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Lagrangian in weak field

  1. Oct 7, 2012 #1
    I just read some basic concepts on General Relativity, and this idea pops up: I know we should use variations of metrics for gravitational field in the Lagrangian. But considering the resemblance of gravitational field(weak-field) to electromagnetic field, can we construct a 4-potential similar to that of the electromagnetic field, say [itex]A_{G} = (ψ(gravitational potential),0, 0, 0)[/itex]. So the Action for the effect of gravitational field would be: [itex]\int -\frac{m}{c} A_{G}dx^{i}[/itex]. Would that be a good approximation for weak field?
  2. jcsd
  3. Oct 7, 2012 #2


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    No. What you wrote down is identically zero. You should consider the weak field limit directly for the point particle action. So, take this action, write the metric as a
    perturbation of e.g. Minkowski spacetime, and perform a Taylor expansion on the square root. Taking static gravity and slowly moving particles should then on its turn give newtonian gravity, as in e.g. Eqn 2.10 of arxiv:1206.5176.

    In the weak field limit GR becomes Fierz-Pauli theory, massless spin 2. A vector potential would mean that gravity is represented by spin-1. In the Newtonian approx. gravity is effectively reduced to a Gallilean scalar field, which could be interpreted as spin 0 (but then with spin defined wrt to the Galilei group).
    Last edited: Oct 7, 2012
  4. Oct 7, 2012 #3
    Eh I was trying to make a parallel comparison to [itex]S_{mf}[/itex], so the integral is actually [itex]\int -\frac{m}{c}ψ d(ct)[/itex] and when v is relatively small the whole expression does degrade into the classical Lagrangian. But it is a very imprudent thought as I only start to scratch the surface of Relativity Theory. Thank you for your reply that I know where my problem is.
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