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Spacetime dilation in special relativity

  1. Apr 10, 2012 #1
    I have recently began to read about special relativity. I think I've got a decent grasp of how it works, but I have been confused on one point. Let's say you are in a spaceship going along at some constant velocity. If you fired a laser beam out into space you would see it moving at 300,000km/sec. Now, in order to observe this, wouldn't you have to be moving at zero speed? It seems to me that space would have to dilate to infinitely long and/or time dilate to infinitely slowly for you in order for you to be able to achieve results like this. I know that the finite speed of light in any reference frame is a proven fact but I just can't seem to visualize it in a scenario that makes sense to me.
     
    Last edited: Apr 10, 2012
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  3. Apr 10, 2012 #2

    jtbell

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    In the reference frame in which the ship is at rest, you are moving at zero speed.
     
  4. Apr 10, 2012 #3

    ghwellsjr

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    How can you see the laser beam in space? Can you see the sun light as it travels in space past the earth at night on its way to the moon? You see the moon lit up by the sun light, but can you see the light traveling on it's way to the moon? If you can't see it, how can you tell how fast it's going?
     
  5. Apr 10, 2012 #4
    Let's say you had something you could reflect the light off and then time how long the light took to get back to you.
     
  6. Apr 10, 2012 #5

    ghwellsjr

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    As long as you could measure how far away from the laser/timer the reflector is with a rigid ruler and its distance remains constant, then no matter how fast your spaceship is going, time dilation and length contraction will work together to make sure you measure the speed of the laser beam as c.
     
  7. Apr 10, 2012 #6
    So, what you're saying is, since speed is simply distance/time, you will actually see the light travel a different distance and/or time to maintain its constant speed of c? I mean different than if you would have measured the distance or time if you were standing still.
     
  8. Apr 10, 2012 #7

    ghwellsjr

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    Speed is distance/time but I'm not saying that you would see a different distance and/or time because if your ruler is contracted due to speed and the distance to your mirror is contracted due to speed by the same amount, you would continue to measure it as the same distance as you did at any other speed. In a similar manner, if your timer slowed down due to speed (so would your heart rate so you'd never notice that either) with the result that the division would come out the same as it would at any other constant velocity. In other words, everything to you would seem as though you were at rest, even though you knew you had previously accelerated up to some great velocity and the earth you left behind is indeed traveling away from you at close to the speed of light.
     
  9. Apr 10, 2012 #8
    I will give a worked example and see if that helps.

    Imagine we have a rocket travelling at 0.8c relative to us on the ground. The rocket has a proper length of 1 light second. As it passes close by we shine a light at the rear of the rocket and as the light hits the rear a clock at the rear of the rocket reads 0 seconds and a clock on the ground nearby also reads 0 seconds. At this point in time we notice that the clock at the front of the rocket is reading -0.8 seconds.

    The light continues towards the front of the rocket but because the rocket is going so fast it takes 3.0 seconds for the light to reach the front of the rocket according to our clocks on the ground. At that event the nose of the rocket is 3 light seconds away from where the light passed the rear, so in the ground rest frame it is easy to see that the speed of light is c because 3/3=1.

    Now we know that clocks on the rocket are ticking slower by a factor of gamma (which equals 0.6 in this case) so 3 seconds in the ground frame is equal to 3*0.6 = 1.8 seconds according to clocks on the rocket. The clock at the nose initially read -0.8 seconds and after advancing by 1.8 seconds it now shows 1 second as the light reaches the front of the rocket. As far as the rocket occupants are concerned the light has taken one second to travel the length of the ship which they know is one light second in length, so they also measure the speed of light to be c.

    In brief, the observers on the ground measure the light signal to take 3 seconds to travel 3 lightseconds in its journey from the rear to the front, while the observers on the rocket measure the same light signal to travel 1 lightsecond in one second, so everyone agrees on the same speed for the light, although they disagree on the time intervals and distances between the same two events.
     
    Last edited: Apr 10, 2012
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