Talk:Stress–energy tensor/Archive 1

Merge reproposed
Hi all, I just put in a merger template and discovered that Lethe had already independently made the same suggestion, but apparently the merger did not occur. Seems to me that this article is about the stress-energy tensor, so it should be called that, but there is already an existing article with some material worth keeping. Conclusion: the thing to do is to merge this article into the existing "stress-energy tensor" article.

FWIW, I am revising the gtr pages and plan to eventually have much better articles on Noether symmetry and all that.---CH (talk) 22:41, 11 August 2005 (UTC)

I added some stuff, but there really ought to be a separate article on plain old stress tensors as opposed to relativistic stress-energy tensors, as I know this is a pretty important subject in engineering, and the relativity stuff is completely irrelevant in that context. We could then refer to the stress tensor article when talking about the space-space components of the stress-energy tensor, which incorporates it. I'd even say that this is more important than beefing up the relativistic content, but unfortunately the relativity stuff is what I know a lot about. Are there any structural engineers out there? --Matt McIrvin 02:39, 17 August 2005 (UTC)

Notation
In the section As a Noether current, I'm unclear on the notation used in the expression
 * $$\nabla_b T^{ab}=T^{ab}{}_{;b}=0$$

I've never seen notation with the subscript ;b and I would have written the expression as
 * $$\partial_\mu T^{\mu\nu}=0$$

Jeodesic 13:16, 23 October 2007 (UTC)


 * The semi-colon indicates a covariant derivative. The covariant derivative is used frequently in general relativity. See mathematics of general relativity for further enlightenment (possibly). :) MP (talk•contribs) 15:27, 23 October 2007 (UTC)
 * I've added the statement $$\partial_\mu T^{\mu\nu}\ne0$$ to the section to make it clearer that we are not talking about the ordinary, non-covariant derivative. --Michael C. Price talk 07:02, 25 October 2007 (UTC)

Some Suggested Improvements
I think this article should link to a new article on conservation laws in general relativity, in which it should be stressed that the divergence law for the stress-energy tensor does not have quite the interpretation expected from flat spacetime. It is not really a conservation law anymore (because it does not account for energy/momentum exchanged between matter or nongravitational fields and the gravitational field itself).

Another point is that the stress-energy tensor plays a role in other metric theories of gravitation besides general relativity, so strictly speaking, I think this article really belongs in the category of things dealing with physical interpretation of Lorentzian manifolds. --CH

The article could certainly use a lot of additional content; some examples, some mention of the interpretation of its various matrix entries and why it's called the "stress-energy tensor", the ambiguity in its definition as a Noether current and the special role of the symmetric form in general relativity, etc. --Matt McIrvin 01:22, 9 August 2005 (UTC)

I do not understand the title of the section As a Noether current. I would expect a derivation of the energy-momentum tensor from a space-time transformation of a field, and I would pose it in the context of field theory, not general relativity. I would name this section under its current content something like "stress-energy momentum conservation in curved spacetime" or something like that. --Daniel 25 feb 2008 —Preceding unsigned comment added by 80.26.142.131 (talk) 18:02, 25 February 2008 (UTC)

relativistic implies homogeneous?
it's not obvious to me. what about a fast bound particle? Also, this isn't particular to field theories, is it? -Lethe | Talk 11:55, August 15, 2005 (UTC)

That a theory is symmetric under certain group doesn't mean that their states are invariant. The group of special relativity is generated by translations, rotations and boosts, but the state of, say, an atom, is obviously not invariant. For your example, translation invariance means that the bound solutions are the same no matter where you place the center of mass. Do not confuse "relativistic" in the sense of "in the framework of special relativity" with in the sense of "moving really fast".-Daniel —Preceding comment was added at 16:42, 2 March 2008 (UTC)

Lagrangian for an idealized fluid
What is the lagrangian for the idealized fluid, which leads to the stress-energy tensor given in the article:


 * $$T^{\alpha \beta} \, = (\rho + {p \over c^2})u^{\alpha}u^{\beta} + p g^{\alpha \beta}$$

--Michael C. Price talk 06:09, 30 July 2008 (UTC)


 * I tried looking for one, but the subject just kept getting more and more complicated. There does not seem to be any one definite idea of a perfect fluid (or ideal gas), but many variations. And the equations that I saw were not only not relativistic, they did not consider gross motion at all. I suspect that relativity is incompatible with the idea of an ideal fluid just as it is incompatible with the idea of an absolutely rigid rod.
 * Why do you need to know? Although one can derive a stress-energy from an action, one is not required to get it that way. If you really must have something, then you could try something like this
 * $$L = f(\sigma) \sqrt{-g} + \mu (c^2 + u^{\alpha} u^{\beta} g_{\alpha \beta}) + \lambda (\sigma u^{\alpha} \sqrt{-g})_{, \alpha} $$
 * where f is an unspecified function, &sigma; is a "particle density", and &mu; and &lambda; are Lagrange multipliers. The density and pressure would have to be determined after the fact by calculating the stress-energy tensor. JRSpriggs (talk) 01:53, 31 July 2008 (UTC)
 * One problem seems to be $$ u^{\alpha} u^{\beta} g_{\alpha \beta} = 1 \,$$. The general proof that  $$ T^{\alpha \beta} _{\ \ \ ; \beta} = 0  \,$$ uses the variational principle which implies that a lagrangian formulation exists.--Michael C. Price talk 13:25, 31 July 2008 (UTC)
 * I use a convention in which $$ u^{\alpha} u^{\beta} g_{\alpha \beta} = -c^2 \,$$ because at rest in an inertial frame, $$g_{0 0} = -c^2 \,$$ and $$u^0 = 1 \,.$$
 * For real fluids, the Lagrangian is probably extremely & prohibitively complex because such fluids are made up of particles and fields of many types in complex arrangements. The best we can do is to use simplifying assumptions, e.g. a fluid in equilibrium is isotropic in its comoving frame.
 * I have worked out the stress-energy for the hypothetical Lagrangian I gave above, if you are interested enough to justify showing you the lengthy derivation. JRSpriggs (talk) 13:47, 31 July 2008 (UTC)
 * I use the same conventions -- it was just a slip of the fingers that left the minus sign off. I'd be interested in seeing the derivation. --Michael C. Price talk 15:30, 31 July 2008 (UTC)

(unindent) First a caveat, even in the simplified thermodynamic model of a working fluid which uses the fundamental thermodynamic relation, there are two independent variables (out of five). Of these, I am using only one &mdash; &sigma; is proportional to 1/V. In other words, I am ignoring temperature and/or entropy. This Lagrangian
 * $$L = f(\sigma) \sqrt{-g} + \mu (c^2 + u^{\alpha} u^{\beta} g_{\alpha \beta}) + \lambda (\sigma u^{\alpha} \sqrt{-g})_{, \alpha} $$

allows the action to depend only on that one variable, &sigma;. The last term ensures that the particle density flows in the indicated direction and is conserved. The middle term ensures that the direction of flow is, in fact, a valid direction. Varying with respect to &mu; gives the constraint equation
 * 1. $$u^{\alpha} u^{\beta} g_{\alpha \beta} = -c^2 \,.$$

Varying with respect to &lambda; gives the constraint equation
 * 2. $$(\sigma u^{\alpha} \sqrt{-g})_{, \alpha} = 0 \,.$$

Varying with respect to &sigma; gives
 * 3. $$f'(\sigma) = \lambda_{, \alpha} u^{\alpha} \,.$$

Varying with respect to u&alpha; gives
 * 4. $$2 \mu \, g_{\alpha \beta} u^{\beta} = \lambda_{, \alpha} \sigma \sqrt{-g} \,.$$

Varying with respect to g&alpha; &beta; gives
 * $$\frac{\sqrt{-g}}{2} T^{\alpha \beta} = f(\sigma) {1 \over 2} g^{\alpha \beta} \sqrt{-g} + \mu u^{\alpha} u^{\beta} - \lambda_{, \gamma} \sigma u^{\gamma} {1 \over 2} g^{\alpha \beta} \sqrt{-g} \,$$

which simplifies to
 * 5. $$T^{\alpha \beta} = f(\sigma) g^{\alpha \beta} + \frac{2 \mu}{\sqrt{-g}} u^{\alpha} u^{\beta} - \lambda_{, \gamma} u^{\gamma} \sigma g^{\alpha \beta} \,.$$

Combining #4 and #3 and #1, we solve for &mu;
 * $$2 \mu \, u^{\alpha} g_{\alpha \beta} u^{\beta} = \lambda_{, \alpha} u^{\alpha} \sigma \sqrt{-g} = f'(\sigma) \sigma \sqrt{-g} \,$$


 * 6. $$\mu = {-1 \over 2 c^2} f'(\sigma) \sigma \sqrt{-g} \,.$$

Substituting #6 and #3 into #5, we get
 * $$T^{\alpha \beta} = f(\sigma) g^{\alpha \beta} - \frac{1}{c^2} f'(\sigma) \sigma \, u^{\alpha} u^{\beta} - f'(\sigma) \sigma g^{\alpha \beta} \,.$$

which can be rearranged to
 * 7. $$T^{\alpha \beta} = (- \frac{1}{c^2} f'(\sigma) \sigma) u^{\alpha} u^{\beta} + (f(\sigma) - f'(\sigma) \sigma) g^{\alpha \beta} \,.$$

Comparing this to the article, we see that the pressure is
 * 8. $$p = f(\sigma) - f'(\sigma) \sigma \,$$

and
 * 9. $$\rho + {p \over c^2} = - \frac{1}{c^2} f'(\sigma) \sigma \,$$

whence the mass density is
 * 10. $$\rho = - \frac{f(\sigma)}{c^2} \,.$$

If we substitute $$f(\sigma) = - c^2 g(\sigma) \,,$$ then we get
 * $$\rho = g(\sigma) \,.$$
 * $$p = c^2 (g'(\sigma) \sigma - g(\sigma)) \,.$$

To describe a fluid which is infinitely compressible, p=0, we can use $$g(\sigma) = \sigma \,.$$ Functions which increase faster than linear give positive pressure. OK? JRSpriggs (talk) 20:21, 31 July 2008 (UTC)
 * Thanks for that. --Michael C. Price talk 07:07, 1 August 2008 (UTC)

Noether
"The stress-energy tensor is the conserved Noether current associated with spacetime translations." Can someone provide (or at least cite) a derivation? Cesiumfrog (talk) 05:29, 10 August 2008 (UTC)


 * There are a several derivations (of varying levels of sophistication and breadth of application) given at Noether's theorem and its external links and references. Also see my derivation at Talk:Stress-energy-momentum pseudotensor. However, bear in mind that the "canonical stress-energy" given by Noether's theorem does not have all the properties we want in a stress-energy tensor. To get those desirable properties, you need to modify it according to Belinfante's method. Unfortunately, we do not have any account of that method yet. JRSpriggs (talk) 16:03, 10 August 2008 (UTC)

Pressure? Viscosity??
I am confused with the interpretation of some of the tensor elements. Pressure and viscosity? Aren't they just macroscopic manifestations of microscopic properties of matter? When you work witl viscous liquid, viscosity is the result of strong (..er than in e.g. water) bonds between molecules, it is not a property of the region of space where that liquid is currently located.

IOW: on microscopic level (atom....proton scale) there is no "viscosity" as a physical reality. Same goes for pressure.

Please, if you can clear up my confusion, please do so somewhere in the article. Thanks. 89.102.207.196 (talk) 00:17, 12 May 2008 (UTC)
 * See clarification in lead -- the s-e tensor describes matter. --Michael C. Price talk 09:25, 12 May 2008 (UTC)


 * Actually, it is the image which is wrong &mdash; it should say "shear stress" where it says "viscosity". Unfortunately, I cannot change the image.
 * As Mike said, the components of the stress-energy tensor measure non-gravitational properties, that is, not properties of spacetime itself but of its contents. The Einstein field equations connect the curvature of spacetime to this attribute of its contents.
 * Stress (both pressure and shear stress) does exist at a micro level. As particles cross the spatial boundary their momentum contributes to it. And the force fields, especially the electromagnetic field, have pressure and shear stress everywhere, even in vacuum. JRSpriggs (talk) 20:01, 12 May 2008 (UTC)
 * I changed "viscosity" to "shear stress" in the image. Bamse (talk) 10:42, 9 October 2008 (UTC)
 * To Bamse: Thank you for fixing the image. JRSpriggs (talk) 13:54, 9 October 2008 (UTC)

Why symmetric?
I came to this article to learn why it is important or useful to have the stress-energy tensor be symmetric, but the author seems to assume that the tensor is always symmetric and ignores the fact that the stress-energy tensor can be asymmetric when defined in terms of a Lagrangian density. See for instance the chapter on continuum mechanics in Goldstein's Classical Mechanics. Tpellman 14:01, 19 October 2006 (UTC)


 * In general relativity the stress-energy tensor has be symmetric to fit into the Einstein field equations. In other fields I think the reason why the asymmetric forms are not studied so much is that any asymmetric stress energy tensor can be symmetrised by the addition of a physically irrelevant term.  --Michael C. Price talk 14:38, 19 October 2006 (UTC)


 * If the stress-energy tensor is the result of varying the action with respect to the metric tensor, then it must be symmetric because the metric tensor is symmetric. Also, one can prove that the symmetry is equivalent to conservation of angular-momentum which is known to be a physical fact. JRSpriggs 03:39, 20 October 2006 (UTC)


 * The points JRSpriggs and myself have made are further elaborated on pages 562/3 of Goldstein's Classical Mechanics. Perhaps this should be mentioned in the article.  --Michael C. Price talk 07:56, 20 October 2006 (UTC)


 * When one defines the tensor $$t^{\mu\nu}$$ as the current leading to the conserved "charges" $$P^\mu$$ there remain an ambiguity since any tensor of the form $$\partial_\sigma A^{\sigma \mu \nu}$$ with $$A^{\sigma \mu \nu}$$ antisymmetric can be added to $$t^{\mu\nu}$$ leading to a divergenceless tensor whose spatial integral is still $$P^\mu$$. The noether-theoretic tensor obtained assuming an invariant lagrangian density is just one of the possible tensors, but it is not the most usefull, because it does not reflect the Lorentz symmetry. When one takes into account the full Poincarè group then the symmetric (Belinfante) tensor arises. Its generalization to curved spaces is the Hilbert tensor, and it is this tensor the one used in general relativity. So I think that the article cannot leave for the end the "many stress-energy tensors" without explaining why, while using them without distinction in the body of the article. It should be stated all of this at the vey beggining of the article. --Daniel —Preceding comment was added at 17:09, 2 March 2008 (UTC)
 * Since Tpellman's issue was thoroughly addressed here on the talk page four years ago, and there has been no further discussion, I'm deleting the sentence that says, "Some people[who?] have speculated that it could be non-symmetric."--75.83.69.196 (talk) 16:48, 21 August 2010 (UTC)

Stress-energy of spinor fields
A discussion (about whether the stress-energy tensor is symmetric) on my talk page led me to the realization that we need to add a new subsection in Stress-energy tensor on the stress-energy of a spinor field (spin 1/2, see Dirac equation) considered as a non-quantum field.

I looked at the Lagrangian (see Dirac equation) for the Dirac electron to see whether I could apply the Hilbert definition of stress-energy to it, but it does not contain the metric tensor explicitly (instead it appears implicitly via the Gamma matrices) so I could not do so. Does anyone know how to find it? JRSpriggs (talk) 04:36, 1 November 2010 (UTC)


 * I believe there is something called a spin connection current we could mention, which is an additional current that accompanies the non-spin current. I'll check my notes. --Michael C. Price talk 09:04, 1 November 2010 (UTC)


 * Since the Dirac equation implies the Klein–Gordon equation, we should be able to apply the Stress-energy tensor to it. Right? On second thought, the Lagrangians are different even though they both satisfy the KG wave equation, and that is what matters. JRSpriggs (talk) 23:28, 3 November 2010 (UTC)

Belinfante-Rosenfeld
Down near the bottom it says "See the article Belinfante-Rosenfeld stress-energy tensor for more details". It sounds as though there was once an article by that name. The article Weinberg-Witten theorem red-links it. The article Einstein–Hilbert action red-links "Belinfante-Rosenfeld tensor" under "See also". Has it been renamed, or deleted, or merged with "The Banana Boat Song"? 24.36.74.15 (talk) 04:55, 6 July 2008 (UTC)


 * I do not know, but I would guess that the article was never created. That subsection was added a long time ago by who has not made any contributions since November 2005. So unfortunately, we probably cannot ask him for a reference. I did see a reference in a paper on ArXiv which said "J.Belinfante, Physica 6 (1939) 887; 7 (1940) 449.". Apparently, Belinfante devised a method for taking non-symmetric stress-energy tensors produced by Noether's theorem and making them symmetric. JRSpriggs (talk) 11:44, 7 July 2008 (UTC)
 * Although he gives no references, J Leite Lopes in Gauge Field Theories An Introduction ISBN 0080265014 pg 54 ff, shows how to generate a symmetric SEM from an unsymmetric tensor -- I guess this is Belinfante's method. --Michael C. Price talk 12:51, 7 July 2008 (UTC)
 * Thanks for the leads. I have some reading to do, but my first impression is that the symmetrization involved is both arbitrary and dated. I wonder whether it would be better to create the article, or work-around/remove the red-linked references to the subject. 24.36.74.15 (talk) 14:03, 7 July 2008 (UTC)
 * The symmetrisation procedure seems forced, not arbitrary.--Michael C. Price talk 16:15, 7 July 2008 (UTC)
 * Yes, of course. "Empirical" or "ad hoc" would have better expressed what I was thinking. 24.36.74.15 (talk) 21:03, 7 July 2008 (UTC)

(unindent) There is nothing magical about the conserved current derived from Noether's theorem. It may not necessarily be the actual stress-energy tensor. In other words, the stress-energy tensor is a conserved current and Noether's theorem gives a conserved current, but there is no reason to think that they are the same unless there is only one conserved current. (Even so you are free to multiply by a non-zero constant factor.) However, since the stress-energy tensor is symmetric, then modifying the Noether current to make it symmetric (while preserving the conservation law) seems like a good idea. JRSpriggs (talk) 21:04, 8 July 2008 (UTC)


 * A Belinfante-Rosefeld article has now been created. Trodemaster (talk) 19:34, 16 June 2011 (UTC)


 * To Trodemaster: Thank you for creating Belinfante–Rosenfeld stress-energy tensor. JRSpriggs (talk) 23:20, 16 June 2011 (UTC)
 * Thanks seconded. Ta! -- cheers, Michael C. Price talk 23:51, 16 June 2011 (UTC)

Yesterday I spent all morning in the dusty bowels of our main library reading the Rosefeld paper. That was was what prompted me to create the  Belinfante–Rosenfeld stress-energy tensor page. I spent part of this morning reading the Belinfante paper, which thakfully is available on line. Because I had learned of the BR tensor from current textbooks, where the Belinfante construction is presented as being  a cookup, I  had not realized that Belinfante actually derived his modified tensor by evaluating the  Hilbert tensor in terms of varying the tetrad and the spin connection separately. Indeed he has the tricky algebra that evaluates the variation of the spin connection as a result of varying the tetrad frame while preserving the torsion-free condition. (I wasted the last  week trying to get this right.) Anyway, as a result I modified the] page so as to give both B and R equal billing for this insight that Belinfante-Rosefeld=Hilbert. Mike Stone (talk) 16:06, 17 June 2011 (UTC) (aka Trodemaster -I  changed mysigfile)

Dimension problem.
The Stress-energy tensor in GR can not have different dimensions for its components. The dimension of all of its components should be [J/m^3]. This is because the Ricci scalar and all components of the Ricci tensor by definition have dimension of [1/m^2]. The problem in this article steams from the wrong definition in the beginning, which states that x0=t. The correct definition should be x0=c.t. —Preceding unsigned comment added by 161.209.206.1 (talk) 22:44, 5 September 2008 (UTC)


 * You are mistaken. There is no necessity that all components of a tensor have the same units; that is a myth. In fact, choosing to use the same units for all components would be inconsistent with with the International System of Units to which I am trying to conform.
 * You are also mistaken about the Ricci tensor. If one chooses to use arbitrary units, i.e. dimensionless numbers, for all of the components of the position vector $$x^{\mu}\!$$ which would be most in accordance with the spirit of general relativity, the Ricci tensor would be dimensionless (units 1) because the Christoffel symbols are dimensionless in that case. (As you said, its trace, the scalar curvature, would have dimensions of inverse meters squared.)
 * If instead one uses SI units for the position vector, then the spatial components of the Ricci tensor would be in inverse meters squared, the time-time component would be inverse seconds squared, and the mixed components would be in inverse meter-seconds. (The units of the scalar curvature do not depend on which units one chooses for the position vector.) JRSpriggs (talk) 00:43, 6 September 2008 (UTC)
 * Nevertheless, why does the article use an apparently idiosyncratic definition of the position Four-vector? The Four-vector article states:
 * A point in Minkowski space is called an "event" and is described in a standard basis by a set of four coordinates such as


 * $$ \mathbf{x} := \left(ct, x, y, z \right) $$
 * Either way, wikipedia should be consistent.
 * --Michael C. Price talk 08:48, 6 September 2008 (UTC)
 * In as much as I am only one person, I cannot be responsible for fixing all articles in Wikipedia which have been afflicted by the myth. Let me put it the other way, why are some of the articles not conforming to SI units even though they often claim that they are? If you look at Classical treatment of tensors, you will see that tensors are very flexible. One can perform a coordinate transformation which rescales (changes the units) of a single component of the position vector without affecting the other components; and this will propagate to all other tensors without any difficulty. Why are people ignoring the fact that tensor-notation essentially incorporates the changing of units? Why are they doing manually what is best done automatically?
 * And since people around here are enamored of references to reliable sources &mdash; please see "Post, E.J., Formal Structure of Electromagnetics: General Covariance and Electromagnetics, Dover Publications Inc. Mineola NY, 1962 reprinted 1997." which explains the right way to handle the units of tensor's components. JRSpriggs (talk) 00:05, 7 September 2008 (UTC)
 * Whether you choose $$ x^0 = ct $$ or $$ x^0 = t $$ is not an SI issue; it is an issue of clarity and aesthetics. There is surely some benefit in giving all the spacetime coordinates the same dimensions, and in making all the metric components dimensionless.--Michael C. Price talk 06:26, 7 September 2008 (UTC)

Hi Mike, I guess, you are correct, as soon as the condition Dim(Rij)=Dim(R.gij)=Dim(k.Tij) holds for each ij-tensor-component involved. However, imagine how beautiful it would be to pool all dimensions and values out of all tensors and leave there just the “orientation”. In other words to “normalize” each tensor in EFE to standard physical dimension of [1] and Determinant=1. The dimension part can be done by using the Planck values (it should work for SI and any other presentation), and the value part is just normalization. As a result the factor of (-1/2) can disappear from the second term, and the coefficient k on the right site can be greatly simplified. Thank you very much for your explanation. Boris. —Preceding unsigned comment added by 72.87.242.181 (talk) 05:40, 9 September 2008 (UTC)
 * To MichaelCPrice: x0 is the time dimension. In practice, one measures time in different units than the three spatial dimensions. You want to compel the user of our equations to do an unnecessary conversion of units (multiplying by c) just to satisfy your aesthetic preferences. Face the fact that the metric is a field which varies from place to place and time to time. Since it is not constant anyway, there is no point in trying to set one of its components to one. Also, it gives the distance (or duration) associated with a pair of events, so its output is not dimensionless. Thus it is unreasonable to expect the components to have to be dimensionless. Again, please read my reference! JRSpriggs (talk) 13:42, 9 September 2008 (UTC)
 * Since it is not constant anyway, there is no point in trying to set one of its components to one. Eh? Misrepresenting my position does yourself a disservice.  Anyway we are wasting our time here; as I said, Wikipedia should be consistent - take the argument to four-vector. --Michael C. Price talk 05:03, 10 September 2008 (UTC)

My two standard reference books, Misner, Thorne & Wheeler (pg 51, Gravitation) Itzykson & Zuber (pg 5, Quantum Field Theory), define $$ x^0 = ct $$.--Michael C. Price talk 05:37, 10 September 2008 (UTC)
 * Add George Sterman, An Introduction to Quantum Field Theory, pg 4, ISBN 0521311322 to the $$ x^0 = ct $$ ref list. --Michael C. Price talk 17:09, 13 September 2008 (UTC)
 * MTW are just saying that "ct" is better than "ict" in the context of special relativity. I do not have Itzykson & Zuber. I do not understand what you mean about me mis-representing you. As for making Wikipedia consistent, I suggest that you change four-vector which is not on my watch list. I am trying to limit myself to fixing 200 or so articles. I cannot do everything. JRSpriggs (talk) 19:00, 11 September 2008 (UTC)
 * Misrepresentation was perhaps too strong a word; nevertheless : the metric can be dimensionless and non-constant.
 * In general relativity the local coordinates carry over from special relativity, so I don't accept your case that MTW's stance is limited to SR. Clue: the book is called "Gravitation"!
 * I shall not be changing four-vector, except to add references that support its current stance of $$ x^0 = ct $$, since that is supported by the literature. --Michael C. Price talk 20:24, 11 September 2008 (UTC)

The components of a tensor do not all need to have the same units in the context of GR. To suggest that they do is simply to show ignorance of GR. As others have correctly noted, GR allows arbitrary coordinate transformations. You can describe spacetime in spherical coordinates such as (t,r,theta,phi), in which case no tensor of rank 1 or higher will have consistent units for all its components. All sources on GR agree on this point. No, Wikipedia should not try to make all articles consistent on this point. In an article on Newton's laws, one should clearly expect that all three components of a vector will have the same units. In an article on GR, one should not -- and in fact the notation in this article would become horrifically unwieldy if one insisted on inserting factors of c all over the place.--75.83.69.196 (talk) 05:30, 4 October 2010 (UTC)
 * Spherical polars are always a pain, but in Cartesian coords there is no excuse for not setting x0=ct. That's where this debate started. Let's end it there as well. --Michael C. Price talk 08:46, 4 October 2010 (UTC)--Michael C. Price talk 08:46, 4 October 2010 (UTC)

I'd just like to point out, that the SI system or imperial units are just arbitrary. We defined the speed of light to be exactly 2.99792458e8 m/s. This is an arbitrarily choosen figure. It has nothing to do with physics, namely that there is a fundamental speed with which actions propagate. The same goes for all the other units we use. Thus the only usefull units to use there are fundamental units, like "time" (in whatever metric we measure it, "length" or "length/time = velocity". But since length is just "time · speed of light" any measure of length would be of the unit "time · c". IMHO this is the only usefull way do go about this. Physics doesn't care about the metric or the imperial system. 2001:4CA0:4103:1:219:99FF:FE58:B11F (talk) 14:06, 4 July 2012 (UTC)


 * If you want to use natural units rather than SI units, then you can mentally replace all the "c"s in the article with 1. However, the standard is SI units, both for historical reasons and because they fit better with human beings and our measuring instruments. And in SI units, the speed of light (c) is 299,792,458 metres per second exactly. So "c" should be kept in the article to leave each reader free to use whatever system of units he prefers. JRSpriggs (talk) 14:21, 4 July 2012 (UTC)

Formula for isolated particle
I believe that the formula given in the text for the stress-energy tensor of an isolated free particle is incorrect. It has a factor of $$ \gamma $$ in it (the relativistic time dilation/length contraction factor), but I believe this factor should be $$ \gamma^2 $$. This is because the Lorentz transformation of a second-rank tensor (which is how the formula is derived--taking the rest-frame stress-energy of $$ T^{00} = m v^0 v^0 = m c^2 $$ and boosting it--involves two factors of $$ \gamma $$, one for each index of the tensor. If desired, I can try to find a reference for this. --PeterDonis (talk) 22:53, 10 September 2009 (UTC)


 * No, you are ignoring the delta-functions which have to be adjusted by a factor of reciprocal gamma. If I had written the formula in a way which was more nearly manifestly invariant, it would be
 * $$T^{\alpha \beta}[t,x,y,z] = m \, \frac{d x^{\alpha}[t]}{d \tau [t]} \, \frac{d x^{\beta}[t]}{d \tau [t]} \;\, \frac{d \tau [t]}{d t} \frac{d \theta [x - x [t]]}{d x} \frac{d \theta [y - y [t]]}{d y} \frac{d \theta [z - z [t]]}{d y} $$
 * where $$\theta \,$$ refers to the Heaviside step function (indefinite integral of the delta function) and $$\tau \,$$ is the proper time measured along the world line of the particle. However, that would be much harder to understand and would still have notational problems. JRSpriggs (talk) 17:59, 11 September 2009 (UTC)

I see that there is some discussion around the importance of dimension and aesthetics but here, for the common reader, the description of $$v^\alpha$$ really does need to be consitent with the equation that is effectively in the same sentance. That is we should have $$v^{0} = c$$ —Preceding unsigned comment added by 62.25.106.209 (talk) 12:05, 24 May 2010 (UTC)


 * To 62.25.106.209: No. In this article, we are using t, not ct, as the time component of position. Thus the velocity vector's time component is 1, not c. JRSpriggs (talk) 23:30, 24 May 2010 (UTC)


 * Note JRSpriggs is incorrect to use $$x^0=t$$ rather than $$x^0=ct$$ ; not only doesn't his usage agree with the sources (e.g. MTW) and we get consequently get tensors of indefinte dimensionality, but it doesn't agree with the other linked articles. This has been explained before, but he doesn't listen. --Michael C. Price talk 10:19, 25 May 2010 (UTC)


 * To Michael C. Price: You are the one who does not understand. There is no requirement that a whole tensor have the same units. Merely that each component of the tensor have some definite units, and that the components fit together to make all the equations work. As I have explained before, the reference gives the best way to handle units in general relativity. JRSpriggs (talk) 16:57, 25 May 2010 (UTC)
 * Michael C. Price, you are quoting MTW as a source. How many pages of MTW have you read? Apparently not more than 20, because on p. 21 we have an expression involving t^2-x^2. This would obviously make no sense if t and x were taken to have different units, and the units were such that c was not equal to 1.--75.83.69.196 (talk) 05:53, 4 October 2010 (UTC)
 * Yes, and so? MTW are using geometrised units (page 38), so that the dimensions of the co-ordinates are all the same. --Michael C. Price talk 08:26, 4 October 2010 (UTC)
 * And, as has been explained before, even that out-dated reference doesn't support your idiosyncratic claims. No one has ever supported your claims here on Wikipedia. Doesn't that tell you something?--Michael C. Price talk 22:03, 25 May 2010 (UTC)
 * If you understand this stuff better than I do, then why is it that when you convert an article (e.g. Special relativity) from the system of units I prefer to the one which you prefer, I have to come in and correct the mistakes you make in applying your own system? JRSpriggs (talk) 20:44, 26 May 2010 (UTC)
 * Did I claim there was any lack of understanding involved? No. Please stop making up accusations and address the technical issues. Your only source does not support your usage, as has been pointed out before. --Michael C. Price talk 21:10, 26 May 2010 (UTC)


 * I assume that you are referring to the section of talk immediately above this one and/or to Wikipedia talk:WikiProject Physics/Taskforces/Relativity. You did not show that my "only source [E.J.Post] does not support" my usage. You merely noticed that he was aware, as I am, that our position is not the most popular one. However, our usage is more consistent with pre-relativistic and non-relativistic usages which are far more common than relativistic usage. Our usage is more consistent with the SI units which are the world standard for units. While your usage is, as far as I can see, merely based on the prejudice that all components of a tensor must use the same units; and forces one to make an unnecessary additional conversion of units before and after using tensors. JRSpriggs (talk) 10:51, 27 May 2010 (UTC)
 * It is a "prejudice" shared by all the sources.--Michael C. Price talk 16:45, 27 May 2010 (UTC)
 * No. For example, see Spacetime and Geometry: An Introduction to General Relativity by Carroll. See his derivation of the Schwarzschild metric, at "From these we get the following nonvanishing components of the Riemann tensor." He gives an expression for R0101 that involves a second derivative of a unitless quantity with respect to t. This has dimensions of inverse time squared. He also gives R0202 as r multiplied by the derivative with respect to r of a unitless quantity, so this would be unitless. R0101 and R0202 have different units, but they're components of the same tensor.--75.83.69.196 (talk) 06:09, 4 October 2010 (UTC)
 * See my response dated 08:38, 4 October 2010. -- cheers, Michael C. Price talk 20:49, 16 June 2011 (UTC)

I do not believe that MTW supports your claim about ct being preferred to t (they are saying that ct is preferable to ict which is true). They use natural units (in which c=1) in most of their book which obscures the issue between us. However, even if you could find a source which explicitly says that all components of a tensor must use the same units, this would be an argument from authority which is logically invalid. The real issue is which system makes more sense, i.e. is more in accord with the reality faced by physicists/engineers who have to use this stuff. We should continue this discussion on that basis. JRSpriggs (talk) 01:06, 29 May 2010 (UTC)
 * For tensors to have definite units obviously makes most sense. --Michael C. Price talk 01:31, 29 May 2010 (UTC)


 * Tensors with mixed units are somewhat awkward and have no advantages, so they should not be used. It would be good to ask other editors what do they think, to reach the consensus. --antiXt (talk) 10:39, 29 May 2010 (UTC)


 * First, it is necessary to understand that the general covariance of tensor equations in general relativity allows us to change the scale of each component of a coordinate system independently. (Indeed, one can do much more such as making the scale vary from place to place with some restrictions or mixing different coordinates together, i.e. oblique coordinates). Such changes of coordinates supersede the older scheme of choosing units for length and duration. In the older scheme, one person might choose to measure a road in statute miles while another person might choose to measure it in kilometers. To avoid confusion, it was useful to attach an indication of the unit chosen to the value. So instead of saying merely "3.8", one might say "3.8 statute miles". The person who prefers kilometers can read this as "3.8 &middot; (1.609344 kilometres)" which is the same as "6.1 kilometers". General covariance allows one use to use either 3.8 or 6.1 (provided all other points have their coordinate similarly adjusted).
 * Thus one's first thought is <>. This would lead to the metric tensor having units of meters2 for each component.
 * However, the next thought is <>. In effect, one is compelled to use the metric tensor (e.g. clocks and measuring rods) to help name the points with respect to which the metric will be measured. So we are back to using units for length and duration in specifying the components of a position.
 * Now observe that (regardless of the supposed inter-convertibility using c) the instruments available for measuring length are distinct from the instruments available for measuring time. So in practice, we cannot use the same unit for both length and duration. Thus in accordance with the International System of Units, I think that we should use the meter for measuring the spatial components of position and the second for measuring the time component of position. As noted above, there is no need to convert to any other system. And I see no advantage in doing so.
 * The only supposed advantage is the ease of raising and lowering indices of a tensor. But that is something which we should not be doing anyway because that hides the use of the metric tensor, and the whole point of general relativity is to make all use of the metric (gravitational potential field) explicit as we do with other fields such as the electromagnetic potential field. JRSpriggs (talk) 21:43, 29 May 2010 (UTC)
 * This is just one half-truth piled on top of another until we end with an absurdity. E.g. "instruments available for measuring length are distinct from the instruments available for measuring time".  Not so, since the SI metre is defined in terms of c and SI seconds. And so on. etc etc --Michael C. Price talk 06:36, 30 May 2010 (UTC)
 * To Michael C Price: This last comment of yours is not an argument. Now who is failing to "address the technical issues"?
 * "Defining" the meter in terms of the second using c is just guidance on how an apparatus could be constructed to calibrate instruments which measure distance. It does not mean that I could directly use a clock to measure how long my car is. JRSpriggs (talk) 23:44, 30 May 2010 (UTC)
 * Yes it does. Since c is defined as an SI conversion factor, simply measure how long it takes light to traverse the distance. --Michael C. Price talk 07:36, 31 May 2010 (UTC)

If you insist, you may change the article to your system of units. But, if you do, make sure that you do so completely and consistently. I expect that the result will be more complicated (not simpler) and less clear (not more clear). JRSpriggs (talk) 18:09, 1 June 2010 (UTC)
 * Thank you. However no promises re completeness and accuracy. A job 1/2 done is better than not done at all, IMO! --Michael C. Price talk 21:29, 1 June 2010 (UTC)
 * A job done without understanding the subject matter is much worse than a job done with an understanding of the subject matter. Michael C Price, different authors writing on the same subject will make different decisions on whether to use units in which c=1. Your attempts to list sources in favor of your position merely demonstrates your lack of awareness that there is no universally observed convention on whether or not to work in natural units. Units with c not equal to 1 are more commonly used in more elementary sources, or those that only touch briefly on relativity. Regardless of whether one takes c=1, it is not desirable or even possible to force all components of any given tensor in GR to have the same units.--75.83.69.196 (talk) 05:39, 4 October 2010 (UTC)
 * No, setting c=1 is not what this argument is about. That is a red herring.  Polar coordinates (as in your Schwarzschild metric example) are always tricky, but JRSprigges was arguing for x0 = t in cartesian coordinates, a practice that went out with Hermann Minkowski, let alone general relativity. --Michael C. Price talk 08:38, 4 October 2010 (UTC)

People are right to criticise the units. They are confusing, and need to be consistent throughout Wikipedia. Why do we have a covariant form in the diagram and a contravariant form in the text? Personally, I think there are some mistakes that need to be ironed out by the people who wrote the page, A.J.Owen — Preceding unsigned comment added by Ajoajoajo (talk • contribs) 16:19, 18 July 2012 (UTC) Ajoajoajo (talk) 16:28, 18 July 2012 (UTC)


 * This is the price you pay for having millions of editors rather than just one. I cannot even get agreement among the people who edit this article. How can I get agreement on what units to use throughout Wikipedia as a whole? As for the diagram, I do not have the ability to create a (possibly better) diagram, so I leave this one there as better than nothing. JRSpriggs (talk) 17:46, 18 July 2012 (UTC)

Wrong dimension of the Stress-energy tensor
The dimension of the Stress-energy tensor (Tab) defined here has dimension of [J/m^3]. It follows that the dimension of Einstein relativity equation in the main article becomes [1/m^2], and this is wrong. Regards, bspasov@yahoo.com — Preceding unsigned comment added by 98.154.17.2 (talk) 00:27, 10 April 2012 (UTC)


 * As I have argued at great length in some other places, the components of a tensor do not all have to have the same units. So to which component(s) of which version (covariant or contravariant or mixed) of the tensor are you referring? What do you mean by the "main article"; are you referring to Einstein field equations or some other article? JRSpriggs (talk) 11:26, 10 April 2012 (UTC)

The main article I am referring to is “Einstein field equations”. I agree that the components of a tensor do not have to have the same units. However, in the “Stress-energy tensor” article, all components of the Stress-energy tensor Tab (it is covariant, like in the main article) have the same dimension of [J/m^3]. When in the main article, substitute the dimension of Tab, the dimension of the field equation becomes [1/m^2]. — Preceding unsigned comment added by 98.154.17.2 (talk) 17:05, 12 April 2012 (UTC)


 * Please see Talk:Einstein field equations/Archive 1 and Einstein field equations.
 * Although some others here violently disagree, I have tried to follow the convention that the purely spatial components of a tensor should have the same units as the corresponding quantity in non-relativistic physics. The components involving time would then have those units multiplied by (second/meter)^(#contravariant_time_indices &minus; #covariant_time_indices). For the (covariant) metric tensor, the spatial components would have units of 1=(meter^2/meter^2).
 * The purely spatial components of the stress-energy tensor are the stress components whose units are the same as pressure, i.e. (kg m / s^2) / m^2 = kg / m s^2. Multiplying by 8piG/c^4 which has units of s^2/mkg gives 1/m^2 for the right hand side of the Einstein equation. On the left hand side, the Christoffel symbols have units of 1/m because they involve a derivative with respect to a spatial coordinate and the metric units cancel since one factor is covariant and the other is contravariant. The Riemann-Christoffel tensor has units of 1/m^2 because it is made from derivatives of the Christoffel symbols and products of two Christoffel symbols. Ricci tensor and thus the Einstein tensor are the same. Thus the equation balances with spatial components on both sides having 1/m^2 as their units.
 * Notice however, that the covariant time-time component of the stress-energy tensor is c^4 times the contravariant time-time component, not merely c^2 times it as you thought. JRSpriggs (talk) 07:07, 13 April 2012 (UTC)

Thank you, Sir. I do appreciate your explanation. — Preceding unsigned comment added by 98.154.17.2 (talk) 23:19, 13 April 2012 (UTC)


 * The above assumes a Cartesian coordinate system and SI units. If you want to use cylindrical or spherical coordinates (e.g. for a Schwarzschild metric), then multiply the above units by (radian/meter)^(#contravariant_angular_indices &minus; #covariant_angular_indices).
 * On the other hand, if you want to use a coordinate system in which each of the coordinates is regarded as a dimensionless number (so called "arbitrary units"), then instead multiply the units of the non-relativistic equivalent of the spatial part of the tensor by (1/meter)^(#contravariant_indices &minus; #covariant_indices). In this case, all of the components will have the same units. This is close to the system preferred by my opponents, except they use natural units instead of SI units. While this may be preferable for theoretical physics, I think that it causes problems for practical applications. In any real world situation, the coordinates themselves will not be arbitrary but must be defined by some physical process. Thus I believe that it is more realistic to expect that the coordinates will have non-trivial units (e.g. meters, seconds, or radians) arising from the process by which they are determined. JRSpriggs (talk) 08:55, 16 April 2012 (UTC)


 * I am sorry to have to disagree--this is really not sensible! Even though in practice you can work that way, it is unnecessary, confusing, and not common practice! When teaching this material--and I do--I would always use $x_0=ct$, and the value of $T_{00}$ is multiplied by $c^2$. I know of no textbook (apart from those working with units in which $c=1$) where definition of this ilk are used. The reason is simple, and can already be seen in special relativity: If one would like a dimensionless Lorentz transformation, then all components of any tensor must have the same dimension. If one wants to be able to think of Lorentz transformations as astract mathematical objects, this is the right thing to do! Nielswalet (talk) 15:40, 13 December 2012 (UTC)


 * Agreed - as far as I can tell most people write all components of 4-vectors/tensors in the same units by convention, but there is no reason to state they could all be different units. I.e. include both statements... currently the article has the tensor components all in different units.... Maschen (talk) 17:12, 13 December 2012 (UTC)


 * Carrying around factors of c2 on the metric seems highly nonstandard. What sources follow this convention? Zueignung (talk) 20:09, 13 December 2012 (UTC)


 * I didn't necessarily mean the metric - I just meant that the entire 4-vector or tensor use the same units. Nevertheless - Penrose (Road to reality) does use factors of c±2 placed in the metric, to remove the factors from the 4-vectors... If that's what you mean. Maschen (talk) 20:44, 13 December 2012 (UTC)

Tensor diagram in lead
Concerning the up-and-coming WP:MOSPHYS, I'll move the tensor diagram back and the GR template lower down. It was raised above (end of section Formula for isolated particle) by Ajoajoajo and JRSpriggs to have contravariant components in the picture instead of the currently covariant components. Shall I create another image like that with the correct indices? (Creating another image rather than just changing the current one allows both covariant and contravariant forms to be used later if needed). M&and;Ŝc2ħεИτlk 13:59, 1 March 2013 (UTC)


 * The diagram is not consistent as it stands. I suggest that you revise it to agree with Stress–energy tensor. If you want a diagram for covariant indices, you should not use the existing one since it is wrong. The labels on the existing diagram might be appropriate for the mixed tensor version (1 covariant index and 1 contravariant index). JRSpriggs (talk) 14:58, 1 March 2013 (UTC)
 * Tij = Tij = Tij = the flux in the j direction of the i component of linear momentum;
 * Ti0 = the density of the i component of linear momentum;
 * T0j = &minus; the flux in the j direction of energy;
 * T00 = &minus; the density of energy;
 * Ti0 = T0i = &minus; the flux in the i direction of energy; and
 * T00 = c2 &times; the density of energy. JRSpriggs (talk) 18:34, 1 March 2013 (UTC)


 * Ok, will get to it. Before creating any diagrams, perhaps we could just use LaTeX like so;
 * $$\begin{align}

T^{\mu\nu} & = \left(\begin{array}{c|ccc} T^{00} & T^{01} & T^{02} & T^{03}\\ \hline T^{10} & T^{11} & T^{12} & T^{13}\\ T^{20} & T^{21} & T^{22} & T^{23}\\ T^{30} & T^{31} & T^{32} & T^{33} \end{array}\right) \\ & = \left(\begin{array}{c|c} T^{00} & T^{0i} \\ \hline T^{i0} & T^{ij} \\ \end{array}\right) \\ & = \left(\begin{array}{c|c} c^{-2} \cdot \text{(energy density)} & \text{component}\,i\,\text{of momentum density} \\ \hline \text{component}\,i\,\text{of momentum density} & \text{flux in}\,j\,\text{direction of momentum component}\,i \\ \end{array}\right) \end{align} $$


 * Although the code is quite complicated, anyone proficient in LaTeX can fix if incorrect. Colours can be added if desired. Unfortunately the full equation large. If compact and colourful labelling is preferred, I'll proceed with the diagram. M&and;Ŝc2ħεИτlk 22:58, 1 March 2013 (UTC)


 * I added a minus sign in the exponent c&minus;2 in your expression above. Whether the time-time component is energy-density multiplied by +c&minus;2, &minus;1, or +c+2 depends on whether it is given as contravariant, mixed, or covariant respectively. JRSpriggs (talk) 08:40, 2 March 2013 (UTC)


 * Ok, thanks - but should we use LaTeX or a diagram? M&and;Ŝc2ħεИτlk 08:44, 2 March 2013 (UTC)


 * Well, I am not up to making a diagram. So it is up to you whether you want to go to that much trouble.
 * The array you gave here has most of the information. Aside from pretty colors, the only extra thing the diagram gives us is the division of stress between shear stress and pressure. JRSpriggs (talk) 08:50, 2 March 2013 (UTC)


 * Apologies for the delay, here is the contravariant type:




 * Any modifications please say. Thanks, M&and;Ŝc2ħεИτlk 09:28, 2 March 2013 (UTC)


 * It looks good to me. Thank you. JRSpriggs (talk) 09:49, 2 March 2013 (UTC)
 * No problem - let's replace the current version. M&and;Ŝc2ħεИτlk 10:01, 2 March 2013 (UTC)

History
Can some one write a section about history of this tensor, who is the first mentioned or invented it? Thank! (Did Einstein first?) 113.160.85.34 (talk) 05:53, 5 January 2013 (UTC)


 * A history section would be nice. Here is a nice pdf from arxiv.org:
 * and another from google:
 * I will write a section in time after looking around more, unless someone beats me to it. M&and;Ŝc2ħεИτlk 13:33, 16 June 2013 (UTC)
 * I will write a section in time after looking around more, unless someone beats me to it. M&and;Ŝc2ħεИτlk 13:33, 16 June 2013 (UTC)
 * I will write a section in time after looking around more, unless someone beats me to it. M&and;Ŝc2ħεИτlk 13:33, 16 June 2013 (UTC)

Conservation of energy and momentum
I have a simple question concerning the conservation law of the energy-momentum tensor, and I cannot figure out where did I make the mistake in the following calculations. Many thanks in advance!

Let us discuss a simple example. Considering a single free particle with 4-momentum $$p^\mu$$ in Cartesian coordinates. In the following, we will try to verify explicitly for a closed surface, one has

I=\int d\sigma_{C\mu} T_C^{\mu 0}= 0 $$ where the closed 3-surface $$\sigma_\mu$$ is defined by $$\tau=\tau_1,\tau_2$$, $$\eta=\eta_1,\eta_2$$, $$x=x_1,x_2$$ and $$y=y_1,y_2$$ with the following definitions

\tau=\sqrt{t^2-z^2} $$

\eta=\frac{1}{2}\ln\frac{t+z}{t-z} $$ Since the energy momentum tensor is evaluated in the Cartesian coordinates, the expression is not covariant, the vanishing contraction corresponds to the energy conservation in Cartesian coordinates. We use the following notations

I=I_\tau+I_\eta+I_x+I_y $$

As shown below, the 3-surface element at $$\tau=const$$ reads

d\sigma_{C\mu}= dxdyd\eta \left( \begin{array}{c} \tau\cosh\eta\\ 0\\ 0\\ -\tau\sinh\eta \end{array}\right) $$ and the integral on the 3-surface $$\tau=\tau_1, \tau_2$$ yields

{I}_\tau=\int_{\tau_1}\tau dxdyd\eta(\cosh\eta T_C^{00}-\sinh\eta T_C^{30})-\int_{\tau_2}\tau dxdyd\eta(\cosh\eta T_C^{00}-\sinh\eta T_C^{30}) $$

Similarly, the 3-surface element at $$\eta=const$$ reads

d\sigma_{C\mu}= dxdyd\eta \left( \begin{array}{c} \sinh\eta\\ 0\\ 0\\ -\cosh\eta \end{array}\right) $$ and the integral on the 3-surface $$\eta=\eta_1, \eta_2$$ yields

{I}_\eta=\int_{\eta_1}dxdyd\tau(\sinh\eta T_C^{00}-\cosh\eta T_C^{30})-\int_{\eta_2}dxdyd\tau(\sinh\eta T_C^{00}-\cosh\eta T_C^{30}) $$

Expressions for the other two 3-surfaces are more intuitive, one has

{I}_x=\int_{x_1}dtdydz T_C^{10}-\int_{x_2}dtdydz T_C^{10} $$ and

{I}_y=\int_{y_1}dtdxdz T_C^{20}-\int_{y_2}dtdxdz T_C^{20} $$

The energy momentum tensor of a point-like particle $$q$$ in Cartesian coordinates possesses the following form

T^{\mu\nu}_C = \frac{p^\mu p^\nu}{p^0}\delta^3(\vec{x}-\vec{x}_q)= \frac{p^\mu p^\nu}{p^0}\delta(x-x_q)\delta(y-y_q)\delta(z-z_q) $$ where

\delta(z-z_q)=\frac{\delta(\eta-\eta_q)}{\tau\sinh\eta}=\frac{\delta(\tau-\tau_q)}{\cosh\eta} $$ depending on whether one deals with 3-surface parameterized in z, $$\eta$$ or $$\tau$$. Substituting into the expressions of $${I}$$, one gets

{I}_x=\int_{x_1}dtdydz T_C^{10}-\int_{x_2}dtdydz T_C^{10}=\int dt(p^1-p^1)=0 $$

{I}_y=\int_{y_1}dtdxdz T_C^{20}-\int_{y_2}dtdxdz T_C^{20}=\int dt(p^2-p^2)=0 $$

{I}_\tau= \int_{\tau_1} d\eta (p^0{\coth\eta} - p^3)\delta(\eta-\eta_q)-\int_{\tau_2}d\eta (p^0{\coth\eta} - p^3)\delta(\eta-\eta_q)=\left\{\begin{matrix} p^0(\coth\eta_a-\coth\eta_b)&\textrm{the world-line goes across both surfaces}\\ 0&\textrm{the world-line goes across neither surface}\\ \textrm{other values}&\textrm{other cases} \end{matrix}\right. $$

{I}_\eta=\int_{\eta_1}d\tau(p^0\tanh\eta - p^3)\delta(\tau-\tau_q)-\int_{\eta_2}d\tau(p^0\tanh\eta - p^3)\delta(\tau-\tau_q)=\left\{\begin{matrix} p^0(\tanh\eta_1-\tanh\eta_2)&\textrm{the world-line goes across both surfaces}\\ 0&\textrm{the world-line goes across neither surface}\\ \textrm{other values}&\textrm{other cases} \end{matrix}\right. $$ where $$\eta_a, \eta_b$$ are the values of $$\eta$$ coordinates when the particle does go across the surfaces $$\tau_1, \tau_2$$. Among the four possible surfaces discussed above, one notes that the particle always goes across two of them. In practice, one may always choose a big enough area of $$\tau_1, \tau_2$$ surface, so that the particle goes across $$\tau_1, \tau_2$$ surfaces and, one ignores the integral on the $$\eta_1, \eta_2$$ surfaces. In a more general case, if the system consists of many point-like particles, one can usually find big enough area of $$\tau_1, \tau_2$$ so that all the particles go through the $$\tau_1, \tau_2$$ area. One sees that the above expression does not vanish as expected from the conservation of energy.


 * corrections: The expression of :$$\delta$$ function above is not correct, it should consider contain the information of the world-line of the particle, but the above expression does not. The correct expression is

\delta(z-z_q)=\frac{\delta(\eta-\eta_q)}{\tau\cosh\eta-v_z\tau\sinh\eta} $$
 * when parameterized in term of :$$\eta$$. The final result is, of course, :$$I=0$$.

Surface element and initial energy
To evalute the surface element, one may transform (the orientation of) a surface element in hyperbolic coordinates $$\tau=const.$$, which is a unitary vector in hyperbolic coordinates, into Cartesian coordinates (here we use :$$B$$ to indicate hyperbolic coordinates and :$$C$$ to indicate Cartesian coordinates.)



e_{B}^{\mu}= \left( \begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right) $$



e_{C\mu}=\eta_{\lambda\mu}e_{C}^{\lambda} =\eta_{\mu\lambda}\frac{\partial x_C^{\lambda}}{\partial x_B^{\nu}}e_{B}^{\nu} =\left( \begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right) \left( \begin{array}{cccc} \cosh\eta & 0 & 0 & \tau\sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \sinh\eta & 0 & 0 & \tau\cosh\eta \end{array}\right) \left( \begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right) =\left( \begin{array}{c} \cosh\eta\\ 0\\ 0\\ -\sinh\eta \end{array}\right) $$

It is straightforward to verify at this point $$|e_{C\mu}|=|e_{B\mu}|=1$$. The area of the surface element can be calculated as following



d\vec{r}_1\equiv \frac{\partial x_C^{\mu}}{\partial \eta}d\eta =\left( \begin{array}{c} \tau\sinh\eta d\eta \\ 0 \\ 0 \\ \tau\cosh\eta d\eta \end{array}\right) $$



d\vec{r}_2\equiv \frac{\partial x_C^{\mu}}{\partial x}dx=\left( \begin{array}{c} 0 \\ dx \\ 0 \\ 0 \end{array}\right) $$ and



d\vec r_3\equiv \frac{\partial x_C^{\mu}}{\partial y}dy=\left( \begin{array}{c} 0 \\ 0 \\ dy \\ 0 \end{array}\right) $$ for completeness, we also write down here



d\vec{r}_0\equiv \frac{\partial x_C^{\mu}}{\partial \tau}d\tau =\left( \begin{array}{c} \cosh\eta d\tau \\ 0 \\ 0 \\ \sinh\eta d\tau \end{array}\right) $$



$$
 * d\sigma_{C\mu}|_{(\tau=const.)}\equiv |d\sigma_{C\mu}|=|d\sigma^{\mu}_{C}|=|d\vec{r}_1||d\vec{r}_2||d\vec{r}_3|=\tau  dxdyd\eta

which had used the fact that the three vectors are mutually orthogonal. In fact, this also implies



$$
 * d\sigma_{B\mu}|= \tau dxdyd\eta \rightarrow |d\sigma_{C\mu}|=\tau dxdyd\eta

where $$d\vec r$$  are the 4-vectors in Cartesian coordinates. One sees that the size of the surface element defined by $$dx, dy, d\eta$$  does not change.

As a side note, in the above, the surface element was calculated by (1) assuming that it s a vector, therefore its components in Cartesian coordinates was obtained by coordinate transformation (2) the module of the surface element was obtained by using the fact the surface $$\tau = const.$$ is parameterized in terms of orthogonal coordinates, therefore all displacements corresponding to each variables ($$\tau,\eta,x,y$$) are orthogonal, the total area of the 3-surface element is simply a product of the module of three displacements.

On the other hand, the surface element can be calculated by using some more general formulae, which reads

d\sigma_{\mu} = \epsilon_{\mu\alpha\beta\gamma}e_1^\alpha e_2^\beta e_3^\gamma d^3y $$ where $$\epsilon_{\mu\alpha\beta\gamma}=\sqrt{-g}[\mu,\alpha,\beta,\gamma]$$ is the Levi-Civita tensor, $$y_1,y_2,y_3$$ parameterize the 3-surface, and $$e_1^\alpha=\partial x^\alpha/\partial y_1,e_2,e_3$$ describe the connection between the coordinates and its parameters. For instance, see the discussion near (3.2.2) of. In our case of the Cartesian coordinates, $$\sqrt{-g}=1$$, therefore $$d\sigma_{C\mu}=[\mu,\alpha,\beta,\gamma]e_1^\alpha e_2^\beta e_3^\gamma d^3y$$. It is straightforward to verify that one obtains the same result, namely,

d\sigma_{C\mu}= dxdyd\eta \left( \begin{array}{c} \tau\cosh\eta\\ 0\\ 0\\ -\tau\sinh\eta \end{array}\right) $$

The energy momentum tensor
Up to this point, we have not discussed the explicit form of energy momentum tensor. According to ,
 * $$\begin{align}

T^{\mu\nu}_C = \sum_i\frac{p^\mu_i p^\nu_i}{p^0_i}\delta^3(\vec{x}-\vec{x}_i) \end{align}$$ Integrating on the $$t=t_0$$ surface, one obtains
 * $$\begin{align}

E_C(\text{IC})=\int_{\sigma_\mu(t=const.)} T_C^{\mu 0}=\int dxdydz T_C^{00}= \sum_i p^0_i \end{align}$$ which is as expected. --Gamebm (talk) 19:05, 28 April 2014 (UTC)


 * Your question is far from simple (at least for someone like me who does not do these calculations often). I have not checked everything you did above, so I cannot testify to its correctness. But I do see one possible source for your problem &mdash; your division by $$p^{0}$$ in the expression for the stress-energy tensor. The article says "In special relativity, ...". This implies that you are using the kind of Cartesian coordinates in which the Lorentz transformation is expressed. But you are not using such a system of coordinates. Accordingly, you should adjust the formula to make the factor of p by which you are dividing orthogonal to the surface in terms of which the delta-functions are expressed, for example, the $$\tau = \tau_1 $$ surface. JRSpriggs (talk) 05:17, 2 May 2014 (UTC)


 * Hi JRSpringgs, I found my mistake and corrected it, it came from the $$\delta$$ function, as a result, the integral cancels out perfectly. I would not say the problem is not simple, it is the energy conservation problem of a free particle, the geodesic of a particle in flat space-time. I was careful about the energy momentum tensor, that part was all right. Again, thanks for the help! --Gamebm (talk) 12:08, 5 May 2014 (UTC)

En-dash vs. hyphen
I see that this article's subject is being spelled with an en-dash throughout the article, rather than a hyphen. My own intuition agrees that this is how it ought to be spelled, but are there any rules, style guides, or authoritative references to corroborate this choice? Thecommexokid (talk) 02:21, 27 May 2014 (UTC)


 * See MOS. "Stress" is not modifying "energy"; rather this is the tensor containing energy-density, momentum-density (or energy-flux), and stress (shear stress and pressure or tension). So an en dash is probably appropriate. JRSpriggs (talk) 03:14, 27 May 2014 (UTC)