0:00

Here you

can see, the

frequency

seems to be starting out about here. And

going out here it's

centred around oh, about 0.2071

cycles per second. So our little

spike turns out to be a set of

frequencies between 0.2068

cycles per second, and

0.2074 cycles per

second. It's not quite constant.

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

corresponds to a period, of 1 over f. Now

these two are different periods right?

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

within the star. We get for this, periods

of about 4.836

seconds to

4.822

seconds.

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

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

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

And guess what?

It does so smoothly, varying back and forth over,

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

days. We have apparently uncovered a classic

case of an eclipsing binary star. But

wait just a cotton picking minute. If every 2.09

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

can we predict anything from this observation?

Sure. If the x-ray emitting object is going

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

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

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

at sunrise and sunset, as the sunlight passes through

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

3:19

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

behind the object and then comes out of eclipse.

And as it goes through the limb of our companion

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

companion star, it is possible that the energy

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

just as the sky changes to red during sunrise and

some, and sunset. And indeed, although the mechanisms

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

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

So we are beginning to put together

a satisfyingly coherent picture of our system.

A source of cosmic x-rays

which somehow, has an internal clock of about 4.8 seconds

4:30

is orbiting another object every 2.09 days.

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

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

Let's examine what we already have and

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

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

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

Since delta f over f equals v over c, our

change in frequency over the frequency

is equal to the speed divided by the velocity

of light, and our range in frequencies

is 0.2074 To 0.2068

Hertz. We end up with, Delta

f over f, equals v over c,

equals 0.2074 minus

0.2068, this is the

change in frequency, divided

by 2, all over the regular

frequency, or the average

frequency, 0.2071.

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

two in there. That means, that v

over c, is equal to and we probably

should put in approximately equal things, just to show

that we're really not exactly 100% accurate

here. This is going to be equal to

0.0003 divided by

0.2071. And

that equals approximately

0.00145.

So when we solve for v, v turns out

to be around 430 Kilometers

per second, if we have a circular

orbit. And in fact, these objects really are

mostly in circular orbits because of other factors

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

size of the orbit. Since the circumference of the orbit

must be equal to 2 pi r, okay, the

object goes around at a radius r

from the companion, it goes once around and

if the speed is constant at 430 kilometers

per second, that distance is nothing more than the

velocity times the amount of time it takes to go once

around, namely 2.09 days.

So now we can figure out what

r is. R is going to be equal to the

velocity times the orbital period divided by 2

pi, and our velocity is 430

kilometres per second times the time,

which is 2.09 days times

86,400 seconds in a day divided by

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

9:21

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

So, these two objects are really close together.

Now we can do something really neat.

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

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

Just take for

its diameter that 40,000 seconds worth

of blackout, and multiply

by the speed that the neutron star is going

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

seconds times 430 kilometers

per second is equal to

1.7 times 10 to

10:22

the 7 kilometers. This would be a crude

estimate of the diameter of our companion

star. It's approximating the arc

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

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

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

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

star. Here's the radius of the orbit and

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

to the 6 kilometers.

11:22

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

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

But it's actually not a really bad approximation.

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

the sun.

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

And we talked about one of those approximations, okay.

What are some of the other assumptions that we made?

And I will leave that to you to figure out.

12:04

So if our theory about the nature of the low state

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

It is that the

companion star should be quite large and massive.

Can we test that theory?

Yes!

Remember when we talked about gravity?

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

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

square of the period of an object in an orbit

is proportional to the cube of the radius

of that orbit or T squared equals

4 pi squared over GM times

the radius of the orbit cubed. Well, now

we know what T is, it's 2.09 days.

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

We can solve for M.