Nonstationary Configurations of a Spherically Symmetric Scalar Field

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Action and stress-energy tensor, Einstein-Klein-Gordon equations, and method for constructing nonstationary configurations of a spherically symmetric scalar field are discussed in this study. The behavior of the characteristic function allows interpretations such as black holes, wormholes, or naked singularities. The metric functions and coordinates play a crucial role in defining the configurations and conditions for Schwarzschild asymptotics.


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  1. Nonstationary configurations of a spherically symmetric massless scalar field Ju.V. Tchemarina , I.M. Potashov Faculty of Mathematics Tver StateUniversity ICPPA 2022

  2. Action and stress-energy tensor We study nonstationary spherically symmetric solutions of the Einstein- scalar field system with a massless scalar field minimally coupled to gravity. We begin with the action ? = 1 ? ?4? . 2? + ??,?? The components of the stress-energy tensor are determined by formula ???= 2?????? (?????????)???.

  3. Einstein and Klein-Gordon equations ??? 1 1 ? ?????? = 0, ? = det ??? , 2????= ??? , ??? A configuration will be called stationary if ? = ? (?), where ? is the radius of the sphere. Otherwise, the configuration will be called nonstationary. This allows us to use the coordinate system ?,?,?,? , at least locally, for any nonstationary configuration.

  4. Method for constructing nonstationary configurations of a spherically symmetric scalar field Characteristic function: ? = ??,?? . Metric in coordinates ?,?,?,? ???? + ??? + ? 1 ??2 ??2= 4?2???2+ ??? ?2??2+ sin2???2. 2 + ? 1 ?? 4? ??? The behavior of function ?(?,?) makes it possible to interpret the solution as a black hole, wormhole, or naked singularity.

  5. Method for constructing nonstationary configurations of a spherically symmetric scalar field The equation for the characteristic function is equivalent to the Klein- Gordon equation, which takes the form 1 ? ?????? = 0 ?? ? ??? ?? + ?? 2?2+ 3?? 3? ?? + 1 ? ?? ? ??? ?2? + ???? +4?2 6? + 2 = 0. (*) Coordinate system (t, ?, ?, ?) : 4?2???2 2 ??2 ??2= ?2??2+ sin2???2, 2 ? + ? 1 ?? 4? ??? ?? = ?? 2?? ?? ??

  6. lasses of metric functions f = I. ( . ) f Form of the equation ? ? 4? ? + 2 = 0 1 2+ ?1?2?+?2? 2?. ? ? = Metric in coordinates ?,?,?,? ??2= ?2(1 2+ ?1?2?+?2? 2?)??2+ 2?(?1?2? ?2? 2?)???? + ( 1 2+ ?1?2?+?2? 2?)??2 = 4 1 16?1?2 ?2??2+ sin2???2, The metric signature ( + - - - ) entails the condition 1 16?1?2>0 .

  7. Next, we revert to the usual coordinates (t,C,,) 1 2+?1?2?+?2? 2? ?1?2? ?2? 2??? ? = ?? . Schwarzschild asymptotics are possible if the following conditions are met: 1 2+ ?1?2?+?2? 2?= 1, 1 16?1?2= 0 - degenerate case. ?1?2? ?2? 2?=0 ? ? = ?1=1 8,?2 = 1 ?1=?2 =1 8 8 ? = ? ? 2? ?2????? ?2?, ? = ? ? ? ? ?, ? = 0 ? =??? ( 2) 2 3,? ? ? 2 ? ~

  8. lasses of metric functions 2 = + II. ( , ) 1 ( ). f C C h Klein-Gordon equation 2 12 ? 2= 0 3 ? ? 2 ? 3 2 3 3?+?2? 2 3 3? h ? = ?1? . ( ) , C , , The exact form of the metric in the coordinates ( ( 1 ) ) 4 2 2 2 2 2 2 = + + dC + ( ) ( ) 3 ( ) , ds h d C h d h d C d 2< 0. = 12 ? 1 + ?2 ? ?2 ?

  9. Coordinates (t,C,,): 8 3 3? ?22 ? 4 3 3?+?2 2 3 3? ? =2?1?2?12? ?1? . 2 3 3??1? ?1=1, ?2= 1, ? = 1 + 8?2? 32 3 3?, 1 ? > 0, ?2> ? < 0 8? 2 3 ? 3 3 3 ? >1 1 ? = 0 ? = ? > 1, 2??? ? ~ 16?2, ? , ? = ?????, 2 3 2 2?

  10. Metric in coordinates ?,?,?,? 4?2???2 2 ??2 ??2= ?2??2+ sin2???2. 2 ? + ? 1 ?? 4? ??? ?? 1 ? > 0, ?2> ? < 0 8? 2 3 ? 3

  11. lasses of metric functions f = III. ( ). f C Scalar field equation ?2? + ?? 2?2+ 3? ? 1 ?? + 4?2 6? + 2 = 0. ??? Under the assumption that the scalar field depends on time only (? = ?), the metric can be written as ?2??2 1 ??? ? ??2 ??2= ?2??2+ sin2???2, ? 1 ??? ? ? > 0.

  12. Direct substitution of the series for the characteristic function into the Klein-Gordon equation allows us to conclude that there are no solutions with Schwarshild asymptotics in this case. Also there are no solutions with de Sitter asymptotics. Numerical solutions

  13. Conclusions: Scalar field equation in coordinate system (?, ?, ?, ?) ??? ?? + ?? 2?2+ 3?? 3? ?? + 1 ? ?? ? ??? ?2? + ???? +4?2 6? + 2 = 0 allows one to obtain both exact and numerical solutions for a massless scalar field. Exact solutions are obtained for a massless scalar field. These solutions are related to characteristic functions of a special kind. Analysis of specific exact solutions can help clarify the general features of nonstationary scalar field configurations. Studying the behavior of the characteristic function contributes to a more correct formulation of the problem of obtaining numerical nonstationary solutions.

  14. Thank you for attention!

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