Talk:Stress–energy tensor/Archive 2

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Archive 1

Archive 2

In general relativity

Latest comment: 10 years ago44 comments6 people in discussion

The article states

"However, when gravity is non-negligible or when using arbitrary coordinate systems, the divergence of the non-gravitational stress–energy may fail to be zero. In this case, we have to use a more general continuity equation which incorporates the covariant derivative"

This is nonsense. The divergence is defined via the covariant derivative and the equation actually says that the divergence vanishes. DvHansen (talk) 21:11, 23 July 2014 (UTC)

I've corrected this in the article. —Quondum 22:33, 23 July 2014 (UTC)

It's still there. The divergence of the stress energy tensor is always zero (this can easily be seen by taking the divergence of the Einstein equation, actually that's how most textbooks motivate the use of the Einstein tensor instead of the Ricci curvature). Also the divergence is defined independent of a choice of coordinates as the contraction of the covariant derivative. The above quoted sentence makes no sense at all. DvHansen (talk) 23:04, 23 July 2014 (UTC)

Huh? It now reads "The divergence of the non-gravitational stress–energy is zero." – which is what you are saying. The equation has been restated as the covariant form, so is independent of coordinates. The partial derivative equation is now qualified to be usable for specific conditions, and for this the term divergence is not used. —Quondum 02:37, 24 July 2014 (UTC)

I'm sorry, I don't know how to quote the article correctly, please reformat if there is a better way. The article still states:

However, when gravity is non-negligible or when using arbitrary coordinate systems, the divergence of the non-gravitational stress–energy may fail to be zero. In this case, we have to use a more general continuity equation which incorporates the covariant derivative

0=T^{\mu \nu }{}_{;\nu }=\nabla _{\nu }T^{\mu \nu }=T^{\mu \nu }{}_{,\nu }+T^{\sigma \nu }\Gamma ^{\mu }{}_{\sigma \nu }+T^{\mu \sigma }\Gamma ^{\nu }{}_{\sigma \nu }

The formula is actually just a complicated way to write "div T = 0". Every statement of the quoted sentence is wrong. See e.g. O'Neill "Semi Riemannian Geometry" Ch. 3 the section "Some Differential Operators" and Ch. 12 section "The Einstein Equation". DvHansen (talk) 12:24, 24 July 2014 (UTC)

To DvHansen (sarcastically): So if I give a baseball kinetic energy by throwing it upwards, then its kinetic energy (which is non-gravitational after all) does not change even as it slows down and stops momentarily at the zenith of its trajectory? JRSpriggs (talk) 11:32, 24 July 2014 (UTC)

You try to use terms of Newtonnian mechanics in a relativistic setting. This must fail. Usually you would model the ball as a test particle, i.e. ignore it's energy/momentum in the sources of the gravitational field. If you want to include it you would have to solve the two body problem earth/ball. And then, yes: div T = 0 applies. This follows directly from the Einstein equations, just calculate. DvHansen (talk) 12:24, 24 July 2014 (UTC)

@DvHansen: My apologies; I was getting confused between two very similar passages. I agree with what you are saying and that the problem remains, and must be fixed. I only fixed the abuse of the term "divergence" in the one section. Feel free to edit the passage yourself (you should be bold anyway), but I'll also look at fixing it at some stage. —Quondum 15:31, 24 July 2014 (UTC)

@Quondum: OK, I think I fixed it. I'm not familar with how physicist do geometry so please have a look if I use the right "slang". And while we are at it: Do physicist really call the Christoffel symbols the gravitational force field? To me this sounds very odd. I think the "force" you would measure in experiments is the curvature (via the Jacobi equation). And the degrees of freedom are the metric. Even if we refer to alternative theories which use the connection as the degrees of freedom it still sounds "wrong" to me to call a coordinate dependend quantity "gravitational force". DvHansen (talk) 17:13, 24 July 2014 (UTC)

I think both of you are wrong to try to ignore coordinate systems. Any actual measurements or calculations have to be done (at some level) in a coordinate system. And I disagree with changing the meaning of "divergence" to be a synonym of "covariant divergence". Divergence is defined in terms of partial derivatives.

The point of my sarcastic comment is that by trying to only work with invariant quantities, you are effectively denying the existence of the gravitational field which cannot be handled that way. JRSpriggs (talk) 20:41, 24 July 2014 (UTC)

@JRSpriggs: Now that you have changed back the article it is clearly wrong and contradicts every geometry textbook available. Your private "definition" of the divergence is not even a map from vector/tensor fields to vector/tensor fields. In fact it's not even a defintion without specifying a chart.

And BTW, the metric is tensor field which is independent of any coordinates and uniquely determines the Levi-Civita connection which in turn uniquely determines the curvature. The gravitational field is completely independent of any choice of coordinates, GR would be falsified if it would depend on a choice of coordinates. DvHansen (talk) 21:27, 24 July 2014 (UTC)

Yeah, I'm afraid that is Einstein's equivalence principle: a "gravitational field" has no physical meaning. An accelerating observer experiences gravitation. Would we call it a gravitational field? In any event, even the calculation of a gravitational field, in the sense of a fictitious force experienced by an observer requires a world line, or family of world lines, to be specified. The Christoffel symbols do not qualify. Let's keep OR out of it. —Quondum 23:30, 24 July 2014 (UTC)

The metric tensor field is the gravitational field in usual terminology, and it certainly has physical meaning. There is a difference between actual gravitational fields and those obtained artificially be coordinate changes. See sections 81-83 in The Classical Theory of Fields, Landau and Lifshitz.

The Christoffel symbols can, by formal analogy (with Newton's second law and Newton's gravitational law), be interpreted as "field intensities" for the reason that they determine the motion of a test particle (via the invariant equation

Du μ = 0

, where

u

is 4-momentum and

D

is the covariant derivative). Here the components of the metric tensor field act as "potentials". See section 87 in The Classical Theory of Fields, Landau and Lifshitz. YohanN7 (talk) 13:25, 27 July 2014 (UTC)

What I can see in Google books looks like part of an interesting discussion on the topic. Your statement "The metric tensor field is the gravitational field in usual terminology, and it certainly has physical meaning" is not particularly helpful in the context. The metric has physical meaning, but this does not contradict what I was saying because it is using the phrase "gravitational field" in a completely different, and particularly sloppy, sense. The Classical Theory of Fields does say "a gravitational field is nothing but a change in the metric of space–time", which makes more sense than your "usual terminology". Your statement about the Christoffel symbols makes sense only with certain constraints on the choice of the coordinate system, which neither you nor the article's lead states. —Quondum 20:04, 27 July 2014 (UTC)

What are these "certain constraints" on the coordinate system that you talk about? YohanN7 (talk) 20:25, 27 July 2014 (UTC)

MTW lists five different candidates for "gravitational field", including the metric tensor and the Christoffel symbols, but adopts none exclusively. YohanN7 (talk) 21:28, 27 July 2014 (UTC)

Just an opinion: The problem with using Christoffel symbols is that they may not vanish even in flat space-time where one would expect that the "gravitational force field" is zero. Trying to get an intuition by identifying certain terms in the geodesic equation as a "force" it's certainly a good idea to use normal coordinates with proper time being the time like basis vector (the momentary rest frame of the observer). But then the Christoffel symbols reduce to (roughly) 0 + curvature time distance + higher order terms. So IMHO if one wants to introduce the term "force field" I would use the curvature. Also this is compatible with the terminology in gauge theories.

But before continuing the discussion: Within this article it doesn't matter. I just asked out of curiosity (and because it smelled fishy). We could just shorten the equation and not mention the Christoffel symbols at all. The interested reader will just have to click the reference to the covariant derivative. DvHansen (talk) 22:09, 27 July 2014 (UTC)

What does matter is pasting factoids uncited into articles. In an exploration of possibilities, I can posit by big toe as a candidate for a gravitational field. Either reference the phrase "which is the gravitational force field" or it must be removed. The onus is not on anyone to prove that it is wrong; I am challenging its notability. —Quondum 22:57, 27 July 2014 (UTC)

I agree with you both. It is fishy, but it remains true that the interpretation is presented often in the literature (therefore notable, either of L&L or MTW alone would be enough). It is based on an analogy with non-relativistic mechanics and gravitation. Then the case of the components (especially

g 00

) of the metric tensor field being gravitational "potentials" is more than a formal analogy, it is the classical limit (if I remember correctly), lending some more credibility to naming the Christoffel symbols "field intensities". (L&L use quotation marks.) I totally agree that the sentence in the article should be reworded (and sourced) or possibly removed. I don't agree with simply deleting unsourced, but sourcable material based on unmotivated guesses that it might be OR. Plenty of our articles would then lose more than 50% of their content. But, perhaps we don't need this in the present article?

@Quondum: What are the "certain constraints on the choice of the coordinate system" you referred to above? There is no mention of this in L&L. They distinguish though between cases where you can make the Christoffel symbols vanish (in an extended region) and when you can't. The latter is one criterium for an "actual" gravitational field. (An actual gravitational field must also solve the field equation, meaning there is a (reasonable!) stress-energy tensor for which the gravitational field (metric, curvature, whatever) is a solution.) YohanN7 (talk) 23:48, 27 July 2014 (UTC)

Where I'm coming from is this: a perfectly reasonable way to define a gravitational field would be what would be measured by a "static" observer. For example, measuring the gravitational field on, in and around the Earth can be done with respect to a (vector) field of "static" 4-velocities, which correspond to observers moving with the Earth. Given such a field, a gravitational field can be defined in a coordinate-independent fashion, and I strongly suspect that the result is a tensor of order 3. If we choose coordinates such that the spatial coordinates are correctly scaled, remain constant locally on and around every world line and do not rotate over time, and the time coordinate matches each observer's proper time, in this coordinate system the components of the gravitational field tensor as defined might match the Christoffel symbols, and we might be tempted to use them directly. This construction requires immense constraints on the coordinate system, and this is what I was referring to, and it is not clear that this coordinatization can be in an extended region, so the Christoffel symbols might be a poor match further afield due to necessary divergence from the constraint. The above construction of the tensor works globally, however, wherever one can define a reference 4-velocity field, or even a world line of a "static" observer. The converse is essentially as stated by DvHansen above: deliberately choose the coordinates so that the deviate from these constraints, as is easy to do even in flat space, and the short cut of using the Christoffel symbols will result in measuring the "gravitational field" of artificially accelerated observers, and will thus be measuring something quite different and coordinate-dependent, and in particular, a non-zero "gravitational field" in flat Minkowski space "generated" by coordinates that do not fit the constraints or the movement of any "static" observer. Thus, the Christoffel symbols are not, for general curvilinear coordinates, useful for interpretation as a gravitational field. But a coordinate-independent approach will yield a useful field with respect to any given "static" set of frames. What you refer to as 'an "actual" gravitational field' is, as DvHansen implies, presumably nothing other than the Riemann curvature tensor, but I'd rather not use the phrase in this sense. —Quondum 01:11, 28 July 2014 (UTC)

I suspected something like this. But, no, what you (and perhaps also I and DvHansen) think ought to be called a gravitational force field is unimportant. What matters is what is in the literature, and there the Christoffel symbols (not constrained to any particular coordinates) are often used for field strength in GR because they (and only they) have formal resemblance with a force field of non-GR theories, like the Newtonian field of gravity and the field strength tensor of EM. That is all there is to it. Whether these forces are real in any sense isn't an issue (for us to argue about in the article). It could be pointed out that they are coordinate dependent. It may be pointed out too that they can be made to vanish at any one point, even if there is an actual gravitational field ≈ non-vanishing Riemannn curvature tensor. YohanN7 (talk) 09:26, 28 July 2014 (UTC)

Note that in this context force field and field strength are not necessarily the same thing. A force field is a thing that when applied to a test particle gives a force. In the case of GR this is given by the Christofel symbols (i.e. you just move the christofel symbols to the right hand side of the geodesic equation). This not a properly covariant thing because because force is not a properly covariant thing in GR. A field strength simply is measure of the strength of a field. This is usually (in gauge theory) defined as the curvature of a connection. I.e. in GR this is the Riemann curvature.T R 12:00, 28 July 2014 (UTC)

Thank you for the clarifying remarks. I should perhaps have written "field strength of the "force field"" with or without some or all quotation marks. I am not trying to make a point that has physical relevance. I'm trying to make the point that the terminology exists in the literature and is based on a formal analogy as seen in the geodesic equation for a test particle, whether or not gravity is actually present (however that last clause is interpreted).

Question: Is it appropriate (albeit perhaps unconventional) to call the metric tensor field the gravitational field? It is certainly "gravitational" and a "field", and all quantities (AFAIK) of any relevance in GR can be derived from it. MTW explicitly avoids making a choice as to what is the gravitational field. I can certainly still live with the Riemann curvature tensor being the gravitational field and its components being the "field strength" of the gravitational field. You indicate that this is (close to?) standard in the mathematical literature. YohanN7 (talk) 13:00, 28 July 2014 (UTC)

No, I would not think that is inappropriate. In many contexts, the metric is treated as "the" dynamical field that gives rise to gravity.

In gauge theories the "force" is given by the curvature of the gauge connection. See David Bleecker: Gauge Theories and Variational Principles theorem 10.1.6. DvHansen (talk) 13:27, 28 July 2014 (UTC)

More conventionally that is called the "field strength" since it does not necessarily produce a force. (For many gauge theories it does though.) T R 14:01, 28 July 2014 (UTC)

Aargh. I'll accept that the term "gravitational force field" gets used in various senses by various authors. But do I need to point out that to define it here with a specialized meaning that is incompatible from author to author and that conflicts with the everyday usage without any form of clarification in an encyclopaedia article is a no-no? I still say we must remove the phrase. —Quondum 14:49, 28 July 2014 (UTC)

No, the meaning of the Christoffel symbols as being the the "gravitational force field" is not ambiguous (being incompatible between authors) as you suggest, see JRSprigg's post below that prints out in fine detail what has been said. It doesn't conflict with everyday usage since it appears to be the everyday usage. There are obviously other interpretations, but these are clearly fringe in the context because physics gets the upper hand in this subject. I can sympathize with your efforts to uniformalize terminology (here and elsewhere), but please, don't be a notational and terminological nazi. YohanN7 (talk) 15:33, 28 July 2014 (UTC)

Really? That equation is incorrect in general. It is only valid when the constraints on the coordinate system that I mentioned are met, which is a point that I've already made. —Quondum 15:49, 28 July 2014 (UTC)

No. It is an invariant equation of general relativity. It is the geodesic equation. YohanN7 (talk) 16:00, 28 July 2014 (UTC)

For clarity, YohanN7, you are saying that this equation is independent of the choice of coordinates; that means it is a tensor equation. The implication is that the Christoffel symbols define a true tensor. Since clearly I must be wrong, I will exit the discussion. After all, I have been told before that I do not know what I'm talking about. —Quondum 21:08, 28 July 2014 (UTC)

~~This comment is deleted out by its author. It unjustifiably puts Quondum (and (justifiably) me for writing it) in a bad light. It's all there in the history if you care.~~YohanN7 (talk) 10:47, 16 August 2014 (UTC)

That's the problem with this point of view. The right hand side of the geodesic equation

\nabla _{\dot {\gamma }}{\dot {\gamma }}=0

is zero. From a geometric point of view there is no force at all. Only if you choose coordinates you get a term that looks like a force but heavily depends on your choice of chart (you can make it vanish in any point). Lets consider a free non relativistic particle on a sphere. The Lagrangian is given by

L[\gamma ]=g({\dot {\gamma }},{\dot {\gamma }})/2

, g being the first fundamental form. In this case: would anyone call the Christofel symbols "force"? — Preceding unsigned comment added by DvHansen (talk • contribs) 16:24, 28 July 2014 (UTC)

The above post is entirely correct, except for the conclusion. The answer is definitely yes as there are sources enough to warrant calling it a gravitational force field. I din't invent the terminology, but I'll defend it on the basis that knowledgeable enough people use it. I also understand their reasons for using the terminology (as explained in detail by JRSpriggs below), so I don't dismiss it for your reasons. It's also not up to you and me. What figures in the relevant literature matters. YohanN7 (talk) 16:40, 28 July 2014 (UTC)

But there are other sources with a completely different point of view

R.M. Wald: General Relativity. "We have no meaningful way of describing gravity as a force" (Chapter 4.3) but notes that tidal force (i.e. curvature) is still meaningful.
F de Felice, C.J.S. Clarke: Relativity on curved manifolds. Distinguishes between three points of view (Chapter 3): The infinitesimal (there is no force), the local point of view (difference between nearby geodesics, i.e. again the tidal force), and the global (if there is a global chart) which has "little or no physical significance".
B. O'Neill: Semi-Riemannian Geometry - With Applications to Relativity. "In this way [Jacobi eqution] curvature, in its role as tidal force, replaces the Newtonian notion of gravitation. In general, an instantaneous observer measures gravity by the tidal force operator." (Ch. 12, section "Foundations").
R. Oloff: Geometrie der Raumzeit. No mentioning of "gravitational force" at all (but describes the Jacobi equation and calls it "Gezeitenkraft" (tidal force in German)). DvHansen (talk) 17:50, 28 July 2014 (UTC)

That there is no no general description of "force" doesn't preclude the literature from inventing one that matches the non-relativistic expressions. Thank you for the lecture. YohanN7 (talk) 18:48, 28 July 2014 (UTC)

The Christoffel symbols are the gravitational force field as can be seen from the equation for force

{\frac {dp_{\mu }}{dt}}=\Gamma _{\mu \nu }^{\lambda }p_{\lambda }{\frac {dx^{\nu }}{dt}}+F_{\mu \nu }q{\frac {dx^{\nu }}{dt}}+{\text{ other forces }}\,.

That is, the rate of change of momentum is the gravitational force field times the momentum times the velocity plus the electromagnetic force field times the charge times the velocity plus other forces. JRSpriggs (talk) 15:00, 28 July 2014 (UTC)

To see that the force equation in my previous comment is actually invariant, move the first term on the right side to the left and multiply by dt/dτ to get the tensor equation:

{\frac {Dp_{\mu }}{d\tau }}=F_{\mu \nu }q{\frac {dx^{\nu }}{d\tau }}+({\text{ other forces }}){\frac {dt}{d\tau }}\,

where "D" represents the covariant derivative along the particle's path.
Also notice that for massive particles

p_{\alpha }=m\,g_{\alpha \beta }\,{\frac {dx^{\beta }}{d\tau }}\,

although the force equation is also true for massless particles like photons. See User:JRSpriggs/Force in general relativity for more details. JRSpriggs (talk) 11:46, 29 July 2014 (UTC)

@JRSpriggs: Question: What is the divergence in spherical coordinates? It sure as hell is not

\partial _{r}+\partial _{\theta }+\partial _{\phi }

. (Which is what you seem to be claiming.)T R 20:26, 26 July 2014 (UTC)

Divergence is characterized by the divergence theorem#Mathematical statement which is stated in terms of partial derivatives. When passing to a more general curvilinear coordinate system, the integral will not be invariant unless the divergence is a scalar density. For the divergence to be a scalar density, the field to which the divergence operation is applied must be a contravariant vector density (e.g. electric current). If

{\mathcal {J}}^{\alpha }

is such a contravariant vector density, then

\operatorname {div} ({\mathcal {J}})={\frac {\partial {\mathcal {J}}^{r}}{\partial r}}+{\frac {\partial {\mathcal {J}}^{\theta }}{\partial \theta }}+{\frac {\partial {\mathcal {J}}^{\phi }}{\partial \phi }}

is indeed the divergence of that vector density in a spherical coordinate system. Although this seems to differ from what is said at divergence#Spherical coordinates, that is because that section is stated in terms of an ordinary contravariant vector. If you use the fact that

{\sqrt {\det {g_{\alpha \beta }}}}=r^{2}\sin \theta \,,

then

{\mathcal {J}}^{\alpha }=r^{2}\sin \theta \,J^{\alpha }

which yields

{\frac {\partial (r^{2}J^{r})}{r^{2}\,\partial r}}+{\frac {\partial (\sin \theta \,J^{\theta })}{\sin \theta \partial \theta }}+{\frac {\partial J^{\phi }}{\partial \phi }}

after converting the result back to an ordinary scalar. JRSpriggs (talk) 05:10, 28 July 2014 (UTC)

In other words, the divergence of a vector is not

(\partial _{r}+\partial _{\theta }+\partial _{\phi }).v

.T R 08:14, 28 July 2014 (UTC)

Moreover, you might want to read the articles you cite. The divergence theorem#Mathematical statement you cite quite clearly defines the divergence as a scalar (not a scalar density). (The volume form "dV" is the form that when integrated over a volume gives the volume of that volume, i.e.

dV=r^{2}\sin \theta drd\theta d\phi

in spherical coordinates not

drd\theta d\phi

) see volume element)T R 11:48, 28 July 2014 (UTC)

The ρ in the volume element is just another name for √|g|. The associative law of multiplication shows that we are saying the same thing

\int \left(f\,\rho \right)dx^{1}\cdots dx^{n}=\int f\left(\rho \,dx^{1}\cdots dx^{n}\right)\,.

The only difference is over whether ρ should be classified as part of the integrand (as I would have it) or part of the integral operator (as you would have it). I think that my version is better because it is more general — even in the absence of a metric (or the presence of 2 or more different metrics) one may be able to identify scalar densities which can be integrated. But your method only works when there is exactly one (or one preferred) metric. JRSpriggs (talk) 05:23, 31 July 2014 (UTC)

The integrand of a n-dimensional integral must always be a n-form. Integration of scalar quantities on any n-dimensional (orientable) manifold is possible as long as one specifies a top level n-form dV. The choice of a volume form also defines duality between the bundle (n-1)-forms and the tangent bundle (and between n-forms a line bundle). The divergence operator is simply the dual of the exterior derivative under this duality. Non of this requires the choice of a metric as you insist.

Moreover, the problem with your "definition" of integration (which amounts to picking the Lebesgue measure on whatever chart you are using) is that is only defined locally and may be ill-defined globally for manifolds with non-trivial topology.T R 09:26, 31 July 2014 (UTC)

Just to pile on: defining the divergence as simply the sum of partials seems exceedingly peculiar. Maybe there is some obscure textbook that insists it be defined that way, but per the principle of least astonishment we should stick to the usual definition: the divergence in curvilinear coordinates is, in general, a complicated expression, and reduces to the sum of partials only in Cartesian coordinates. Zueignung (talk) 15:43, 4 August 2014 (UTC)

Electromagnetic stress–energy tensor

Latest comment: 10 years ago32 comments4 people in discussion

The formula

T^{00}={\epsilon _{0} \over 2}\left({E^{2} \over c^{2}}+B^{2}\right)

seems, as User:Issaeghdami points out, wrong, whether you are supposed to divide by $c 2$ or not. Correct versions for this tensor component is given in Electromagnetic stress–energy tensor for SI units (see Wangsness) and for CGS units (see Landau & Lifshitz or MTW). YohanN7 (talk) 18:06, 30 August 2014 (UTC)

It is very tiresome having this same kind of argument over and over again, but here we go ...

MTW uses c=1 which I am not doing here, so their formula is irrelevant.

I am following the convention that the units of the purely spatial part of a tensor should agree with the SI units for corresponding quantity in classical physics. In this case, this means pressure (compressive stress) which has units of Newtons/square meter (which is the same as Joules/cubic meter). For each contravariant index which is temporal, one must multiply the units by a second/meter; and for each covariant index which is temporal, one must multiply the units by a factor of meter/second. Since T^{00} has two contravariant temporal indices, we must multiply the units by (second/meter)^2. Hence it is necessary to divide the energy density by c^2 to get the units right. In my opinion, this is the only logical way to do it.

However, I cannot be responsible for keeping all articles consistent with this convention against the desire of innumerable editors to change them to agree with the units used in their favorite reference. JRSpriggs (talk) 12:29, 31 August 2014 (UTC)

How is the choice "the purely spatial part of a tensor should agree with the SI units for corresponding quantity", the only logical way to do it? That choice is quite arbitrary, and other choices can be (and are) made which are just as consistent. The most obvious being "the purely temporal part of a tensor should agree with the SI units for corresponding quantity"T R 09:16, 1 September 2014 (UTC)

No. No argument. I made the simplest algebraic mistake imaginable when I divided my version by

c^{2}={\frac {1}{\mu _{0}\epsilon _{0}}}.

Well, I was tired

.

This raises a broader issue: consistent adoption of SI when we say "SI units". I regard measuring time in units of metres as being non-SI, so I think we need to pull a number of articles into alignment with what JRSpriggs says, including Electromagnetic stress–energy tensor. IMO, this should be put into MOS. —Quondum 14:06, 31 August 2014 (UTC)

Either way you write the tensor (with or without division by

c 2

), everything is in SI units. There is nothing wrong anywhere, so there is no need to change anything. Apparently this varies in the literature, so we don't have to change anything. I'm neutral in the matter, just saying it would be a waste of energy to do it (or mass if we divide by

c 2

:D) YohanN7 (talk) 16:25, 31 August 2014 (UTC)

True enough, no need; it is after all a matter of convention, and that has not been agreed on WP. As to it being in SI units, the same statement could be applied to measuring weight in kilograms, or energy in watts. —Quondum 17:26, 31 August 2014 (UTC)

It is not enough that each component be given in SI units. The components must work together so that the object transforms as a tensor.

To TimothyRias: The reason that we should go with the spatial part is as follows: When going from the 3+1 dimensional viewpoint to the 4 dimensional viewpoint, it is more natural to regard the time dimension as a special space dimension than to regard the space dimensions as special time dimensions. Also it is much more common for there to be a unique quantity representing the spatial part of the tensor than to have a unique quantity for the temporal part, so it reduces the ambiguity. For example, there is just linear momentum on one side, but one has to choose between energy and mass on the other side. JRSpriggs (talk) 11:05, 1 September 2014 (UTC)

I hope you are not saying that multiplication of a tensor by a constant renders it a non-tensor. YohanN7 (talk) 11:16, 1 September 2014 (UTC)

Whoa. Maybe I've missed something. JRSpriggs, the way I understood you, a displacement tensor would have units of metres for the spatial components and seconds for the time component. YohanN7, no this does not render it into a non-tensor: it still transforms correctly. I have seen many treatments that use this approach; the metric tensor is

diag(c 2,-1,-1,-1)

or similar. A displacement vector then has the form

(t, x, y, z)

, with everything perfectly SI. Let us not tolerate discrimination between time and space! —Quondum 14:58, 1 September 2014 (UTC)

To YohanN7: No, do not be absurd. The Maxwell stress tensor is the negative of the spatial part of the Electromagnetic stress–energy tensor. The Maxwell stress tensor has units of pressure (Newtons/meter^2). So if the time-time component of the Electromagnetic stress–energy tensor does not have units of mass density (kilograms/meter^3), it will get mangled into garbage when you transform to a different reference frame.

To Quondum: In an inertial frame of reference (free-falling, non-rotating Cartesian coordinates), the metric tensor is

diag(- c 2,+1,+1,+1)

because the spatial part is just the Kronecker delta in three dimensional space (which is why it does not appear explicitly in classical physics). The c² is there to compensate for the fact that time is measured in seconds rather than meters, and the minus sign is because of the special nature of time. Notice that this fits with what I said above, the units of the spatial part are 1, so the units of covariant time-time part are that multiplied by (meters/second)^2.

We can think of the metric as a machine which inputs two displacements (in whatever units they are given) and outputs the square of a distance (meters^2). JRSpriggs (talk) 19:54, 1 September 2014 (UTC)

I don't follow. Are you saying that we have "4-tensor" with a "3-tensor" inside and that the "3-tensor" will crack if we Lorentz transform (with the other convention)? YohanN7 (talk) 20:47, 1 September 2014 (UTC)

Sorry JRSpriggs, I see no logic there only personal preference. Preference for the one or the other, is similar to preferring -+++ or +--- for the metric signature. (Coincidentally, +--- would also suggest expressing the metric in units of time). Each has its own logic.T R 20:39, 1 September 2014 (UTC)

To YohanN7: Please do not insert your comments in the middle of my comments. I do not know what you mean by "crack".

To TimothyRias: (Sarcastically) Classical physicists will love your proposal to change the definition of the dot-product of two vectors by putting in a factor of -c^-2. And we will also be glad to change the definitions of gradient, cross-product, etc. as necessary to comply with your new metric tensor. JRSpriggs (talk) 12:52, 3 September 2014 (UTC)

By "crack" I mean the same thing (probably) as when you say "mangled into garbage". I'm asking, is there anything wrong with the other convention? You indicate that it is vital to have mass-density in the

T 00

-component, otherwise something will get mangled (crack) upon transformation of coordinates. Not clear if this is the

T 00

-component or the space-space part that is useless afterwards - or the whole tensor (in which case it isn't a tensor). I'm asking because I don't know, I'm not arguing one way or the other. YohanN7 (talk) 13:08, 3 September 2014 (UTC)

To be clear, I want to know if this whole thing is just another convention jihad (have seen a few around here by now) or if there are undisputed reasons to say that either convention is wrong in that it leads to incorrect results. YohanN7 (talk) 13:28, 3 September 2014 (UTC)

I think JRSpriggs merely explained that once you have chosen the ratio of units between space and time components for vectors, the unit ratios between components for all tensors are determined. Thus, displacement vectors being time-and-space forces the momentum four-vector to be kg-and-kg·m·s⁻¹, but also the momentum covector units being J-and-kg·m·s⁻¹ (each up to a total factor of units). If one does not stick to consistent unit ratios, things do not hang together. I think comments about sign conventions are confusing things; let's not side-track onto those. —Quondum 15:34, 3 September 2014 (UTC)

Sign-conventions? YohanN7 (talk) 15:37, 3 September 2014 (UTC)

diag(−c²,1,1,1) vs. diag(c²,−1,−1,−1) —Quondum 16:18, 3 September 2014 (UTC)

I didn't bring those up. YohanN7 (talk) 16:30, 3 September 2014 (UTC)

No, you didn't, but you were responding to a section in which J and T were including that. I think they agree on everything else here (but I could be wrong!). —Quondum 17:10, 3 September 2014 (UTC)

The relevant thing is that when you perform a coordinate transformation the components that get added together have the same units. Consequently, once you have determined the units of your coordinate system(s), this determines the relation between the units of different components of a tensor. If these relations are not obeyed, the thing is not tensor. Other than that you are free to apply an overall scaling to the tensor to get it in whatever units you like/are convenient. There is not even a need for a convention that specifies that the purely spacelike or timelike components should have the units of the corresponding classical quantity. In fact, MHO is that trying to impose such an artificial convention is just silly.T R 15:50, 3 September 2014 (UTC)

Yes, Q & TR explained it better than I did. When you apply the transformation matrix, you must apply the same matrix on each contravariant index (and its inverse to each covariant index). If the units of the products of factors which are part of the same sum are not expressed in the same units, then the resulting numbers are meaningless.

If I am not mistaken, TR is advocating that the metric tensor in an inertial reference frame be diag(+1, −c⁻², −c⁻², −c⁻²) [or possibly diag(+1, −1, −1, −1)]. Is that not correct? JRSpriggs (talk) 20:49, 3 September 2014 (UTC)

I am not advocating anything. I am point out that those are are equally valid conventions that are also commonly used. For the record, my personal preference is -+++ signature and using geometrized units (i.e. G=c=1). Actually, as I work in black hole perturbation theory, my units are also adapted to the length scale set by the black hole being studied. Preferences, however are just that preferences.T R 21:14, 3 September 2014 (UTC)

It feels as though there's been a lot of talk without a real disagreement, possibly people missing each other on the odd point. I conclude with respect to the original point that it comes down to: check what convention is in use (both units and signs), since there are several. —Quondum 00:01, 4 September 2014 (UTC)

I think it just confuses most people to have a host of different "conventions" instead of one conventional convention. And I think that for most purposes relativity should be a set of (usually minor) corrections to classical physics. That is why I think we should use the convention I suggested. It is the only one which does not require one to be constantly changing the scale of the numbers depending on what one is doing. JRSpriggs (talk) 00:59, 4 September 2014 (UTC)

I've got a feeling that your preference is the same as mine. Under a claim of being SI units, we would be restricted to something very close to this, and c = 1 or G = 1 would not apply. Under SI, mostly one would also be restricted to normalized basis vectors, which would be typical only in SR. If I understand you, a displacement vector would have units of s and m, the momentum four-vector would be kg and kg⋅m/s, and 4-velocity would be 1 and m/s? For clarity, I'm looking only at what falls under an explicit claim of "in SI units". —Quondum 03:06, 4 September 2014 (UTC)

It does confuse people to have different conventions around, but there is nothing to do about it. I even argue that we should definitely not do anything about it. Different units/conventions serve different purposes. In the context of engineering, the SI units are probably well suited. In the context of fundamental physics, they are probably not optimal. We just have to keep the articles internally consistent. Forcing one convention all over Wikipedia wouldn't reflect what is used in the real world.

What confuses people more, (and is such a total waste of everybody's time) is to argue, not only that "my convention is right", this is fine, but to argue that the other conventions are wrong. YohanN7 (talk) 07:58, 4 September 2014 (UTC)

It is inherent in the way WP works that we will have different conventions in different articles, and internal consistency per article is all that we can strive for – as we should well know. I think the more modest goal of identifying what is meant by a specific convention in the context of tensors would be more useful. If we divert onto the idea of uniformity across articles without even know what the convention in use is, we can be sure of chaos. In the above, I am trying to see whether there is something that can be called "tensor components expressed in SI units", but no-one seems interested in that. —Quondum 13:30, 4 September 2014 (UTC)

The answer to the question is no. (other of course than the components being expressed in SI units) The SI will not tell you whether the 00 component of the stress energy tensor should have units of mass or energy. T R 19:18, 4 September 2014 (UTC)

True, there is some degree of freedom, considered in isolation. Part of the apparent freedom comes from the fact that we are free to use mass or energy, which are both concepts in SI, and should be named accordingly (an energy of 5 picogram is not SI). However, for a consistent set of components of multiple tensors in physics that predominantly make sense in terms of SI units, one's choices become highly constrained, including this one. —Quondum 20:19, 4 September 2014 (UTC)

No, not really. The only constraint is that the units for the component for each individual tensor are consistent.T R 21:29, 4 September 2014 (UTC)

Once the units have been chosen for the metric tensor, the ratio of units between components within a tensor is determined for every tensor. Within this constraint, it remains to choose what the tensor represents, so that the components represent a real quantity, otherwise the label of "in SI units" does not apply: it does not suffice that the components have SI units. For all I can tell from your description, an energy–momentum vector with units of newton-candelas would be "consistent". —Quondum 01:28, 5 September 2014 (UTC)

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