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Two people in a fast ship

  1. Oct 24, 2006 #1
    Lets say we have a ship 100 meters long with one person up front and one in the back, call it a 100 meter long hallway. Let us say it has attained a speed within a centimeter of the speed of light per second. Let us say that each person heads to the front or back of the ship. Wouldn't the one heading to the back of the ship have no problem getting there, while the one heading to the front be limited to a speed no more than one centimeter per second? It is said that all things would seem normal to a person approaching the speed of light, but how can it be normal for the dude in the back of the ship? This person would expect to see things like as if he were traveling at walking speed as he heads to the front of this ship, but that speed is not possible.
     
  2. jcsd
  3. Oct 24, 2006 #2

    DaveC426913

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    Relativistic Speeds are not additive. Use the formula

    v = 1 / (1 - (v^2 / c^2) )^.5

    You will see that v never exceeds c.
     
    Last edited: Oct 24, 2006
  4. Oct 25, 2006 #3
    They're not additive, but "v=f(v)" doesn't seem helpful.

    The basic physics is that time slows on the ship (plus the whole ship gets shorter), so that your extra 1cm/s seems like the speed of light to them.
     
  5. Oct 25, 2006 #4
    Relative to what?
    Remember that speeds are always relative to other objects that have mass.
    The speed of an object of mass relative to light is aways 0.
     
  6. Oct 25, 2006 #5
    What a bizarre way to think about it! :wink: Is that also a contradiction?
     
  7. Oct 25, 2006 #6
    If v is the ship's speed and u the person's speed (walking in the same direction and sense of the ship), then his resultant speed is:

    V = (v+u)/(1+v*u/c^2).

    This is the famous relativistic rule for speeds addition (so, it's not v+u, as it was already pointed out).
    If you try to substitute some values to v or u or both, you will see that V never exceeds c. For example, if you put v = c:

    V = (c+u)/(1+c*u/c^2) = (c+u)/(1+u/c) = c*(c+u)/(c+u) = c.

    You can enjoy substituting other values and see what happens!
     
  8. Oct 25, 2006 #7
    There is no contradiction.
    Absolute speed does not exist, it is always relative to something, and that something must have mass.
     
  9. Oct 25, 2006 #8

    Garth

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    Agreed
    Surely, as photons are massless, for consistency you mean the speed of light relative to an object is aways c?

    Garth
     
  10. Oct 25, 2006 #9
    Correct.

    Light always travels at c for a given object O thus it follows that it's speed relative to light must be 0.
    From the perspective of an object O' that is in relative motion with O, O travels at a speed relative to c and as a consequence O's length is observed as contracted and O's time is observed as dilated.
     
  11. Oct 25, 2006 #10

    Garth

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    But by your statement
    you cannot have a speed relative to light, how could you measure it, except as a speed of light relative to the object, in which case the speed is c?

    Garth
     
  12. Oct 25, 2006 #11
    Regardless of your relative movement with something else that has mass you will always measure the speed of light as c.
     
  13. Oct 26, 2006 #12
    Castlegate,

    The ship passengers get to the ends of the vessel simultaneously per them, and they do it just as they would on earth. Nothing seems out of the ordinary per them. A stationary earthling observes the aftward passenger to get to the back of the ship first, then later the foreward passenger gets to the front of the vessel. The faster the ship travels, the greater the seperation in time between the 2 passengers arriving at ship's end. At near c, this temporal seperation may be many eons. One may say that the guy walking to the front of the ship was more massive, and so it took him longer to get there per the stationary earthling, given they each expended the same energy to walk in their own frame. Yet, if you could see their wrist watches, the time readouts would be identical if you considered only the time when a ship passenger got to a ship's end.

    pess
     
  14. Oct 26, 2006 #13

    JesseM

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    Well, it couldn't really be eons unless the ship was thousands of light-years long--if you have clocks at either end of the ship which are synchronized in the ship's frame, and the distance between them in the ship's frame is L, then in a frame where the ship is moving at speed v, the clocks will be out-of-sync by Lv/c^2. So, even if v gets arbitrarily close to c, the clocks can never be out-of-sync by more than L/c, which means the ship would have to be at least 1000 light-years long for the clocks to be out-of-sync by 1000 years or more. And as you say, the two reach either end simultaneously in their own frame, so clocks at either end of the ship should show the same time at the moment each person reaches them.
    I don't think this argument makes sense, since if you're talking about the energy expended in the guy's own frame, you should also be talking about the relativistic mass in his own frame, which is just the same as his rest mass. Anyway, time dilation doesn't really have anything to do with relativistic mass, two clocks of different masses which are moving at the same velocity will always tick at the same rate.
     
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