GR Lagrangian (Part 2)

  • Thread starter Malamala
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Homework Statement


This is a continuation of this problem. I will rewrite it here too:
The Lagrangian density for the ##h=h^{00}## term of the Einstein gravity tensor can be simplified to: $$L=-\frac{1}{2}h\Box h + (M_p)^ah^2\Box h - (M_p)^b h T$$ The equations of motion following from this Lagrangian looks roughly like (I didn't calculate this, they are given in the problem): $$\Box h = (M_p)^{a}\Box(h^2)-(M_p)^bT$$ For a point source ##T=m\delta^3(x)##, solve the equation for h to second order in the source T, with ##M_p=\frac{1}{\sqrt{G_N}}## and calculate the correction to the Mercury angular frequency orbit around the sun compared to the first order approximation.

Homework Equations




The Attempt at a Solution


Continuing the same logic as before I write $$h=h_0+h_1+h_2...$$ where ##h_0## is first order in T, ##h_1## is second order and so on. Before, I got for the first order $$h_0=-\frac{M_p^b m }{4 \pi r}$$ Now if we go to the second order, the equation we need to solve is: $$\Box h_1 = (M_p)^a\Box(h_0^2)$$ which is equivalent to $$\Box (h_1 - (M_p)^a h_0^2) = 0$$ $$h_1 = (M_p)^a h_0^2 + f(x)$$ where ##f(x)## is such that ##\Box f = \nabla f = 0## (I replaced ##\Box## with ##\nabla## as the source is time independent). Is it ok up to now? Now first thing I am confused about, can I discard this ##f##? in the equation of motion it seems like ##h## appear with ##\Box## so I think f will not affect the equations, but I am not 100% sure this is true. Now for the orbit, to first order I used $$m_{Mercury}\omega^2 r = \frac{m_{Mercury}}{M_p}(-\nabla h_0)$$ where that ##M_p## comes from the normalization (as I was told in the first post). And from here I got the first order approximation for ##\omega##. For the second order, I tried the same thing (this is QFT class, so I don't think I am expected to use GR, and they ask for a rough approximation, anyway) $$m_{Mercury}\omega^2 r = \frac{m_{Mercury}}{M_p}(-\nabla (h_0+h_1))$$ However something is wrong as the correction should be very small, however, ##h_1## going like ##h_0^2##, contains ##m^2## which here m is the mass of the sun so overall, the value of ##h_1## is bigger than ##h_0## which doesn't make sense for a second order correction. What is wrong with my calculations? Thank you!
 

Answers and Replies

  • #2
Orodruin
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Since ##h_0 \ll 1##, clearly ##h_0^2 \ll h_0##.
 

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