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Justify this!

  1. Jul 25, 2004 #1
    Einstein said that the speed of light will be constant irrespective of the speed of its source..

    So, suppose your are onboard a supersonic car (imagine!) with the speed of 8 Mach, and you throw a ball in the same direction with a speed of say 1500m/s, the ball will actually have the velocity of 1500 M/s + 8 Mach..

    Now instead of the ball, take a torch & switch it on..Einstein says that Light wud still travel at c m/s and not c + 8 Mach m/s...

    Can anyone justify this? Is there any proof of this??

  2. jcsd
  3. Jul 25, 2004 #2

    Tom Mattson

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    No, the ball will not have that speed exactly. It will have a speed that is slightly less.

    Yes, there is plenty of experimental evidence that the lighspeed barrier cannot be broken. Have you looked into it yourself?
  4. Jul 25, 2004 #3
    Yeah actually, I have read a lot on light speed barrier...Infinte mass & infinte force, rate of time being affected etc etc...But I just dun get this thing, CANT EVEN LIGHT OUTSPEED LIGHT?
  5. Jul 25, 2004 #4


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    Reemeber Einstein was hradly a dunce, he certainly would not of missed something as simple as this (and that also goes for the millions of people who have learnt the theory of special relativty since it was conceived). There's a good rule of thumb for this situation, if you find a trivial error in a simple, well-known theory that has been around for a century it is almost certainly you who have made the error (thats's not aimed at you in particular more at certain other postes who will remain nameless *cough*geistkeisel*cough*).

    In Newtonian physics the velocity is subject to a simply vecor addition, that is to say if you were standing on the ground and spersonic car wizzed past you with velocity u and threw the occupant of the car threw a ball with velocity v (relative to the car), you would measure the velocity of the ball w as:

    w = u + v

    N special relativity things are not so simple as the above formula becomes:

    [tex]w = \frac{u + v}{1 - \frac{uv}{c^2}}[/tex]

    Now you can see from the formula when u and v are much smaller than c,

    w = u + v

    is a very, very good approximation.
  6. Jul 25, 2004 #5
    Hmm..I think I have got it now..Thanx for help!!
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