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csmyth3025
2011-Mar-02, 08:41 PM
I realize this thread is 3 years old, but it's the first one in the search list that specifically references the subject of my question, as follows:

I was reading about PSR J0737-3039 - a binary pair of pulsars - and I came across this table in Wikipedia:

Property..........................Pulsar A.............................Pulsar B

Spin period...................23 milliseconds..................2.8 seconds

Mass..........................1.337 solar masses..........1.250 solar masses

Orbital period 2.4 hours

(ref. http://en.wikipedia.org/wiki/PSR_J0737-3039#Known_double_pulsars)

I was wondering what effect frame dragging would have on this system. Do they appear to a distant observer (us) to be orbiting each other faster than their masses would otherwise indicate if they weren't rotating at such high rates?

I ask this because both components are massive, compact, have high spin rates (especially pulsar A) and are very close to each other (~about 800,000 km according to a related article here: http://www.skyandtelescope.com/news/3310106.html?page=1&c=y )

I tried using the formula for orbital periods to see what the orbital period without any frame dragging would be:

The mass would be (1.25+1.337)=~2.587 solar mass

The semi-major axis would be 400,000,000 meters

When I plug these numbers into the WolframAlpha site, I get 45.21 min.

(ref. http://www.wolframalpha.com/input/?i=orbital+period&a=*C.orbital+period-_*Formula.dflt-&a=*FS-_**KeplersThirdLaw.T-.*KeplersThirdLaw.a-.*KeplersThirdLaw.m1--&f3=4x10%5E8+m&f=KeplersThirdLaw.a_4x10%5E8+m&f4=2.659x10%5E30+kg&f=KeplersThirdLaw.m1_2.659x10%5E30+kg&f5=2.486x10%5E30+kg&f=KeplersThirdLaw.m2_2.486x10%5E30+kg&x=4&y=10 )

When I make this calculation on paper I come up with ~2700 seconds (~45 min.).

There's a bit of a difference between my calculation and the published orbital period :eek:

Can anyone tell me what I'm doing wrong?

This is my first post here, so if I've put it in the wrong place or made some other miss-step, please let me know.

Thanks,
Chris

Jim
2011-Mar-03, 01:54 AM
Post moved from that 3yo thread to here.

Oh, and welcome to BAUT!

antoniseb
2011-Mar-03, 04:18 PM
When I plug these numbers into the WolframAlpha site, I get 45.21 min. ...

When I make this calculation on paper I come up with ~2700 seconds (~45 min.).

There's a bit of a difference between my calculation and the published orbital period :eek:

Can anyone tell me what I'm doing wrong?
...

It's an interesting question whether frame-dragging has been included here. I assume that since gravitational radiation is being assigned as the means of orbital energy loss that frame-dragging must be a factor, but it should be straight forward to use some formulae concerning the Kerr Metric to see if the scale of the frame dragging accounts for your 16 seconds difference. I'm a lousy guesser about such things but I'm guessing the effect in this case would be much smaller than that. Since I *am* guessing, I will be delighted to see someone supply a real analysis.

csmyth3025
2011-Mar-03, 08:42 PM
It's an interesting question whether frame-dragging has been included here....but it should be straight forward to use some formulae concerning the Kerr Metric to see if the scale of the frame dragging accounts for your 16 seconds difference. I'm a lousy guesser about such things but I'm guessing the effect in this case would be much smaller than that. Since I *am* guessing, I will be delighted to see someone supply a real analysis.
The 16 second difference (from the WolframAlpha calculation) is the result of rounding by me in my paper calculation. I'm not worried at all about that.

What I am worried about is the 99 minute difference between the orbital period calculation I've made and the published orbital period of this binary system.:confused:

Chris

antoniseb
2011-Mar-03, 11:47 PM
I'm guessing that you didn't take into account that the semi-major axis is to the center of gravity, and that the SMAs are 424,000 and 453,000 km respectively, so the actual distance between the masses is about double, and the gravitational force is about 1/4 of what you were assuming, and so the period will be much longer.

I have not worked this out myself, that's just where I would start looking.

StupendousMan
2011-Mar-04, 12:00 AM
The mistake may be your choice of 4 x 10^(8) m for the semi-major axis. What happens if you use the separation (2.9 light seconds = 8.7 x 10^(8) m) instead?

csmyth3025
2011-Mar-04, 02:04 AM
The mistake may be your choice of 4 x 10^(8) m for the semi-major axis. What happens if you use the separation (2.9 light seconds = 8.7 x 10^(8) m) instead?
4x10^8 meters for the semi-major axis is not my choice. The published literature stipulates that these pulsars are separated by a distance of 800,000 km.
(ref. http://www.skyandtelescope.com/news/3310106.html?page=1&c=y )

You're correct about changing the semi-major axis. I back-calculated the semi-major axis using the published orbital period and the published masses of these pulsars and arrived at a semi-major axis of 866,000 km.
(ref. http://www.wolframalpha.com/input/?i=orbital+period&a=*C.orbital+period-_*Formula.dflt-&a=*FS-_**KeplersThirdLaw.T-.*KeplersThirdLaw.a-.*KeplersThirdLaw.m1--&f3=8.66x10%5E8+m&f=KeplersThirdLaw.a_8.66x10%5E8+m&f4=2.659x10%5E30+kg&f=KeplersThirdLaw.m1_2.659x10%5E30+kg&f5=2.486x10%5E30+kg&f=KeplersThirdLaw.m2_2.486x10%5E30+kg&x=9&y=8 )

This puts the separation between these pulsars at 1,732,000 km for an orbital period of 2.4 hr.

Therein lies the problem: If I calculate the orbital period using the published separation between these pulsars (800,000 km), I get an orbital period of ~45 min. - which doesn't agree with the published orbital period of 2.4 hr. If I calculate the orbital separation using the published orbital period of these two pulsars, I get 1,732,000 km - which doesn't agree with the published orbital separation of 800,000 km.

Chris

csmyth3025
2011-Mar-04, 02:14 AM
I'm guessing that you didn't take into account that the semi-major axis is to the center of gravity, and that the SMAs are 424,000 and 453,000 km respectively, so the actual distance between the masses is about double, and the gravitational force is about 1/4 of what you were assuming, and so the period will be much longer.

I have not worked this out myself, that's just where I would start looking.
If I tailor the orbital calculation to produce an orbital period of 2.4 hr, I get a semi-major axis of 866,00 km.
(ref. http://www.wolframalpha.com/input/?i=orbital+period&a=*C.orbital+period-_*Formula.dflt-&a=*FS-_**KeplersThirdLaw.T-.*KeplersThirdLaw.a-.*KeplersThirdLaw.m1--&f3=8.66x10%5E8+m&f=KeplersThirdLaw.a_8.66x10%5E8+m&f4=2.659x10%5E30+kg&f=KeplersThirdLaw.m1_2.659x10%5E30+kg&f5=2.486x10%5E30+kg&f=KeplersThirdLaw.m2_2.486x10%5E30+kg&x=9&y=8 )

This puts the separation between these two pulsars at 1,732,000 km - a figure which is at odds with the published value of 800,000 km.
(ref. http://www.skyandtelescope.com/news/3310106.html?page=1&c=y )

Chris

StupendousMan
2011-Mar-04, 02:59 AM
Ahem. My earlier hint apparently was too subtle. You are misunderstanding the meaning of the semi-major axis of the orbit of two objects of (roughly) equal mass. It is _not_ half the separation.

It is the separation.

csmyth3025
2011-Mar-04, 06:46 AM
Ahem. My earlier hint apparently was too subtle. You are misunderstanding the meaning of the semi-major axis of the orbit of two objects of (roughly) equal mass. It is _not_ half the separation.

It is the separation.
This is, indeed, a misunderstanding on my part. I was relying on the following definition:

The major axis of an ellipse is its longest diameter, a line that runs through the centre and both foci; its ends being at the widest points of the shape. The semi-major axis is one half of the major axis, and thus runs from the centre, through a focus, and to the edge of the ellipse; essentially, it is the measure of the radius of an orbit taken from the points of that same orbit's two most distant points. For the special case of a circle, the semi-major axis is the radius. One can think of the semi-major axis as an ellipse's long radius...
and
...Note that for a given amount of total mass, the specific energy and the semi-major axis are always the same, regardless of eccentricity or the ratio of the masses. Conversely, for a given total mass and semi-major axis, the total specific energy is always the same. This statement will always be true under any given conditions
(ref. http://en.wikipedia.org/wiki/Semimajor_axis )

In the case of these pulsars I used this definition when I entered the semi-major axis in my calculation.

Can you direct me to a source that more clearly defines the semi-major axis when the orbiting bodies are of (roughly) equal mass?

Chris

csmyth3025
2011-Mar-04, 10:39 AM
I've tried to research the applicable equations for the "two body" problem. The math is beyond my capabilities. I've been able to figure out that I need to use the reduced mass in the equation, which works out to be 1.285x10^30 kg.
(ref. http://www.wolframalpha.com/input/?i=reduced+mass&a=*FS-_**ReducedMass.m12-.*ReducedMass.m2-.*ReducedMass.m1--&f2=2.659*10%5E30+kg&f=ReducedMass.m1_2.659*10%5E30+kg&f3=2.486*10%5E30kg&f=ReducedMass.m2_2.486*10%5E30kg&x=8&y=10 )

I've studied the equations immediately available to me but I can't figure out how to use them.
(ref. http://scienceworld.wolfram.com/physics/Two-BodyProblem.html )

Any help in how to use these equations for determining the orbital distance and orbital period of these two pulsars would be greatly appreciated.

Chris

IsaacKuo
2011-Mar-04, 03:28 PM
Can you direct me to a source that more clearly defines the semi-major axis when the orbiting bodies are of (roughly) equal mass?

It is the same definition. The thing that you might be missing is the position of the bodies along the orbital path. If one body is much heavier than the other--like the Sun and Earth--then the heavy body stays still in the center while the other body travels along an orbital path. So, the semi-major axis is equal to the separation in this case.

But if both bodies are roughly equal mass, then they BOTH travel along orbital paths. If they're the same mass, then they may travel along the same path, staying on opposite ends. So, the semi-major axis is equal to half the separation in this case.

csmyth3025
2011-Mar-04, 04:00 PM
...But if both bodies are roughly equal mass, then they BOTH travel along orbital paths. If they're the same mass, then they may travel along the same path, staying on opposite ends. So, the semi-major axis is equal to half the separation in this case.
I thought I was starting to get this thing figured out - but now I seem to have gotten some conflicting information.

When I use half the published separation of 800,000 km to calculate the orbital period of these pulsars I get ~45 minutes. The published orbital period is 2.4 hr.

StupendousMan tells me that the semi-major axis is not half the separation - it should be the whole separation for two similar mass objects.

The Wikipedia article on the two-body orbital calculation includes several equations which I'm having a great deal of difficulty understanding. These equations are not the same as the equation for calculating "normal" orbits (the orbit of the Earth around the sun, for instance) - although I suspect that once I understand them I'll find that they produce the same results as the so-called normal calculation for the Earh's orbit.

It would be helpful if you could calculate the orbital period of these pulsars based on their mass and separation and let me know what you come up with.

Thanks,
Chris

StupendousMan
2011-Mar-04, 05:04 PM
The information from this lecture

http://spiff.rit.edu/classes/phys440/lectures/reduced/reduced.html

may help you. The basic idea is that one can take a real case of stars with masses m1 and m2, each of which is moving in an orbit, and convert it to an equivalent system -- the reduced mass system -- in which

one star, "A", has mass M = m1 + m2
the other star, "B" has mass mu = (m1*m2)/(m1 + m2)

star "A" sits motionless at the origin
star "B" orbits star "A" so that the separation of the two is always the same as the separation of the two real stars

Why make all these conversions? It turns out that the mathematical solution of the real orbital problem is identical to the mathematical solution of the fictitious, "reduced mass" system. And, because in the reduced mass system, only one object is moving, it's easier to visualize that motion, and to solve the equations of motion, too.

In this particular case, you can apply Kepler's Third Law to the reduced mass problem and get the same result as Kepler's Third Law applied to the real situation. You know the orbital period, and the total mass, so you can solve for the separation of the two stars using Kepler's Third Law. That separation will be the the distance between the two stars in real life, which is in this case about 2.9 light-seconds.

Contrary to the assertion above by IsaacKuo, the semi-major axis of the fictitious reduced-mass orbit is the separation of the two stars in the real orbit, not _half_ of that separation. That is the same mistake the OP made to start this thread. When two stars are approximately equal in mass, the semi-major axis of their mutual orbit (for purposes of Kepler's Third Law) is the separation of the two stars, as counterintuitive as that may sound.

csmyth3025
2011-Mar-04, 10:27 PM
,,,Contrary to the assertion above by IsaacKuo, the semi-major axis of the fictitious reduced-mass orbit is the separation of the two stars in the real orbit, not _half_ of that separation. That is the same mistake the OP made to start this thread. When two stars are approximately equal in mass, the semi-major axis of their mutual orbit (for purposes of Kepler's Third Law) is the separation of the two stars, as counterintuitive as that may sound.
I finally had that AHA! moment. I now understand what you were dropping a subtle hint about in your original post. If one were to assume that the Earth's orbit around the sun is circular (for simplicity's sake), the major axis is the diameter of the Earth's orbit and the semi-major axis is half that (the radius of the orbit) or, in other words, the distance separating the Earth and the sun.

If I apply the published orbital distance (800,000 km) to Kepler's third law for orbital period, I get 127.9 minutes (2.131 hr).
(ref. http://www.wolframalpha.com/input/?i=orbital+period&a=*C.orbital+period-_*Formula.dflt-&a=*FS-_**KeplersThirdLaw.T-.*KeplersThirdLaw.a-.*KeplersThirdLaw.m1--&f3=8x10%5E8+m&f=KeplersThirdLaw.a_8x10%5E8+m&f4=2.659x10%5E30+kg&f=KeplersThirdLaw.m1_2.659x10%5E30+kg&f5=2.486x10%5E30+kg&f=KeplersThirdLaw.m2_2.486x10%5E30+kg&x=6&y=6 )

If I similarly use (m_1+m_2) for mass A and (m_1*m_2)/(m_1+m_2) for mas B I get 114.4 minutes (1.906 hr).
(ref. http://www.wolframalpha.com/input/?i=orbital+period&a=*C.orbital+period-_*Formula.dflt-&a=*FS-_**KeplersThirdLaw.T-.*KeplersThirdLaw.a-.*KeplersThirdLaw.m1--&f3=8x10%5E8+m&f=KeplersThirdLaw.a_8x10%5E8+m&f4=5.145x10%5E30+kg&f=KeplersThirdLaw.m1_5.145x10%5E30+kg&f5=1.285x10%5E30+kg&f=KeplersThirdLaw.m2_1.285x10%5E30+kg&x=5&y=5 )

As you can see, both calculations are closer to the published orbital period of 2.4 hr.

We're not quite there yet, so I'm still not using the correct calculation.

Any suggestions?

Chris

Solfe
2011-Mar-05, 03:57 AM
***Edit - never mind, Stupendousman and you already said what I posted. Unedited post: "A semi-major axis of 8.6605x10^8 m gives exactly 2.4 hours in the first link."

csmyth3025
2011-Mar-06, 03:03 AM
Although I may seem a bit obsessed with this topic, I continue my quest to arrive at an orbital calculation that matches the published data for PSR J0737-3039 (as follows)

Property..........................Pulsar A.............................Pulsar B

Spin period...................23 milliseconds..................2.8 seconds

Mass..........................1.337 solar masses..........1.250 solar masses

Orbital period 2.4 hours

(ref. http://en.wikipedia.org/wiki/PSR_J0737-3039 )

The separation of these pulsars is given as 800,000 km.

(ref. http://www.skyandtelescope.com/news/3310106.html?page=1&c=y )

Using a mass of 1.989 x 10^30 kg for one solar mass, I get a mass of 2.659 x 10^30 kg for component A and a mass of 2.486 x 10^30 kg for component B.

If I use the Kepler orbital calculation and substitute (m_1 + m_2) for component A (5.145 x 10^30 kg) and (m_1 x m_2)/(m_1 + m_2) for component B (the reduced mass, which equals 1.285 x 10^30 kg) in accordance with what I understand is the solution to the classic two-body problem, I get an orbital separation of 937,200 km based on an orbital period of 2.4 hr.

(ref. http://www.wolframalpha.com/input/?i=orbital+period&a=*C.orbital+period-_*Formula.dflt-&a=*FS-_**KeplersThirdLaw.T-.*KeplersThirdLaw.a-.*KeplersThirdLaw.m1--&f3=9.327x10%5E8+m&f=KeplersThirdLaw.a_9.327x10%5E8+m&f4=5.145x10%5E30+kg&f=KeplersThirdLaw.m1_5.145x10%5E30+kg&f5=1.285x10%5E30+kg&f=KeplersThirdLaw.m2_1.285x10%5E30+kg&x=7&y=7 )

If I make the same calculation to obtain an orbital period based on the published separation of 800,000 km., I get 1 hr 54.4 min. (1.906 hr)

(ref. http://www.wolframalpha.com/input/?i=orbital+period&a=*C.orbital+period-_*Formula.dflt-&a=*FS-_**KeplersThirdLaw.T-.*KeplersThirdLaw.a-.*KeplersThirdLaw.m1--&f3=8x10%5E8+m&f=KeplersThirdLaw.a_8x10%5E8+m&f4=5.145x10%5E30+kg&f=KeplersThirdLaw.m1_5.145x10%5E30+kg&f5=1.285x10%5E30+kg&f=KeplersThirdLaw.m2_1.285x10%5E30+kg&x=4&y=5 )

Does the orbital calculation for the two body problem take a different form than that of the Keppler orbital calculation?

Chris

StupendousMan
2011-Mar-06, 03:52 AM
I think that if you want to perform calculations at the level of 10 percent or better, you ought to leave the press releases behind and read the primary literature. Here are some papers about the double pulsar system -- read them and check your numbers against theirs. You should be able to access the arXiv preprints for each one, even if the official journal articles aren't accessible to you.

csmyth3025
2011-Mar-06, 08:58 AM
I think that if you want to perform calculations at the level of 10 percent or better, you ought to leave the press releases behind and read the primary literature. Here are some papers about the double pulsar system -- read them and check your numbers against theirs. You should be able to access the arXiv preprints for each one, even if the official journal articles aren't accessible to you.

Thanks for that advice and your links. As it turns out the first paper you cited (B.B.B. Perera, et al) does, indeed, cite an orbital separation of 2.9 lt-s as your March 3 post suggested. :clap:

This more exact orbital separation (869,000 km) makes the Kepler orital period calculation work out to very nearly 2.4 hr.

Henceforth I'll be more skeptical about the numbers I read in popular science articles.

I'm thinking that the equations for the two-body problem should give (approximately) the same result as the Kepler orbital calculation. Unfortunately,I don't have the math skills to work through the equations for the two-body problem (in short, I don't understand them).

Although I started this thread with a question about the effect of frame dragging on this binary pulsar, I've now redirected my attention to just understanding "normal" orbital calculations. It will definitely be a learning experience for me.

Thanks again

Chris

Hornblower
2011-Mar-06, 01:29 PM
I did the calculation using the masses given above and an assumed separation of 800,000 km, and found a period of about 2.12 hours. If my number crunching was accurate I can only conclude that at least one of the given numbers was only a rough approximation.

I find it odd that the masses were given to four significant figures, while only rough figures were given for the period and separation. Surely the period is known very precisely, and the orbital velocities were precisely determined from the Doppler effect on the pulses. That should enable very accurate calculation of the separation. I cannot see any other way to calculate such precise numbers for the masses, not to mention the ability to observe the relativistic effects.

Here is my understanding of Kepler's formula and its generalizations:

1. P2 = a3, Kepler's 3rd law for planets orbiting the Sun, where P is the period in years and a is the semimajor axis of the planet's orbit in astronomical units. The planet has vanishingly low mass in this case.

2. P2 = a3/M, Kepler's formula for another star, with M in units of solar masses.

3. P2 = a3/(M + m), the general case for massive bodies where neither is vanishingly lightweight compared to the other. Here a is the mean separation of the bodies, with M and m in solar masses. In this case each body follows an elliptical orbit around the system barycenter, and the sum of the semimajor axes of these orbits equals a.

I understand the reduced mass method illustrated earlier, but for the purpose of this exercise equation 3 was less work for me to crunch. My old Texas Instruments calculator was dead, and I did not feel like setting up the equations in my spreadsheet for such a simple job, so I used pencil and paper and a slide rule.

Tobin Dax
2011-Mar-07, 12:13 AM
3. P2 = a3/(M + m), the general case for massive bodies where neither is vanishingly lightweight compared to the other. Here a is the mean separation of the bodies, with M and m in solar masses. In this case each body follows an elliptical orbit around the system barycenter, and the sum of the semimajor axes of these orbits equals a.

I understand the reduced mass method illustrated earlier, but for the purpose of this exercise equation 3 was less work for me to crunch. My old Texas Instruments calculator was dead, and I did not feel like setting up the equations in my spreadsheet for such a simple job, so I used pencil and paper and a slide rule.

Hornblower, your third equation is derived using the reduced mass of the two bodies. That is why it is the general case for such situations.

Using the reduced mass method, the central mass is assigned a mass of (M+m) while the second body is assigned the negligible mass of (Mm)/(M+m). By replacing the semi-major axis in your equation 2 with the mean separation between the two bodies and replacing the mass of the central star with (M+m), we arrive at your equation 3.

Hornblower
2011-Mar-07, 01:12 AM
Hornblower, your third equation is derived using the reduced mass of the two bodies. That is why it is the general case for such situations.

Using the reduced mass method, the central mass is assigned a mass of (M+m) while the second body is assigned the negligible mass of (Mm)/(M+m). By replacing the semi-major axis in your equation 2 with the mean separation between the two bodies and replacing the mass of the central star with (M+m), we arrive at your equation 3.

Thanks for the words of wisdom. I am here to learn as well as to respond to others.

Tobin Dax
2011-Mar-07, 03:54 AM
Thanks for the words of wisdom. I am here to learn as well as to respond to others.

Let me also say that's it's been a while since I've done such things, so I had to spend about ten minutes making sure that I knew what I was talking about. The "ah-ha moment" in my second paragraph was something I had forgotten. (Assuming I got it at some earlier point. Who knows these days.) :)

csmyth3025
2011-Mar-08, 05:31 AM
The mistake may be your choice of 4 x 10^(8) m for the semi-major axis. What happens if you use the separation (2.9 light seconds = 8.7 x 10^(8) m) instead?
I followed your suggestion and studied the primary documents related to this system. I've tried three times to post a reply with citations and refernces but each time my log-in had expired by the time I was ready to post the reply. Naturally, when I logged back in my reply was wiped out by some mysterious computer logic.

Suffice it to say that using mass data to four significant digits and an orbital period given in the literature as 8834 seconds I arrived at a separation of ~879,000 km.
(ref. http://www.wolframalpha.com/input/?i=orbital+period&a=*C.orbital+period-_*Formula.dflt-&a=*FS-_**KeplersThirdLaw.T-.*KeplersThirdLaw.a-.*KeplersThirdLaw.m1--&f3=8.7902x10%5E8+m&f=KeplersThirdLaw.a_8.7902x10%5E8+m&f4=2.6593x10%5E30+kg&f=KeplersThirdLaw.m1_2.6593x10%5E30+kg&f5=2.4882x10%5E30+kg&f=KeplersThirdLaw.m2_2.4882x10%5E30+kg&x=6&y=9 ).

This works out to 2.932 light-seconds separation (close to your suggested value). The 2.9 light-second separation you mentioned is in the literature containing the data I used.

Chris.

StupendousMan
2011-Mar-08, 03:12 PM
Good work. I would say that you've taken this as far as you can, and learned quite a bit in the process.

Sorry about the lost posts -- that's happened to me, too. I've learned that after I've written a long
screed, I should first cut-and-paste the text into some temporary place on my machine before
I press the "Post" button. If the post fails, I can just copy the text back for a second try.

csmyth3025
2011-Mar-19, 07:23 AM
I'm resurrecting this thread because of a recent post: "In GR, can objects and light exceed C?". In that thread Macaw provided a link to a previous thread in which he presented a formula for the Kerr metric. I have no idea how to use this formula.

I'm hoping someone can tell me how to apply it to the binary pulsars that are the subject of this thread.

Originally, I got sidetracked on the question of the orbital separation of these objects - which I finally managed to muddle through after 24 replies. Unfortunately, my original question got lost in the shuffle:

I was reading about PSR J0737-3039 - a binary pair of pulsars - and I came across this table in Wikipedia:

Property..........................Pulsar A.............................Pulsar B

Spin period...................23 milliseconds..................2.8 seconds

Mass..........................1.337 solar masses..........1.250 solar masses

Orbital period 2.4 hours

(ref. http://en.wikipedia.org/wiki/PSR_J0737-3039)

I was wondering what effect frame dragging would have on this system. Do they appear to a distant observer (us) to be orbiting each other faster than their masses would otherwise indicate if they weren't rotating at such high rates?

I ask this because both components are massive, compact, have high spin rates (especially pulsar A) and are very close to each other (about 880,000 km).

Chris

csmyth3025
2011-Mar-21, 01:25 AM
macaw

Light being dragged around by rotating black holes

In general relativity, the Kerr metric (or Kerr vacuum) describes the geometry of spacetime around a rotating massive body. According to this metric, such rotating bodies should exhibit frame dragging, an unusual prediction of general relativity. Roughly speaking, this effect predicts that objects coming close to a rotating mass will be entrained to participate in its rotation, not because of any applied force or torque that can be felt, but rather because of the curvature of spacetime associated with rotating bodies. At close enough distances, all objects — even light itself — must rotate with the body; the region where this holds is called the ergosphere. The net effect is that coordinate light speed as measured by a distant viewer can exceed c. The fact is mentioned in a few textbooks [1].
Indeed, in the vicinity of a rotating black hole, the photons' equation of motion, as derived from the Kerr metric is:

0=a_1(cdt)^2-a_2dr^2+a_3dt(d\phi)

Dividing by dt^2:

a_2(dr/dt)^2=a_1*c^2+a_3*d(\phi)/dt=c^2(a_1+a_3/c^2 \omega)

where \omega is the angular speed of the rotating black hole and dr/dt=c_coord is the coordinate light speed from the perspective of a distant observer while c is the light speed in vacuum

Therefore:

c_coord=c*sqrt(a_1/a_2+a_3/(a_2*c^2)\omega)

The term a_3/(a_2*c^2)\omega expresses the "dragging" effect. The larger \omega is, the larger the "dragging" which is quite intuitive. Depending on the value of \omega c_coord can be smaller or larger than c

[1] An Introduction to the Theory of Relativity, W.Rosser p460
Last edited by macaw; 16-March-2011 at 01:35 AM.
I've been trying to decipher your equations for the more massive of these two pulsars (1.337 solar mass). Using the Wikipedia cited value of ~1.989 x 10^30 kg for one solar mass I obtain a figure of 2.659 x 10^30 kg for this object. Applying this number to the equation for the Schwartzchild radius I get a radius of 3.948 km and an angular momentum of 5 x10^40 Joule seconds.

The actual radius* of this object is probably on the order of 10-12 km. Also, the formula used is for a non-rotating black hole. This larger pulsar is rotating at the rate of one rotation every 23 milliseconds (~43.5 times per second).

I'm hoping that you - or someone else here - can help me translate these numbers into the frame-dragging effect this object would have on an object orbiting at a distance of ~880,000 km. I expect that at this distance the effect would be very small - but not insignificant.

Chris

csmyth3025
2011-Mar-21, 08:22 AM
I figured out that the WolframAlpha calculation in my last post used generic values which don't relate to the problem presented in this thread.

The angular velocity of the larger pulsar=43.5 revolutions per sec=273.3 rad/s.

The moment of inertia (assuming a radius of 12 Km - 12,000 meters)=1.532×10^38 kg m^2 (ref. http://www.wolframalpha.com/input/?i=moment+of+inertia&a=*C.moment+of+inertia-_*Formula.dflt-&a=FSelect_**MomentOfInertiaSphere-.MomentOfInertiaPointMass.MomentOfInertiaCone.Mome ntOfInertiaCuboid.MomentOfInertiaCylinder.MomentOf InertiaDisk.ParallelAxisTheorem-&a=*FS-_**MomentOfInertiaSphere.I-.*MomentOfInertiaSphere.m-.*MomentOfInertiaSphere.r--&f4=2.659x10%5E30+kg&f=MomentOfInertiaSphere.m_2.659x10%5E30+kg&f5=12000m&f=MomentOfInertiaSphere.r_12000m&x=4&y=8 )

The angular momentum of this object would be 4.187×10^40 J s (joule seconds)

I don't know if any of this helps in calculating the frame-dragging effect. Right now I'm just taking shots in the dark and hoping that I'll stumble across something I can use.

Chris

csmyth3025
2011-Mar-21, 08:29 AM
The term a_3/(a_2*c^2)\omega expresses the "dragging" effect.

Can anyone explain the terms used in this formula to me?

Chris

csmyth3025
2011-Mar-22, 09:47 AM
In my quest to calculate the frame dragging effects of PSR J0737-3039 A only, I've been trying to decipher the following Wikipedia passage:

Frame-dragging may be illustrated most readily using the Kerr metric, which describes the geometry of spacetime in the vicinity of a mass M rotating with angular momentum J

where rs is the Schwarzschild radius

and where the following shorthand variables have been introduced for brevity

(other unnecessary variables ommited)
..................................................
...This metric is equivalent to a co-rotating reference frame that is rotating with angular speed Ω that depends on both the radius r and the colatitude θ

In the plane of the equator this simplifies to:

Thus, an inertial reference frame is entrained by the rotating central mass to participate in the latter's rotation; this is frame-dragging.
(ref. http://en.wikipedia.org/wiki/Frame_dragging )

Per my previous post, the Swarzschild radius r_s for object A is ~3950 m. The mass M is 2.659x10^30 kg. The angular momentum J is 4.187 x 10^40 J-s (joule seconds). The orbital radius r is estimated to be 8.8x10^8 m.

According to the above equivalence, a=(4.187x10^40 kg*m^2*s^-1)/[(2.659x10^30 kg)(3x10^8 m*s^-1)]=52.49 meters
(ref. http://www.wolframalpha.com/input/?i=%284.187x10%5E40+kg*m%5E2*s%5E-1%29%2F%5B%282.659x10%5E30+kg%29%283x10%5E8+m*s%5E-1%29%5D )

(ref. http://www.wolframalpha.com/input/?i=%283950m%29%2852.49m%29%283*10%5E8m%2Fs%29%2F%5 B%288.8*10%5E8m%29%5E3%2B%2852.49m%29%5E2%288.8*10 %5E8m%29%2B%28%283950m%29%2853.49m%29%5E2%29%5D )

Is this 5.7x10^-13 rad/s the amount that the orbiting body is "speeded up" by the frame-dragging effect?

Chris

csmyth3025
2011-Mar-23, 03:04 AM
Based on an orbital separation of 8.8x10^8 m and an orbital period of 8834 s, the orbital velocity of this binary pulsar system is 625,900 m/s (625.9 km/s)
(ref. http://www.wolframalpha.com/input/?i=%282*pi*8.8*10%5E8+m%29%2F8834+s )

If 5.7x10^-13 rad/s is the frame-dragging effect of pulsar A at an orbital distance of 880,000 km, then the increase in the orbital speed of this system would be 5.016x10^-4 m/s (about 0.5 mm/s).