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PC cpu limitation ruining our movie promise

  1. Jun 22, 2011 #1
    Hi, I'm new here.

    My colleague and I are working on a sci-fi movie script. It's almost finished and it's going very well, thanks. At the beginning we made a promise. Although there inevitably has to be some small fantasy elements to any sci-fi story, we promised each other we wouldn't break Einsteinian physics, and there's no darn sound in space! Now, we know that normally these rules would lead, probably, to a kind of boring film, but we think we've got a good story, so please don't worry about that!

    There is a scene where, because of a freak incident (the fantasy part), a ship is unexpectedly accelerated to a speed that is over 99% of SOL. More fantasy parts are that the passengers are not pasted like jelly on the aft wall of the ship's bridge, and space dust doesn't puncture holes in the hull. What was that promise again?

    Because of an important subplot, it's important that they travel at least 30 million years into the future 'cause of time dilation. Now, I THINK I understand the Lorentz factor. I worked out that if c = 1 ly, and v = 0.9 (etc) with 15 9's after the decimal, that in one year the ship would travel 15.8 million years into the future. Not only is that not enough years, but we want the trip to only be a few days at most inside the ship.

    Now, as I understand it, in theory, if the freak incident made them go so fast that there were more than just 15 9's after the decimal point, then we could achieve what we're looking for. Trouble is, it would appear that a PC isn't capable of doing that! At least that's what we think is going on. So, could anyone kindly confirm this, and also let me know if there's any kind of workaround that would allow us to determine the speed it would take to do the 30 million in just a few days? Would it be something like 0.9 (etc) with more like 25 or 30 9's after the decimal?

    I know, kinda bizarre, but hey, I think it's fun!

    Thanks, Hal Dace
     
  2. jcsd
  3. Jun 22, 2011 #2
    The formula for figuring out what percent of the speed of light you would have to go is:

    [tex]Sqrt(1-v^2)[/tex]
    v = (number of days the trip takes)/(Number of days in 30 million years)
    So a 3 day trip makes: v=0.000000000456621004566210045662100456621 (not counting leap year)

    The calculator needs more than 15 significant digits to calculate it. What you can do for a scene is have a character try to calculate it and not have a calculator that can do it. So maybe they are not sure how close to light speed they are. They just know it cannot be very many days at that speed, which they very roughly know by how fast they get by known star systems.

    Of course you can also download a calculator from the internet with enough digits to pull it off.
     
  4. Jun 22, 2011 #3

    jtbell

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    Staff: Mentor

    Using the binomial approximation (which is an excellent approximation in this case):

    [tex]\sqrt{1-v^2} = (1-v^2)^{1/2} \approx 1-\frac{1}{2}v^2[/tex]

    Using your value of v, I get 1 - 0.000000000000000000104 = 0.999999999999999999896. The last step is easy enough to do without a calculator. For me it is, at least.
     
  5. Jun 22, 2011 #4

    PAllen

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    Science Advisor
    Gold Member

    Assuming a few is 3, the gamma factor to time dilate 30 million years into 3 days is 3.65 * 10^9, ignoring leap years, etc. For large gamma, it is straightforward to compute that the following is a very good approximation (in units of c):

    v = 1 - 1/(2*gamma^2)

    Thus v (appx) = 1 - 3.75*10^-20 =
    .9999999999999999999625 (19 nines)
     
  6. Jun 22, 2011 #5
    To 40 decimal places, taking 3 days as 0.008214 years, Mathematica gives, for the exact formula,

    N[Sqrt[1 - ((365230*10^(-5))*(10^9))^(-2)], 40]
    = 0.9999999999999999999625167605087179983996,

    and for the suggested approximation (2nd order Taylor series expansion about 0),

    N[1 - (1/2)*((365230*10^(-5))*(10^9))^(-2), 40]
    = 0.9999999999999999999625167605087179984003.

    You can also compare these methods online, using Wolfram Alpha:

    http://www.wolframalpha.com/

    Exact formula: Sqrt[1-(3.65230*10^9)^(-2)]
    Approximation: 1-(1/2)*(3.65230*10^9)^(-2)
     
  7. Jun 22, 2011 #6
    So three days equals 30 million years if you're traveling at 0.9 (etc) of c with 19 nines after the decimal (etc). Thank you so much for your help guys!!!!! We really appreciate it.

    Yes, the characters in the story have almost no food and little water, so 3 days are about right. Now we can get the dialog right. Thanks again.
     
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