*A pulsar discovered last April is helping astronomers measure the magnetic field surrounding our galaxy’s central black hole.*

The supermassive black hole hiding in Milky Way’s core just got a new neighbor.

After decades of searching, astronomers discovered a young pulsar in the galactic center a few months ago. Now Ralph Eatough (Max Planck Institute for Radio Astronomy, Germany) and his colleagues are putting the pulsar to work spying on the behemoth next door. In this week’s Nature, they announce that the strong magnetic field they find surrounding the black hole might put the beast on a permanent diet.

Like L. A. smog, thick clouds of gas and dust hang over our galaxy’s downtown, hiding millions of stars from view. Only very long and very short wavelengths penetrate the veil. A flash of X-rays did just that on April 24th (which by cosmic coincidence also happens to be my birthday). Follow-up observations showed X-ray and radio-wave pulses arriving every 3.76 seconds, confirming that astronomers had discovered a bona fide pulsar in the galactic center.

Such pulsars shouldn’t be uncommon — theorists think there ought to be more than a thousand within spitting distance of the black hole — but somewhat mysteriously, this pulsar is the first find after a three-decade hunt.

The emission from this pulsar, named PSR J1745–2900, is almost all linearly polarized, meaning the wavelengths tend to oscillate along a single direction. That’s an advantage astronomers can use, because magnetic fields between the pulsar and Earth twist the waves as they travel, rotating the direction that they oscillate. The effect is known as Faraday rotation.

So, measure the twist, measure the magnetic field? Well, almost. How much the waves rotate also depends on the amount of hot, ionized gas between the pulsar and Earth. And that’s a little harder to measure since, surprisingly enough, we don’t know exactly where the pulsar is.

Though the pulsar was found just 3 arcseconds away from the black hole on the plane of the sky, the two could still be quite far apart in space. “[Distance] is always a problem for pulsars,” explains Eatough. “We cannot get accurate [direct] distance measurements without a parallax.” And parallax is impossible to measure for a pulsar so far away.

So Eatough’s team used indirect measurements instead. For example, ionized gas between the pulsar and Earth causes pulses at lower frequencies to lag slightly behind pulses at higher frequencies. This so-called dispersion measure is higher than that measured for any other known pulsar and suggests the pulsar lies within 33 light-years of the black hole.

Eatough and his colleagues also compared the pulsar’s location to two nearby pockets of ionized gas. In front of the black hole, hot gas glows bright in the X-rays; in back, a cooler gas stream emits radio waves. One or both of these gas pockets could affect the pulsar’s light as it travels through. The gas stream (technically known as the Northern Arm of Sagittarius A West) is especially well studied, and its polarized emission is far more rotated than the pulsar’s. So the pulsar probably lies in front of the cooler gas stream.

That means it’s the hot gas that twists the pulsar’s polarized light, gas that will ultimately become the black hole’s meal. Eatough’s team estimates the magnetic field threading the ionized gas is roughly 8 milligauss. That’s about 1,000 times stronger than the fields crisscrossing the rest of the galaxy.

A strong magnetic field would have important consequences for how the black hole feeds. Although there’s enough gas in the black hole’s neighborhood to feed it the equivalent of 30 Earths every year, only 0.003 Earth’s worth of gas makes it past the event horizon. A strong magnetic field might help the black hole diet by holding the gas back and preventing it from crossing the point of no return.

Fred Baganoff (MIT), who measured the hot gas reservoir surrounding the black hole a decade ago, cautions that a strong magnetic field isn’t necessary to explain why the black hole fasts. Despite gravity’s strong pull, it’s harder than you might think to swallow matter away into eternity. Much of the gas might never reach the black hole, surging outwards instead in a wind or a jet.

“I think we need theoretical studies now,” explains Heino Falcke (Radboud University, the Netherlands). “We have the density of the material [around the black hole], its temperature, and the strength of the magnetic field. That’s the only accretion disk of all accretion disks where you can say that. Now based on the observations, we can say, look guys, . . . this is what you should start with, now tell us what comes out of your simulations.”

## Comments

Peter

August 15, 2013 at 6:56 pm

Consider two particles of dark matter, each of mass M, initially at rest and infinitely far apart. According to special relativity (SR), they possess an innate energy, equal to 2Mc^2. Let them start falling towards each other. This “rest energy” will be converted into kinetic energy, in an amount equal to the gravitational energy lost, or GM^2/r, where r is their now-finite separation. The process of converting potential energy into kinetic energy must STOP, however, at some minimum separation, call it r(s), at which point the rest energy is all used up. This happens when: 2Mc^2 - GM^2/r(s) = 0. Solving for r(s), we find,

r(s) = GM/2c^2, which some of you will recognize as the Schwarzschild radius! In other words, the event horizon is not just a point-of-no-return. It is the point at which all the rest-mass-energy has been converted to either kinetic or radiant energy. Picture matter approaching the event horizon: it is moving almost the speed-of-light, and has radiated away immense quantities of energy as it spiraled down. We imagine it crossing the event-horizon, going even faster, but that would imply more energy than it started with. So what happens? The event horizon crosses it, not the other way around. Got it? A quantity of matter m cannot cross the event horizon without violating conservation-of-energy principles; ergo, it does not cross it. Instead, the event horizon expands by a distance of Gm^2/2c^2.

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Mike W. Herberich

August 16, 2013 at 9:20 pm

Dear Peter, what a beautiful thought experiment (like Einstein's "Gedanken-Experimente"!). I wasn't able to follow all of it, though, which might well be due to my ignorance, I suspect. 1., a formal question: your last term, Gm^2/2c^s has the dimension of [kg*m] which is a distance times a mass, as opposed to your calling it a mere distance. I suspect that is a slip of the pen, since in the middle of your article you rightfully call r(s)=GM/2c^2 a distance. Should it read Gm/2c^s instead?

Another question puzzling me for the first time, really, is: is the event horizon one and the same thing as the Schwarzschildt radius?

A third question would be: looking at a particle falling from afar into a black hole, could one use your primary thread of thought and say: r(s) = GMm/((M+m)*c^s), where M is the black hole's mass and m any particle falling into it?

If so, the Schwarzschildt formula would look pretty awkward now.

If I'm not totally out on a wrong limb I'd like you to elaborate a bit on either my questions or, in general, on the, say, last third of your post. Thank you very much in advance.

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Peter

August 17, 2013 at 10:01 am

Thanks, Mike. Yes, that’s a typo, should be: “the event horizon grows by a distance of Gm/2c^2.” I was astounded at the result of the this simple calculation. Why is it not in every physics textbook? It 1) adds confidence that the calculation of the Schwarzschild radius/event horizon is correct, because using a totally different approach yields the same answer, and 2) it uses conservation-of-energy to resolve the divide-by-zero conundrum. Elaborating a little: you may have read something like, “As much as 10% of the rest-mass energy of matter falling into a BH gets converted to radiant energy.” Throw 10 kg into a BH, and it only gains 9 kg, the other 1 kg being converted into radiant energy, mostly gamma and x-rays. As it nears the event horizon, it is circling at nearly the speed-of-light. But because of the effect of velocity on inertia, according to special relativity, the picture is not one of 9 kg moving at nearly c, which would have an inertia of much more 9 kg. Instead, the picture is one of an infinitesimal mass, traveling at a velocity infinitesimally less than c, having a non-infinitesimal inertia of…9 kg! The prediction is that BHs should be shaped like a ring, not a point-mass.

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Gordon Nanninga

August 17, 2013 at 12:05 pm

For a rotating emitter it is also possible to determine the distance by the apparent change in rotation by distance and relative motion. This takes fitting the pattern to a lot of observations and really stable time source. Having to wait half a year puts further requirements on the time source. At this distance it would probably have to be done at NIST.

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Mike W. Herberich

August 20, 2013 at 7:03 pm

Dear Peter, thank you for answering and elaborating. When you say "inertia" (in connection with special relativity) are you talking about m=m0*gamma, where m is any mass, m0 its rest mass and gamma=1/sqrt(1-(v^2/c^2)) ? If I get this correctly, shouldn't that result in much more than 10% of radiation, taken from the original "slow" mass (10 kg in your example)? Accelerating whatever is left over of the mass at any time (as opposed to a moment earlier) enlarges that mass by gamma. Having reached c (at the event horizon?), there can't be any "mass" left (which would be infinitely large anyway), meaning, that everything would have to be radiated away. Are these thoughts correct? --- Also, I am constantly wondering how much of a "mass" such a speedy one could remain, as opposed to flying apart, torn to bits by acceleration and/ or tidal effects! Is it imaginable that a mass torn to tiniest bits, maybe even only of molecular size could be torn apart even further, into atoms or even constituent parts thereof? Or are even the largest gravitational/ tidal forces much too weak against electrochemical, let alone strong molecular forces?

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miler

August 21, 2013 at 3:23 am

Precisely how have you figure all of this out relating to this topic? I enjoyed scanning this, Ill must visit other pages on your own site straight away.

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Peter

August 21, 2013 at 10:51 am

Great question, Mike. It’s a tricky subject. Whats glossed over in the standard picture is the source of energy. Consider our 10 kg fired as a projectile from an artillery. Imagine it is fired level, and we consider only the moment it leaves the barrel, with a velocity v. Is its inertia now greater than 10 kg by 1/sqrt(1-(v^2/c^2)? Yes, and there is no conflict with conservation-of-energy (CE), because the supply of energy was external, specifically, the powder-charge in the shell that drove it. But now let’s drop that same 10 kg from the tower of Piza. Now there is no external source of energy. Consider the moment before impact, traveling at a velocity v. Is its inertia greater than 10 kg by 1/sqrt(1-(v^2/c^2)? If so, we have a CE problem, because its kinetic and potential energy have both increased! To conserve energy, we must assume its increase in kinetic energy is offset by a loss in potential energy. The projectile has less potential energy at the bottom of the tower than at the top. Its rest-mass must be lower at the bottom. Its rest-mass has to have decreased by something like sqrt(1-(v^2/c^2). Returning to the original subject: throw 10 kg into a BH and the system’s inertia cannot increase by more than 10 kg. Therefore, the increase in kinetic energy as the material accelerates must be offset by a decrease in its potential-energy (rest-mass/inertia). And when all 10 kg of potential energy has been converted to kinetic energy, it is at the event horizon. It cannot go any faster, because doing so would violate CE.

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Joe S

August 25, 2013 at 3:14 pm

While I very much enjoyed Peter's idea and the follow-up discussion, I have to be the nay-sayer. The kinetic energy that grows as the mass falls toward the BH does not come at the expense of the rest mass. It comes from the potential energy. The zero-point of potential energy can be chosen at any location. If you choose it to be zero at infinity, then as the mass falls, the potential energy decreases (becomes negative) and the kinetic energy grows so that the sum is constant (and in fact is zero). These ideas are classical, but they can be generalized to use special relativity. The rest mass of an electron, for example, never changes (all electrons are identical); so dropping one towards a BH cannot affect its rest mass.

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Mike W. Herberich

August 27, 2013 at 7:05 pm

Dear Joe S (and Peter), I (and probably Peter) understand and agree that the growth in kinetic must be feed through potential energy, at all times of the path in the thought experiment, and therefore be constant in sum at all times, too. I would suggest though that the sum should not be zero in the relativistic case, but rather m0*c^2, the rest energy of the sample body at infinity: -GM*m0/r(=+infinity), the potential energy at infinity = 0, + the then still 0 kinetic energy, + the above rest energy, = in sum m0*c^2. (G = gravitational constant; c = speed of light; M = central mass spanning up the gravitational field GM/r^2 at every point in space around the black hole in our case). At any later time this formula then becomes -GM*m0*gamma/r(t) + m0*gamma*c^2 + Erad, where gamma = 1/sqrt(1-(v^2/c^2)), r(t) = distance (vector) positioned at center of mass M, at any time t, Erad a (hypothetical?) "forgotten" energy, such as the one radiated away (the 2nd term contains the rest plus the inketic energy now). I am very insecure about the factor gamma in the first term for the potential energy, though (maybe, because of what Peter said above)! --- What I do really struggle with is: -gamma (+ Erad) = 1, assuming r(s) (Schwarzschildt) = GM/2c^2. Ignoring Erad, -gamma = 1 doesn't make sense at all; at best it could mean v = 0 which is surely wrong: v should be VERY high at r(s), if not outright = c yet! Including Erad could result in 0 > Erad > 2(m0*c^2) which entails halfway sensible values for v/c (>= 0, =< 1). Erad > 2, though means MORE THAN 2 rest energies were radiated away! Erad < 0 means energy got ADDED!? And, in any case, we won't arrive at v = c, ever. Where is the fault, are the faults? In the setup? In the calculation? In interpretation? Please, help me out!

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Mike W. Herberich

August 30, 2013 at 9:25 am

"... (+ Erad) = 1", of course, should read "... Erad/m0*c^2 ...". Come to think of it, Erad < 0 doesn't look THAT bad anymore to me. It means the + sign in front of it in the formula is actually a minus sign (radiated AWAY, rather than gained). Also, v/c = sqrt[1-{1/((Erad/m0*c^2)-1)^2}] that way results in "viable" values: 0 < v/c < 1. At Erad = -1m0*c^2 --> v/c = .866, already. At Erad = -2m0*c^2 --> v/c = .943. That seems to say that c can not be reached at the event horizon at all, ever. That seems to make sense in that, BEFORE event horizon there is always a small remnant of mass, material ... which by special relativity can not reach c.

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