Meanwhile, on the optical frontier, the hunt was on for the optical counterpart of our x-ray source. The difficulties were enormous. Centaurus is in a heavily obscured part of the Milky Way galaxy, and remember, our early error boxes for locations of sources were quite crude. So despite our detailed understanding of the orbit based on the x-ray data, it was three years before we found this object optically. But the discovery was unmistakable. A Polish astronomer Voytek Krisminsky, found a hot, massive star that changed its brightness ever so slightly every 2.09 days, in lockstep with Cen X-3's x-ray clips. Why such a small variation? Well maybe Cen X-3 is very small, so during its orbit, Krisminsky's star's brightness doesn't change very much. Indeed, the situation is somewhat more complicated, but it's a good starting point, based on our understanding that GK Per had a white dwarf for it's x-ray emitting component. Come to think of it, let's check to see if a white dwarf can be responsible for Cen X-3 x-ray output. If it's a white dwarf spinning around every 4.8 seconds, we know that its speed on the equator must be v equals 2 pi r over T. Where now, r, let's make this look a little bit more like an r, in fact, let's do that. Where now r is the radius of a white dwarf, about 6 times 10 to the 8 centimeters, 6 times 10 to the 8 centimeters, and T now, is 4.8 seconds. Right, it goes once around on the equator, traveling this distance, and it does so in 4.8 seconds, so the velocity just distance divided by time. And in other words, we have an Earth sized object, spinning around every 4.8 seconds. If we solve this for V, we get V equaling about 8 times 10 to the 8th, centimeters per second. So the acceleration we would feel, trying to pull us off the star, due to it's rotation, would be a equals v squared over r. We know what v is, we know what r is, so we get about 64 times 10 to the 16 divided by 6 times 10 to the 8, and the units on that are centimeters per second squared. This is an acceleration of about 10 to the 9 centimeters per second squared. Is gravity up to this task? Can it pull us in from the surface of a white dwarf with enough acceleration so that we won't fly off due to the acceleration that is trying to get us to go away from the center of the star? Let's see what a one solar mass white dwarf has for its gravitational acceleration. So we have a, due to gravity equalling GM over r squared. And that's about equal to 6.7 times 10 to the minus 8 times, that's G. Our one solar mass white dwarf has 2 times 10 to the 33 grams associated with it. And our distance is 6 times 10 to the 8 centimeters squared. And again, our units are going to be centimeters per second squared. We do this math, and we get a number 4 times 10 to the 8 centimeters per second squared. Uh-oh, we're in trouble. This number is smaller than this number. Gravity cannot prevent a white dwarf from just flying apart due to its spin. What do we do now? Well, fortunately, since Jocelin Bell's discovery of the radio pulsars a few years before, we had a fairly good idea about what the solution might be. Neutron stars, a new type of object in the cosmos, or at least our understanding of it, would fit the bill. This story actually begins with the white dwarfs, and with Subrahmanyan Chandrasekhar, a truly wonderful man with whom I was fortunate enough to spend several days when I was a graduate student at Princeton, and incidentally after whom the Chandra x-ray Satellite was named. In 1930, Chandra, at the age of 19, discovered that there was a limit to the size of a white dwarf. If such a star had a mass in excess of about 1.4 solar masses, it simply could not resist the pull of gravity, and must collapse. It had no choice. This was initially met with derision on the part of most astronomers. Where would the star go? Ha, ha. Would it disappear? Hah, ha, ha. Well the universe seems to be ultimately impervious to prejudice. Just four years after Chandra put forward his bold idea, Walter Baade and Fritz Zwicky proposed the existence of a neutron star. If the white dwarf had to collapse, the electrons would be the first to go, and the star would consist essentially of a giant nucleus. The neutron had just been discovered a year earlier by James Chadwick, so slamming an electron into a proton to form a neutron seemed at least plausible. But, wasn't this preposterous? Nuclear densities without the support of the vast empty space between it and the electrons would mean that a teaspoon full of this star, remember our teaspoon? Remember our white dwarf? A teaspoonful of this star would weigh about 5 times 10 to the 11 kilograms. This is just a number. Teaspoon. What does it mean? It's a big number. Well, a typical African elephant weighs about 5,000 kilograms, so our teaspoon-full of this neutron star stuff has the same mass as about 100 million African elephants. Absolutely astounding. But how does this solve our Cen X-3 problem? Well, the radius of this star, according to a calculation that you can and should do by simply scaling up a neutron to the mass of our star, would be about 10 kilometers, about half the length of Manhattan island. Oh, I know that Manhattan isn't circular or spherical, but you get the idea, right? So let's imagine a two solar mass neutron star, the radius of which is 10 kilometers. Would gravity be sufficient to hold this star together? Let's find out. With a period of 4.8 seconds and a radius of 10 kilometers, the speed of a test mass on the equator of the star would have the following velocity: V equals 2 pi r over T, where now T is 4.8 seconds, and r is 10 kilometers. 2 pi is about six. And that turns out to be 1.3 times 10 to the 6 centimeters per second. So, our acceleration, v squared over r, would be 1.3 times 10 to the 6, squared, divided by 10 to the 6 centimeters for the radius of the star in centimeters per second squared. And that equals 1.7 times 10 to the 6 centimeters per second squared. Now, once again, we compare this to the acceleration due to gravity, GM over r squared, 6.7 times 10 to the minus 8 times 4 times 10 to the 33 grams, 2 solar masses, divided by r squared which is 10 to the 12 also, centimetres per second squared. And this turns out to be approximately 2.7 times 10 to the 14 centimetres per second squared! Yes, no problem. The acceleration due to gravity is much, much stronger than the acceleration trying to get the star to fly apart. This is no problem, but an astonishing number anyway. What this means is that if an object were dropped at a distance of one meter from the surface of such a neutron star, it would hit the surface in about a millionth of a second, and it would arrive at the surface traveling almost 3,000 kilometers per second. Calculate what that is in miles per hour. But what exactly is it that provides our clock? It turns out that the most likely possibility is similar to what causes aurorae on the Earth. In that case, charged particles from the sun are funneled into the magnetic polar regions of our planet where they collide with each other and give rise to spectacular light show. We believe that this is likely to happen in the outer reaches of a neutron star's environment as well. Material form the companion star is concentrated to, by the magnetic field of the neutron star and this, coupled with the intense gravitational field, heats up the gas to incredible temperatures. If the rotational axis is inclined to the magnetic field, similar to the way that the Earth's magnetic poles are not coincident with the geographic North and South Poles, we have a natural way to provide a beacon of light that we can see every 4.8 seconds. Essentially, the magnetic poles become hot spots that radiate copious amounts of x-rays because of the tremendous amount of gravitational energy that can be converted into light. And you can see in a little cartoon what that's likely to look like. Now, let's go one step further. Let's revisit Cen X-3, 15 years after the EXOSAT observation we just examined, and see what Chandra can add to the picture.