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Time dilation formula explanation?

  1. Aug 20, 2014 #1
    Can anyone explain how the time dilation formula works?

    T=t/(√(1-v^2)/c^2))...

    ... What does the 1 stand for? Why include it?

    Can you write down the process of the formula?... How does ((C*t)^2))/c^2-v^2
    lead to the time dilation formula? Please explain the process.
     
  2. jcsd
  3. Aug 20, 2014 #2

    BiGyElLoWhAt

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    It comes from pythagoreans theorem. I'd have to go through and try to set it up and solve it again.

    Basically you need to have some observer which you use as the rest frame, then you need some observer moving at a velocity v_0 relative to the stationary observer. You then have them observe the same photon.

    The most important thing that, at least for me, wasn't all that intuitive at first, you have to assume that no matter how fast or slow an observer is moving, via ideal experimental trials, if they measure the speed of a photon, it will always be c.

    That should be enough info to get you started on deriving the formula for time dialation.
     
  4. Aug 20, 2014 #3
  5. Aug 20, 2014 #4
    So if I'm moving relative to the speed of light (e.g 3*10^5 m/s), does that mean that the speed of light still will be moving faster than me, NOT have the speed v=0 relative to me, or just stay C? :confused:
     
  6. Aug 20, 2014 #5

    Orodruin

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    One of the basic assumptions in special relativity is that light will always move at speed c relative to any observer. This means that no matter what your velocity is relative to a light source, you will always see the light moving with this speed.
     
  7. Aug 20, 2014 #6

    BiGyElLoWhAt

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    Say you're moving parallel in line with a photon, but you're moving at .5c, the photon is obviously moving at c, but, contrary to say 2 cars moving down the road, if you could somehow measure the speed of that photon, you would still measure c, not .5 c like expected.
     
  8. Aug 20, 2014 #7

    BiGyElLoWhAt

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    It will just stay C.

    If you move 99% C in the same direction as a photon, you would expect it to only seem to be moving 1% C, right? The fundamental assumption of Relativity is that you actually see the photon moving at 100% C, not 1% C like expected.
     
  9. Aug 20, 2014 #8

    Orodruin

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    I would not use the expression "as expected" in this context. Anyone familiar with relativity will expect the photon to move with speed c ... If just going by everyday experience we do not know what to expect since the speed of light is so far removed from this that c is practically infinite for most phenomena that occur in our everyday lives.
     
  10. Aug 20, 2014 #9

    BiGyElLoWhAt

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    I see your point Orodruin, but to the average observer who is NOT familiar with relativity, as I'm assuming the op isn't, would expect velocities to sum non relativistically, such as my car metaphor.

    If I'm doing 45 mph behind a car doing 55 mph, they're moving 10mph w.r.t. me; whereas if the car were a photon it would still move at c w.r.t. me, not (c-45mph). That's what I mean by as expected. If my wording caused any confusion, I apologize.
     
  11. Aug 21, 2014 #10
    Wait... in that case, isn't the speed of light infinite as it never changes?
     
  12. Aug 21, 2014 #11

    Orodruin

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    No, the speed of light is finite, 299792458 m/s. The fact that any observer will see the same light speed regardless of how the observer moves is one of the foundations of relativity and ultimately leads to the fact that space and time cannot be considered independent.
     
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