6.1 Introduction

   In the laboratory, we are familiar with electrical currents, and with magnetic fields caused by the currents. Solenoids play an important role in technology. A well known cosmic analogy to solenoids are sunspots. There, we measure the magnetic field (by the Zeeman splitting of the spectrum of light from sunspots). We infer the electrical currents, and we find that a sunspot is much like a solenoid in the laboratory (see problem 6.2).

   In the 1940's, new observations showed that magnetic fields must be truly cosmic: radio astronomers showed that much cosmic radio radiation is polarized, and optical astronomers showed that light from the stars in the Milky Way becomes polarized as it travels toward us. Only magnetic fields can explain the preferred directions inherent in the polarization of the radiation.

   Magnetic fields can be detected everywhere, in and near planets, in and near stars, between stars, and throughout galaxies. The associated electrical currents and I x B forces must be inferred, with much uncertainty. Given that uncertainty, theory has emphasized rather simple concepts that are equivalent to mechanics and gas dynamics: magnetic energy can do work like any other reservoir of energy (problem 6.3), magnetic pressure acts much like a gas pressure (problems 6.4, 6.5), and magnetic tension acts like the tension on a violin string. Moreover, the concept of magnetic induction can be used to visualize magnetic fields and to convert the magnetic fields from a mathematical construct to what is practically a physical quantity (problem 6.6). Thus the topic of cosmic magnetic fields can be used in physics courses as example of energies, pressures, waves, and Faraday's law of induction.

6.2 Sunspots and their tera-Amperes
      solenoid - magnetostatics - Zeeman effect

The problem: Treat a sunspot as a solenoid much longer than its diameter: The spot magnetic field, vertical relative to the surface of the Sun, is B = 0.15Tesla. Electrical current, horizontal relative to the surface of the Sun, encircles the magnetic field. Find the current per unit length of the solenoid (in Amp/m). Find the total current (in Amp) if the depth of the sunspot is 3x104km.


Credit: Lockheed Research
Laboratories and NASA


The setting: Most of the solar surface is at a temperature of about 5800oK. A photograph of the Sun shows "dark" spots. The fairly uniformly "dark" central portion, typically about 104km in diameter, has temperatures as low as about 4000oK. Thus it is less bright than most of the surface according to the Stefan-Boltzmann law, but the so-called darkness is only relative to the surroundings. A spectrum of the light from a sunspot shows spectrum lines split by the Zeeman effect. Typically, the magnetic field is vertical relative to the surface of the Sun, it has a relatively sharp boundary, roughly 103km thick, and the value of the field strength in most spots is about 0.1 to 0.2 Tesla. Since all the Sun is gaseous, there cannot be any solid magnetic material. The magnetic field must be due to an electrical current running through the solar gas. The gas is sufficiently ionized so that free electrons can easily carry the current.

   A model of the sunspot is that of a solenoid: wire wrapped tightly in the form of a cylinder. The "wire" is the gas in the boundary, proportional 103km thick, and the cylinder is the sunspot, with diameter proportional 104km. The length of the solenoid corresponds to the depth to which the sunspot reaches inside the Sun.

   Strictly, the magnetic field inside a solenoid is uniform only if the solenoid is much longer than its diameter. For our estimate, we shall assume this is true. In reality, however, we know only that a typical spot is at least 104km deep (based on the details observed at the surface), probably about 3x104km deep (based on observations of sound waves interacting with the sunspot), and less than 105km (=1/7 Ro) deep (based on the details of sunspots emerging from the interior).

The solution: The formula for the magnetic field inside an "infinitely" long solenoid wound by N turns of wire per meter and carrying current I is B = muNI where mu =4pi x 10-7 if B is in Tesla and I in Amperes. Given B = 0.15Tesla, we find NI = 1.2x105amperes/meter. This is the current encircling the solenoid for each meter of length. (The value of N is irrelevant for continuous gas.) For a depth of 3x104km, the total current encircling the sunspot is 4x1012Ampere!

Interpretation: Compared to electrical currents running in our cities, the currents running in the Sun are enormous. Quite generally, astronomical currents are well beyond our imagination. For instance, the magnetic field between the stars is roughly 3x10-10Tesla; it is arranged in magnetic structures (not necessarily in the shape of a solenoid) with dimensions of the order of 10 light years = 1017m; therefore, currents of the order of 3x1013Amperes must commonly run between the stars. Astrophysical gases must be thought of as giant electromagnets. Moreover, because the "wires" are so thick (103km for sunspots, 1014km between the stars), currents are not turned into heat (problem 6.6). In this sense, astrophysical gases are rather like superconducting electromagnets.

Didactics: Because there are no solid magnets in astrophysics and all the currents in gases are accounted for explicitly, there is no need to consider separately the magnetic field H and the magnetic induction or flux density B . In astrophysics, B is referred to as the magnetic field.

6.3 Solar coronal mass ejections
      conservation of magnetic and kinetic energies

Physics introduction: In the laboratory, a wire carrying current I in a magnetic field B has a force exerted on it which is IxB per unit length of wire. In astrophysics we deal with continuous gases. We replace I by the current flowing through a unit cross-section, j. Then the force exerted on the gas is jxB per unit volume. In astrophysics, j is not observed, but B is observed, and B can be easily thought of in terms of magnetic lines of force : Lines of force are familiar in terms of iron filings around a bar magnet and in terms of the Van Allen belts in the Earth's magnetosphere. Therefore, simple ways of thinking have been developed that allow us to focus on B and on field lines rather than on j.

   Simple ways of thinking always start with energy. We know from theory that magnetic fields involve energy, specifically an energy density B2/2mu j/m3. When jxB forces accelerate gas, they do work on the gas, and the kinetic energy of the gas comes from B2/2mu.

The setting: If we cover up the bright solar disk and observe the edge of the Sun against the dark sky, we see many features that look like loops standing over the Sun. Their "feet" are on the solar surface, in regions with sunspots. The shape of the loops reminds us of magnetic field lines near magnets, consistent with sunspots. We recognize the loops because they contain somewhat denser gas than the surroundings.
   Photo Credit: National Optical Astronomy Observatories

   Many of these loops remain nearly constant for days. Why do the gases not fall to the surface? The shape of the loops suggests that the gases are held up against gravity by j x B forces. Electrical currents must be running there, but we visualize only the magnetic field.

   When we image the Sun using x-rays, that is, when we look at hot coronal gas, then we also see many loops standing on the Sun. These, also, usually last many days. But almost daily, one or another of these loops is suddenly blown away from the solar corona, achieving speeds in the range 400 to 1000 km/s. The ejected gases involve a mass typically of the order of 1012kg. How does the gas acquire so much kinetic energy? Apparently the magnetic fields in the solar corona, previously steadily holding up the embedded gas with j x B forces, suddenly lose their equilibrium. Then the j x B forces accelerate the gas.

   These phenomena are called "coronal mass ejection", abbreviated as CME. When a CME hits the Earth's magnetosphere, the resulting electric currents may damage satellite electronics, disturb communications, cause electrical power failures, and more. This problem concerns the predicted speed of the CME.

The Problem: Assume that ionized hydrogen gas in the solar corona has a density n = 1015 protons m-3 (and the same number density of electrons) and a magnetic field of strength 10-3Tesla. If all the magnetic energy in a cubic meter is converted into kinetic energy of the gas initially in that cubic meter, what speed does this gas attain?

The Solution: Setting 1/2 rho v2 = B2/2mu yields v = B/(murho)1/2 = 700km/s.



Credit: NASA



Interpretation: The estimate made in the problem is actually a fairly realistic estimate. The gas density is derived from the intensity of x-radiation. The magnetic field is measured at the surface (Zeeman splitting) and extrapolated. The predicted speed fits well into the range of observed speeds, 400 to 1000 km/s. The effect of these gases arriving at Earth appears in problem 6.4.

Didactics: The treatment of cosmic electrical currents and magnetic fields, when done in detail, involves nonlinear partial differential equations, which are very difficult to solve. Thus most of the theory of the last forty years has dealt primarily with a few simple soluble problems, such as energy estimates, as in this problem, or forces in very simple magnetic geometries, as in the next two problems. So far, no computer program can start with a steady loop in the corona, change it so as to lose its equilibrium, compute its acceleration, and follow the CME into space. But cameras on the space craft SOHO, which orbits permanently between Earth and Sun (at the LaGrangian point), can now warn us when CME's are approaching the Earth.

6.4 The solar wind and the Earth's magnetosphere
      magnetic and dynamic pressures

Physics Introduction: In Maxwell's equation, we neglect the term c-2partial_derivativeE/partial_derivativet term which was introduced by Maxwell. The physical approximation made by this neglect is outlined below. What remains of Maxwell's equation, curlB = muj, mu = 4pix10-7, is used to express the current density j in terms of B . The force on a unit volume of gas can now be written jxB = curlB x B/mu = -grad(B2/2mu) + (Bdotgrad)B/mu . In the first term, B2/2mu is identified as magnetic pressure, quite in analogy to the normal gas pressure. The second term corresponds to a tension (force per unit area) of B2/mu along the lines of force. Since pressure and tension are familiar quantities from the laboratory, it is relatively easy to think of magnetic pressure and magnetic tension.

   The tension acting on magnetic field lines is quite analogous to the tension on a violin string: The wave speed along the string is (tension / mass per unit length) 1/2. For a continuous gas, we replace tension by force per unit area, B2/mu, and we replace mass per unit length by mass per unit volume, rho. Then the wave speed along the magnetic field line is (B2/murho)1/2. The wave is called an Alfvén wave, traveling at a speed called the Alfvén speed.

   Here we deal with magnetic pressure.

The problem: Estimate the distance from the Earth where the solar wind meets the Earth's magnetosphere. Specifically, how far from the Earth (r in units of Earth's radius rE) at the magnetic equator is the magnetic pressure B2/2mu of the Earth's dipole magnetic field equal to the dynamic pressure rhov2 exerted by the solar wind? Use B = 10-4T at the Earth's surface, and for wind parameters use pure ionized hydrogen, np = 107m-3, v = 500 km/s.

The setting: The Earth's magnetic field shields the Earth's surface from the direct impact of the solar wind (or of a comet's poisonous tail). The volume controlled by the Earth's magnetic field, called the magnetosphere, has a tear-drop shape. It is round where it faces the solar wind. Its tail is very long and literally flaps like a flag in the solar wind. In first approximation, the magnetosphere is simply a blunt obstacle to the wind. At the sunward stagnation point, the pressure imparted by the wind must be balanced by magnetic pressure due to magnetospheric fields. This problem asks for an estimate on how far from the Earth the magnetic pressure can balance the wind pressure.




The solution: The strength of the Earth's magnetic field falls off as a dipole, proportional r-3. Take B = BE (rE/r) 3. We need B2/2mu = 4x10-3 (rE/r) 6 = rhov2 = 4x10-9. Thus r = 10rE.

Interpretation: Obviously this is only an estimate. A dipole field carries zero electrical current and thus cannot produce a force. However, electrical currents are induced by the impact of the wind. The currents alter the Earth's magnetic field such that the new forces stop the wind. In first approximation, the currents run on the surface and the dipole field is terminated abruptly. Then magnetic pressure is the relevant force, balancing the external pressure of the wind.. The estimate for the standoff distance is really quite good. However, the details of the boundary and its interactions with the wind are still not quite settled, even after three decades of satellite exploration, because the solar wind carries its own magnetic field. When a coronal mass ejection (CME, see problem 6.3) suddenly hits the magnetosphere, the front of the magnetosphere is compressed to a smaller radius. New electrical currents are induced throughout the magnetosphere, where they endanger satellite electronics, and in the ionosphere, where they interrupt radio communications, and on long conductors on Earth surface, where they can interrupt electrical power systems serving millions of people. CME's can now be watched by satellites all the way from their launch at the Sun to their arrival at Earth, but they can still provide surprises. One CME, in January 1997, seemed to pass quietly until an order-of-magnitude pressure pulse arrived at the end. That pulse compressed the magnetosphere so power-fully that the resulting currents short-circuited and destroyed a new communications satellite valued at 400 million dollars.

Didactics: Maxwell added the last term to what is now called Maxwell's equation, curlB = muj + c-2partial_derivativeE/partial_derivativet. Now we omit this term. Thus we cannot deal with vacuum electromagnetic radiation : In the vacuum, necessarily j = 0, but curlB is finite, so partial_derivativeE/partial_derivativet must be finite. The relation curlB = muj implies divj = div curlB = 0 (because divB = 0). Thus we can deal only with problems where there is no accumulation of electric charge. The equation cannot describe most waves invoked in plasma physics. It can, however, describe most phenomena that happen on long time scales, that is, much longer than any plasma or cyclotron frequencies. This is the subject of magnetohydrodynamics, usually abbreviated MHD, a subject begun about 50 years ago by H. Alfvén. It could have been investigated even before Maxwell.

6.5 Sunspots and the Earth's climate
      magnetic pressure

The Problem: A cylindrical sunspot is in pressure equilibrium, such that the internal pressure nikTi + B2/2mu equals the external pressure nokTo. The magnetic field strength outside the spot is zero, inside it is B = 0.15 Tesla. The temperatures are To =5800oK and Ti = 4800oK. The density outside is no = 2x1023H atoms m-3. Find the density inside, ni. Compare nito no.

The setting: Sunspots "live" some days or weeks. As the Sun turns, the spots move with the Sun. Seen from earth, the spots apparently move across the disk of the Sun. They become invisible before they are turned fully toward the edge of the Sun. Apparently, the surface of the sunspots is lower than the normal solar surface. Therefore, the gas at any one height must be more transparent within the sunspot than outside it. Why is that? Because of the magnetic pressure, the gas pressure inside the sunspot is much smaller than outside, as shown in the problem, and so is the gas density.

The solution: Outside of the sunspot, nokTo = 1.7x104, inside B2/2mu = 0.9x104, so that inside nikTi =0.8x104, and ni =1.2x1023m-3 = 0.6no.

Interpretation: In the solar atmosphere, gas is stratified vertically according to the equation of hydrostatic equilibrium (problem 5.2). But the stratification can be different outside and inside a sunspot because of the spot magnetic field. At the height of the normal photosphere, i.e. from where most solar radiation can leave into space, the sunspot is relatively empty and transparent. The spot surface resides further down, where the density is more like that of the normal photosphere. Since the scale height of the pressure distribution in the solar surface is about 180 km (problem 5.2), the spot surface is about 200km below the solar surface. This roughly fits the observations.

Sunspots and Earth's climate: A sunspot is less bright than the surrounding photosphere because it is less hot. Why is it less hot? In the surroundings, convection brings up heat which must then be radiated away at the same rate. The energy flow determines the temperature at the solar surface (Stefan-Boltzmann law). The spot's magnetic field resists the convective motions, so that less energy is convected up from the interior within the spot. The arriving heat can be radiated away at a lower temperature.

   The number of sunspots varies in a cycle of approximately 11 years. One might expect that Earth receives less sunlight in years when there are many spots. Indeed, when any one spot appears, we do receive less light than just before. Surprisingly, according to satellite measurements, we receive roughly 0.1% more energy on average during years with many spots, and less during years with very few spots. Indeed, historical periods with long absences of sunspots are times when the Earth's climate was relatively cool.

   Why is the observation opposite to the prediction? In times when there are many spots, detailed observations of the solar surface also show very many very thin bundles of magnetic fields with B proportional 0.15Tesla. These are solenoids similar to sunspots but merely some 100 km in diameter. Like sunspots, they have a depressed surface. However, they are so thin that most of their radiation is lost from the sides rather than from the depressed surface. We observe the walls of the solenoids. They are hotter than the solar surface, and so we receive more energy.


  6.6 Magnetic fields of white dwarfs and neutron stars
      Faraday's law of induction

Physics Introduction: In the laboratory, if a wire heats up because of electrical resistance, one uses a thicker wire. Cosmic magnetic fields imply electrical currents flowing in extremely thick gaseous "wires", so thick that dissipation of the currents into heat is negligible over time scales of interest. Effectively, there is no electrical resistance. Therefore, the electrical field in the frame of the ions vanishes. The electrical field seen in any other frame of reference is E = -vxB, and Faraday's law of induction becomes deltaB/deltat = -curlE = curl(vxB). The integral form of Faraday's law is delta/deltatintegralBdotds = integralvxBdotdl where the integrals are over any fixed surface and around the edge of that surface. On interchanging dot and x in the line integral, one can show geometrically that Faraday's law reduces to d/dtintegralBdotds = 0 , where now the integral is over any surface moving with the gas. The magnetic flux through any surface moving with the gas is conserved.

   Imagine two elements of gas connected by a field line at some time. Draw a surface, generally not flat, which includes this field line and which also includes neighboring field lines. This surface has zero magnetic flux through it. Let this surface move with the gas. By Faraday's law without current dissipation, the surface must continue to have zero magnetic flux through it. Therefore, the same field line still connects the two elements of gas. In effect, the field lines are attached to the gas. One says the magnetic field is "frozen into" the gas. Although magnetic field lines are mathematical constructs, in MHD they take on a real physical property. That is why, for instance, the vibration of field lines and gas participating in an Alfvén wave is completely analogous to the vibration of a string with tension as in a violin.

The problem: Imagine that the Sun, in its interior, contains a magnetic field encircling the axis of the Sun, parallel to the solar equator, much as if a giant solenoid were stretched around the Sun at its equator. Now imagine making the Sun smaller, keeping the same structure of density and frozen-in magnetic field. If the field strength B changes in proportion to Rn, what is n? If the field strength within the Sun is 10 Tesla, what is it after the Sun shrinks to the size of a white dwarf, and after the Sun shrinks to the size of a neutron star? For the stellar radii, use R = 106km, 104km, and 10 km, respectively.

The solution: It is simplest to consider any part of a plane through the axis of the star, that is, normal to the magnetic field, for instance a circle within that plane. Follow the motion of the gas on the boundary of that circle as the circle shrinks. The magnetic flux through the circle must remain constant. But the area of the circle, attached to the gas, shrinks as R-2. Thus n = -2, and B/Bo = (Ro/R) 2. The white dwarf then is expected to have B = 105Tesla, the neutron star B = 1011Tesla.

Interpretation: The strongest magnetic field observed on a white dwarf is 5x104Tesla. Magnetic fields of neutron stars observed as radio pulsars have values up to 4x109Tesla. The highest field deduced in an x-ray pulsar is 1.6x1011Tesla. So the estimates are quite satisfying. One can really expect only order-of-magnitude agreement because the solar B = 10 Tesla is estimated only rather indirectly, from the manner how sunspots emerge from the deep solar interior; because the white dwarf fields are left in the star after the star has shed its outer layers; and because the neutron star fields are left after the star exploded as a supernova. Moreover, it is possible that powerful convection during the supernova explosion amplified the field in the neutron star. Nevertheless, it seems reasonable that the observed fields on white dwarfs and neutron stars derive from the compression of magnetic fields that were within the stars when these stars were main sequence stars.

   Exceptions to the frozen-in approximation occur at very powerful events, such as solar flares (whose explosive energy must derive from powerful localized electric currents suddenly dissipated in the solar corona), the turbulence during a supernova explosion, the interaction of the Crab Nebula with the pulsar at its center (since the magnetic fields in the nebula now are far too strong as to have expanded in frozen-in form from the original supernova and its progenitor star), and the turbulence of the gases between the stars (which must have created the observed interstellar galactic magnetic field), and at singular geometries such as the front and back of our magnetosphere (diagram in problem 6.4).

Didactics: When electrical currents are negligibly dissipated into heat, this is often interpreted as "zero resistance" or "infinite conductivity", inspiring the analogy with superconductors. However, the conductivity of many astrophysical gases is within an order of magnitude of the conductivity of copper in Earth's laboratories. Negligible dissipation in astrophysics is literally due to the extreme thickness of the "wires" (mathematically, the very large value of the length B/|curlB|).

   Possible students' problem. (Students should discuss each part and agree on the answer before moving on). Suppose the magnetic field now in the Crab Nebula, B = 4x10-8Tesla, were truly frozen into the expanding nebular gas, approximately a sphere of radius r = 3 light years. Earlier, r was smaller and B was higher. With what exponent does B depend on r? (Answer: B is proportional to r-2, as for the stars.) How do the magnetic energy density and the total magnetic energy in the nebula depend on r? (Answer: r-4 and r-1, respectively). Estimate backwards in time: what was the total magnetic energy when r equaled the radius of an about-to-explode star, say r = 150Ro = 1011m ? (Answer: Now the magnetic energy density = 2/pi x 10-9j m-3, the volume = 3.6 pi x1049m3, the total energy = 7.2x1040j; scaled by a factor 3x1016-11, the total magnetic energy then = 2x1046j.) Compare to the gravitational energy of the original star, assuming M = 10Mo, and interpret. (Answer: GM2/R proportional 2.6x1041j << 2x1046j. No stable star can exist with negligible gravitational energy.) [The Virial theorem extended to MHD has the total kinetic energy K replaced by K + total magnetic energy.]