0:00

Here you

Â can see, the

Â frequency

Â seems to be starting out about here. And

Â going out here it's

Â centred around oh, about 0.2071

Â cycles per second. So our little

Â spike turns out to be a set of

Â frequencies between 0.2068

Â cycles per second, and

Â 0.2074 cycles per

Â second. It's not quite constant.

Â And if we follow the values of our new clock, which

Â corresponds to a period, of 1 over f. Now

Â these two are different periods right?

Â This is a two day period, and this is going to correspond to something else

Â within the star. We get for this, periods

Â of about 4.836

Â seconds to

Â 4.822

Â seconds.

Â And let's see, this is the larger frequency, so it corresponds

Â to the smaller period. I think I might even have that right, but

Â you can check me on that. It's changing its frequency.

Â And guess what?

Â It does so smoothly, varying back and forth over,

Â what do you think the time period is? Right, 2.09

Â days. We have apparently uncovered a classic

Â case of an eclipsing binary star. But

Â wait just a cotton picking minute. If every 2.09

Â days we really see an eclipse that lasts about 40,000 seconds,

Â can we predict anything from this observation?

Â Sure. If the x-ray emitting object is going

Â in and then coming out from behind another star, as it

Â does so, maybe we should see a change in the x-ray energy

Â spectrum, similar to the way the sun and sky changes color

Â at sunrise and sunset, as the sunlight passes through

Â more and more of the Earth's atmosphere. So, the picture is this.

Â 3:19

We have an object and we have our x-ray source that goes

Â behind the object and then comes out of eclipse.

Â And as it goes through the limb of our companion

Â star, and as it comes out from the other limb of the

Â companion star, it is possible that the energy

Â that we see from the x-rays will change its composition,

Â just as the sky changes to red during sunrise and

Â some, and sunset. And indeed, although the mechanisms

Â are very different for x-rays, the result is that we do

Â see the source change as it goes into and out of eclipse.

Â So we are beginning to put together

Â a satisfyingly coherent picture of our system.

Â A source of cosmic x-rays

Â which somehow, has an internal clock of about 4.8 seconds

Â 4:30

is orbiting another object every 2.09 days.

Â But what are these objects, and what is the 4.8 second periodicity due to?

Â Well, let's roll up our sleeves and get back to work.

Â Let's examine what we already have and

Â where it can get us. First, it appears that the

Â regularity of our range of different periods is due to the Doppler

Â shift of the x-ray emitting star travelling around its companion.

Â Since delta f over f equals v over c, our

Â change in frequency over the frequency

Â is equal to the speed divided by the velocity

Â of light, and our range in frequencies

Â is 0.2074 To 0.2068

Â Hertz. We end up with, Delta

Â f over f, equals v over c,

Â equals 0.2074 minus

Â 0.2068, this is the

Â change in frequency, divided

Â by 2, all over the regular

Â frequency, or the average

Â frequency, 0.2071.

Â See if you can understand, why there's a factor of

Â two in there. That means, that v

Â over c, is equal to and we probably

Â should put in approximately equal things, just to show

Â that we're really not exactly 100% accurate

Â here. This is going to be equal to

Â 0.0003 divided by

Â 0.2071. And

Â that equals approximately

Â 0.00145.

Â So when we solve for v, v turns out

Â to be around 430 Kilometers

Â per second, if we have a circular

Â orbit. And in fact, these objects really are

Â mostly in circular orbits because of other factors

Â of their orbital circumstances. But now, we can find the

Â size of the orbit. Since the circumference of the orbit

Â must be equal to 2 pi r, okay, the

Â object goes around at a radius r

Â from the companion, it goes once around and

Â if the speed is constant at 430 kilometers

Â per second, that distance is nothing more than the

Â velocity times the amount of time it takes to go once

Â around, namely 2.09 days.

Â So now we can figure out what

Â r is. R is going to be equal to the

Â velocity times the orbital period divided by 2

Â pi, and our velocity is 430

Â kilometres per second times the time,

Â which is 2.09 days times

Â 86,400 seconds in a day divided by

Â about 6. And that's going to be in kilometers.

Â 9:21

This is about one quarter the size of Mercury's orbit around the Sun.

Â So, these two objects are really close together.

Â Now we can do something really neat.

Â Since, the x-ray source is eclipsed for about 40,000 seconds

Â every orbit, we can estimate crudely the size of the other object.

Â Just take for

Â its diameter that 40,000 seconds worth

Â of blackout, and multiply

Â by the speed that the neutron star is going

Â as it goes around in its orbit. So 40,000

Â seconds times 430 kilometers

Â per second is equal to

Â 1.7 times 10 to

Â 10:22

the 7 kilometers. This would be a crude

Â estimate of the diameter of our companion

Â star. It's approximating the arc

Â of a circle with a straight line. So the companion star's radius

Â is about half that, or 8.5 times 10 to the 6

Â kilometers. Thus, not only is the orbit small, the

Â x-ray source must be very close to the surface of the other

Â star. Here's the radius of the orbit and

Â the radius of the star, r star, is about 8.5 times 10

Â to the 6 kilometers.

Â 11:22

So what we're doing is we're imagining that we have a star and that

Â the object is just moving behind it but in a straight line instead of a circle.

Â But it's actually not a really bad approximation.

Â But what it means is the radius of the star is over 10 times the size of

Â the sun.

Â Now, I do have to say we did make some approximations here.

Â And we talked about one of those approximations, okay.

Â What are some of the other assumptions that we made?

Â And I will leave that to you to figure out.

Â 12:04

So if our theory about the nature of the low state

Â in Cen X-3's light curve is correct, we have a prediction.

Â It is that the

Â companion star should be quite large and massive.

Â Can we test that theory?

Â Yes!

Â Remember when we talked about gravity?

Â We derived an equation relating the period of an orbit to its size, or radius.

Â Better known as Kepler's Third Law, it states that, the

Â square of the period of an object in an orbit

Â is proportional to the cube of the radius

Â of that orbit or T squared equals

Â 4 pi squared over GM times

Â the radius of the orbit cubed. Well, now

Â we know what T is, it's 2.09 days.

Â We know what r is, it's 1.2 times 10 to the 7 kilometers.

Â We can solve for M.

Â