Spatially Homogeneous Scalar Field on Spacetime: Showing $\nabla^2 f$

  • Context: Undergrad 
  • Thread starter Thread starter ergospherical
  • Start date Start date
  • Tags Tags
    Mean
Click For Summary
SUMMARY

The discussion focuses on demonstrating that for a spatially homogeneous scalar field ##f## on spacetime described by the metric ##ds^2 = dt^2 - a^2(t) \delta_{ij} dx^i dx^j##, the Laplacian operator yields the result ##\nabla^2 f = \ddot{f} + 3H \dot{f}##. The scalar field ##f## is defined as a function of time only, leading to the conclusion that only time derivatives contribute to the Laplacian. The derivation involves the use of the covariant derivative and the trace of the Christoffel symbols, specifically ##\Gamma_{0i}^{j} = H\delta^{j}_{i}##, where ##H## is the Hubble parameter.

PREREQUISITES
  • Understanding of scalar fields in general relativity
  • Familiarity with covariant derivatives and Laplacians
  • Knowledge of the Friedmann-Lemaître-Robertson-Walker (FLRW) metric
  • Basic concepts of cosmological parameters, particularly the Hubble parameter
NEXT STEPS
  • Study the derivation of the Laplacian in curved spacetime using the covariant derivative
  • Explore the implications of spatially homogeneous fields in cosmology
  • Learn about the role of Christoffel symbols in general relativity
  • Investigate the relationship between scalar fields and cosmological evolution
USEFUL FOR

Researchers and students in theoretical physics, particularly those focusing on cosmology, general relativity, and scalar field theory.

ergospherical
Science Advisor
Homework Helper
Education Advisor
Insights Author
Messages
1,100
Reaction score
1,387
If ##f## is a "spatially homogeneous" scalar field on spacetime ##ds^2 = dt^2 - a^2(t) \delta_{ij} dx^i dx^j## then show that ##\nabla^2 f = \ddot{f} + 3H \dot{f}##. Should be easy if I knew what the condition on ##f## is, i.e. ##\nabla^2 f = \partial_{\mu} \partial^{\mu} f = \ddot{f}- a^{-2}(t) \delta^{ij} \partial_i \partial_j f = \dots##?
 
Physics news on Phys.org
Spatially homogeneous = does not depend on spatial position. In other words, ##f## is a function of ##t## only.
 
  • Like
Likes   Reactions: FactChecker, vanhees71 and ergospherical
I messed up the double covariant, should be:\begin{align*}
\nabla^2 f &= \nabla_{\mu}(\partial^{\mu} f) \\
&= \partial_{\mu} \partial^{\mu} f + g^{\nu \rho} \Gamma_{\mu \nu}^{\mu} \partial_{\rho} f \\
&= \partial_t^2 f + 3H \partial_t f
\end{align*}(##\Gamma_{0i}^{j} = H\delta^{j}_{i}##)
 
  • Like
Likes   Reactions: vanhees71
Indeed. Only the time derivatives survive since the field only depends on t. Then it is just a matter of finding the appropriate trace of Christoffel symbols.
 
  • Like
Likes   Reactions: vanhees71

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 53 ·
2
Replies
53
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
591
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K