Talk:Tensor density/Archive 1

Requested move

 * The following discussion is an archived discussion of a requested move. Please do not modify it. Subsequent comments should be made in a new section on the talk page. No further edits should be made to this section. 

The result of the move request was: page stays at Tensor density, per discussion below. - GTBacchus(talk) 17:53, 10 July 2011 (UTC)

Tensor density → relative tensor – This is what this move discussion is genuinely about. Anthony Appleyard (talk) 04:30, 27 June 2011 (UTC)
 * Note that the responses below actually responded to the opposite request. I.e. responses starting with Ýes, intend the article to be at tensor density.TR 07:58, 27 June 2011 (UTC)


 * Because controversiality is now established, I'm submitting this very recent page move to the formal process. Until very recently this page was titled "relative tensor" and now I am formally proposing that it be maintained under the recently acquired title "tensor density".  Reasons, including those that predate this formal posting are as follows.  Thanks — Q uantling (talk &#124; contribs) 14:57, 10 June 2011 (UTC)

I must have missed the discussion, if there was one. Tensor density is the more common term (by 4:1 on ghits). Suggest we move back to the more prevalent term. Also, I am not aware that tensor densities have to have weight 1 - that's not what I was taught. -- cheers, Michael C. Price talk 09:32, 9 June 2011 (UTC)


 * Yes, I never even heard of the term "relative tensor" until moved this article to that name. [Explanation added later &mdash; For most of its existence this article covered tensor densities of all weights and called them tensor densities ("relative tensor" was not mentioned). Recently Jason Quinn moved the article to "Relative tensor" and introduced that term into the article. He did so without consultation with other editors. After a while, some of us undid some of his changes, and Michael C Price moved it back to "Tensor density". Then Jason Quinn began arguing to move it again to "Relative tensor". This is that discussion.]
 * My understanding was that if the weight of a "tensor density" is not specified, then it is assumed to be +1 (which is the most common weight other than 0). JRSpriggs (talk) 10:14, 9 June 2011 (UTC)


 * I wasn't aware of that convention, although it seems plausible. More concretely, though, MTW, page 501, are clear that tensor densities can have any weight, although they imply that weight=+1 is most common usage (after W=0, of course). -- cheers, Michael C. Price talk 10:32, 9 June 2011 (UTC)


 * Yes. I too was unfamiliar with the term "relative tensor" until I found this article; I have always used "tensor density". Personally, I use "tensor density of weight W" to mean a tensor density of a weight W, and I use "tensor density" without a specification of the weight to mean a weight of one.  But among Weinberg, MTW, or Ohanian and Ruﬃni, I do not recall how I got to that state.  — Q uantling (talk &#124; contribs) 13:32, 9 June 2011 (UTC)

Okay, I've moved it back [to tensor density]. -- cheers, Michael C. Price talk 19:11, 9 June 2011 (UTC)

No "Relative tensor" seems to be the term used by books that actually discuss the concept in full generality (that is with weights 2 and more). In fact, I don't recall a text that *doesn't* use the term when discussing these higher weight tensors. I don't recall the extent of the MTW discussion on tensor densities but if it was just a footnote about other weights, that doesn't count. I'm away from my personal library right now but the most influential source I can recall that uses "relative tensor" is the classic Lovelock and Rund Dover text. Using Google hits to support the title of "tensor density" over "relative tensor" is a poor idea. If you do a search for "vector" vs "tensor", you will also find that "vector" is more popular even though the latter is just a special case. Jason Quinn (talk) 01:22, 10 June 2011 (UTC)

Regarding the my initial move. I was bold and did it. I didn't expect the move to be controversial. The move was perfectly in accord with Wikipedia practices. I am somewhat annoyed by the speed of the move back to the original title. No sooner do I receive a message mentioning the discussion (thanks Quantling) but I find the revert has already been done, all within like half a day of the call discussion beginning. There's no reason for such haste on such a low-traffic article. Jason Quinn (talk) 01:38, 10 June 2011 (UTC)

Regarding the 4:1 ratio on Google hits. It wouldn't surprise me if the ratio of mathematicians who have been exposed to tensors densities of weight 2 or more to mathematicians who have been exposed only to tensor densties of weights 0 and 1 is roughly about 4:1. In that sense, interpreting the Google hits makes a strong argument that they SUPPORT using the title of "relative tensor" for the article. Jason Quinn (talk) 01:44, 10 June 2011 (UTC)
 * The MTW mention is not a footnote, but where they define tensor density. And the Kuptsov reference at the end of the article  also supports the use of tensor density.  (It's also how I was taught it by Michael Duff.)  Perhaps this is a case of different useage by physicists and mathematicans. -- cheers, Michael C. Price talk 04:49, 10 June 2011 (UTC)


 * That could be true that there's a physicist vs mathematician difference. I'm reluctant at the moment to look into this due to busy real life stuff but at some point I could try to estimate the ratio of books that use the one terminology over another. Using the adjective "density" in the general case is sort of a misnomer and I have a bias against it. Jason Quinn (talk) 09:16, 10 June 2011 (UTC)

Here are just a small sample of some books found using with Google Books that use "relative tensor": There are many more. I reject the notion that "relative tensor" is an obscure term. I also do not wish for the article title to become "frozen" at "tensor density" due to a small sample-size discussion. In fact, as the article was moved from "relative tensor" after the controversy was first discovered, I believe it ought to be the default article title in the face of an inconclusive discussion. I would like to see a list of books that use "tensor density" for weights other than 0 or 1. Apparently MTW is one (I still wish to see it in context first). What others are there? If the terms have roughly equal usage in the literature, it only makes logical sense if "relative tensor" is used because "tensor density" is a misnomer in general. Jason Quinn (talk) 06:27, 14 June 2011 (UTC)
 * Mathematical methods for physics and engineering by Riley, Hobson, and Bence (defined on p.964)
 * Tensors, differential forms, and variational principles by Lovelock and Rund (defined on p. 105)
 * Tensor calculus by Synge, Schild, and Schild (defined on p. 240)
 * The differential invariants of generalized spaces by Yerkes-Thomas (define p. 193)
 * Differential Geometry And Tensors by Dube (defined p. 268)
 * An Introduction to RIEMANNIAN GEOMETRY AND THE TENSOR CALCULUS by Weatherburn (defined p. 33)
 * Alas, no preview for MTW, AFAICS.
 * When g-searching for "tensor density" I get more book references than for "relative tensor". Some people define relative tensors as you say, some people (apparently more people) define tensor densities the same way.  So I see no objective way of deciding the issue, except by head count. -- cheers, Michael C. Price talk 08:56, 14 June 2011 (UTC)
 * I think my previous comment, that using Google searches to determine this is a bad idea, has merit. If you Google search for "vector" you get 310 million results compared to 13 million for "tensor". Should we merge tensor into the vector article? Of course not. By the very nature of a generalization, the inspiring concept will almost always be a more common term. The concept of a tensor density (meaning only those of weight 1) is often presented in low level physics, math, or engineering courses. Most of those students never learned or thought about the higher weight concept. If this is true, then is explains why "tensor density" out-googles "relative tensor". It's the same reason that "absolute value" would out-google the p-adic norm. Google hits are only a guide to help solve a problem. Here there are potentially important bias factors at work that I am suggest makes them misleading. For that reason, I cannot accept google hits as a valid argument to support the claim that "tensor density" is a better title than "relative tensor" and I am dismissing that line of reasoning. I also do not accept the premise that "more people" are using "tensor density" than "relative tensor" when explicitly talking about the general concept. If the premise is true, please try to convince me. But just because you didn't previously hear the term "relative tensor" doesn't make it so. I admit that it appears both conventions for the definition must be in use. What matters now is the ratio of those conventions. Besides a couple anecdotal stories and some (biased) use of Google hits, I have not seen a compelling argument that "tensor density" is the prevailing definition for general weights. In fact, my anecdotal story conflicts with that. I would like to see references now in addition to MTW to support the case. Jason Quinn (talk) 11:06, 14 June 2011 (UTC)


 * No. Jason Quinn (talk) makes a convincing argument.  Barring a convincing counterargument, I now prefer "relative tensor". — Q uantling (talk &#124; contribs) 15:27, 15 June 2011 (UTC)
 * Yes-ish Agree that Jason makes some good points - although the tensor/vector analogy fails. We all agree that "both conventions for the definition must be in use. What matters now is the ratio of those conventions." but the MTW reference carries much weight IMO, since MTW is the standard iconic ref'. I also looked up my Ray d'Inverno notes and he agrees with MTW. -- cheers, Michael C. Price talk 09:01, 16 June 2011 (UTC)

Tensor density: -- cheers, Michael C. Price talk 09:35, 16 June 2011 (UTC)
 * 1) Lectures on general relativity by Achilleus Papapetrou, defined,  page 12
 * 2) Introduction to the theory of relativity by Peter Gabriel Bergmann, defined Page 57
 * 3) Relativity: modern large-scale spacetime structure of the cosmos by Moshe Carmeli, defined page 99
 * 4) An introduction to general relativity and cosmology by Jerzy Plebański, Andrzej Krasiński, defined page 18
 * 5) Space, Time, Matter by Hermann Weyl, defined page 109
 * 6) Advanced Mechanics and General Relativity by Joel Franklin, defined page 184
 * 7) Space-time structure by Erwin Schrödinger, used repeatedly:


 * So, what do we do now? We have about 25 hours left on the one-week clock usually given for this debate.  I suppose we can always extend, if we think that that will help resolve things….


 * On a somewhat related topic, if we have a tensor and want to emphasize that it is not a relative tensor (of any non-zero weight) we can call it an absolute tensor. On the other hand, if we have a tensor and want to emphasize that it is not a tensor density (of any non-zero weight) then what do we call it; a tensor non-density?  — Q uantling (talk &#124; contribs) 13:14, 16 June 2011 (UTC)


 * We can ignore any artificial time constraints. I see no reason why the resolution of this has to be done quickly. All that is important to me is that we make a good decision eventually. Sooner or later new people will chime in with their opinions and that may help us. That may take weeks, months, or even years. It does seem as if a physics/math community split is at least somewhat at work here as Micheal suggested. As this is a pure mathematical concept perhaps deferring to the mathematicians is a good idea, especially in light of the slightly better name. PS In practice, if that situation arose, I would say or write a tensor of weight not equal to one with "relative" implied. "Tensor non-density" just sounds strange. Jason Quinn (talk) 12:35, 17 June 2011 (UTC)
 * Hi. I'm here from Requested Moves, and I would concur that there's no real time constraint. Take your time, and when there's a consensus worked out, we'll either move the page or not. -GTBacchus(talk) 23:12, 18 June 2011 (UTC)

I suggest that the article be kept at tensor density. This seem by far the most common name for these objects in physics. (To Mischael's list above, we can also add Sean Carroll's "Spacetime and geometry" on of the most common currently used graduate textbooks for general relativity.) In mathematics, the term "relative tensor" seems to be commonly used, however whenever it is used the term tensor density is used for a special case. As a result a mathematician looking up "relative tensor"and being redirected to "tensor density" will be less surprised than a physicist looking up tensor density and ending up at "relative tensor". Also, "tensor density"+ weight outnumbers "relative tensor" by 5:1 on gbooks, and 2:1 on gscholar. (adding the + weight to the search should filter out cases that only discuss the weight one case.) So, WP:COMMONNAME also seems to apply.TR 07:45, 20 June 2011 (UTC)
 * I agree with Micheal Price and TR (TimothyRias). These two have presented the best arguments for keeping the title "Tensor density". This appears to be the convention for physics. Steve Quinn (talk) 05:41, 5 July 2011 (UTC)
 * The above discussion is preserved as an archive of a requested move. Please do not modify it. Subsequent comments should be made in a new section on this talk page. No further edits should be made to this section.

Even vs. odd vs. pseudo
It seems to me that there are four ways a tensor density of weight W could transform. For instance if &phi; is a scalar density of weight W then we could have any of Is it that these are examples of, in order, an even scalar density, an odd scalar density, and even pseudoscalar density, and an odd pseudoscalar density? The article's present text on "odd" &mdash; specifically, "A distinction is made between odd tensor densities, in which a factor of the sign of the Jacobian determinant is also in the product, and even tensor densities which omit the sign from the transformation rule." &mdash; does not make it clear to me which case is "odd"; is it the second or third? FWIW, the first and third are apparently defined for any real-valued W, but the other two are apparently defined only when W is an integer. Also, when W is even, the first and second are the same as are the third and fourth; but when W is odd, it is the first and fourth that are the same and likewise for the second and third. It would be nice if the article made this all clear. Thanks — Q uantling (talk &#124; contribs) 15:37, 10 June 2011 (UTC)
 * $$\phi = \left|\det \frac{\partial \hat{x}^\mu}{\partial x^\nu}\right|^W \hat{\phi}$$
 * $$\phi = \left(\det \frac{\partial \hat{x}^\mu}{\partial x^\nu}\right)^W \hat{\phi}$$
 * $$\phi = \sgn\left(\det \frac{\partial \hat{x}^\mu}{\partial x^\nu}\right) \left|\det \frac{\partial \hat{x}^\mu}{\partial x^\nu}\right|^W \hat{\phi}$$
 * $$\phi = \sgn\left(\det \frac{\partial \hat{x}^\mu}{\partial x^\nu}\right) \left(\det \frac{\partial \hat{x}^\mu}{\partial x^\nu}\right)^W \hat{\phi}$$


 * You cannot raise a negative number to a non-integer power unless you want to go into the complex numbers (which I do not). Notice that
 * $$\left(\det \frac{\partial \bar{x}^\mu}{\partial x^\nu}\right) = \sgn\left(\det \frac{\partial \bar{x}^\mu}{\partial x^\nu}\right) \left|\det \frac{\partial \bar{x}^\mu}{\partial x^\nu}\right| \,.$$
 * So the only rules which makes any sense are your first and third rules, and they are sufficient. JRSpriggs (talk) 00:31, 11 June 2011 (UTC)

Ah, maybe it is that the second formula describes a tensor density of weight W, the fourth formula describes a pseduotensor density of weight W, the first formula describes an even tensor density of weight W and the third formula describes an odd tensor density of weight W. In particular, if W is an even integer (e.g., zero) then even tensor density (formula #1) and tensor density (formula #2) are interchangeable as are odd tensor density (formula #3) and pseudotensor density (formula #4); but when W is an odd integer then it is odd tensor density (formula #3) and tensor density (formula #2) that are interchangeable, and even tensor density (formula #1) and pseudotensor density (formula #4) that are interchangeable. — Q uantling (talk &#124; contribs) 20:18, 13 June 2011 (UTC)

My confusion — are the densities even or odd?
I seem to have arrived at a contradiction which I cannot yet resolve.
 * On the one hand, clearly all terms in the true Lagrangian (whatever it is) should have the same character, either all even scalar densities of weight +1 or all odd scalar densities of weight +1. The Lagrangians at Einstein–Hilbert action and Maxwell's equations in curved spacetime are both even scalar densities of weight +1. This implies that $$\mathfrak{J}^{\mu}, \, \mathfrak{D}^{\mu \nu}, \, \mathfrak{M}^{\mu \nu} \, \mathfrak{f}_\mu \,$$ (current, D & H, polarization & magnetization, and Lorentz force in a continuous medium) are also even tensor densities of weight +1.
 * On the other hand, the characterizations of current density and Lorentz force density given at Tensor density can be expanded to:
 * $$\int \mathfrak{J}^{\mu} \, \varepsilon_{\mu \alpha \beta \gamma} \, d x^{\alpha} d x^{\beta} d x^{\gamma} $$ is the net electric charge crossing the 3-volume in the positive direction; and
 * $$\int \mathfrak{f}_{\mu} \, \varepsilon_{\alpha \beta \gamma \delta} \, d x^{\alpha} d x^{\beta} d x^{\gamma} d x^{\delta} $$ is the linear momentum (4-vector) transferred from the electromagnetic field to matter within the 4-volume. This implies that these tensor densities are all odd (since the Levi-Civita symbol is odd).

So what am I doing wrong? JRSpriggs (talk) 19:07, 13 June 2011 (UTC)


 * You are using \mathcal rather than \mathfrak, so I'm not sure whether we're talking apples and oranges, but I write
 * $$\mathfrak{J}^\mu = J^\mu \sqrt{-g}$$
 * so $$\mathfrak{J}$$ is an even tensor density because $$\sqrt{-g}$$ is. — Q uantling (talk &#124; contribs) 19:33, 13 June 2011 (UTC)


 * But, in general, there is also some ambiguity between tensors and pseudotensors (or between even and odd) when it comes to integrands. An example in one dimension, when $$y=-x$$ and $$a < b$$ is:
 * $$\int_a^b f(x) \, dx = \int_{-a}^{-b} f(-y) \frac{dx}{dy} \, dy = \int_{-b}^{-a} f(-y) \left| \frac{dx}{dy}\right| \, dy\,.$$
 * The first integral is equal to either of the other two integrals. In the second integral the integrand transforms like a tensor (or odd tensor) density, but the integration limits are no longer from lower to higher.  In the second integral the integrand transforms like a pseudotensor (or even tensor) density because we have kept the integration limits in proper order, −b < −a.  The usual convention I've seen is to keep the integration limits from low to high and to transform the integrand as a pseudotensor density, but the other approach is equally valid. — Q uantling (talk &#124; contribs) 19:52, 13 June 2011 (UTC)
 * I had been copying the notation from Maxwell's equations in curved spacetime where I had not chosen to use mathfrak because it is harder to read and it suggests incorrectly that these are not the ordinary notions used in Maxwell's equations. I have now changed it in my comment above to allow us to communicate more clearly here.
 * What do you mean by $$J^\mu \,$$? The only way that I could define it is by
 * $$J^\mu = \frac{\mathfrak{J}^\mu}{\sqrt{-g}} \,$$
 * which renders your definition circular.
 * While assuming that the limits of integration are chosen in a way that allows one to use the absolute value of the Jacobian rather than the Jacobian itself might be acceptable for the Lorentz force where one is integrating over a 4-volume (where one might suppose that there is a preferred orientation), that will not work for a 3-volume where there are two equally valid orientations. JRSpriggs (talk) 12:29, 14 June 2011 (UTC)

If we cannot find an error in what I said in my first comment, then the only apparent ways out would be: Neither of these possibilities seems very satisfactory. What say you? JRSpriggs (talk) 12:29, 14 June 2011 (UTC)
 * Assume that the laws of physics only apply in right-handed coordinate systems, so that there is no distinction between even and odd to worry about. This would mean that we would be in a pickle if the manifold which is our spacetime turned out to be non-orientable. And it seems inconsistent with the common understanding that only the weak force (not gravity or electromagnetism, as here) fails to be invariant under spatial inversion (P-symmetry).
 * Or we could re-define $$\sqrt{-g} \,$$ to be an odd density by saying that it is positive in right-handed coordinates and negative in left-handed coordinates. This would probably also cause trouble if the universe is non-orientable.


 * I would say that $$J^\mu\,$$ is a tensor, that $$\mathfrak{J}^\mu = J^\mu \sqrt{-g}$$ is an even tensor density of weight one and that $$\mathcal{J}_{\alpha\beta\gamma} = \mathfrak{J}^\mu \varepsilon_{\mu\alpha\beta\gamma} = J^\mu \sqrt{-g} \varepsilon_{\mu\alpha\beta\gamma}$$ is a pseudotensor (or an odd tensor of weight zero). Conservation of current is written as the vanishing of $$\nabla_\mu \, J^\mu$$ or $$\partial_\mu \mathfrak{J}^\mu$$ or $$\partial_\alpha \mathcal{J}_{\beta\gamma\delta} - \partial_\beta \mathcal{J}_{\alpha\gamma\delta} + \partial_\gamma \mathcal{J}_{\alpha\beta\delta} - \partial_\delta \mathcal{J}_{\alpha\beta\gamma}$$.
 * The "AJ" contribution to the action is written as any of
 * $$\int A_\mu J^\mu \sqrt{-g} \, \varepsilon_{\alpha\beta\gamma\delta} \, dx^\alpha dx^\beta dx^\gamma dx^\delta$$
 * $$\int A_\mu \mathfrak{J}^\mu \, \varepsilon_{\alpha\beta\gamma\delta} \, dx^\alpha dx^\beta dx^\gamma dx^\delta$$
 * $$\int A_\mu \mathcal{J}_{\pi\rho\sigma} \, \frac{\varepsilon^{\mu\pi\rho\sigma}}{3!} \, \varepsilon_{\alpha\beta\gamma\delta} \, dx^\alpha dx^\beta dx^\gamma dx^\delta\,.$$
 * The shared integrand $$\nu_{\alpha\beta\gamma\delta} = A_\mu J^\mu \sqrt{-g} \, \varepsilon_{\alpha\beta\gamma\delta}$$ is a pseudotensor (or an odd tensor density of weight zero).
 * I don't see the distinction you are making between three-dimensional surface integrals and four-dimensional volume integrals. In either case the usual convention is to keep the integral limits in proper order and to use the absolute value of the Jacobian determinant — i.e., to have an integrand that is a pseudotensor.  Sure, with a volume integral there is the usual convention that the volume element is non-negative, and with a surface integral there is the usual convention of outward orientation, but I think that's a mostly orthogonal discussion.  — Q uantling (talk &#124; contribs) 20:11, 14 June 2011 (UTC)


 * You say $$\mathfrak{J}^\mu \,$$ is even. If so, then would that not imply that
 * $$\int \mathfrak{J}^{\mu} \, \varepsilon_{\mu \alpha \beta \gamma} \, d x^{\alpha} d x^{\beta} d x^{\gamma} $$
 * changes sign when a coordinate is negated? And does that not imply that the electric charge crossing that 3-surface changes sign? But how can it, since it is a physical thing which should not be affected by our choice of coordinates? JRSpriggs (talk) 06:17, 15 June 2011 (UTC)


 * No, the integral won't change sign. Because $$\mathfrak{J}^\mu$$ is even and of weight one, $$\mathfrak{J}^\mu \varepsilon_{\mu \alpha \beta \gamma}$$ will be odd and of weight zero; that is, it will be a pseudotensor.  If we perform a coordinate transformation that serves to flip the $$x^\alpha$$ coordinate, we we will pick up a sign flip as we transform the $$\alpha$$ index of $$\mathfrak{J}^\mu \varepsilon_{\mu \alpha \beta \gamma}$$ and then we will flip back because our integrand is a pseudotensor.  So the integrand keeps its sign.  And we don't have to think about whether $$dx^\alpha$$ keeps its sign because any potential sigh flip is canceled out by our convention of always integrating each coordinate from a lower value to a higher value.


 * Or perhaps another way of thinking of it is to say, by virtue of our prejudice that each coordinate should be integrated from a lower value to a higher value, that the integral sign is a pseudotensor (aka, an odd tensor of weight zero). Then it cancels with the pseudoscalar nature of $$\mathfrak{J}^{\mu} \, \varepsilon_{\mu \alpha \beta \gamma} \, d x^{\alpha} d x^{\beta} d x^{\gamma}$$ to make the integral be a scalar as desired. — Q uantling (talk &#124; contribs) 16:05, 15 June 2011 (UTC)


 * I do not think that the limits of integration can save this. Suppose the 3-surface is defined as the image in our spacetime manifold of a function with domain [0,1]&times;[0,1]&times;[0,1] which is parameterized by . The the integral becomes
 * $$q = \int_{w=0}^{w=1} \int_{v=0}^{v=1} \int_{u=0}^{u=1} \mathfrak{J}^{\mu} \, \varepsilon_{\mu \alpha \beta \gamma} \, \frac{\partial x^{\alpha}}{\partial u} \frac{\partial x^{\beta}}{\partial v} \frac{\partial x^{\gamma}}{\partial w} \, d u \, d v \, d w \,.$$
 * Changing from the x coordinate system to the y coordinate system for our spacetime does not affect u,v,w. The partial derivatives transform as ordinary contravariant vectors. The Levi-Civita symbol is unchanged which can be regarded as transforming as an odd tensor density of weight -1. Thus the current $$\mathfrak{J}^{\mu} \,$$ must transform as an odd contravariant vector density of weight +1 in order for q to be unaffected. Is this not correct? JRSpriggs (talk) 23:56, 16 June 2011 (UTC)


 * But this not an integral over coordinates. In this case, I agree with you that we want the integrand $$\mathfrak{J}^\mu \varepsilon_{\mu \alpha\beta\gamma}$$ to be an ordinary tensor, not a pseudotensor, and, furthermore, if we somehow put this in a form that used $$\det[\partial{y^\kappa}/\partial{x^\lambda}]$$, we would not use its absolute value.  The difference between this case and the earlier cases is precisely that, in this case, orientation-reversing coordinate transformations don't mess with the order of the limits of integration.


 * This case is more like the case
 * $$\int U^\alpha U^\beta g_{\alpha\beta} \, d\tau\,,$$
 * where the integrand is even more clearly a tensor, not a pseudotensor.


 * The remark of mine that an integral sign could be interpreted as a pseudotensor doesn't work; as your latest example clearly illustrates. At the very least, we'd have to add the caveat that this "rule" works only with integrals over coordinates.  — Q uantling (talk &#124; contribs) 01:35, 17 June 2011 (UTC)


 * If the current is odd in the case which I mentioned in my last comment, then it must be odd in any case.
 * Similarly to that argument, one can consider an integral which gives the action (physics). Suppose the 4-volume is defined as the image in our spacetime manifold of a function with domain [0,1]&times;[0,1]&times;[0,1]&times;[0,1] which is parameterized by . The the action integral is
 * $$\mathcal{S} = \int_{z=0}^{z=1} \int_{w=0}^{w=1} \int_{v=0}^{v=1} \int_{u=0}^{u=1} \mathfrak{L} \, \varepsilon_{\alpha \beta \gamma \delta} \, \frac{\partial x^{\alpha}}{\partial u} \frac{\partial x^{\beta}}{\partial v} \frac{\partial x^{\gamma}}{\partial w} \frac{\partial x^{\delta}}{\partial z} \, d u \, d v \, d w \, d z \,$$
 * where $$\mathfrak{L} \,$$ includes terms proportional to these
 * $$R_{\alpha \beta} \, g^{\alpha \beta} \, \sqrt{-g} \,$$
 * $$F_{\alpha \beta} \, F_{\gamma \delta} \, g^{\alpha \gamma} \, g^{\beta \delta} \, \sqrt{- g} \,$$
 * $$A_{\alpha} \, \mathfrak{J}^{\alpha} \,.$$
 * Thus all of these terms in $$\mathfrak{L} \,$$ must be odd scalar densities of weight +1. This is consistent with the fact that $$A_{\alpha} \,$$ is an ordinary covariant vector. But the other two terms appear to be even scalar densities, unless we re-define $$\sqrt{-g} \,$$ to be odd instead of even (as I suggested above). JRSpriggs (talk) 09:59, 18 June 2011 (UTC)


 * That is not the action integral for electromagnetism. The electromagnetism action integral is over coordinates.  It requires that $$\mathfrak{L}$$ be a pseudoscalar density (a.k.a., an even tensor density of weight +1), which it is, because $$\sqrt{-g}$$ and $$\mathfrak{J}^\mu$$ are both pseudotensors densities (a.k.a., even tensor densities of weight +1).  — Q uantling (talk &#124; contribs) 14:02, 20 June 2011 (UTC)


 * If
 * $$x^1 = u; x^2 = v; x^3 = w; x^4 = z \,$$
 * then
 * $$\varepsilon_{1 2 3 4} = +1 \,$$
 * and the integral which I gave in my last message simpifies to
 * $$\mathcal{S} = \int_{z=0}^{z=1} \int_{w=0}^{w=1} \int_{v=0}^{v=1} \int_{u=0}^{u=1} \mathfrak{L} \, d u \, d v \, d w \, d z \,$$
 * which is (I presume) the form which you consider correct. The reason I used the more complex form was to make it easier to address the question of the transformation law for $$\mathfrak{L} \,$$; and in particular to make it clear that the convention used for limits of integration is not controlling on this issue. JRSpriggs (talk) 06:10, 21 June 2011 (UTC)

Even or odd? (continued)
No, I don't think of that last formula as the action integral. In
 * $$S = \int_{z=0}^{z=1} \int_{w=0}^{w=1} \int_{v=0}^{v=1} \int_{u=0}^{u=1} \mathfrak{L} \, \varepsilon_{\alpha \beta \gamma \delta} \, \frac{\partial x^{\alpha}}{\partial u} \frac{\partial x^{\beta}}{\partial v} \frac{\partial x^{\gamma}}{\partial w} \frac{\partial x^{\delta}}{\partial z} \, d u \, d v \, d w \, d z \,,$$

I would say that the tensor $$\mathfrak{L}$$ is an authentic tensor with respect to $$(x^0, x^1, x^2, x^3)$$ and is a pseudotensor with respect to $$(u, v, w, z)$$ because the former can be transformed without messing with the integration limits but the latter messes with the integration limits.

In
 * $$S = \int \mathfrak{L} \, \varepsilon_{\alpha\beta\gamma\delta}\,dx^\alpha\,dx^\beta\,dx^\gamma\,dx^\delta\,,$$

especially with terms like $$R\sqrt{-g}$$ contributing to $$\mathfrak{L}$$, it appears to all boil down to whether $$\sqrt{-g}$$ is an even tensor density or an odd tensor density. Somewhere we should be able to find a reliable source that answers that question!

Thanks for the pointer to ! — Q uantling (talk &#124; contribs) 13:09, 21 June 2011 (UTC)


 * I think it would be very difficult to find a source (reliable or not) which addresses this issue because physicists are usually focused on calculating the equations of motion, not on fine details of the coordinate transformation laws. If you look at Einstein–Hilbert action, you will see that the square-root cancels out in the course of the calculation, so whether the square-root is the positive root or the negative root does not affect the equations of motion.
 * Our question is whether $$\sqrt{-g}\,$$ is an even scalar density of weight +1 (i.e. a pseudo scalar density) or an odd scalar density (i.e. an authentic scalar density). One could choose it to be either one since $${\sqrt{-g}\;}^2 = - g \,$$ is an even (authentic) scalar density of weight +2 in either case. Indeed, one might want to use one version for some purposes and the other version for other purposes. For example, for its use in this article, it is more convenient that it be even (although we could put absolute value signs around it, if it is odd).
 * Where it does matter is in deriving Maxwell's equations in curved spacetime, specifically
 * $$\mathcal{D}^{\mu\nu} \, = \, \frac{1}{\mu_{0}} \, g^{\mu\alpha} \, F_{\alpha\beta} \, g^{\beta\nu} \, \sqrt{-g} \,.$$
 * If the current density is odd, then $$\mathcal{D}^{\mu\nu} \,$$ (the electromagnetic displacement tensor density which combines the electric displacement field (D_x,D_y,D_z) and the magnetic field intensity (H_x,H_y,H_z) ) must also be odd which appears to contradict this equation unless the square-root is also odd. JRSpriggs (talk) 08:05, 22 June 2011 (UTC)

Hmm. Where Maxwell's equations in curved spacetime writes:
 * $$\begin{align}

F_{\alpha \beta} & = \partial_{\alpha} A_{\beta} - \partial_{\beta} A_{\alpha} \\ \mathcal{D}^{\mu\nu} & = \frac{1}{\mu_{0}} g^{\mu\alpha} F_{\alpha\beta} g^{\beta\nu} \sqrt{-g} \\ J^{\mu} & = \partial_\nu \mathcal{D}^{\mu \nu} \\ f_\mu & = F_{\mu\nu} \, J^\nu \,, \end{align}$$ I would write
 * $$\begin{align}

F_{\alpha \beta} & = \partial_{\alpha} A_{\beta} - \partial_{\beta} A_{\alpha} \\ D^{\mu\nu} & = \frac{1}{\mu_{0}} g^{\mu\alpha} F_{\alpha\beta} g^{\beta\nu} \\ J^{\mu} & = \nabla_\nu D^{\mu\nu} = \frac{1}{\sqrt{-g}} \partial_\nu D^{\mu\nu} \sqrt{-g} \\ f_\mu & = F_{\mu\nu} J^\nu \end{align}$$ and
 * $$\begin{align}

\mathfrak{D}^{\mu\nu} & = D^{\mu\nu} \sqrt{-g} \\ \mathfrak{J}^{\mu} & = J^\mu \sqrt{-g} = \partial_\nu \mathfrak{D}^{\mu \nu} \\ \mathfrak{f}_\mu & = f_\mu \sqrt{-g} = F_{\mu\nu} \mathfrak{J}^\nu \end{align}$$ and
 * $$\begin{align}

\omega_{\alpha\beta\gamma\delta} & = \sqrt{-g} \varepsilon_{\alpha\beta\gamma\delta} \\ \mathcal{D}_{\gamma\delta} & = \frac 12 \mathfrak{D}^{\alpha\beta} \varepsilon_{\alpha\beta\gamma\delta} = \frac 12 D^{\alpha\beta} \omega_{\alpha\beta\gamma\delta}\\ \mathcal{J}_{\beta\gamma\delta} & = \mathfrak{J}^{\alpha} \varepsilon_{\alpha\beta\gamma\delta} = J^{\alpha} \omega_{\alpha\beta\gamma\delta} = \partial_{\beta} \mathcal{D}_{\gamma\delta} + \partial_{\gamma} \mathcal{D}_{\delta\beta} + \partial_{\delta} \mathcal{D}_{\beta\gamma} \,, \end{align}$$ where $$A\,$$, $$F\,$$, $$D\,$$, $$J\,$$, and $$f\,$$, are (even) authentic absolute tensors, where the mathfraks, $$\sqrt{-g} \,$$, $$\mathfrak D \,$$, $$\mathfrak J \,$$, and $$\mathfrak f \,$$, are (even) pseudotensor densities, and where the mathcals, $$\omega \,$$, $$\mathcal D \,$$, and $$\mathcal J \,$$ are (odd) absolute pseudotensors. — Q uantling (talk &#124; contribs) 13:29, 22 June 2011 (UTC)


 * I edited the article. Please comment and/or edit.  — Q uantling (talk &#124; contribs) 13:16, 23 June 2011 (UTC)

Even or odd? (continued2)

 * Let me set aside the question of even versus odd for a moment, and instead address the question of whether electric current is an (odd or even) contravariant vector density of weight +1 or an ordinary contravariant vector. I will do this by converting the equations for conservation of electric charge back into classical physics notation.
 * A (locally) inertial frame is a free-falling, non-rotating, Cartesian coordinate system. A laboratory sitting on the surface of the Earth is undergoing two accelerations relative to an inertial frame: (1) the Earth is holding it up with $$- \vec{g} \,$$ against gravity minus centrifugal force; and (2) the Earth is carrying it along in its daily rotation at $$\vec{\omega} \,.$$
 * Consequently, in the natural coordinate system $$\langle t, \vec{r} \rangle \,$$ of the laboratory the metric is given by
 * $$(ds)^2 = - (c^2 - 2 \vec{r} \cdot \vec{g}) (dt)^2 + (d\vec{r} + (\vec{\omega} \times \vec{r}) dt)^2 + O(\vec{r}\;^2) \,.$$
 * Thus
 * $$\sqrt{\frac{g}{\eta}} = \sqrt{1 - \frac{2\vec{r}\cdot\vec{g}}{c^2} + O(\vec{r}\;^2)} = 1 - \frac{\vec{r}\cdot\vec{g}}{c^2} + O(\vec{r}\;^2) \,.$$
 * So if the classical $$\langle \rho, \vec{j} \rangle \,$$ is the vector density $$\mathfrak{J} \,,$$ then the conservation law is
 * $$0 = \mathfrak{J}^\nu_{; \nu} = \mathfrak{J}^\nu_{, \nu} = \frac{\partial \rho}{\partial t} + \operatorname{div}{\vec{j}} \,.$$
 * On the other hand, if the classical current is the ordinary vector $$J \,,$$ then the "conservation" law is
 * $$0 = J^\nu_{; \nu} = \frac{\partial \rho}{\partial t} + \operatorname{div}{\vec{j}} - \frac{\vec{j}\cdot\vec{g}}{c^2} + O(\vec{r}\;^1) \,.$$
 * Is there any evidence for the gravitational term in this formula? No! JRSpriggs (talk) 07:33, 4 July 2011 (UTC)
 * Sorry, but why exactly do you think that the conservation law would even depend on the quantity being a density or not? The covariant derivative commutes with the determinant of the metric and all that.TR 13:30, 4 July 2011 (UTC)

I'd say $$\mathfrak{J} = (\rho, \vec{j})$$, so the metric does not come into play when talking about conservation. If we write $$\mathfrak{J} = \sqrt{-g} J$$, and note that $$\sqrt{-g}$$ sure looks like a pseudotensor then we have the problem that we can't have both $$\mathfrak{J}$$ and $$J\,$$ be authentic tensors. I agree that I'd want them to both be authentic tensors. However, by my thinking the above divergence criterion for conservation is really shorthand for our concept of conservation as a flux through a surface, with an equivalence of the two approaches via Stokes' Theorem (Divergence Theorem). That is, to have current conserved we're going to want to look at an integral of $$\mathcal{J}_{\alpha\beta\gamma} = \mathfrak{J}^\mu\varepsilon_{\mu\alpha\beta\gamma}$$ through a closed surface. And the change of variables theorem sure makes it look like integrands are pseudotensors. So, there is at least some reason to treat $$\mathcal{J}$$ (and $$\mathfrak{J}$$) as a pseudotensor. But I agree that I also have "intuitive" motivations for wanting $$(\rho,\vec{j})$$ to be an authentic tensor density. — Q uantling (talk &#124; contribs) 14:26, 4 July 2011 (UTC)


 * To TimothyRias: I think that the equations are different because that is what I got when I did the differentiation. Why do you think that they should be the same? In either case (vector density or ordinary vector), I hope we agree that the equation
 * $$0 = \frac{\partial \rho}{\partial t} + \operatorname{div}{\vec{j}} \,$$
 * holds in an inertial frame of reference. If you transform that to a non-inertial frame such as that of a laboratory sitting on the surface of the Earth, then you should get the equations I showed above, depending on the transformation law (which is the difference between a vector density and an ordinary vector).
 * To Quantling: Since you agree that the vector density is the real deal, then why worry about the ordinary vector? JRSpriggs (talk) 00:09, 5 July 2011 (UTC)
 * What you should get after the transformation is:
 * $$D_\mu (\sqrt{|g|}^n J^\mu) = 0,$$
 * where n depends on the weight of J. Since the covariant derivative commutes with the metric, this is equivalent to
 * $$D_\mu J^\mu = 0.$$
 * So, if you are getting that the conservation law after transformation depends on the weight, you are doing something wrong.TR 08:09, 5 July 2011 (UTC)


 * Although the covariant derivative commutes with multiplication by the square-root of the determinant of the metric, the same is not true for the partial derivative. That is the source of your error. JRSpriggs (talk) 08:56, 5 July 2011 (UTC)
 * But after the transformation you should be getting a covariant derivative. You seem to be forgetting to transform the derivatives as well.TR 09:18, 5 July 2011 (UTC)

By definition the electric current (as defined in classical physics), $$J^{\alpha} = \langle \rho, \vec{j} \rangle \,,$$ satisfies
 * $$J^{\mu}_{, \mu} = \frac{\partial \rho}{\partial t} + \operatorname{div}{\vec{j}} \,$$

in any coordinate system. And we believe that
 * $$0 = J^{\mu}_{; \mu} \,$$

is true in any coordinate system. In a coordinate system where $$g_{\alpha \beta} = \eta_{\alpha \beta} \,$$ and $$g_{\alpha \beta, \gamma} = 0 \,$$ at an event, we have that
 * $$J^{\mu}_{; \mu} = J^{\mu}_{, \mu} \,$$

holds at that event and thus
 * $$0 = \frac{\partial \rho}{\partial t} + \operatorname{div}{\vec{j}} \,$$

also holds at that event. However, we need to see what it becomes when transformed to the natural coordinate system of a laboratory sitting on the Earth. Now suppose for the sake of argument (reductio ad absurdum) that the electric current $$J^{\alpha} \,$$ is an ordinary contravariant vector. Then the covariant derivative in an arbitrary coordinate system is
 * $$J^{\mu}_{; \mu} = J^{\mu}_{, \mu} + \Gamma^{\mu}_{\nu \mu} J^{\nu} \,.$$

Conveniently, we know that
 * $$\Gamma^{\mu}_{\nu \mu} = \frac{(\sqrt{-g})_{, \nu}}{\sqrt{-g}} \,.$$

And for the laboratory, this gives
 * $$\Gamma^{\mu}_{\nu \mu} = \frac{(1 - \frac{\vec{r}\cdot\vec{g}}{c^2} + O(\vec{r}\;^2))_{, \nu}}{1 - \frac{\vec{r}\cdot\vec{g}}{c^2} + O(\vec{r}\;^2)} \,$$

which simplifies to
 * $$\Gamma^{\mu}_{\nu \mu} = \langle 0, - \frac{\vec{g}}{c^2} + O(\vec{r}\;^1) \rangle \,.$$

Putting this all together gives
 * $$0 = J^\mu_{; \mu} = \frac{\partial \rho}{\partial t} + \operatorname{div}{\vec{j}} - \frac{\vec{j}\cdot\vec{g}}{c^2} + O(\vec{r}\;^1) \,$$

which contradicts what we know of electricity. Thus our supposition that electric current is an ordinary vector must be false. JRSpriggs (talk) 10:32, 5 July 2011 (UTC)


 * I agree that the settings where $$(\rho, \vec{j})$$ is most relevant tend to be one and the same as the settings where $$\mathfrak{J}^\mu = (\rho, \vec{j})$$ is more relevant than $$J^\mu = \mathfrak{J}^\mu / \sqrt{-g},$$. And I agree that $$\mathfrak{J}^\mu$$ is a density of weight +1.  And I agree that $$\partial_\mu \mathfrak{J}^\mu$$ is a proper expression for conservation of charge.  But how does any of this help us decide which of $$J\,$$ and $$\mathfrak{J}$$ are authentic, pseudo, even, or odd?  — Q uantling (talk &#124; contribs) 19:05, 5 July 2011 (UTC)
 * We should not be deciding anything :), you should find sources that decide it for you. How about starting with finding a source that actually simultaneously defines pseudo vs. authentic and even vs. odd? TR 16:13, 6 July 2011 (UTC)
 * Agreed, we need sources. Unfortunately, my books are not clear on whether $$\mathfrak{J}$$ and $$\sqrt{-g}$$ are psuedo or not.  Yours?  — Q uantling (talk &#124; contribs) 17:00, 6 July 2011 (UTC)

In view of Integration by substitution and after further consideration, I realize that
 * $$\mathcal{S} = \int_{z=0}^{z=1} \int_{w=0}^{w=1} \int_{v=0}^{v=1} \int_{u=0}^{u=1} \mathfrak{L} \, \varepsilon_{\alpha \beta \gamma \delta} \, \frac{\partial x^{\alpha}}{\partial u} \frac{\partial x^{\beta}}{\partial v} \frac{\partial x^{\gamma}}{\partial w} \frac{\partial x^{\delta}}{\partial z} \, d u \, d v \, d w \, d z \,$$

is incorrect because the part involving u,v,w,z is not invariant under an orientation reversing transformation of those coordinates. Perhaps
 * $$\mathcal{S} = \int_{z=0}^{z=1} \int_{w=0}^{w=1} \int_{v=0}^{v=1} \int_{u=0}^{u=1} \mathfrak{L} \, \left\vert \, \varepsilon_{\alpha \beta \gamma \delta} \, \frac{\partial x^{\alpha}}{\partial u} \frac{\partial x^{\beta}}{\partial v} \frac{\partial x^{\gamma}}{\partial w} \frac{\partial x^{\delta}}{\partial z} \, \right\vert \, d u \, d v \, d w \, d z \,$$

might be correct. If so, this would support Quantling's position that $$\mathfrak{L} \,$$ is even. JRSpriggs (talk) 07:32, 9 July 2011 (UTC)
 * To TimothyRias: Do you now see your error? The covariant derivative of the vector density is
 * $$\mathfrak{J}^{\alpha}_{; \beta} = \mathfrak{J}^{\alpha}_{, \beta} + \Gamma^{\alpha}_{\gamma \beta} \mathfrak{J}^{\gamma} - \Gamma^{\delta}_{\delta \beta} \mathfrak{J}^{\alpha} \,$$
 * where the second term on the right hand side is because the current is a contravariant vector and the third term is because it is a density of weight +1. If we then contract to get the divergence, those terms cancel out, so
 * $$\mathfrak{J}^{\mu}_{; \mu} = \mathfrak{J}^{\mu}_{, \mu} + \Gamma^{\mu}_{\gamma \mu} \mathfrak{J}^{\gamma} - \Gamma^{\delta}_{\delta \mu} \mathfrak{J}^{\mu} = \mathfrak{J}^{\mu}_{, \mu} \,$$
 * due to the symmetry of the Christoffel symbol in the lower indices. Thus the problematic gravitational term in the conservation law will not arise, if current is a vector density. JRSpriggs (talk) 07:32, 9 July 2011 (UTC)

Need sources for definitions
The definitions of "pseudotensor density", "authentic tensor density", "odd tensor density" and "even tensor density", need references. As they are currently given they appear to be at odds with the definitions given in the general references given for the article.

For example, Encyclopaedia of Mathematics, defines (citing Spivak) an odd relative tensor as what is defined as an "authentic tensor density" in the article. It defines an even relative tensor as what is here defines as a "pseudotensor density".

Moreover, it seems to me unlikely that any source would define these concepts at the same time. It just seems that there are varying conventions as to what is "even" and what is "odd". If that is truly the case the article should not so, and pick one convention to use in the rest of the article. The current representation using the terminology side by side seems to be unnecessarily confusing and a little misleading.TR 13:49, 23 June 2011 (UTC)


 * I haven't looked at your example, Encyclopaedia of Mathematics citing Spivak, so I may regret saying: those definitions are consistent with our current Wikipedia article if by "relative tensor" they implicitly are fixing the weight at +1.


 * I agree we should mention that it is unclear whether everyone in the literature agrees on these conventions. I'll see what I can do.  I also agree we need citations.  — Q uantling (talk &#124; contribs) 18:30, 23 June 2011 (UTC)
 * No they are not fixing the weight at one. (when on earth would calling it a relative tensor imply that anyway?)TR 19:55, 23 June 2011 (UTC)
 * I was correct that I would regret saying that! :-)  — Q uantling (talk &#124; contribs) 12:40, 24 June 2011 (UTC)

Well-definition
The covariant derivative of tensor densities defined here isn't compatible with the definition of the Levi-Civita connection. First and foremost, the determinant of the metric is a scalar. Under a coordinate change, $$ \det{g}(x) \rightarrow \det{g} (f^{-1}(x))=\det{J(f,x)}\det{g'(x)} $$. Thus, while being a scalar density, it is also a scalar. And for scalars, we have already a definition for the covariant derivative (using the Levi-Civita connection), namely $$\nabla_a \det{g}=\partial_a \det{g} $$. Thus, $$ \nabla_a \det{g} \neq 0 $$ in general. Take one simple example. We use the sphere with the metric $$ g=diag(1,\sin(\theta)^2) $$. We define a scalar function $$ f(\theta)=\sin(\theta) $$. Obviously, $$ \sqrt{\det{g}}-f=0 $$ everywhere. However, applying the covariant derivative to the term yields (IF the wrong definition was true) $$ \nabla_a 0=\nabla_a (\sqrt{\det{g}}-f)=-\nabla_a f=-\partial_a f $$, which is simply painful to even look at. Cretu (talk) 22:07, 26 October 2011 (UTC)
 * The true lesson there is that one should not add tensor densities of different weight.TR 06:00, 27 October 2011 (UTC)
 * Any tensor density is still a tensor as under a coordinate transformation the determinant prefactor is a smooth function. So the addition of tensors of different weight would simply be a tensor addition, which is well defined. Also, in one dimensions, $$\det{g}=g_(e_1,e_1)$$. From the definition of the Levi-Civita connection it follows that $$\partial_1 (g(e_1,e_1))=\nabla_1 (g(e_1,e_1))$$. So we would have $$0=\nabla_1 \det{g}=\nabla_1 (g(e_1,e_1))= \partial_1 (g(e_1,e_1))$$ which is simply non true for arbitrary positive functions $$ g(e_1,e_1) $$. Something can't be right about the tensor weight differentiation.. Cretu (talk) 06:52, 27 October 2011 (UTC)
 * I am sorry, but you are mistaken. You said "First and foremost, the determinant of the metric is a scalar. Under a coordinate change, $$ \det{g}(x) \rightarrow \det{g} (f^{-1}(x))=\det{J(f,x)}\det{g'(x)} $$. Thus, while being a scalar density, it is also a scalar.". No, you are misinterpreting this. If the determinant of the metric were a scalar, then the transformation would be $$ \det{g}(x) \rightarrow \det{g'(x)} $$, but as you correctly noted it is not. JRSpriggs (talk) 07:34, 27 October 2011 (UTC)
 * It is a scalar on the tangent space. Not a "physical" scalar. Also, it is a smooth function. And it obeys $$\det{g}(x)=\det{g}(f^{-1}(y))$$. And as such, the covariant derivative is already defined for it as the normal partial derivative. Besides, you missed the point about the thing not being well defined in 1d. Also, as I said, tensor densities are tensors. One can define a tensor density via $$\mathfrak{T}^\alpha_\beta (g(x),x)=\det{J}^W \frac{\partial_x^\alpha}{\partial_y^\gamma} \frac{\partial_y^\delta}{\partial_x^\beta} \bar{\mathfrak{T}}^\gamma_\delta (g'(y),y)=\frac{\partial_x^\alpha}{\partial_y^\gamma} \frac{\partial_y^\delta}{\partial_x^\beta} \bar{\mathfrak{T}}^\gamma_\delta (g(f^{-1}(y)),y)= $$. Thus, it is still a tensor and the covariant derivative acts on it like one. The new definition is in violation of the old one. Thus it is ill defined.Cretu (talk) 11:05, 27 October 2011 (UTC)
 * The whole point of tensor densities is that they are not tensor fields. They do not transform as tensor. They are not sections of a tensor bundle. Etc. You can claim that 1 = 2 all you want, but you should not be surprised if that leads to the conclusion that the sum 2+1 is ill-defined.TR 11:26, 27 October 2011 (UTC)
 * Taken from here: Definition: A tensor field of type (2, 0) on the n-dimensional smooth manifold M associates with each chart x a collection of n2 smooth functions Tij(x1, x2, . . ., xn) which satisfy the transformation rules shown below. Similarly, we define tensor fields of type (0, 2), (1, 1), and, more generally, a tensor field of type (m, n). Now let's check. $$\mathfrak{T}^\alpha_\beta (g(x),x)=\frac{\partial_x^\alpha}{\partial_y^\gamma} \frac{\partial_y^\delta}{\partial_x^\beta} \bar{\mathfrak{T}}^\gamma_\delta (g(f^{-1}(y)),y)$$. The transformation rule is satisfied. It's also smooth. Thus it is a tensor field. Cretu (talk) 11:41, 27 October 2011 (UTC)
 * Actually, $$\mathfrak{T}^\alpha_\beta (g(x),x)$$ doesn't satisy the definition rule, since the dimensionality of the indices does not match the dimensionality of the argument of the functions. (i.e. the indices run from 1 to d, where d is the dimension of the manifold, yet each function has d+ 1/2d(d+1) arguments).TR 13:52, 27 October 2011 (UTC)
 * Also, may I remind you that Wikipedia is not a forum. This is not the place to discuss your thoughts on tensor densities.TR 13:56, 27 October 2011 (UTC)
 * This doesn't change my original point. Given a metric in a Riemannian manifold, the indices match the number of dimensions and the arguments, as a metric depends on the position only. On the manifold, it is a simple smooth function. And as my example stated, given the definition in the article for the covariant derivative of the metric, it is ill defined in 1d. If you would be so kind to point the mistakes out here, I'm happy to agree with you. Cretu (talk) 14:51, 27 October 2011 (UTC)
 * $$\det(g_{\alpha \beta})$$ is a scalar field density of weight +2, not of weight 0. If you define $$f$$ to be a scalar field (of weight 0) than you cannot reasonably compute their difference.  The article looks right to me.  — Q uantling (talk &#124; contribs) 17:09, 27 October 2011 (UTC)
 * I've derived what I worry about in a rather proper fashion on my user page: http://en.wikipedia.org/wiki/User_talk:Cretu. Please take a look at it and tell me where I am mistaken. If not, there must be an error in the definitions. Cretu (talk) 17:51, 27 October 2011 (UTC)

To Cretu: You are using a language which is not used by us physicists, so it is hard for me to know what you mean. However, right at the beginning of your argument, you assume that the coordinate transformation is an isometry. If you are saying that the components of the metric tensor do not change when the coordinates are transformed, then that is false in general. Also, the distinction between tensors and tensor densities disappears in a one-dimensional space. However, we are not interested in one-dimensional space. We only care about four-dimensional (or possibly higher dimensional) spaces because that is where general relativity physics is done. JRSpriggs (talk) 05:25, 28 October 2011 (UTC)
 * An isometry means precisely what it states there. The metric transforms as a tensor and is given by the transformation formula there, which is just what you assume anyway. The language is mathematically precise, saying "transforms as" can really mean anything, however, the definition for the transform given there is just what is in the article simply implies without stating. Namely, a tensor density is an object depending on the metric and the coordinates. Under change of both it transforms as a tensor multiplied with the determinant of the transformation. However, the defining property of a tensor is just the way it transforms under a simple coordinate change as the metric determinant is a smooth function. Thus, the tensor density is in essential a tensor and needs to be treated as one, which is also the problem arising in 1d. The covariant derivative is an intrinisic derivative. It doesn't depend on how something transforms. Cretu (talk) 08:47, 28 October 2011 (UTC)


 * Your formula

(\mathbf{f}^* \bar{\mathbf{T}}_{ij})(x)=\bar{\mathbf{T}}_{kl}(\bar{g}(f(x)),f(x)) J^k_i(x) J^l_j(x) =\det{J}^{-W}(x)\mathbf{T}_{ij}(g(x),x)=\mathbf{T}_{ij}(\bar{g}(f(x)),x) $$
 * "defining" tensor densities appears to be wrong. There is no dependence on g. The formula should be

\bar{\mathbf{T}}_{kl}(f(x)) J^k_i(x) J^l_j(x) (\det{J}(x))^{W} = \mathbf{T}_{ij}(x) $$
 * instead.
 * The covariant derivative is not "intrinsic"; it is just a way of patching up the partial derivative so that it becomes a tensor (or tensor density, in our case). So it certainly does depend on how the tensor or tensor density being derived transforms. JRSpriggs (talk) 09:21, 28 October 2011 (UTC)
 * Multiply both sides with the determinant to obtain your formula from my formula. Thus they are equal. On a fixed chart, the determinant is a smooth function. And for smooth functions on a fixed chart, the covariant derivative is already defined and coincides with the partial derivative. Imagine yourselves to live on a chart. You're being given two number fields. One is a function whose value is the determinant. Another one is the scalar density determinant. You can't separate these two. Yet their covariant derivative should differ? Cretu (talk) 13:01, 28 October 2011 (UTC)
 * By your logic the covariant derivative of any component of a tensor should be just the normal partial derivative (i.e. the covariant derivative should always be equal to the partial derivative) because on a single chart you cannot distinguish a scalar function from a component of a tensor. NOFI but your basic understanding of differential geometry seems to be lacking here.TR 13:12, 28 October 2011 (UTC)
 * In fact, for any tensor, $$\nabla_i( T(e_j,e_k))=\partial_i( T(e_j,e_k))$$. Component wise, this is indeed just the partial derivative. What is commonly written as $$\nabla_i T_{jk}$$ is in fact $$\nabla_i T_{jk}=(\nabla_i T)(e_j,e_k)$$ i.e. the components of the tensor $$(\nabla_i T)$$. Cretu (talk) 13:22, 28 October 2011 (UTC)
 * That is only true if the e_i are proper vector fields. If they represent the coordinate frame, then they are not vectors fields.TR 13:48, 28 October 2011 (UTC)

The simple case of zero time dimensions and two spatial dimensions, when examined in Cartesian coordinates, gives $$\det(g_{\alpha\beta}) = 1$$ everywhere (in the chart). Yet, when examined in polar coordinates, $$\det(g_{\alpha\beta}) = r^2$$ everywhere. Because the right-hand sides of these two formulae are not the same, I surmise that $$\det(g_{\alpha\beta})$$ is not transforming like a scalar field. Rather, it turns out, that it is transforming like a scalar field density of weight +2. — Q uantling (talk &#124; contribs) 15:26, 28 October 2011 (UTC)