Talk:Tensor density

• Talk:Tensor density/Archive 1

Definition from elsewhere than physics?

Is there a definition of the thingies in the article coming from a pure math source, like a definition that does not relate to transformation properties? Physicists tend to dislike to formally define the concepts they are using, and when they (most reluctantly) do define them, they don't always agree among themselves. YohanN7 (talk) 14:06, 29 November 2013 (UTC)

For example, a tensor (w/o all pseudo, etc qualifiers) may be defined without reference to transformation properties. YohanN7 (talk) 19:23, 28 November 2013 (UTC)

I have a possibly different, but essentially related issue – the chosen wording. The concept of a tensor used in this article is the matrix of coefficients with respect to a chosen basis, and it excludes the coordinate-free concept of a tensor. I've seen this concept mathematically formalized somewhere in WP as a matrix function of the space of basis vectors, thus formally capturing the transformation requirement. In this context (whether formally or in the less formal physicists' approach), tensor densities are easily defined. In a coordinate-free approach to tensors, or even with wording that is agnostic to which approach is to be used, tensor densities don't work. For example, the concept of a determinant of a tensor in the coordinate-free formulation only makes sense in the case of a linear transformation (necessarily a type-(1,1) tensor; see Determinant#Abstract_algebraic_aspects). This article (to some extent unlike the article Tensor), in its wording implicitly denies the alternative abstract definition of tensors. For example, the phrase "the determinant of the metric tensor" does not make sense using the abstract geometric concept of tensor, where what is meant is "the determinant of the metric tensor coefficient matrix [with respect to a chosen basis]". By way of establishing the context in which a reader is to interpret this article, it should at least point out at the start that it is using the formalism in which the matrix of coefficients representing a tensor with respect a basis, and that here this matrix is referred to as the tensor.

A second point, related in that it suffers from the same encyclopaedic style problem, namely that it makes the implicit assumptions of a certain approach without stating the assumptions, is that the definitions assume a holonomic basis. A coordinate-based approach still makes sense with a nonholonomic basis, and thus tensor densities can defined in the same way in terms of the determinant of the transformation matrix, but the transformation matrix cannot be expressed as the matrix of partial derivatives as assumed in this article.

Am I being reasonable in saying that this article simply does not adequately outline the context and assumptions sufficiently before launching into the detail? —Quondum 14:28, 14 June 2014 (UTC)

See Spivak's book listed in the references. (It should be easy to find a digital copy using google)TR 09:00, 3 July 2014 (UTC)

Sign of the weight of the determinant of the metric tensor

I have been in an edit-war with an unregistered user with a constantly changing IP address. This makes for a very difficult situation as it is almost impossible to communicate with such people.

Although there are several things I dislike about his edit, the main problem is that he is getting the sign of the weight of the determinant of the metric tensor wrong. Essentially he has reversed the definition of weight given in the first section of the article (without changing it there). I think that changing that definition would be wrong as it would not be the usual definition and would conflict with the usage of the word "density" in ordinary speech. A weight of +1 should be used for "density" in the ordinary sense.

The integral of a scalar field over a hyper-volume of space-time is not an invariant. To make an invariant integral, one has to integrate a scalar density instead. How does this scalar density transform? For simplicity, let us consider a change of coordinate system in which each of the coordinates is doubled. Then the hyper-volume of the region over which one is integrating would seem to increase by a factor of 16 (if one ignores the metric tensor). So to compensate for that and make the integral invariant, we must multiply the scalar density by 1/16. Under this doubling of coordinates, an arrow (contravariant vector) would have its components double. So the metric tensor (a covariant matrix) would have its components multiplied by 1/4. The determinant of the metric would be multiplied by 1/256. So the square-root of its absolute value would be multiplied by 1/16. This is just the factor needed to change an ordinary scalar into a scalar density. So that square-root should have weight +1. JRSpriggs (talk) 05:54, 1 July 2014 (UTC)

Page protection?

It would force him/her to either quit or get an account and discuss. YohanN7 (talk) 13:35, 1 July 2014 (UTC)

It does not seem necessary or useful to apply page protection. I support JRSpriggs's perspective, and hope that the IP will choose to discuss here. Using informative edit summaries and the opening of this thread may be helpful in this direction. —Quondum 14:32, 1 July 2014 (UTC)

The IP(s) doesn't seem to be interested in discussion. This has been going on for some time. Therefore page protection is both necessary and useful (useful in that he/she(/they) is forced to discuss). YohanN7 (talk) 15:52, 1 July 2014 (UTC)

He reverted. This guy is not here to discuss. Semi-protect! YohanN7 (talk) 20:41, 1 July 2014 (UTC)

Hi, This is the unregistered user with the changing IP address. I only found the 'editing talk' section now and I do apologize for not writing here sooner. There are two main resons why (as you said) I 'reverse the sign of the weight'.

1) When I took a General Relativity (GR) course we used the GR textbook by Sean Carroll. (http://arxiv.org/pdf/gr-qc/9712019v1.pdf) In that book (and course) we went over tensor densities. One thing we showed/learned was that the square-root of the determinant of the metric is a scalar density of weight -1. (Sean Carroll shows it on p.52).

2) The volume element is a (0,n) tensor density of weight +1. (http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll2.html, equation 2.44). Now, in curved space-time we want the expression

$\int \phi (x^{\mu })Ad^{n}x$

to be a intergral of a scalar over a proper n-form. Here $A$ is some constant to be determined. The scalar is a type (0,0) tensor density of weight 0 (i.e. a scalar), so that is good. Since the volume element is a tensor density of weight +1 we require $A$ to be a object of weight -1 so that the combination is a (0,0) tensor of weight 0. It so happens that ${\sqrt {\vert g\vert }}$ fulfills this requirement as seen by (http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll2.html, equation 2.41).

As further proof that this is correct, consider a famous action, say the Einstein-Hilbert (EH) action. For simplicity lets look at the EH-action with zero cosmological constant and no matter. The EH action is $S\propto \int R{\sqrt {-g}}d^{4}x$. Do you deny this is correct? According to you ${\sqrt {-g}}$ has weight +1 which (since the volume element has weight +1) means that the intergral has overall weight $+2\not =0$. But it is a known fact that the EH-action is a true scalar. Or if you insist that it has weight 0 then you are saying that the EH-action should be $S\propto \int {\frac {R}{\sqrt {-g}}}d^{4}x$ which, obviously is wrong. QED

P.s. I use absolute value on g to ensure positivity even in a Riemannian manifold. — Preceding unsigned comment added by 206.45.187.125 (talk) 21:58, 1 July 2014 (UTC)

Hi, and welcome to the talk page. YohanN7 (talk) 22:42, 1 July 2014 (UTC)

Three things you need to know (for a start):

• You need to update the article with references supporting your claims (or you are likely to be reverted).
• Lecture notes do generally not constitute acceptable references. Nor does your own arguments - even if they are correct.
• It is customary here at Wikipedia that when an edit war occurs, then the matter should be settled on the talk page before any changes are made.

YohanN7 (talk) 00:09, 2 July 2014 (UTC)

Hi Yohan, thank you for your reply.

In reference to your three items...

i) I did support my claims both on the talk and the article.

ii) The reference I used were from a textbook called 'Spacetime and Geometry: An Introducton to General Relativity' by Sean Carroll which is a graduate level textbook and widely accepted in physics today. It so happens that Carroll made lecture notes as well (which I referenced). However his 'lecture notes' are identical to his textbook. Here is his blog: http://preposterousuniverse(dot)com/spacetimeandgeometry/

iii) That I did not know and I apologize! — Preceding unsigned comment added by 205.200.233.195 (talk) 01:00, 2 July 2014 (UTC)

It looks like things are going to work out just fine. I'm not an expert on the subject, so I'll leave it to you guys who are to sort it out. But I have a hunch that it is just a matter of definition or convention. If this is the case, then both conventions can be mentioned in the article, with the most common one being the "main" definition to be used throughout. YohanN7 (talk) 01:54, 2 July 2014 (UTC)

To anonymous:

Your teacher was mistaken, if he equated what Carroll called the "naive volume element" with a density of weight +1. In ordinary language density is characterized by

$Density={\frac {d\,Mass}{d\,Volume}}$

So density is inversely related to volume, not proportional to volume. That is, your basic assumption is wrong.

Also, density is used in lieu of four anti-symmetric covariant indices. So you should not combine both a notion of weighted density and tensor indices in your volume element. JRSpriggs (talk) 06:22, 2 July 2014 (UTC)

Saying that people adhering to a different convention are wrong is silly.TR 10:49, 2 July 2014 (UTC)

To JRSpriggs

Carroll SHOWS that ${\sqrt {\vert g\vert }}$ is a scalar density of -1!!! Furthermore (Carroll also shows this) the volume element transforms under a co-ordinate transformation with a determinant as such: $d^{n}x'=\left\vert {\frac {\partial x'}{\partial x}}\right\vert d^{n}x$ Thus it is a density of +1.

So the combination ${\sqrt {\vert g\vert }}d^{n}x$ has total weight $-1+1=0$. So it is a ordinary $(0,0)$ tensor.

Also I highly doubt that both Carroll and my professors are wrong...espcially that they have a PhD in physics, specilizing in GR. — Preceding unsigned comment added by 205.200.233.195 (talk) 08:06, 2 July 2014 (UTC)

It is useful to realize that there are different conventions for the definition of weight of a tensor density that differ by a sign. Carroll (like for example Weinberg) use a convention in which the Levi-Civita symbol with lower indices has weight +1. Consequently, the square root of g has weight -1 in that convention. The definition currently appearing in the article uses the opposite convention, where the square root of g has weight +1. This is not a matter of right or wrong, but a matter of convention, and whatever convention is used in the article should be used consistently.TR 10:49, 2 July 2014 (UTC)

To JPSpriggs

Ok, this lasted too long and I am sick of it. You know what? I give up. I raise the white flag. YOUR RIGHT AND I (with my tiny miniscule insignificant pee brian) AM WRONG!!!! HAPPY?

FYI, Personally I will keep believing what I believe but you probably don't care about that. I will continue physics and continue with my convention and if I fail university and life and become a hobo on the streets because of it, then SO BE IT, (You can come by and laugh at my stuipidity then). By the looks of it, my miniscule insignificant pee brain and I deserve it. (and according to you it will happen)

At the beginning you said there were other things you didn't like about my post. I was going to ask you what but I don't need to because I can guess...it is EVERYTHING.

Have a good life, this future stupid IDIOTIC loser (aka me) won't bother you again!!!! — Preceding unsigned comment added by 206.45.29.3 (talk) 19:08, 2 July 2014 (UTC)

Ok I just realized that my argument was valid if you defined the tensor transformation as ${\mathfrak {T}}_{\nu '_{1}\cdots \nu '_{m}}^{\mu '_{1}\cdots \mu '_{n}}=\left|{\frac {\partial x'}{\partial x}}\right|^{w}{\frac {\partial x^{\mu '_{1}}}{\partial x^{\mu _{1}}}}\cdots {\frac {\partial x^{\mu '_{n}}}{\partial x^{\mu _{n}}}}{\frac {\partial x^{\nu _{1}}}{\partial x^{\nu '_{1}}}}\cdots {\frac {\partial x^{\nu _{m}}}{\partial x^{\nu '_{m}}}}{\mathfrak {T}}_{\nu _{1}\cdots \nu _{m}}^{\mu _{1}\cdots \mu _{n}}$

(then g has weight -2)

BUT you defined it as

${\mathfrak {T}}_{\nu '_{1}\cdots \nu '_{m}}^{\mu '_{1}\cdots \mu '_{n}}=\left|{\frac {\partial x}{\partial x'}}\right|^{w}{\frac {\partial x^{\mu '_{1}}}{\partial x^{\mu _{1}}}}\cdots {\frac {\partial x^{\mu '_{n}}}{\partial x^{\mu _{n}}}}{\frac {\partial x^{\nu _{1}}}{\partial x^{\nu '_{1}}}}\cdots {\frac {\partial x^{\nu _{m}}}{\partial x^{\nu '_{m}}}}{\mathfrak {T}}_{\nu _{1}\cdots \nu _{m}}^{\mu _{1}\cdots \mu _{n}}$

(then g has weight +2)

So I again apoligize. Truthfully, looking back, I am appalled at how stupid and mentally challenged I am. I really should be in a mental institution come to think of it. I know I graduated with honors but, honestly, I am SHOCKED that I passed elementry school with my mentally challanged brain.

Have a good life! — Preceding unsigned comment added by 206.45.181.160 (talk) 02:30, 3 July 2014 (UTC)

Hi, we all make mistakes. An important thing on Wikipedia is to follow etiquette so that you do not make life difficult for other editors. I know this because I found out the hard way! — Cheers, Steelpillow (Talk) 08:29, 3 July 2014 (UTC)

I'm not sure what the current state of this discussion is. I only just became aware of it after editing Tensor#Tensor densities because the convention used there seemed to me to be wrong (or at least inconsistent with most conventions). Let me explain why JRSpriggs is "right" and other conventions are "wrong". I will just discuss scalar densities. (Here by "density" I just mean "1-density"). A density is a quantity on a manifold that is integrable. Examples are charge densities, mass densities, population densities, (even "volume densities", although if we think of this, we are likely to confuse ourselves). The main point is that a density is something that can be integrated over small pieces of the manifold (in a coordinate-independent manner), producing a single number (scalar). The Jacobian change of variables formula dictates what the transformation law for a density must be under a change in coordinates. It must vary by the inverse of the Jacobian of the coordinate change. That is,

$f(y)=f(x)|\det(\partial y/\partial x)|^{-1}.$

This is the convention adopted in the present article, and in JRSpriggs' comment above.

Put another way, a density is something that defines a scalar field when it is multiplied by the wedge product of the coordinate functions. Thus if the wedge product of the coordinate functions has units of volume, densities have units of 1/volume (or (whatever unit)/volume). Sławomir Biały (talk) 13:32, 3 July 2014 (UTC)

@Sławomir Biały: I'm slightly confused. Both the Tensor article and the present article follow the mnemonic "new variables downstairs in weighting factor" and they are presumably using the "correct" convention. Your change of variables formula above have "new variables upstairs". YohanN7 (talk) 15:50, 3 July 2014 (UTC)

yes, you're right. It should be the inverse of the jacobian. Sławomir Biały (talk) 16:28, 3 July 2014 (UTC)

The long-term edit war (which I think justifies semiprotection), has spilled over to the Tensor article as well, fwiw. Sławomir Biały (talk) 19:38, 3 July 2014 (UTC)

The convention where the weight of g is -2 is NOT I repeat NOT wrong...it is just a convention (Which I prefer) — Preceding unsigned comment added by 142.132.71.246 (talk) 21:29, 3 July 2014 (UTC)

Fine. Except in that convention, the word "density" is unrelated to its usual mathematical meaning. Sławomir Biały (talk) 22:35, 3 July 2014 (UTC)

To JRSpriggs

For your information I spoke to THREE different university professors who all have a PhD IN THEORETICAL PHYSICS specializing in GRAVITY. They ALL said that the transformation of tensor densities defined as:

${\mathfrak {T}}_{\nu '_{1}\cdots \nu '_{m}}^{\mu '_{1}\cdots \mu '_{n}}=\left|{\frac {\partial x'}{\partial x}}\right|^{w}{\frac {\partial x^{\mu '_{1}}}{\partial x^{\mu _{1}}}}\cdots {\frac {\partial x^{\mu '_{n}}}{\partial x^{\mu _{n}}}}{\frac {\partial x^{\nu _{1}}}{\partial x^{\nu '_{1}}}}\cdots {\frac {\partial x^{\nu _{m}}}{\partial x^{\nu '_{m}}}}{\mathfrak {T}}_{\nu _{1}\cdots \nu _{m}}^{\mu _{1}\cdots \mu _{n}}$

is a convention that is RIGHT. By THIS convention $g$ is a scalar density of $-2$. This is NOT WRONG. It is just a convention. So by saying (and I quote) "Your teacher was mistaken, if he equated what Carroll called the "naive volume element" with a density of weight +1. ...That is, your basic assumption is wrong." your simply telling everyone that you are arrogant!

Good day! — Preceding unsigned comment added by 207.161.69.254 (talk) 23:11, 3 July 2014 (UTC)

Fine. If three PhDs agree that a "mass density" or "charge density" is not a density at all, clearly they must be right. Sławomir Biały (talk) 23:22, 3 July 2014 (UTC)

To be honest, I'm reacting a bit against the right-or-wrong formulation. Chose convention as suits the occasion. It is little more (actually no more) than choosing a basis in Rⁿ. Sometimes you don't even want it to be orthogonal because it would complicate things. YohanN7 (talk) 00:41, 4 July 2014 (UTC)

In the end, any vocabulary is a matter of convention. But not all vocabularies are equally useful. In this case, calling something a "density" should agree with the premathematical use of the term, in my opinion. The convention should be chosen so that mass, charge, population, etc, are densities. Wouldn't you agree? Choosing the opposite convention therefore seems arbitrary and unmotivated. Also it is not a common one in my experience. Sławomir Biały (talk) 01:41, 4 July 2014 (UTC)

The premathematical (in the sense of me not properly understanding the detail) use would lead me to expect that a scalar density (which I'd assume would be said to have weight +1) would be one that when integrated over a region (using the volume element), yields a true scalar. Charge and mass densities do not qualify, since they are components of tensors, not scalars, unlike the trace of the stress–energy tensor or the Ricci scalar. Here there is room to define what "of weight w" means, but I guess this'd be the most natural. Is this the sense in which it is currently used? —Quondum 02:14, 4 July 2014 (UTC)

@Sławomir, I agree of course. I was speaking in general terms. I don't have a problem at all with the Wikipedia articles making internal choices that could be said are natural. But other choices in common existence could (and should imo) be mentioned, and not be called wrong, and, likewise, the Wiki-choice shouldn't be called right, just natural or appropriate or whatever, preferably with an explanation. It is a potential blunder to call, say, Weinberg, wrong. It is hard to find even the slightest typo is his books, errors in equations are almost unheard of, and would he make gross conceptual errors? (Having said that, his mathematical taste is horrendous )

So, I suggest, in the present case, that we motivate briefly our choice of definition (so that it goes into the top part of the article), and there also mention the alternative, with pop-up citations. This would increase the quality of the article, and keep the peace longer. YohanN7 (talk) 03:26, 4 July 2014 (UTC)

I think it is misleading to think of the volume form as a density in this context, since the "volume form" might refer to a scalar or a density, depending on what "volume form" means. If "volume form" means the wedge product of the differentials of the coordinate functions, then it is not coordinate invariant (it picks up a factor of the Jacobian). But of course, it does not do so in a way that results in an invariant integral. For that, one needs an actual scalar. This scalar is obtained by taking the product of this "volume form" with a density! This is what the slogan "densities have a well-defined integral" means.

I don't mean to say that Weinberg is mathematically wrong in his choice. I have been using "wrong" in quotes, because it is after all only my opinion. But no one so far seems to have produced a clear motivation for the opposite convention, aside from just appeal to authority (an authority which I think is fairly limited as far as it goes). Maybe Weinberg did have good reasons, but Weinberg is somewhat unconventional in his entire approach to GR, and he often uses the opposite conventions of others working in that area (cf. Penrose, Hawking, for example). I have no doubt that he had his reasons for choosing these conventions, but so far no one has been able to give a compelling account of those reasons. Sławomir Biały (talk) 12:45, 4 July 2014 (UTC)

I'm not getting my point through. I am not defending unconventional choices, nor do I say every choice is as good as any other (though I wrote something that could be misinterpreted to that effect above).

The point is this: The articles are supposed to reflect what is in the literature, not what we think ought to be in the literature. When there are multiple conventions around, the article should make its choice. At that point a motivation could well be given if there (like in our case) is a compelling reason for making the choice. Then, in the interest of completeness, list some other common choices (naturally without motivation unless there is one), and then get on with the article. This pattern reflects what it looks like in (good) textbooks as well. Alternatives are mentioned. But don't misunderstand me as proposing that every fringe convention should be mentioned. Is Weinberg's authority enough to make his choice of convention get mention in this case? Don't know. I haven't read his work on GR, just his QFT vol 1,2,3. (Yes, these were totally unconventional - and great.) Just trying to reason in general terms here. YohanN7 (talk) 15:31, 4 July 2014 (UTC)

I agree with Yohan completely. As an aside: I have been thinking and (I will probably be criticised severely for this), I think (personally) the name 'Tensor density' is a misnomer. The reason being is, when one hears the word 'density' or the words '____ density', our minds automatically think that this is something that (once integrated) is a physical thing, e.g. can be visualized or picked up...in other words one can interact with it, be it in implicit way.

For example mass density can be integrated to give a mass of a object that we can pick up...charge density can be integrated to give charge of an object which we can see (some of it) for example electric charge and electricity, etc.

But Tensors are purely abstract and mathematical concepts which can be written on paper and were INVENTED to understand GR yes, but one can't see tensors, pick them up, or interact with them at all in any way. (Like numbers for example...you cannot pick up 2)...[FYI I don't consider working with them on paper as interacting lol]

So by calling something a 'tensor density' one assumes that one can get a tensor which is a physical thing. But it's NOT. It makes more sense for it to be called just a 'weighted tensor'

What do you guys think? Or am I off my rockers? — Preceding unsigned comment added by 206.45.181.185 (talk) 04:13, 4 July 2014 (UTC)

Often it is said that definitions and conventions are arbitrary and that it does not matter which version one uses. This is wrong. It is important to make them as intuitive as possible, especially for people just learning the subject. Just imagine if when you swept your finger to the right over a cell phone, the image moved upward or spun around or shrank, rather than moving to the right.

Tensor densities (of weight +1) are called "densities" precisely because they are used as the mathematical representation of things which are called densities in ordinary life (classical physics). Sławomir Biały is right about that. Action is a scalar density. The density and current of electric charge form a vector density. It could be represented either as a contravariant vector density with 4 components or equivalently as a anti-symmetric three-times-covariant ordinary tensor with 64 (mostly redundant) components. JRSpriggs (talk) 04:36, 4 July 2014 (UTC)

Sorry but I disagree with you JRSpriggs: 1) Conventions ARE arbitrary and just because person A uses a different convention from person B doesn't make it wrong. As long as they are consistent with their conventions the physics that the two are describing are COMPLETELY EQUIVALENT. The math will be different but the answer/interpretation will be the same.

Consider the sign convention for the metric (-,+,+,+) vs (+,-,-,-). I personaly use the mostly plus convention, but as long as people are consistant they are both right. By saying that "Often it is said that definitions and conventions are arbitrary and that it does not matter which version one uses. This is wrong." you are saying that either ALL particle physicists are wrong or ALL relativity physicists are wrong, which (if you ask me) is arrogant.

Math IS arbitrary, numbers are arbitrary...Just look at complex numbers for example. You won't see a '2+i' walking down the street and smiling at you now would you? But we use numbers to describe the world, so how can it be wrong?

2) Action is not a density but rather a functional. You probably meant to say Lagrangian density. — Preceding unsigned comment added by 206.45.181.185 (talk) 05:07, 4 July 2014 (UTC)

Perhaps we should avoid the use of the terms "right" and "wrong" in this discussion. And I interpret JRSpriggs differently – he seems to be saying that we do ourselves a disservice if we choose to completely sweep the benefits of selecting certain conventions under the rug, with which I agree. Essentially, since the word "density" is used, it will trigger certain natural associations as its selection was presumably intended to do. Thus, I would expect the components a tensor density of weight +1 to scale exactly as the components of something that I would call a density, and hopefully this would suggest a convention.

This is where I fall off the wagon, though, since it seems to me that scalar densities (in the everyday sense) are invariant, and all tensors transform with weight 0. Thus, the concept of a "tensor density" as described in this article is an artefact of retaining a form of integration that is coordinate-dependent, and thus (quite unnecessarily) undermines the fundamental motivation of tensors: coordinate-independence. It leads to defining densities in the sense of this article (of weight +1?) as entities that have been scaled to allow integration (using what I would call a flawed form of integration, or more specifically, volume element) over a hypervolume region to yield an invariant quantity. This article describes tensor densities, but fails to explain why they have any value, or what motivates them. As far as I can tell, the motivation will relate to a specific form of integration, where it was presumably decided to retain a form that rescaled the units with a change of coordinates, and dumped the responsibility for compensating this on the quantity being integrated. I would encourage a knowledgeable editor to insert such a motivation into the article; surely it must be sourceable. —Quondum 18:48, 4 July 2014 (UTC)

Maybe I am interpreting JRSpriggs incorrectly, but it sounds like he is trying to state that there is the one and only convention. (Whatever that convention is) and all other conventions and by extention people who use these "wrong" conventions are stupid/wrong... (In analogy it sounds like religion - there is only one God and all other Gods are false Gods and if you belive in them then you go to hell). There are MANY physicists who work in the same feild (eg Gravity) who use different conventions.

• Steven Weinburg in his book Gravitation and Cosmology: Principles and Applications of the General Theory of Relativity uses the $(-,+,+,+)$ signature and says that g has weight $-2$
• Misner, Thorne, Wheeler in their book Gravitation uses the $(-,+,+,+)$ signature and says that g has weight $+2$
• Sean Carroll in his book Introduction to General Relativity Spacetime and Geometry uses the $(-,+,+,+)$ signature and says that g has weight $-2$
• Ray D'Inverno in his book Introducing Einsteins's Relativity uses the $(+,-,-,-)$ signature and says that g has weight $+2$

Now all these physicists understand and know GR and (more to the point) come to the same conclusions. I highly doubt anyone would say that any of these people are wrong simply because they use a convention that someone dislikes. Yet by JRSpriggs argument he is basically saying that people like me, Weinburg and Carroll who use the $(-,+,+,+)$ signature and uses the definition that g has weight $-2$ are wrong/stupid/etc... — Preceding unsigned comment added by 207.161.172.197 (talk) 02:32, 5 July 2014 (UTC)

I suggest you mellow your interpretation; it is a principle in Wikipedia to assume good faith on the part of other editors. I've given an interpretation that differs from yours and is compatible with none of these authors, conventions or you being slighted. In any event, Wikipedia does not go by what is "right" or "true", but rather by what is published by notable sources. Given that both conventions are clearly used by notable sources, you can proceed from the assumption that they should both be mentioned in the article. For internal consistency, however, the article does need to select one only of the prevailing conventions for its presentation of the details. A further principle is not to change the convention in use in an article without consensus. This essentially means that arguments about the best convention have little point other than to sway the consensus prior to any change. JRSpriggs and Sławomir have both expressed their preference, YohanN7 is essentially saying the same as I've said now, and as you will gather from what I've said above, I fail to grasp the utility of the concept of tensor densities, nor do I see a natural preference for either convention. Simply put, if you wish to change the article, you will need to persuade the other involved editors to agree to the changes that they are presently opposed to, and taking a combative stance is not going to help with that. Editing works best through mutual understanding and cooperation. As you settle in as a Wikipedia editor, I expect you'll learn to enjoy being a part of a mutual, cooperative endeavour that we all find rewarding. Your attention to detail and energy for researching specifics are a plus here. You may find Wikipedia:Five pillars and the numerous articles and essays that it links to illuminating. —Quondum 04:17, 5 July 2014 (UTC)

@Anonymous: The choice of signature is AFAIK arbitrary, while the choice of convention for tensor density isn't completely arbitrary. In the latter case there is a natural choice. The other choice isn't wrong because it will produce identical results (apart from a sign switch for weights). But it goes counter the intuitive when it comes to the word "density". The convention doesn't match the terminology. That is all there is to it. The reason this is so is made clear in this thread, especially in JRSprigg's posts.

There's nothing stopping you from editing the article and mentioning the other convention. I can't promise it wont be reverted, but you might as well try. But don't get the idea that there should be a parallel development of the two conventions (I'd be quick to revert) or anything like it. I'd try to describe it in words in one sentence with one reference (hint: Weinberg has higher weight than Carroll), the more economical, the bigger chance it lasts, but that's really up to you. YohanN7 (talk) 09:56, 5 July 2014 (UTC)

Actually, upon closer inspection, the article was already mentioning the other convention before the definitions. All that was needed was a reference. I added Weinberg (and got rid of a reference needed tag in the process). YohanN7 (talk) 11:45, 5 July 2014 (UTC)

I also added a brief motivation for the choice of convention in the article along with the difference in the other convention. The latter is contained in the popup reference.

Correct me if I'm wrong, but there is little conceptual difference and no substantial difference at all between the two conventions. The conceptual difference is that "ordinary" densities get weight +1 with our convention, and -1 with the other. Both are equally invariantly integrable since while our convention assigns the weight -1 to the "coordinate volume element" (wedge of coordinate differentials), the other assigns +1; hence the net result is a true scalar (vector, tensor) when multiplied together with the density.

Ironically, there is one possible convention that seems to me to be better than the present choice of ours. That would be to use the other convention, but with an additional flip in the sign of the weight. That way Jacobian determinants of the inverse transformation are kept out of sight, and ordinary densities are assigned weight +1. (Admittedly, this is our convention, but with the defining formula reformulated in a more natural way (Jacobian determinant of forward transformation.))

At the risk of making myself more unpopular than I already am, I'll (hesitantly) say that there seems to be very little real substance in what has been said (with close to religious conviction) in this thread. What it seemingly boils down to is where to put minus-signs and inverses. I feel little religiosity about having +1 for weight for an ordinary density though I prefer it before -1. A "weight" is anyhow something that is over, above and beyond the concept of a classical density.

So now I have screwed up badly YohanN7 (talk) 14:41, 5 July 2014 (UTC)

Not religious conviction, but usually on the literature the bare word "density" means "1-density". From this basic premise, the rest follows. Sławomir Biały (talk) 16:21, 5 July 2014 (UTC)

Ah, but that is far away from intuitive notions about classical densities and, furthermore, already has a choice of convention (ours) built in. The motivation would then be commonality, not naturality. But let us not argue . Article ok (last sentence in lead + popup)? YohanN7 (talk) 16:41, 5 July 2014 (UTC)

Sławomir, I can't shake the feeling that different concepts are being conflated under a shared name, causing a confusion. Looking at Density on a manifold, it seems to me that the concept is best understood in the absence of a metric tensor, and the crucial feature appears to be that a density is a quantity that can be integrated (intrinsically) over a region of the manifold. I deduce that this is a coordinate-independent concept, which puts it at odds with this article. If one defines a volume form in terms of the coordinates as say dx ∧ dy ∧ dz, one has an inherently coordinate-dependent volume form, but if volume is to be a coordinate-independent concept, one is forced to multiply this expression by the Jacobian determinant J, thus arriving at an expression v = J dx ∧ dy ∧ dz that is a coordinate-independent "density of volume" (though one needs an initial frame for a density reference). Now I am guessing here, but this invariant "density of volume" is a 3-form, and thus an ordinary tensor; this is deserving of the name "density". This is, however, not the concept represented by a "tensor density" of this article; this article considers the quantity dx ∧ dy ∧ dz as a density of weight ±1, and v as a density of weight 0. Ergo: the word "density" is being used with two distinct meanings. In light of this, I do not follow you when you say "the rest follows". —Quondum 18:54, 5 July 2014 (UTC)

Do you still have access to Lee's book? You can find that kind of density in chapter 14 (old ed). It allows you to do invariant integration on non-orientable manifolds. Sorry for interrupting. YohanN7 (talk) 19:06, 5 July 2014 (UTC)

Hi Yohan

Ok I will put at the bottom of the article a side note about the other convention and *hope* that it doesn't get deleted. (Though I suspect it will) Does that sound ok?

The article is updated already. YohanN7 (talk) 19:20, 5 July 2014 (UTC)

Yohan

I added a secton titled "Note about conventions" Is this fine? I hope it won't get deleted...though I bet JRSpriggs will delete it...lol

To Anon

Do you remember my item 3 above? You wrote "That I did not know and I apologize!" in response too it. You seem to understand the principle, but just don't care about following it.

Please don't do any more edits to the article before consensus is reached. YohanN7 (talk) 13:53, 2 July 2014 (UTC)

I'm not saying you are wrong (or right). As I suspected, this is much a matter of definition and convention. Even so, since you have been edit warring, the original version should stand as long as no consensus is reached. YohanN7 (talk) 14:43, 2 July 2014 (UTC)

To Yohan

Yes and I am sorry about that. This did last too long and should have stopped....I am sorry and I'll stop contributing to this page.

Now you screw up pretty badly. I had written "I'm not saying you are wrong (or right)." You changed that to "I'm not saying you are right." Here's diff: [1]

Do not ever edit what other people have written on a talk page! YohanN7 (talk) 19:18, 2 July 2014 (UTC)

Fine, I am sorry I'll stop contributing. — Preceding unsigned comment added by 206.45.181.160 (talk) 21:15, 2 July 2014 (UTC)