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In astrophysics there is a lot going on about strong, large scale magnetic fields: in stars (prominences), magnetic dynamos, compact accretors collimating jets, etc. There's even a special computational formalism called magnetohydrodynamics (MHD), which allows to deal with space plasma.

Yet I've never read about large scale electric fields. I know that most of the matter we model in astrophysics is plasma but, naively, one would assume that this introduces both $\mathbf{E}$ and $\mathbf{B}$ fields on an equal footing.

So where does this asymmetry come from?

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I work in astrophysics and have wondered this from time to time myself. I like the first answer; maybe it also has to do with special relativity telling us that $E$ and $B$ are essentially the same entity? (Just change your reference frame and one becomes another). –  o0BlueBeast0o Jul 8 at 12:34

4 Answers 4

up vote 26 down vote accepted

Many astrophysical plasmas are well modeled as perfect conductors. Ideal MHD assumes this limit. As a result, there is no electric field in the fluid's rest frame. In other frames, we generally have $\vec{E} = -\vec{v} \times \vec{B}$, so there is an electric field. However, the perfect conductivity constraint means we don't have to model the electric field - if we evolve just the magnetic field (and the other properties of the fluid like its velocity and density), then we have the complete picture.

The natural followup question is, "Why can we assume infinite conductivity?" Most people's intuition about space is that it is mostly vacuum, and vacuum seems like as good an insulator as one can find. The thing about vacuum is that even though there are few charge carriers per unit volume, what charge carriers there are can proceed uninterrupted and respond to any electric field.

The book Physics of the Interstellar and Intergalactic Medium (Bruce Draine) gives some equations to quantify this. In eq. 35.48 it gives the conductivity of a pure hydrogen fully ionized plasma at temerature $T$ as $$ \sigma = 4.6\times10^{9}\ \mathrm{s}^{-1} \left(\frac{T}{100\ \mathrm{K}}\right)^{3/2} \left(\frac{30}{\log\Lambda}\right) $$ (CGS units), where kinetic effects and the Debye length are approximately captured by the Coulomb logarithm $$ \log\Lambda = 22.1 + \log\left(\left(\frac{E}{kT}\right) \left(\frac{T}{10^4\ \mathrm{K}}\right)^{3/2} \left(\frac{n_e}{\mathrm{cm}^{-3}}\right)^{-1}\right) $$ (eq. 2.17). Here $E$ is the particle kinetic energy, and $n_e$ is the number density of electrons.

To give some sense to these numbers, take a look at the conductivities in this table on Wikipedia. Copper has a conductivity of about $6.0\times10^7\ \mathrm{S/m} = 5.4\times10^{17}\ \mathrm{s}^{-1}$, so a $100\ \mathrm{K}$ hydrogen plasma is not nearly as conductive. However, drinking water has a conductivity of no more than $5\times10^{-2}\ \mathrm{S/m} = 4.5\times10^{8}\ \mathrm{s}^{-1}$, and air's conductivity is at most $8\times10^{-15}\ \mathrm{S/m} = 7\times10^{-5}\ \mathrm{s}^{-1}$. Thus astrophysical plasmas are not particularly insulating.

Bruce Draine's book also quotes a timescale for a magnetic field to decay over a lengthscale $L$: $$ \tau = 5\times10^{8}\ \mathrm{yr}\ \left(\frac{T}{100\ \mathrm{K}}\right)^{3/2} \left(\frac{30}{\log\Lambda}\right) \left(\frac{L}{\mathrm{AU}}\right)^2 $$ (eq. 35.49). Thus if the smallest length scales in your problem are at least $10\ \mathrm{AU}$ (and you are working around $100\ \mathrm{K}$), the magnetic field decay time due to the finite conductivity of the plasma (due in turn to e.g. ion collisions) is well over the current age of the universe. On smaller scales you may have to model such effects, and indeed many astrophysicists do just that.

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The lack of the electric field in modeling plasmas stems from the Lorentz force, $$ \mathbf F=q\mathbf E+q\boldsymbol\beta\times\mathbf B $$ where $\boldsymbol\beta=\mathbf v/c$. For most astrophysical plasmas, the force is zero, so we have that $$ \mathbf E=-\boldsymbol\beta\times\mathbf B $$ So any time we see an electric field, we can simply replace it with the above cross product. This happens in Faraday's law: $$ \frac{\partial\mathbf B}{\partial t}=-\nabla\times\mathbf E=-\nabla\times\left(-\boldsymbol\beta\times\mathbf B\right)=\nabla\times\boldsymbol\beta\times\mathbf B $$

The justification for $\mathbf F=0$ is as Chris White says: we assume perfect conductivity.

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Probably because matter on a large scale is electrically neutral and therefore the electric effects cancel out .This asymmetry arises from the fact that atom as a whole is electrically neutral.

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I think this answer is only partially correct. A star is neutral because the gravitational force is to weak to overcome the electrostatic force between to equally charged particles, not because an atom is neutral. –  Noldig Jul 8 at 12:36
    
voted -1 for it doesn't address the question. It doesn't explain why magnetic effects take over; one could equally as well say that since there are no magnetic monopoles, the atom as a whole is also magnetically neutral. –  Arash Arabi Jul 8 at 16:24
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The atom being electrically neutral ($n_p=n_e$) leads to the large scale neutral state because the plasmas are generally hot enough to consist of only ions & electrons. Note also that local charges can (and do) build up (see any "particle in cell" (PIC) simulation). –  Kyle Kanos Jul 9 at 14:25

There exists a basic asymmetry: there are no magnetic monopoles of the number and dimensions that the electric monopoles exist ( there are theories with magnetic monopoles and people are looking for them but we are talking of masses much larger than electrons and quarks) .

Hand waving, (because I have not checked the math just extending the symmetry that would have to be imposed in maxwell's equations) , if there existed magnetic monopoles of the size/strength of electric ones ( electrons, quarks) then an alternate state of neutral (magnetically neutral) matter could have appeared, where the spill over electric dipole forces would dominate also over magnetically neutral matter the way magnetic forces do over electrically neutral matter, symmetrically. Then it would be relevant for astrophysics plasma observations, but this is science fiction.

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voted -1 for it doesn't address the question –  Arash Arabi Jul 8 at 16:21
    
@ArashArabi thats fine, but it is complementary to the lorenz force answer of Kyle. –  anna v Jul 8 at 17:22

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