Einstein-Hilbert term and mass, gravity

[Tomahoc]

It is said that the Einstein-Hilbert term in the following couple mass and gravity.

Lagrangian = R/(16*pi*G_N) [Einstein-Hilbert term] + L (nongravitational)

How is the above derived? Is it enough to prove that there should be a theory of quantum gravity since mass and gravity are coupled together in that term?

 

[Mentz114]

That Lagrangian does not have the form L = L_grav + L_matter + L_interaction which is usually necessary. If there was such a term it would be something like θ^abT_ab where θ is the field and T is the matter SET as in field theory gravity (FTG). I understand that this term gives rise to the spin-2 carrier bosons. I'm no expert though.

 

[PeterDonis]

It is said that the Einstein-Hilbert term in the following couple mass and gravity.

Can you give a specific reference where this is said? The usual interpretation that I'm familiar with is that the Einstein-Hilbert term only contains spacetime curvature (i.e., "gravity"), and the other term, which you call L(nongravitational), only contains the "matter", which means all non-gravitational fields. The only coupling between the two, as far as classical GR is concerned, comes from the Einstein Field Equation, which is obtained by varying the Lagrangian with respect to the metric.

 

[Tomahoc]

Can you give a specific reference where this is said? The usual interpretation that I'm familiar with is that the Einstein-Hilbert term only contains spacetime curvature (i.e., "gravity"), and the other term, which you call L(nongravitational), only contains the "matter", which means all non-gravitational fields. The only coupling between the two, as far as classical GR is concerned, comes from the Einstein Field Equation, which is obtained by varying the Lagrangian with respect to the metric.

What I heard is just like what you described, that the coupling between them comes from the Einstein Field Equation, which as you said is "obtained by varying the Lagrangian with respect to the metric". Can you please describe what it means to "vary the Lagrangian with respect to the metric"? Is varying the Lagrangian natural or kinda forced?

 

[haushofer]

That's the whole idea behind the action principle; you regard the metric as fundamental field. The action is then a functional of the metric, and its extremum gives the EOM.

You can look at e.g. Carroll's notes for a nice motivation for this action: it is "simple" and contains up to second order derivatives of the metric.

Btw, it's not just mass, but energy in general which couples to gravity. An electromagnetic field curves spacetime, deflecting even electrically neutral particles.

 

[PeterDonis]

What I heard is just like what you described, that the coupling between them comes from the Einstein Field Equation

That's not the same as saying "the Einstein-Hilbert term couples mass and gravity", which is what you said in the OP. That's why I wasn't sure what you were referring to.

Can you please describe what it means to "vary the Lagrangian with respect to the metric"?

As haushofer said, you find the extremum of the action (the action is just the integral of the Lagrangian over all of spacetime), and that gives you the equation of motion (which is what the Einstein Field Equation is: it's the "equation of motion" for gravity).

Finding the extremum just means taking the derivative (of the action) and finding where it is zero. Varying "with respect to the metric" just means the metric is what you take the derivative of the action with respect to.

Is varying the Lagrangian natural or kinda forced?

I'm not sure what you would consider "natural" vs. "forced". Finding the extremum of a function by finding where its derivative is zero is an elementary operation in calculus.