Why is 1/G Used in GR Lagrangian?

  • Context: Graduate 
  • Thread starter Thread starter nickyrtr
  • Start date Start date
  • Tags Tags
    Gr Lagrangian
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
4 replies · 2K views
nickyrtr
Messages
93
Reaction score
1
I have started reading about the Lagrangian in General Relativity, in relation to the Einstein-Hilbert action, and there is something that does not make sense to me. The Lagrangian is split into two pieces, one derived from the Ricci curvature and the other labeled L_matter, so far so good.

What I find odd is that the Ricci curvature part of the Lagrangian is proportional to 1/G. This seems a contradiction with the way gravity is always described as much weaker than the other forces, as represented by a very small value for G. Why then is the gravitational part of the Lagrangian proportional to 1/G, which would be a very large number if G is small. I would have expected the gravity contribution to the Lagrangian to be smaller than the other forces' contribution. What am I missing?
 

Discussion

Physics news on Phys.org
You need to look at the overall picture. In general, multiplying the Lagrangian by a constant will not change the EoMs. The term with the Ricci scalar is not going to affect the EoMs of the matter fields simply because it does not depend on them. In a world without matter, the G does not matter for the EoM of the metric and therefore the matter interaction terms arise from the matter Lagrangian. For these terms to be suppressed for small G, you need the matter Lagrangian to be suppressed by G wrt the Ricci scalar term.
 
A Lagrangian has units of energy (or energy density), so this question about the Lagrangian is equivalent to the following. Suppose we write an expression for the energy density of a field in terms of the field strength. Expressions like this are always of the form (energy density)=(1/k)(field)2, where k is a coupling constant. Why is it the inverse of the coupling constant that appears here? The answer is that for a field created by a given source, the field strength is proportional to k. Therefore if we have a fixed source, the energy of its field is actually proportional to ##(1/k)(k^2)=k##.
 
bcrowell said:
A Lagrangian has units of energy (or energy density), so this question about the Lagrangian is equivalent to the following. Suppose we write an expression for the energy density of a field in terms of the field strength. Expressions like this are always of the form (energy density)=(1/k)(field)2, where k is a coupling constant. Why is it the inverse of the coupling constant that appears here? The answer is that for a field created by a given source, the field strength is proportional to k. Therefore if we have a fixed source, the energy of its field is actually proportional to ##(1/k)(k^2)=k##.

OK, that helps. If you double G, then (1/G) is cut in half, but the Ricci curvature is four times stronger, so the Lagrangian is doubled. Is that right?
 
nickyrtr said:
OK, that helps. If you double G, then (1/G) is cut in half, but the Ricci curvature is four times stronger, so the Lagrangian is doubled. Is that right?

Right.