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A typo or I don't understand it?

  1. Aug 28, 2012 #1
    Hi

    First, sorry for a horrible title for this thread.
    Second, sorry if this was is the wrong section.

    However I recently read this article (below). I'm probably missing something with my knowledge, I doubt it being a typo.

    http://mcdonaldobservatory.org/news/releases/2012/0828

    I quote from the article:

    I don't get it. They asy the orbital period will shrink with about 0.25 ms each year, that's 1 ms after 4 years. The way I have perceived this, is that we can expect the star eclipsing the other star 1ms earlier for each 4 years.
    However, later in this quote, it says 6 seconds compared to a study performed about 14months earlier. 6 seconds? And they say 0.25 ms/year?

    Something is wrong, can you please help me out. I'd be suprised if it's a typo. I feel there is something that I have misunderstood, or something I did not understand at all.

    Best Regards
    Robin Andersson
     
  2. jcsd
  3. Aug 28, 2012 #2

    Bill_K

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    They say the orbital period is about 13 minutes, so it makes 365 x 24 x 60 / 13 = about 40,000 orbits per year. If the orbital period loses 0.25 milliseconds over the course of a year, the system will be 40,000 x 0.25 msec = about 10 seconds early. Well this is order of magnitude, they say 6 seconds after 14 months.
     
    Last edited: Aug 28, 2012
  4. Aug 28, 2012 #3

    ghwellsjr

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    It's 0.25 milliseconds per orbit. You need to count up how many orbits there were in 14 months where each orbit takes less than 13 minutes.

    I take 60 * 24 * 30 * 14 to get 604800 minutes in 14 months. Divide that by 13 to get 46523 total orbits. Multiply that by 0.00025 seconds per orbit to get 11.6 minutes. But now we have to divide that by 2 to get the average loss over the 14 months or just under 6 seconds. If we knew the actual time of the orbit (less than 13 minutes) we would get closer to 6 seconds.
     
  5. Aug 28, 2012 #4
    Oh yes! Sorry, I totally forgot that the orbital period was 13 minutes! Thanks a lot. Gosh, sometimes the obvious is the problem.
    I only had the orbital period equal to 1 year in my head.

    Thanks!
     
  6. Aug 28, 2012 #5

    Bill_K

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    The J0651 system is a very important discovery, thanks for calling attention to it! A binary system consisting of two white dwarfs, that gives us a very clean picture of the effect of gravitational wave emission.
     
  7. Aug 29, 2012 #6
    Hmm, I've just read the piece in question, and it says 0.25ms per YEAR, twice. Looks like a badly written article . . . ?
     
  8. Aug 29, 2012 #7
    Yes it's a bit sloppy - they obviously mean binary star year, but forgot to indicate that.
     
  9. Aug 29, 2012 #8
    Still, I'd like to see some bigger guns pointed at this pair. From the article, it would appear that these bodies will actually merge sometime in the next few years (20s (or is it 40, 2 eclipses per revolution?) reduction by May next year(sic) seems like a huge change).
     
  10. Aug 29, 2012 #9

    phyzguy

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    I think what they are saying in the article is correct. The orbital period is changing by 0.25 milliseconds per year. It's just that this causes an accumulating change in the time of eclipse. This time starts out at 0 milliseconds per orbit, and reaches 0.25 milliseconds per orbit one year later, so it is an average time of 0.25/2 milliseconds per orbit over the course of a year, as ghwellsjr correctly calculated.
     
  11. Aug 29, 2012 #10
    ghwellsjr said "It's 0.25 milliseconds per orbit", and I agree that what he says makes sense of the article.

    It seems to me that you are merely compounding the confusion in the original article by using the word "year" to refer to time intervals in two independent and widely separated non-inertial frames. Big no-no, no?

    Unless I still don't get it . . .
     
  12. Aug 29, 2012 #11

    phyzguy

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    I think you still don't get it. The orbital period is changing by 0.25 milliseconds/year, but the accumulated delay is much more than this. Suppose you have a watch that loses 1 second per day. After 10 days, how far is it off? By your logic in post #1, it would be off 10 seconds (1 second/day * 10 days), but in fact it is off by 50 seconds, since it is off by 1 second the first day + 2 seconds the second day + 3 seconds the third day....
     
  13. Aug 30, 2012 #12
    If planet A has an orbital period of 8,766 hours then its year is 8,766 hours. If planet B has an orbital period of 13 minutes then its year is 13 minutes long. The only possible confusion would be to mistakenly think a year is the same for every planet.
     
  14. Aug 30, 2012 #13

    ghwellsjr

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    There's only one year of concern in this thread, that of an Earth year. The 13-minute "year" of the binary star system is not the year the article was referring to. The article correctly said that the 13-minute orbital period is shrinking by 0.25 milliseconds per earth year. That means that in April 2011, if the orbital period was exactly 2340.00000 seconds, then in April 2012 it would have been 2339.99975 seconds. The average period over that one-year interval would be 2339.999875 seconds. This is an average loss of 0.125 milliseconds from April 2011. There were 60 times 24 times 365 divided by 13 or 40430 orbits during the year. Multiply 40430 by 0.125 milliseconds and you get 5.05 seconds for the year. To get the [STRIKE]loss[/STRIKE] gain for 14 months we would multiply that by the ratio 14/12 or 5.896 seconds.

    Now the article is in error in their prediction that by April 2013 the [STRIKE]loss[/STRIKE] gain will grow to 20 seconds. They neglected to divide by 2 and the correct answer is 10 seconds.

    EDIT: I changed my wording from "loss" to "gain" because as a clock, the binary system is running faster and we normally think of a fast running clock as gaining time, not losing it.
     
    Last edited: Aug 31, 2012
  15. Aug 30, 2012 #14

    ghwellsjr

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    It will take much longer than a few years for the two bodies to merge. If the loss per year stayed the same, it would take 13 * 60 / 5.05 or 154 years but of course that isn't exactly right.

    EDIT: Well, I sure screwed that one up. I calculated how long it would take (at the current rate) for the accumulated orbit time to have gained 13 minutes, not how long it would take for the period of the orbit to reach zero. That calculation would be 13 * 60 / 0.00025 which is over three million years.

    As the article says, "Every six minutes the stars in J0651 eclipse each other as seen from Earth, which makes for an unparalleled and accurate clock some 3,000 light-years away,"

    Now an accurate clock should take the same amount of time per "tick". But this clock keeps running faster and faster. The times that we are calculating are how much earlier the accumulated time would be compared to a clock that maintained the same tick rate.
     
    Last edited: Aug 31, 2012
  16. Aug 31, 2012 #15
    Good post - indeed, it's an acceleration of the rotation, funny that I did not get that. :redface:
     
  17. Aug 31, 2012 #16
    Hi
    Wouldn't the conservation of angular momentum dictate that the rate the periodicity increased in frequency would also increase as the orbit decreased?
    In fact it seems like the period would decrease exponentially along with the orbital decay.
    Does this sound right?
     
    Last edited: Sep 1, 2012
  18. Sep 1, 2012 #17

    mfb

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    The orbital decay will be quicker in the future, but this has nothing to do with angular momentum conservation - it is just a result of general relativity, as the velocity and acceleration (around their common center of mass) of the objects will increase.

    There is nothing exponential here.
     
  19. Sep 1, 2012 #18

    Bill_K

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    MTW p988 discusses gravitational radiation for a binary star system. The rate of energy loss given there is E ~ a-5 where a is the usual semimajor axis. The time evolution is a ~ a0(1 - t/τ0)1/4, and by Kepler's Law, ω(t) ~ a(t)3/2 ~ (1 - t/τ0)3/8.
     
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