Timeline for When does $T^{ij} = T_{ij}$?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Mar 12, 2013 at 22:25 | comment | added | joshphysics | @JerrySchirmer Thanks; I just changed it to if $g_{ij} = \delta_{ij}$ so as to avoid confusion. | |
| Mar 12, 2013 at 22:24 | comment | added | joshphysics | @Vibert Yeah I've seen that notation used quite a bit as well; it's simply a matter of notational taste it seems. | |
| Mar 12, 2013 at 22:23 | history | edited | joshphysics | CC BY-SA 3.0 |
added 73 characters in body
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| Mar 12, 2013 at 21:42 | vote | accept | user12345 | ||
| Mar 12, 2013 at 21:28 | comment | added | Zo the Relativist | slight correction to your last line--if the metric is Euclidean <b>and the coordinates are Cartesian</b>, then it is true that... | |
| Mar 12, 2013 at 21:22 | comment | added | David Z | I guess it varies by field. (Personally, I'm also used to seeing $(\cdot)^2$ used to indicate a trace over all written indices, $(T_{ij})^2 = T_{ij}T^{ij}$.) | |
| Mar 12, 2013 at 21:20 | comment | added | Vibert | In general, I would understand summation in $(T_{ij})^2$. In gauge theory, for example, it's almost standard to write $S = \int -\frac{1}{4} (F_{\mu\nu})^2.$ What would it be, otherwise? | |
| Mar 12, 2013 at 21:08 | history | answered | joshphysics | CC BY-SA 3.0 |