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By using some axioms people derives equation for Lorentz force and, then, Maxwell's equations from the Coulomb's law and Lorentz transformations. When I used analogical methodology for Newton's law of universal gravitation, I derived some equations like gravitomagnetic equations deriving from the GR equations in the weak field limit. But I lost factor 2, which multiplies "magnetic" component. It's binded with representation of gravitation field as the tensor. Can you explain me, how to, using methodology described above and representative of gravitation field by tensor, get factor 2?

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    $\begingroup$ Hi Maxim_Ovchinnikov, and welcome to Physics Stack Exchange! We really can't do anything to address this question unless you show the details of the procedure you have used to do these derivations. $\endgroup$ – David Z Apr 22 '12 at 0:35
  • $\begingroup$ He's asking what happens if you boost to the rest frame in a weak-gravitational field situation, should you reproduce the Newtonian force. The problem is probably the factor of 2 in the definition of the metric connection, which you would misguess if you started using the Newtonian potential which does not come with a factor of 2. $\endgroup$ – Ron Maimon Apr 22 '12 at 1:10
  • $\begingroup$ Perhaps, but it's quite a stretch to get that from the question as written. $\endgroup$ – David Z Apr 22 '12 at 3:38
  • $\begingroup$ @DavidZaslavsky: Except I had similar confusions a bazillion years ago. But it might be a different factor of 2, hard to tell, but I would bet it's the one I said at 60% confidence. $\endgroup$ – Ron Maimon Apr 22 '12 at 4:07
  • $\begingroup$ Derivation of Loternz force and Maxwell equations using Coulomb law and Lorentz transformations are like derivation below: docs.google.com/… . Analogical derivation I used for derivation the equals of gravitomagnetism (like in a ref.): en.wikipedia.org/wiki/Gravitomagnetism . $\endgroup$ – user8817 Apr 22 '12 at 10:18
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What's going on here is that the relativistic force is

$$ {d^2x^\mu \over d\tau^2} = - \Gamma^\mu_{\alpha\beta} \dot{x}^\alpha\dot{x}^\beta$$

Where the Christoffel symbol is

$$ \Gamma^\mu_{\alpha\beta} = -{1\over 2}g^{\mu\nu}(g_{\alpha\nu,\beta} + g_{\beta\nu,\alpha} - g_{\alpha\beta,\nu} ) $$

The factor of 2 in your derivation almost cetainly the half in front of the right hand side. The metric tensor is the square of the length of a vector, and the preservation of length in the parallel transport is what requires the factor of 1/2, the weak field metric tensor is twice the Newtonian potential.

The weak field limit has $g_{\mu\nu}= \eta_{\mu\nu} + h_{\mu\nu}$, and the weak field gravitational force field $\Gamma$ is

$$ \eta_{\mu\nu} \Gamma^\mu_{\alpha\beta} = -{1\over 2} h_{\alpha\nu,\beta} -{1\over 2} h_{\beta\nu,\alpha} + {1\over 2} h_{\alpha\beta,\nu} $$

You want to expand this noncovariantly. In noncovaraint equations (like the Lorentz force law) you write the derivative of the velocity and the velocity using time, not proper time. To do this, separate out the 0 index, and write the force law (using ${d\over d\tau}=\gamma{d\over dt}$ as:

$$ \gamma {d \over dt }(\gamma v^i) = - \Gamma^i_{00} + 2\gamma \Gamma^i_{0j}v^j + \gamma^2 \Gamma^i_{jk} v^j v^k $$

Where $\gamma={1\over \sqrt{1-v^2}}$ (there's a reason nobody writes weak field GR in nonrelativistic notation). The result in terms of the weak field metric is:

$$ {d\over dt}{\gamma v_i} = {1\over 2\gamma} h_{00,i} -{1\over \gamma} h_{0i,0} - {1\over 2} v^j (h_{i0,j} + h_{j0,i} - h_{ij,0}) - {\gamma\over 2} v^j v^k (h_{ji,k} + h_{ki,j} - h_{jk,i}) $$

To get the Newtonian limit, set v to zero, and the time derivative of h to be small (in Newtonian mechanics, everything is changing slowly). Then the leading term leads to to identify the Newtonian potential as half the 00 component of h.

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  • $\begingroup$ Your guess where the wrong factor of 2 could have arisen may be right or wrong but I still think that if you show the whole derivation which is right, the OP may still find an error wherever it is, assuming that he or she followed a similar strategy, +1. $\endgroup$ – Luboš Motl Apr 22 '12 at 3:57
  • $\begingroup$ Thanks. But I have one another question: how I can "recover" factor 2 after using derivation below (representated in a text from a reference) for the Newtonian law? docs.google.com/… $\endgroup$ – user8817 Apr 22 '12 at 11:14
  • $\begingroup$ @Maxim_Ovchinnikov: You can't get Einstein's equations using the method in this source--- it is specific to a miracle of the linear Maxwell equations--- that the E field in any frame is the instantaneous Coulomb field up to radiative stuff, so that if things are moving in straight lines at constant velocities, you can compute E. You can't do this in GR or linear GR in any way I know--- you need to know at least the on-diagonal linear field from a point mass in some gauge, and include pressure. The h-field is like the vector potential in EM. $\endgroup$ – Ron Maimon Apr 22 '12 at 16:33
  • $\begingroup$ @Maxim_Ovchinnikov: I don't mean to discourage you --- it might be possible--- but then you should use weak-field Schwartzschild for the point-mass solution which you superpose, not just Newton (which is weak field and slow source velocity and slow test-mass velocity), and then you need to check if the integrated "Coulomb law" of weak field Schwartzschild is valid in all frames (I don't see why--- Dirac gauge is a miracle of EM). It is easier to just boost weak-field Schwartzschild, and the result is as above--- really complicated. $\endgroup$ – Ron Maimon Apr 22 '12 at 16:37

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