GR in Newtonian Limit: Understanding Weak Fields & Inequalities

  • Context: Undergrad 
  • Thread starter Thread starter Silviu
  • Start date Start date
  • Tags Tags
    Gr Limit Newtonian
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 2K views
Silviu
Messages
612
Reaction score
11
Hello! I am reading A first course in General Relativity by Schutz and at a point he proves that for a weak gravitational field and assuming ##\Lambda = 0## we have ##\Box \bar{h}^{\mu \nu} = -16\pi T^{\mu \nu}##. Leaving the notations aside, he says that for a weak gravitational field (and non-relativistic speeds) we have ##|T^{00|}>>|T^{0i}|>>|T^{ij}|## and this implies ##|\bar{h}^{00}|>>|\bar{h}^{0i}|>>|\bar{h}^{ij}|##. Can someone explain to me why do we have this last inequality? This is Chapter 8.3 in the second edition. Like for example, taking ##T^{0i}=0##, we get ##\Box\bar{h}^{0i}=0##. Why does this implies in any way that ##\bar{h}^{0i}## is very small? Thank you!
 
Physics news on Phys.org
Orodruin said:
The right-hand side is the source term. Without a source, your solution is identically ##\bar h^{0i} = 0## (assuming homogeneous boundary conditions).
Sorry I am a bit confused, if ##\bar{h}^{0i}## is a constant, doesn't the equality still holds?