Timeline for Are covariant components of 4-gradient of scalar product made of contravariant components of vector?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Feb 2, 2014 at 6:58 | comment | added | Ruslan | @gj255 indeed, I didn't see the obvious possibility to lower the indices on the RHS and get the simple relation. Have to do more index gymnastics to get used to it :) | |
| Feb 1, 2014 at 20:19 | vote | accept | Ruslan | ||
| Feb 1, 2014 at 19:23 | answer | added | Valter Moretti | timeline score: 1 | |
| Feb 1, 2014 at 13:12 | comment | added | gj255 | The vector on the RHS may involve contravariant components of the 4-momentum, but it is certainly not equal to $$ (p^0,p^1,p^2,p^3) \,,$$ which are the contravariant components of the 4-momentum. Using the relation $$p_\mu = \eta_{\mu \nu} p^\nu $$ we see that the list of components on the right are indeed the covariant components of the 4-momentum. The covariant and contravariant components of an object are closely related, and so it is natural that we could write them in terms of each other --- we have $p_1 = -p^1$, for instance. | |
| Feb 1, 2014 at 13:02 | history | asked | Ruslan | CC BY-SA 3.0 |