Timeline for Constant term in action term in general relativity?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 6, 2020 at 15:37 | vote | accept | More Anonymous | ||
| Dec 6, 2020 at 12:50 | answer | added | Qmechanic♦ | timeline score: 3 | |
| Dec 6, 2020 at 11:48 | comment | added | Eletie | The calculation just goes about varying the metric field, in the standard variational way. Note this includes varying the metric determinant. It's in almost all GR textbooks but is a tedious calculation, so I'll link you to the wiki page en.wikipedia.org/wiki/Einstein%E2%80%93Hilbert_action. It also includes the cosmological constant term too. Basically the variation of the metric determinant leads to a term proportional to $g_{ab}$ in the field equations, so the 'constant term' shows up in the field equations too. | |
| Dec 6, 2020 at 11:33 | comment | added | More Anonymous | @eletie can you show the calculations? So I can accept your answer. Or comment including page showing the same? | |
| Dec 6, 2020 at 11:28 | comment | added | Eletie | If you just wrote the Einstein Hilbert action and then the $c$, like in your first equation, it wouldn't contribute to the field equations. But that wouldn't make sense anyway, because in GR we're dealing with curved space so the volume form needs to be $\sqrt{-g} d^4 x$ in order to transform as a scalar. | |
| Dec 6, 2020 at 11:23 | comment | added | Eletie | The reason for the difference in the two calculations, is that in GR the metric is dynamical. Explicitly, the volume element, which contains the metric determinant $g$, means even a constant in the Lagrangian shows up in the field equations. That constant is the Lambda you wrote in the field equations. | |
| Dec 6, 2020 at 9:30 | history | asked | More Anonymous | CC BY-SA 4.0 |