Talk:Stress–energy tensor

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comment from energy-momentum density

I added a whole bunch of stuff to this article. I think this article should be merged with Stress-energy tensor. I was also going to move this article to Stress tensor (which I will still consider doing, after I have merged), but stress tensor is a redirect to Stress (physics) so i wonder if i should. -Lethe | Talk 00:03, Jan 20, 2005 (UTC)

Care to elucidate Noether procedure? --ub3rm4th 16:34, 23 Feb 2005 (UTC)

Pleast note the spelling of 'Belinfante'. It is incorrect here, I believe.

Convention for 0th component

In relativity in Minkowski coordinates there is a choice of convention as to whether $x^{0}=ct$ or $x^{0}=t$ . Either is valid, so long as all tensors abide by the same convention. There has been some recent edit-warring over this in this article. The article needs to be self-consistent, so any change in the value of $x^{0}$ at the beginning of the article needs to be accompanied by appropriate changes to $T^{0i},g^{00},u^{0}$ , etc, throughout the article.-- Dr Greg talk 23:40, 21 September 2018 (UTC)[reply]

Thank you, Dr Greg.

To 41.37.74.160 and Feynman 0511: I would like to point out that this issue has been debated at great length already. Please see my comments at: Talk:Stress–energy tensor/Archive 1#Dimension problem., Talk:Stress–energy tensor/Archive 1#Wrong dimension of the Stress-energy tensor, Wikipedia talk:WikiProject Physics/Taskforces/Relativity/Archive 1#Agreeing on a consistent metric for SR, and User:JRSpriggs/Conventions for general relativity. And as an example, User:JRSpriggs/EM in GR. JRSpriggs (talk) 03:20, 22 September 2018 (UTC)[reply]

Thank you Dr Greg, JRSpriggs.

However, the convention the whole article uses is unheard-of and rather arbitrary. So I suggest rewriting the article in one of the common conventions, or better still, use natural units. You are using the convention that you like, making the page look like it's only yours. I'm starting to recoil from Wikipedia as my primary source because of arguments like this. So sad! Feynman 0511 15:09, 22 September 2018 (UTC)

It is not arbitrary. It is carefully designed to make things as easy as possible for the user of classical physics who is merely trying to see how to correct his formulas for relativistic effects. JRSpriggs (talk) 20:03, 22 September 2018 (UTC)[reply]

Neither convention is consistent with the adjoining figure, and no convention can be made consistent with the figure.

First, the dimensions of energy density, energy flux, momentum density and momentum flux are, respectively, M/(LT²), M/T³, M/(L²T) and M/(LT²); where M, L and T respectively denote the dimensions of mass, length and time duration. In particular, the dimensions for energy flux and momentum density are different, making T^{i0} and T^{0j} of different dimensions. No symmetric rank (2,0) or rank (0,2) tensor can have that property. Only a rank (1,1) tensor can, and only with the convention x⁰ = t. If you use x⁰ = ct, then all the components will have the same dimensions, presumably that of the diagonal entries, M/(LT²).

Second, the description in the figure does not apply to this or any tensor at all; but (as made clear by the names themselves) to a tensor density - and that's the stress tensor density.

The correct dimensional analysis is done by going back to basics. The stress tensor density 𝔗^ρ_ν and canonical stress tensor density have the same dimensions, which are (for the diagonal components) action divided by space-time volume - the same as the dimensions for a Lagrangian density. That is either (ML²/T)/(L³T) = M/(LT²) with the convention x⁰ = t, or (ML²/T)/L⁴ = M/(L²T) with the convention x⁰ = ct. The stress tensor - as a rank (1,1) tensor - is obtained from this by dividing out T^ρ_ν = 𝔗^ρ_ν/√|g|, where g is the determinant of the metric, whose components are g_μν. Their dimensions, in turn, are determined from the dimensions of the line element g_{μν} dx^μ dx^ν, which is conventionally taken to be L². Therefore, √|g| has dimensions L⁴/([0][1][2][3]), where [0], [1], [2], [3] are respectively the dimensions of x⁰, x¹, x² and x³. With the convention x⁰ = t, that is L/T, otherwise with the convention x⁰ = ct, √|g| is dimensionless. In both cases, the diagonal components of T^ρ_ν have dimension M/(L²T), which does not match the figure.

Only the stress tensor density accords with the figure - and only with the convention x⁰ = t, not with x⁰ = ct.

Flux

From wikii on stress mass energy tensor: "The stress–energy tensor is defined as the tensor Tαβ of order two that gives the flux of the αth component of the momentum vector across a surface with constant xβ coordinate. "

From wikii on flux: "Only the parallel component contributes to flux because it is the maximum extent of the field passing through the surface at a point, the perpendicular component does not contribute."

Something seems wrong with the stress mass energy definition since more than the perpendicular components contribute. So, for example, a stress that is pure shear has no flux as it's components are at right angles to the normal of the side being considered. So the stress mass energy tensor cannot be defined using flux solely. Must include parallel components.

Cannot make the edit because too inexperienced but am convinced something is wrong.

Long live Wikipedia! Justintruth (talk) 17:47, 9 March 2019 (UTC)[reply]

Either the article flux is wrong on this point or it is referring to a special case which does not include the stress-energy tensor. JRSpriggs (talk) 11:32, 10 March 2019 (UTC)[reply]

Isolated Particle

Shouldn′t

(v^{\alpha })_{\alpha =0,1,2,3}=\left(1,{\frac {d\mathbf {x} _{\text{p}}}{dt}}(t)\right)

read

(v^{\alpha })_{\alpha =0,1,2,3}=\left(c,{\frac {d\mathbf {x} _{\text{p}}}{dt}}(t)\right)

instead? That would make it consistent with the result one obtains by setting

\rho =m\cdot \delta \left(\mathbf {x} -\mathbf {x} _{\text{p}}\left(t\right)\right)

and $p=0$ in the formula for a perfect fluid. (It is also necessary so that all components of $v$ have the same units.) --77.0.137.249 (talk) 15:43, 30 June 2019 (UTC)[reply]

As I have said before at excruciating length, the idea that all components of a tensor must have the same units is a myth, that is, it is false. This is obvious if you simply consider the units of components of tensors in Schwarzschild coordinates.

In this article we use the International System of Units which is the normal standard for classical physics. In a Cartesian coordinate system, time has units of seconds and distance has units of meters. From this it follows that the time component of (ordinary, not "proper") velocity has units of 1 (dimensionless) while the spatial components have units of meters per second. JRSpriggs (talk) 07:58, 1 July 2019 (UTC)[reply]