Talk:Pseudotensor

Tensor densities?

This entry needs some good examples.

Done --Michael C Price 11:34, 10 June 2006 (UTC)[reply]

Aren't pseudotensors more general than tensor densities? Phys 22:05, 17 October 2005 (UTC)[reply]

Yes! The terminology is muddled in the literature, but pseudotensor is often used in gtr to mean a tensorial like object which is not generally covariant, but which only makes sense in certain coordinate systems, such as pseudo-cartesian systems. ---CH 20:19, 24 December 2005 (UTC)[reply]

The terminology is more than muddled, it's complete rubbish. The article is talking about tensor densities not pseudotensors. I've changed it to reflect the distinction -- at some stage the two concepts should be given their own pages. --Michael C Price 11:34, 10 June 2006 (UTC)[reply]

I agree. Please do it. (I could, but too busy trying to control bad edits in other physics/math topics.)---CH 23:14, 13 June 2006 (UTC)[reply]

Already done. It was a learning experience :-(, but worth it.--Michael C Price 02:56, 14 June 2006 (UTC)[reply]

Math or physics?

Math. BTW, Michael, WP discussion of this kind of confusion has been previously noted but none of the experts has taken the time to clear it up, so feel free to have a go. I think it is important to take advantage of existing math articles and to distinguish between math (tensor algebra/calculus) and applications to physics.---CH 23:13, 13 June 2006 (UTC)[reply]

I spent some time looking over the maths and physics categories and the only firm conclusion I came to was that I wasn't going to be happy with which ever one I chose! Still, I think it better to produce half-finished and inappropriate stubs than none at all....--Michael C Price 02:56, 14 June 2006 (UTC)[reply]

Sign flip

I believe that the term pseudotensor means different things in different contexts: in particle physics they may well gain a sign flip under an inversion but I don't believe this is true in general relativity. The article is currently inconsistent on this issue: I don't think the Landau-Lifshitz pseudotensor, for example, flips signs under an inversion. --Michael C. Price ^talk 08:12, 11 April 2007 (UTC)[reply]

Pseudotensors flip sign also in general relativity. This is true for the Landau-Lifshitz pseudotensor, as well as for the Levi-Civita symbol. The issue is much more complex though. Pseudotensors are generalisation of axial vectors, just as ordinary tensors are generalisation of polar vectors. I am doing an extensive rewrite of the pseudotensor article, in which I will explain this issue through matrix algebra and show how the determinant of a pseudotensor can become -1 or 1 depending on the rotation (improper or proper). I also plan to rewrite the article on Landau-Lifshitz pseudotensor, and the empty entry Stress-energy-momentum pseudotensors (note the plural). However, I must first add some physics meaning in the article Covariant derivative because in the moment it is too math oriented and lacks some definitions which are helpful for explaining the energy conservation in general relativity. --Lantonov 08:43, 11 April 2007 (UTC)[reply]

I don't see why the LL pseudotensor flips sign under a spatial inversion. IOW readers badly need your explanation. --Michael C. Price ^talk 07:48, 13 April 2007 (UTC)[reply]

I am working on the explanation. As an example, this will include a simple pseudotensor, to show its general properties in a manner that is easy to grasp. As for the LL pseudotensor, I don't know whether any modern computer is up to the task of calculating its determinant (in 4D). Only after the determinant is calculated, it can be shown that it flips sign under improper rotation. My slowness is due to the fact that it is not easy for me to find what is explained already in wiki and where. I am afraid that I will duplicate material found in other articles and it is not easy for me to decide what to include in the explanation. --Lantonov 11:36, 13 April 2007 (UTC)[reply]

I'm guessing here -- as you say, direct calculation would be very difficult -- but isn't there a good chance that the determinant sign flip will cancel since det appears both above and below? I realise that it may not be this simple, since one appears inside the bracket and is twice differentiated, whilst the other is outside. --Michael C. Price ^talk 12:04, 13 April 2007 (UTC)[reply]

I think that we are talking about 2 different determinants here. Maybe you have in mind g, the determinant of the of the metric tensor g^αβ. g is fairly easy to calculate depending on the complexity of the metric of the space it is calculated in. However, g never flips sign because it is a det of true tensor. In LL g is defined to be always negative because all spaces can be made through transformations locally Galilean and Galilean frame in LL gives g = -2. J. Synge in his book makes another definition of the Minkowski space which differs in sign from LL, so Synge's g is always positive. The det of LL pseudotensor is something infinitely more complicated. We will have a 16 x 16 matrix, in which each of the 256 components is expressed by the formula (96,8) or (96,9) in the LL book. If we take into account the summation for the dummy indexes, the expression for each component may well expand to 2 pages. Think about calculating the determinant for such a monster where we have to multiply all the components. I feel dizzy just by the thought of it. As for cancelling the sign of g, yes, it does cancel because we can always take (-1) out of the differentiation operator. LL operates with (-g) instead of g because they have defined g to be negative. The same expression in Synge is with g in the place of (-g). --Lantonov 15:04, 13 April 2007 (UTC)[reply]

You're correct, we're talking about two different determinants. My error. --Michael C. Price ^talk 20:06, 13 April 2007 (UTC)[reply]

When I slept on it, I figured that if I find a very simple metric for which $g^{\alpha \beta },_{\mu }\neq 0$ and make the calcs in 3D with a computer I might be able to prove the sign flip of LLP under improper rotation. --Lantonov 07:00, 14 April 2007 (UTC)[reply]

Since the LL pseudotensor vanishes locally in any frame where the first derivatives of the metric vanish (which by the equivalence principle is always possible anywhere), doesn't it also follow that the sign flip can't encapsulate the definition of a pseudotensor? Adding a sign flip to a pseudotensor's transformation would not allow it to vaninh anywhere in anyframe.--Michael C. Price ^talk 06:02, 18 April 2007 (UTC)[reply]

You are perfectly right on this. However, the definition of pseudotensor is rigid and made in mathematics, not in physics, on the basis of coordinate transformations, contractions and relationships with pseudovectors (axial vectors) and pseudoscalars. Pseudotensors are also pivotal for the concept of dual tensors, vector products, and such. The more I read, the more I come to the conclusion that pseudotensor is a misnomer when applied to the grav SEM quantities like LL pseudo. All that Lan-Lif say on this is (ch. 96, after #96,8): "An important property of the quantities $t^{ik}$ is that they do not form a tensor; it can be seen by the fact that $\partial {h^{ikl}}/\partial {x^{l}}$ are composed of ordinary but not of covariant derivatives. However, $t^{ik}$ are expressed by the quantities $\Gamma _{kl}^{i}$ and those behave like tensors in respect to linear coordinate transformations (see ch. 85); it follows that the same is true in respect to $t^{ik}$ ." Now, the transformation rule for $\Gamma _{kl}^{i}$ is (LL,#85,15) and differs from the tensor rule by the presence of a monom with a second derivative (which disappears in linear transformations). Therefore, $\Gamma _{kl}^{i}$ is not a tensor but it is not a pseudotensor either because this transformation rule is different from that for a pseudotensor. There is no reason to expect that $t^{ik}$ will transform somehow differently than the Christoffel symbols. I think that the term pseudotensor is given here only to designate the fact that LLP is not a true tensor. In the newer literature, the SEM quantities for gravitation are called quasitensors; eg. Landau-Lifshitz quasitensor. I recently read an article with exactly this term, and the other SEM pseudos were also called quasitensors. I'll post a reference to this article when I find it in my messy folders. I still haven't read anything rigorous about quasitensors (definitions, properties etc) so I don't know if this is not another misnomer. However, I am almost certain that LLP is a very loose application of the term pseudotensor. --Lantonov 07:41, 18 April 2007 (UTC)[reply]

You may well be right about the change in usage of the term "pseudotensor" from the days of Landau et al. I suspect that the new usage is more prevalent in QFT/particle physics than GR. I have never heard of quasitensors, but the distinction you make is logical.--Michael C. Price ^talk 09:48, 18 April 2007 (UTC)[reply]

Quasitensors

I read something further on quasitensors. They seem to be related to dual quasialgebras and quasi Hopf-algebras, and yes, some SEM (quasi?)tensors, eg Bergmann-Thompson quasitensor is mentioned in this respect. Need some more time to dip in this. --Lantonov 08:17, 18 April 2007 (UTC)[reply]

Mostly they avoid the 'p' word in recent papers, preferring to speak about 'energy-momentum complexes' or 'energy-momentum formulations' of LL, BT, Møller, Papapetrou or whoever. --Lantonov 13:46, 18 April 2007 (UTC)[reply]

A Google web search gets the following hit counts:

"Landau-Lifshitz pseudotensor" = 455
"Landau-Lifshitz pseudo-tensor" = 40
"Landau-Lifshitz quasi-tensor" = 3
"Landau-Lifshitz quasitensor" = 0

--Michael C. Price ^talk 11:50, 19 April 2007 (UTC)[reply]

Covariant vs Invariant

I changed the term covariant to invariant. Invariant is a more general term which refers to both covariant and contravariant tranformations. Although often covariant is used as a synonym of invariant this use is informal (see Covariance and contravariance of vectors#Informal usage:invariance). Lantonov 10:06, 23 July 2007 (UTC)[reply]

Invalid claim about volume element?

This article contains the statement "On non-orientable manifolds, one cannot define a volume form globally due to the non-orientability, but one can define a volume element". However, this is a complete swizz (i.e. it is a fallacy), because the volume form can only be defined in terms of a basis, and usually a coordinate basis is assumed. No continuous basis field (whether this is a coordinate basis or not) exists globally on a non-orientable manifold, and consequently on cannot define a volume form globally on a non-orientable manifold. I am ware that a volume form such as the Levi-Civita symbol takes the same component form regardless of the choice of basis, but this does not change the fact that it is only defined where there is something to reference it to – either a basis, or a volume form, neither of which exist globally (in a continuous form) on a non-orientable manifold. —Quondum 04:15, 22 May 2015 (UTC)[reply]

I think that the article is trying to indicate that, e.g., a Möbius strip has a definable total area. Furthermore, despite the lack of a volume form, one could define a mass density on a non-orientable manifold and then integrate it to get a total mass. Whether the article is conveying these ideas properly is a separate question. Leegrc (talk) 17:10, 22 May 2015 (UTC)[reply]

Understood that defining an integrable quantity has value, but I'm not convinced that this construction is mathematically consistent: the definition of integration itself needs to be modified to ignore orientation of the volume element to achieve consistency. The problem becomes more obvious if you consider something like a charge density (i.e. something that can be positive and negative), and then try to integrate it globally. A simpler and more natural construction is to integrate an n-form defined on the double cover of the manifold. My concern is to avoid the claim that a volume element is defined globally while a volume form is not. The examples should also be separated into the two types (as per the lead's two definitions). This is a tricky area, however. —Quondum 17:38, 22 May 2015 (UTC)[reply]

I think that charge density is not any harder than mass; I can imagine a Möbius strip with positive and negative charges on it and ask what is its net charge. I agree that the double-cover approach allows one to side-step this business of volume form vs. volume element, but I am still hopeful that there is a textbook way to accurately describe the volume element approach too. Leegrc (talk) 19:09, 22 May 2015 (UTC)[reply]

Sorry, I was overly cryptic. In a non-orientable Möbius strip universe (with a time dimension), switching the orientation by traveling around the loop changes the sign of the charge from the perspective of the flipped observer, so the net charge is inherently zero once you integrate twice round the strip. This holds for any true density, with or without a time dimension. I don't think there is any way around the conundrum, but would be happy to see a coherent account refuting this. Or even a well-sources claim that a volume element can be integrated globally on a non-orientable manifold (in a way that does not always give zero). —Quondum 20:40, 22 May 2015 (UTC)[reply]

A pseudo-vector (aka axial vector) is not a pseudo-tensor

In fact, they are quite different.

a pseudo-tensor gains an additional sign flip under some passive transform (e.g. a basis inversion),
a pseudo-vector gains an additional sign flip under some active transform (e.g. a space inversion).

Therefore, a pseudo-scalar (e.g. a scalar product of an axial vector by a polar vector) is not a pseudo-tensor as well.

If I am right, the sentence “This is a generalization of a pseudovector.” must be deleted. KharanteDeux (talk) 09:08, 11 May 2021 (UTC)[reply]

Why det(A) and not det(B) ?

This may be a n00b question, but...why does the parity of the transformation depend only on det(A), "the transition matrix for the contravariant indices" and not similarly on det(B) for the covariant components?ScriboErgoSum (talk) 22:48, 8 December 2024 (UTC)[reply]