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Modern Physics Question?

  1. Aug 26, 2004 #1
    Say a jet is flying at 300mph. How much time must pass before a clock in the airplane and one on the ground differ by 1.0 s? I know that this deal with time dilation and the special theory of relativity, yet I have no idea on how to get started.
     
  2. jcsd
  3. Aug 27, 2004 #2

    selfAdjoint

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    The forumula for comparing times of frames with a relative speed between them is [tex] t' = t/\gamma, where \gamma = \frac {1}{\sqrt{1 - \frac {v^2}{c^2}}} [/tex]. v is the relative speed, which is 300mph in your example. You have to get its ration to c, the speed of light (be careful they are in the same units!) then square that ratio, subtract the square from 1, and take the square root of the difference. Then because the square root is in the denominator of gamma, and gamma is in the denominator of the time formula, you can just multiply the airplane time by the square root to see what the differnce is. Since the speed of the plane is constant, this value will never increase, so it will never reach the limit of 1 second you are looking for.
     
  4. Aug 28, 2004 #3
    Need some education for the super mentors?

    Exactly what are the qualifications for a "Super Mentor"? The man asked, "How much time must pass before a clock in the airplane and one on the ground differ by 1.0 s?" And selfAdjoint gave the answer above!!

    Oh, he is ok as far as he goes but he makes a very major error: he confuses apparent rate with reading on the clock, two very different issues. Yes, the rate difference is given by the formula Adjoint quotes but the readings on the clock change with time and the man asked how long it would take for the difference to be one second. If the answer were "never" then the clocks would have to go at the same rate! Plus that, out "super mentor" has failed to take the opportunity to point out some very interesting things.

    The speed of light is roughly 3 x [itex]10^8[/itex] meters/sec which is about 666 x [itex]10^6[/itex]miles/hour (a number easy to remember). So, for the fun of it why don’t we make life simple and let the jet be going 666 miles/hour (closer to what jets run too). Then v/c is simple, it's just [itex]10^{-6}[/itex]; square it and we have [itex]10^{-12}[/itex]. That is a very small number! Subtract it from one and the result is 0.999999999999, awfully close to "one". We now have to take the square root of that. Clearly the answer is going to very close to one (if you try and do it on a calculator, you will probably get one but that is the wrong answer).

    The trick is to look at the fact that the answer will differ from one by a very small amount, x = 1-[itex]\delta[/itex]. We have to find out what [itex]\delta[/itex] is. Well [itex]x^2\,=\,1-2\delta+\delta^2[/itex]. Since [itex]\delta[/itex] is very small, [itex]\delta^2[/itex] is much much smaller and can be (at least initially) be ignored. It is then simple to conclude that [itex]\delta[/itex] is very close to .5 x [itex]10^{-12}[/itex] with an error of about .25 x[itex]10^{24}[/itex] (the [itex]\delta^2[/itex] term).

    In fact, it is the [itex]\delta[/itex] term which is of interest here, not the [itex]\gamma[/itex] Ajoint concludes with. Yes, [itex]t'\,=\, t/\gamma[/itex] (or [itex]\gamma t' \,=\,t[/itex]) but it is (t-t') which the Rocker is asking about. That's just [itex]\delta t[/itex] and it become quite clear that, in order to get an answer of one second, we have to wait 2 x [itex]10^{12}[/itex] seconds or about 11.12 x [itex]10^8[/itex] hours. That is something close to one hundred and thirty thousand years (about 126,800 years actually)!

    Now remember, this is only true for uniformly moving coordinate systems (special relativity, not general relativity), which means the plane (maybe we should change it to a rocket ship) has to keep that speed in a straight line for close to one hundred and thirty thousand years. At that rate it will be some 740 billion miles away. That's a long way away; however, light could make the same trip in just over 46 days so it is only a fraction of a light year (about an eighth). So what's big and what's little depends on your point of view.

    And, speaking of "point of view", if you are going to look at that clock on the plane/rocket ship with a telescope, the answer above is dead wrong! In that case, you have to include the fact that every second the plane is about 977 feet further away (I am presuming here that the clocks agreed as the plane passed you). That means the light you are using to look at that clock had to travel 977 feet more than what you saw a second ago. It follows that, even if there were no relativistic effect (and at 666 mph there practically isn't one), the clocks would "appear" to differ by one second after the ship has traveled for a mere 277 hours (11.5 days, considerably smaller than 130,000 years). This is called the Doppler effect; the frequency appears to be higher for things approaching you and lower for things going away.

    If I have made a numerical error in the above, I apologize; I will stand behind the ideas but I often make little math errors.

    Take a look at

    http://home.jam.rr.com/dicksfiles/StarCurv.htm

    you might find it interesting -- Dick
     
    Last edited: Aug 28, 2004
  5. Aug 28, 2004 #4
    Take it easy Dick. Self-adjoint made a little mistake probably because he did not read the question right. I think it is admirable making so little mistakes in so many posts.

    And ofcourse you're right in your corrections... :smile:
     
  6. Aug 28, 2004 #5
    I agree with doctordick that it seems SelfAdjoint misinterpreted the question. CollectiveRocker, remember that because this question involves time dilation, the amount of time that must pass will be different depending on which perspective you choose- one less second must pass for the people on the airplane than for the people on the ground for the clocks to differ by one second. We can calculate for both perspectives. Using the numbers CollectiveRocker provided, the answer can be calculated like this:
    [tex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
    Relativistic gamma is the ratio of the amount of time that passes for the people on the ground to the amount of time that passes for the people on the plane. In this case, it equals about 1.0000000000001004. As you can see, the difference between how much time passes for the people on the ground and on the airplane is very small due to the plane's low speed. Because t = t' / gamma, where t' is the proper time (the time measured by the airplane people), we're looking for a value for t' that satisfies:
    1.0000000000001004*t' - t' = 1
    therefore:
    t' = 1 / (1.0000000000001004 - 1)
    t' = 9960159362549.8008s
    Multiplying that by relativistic gamma should result in a number 1 second larger.
    t = t'*gamma
    t = 9960159362550.8008s
    t - t' = 1, so we have our answer:
    After about 9960159362549.8008s have passed for the people on the airplane, or 9960159362550.8008s have passed for the people on the ground, a clock keeping perfect time on the plane will be one second behind a clock keeping perfect time on the ground. In other words, it'll take about 315,834.6 years.
     
  7. Aug 28, 2004 #6
    CollectiveRocker, if you have any more questions of this nature, PF has a forum specifically for relativity here.
     
  8. Aug 28, 2004 #7
    I would agree with you 100% except for one small issue: when Adjoint wrote down the comment "it will never reach the limit of 1 second you are looking for" some bells should have gone off in his head no matter what the question was. This comment makes a statement concerning a time measurement (the units of measure are explicity seconds) and to say two time measurements can never differ by one second is a little to broad a statment to make. He should have realized something was amiss.

    But I will excuse him as I have made some pretty stupid comments too.

    Have fun -- Dick
     
  9. Aug 28, 2004 #8
    Thank you everybody. That really helps.
     
  10. Aug 28, 2004 #9

    cronxeh

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    the best movie i could come up with.. Back to the Future..

    Doc Brown sends Einstein (his dog) in the time machine.. 1 minute into the future.. and their clocks differ by precisely one minute. sure the car didnt travel at the speed of light or use wormholes.. but hey.. back in the day you couldnt buy a flux capacitor off of ebay
     
  11. Aug 29, 2004 #10
    True, but there's no need to attack him the way you did. But anyway he should be thankful for someone pointing out his errors :smile:
     
  12. Sep 1, 2004 #11
    And, they're still going to lose your luggage...THAT too is another constant.
     
  13. Sep 2, 2004 #12

    Chronos

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    DrD is a very good physicist, just a lousy philosopher. SA is also a very good physicist. He is also a lousy philosopher. Physics and philosophy do not mix well. I am a lousy physicist and lousy philosospher.. go figure. You can, however, learn a lot from the former two guys. They know their stuff.
     
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