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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)
In the section As a Noether current, I'm unclear on the notation used in the expression
I've never seen notation with the subscript ;b and I would have written the expression as
Jeodesic 13:16, 23 October 2007 (UTC)
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)
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)
What is the lagrangian for the idealized fluid, which leads to the stress-energy tensor given in the article:
-- Michael C. Price talk 06:09, 30 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 — σ is proportional to 1/V. In other words, I am ignoring temperature and/or entropy.
This Lagrangian
allows the action to depend only on that one variable, σ. 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 μ gives the constraint equation
Varying with respect to λ gives the constraint equation
Varying with respect to σ gives
Varying with respect to uα gives
Varying with respect to gα β gives
which simplifies to
Combining #4 and #3 and #1, we solve for μ
Substituting #6 and #3 into #5, we get
which can be rearranged to
Comparing this to the article, we see that the pressure is
and
whence the mass density is
If we substitute then we get
To describe a fluid which is infinitely compressible, p=0, we can use Functions which increase faster than linear give positive pressure. OK? JRSpriggs ( talk) 20:21, 31 July 2008 (UTC)
"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)
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)
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)
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#Stress-energy in special situations 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#Adjoint equation and Dirac current) 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)
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)
(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)
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)
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)
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 “orientationâ€. 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)
My two standard reference books, Misner, Thorne & Wheeler (pg 51, Gravitation) Itzykson & Zuber (pg 5, Quantum Field Theory), define .-- Michael C. Price talk 05:37, 10 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)
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)
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 in it (the relativistic time dilation/length contraction factor), but I believe this factor should be . 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 and boosting it--involves two factors of , 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)
I see that there is some discussion around the importance of dimension and aesthetics but here, for the common reader, the description of really does need to be consitent with the equation that is effectively in the same sentance. That is we should have —Preceding unsigned comment added by 62.25.106.209 ( talk) 12:05, 24 May 2010 (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)
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)
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)
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)
The main article I am referring to is “Einstein field equationsâ€. 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)
Thank you, Sir. I do appreciate your explanation. — Preceding unsigned comment added by 98.154.17.2 ( talk) 23:19, 13 April 2012 (UTC)
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∧Ŝ c2ħε Иτlk 13:59, 1 March 2013 (UTC)
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)
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 in Cartesian coordinates. In the following, we will try to verify explicitly for a closed surface, one has
where the closed 3-surface is defined by , , and with the following definitions
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
As shown below, the 3-surface element at reads
and the integral on the 3-surface yields
Similarly, the 3-surface element at reads
and the integral on the 3-surface yields
Expressions for the other two 3-surfaces are more intuitive, one has
and
The energy momentum tensor of a point-like particle in Cartesian coordinates possesses the following form
where
depending on whether one deals with 3-surface parameterized in z, or . Substituting into the expressions of , one gets
where are the values of coordinates when the particle does go across the surfaces . 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 surface, so that the particle goes across surfaces and, one ignores the integral on the surfaces. In a more general case, if the system consists of many point-like particles, one can usually find big enough area of so that all the particles go through the area. One sees that the above expression does not vanish as expected from the conservation of energy.
To evalute the surface element, one may transform (the orientation of) a surface element in hyperbolic coordinates , which is a unitary vector in hyperbolic coordinates, into Cartesian coordinates (here we use : to indicate hyperbolic coordinates and : to indicate Cartesian coordinates.)
It is straightforward to verify at this point . The area of the surface element can be calculated as following
and
for completeness, we also write down here
which had used the fact that the three vectors are mutually orthogonal. In fact, this also implies
where are the 4-vectors in Cartesian coordinates. One sees that the size of the surface element defined by 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 is parameterized in terms of orthogonal coordinates, therefore all displacements corresponding to each variables () 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
where is the Levi-Civita tensor, parameterize the 3-surface, and describe the connection between the coordinates and its parameters. For instance, see the discussion near (3.2.2) of [1]. In our case of the Cartesian coordinates, , therefore . It is straightforward to verify that one obtains the same result, namely,
Up to this point, we have not discussed the explicit form of energy momentum tensor. According to [2],
Integrating on the surface, one obtains
which is as expected. -- Gamebm ( talk) 19:05, 28 April 2014 (UTC)
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)
![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 | Archive 2 |
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)
In the section As a Noether current, I'm unclear on the notation used in the expression
I've never seen notation with the subscript ;b and I would have written the expression as
Jeodesic 13:16, 23 October 2007 (UTC)
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)
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)
What is the lagrangian for the idealized fluid, which leads to the stress-energy tensor given in the article:
-- Michael C. Price talk 06:09, 30 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 — σ is proportional to 1/V. In other words, I am ignoring temperature and/or entropy.
This Lagrangian
allows the action to depend only on that one variable, σ. 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 μ gives the constraint equation
Varying with respect to λ gives the constraint equation
Varying with respect to σ gives
Varying with respect to uα gives
Varying with respect to gα β gives
which simplifies to
Combining #4 and #3 and #1, we solve for μ
Substituting #6 and #3 into #5, we get
which can be rearranged to
Comparing this to the article, we see that the pressure is
and
whence the mass density is
If we substitute then we get
To describe a fluid which is infinitely compressible, p=0, we can use Functions which increase faster than linear give positive pressure. OK? JRSpriggs ( talk) 20:21, 31 July 2008 (UTC)
"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)
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)
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)
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#Stress-energy in special situations 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#Adjoint equation and Dirac current) 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)
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)
(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)
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)
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)
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 “orientationâ€. 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)
My two standard reference books, Misner, Thorne & Wheeler (pg 51, Gravitation) Itzykson & Zuber (pg 5, Quantum Field Theory), define .-- Michael C. Price talk 05:37, 10 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)
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)
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 in it (the relativistic time dilation/length contraction factor), but I believe this factor should be . 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 and boosting it--involves two factors of , 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)
I see that there is some discussion around the importance of dimension and aesthetics but here, for the common reader, the description of really does need to be consitent with the equation that is effectively in the same sentance. That is we should have —Preceding unsigned comment added by 62.25.106.209 ( talk) 12:05, 24 May 2010 (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)
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)
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)
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)
The main article I am referring to is “Einstein field equationsâ€. 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)
Thank you, Sir. I do appreciate your explanation. — Preceding unsigned comment added by 98.154.17.2 ( talk) 23:19, 13 April 2012 (UTC)
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∧Ŝ c2ħε Иτlk 13:59, 1 March 2013 (UTC)
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)
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 in Cartesian coordinates. In the following, we will try to verify explicitly for a closed surface, one has
where the closed 3-surface is defined by , , and with the following definitions
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
As shown below, the 3-surface element at reads
and the integral on the 3-surface yields
Similarly, the 3-surface element at reads
and the integral on the 3-surface yields
Expressions for the other two 3-surfaces are more intuitive, one has
and
The energy momentum tensor of a point-like particle in Cartesian coordinates possesses the following form
where
depending on whether one deals with 3-surface parameterized in z, or . Substituting into the expressions of , one gets
where are the values of coordinates when the particle does go across the surfaces . 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 surface, so that the particle goes across surfaces and, one ignores the integral on the surfaces. In a more general case, if the system consists of many point-like particles, one can usually find big enough area of so that all the particles go through the area. One sees that the above expression does not vanish as expected from the conservation of energy.
To evalute the surface element, one may transform (the orientation of) a surface element in hyperbolic coordinates , which is a unitary vector in hyperbolic coordinates, into Cartesian coordinates (here we use : to indicate hyperbolic coordinates and : to indicate Cartesian coordinates.)
It is straightforward to verify at this point . The area of the surface element can be calculated as following
and
for completeness, we also write down here
which had used the fact that the three vectors are mutually orthogonal. In fact, this also implies
where are the 4-vectors in Cartesian coordinates. One sees that the size of the surface element defined by 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 is parameterized in terms of orthogonal coordinates, therefore all displacements corresponding to each variables () 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
where is the Levi-Civita tensor, parameterize the 3-surface, and describe the connection between the coordinates and its parameters. For instance, see the discussion near (3.2.2) of [1]. In our case of the Cartesian coordinates, , therefore . It is straightforward to verify that one obtains the same result, namely,
Up to this point, we have not discussed the explicit form of energy momentum tensor. According to [2],
Integrating on the surface, one obtains
which is as expected. -- Gamebm ( talk) 19:05, 28 April 2014 (UTC)
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)