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We can think that the electric field and the gravitational field operate similarly in the sense that the forms of their governing laws (namely, Coulomb's law and Newton's law respectively) are strikingly similar. The only difference one can point out is that while the electric charges come in two flavors, the gravitational masses come in just one.

Now, I have read that when a charged particle moves, the electric field lines associated with are distorted in a fashion because of the finite time required for the information about the change in the position of the charge to get propagated. And, I have led to the understanding that this is the cause of the existence of the magnetic field (and that if calculus is used it can be proven mathematically).

So (if this is true then) why doesn't the same thing happen to the gravitational field? Why is there nothing like a gravitational magnetic field? Or, is there?

Note

I have changed the language and the tone of the question massively. Although the question was fairly well received, I believe it was really ill-posed. As pointed out by ACuriousMind in the comment, the ''reason''described here behind the existence of the magnetic field is something that cannot be found a good support for. But still, due to the similarity between the equations describing their static behavior of electric and gravitational fields, one can still ask as to whether a boost would create some sort of a gravitational magnetic field if the original frame only had a static gravitational field. As the accepted answer points out, the answer is, roughly, a yes but the Maxwell-type equations for gravity aren't as well-behaved as are the original Maxwell equations of electromagnetism--one should note. In particular, the equations for gravity take the Maxwell-like form in appropriate weak limits only in some appropriately chosen gauges--and aren't Lorentz covariant.

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    $\begingroup$ Can you provide a reference that really says that magnetic fields are caused by finite propagation and such? For me, electric and magnetic fields are just equal components of the electromagnetic field strength tensor, none is "the cause" of the other. Also, can you really write down what the "gravitational field" is? (That is also a point of contention between different approaches to a gauge view on GR) $\endgroup$ – ACuriousMind Jul 29 '14 at 17:22
  • $\begingroup$ Now I am not finding any source proving the points that you want to be proved...I had a confusion regarding the generation of magnetic field due to the motion of a charge...facebook.com/events/539159816188874 In context of this question I found the points that I have talked here... $\endgroup$ – Dvij Mankad Jul 29 '14 at 17:34
  • $\begingroup$ More on gravitoelectromagnetism. Related: physics.stackexchange.com/q/944/2451 , physics.stackexchange.com/q/15990/2451 , physics.stackexchange.com/q/54942/2451 and links therein. $\endgroup$ – Qmechanic Jul 29 '14 at 18:39
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There is a sort of analog called gravitomagnetism (or gravitoelectromagnetism), but it is not discussed that often because it applies only in a special case. It is an approximation of general relativity (i.e. the Einstein Field Equations) in the case where:

  • The weak field limit applies.
  • The correct reference frame is chosen (it's not entirely clear to me exactly what conditions the reference frame must fulfill).

In this special case, the equations of GR reduce to:

$$ \begin{align} \nabla\cdot \vec{E}_g &~=~ -4\pi G \rho_g \\[5px] \nabla\cdot \vec{B}_g &~=~ 0 \\[5px] \nabla\times \vec{E}_g &~=~ -\frac{\partial \vec{B}_g}{\partial t} \\[5px] \nabla\times \vec{B}_g &~=~ 4\left(-\frac{4\pi G}{c^2}\vec{J}_g+\frac{1}{c^2}\frac{\partial \vec{E}_g}{\partial t}\right) \end{align} $$

These are of course a close analogy to Maxwell's equations of electromagnetism.

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    $\begingroup$ Can't we predict gravitational magnetism using classical mechanics only...( without using concepts of relaivity and spacetime ? ) $\endgroup$ – Dvij Mankad Jul 29 '14 at 17:44
  • $\begingroup$ @Dvij I don't think so... I can't think of any way in classical mechanics to get the angular momentum of one body (gravitational analogue of the dipole moment) to couple to the mass of another. $\endgroup$ – Kyle Oman Jul 29 '14 at 17:51
  • $\begingroup$ @Kyle: Try assuming the coupling is nonzero and crunch. I proved L of a photon is independent of wavelength using no QM or Rel assumptions that way. $\endgroup$ – Joshua Jul 29 '14 at 18:06
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    $\begingroup$ @Dvij: you can't predict normal magnetism using classical mechanics only. To see that you need magnetism if you have an electrical force, you have to make an appeal to Lorentz invariance. $\endgroup$ – Jerry Schirmer Jul 29 '14 at 18:09
  • $\begingroup$ @Joshua hmmm, can you still call it classical mechanics in that case? $\endgroup$ – Kyle Oman Jul 29 '14 at 18:15
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There is a gravitational analogue of the magnetic field. See gravitoelectromagnetism and frame dragging on Wikipedia.

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    $\begingroup$ I don't know GR...Can you explain only significance without using terminology of GR ? $\endgroup$ – Dvij Mankad Jul 29 '14 at 17:41
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    $\begingroup$ @Dvij See the Wikipedia gravitoelectromagnetism article. Gravitoelectromagnetism is wholly analogous to Maxwell's equations together with an ever-so-slightly different Lorentz force law. GEM is a known weak field approximation to the Einstein field equations; it's really like Newtonian gravity with a time delay - so kind of a minimum effort to marry Newtonian gravity with the special relativistic ruling out of faster-than-light signalling. Note that GEM is not Lorentz-covariant (even though Maxwell's equations are), since there is no obvious way to make the source (mass and its current) ... $\endgroup$ – WetSavannaAnimal Jul 3 '16 at 6:02
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    $\begingroup$ @Dvij .... into a four-vector, unless you use rest mass and its current (which would leave known gravitational effects unexplained). $\endgroup$ – WetSavannaAnimal Jul 3 '16 at 6:03
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The reason that magnetogravitational fields don't appear in purely Newtonian gravitation is that magnetism is actually a relativistic effect. If you use the CGS system of units, you'll see that only the quantity $B/c$ appears in the Lorentz force law. The nonrelativstic (Newtonian) limit is equivalent to the limit $c \to \infty$, so in this limit the magnetic fields drop out entirely.

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protected by AccidentalFourierTransform Jun 6 '18 at 1:24

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