ask which "location" in the frame corresponds to the point you picked; none of them do since the frame is not describing "locations", it's describing vectors. To the extent you can label any "point" in the frame as corresponding to the chosen point in the spacetime, it would have to be the origin, since that is the "base" of all the vectors.

If you are talking about placing a particular event at a particular point in "local coordinates", the "local coordinates" would be Riemann normal coordinates, which are

This is true in an inertial chart on flat spacetime in Cartesian coordinates. But as the above makes clear, that ishaushofer said:You can always perform a constant translation as general coordinate transformation which doesn't change your connection coefficients

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[tex]

\ddot{x}^i + \Gamma^i_{00} = 0 \ \ \ \ \ (1)

[/tex]

With the group of Galilei transformations, this equation transforms as a tensor (or: vector equation). Physically, this means no Galilei transformation will introduce a fictitious force. But if you e.g. apply a time-dependent rotation,

[tex]

x^{'i} = R^i{}_j (t)x^j \ \ \ with \ \ \ R^i{}_j (t)\ \ \in SO(3) \ \ \forall t

[/tex]

then eqn.(1) does not transform as a tensor anymore, but inhomogenously. The inhomogeneous terms are the fictitious forces. I don't see the problem to speak about eqn.(1) this way, and why it would be so bad to call eqn.(1) a tensor equation under the group of Galilei transformations.

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Yes. But in a curved spacetime, these inertial coordinate systems are on the tangent space at a particular point, not on the spacetime itself. See post #30. So "translations" in the usual sense of that term make no sense. A "translation" in the context of this thread would be, as I said in post #30, a transformation between Riemann normal charts on a local patch of the spacetime, which is not a Poincare transformation.haushofer said:Poincaré transformations connect different inertial coordinate systems

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Is this true for translations in spherical coordinates? I don't think such a transformation is linear.haushofer said:once you've chosen your spatial coordinates (Cartesian, spherical,...), you can calculate your christoffel symbol. If you then apply a Poincaré transformation (Lorentz transformation, rotation or spacetime translation), this symbol you've calculated transforms tensorially (because the inhomogeneous term in the transformation drops out).

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I agree.PeterDonis said:Yes. But in a curved spacetime, these inertial coordinate systems are on the tangent space at a particular point, not on the spacetime itself. See post #30. So "translations" in the usual sense of that term make no sense. A "translation" in the context of this thread would be, as I said in post #30, a transformation between Riemann normal charts on a local patch of the spacetime, which is not a Poincare transformation.

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Hmm, is that true? I have never tried that, and I am not even sure exactly how to do it, but I am rather skeptical about it.haushofer said:So once you've chosen your spatial coordinates (Cartesian, spherical,...), you can calculate your christoffel symbol. If you then apply a Poincaré transformation (Lorentz transformation, rotation or spacetime translation), this symbol you've calculated transforms tensorially (because the inhomogeneous term in the transformation drops out).

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