- #1
Tomas Vencl
- 66
- 13
I would like to ask if the function M(u) in Vaidya metrics must fulfil any special conditions, or is it completely free ?
https://en.wikipedia.org/wiki/Vaidya_metric#Outgoing_Vaidya_with_pure_Emitting_fieldIn other words, when the outgoing Vaidya metrics describes the metrics of radiating body (for example Hawking radiation), probably the M(u) must have some special form, conditions etc. to be a physical situation description.
Does anyone know some details about this topic ?
Thank you.
https://en.wikipedia.org/wiki/Vaidya_metric#Outgoing_Vaidya_with_pure_Emitting_fieldIn other words, when the outgoing Vaidya metrics describes the metrics of radiating body (for example Hawking radiation), probably the M(u) must have some special form, conditions etc. to be a physical situation description.
Does anyone know some details about this topic ?
Thank you.