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Postulates of STR

  1. Aug 24, 2014 #1
    The speed of light is constant in all inertial frames of reference

    Imagine a thought experiment, an observer at rest sending a light pulse towards a spaceship, he measures the speed of light to be 3x10^8 m/s, now assume that the spaceship also travels at the speed of light and an astronaut in spaceship also measures the speed of light pulse is 3x10^8 m/s.
    my question is according to him both him and light travels at the same speed so actually he could measure the speed of light to be zero (i.e, light pulse is at rest) right?, but according to postulates of special relativity the astronaut also measures the speed of light to be 3x10^8 m/s, how?
     
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  3. Aug 24, 2014 #2

    Nugatory

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    Google for "relativistic velocity addition" - speeds do not add the way you're assuming, although the effect is too small to notice in anything in our daily existence.

    Also, in the interests of being precise... Your example is physically impossible because no spaceship (or any other massive object) can move at the speed of light. However, we can fix your example just by saying that the spaceship is traveling at, say, 99% of the speed of light and then your question is good.
     
  4. Aug 24, 2014 #3

    ghwellsjr

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    Have you thought about how the observer can actually measure the speed of the light pulse that he sent out? Please describe in detail exactly what you have in mind.

    As Nugatory pointed, the astronaunt cannot travel at the speed of light, but assuming he is traveling at some lesser speed, how does he measure the speed of the same light pulse?

    That's not an accurate statement of the postulate according to Einstein. Can you look it up and see the difference?
     
  5. Aug 24, 2014 #4
    Say you are at "rest" in the lab frame and the astronaut is traveling in the "primed" frame at .99c, as Nugatory suggested. If you in the lab frame read the same time on your clock as you read on the astronaut's clock, then you'd be right, it would appear as if the astronaut would read the speed the light to be very close to zero. However, you in the lab frame do not read the same time on both clocks. The astronauts clock according to your frame is running much, much, slower. So slow, in fact, the the very small velocity difference in the speed of light versus the astronauts speed that you observe in the lab frame is actually observed to be exactly the speed of light, c, as observed by the astronaut, whose perception of the transit of time is much slower than yours as measured from your lab frame.
     
  6. Aug 26, 2014 #5

    ghwellsjr

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    I don't think you meant what you said. I think you meant you are at "rest" in the lab frame and the astronaut is traveling in the lab frame at .99c. You could then say that the astronaut is at "rest" in the "primed" frame and you are traveling at -0.99c in the "primed" frame. Isn't that what you mean?

    Here is a spacetime diagram for the lab frame showing you as the thick blue line at rest and the astronaut in red traveling at 0.99c. The light pulse that you send out at time zero is shown as the thin blue line. The dots mark off 1 second increments of time. Notice how the astronaut's clock is running about seven times slower than yours:

    attachment.php?attachmentid=72448&stc=1&d=1409058951.png

    Now we use the Lorentz Transformation process to transform to the "primed" frame in which the astronaut is at rest and you are traveling at -0.99c but the light pulse is still traveling at c:

    attachment.php?attachmentid=72449&stc=1&d=1409058951.png

    Notice how in the "primed" frame, it is your clock that is running seven times slower than the astronaut's.

    I think it is important for the OP to understand that neither observer is actually measuring the speed of light to be c, rather the speed of light is defined to be c "in all inertial frames of reference".
     

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