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Speed of light

  1. Feb 27, 2008 #1
    I read somewhere that relative speed of light always remains constant. That is there is no difference between relative speed and absolute speed of light.
    If we move with 99% velocity of light in the direction of light , the speed of light is NOT 1% of actual speed of light , It still remains 100% speed of light.

    I would like to if above statement is correct.
     
  2. jcsd
  3. Feb 27, 2008 #2

    Doc Al

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    All speeds are relative to something; an absolute speed has no meaning. But it's true: No matter what your speed, you will always measure the speed of light with respect to yourself to be the same.

    If you are moving at 99% of lightspeed with respect to me, and you shine a beam of light in the direction of your motion, you and I will both measure that beam to move at speed c with respect to ourselves.
     
  4. Feb 27, 2008 #3
    I agree , but this is the same case with all other motion also.
    Like .. I am travelling on moving body of speed X and i am throwing an object(A) with speed X in the same direction of my motion then the speed of A with respect to myself is X.

    I think i need to put my question with more clarity.

    Lets say , There are light beam L1 and spaceship S1 and an Observer O.
    S1 is 99% speed of Light, both L1 and S1 starts at same instance of time

    With respect to O , S1 travels 99% speed of L1
    With respect to person in spaceship L1 is 1% greater speed than S1

    What I understood from the book was L1 is always 100% speed of light relative to S1.
    I know this is not logically sound.
     
  5. Feb 27, 2008 #4

    Doc Al

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    That's true, by definition--since you told us that you threw the object with speed X with respect to you. But it's not true that the speed of the object is X with respect to anyone, unless it's moving at light speed.
    OK.
    With respect to the spaceship, the speed of the S1 is zero and the speed of L1 is c.

    That's true.
    Why is that?
     
  6. Feb 27, 2008 #5
    .

    Yes , WRT to spaceship speed of S1 is zero. But i dont understnad why L1 is c with respect to spaceship

    Lets say speed of A = 100 and B = 90 then relative speed of A WRT B is 10 and speed of B with respect to B =0 is that not the case? , cant we substitute A with c and B with speed of S1.

    am i missing something?
     
  7. Feb 27, 2008 #6

    Doc Al

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    What you're missing is that for high speeds (where relativistic effects cannot be ignored), velocities do not simply add like they do for low speeds.

    For low speeds, velocities add like this:
    [tex]V_{a/c} = V_{a/b} + V_{b/c}[/tex]

    But for high speeds, we must use relativistic addition of velocities:
    [tex]V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}[/tex]
     
  8. Feb 27, 2008 #7

    jtbell

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    Just to nit-pick a bit: the second formula is always valid (both high and low speeds). The first formula is merely a very good approximation to the second one, for low speeds. jjaji, try calculating a few examples if you're not convinced of this.
     
  9. Feb 27, 2008 #8

    Doc Al

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    I wouldn't call that a nit-pick, but an important clarification! :wink:
     
  10. Feb 27, 2008 #9
    How do we arrive at this expression ? Can you point me the theory behind it ? Pls post weblink if possible.
     
  11. Feb 27, 2008 #10

    Doc Al

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  12. Feb 27, 2008 #11
    yes for low speed, denominator is almost equals to 1. I am curious how this formula is arrived at.

    btw , I am working as a software engineer ,(had physics at graduate level) i just got very curious about SR and QM. so sometime i might ask some basic questions
     
  13. Feb 27, 2008 #12

    jtbell

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    That depends on where you're starting from, logically speaking. It can be derived from the Lorentz transformation:

    http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/veltran.html#c1

    Basically, all of relativistic kinematics (length contraction, time dilation, relativity of simultaneity, velocity addition) can be derived from the Lorentz transformation.
     
    Last edited: Feb 27, 2008
  14. Feb 27, 2008 #13

    robphy

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    The best way to think about the formula [in the case of motion in one spatial dimension]
    is in terms of adding "angles" in Minkowski geometry (i.e. arc-length of the unit "circle" in Minkowski geometry) and interpreting "spatial velocity" as the hyperbolic-tangent of the "angle" multiplied by c. The formula is essentially an identity for the hyperbolic-tangent of a sum of two angles (as seen in Doc Al's link).

    [In the more-general situation, it really amounts to trigonometry on the unit-hyperboloid.]

    The methods above give a geometrical picture that underlies the algebraic formulas that are often (arguably, over-) emphasized in textbooks.
     
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