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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)
In relativity in Minkowski coordinates there is a choice of convention as to whether or . 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 at the beginning of the article needs to be accompanied by appropriate changes to , etc, throughout the article.--  Dr Greg  talk 23:40, 21 September 2018 (UTC)
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 xⰠ= t. If you use xⰠ= 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 đ^Ď_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)/(LÂłT) = M/(LT²) with the convention xâ° = t, or (ML²/T)/Lâ´ = M/(L²T) with the convention xâ° = ct. The stress tensor - as a rank (1,1) tensor - is obtained from this by dividing out T^Ď_ν = đ^Ď_ν/â|g|, where g is the determinant of the metric, whose components are g_Ον. Their dimensions, in turn, are determined from the dimensions of the line element g_{Ον} dx^Îź dx^ν, which is conventionally taken to be L². Therefore, â|g| has dimensions Lâ´/([0][1][2][3]), where [0], [1], [2], [3] are respectively the dimensions of xâ°, xš, x² and xÂł. With the convention xâ° = t, that is L/T, otherwise with the convention xâ° = ct, â|g| is dimensionless. In both cases, the diagonal components of T^Ď_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 xâ° = t, not with xâ° = ct.
From wikii on stress mass energy tensor: "The stressâenergy tensor is defined as the tensor Tιβ of order two that gives the flux of the Îąth 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)
Shouldnâ˛t
read
instead? That would make it consistent with the result one obtains by setting
and in the formula for a perfect fluid. (It is also necessary so that all components of have the same units.) -- 77.0.137.249 ( talk) 15:43, 30 June 2019 (UTC)
![]() | This ![]() It is of interest to the following WikiProjects: | |||||||||||||
|
This page has archives. Sections older than 365 days may be automatically archived by Lowercase sigmabot III when more than 5 sections are present. |
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)
In relativity in Minkowski coordinates there is a choice of convention as to whether or . 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 at the beginning of the article needs to be accompanied by appropriate changes to , etc, throughout the article.--  Dr Greg  talk 23:40, 21 September 2018 (UTC)
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 xⰠ= t. If you use xⰠ= 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 đ^Ď_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)/(LÂłT) = M/(LT²) with the convention xâ° = t, or (ML²/T)/Lâ´ = M/(L²T) with the convention xâ° = ct. The stress tensor - as a rank (1,1) tensor - is obtained from this by dividing out T^Ď_ν = đ^Ď_ν/â|g|, where g is the determinant of the metric, whose components are g_Ον. Their dimensions, in turn, are determined from the dimensions of the line element g_{Ον} dx^Îź dx^ν, which is conventionally taken to be L². Therefore, â|g| has dimensions Lâ´/([0][1][2][3]), where [0], [1], [2], [3] are respectively the dimensions of xâ°, xš, x² and xÂł. With the convention xâ° = t, that is L/T, otherwise with the convention xâ° = ct, â|g| is dimensionless. In both cases, the diagonal components of T^Ď_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 xâ° = t, not with xâ° = ct.
From wikii on stress mass energy tensor: "The stressâenergy tensor is defined as the tensor Tιβ of order two that gives the flux of the Îąth 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)
Shouldnâ˛t
read
instead? That would make it consistent with the result one obtains by setting
and in the formula for a perfect fluid. (It is also necessary so that all components of have the same units.) -- 77.0.137.249 ( talk) 15:43, 30 June 2019 (UTC)