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Gravitational Magnetism

  1. Jun 29, 2011 #1
    I was working with Maxwell's equations recently when I remembered Gauss' Law for gravity. I couldn't help but wonder if there was some sort of parallel between gravity and electricity, in the sense that one might be able to construct "Maxwell's Equations" (obviously they wouldn't really be his) for gravity and gravitational magnetism (I know that magnetism is just a relativistic effect of electricity, so I suspect that there should be a similar relativistic effect for gravity).

    This was my guess as to what they would look like:

    [tex]\nabla \cdot \mathbf{g} = -4\pi G\rho [/tex]
    [tex]\nabla \cdot \mathbf{b} = 0[/tex]
    [tex]\nabla \times \mathbf{g} = -\frac{\partial \mathbf{b}}{\partial t}[/tex]
    [tex]c^2~\nabla \times \mathbf{b} = \frac{\partial \mathbf{g}}{\partial t}-4\pi G\mathbf{J}[/tex]

    where g is the standard gravitational field, b is the "gravitational magnetic field," and J is the "mass current density" (though I have no idea what that would be).

    Anyone have any thoughts?
  2. jcsd
  3. Jun 29, 2011 #2


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    Gauss' Law is actually a general mathematical tool - it is not unique to either gravity or electromagnetism. Also, magnetic fields are not relativistic effects. They can clearly be seen at non-relativistic energies.

    Your first equation actually does exist. It is Newton's Law of Gravity! You typically see it as [itex]\nabla^2 \phi = 4\pi G \rho[/itex]. I don't think there is any reason to believe there is an analogue to the magnetic field in Newtonian gravity, however.
  4. Jun 29, 2011 #3
    Are you sure you aren't thinking of Gauss' Divergence Theorem?

    I don't know how to say it in any way other than just being frank - you're wrong. Magnetic effects do (obviously) occur at lower energies, but they can still be thought of as a consequence of relativistic effect (mainly length contraction).

    http://physics.weber.edu/schroeder/mrr/MRRhandout.pdf#search="purcell simplified"

    I realize that the first equation exists (it's Gauss' Law for gravity) - this post was mainly about the other three.
    Last edited: Jun 29, 2011
  5. Jun 29, 2011 #4


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    No. What you basically are looking at is Laplace's equation with a source term.

    When one says "X is a relativistic effect of Y", one usually means to say that X is not observed (or explained properly) unless relativity is taken into account (ie hyperfine splittings, Thomas procession, etc) or you're talking about high energy situations. Relativity is a consequence of electromagnetism so it is no surprise that you can speak of one giving rise to the other. In fact, you can actually make a couple very simple postulates and assume a linear transformation (this is a problem in Jackson) and out comes the Lorentz symmetry. Take it a step further and you can derive e/m (and thus Maxwell's equations) based off that symmetry alone using gauge fields.

    The problem with Newtonian gravity is that the Lorentz symmetry is not a symmetry of it. If you could actually make a sort of "Newton's equations" like you're trying, you would be able to identify a Lorentz symmetry with it and we know that's not possible. However, you CAN actually linearize Einstein's Equations to look like Maxwell's Equations. It's a weak-field assumption, though.
  6. Jun 29, 2011 #5


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  7. Jul 10, 2011 #6
    Unfortunately, that Gravitomagnetism article on the Wikipedia is about as good as you can hope to get searching the internet -- unfortunately, there seems to be a lot of quackery surrounding the subject.

    Oleg Jefimenko wrote a lot on the topic, and while he has gained a bit of a reputation for stoking the aforementioned quackery, he at least provides one valuable resource that I wasn't able to find elsewhere -- and seems to have since been removed. It's Heaviside's discussion of the same topic from around the turn of the (20th) century; luckily, archive.org seems to have scraped a link:

  8. Nov 18, 2011 #7
    Could magnetism from very large metallic objects such as iron cores in planets be mislabeled gravity? Can magnetism attract non metallic matter if it is large enough? Are there papers or experiments to validate/invalidate this hypothesis?
  9. Nov 18, 2011 #8


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    Why not?

    The classical electromagnetic effect is perfectly consistent with the lone electrostatic effect but with special relativity taken into consideration.

    The simplest hypothetical experiment would be two identical parallel infinite lines of charge (with charge per unit length of [itex] \lambda \ [/itex] and some non-zero mass per unit length of [itex] \rho \ [/itex] separated by some distance [itex] R \ [/itex]. If the lineal mass density is small enough that gravitational forces can be neglected in comparison to the electrostatic forces, the static non-relativistic repulsive (outward) acceleration (at the instance of time that the lines of charge are separated by distance [itex] R \ [/itex]) for each infinite parallel line of charge would be:

    [tex] a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} [/tex]

    If the lines of charge are moving together past the observer at some velocity, [itex] v \ [/itex], the non-relativistic electrostatic force would appear to be unchanged and that would be the acceleration an observer traveling along with the lines of charge would observe.

    Now, if special relativity is considered, the in-motion observer's clock would be ticking at a relative rate (ticks per unit time or 1/time) of [itex] \sqrt{1 - v^2/c^2} [/itex] from the point-of-view of the stationary observer because of time dilation. Since acceleration is proportional to (1/time)2, the at-rest observer would observe an acceleration scaled by the square of that rate, or by [itex] {1 - v^2/c^2} \ [/itex], compared to what the moving observer sees. Then the observed outward acceleration of the two infinite lines as viewed by the stationary observer would be:

    [tex] a = \left(1 - v^2 / c^2 \right) \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} [/tex]


    [tex] a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} - \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} = \frac{ F_e - F_m }{\rho} [/tex]

    The first term in the numerator, [itex] F_e \ [/itex], is the electrostatic force (per unit length) outward and is reduced by the second term, [itex] F_m \ [/itex], which with a little manipulation, can be shown to be the classical magnetic force between two lines of charge (or conductors).

    The electric current, [itex] i_0 \ [/itex], in each conductor is

    [tex] i_0 = v \lambda \ [/tex]

    and [itex] \frac{1}{\epsilon_0 c^2} [/itex] is the magnetic permeability

    [tex] \mu_0 = \frac{1}{\epsilon_0 c^2} [/tex]

    because [itex] c^2 = \frac{1}{ \mu_0 \epsilon_0 } [/itex]

    So you get for the 2nd force term:

    [tex] F_m = \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} = \frac{\mu_0}{4 \pi} \frac{2 i_0^2}{R} [/tex]

    which is precisely what the classical E&M textbooks say is the magnetic force (per unit length) between two parallel conductors, separated by [itex] R \ [/itex], with identical current [itex] i_0 \ [/itex].

    Why cannot the same thought experiment be used to obtain a gravitomagnetic effect for two parallel lines of mass and the inverse-square gravitational effect?
  10. Jun 13, 2012 #9
    A lot of scientists have found parallels between electromagnetism and gravity.
    Some of the differences between electromagnatism and gravity arise from the fact that there cannot be negative mass and that like masses attract rather than repulse.
    I will point to some links

    While General relativity is a very successful theory, any progress would come only by studying it at its breaking points. Gravitomagnetic effects need much more scruitiny I guess
  11. Jun 13, 2012 #10
    The reason gravity is always attractive (and the reason "antigravity"/"gravitational propulsion" would appear to be bogus) has nothing to do with mass/energy always being positive, where charge can be positive or negative; if that analogy held, wouldn't all gravitational forces be repulsive then (like repels like)?
    My understanding is that forces mediated by particles with even spin - like gravity (spin 2) and the inter-nucleon pionic yukawa force (spin 0)) - are always attractive.

    Just my two cents.
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