Investigations into a heuristic Lagrangian of graviton field

  • #1
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The following is taken from page 40 of Matthew Schwartz's "Introduction to Quantum Field Theory."

The Lagrangian for the graviton is heuristically ##\mathcal{L}=-\frac{1}{2}h\Box h + \frac{1}{3}\lambda h^{3}+Jh,## where ##h## represents the gravitational potential. We are ignoring spin and treating gravity as a simple scalar field theory. The ##h^3## term represents a graviton self-interaction, which is present in general relativity and so ##\lambda \sim \sqrt{G_N}##. The equations of motion are ##\Box h -\lambda h^{2}-J=0##.

Why the ##h^{3}## term represent the graviton self-interaction? What does self-interaction mean anyway? Does it mean the interaction among the various excitations of the graviton field?
 

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  • #2
haushofer
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Well, draw the Feynmandiagrams for this theory, and you see that the h^3 term brings together three gravitons. As such it describes an interaction among gravitons themselves, hence "self-interaction". So to answer your last question: yes. You can compare it with a simple phi to the fourth theory :)
 
  • #3
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Thanks!

I was also wondering why the fact that the ##h^3## term is present in general relativity implies that ##\lambda \sim \sqrt{G_N}##?

Where in general relativity is the ##h^3## term present anyway?
 
  • #5
haushofer
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Thanks!

I was also wondering why the fact that the ##h^3## term is present in general relativity implies that ##\lambda \sim \sqrt{G_N}##?

Where in general relativity is the ##h^3## term present anyway?
In GR you have an expansion which gives you an infinite number of self-interactions in the Lagrangian; this is just a matter of expanding the Lagrangian around a vacuum (e.g. Minkowski). A quadratic term is a mass-term, and since the graviton is massless in GR this term is absent. The (mass) dimension of lambda can be derived from considering the (mass) dimension of the Lagrangian or action. This is a nice exercise for you to do; if you want help, consult e.g. Zee's QFT book.
 

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