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Need help in a time dilation question (high school)

  1. Jul 25, 2013 #1
    So basically im in high school and have done my major exam for the end of the year. It was a good test except for one question that i got totally stumped on. The question said that a planet was 6.7 light years away from earth. A space craft has a maximum speed of 0.9c, how long would a clock on the spacecraft measure once it had reached the planet. So the bit that confused me is the fact there is no relativity going on here, or is there? light years are a measurement of distance not time, thus i should be able to restate the question as, point a is x distance away, traveling at y speed (in this case 0.9c) how long will it take to get there? where does special relativity come in? there is no point of reference? I thought there would be no time dilation? I think it is a badly thought out question but i'm afraid i will loose marks because of it. Also assuming there is time dilation, then the maximum speed of 0.9c should be dependent on the frame of reference right? If u measure the speed in the spaceship and the time is dilated then shouldn't this increase the speed as you are traveling the same distance in a less amount of time, or does distance contract, ahhhh im so confused. Any way, hope you guys can help me scab some marks and also solve the question that's been playing on my mind for a while now.

    All help is greatly appreciated, thanks :)
     
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  3. Jul 25, 2013 #2

    phinds

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    Time dilatation is an artifact of remote observation. The spaceship itself experiences no time dilation.

    For example, you yourself, right now as you read this are traveling at .9999c from some frame of reference and from that frame of reference you are massively time dilated. Feel any different?
     
  4. Jul 25, 2013 #3
    Hi Danny20051,

    Your teacher probably could have been a little more careful in specifying which frame of reference he was using when indicating the distance to the planet was 6.7 light-years. However, don't you think the context makes it fairly clear that he was referring to the earth's frame? If so, keep in mind that you are working with two different frames, the earth's and the spacecraft's. Think your problem through again, keeping those two frames in mind (we will assume the planet's motion relative to the earth is not significant). You were almost onto it, then were so confused.
     
  5. Jul 25, 2013 #4
    Hi Danny20051. Welcome to Physics Forums.

    This problem can be done in two different ways, both giving the same answer. It can be done using time dilation, and it can be done using length contraction. Length contraction is a little easier. From the frame of reference of the spacecraft, the distance between the earth and the planet is reduced. So, with a relative velocity of 0.9c, the elapsed travel time measured on the spacecraft will be less. From the frame of reference of the spacecraft, the clocks in the earth-planet frame of reference are running faster than on the spacecraft. So, here again the elapsed travel time measured on the spacecraft is less.

    Chet
     
  6. Jul 25, 2013 #5

    ghwellsjr

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    There's nothing wrong with the question and there is relativity going on and you have pretty much gotten everything right. If we consider the frame of reference in which the earth and the planet are at rest, then the clock on the space craft, traveling at 0.9c will be time dilated and will tick at 0.437 times the rate of the Coordinate Time of the reference frame. Since the space craft will take 6.7 / 0.9 or 7.444 years of Coordinate Time to get there, the clock will progress through 7.444 x 0.437 or 3.253 years during the trip.

    This does not mean that the space craft is traveling faster than the speed of light since from the reference frame of the space craft, the distance between the earth and the planet is Length Contracted by the same factor which is 6.7 x 0.437 or 2.928 light years and the planet is moving towards the space craft at 2.928 / 3.252 or 0.9c.

    It seems like you almost had all these ideas correct, you just didn't go far enough with them.
     
  7. Jul 25, 2013 #6
    Thanks for the quick responses guys, really appreciate it, but still a tiny bit confused, i understand what you guys are saying (i think), but i just want to ask one more question, lets say speed of light is 100m/s and the distance is 1 light second away (100m) and i travel at 0.9 c (90m/s) why am i not able to just go 100/90 as the time, why should i have to introduce another frame of reference in order to determine the time dialation and hence the time to get there. The derivation of the time dilation from memory was using trigonometry, for example a person on a train that is moving at a relativistic speed with light bouncing up and down vertically on mirrors, will see the light just moving up and down, however a person outside the frame of reference will see a triangular path and thus a longer time. how ever this time dilation requires an out side frame of reference to be measured, where as in my question there is no outside observer just the one person. hopefully that made some sense to you guys, im not very good at explaining my self lol.

    thanks
     
  8. Jul 25, 2013 #7

    ghwellsjr

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    One of the best ways to comprehend Special Relativity is to define a scenario in an Inertial Reference Frame (IRF) and draw a spacetime diagram depicting the locations and motions of the various objects. I like to use dots to indicate the passage of time for each object which are placed farther apart when the object is moving according to the factor of gamma (1/√(1-v2)) where v is expressed as a fraction of the speed of light. At a speed of 0.9c, gamma is 2.294. So for your scenario it would look like this where earth is shown in blue, the planet in red and the space craft in black:

    attachment.php?attachmentid=60493&stc=1&d=1374763257.png

    Now we can use the Lorentz Transformation process to calculate the coordinates of all the events (the dots) for a new IRF moving at 0.9c with respect to the original IRF which would be the rest frame of the space craft. Here is what the diagram for that IRF looks like:

    attachment.php?attachmentid=60494&stc=1&d=1374763257.png

    You can see at the Coordinate Time of 0, the distance between the earth and the planet is slightly less than 3 light-years and the planet and the earth are the ones moving at 0.9c while the space craft is stationary.

    Now we can also transform the scenario into any other IRF moving at any speed (short of c) with respect to the defining IRF and even though the objects can have new time dilations and new distances between them and new speeds, they will all show that the clock on the space craft progresses through 3.253 years during the course of the trip. For example, I picked another IRF where the Time Dilations of all three objects are identical. This happens with the IRF moving at 0.627c with respect to the first one:

    attachment.php?attachmentid=60495&stc=1&d=1374763257.png
     

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  9. Jul 25, 2013 #8

    ghwellsjr

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    You don't have to introduce more than one Inertial Reference Frame as I showed in my previous post, any IRF will do, but you have to use Time Dilation for any object that is moving, otherwise when you transform to the IRF in which that object is at rest, its clock will not be ticking at the same rate as the Coordinate Time.

    Also, you should not think of some observers or objects as being in particular IRF's. Everything is in every IRF. You should also not think of any particular IRF as being preferred. Of course, you need to start with one IRF so that you can define your scenario, but after that, you can transform to any other IRF. No IRF is better than any other IRF.
     
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