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Speed of light and position

  1. Jun 10, 2010 #1
    Be gentle, I have physics textbooks on order ;) which should satisfy my cravings, until then..


    I would like to try and stick a pole-in-the-ground so to speak so I can say 'all movement is an offset from here'.

    I've read up a little on the Michelson-Morley experiment.

    I believe light can travel both through a medium (when on Earth) and without a medium (when in space).

    I think I mean using the speed of light outside a medium. I am currently believing that nothing in the universe that emits light is in exactly the same place after it's started emmitting. Wherever the centre of the expanding sphere of light is, is the pole-in-the-ground, and is not where the source is anymore.

    If you take 6 light detectors and arrange them somewhere in space in a sphere, or 12 (is there a name for the number series that describes increasing the points on a sphere symmetrically?). In the centre of this sphere is a light source and you switch it on, measure when the front surface of the light wall hits each detector, then switch it off and measure when the back surface of the light wall hits the detectors.

    If light, either because of its speed or something else about it, behaves independently of anything elses movement once it's been emitted, would this be be picked up as a delay in the readings hitting some of the sensors?

    I am wondering what would be the minimum speed of our solar system/galaxy/planet/group-of-detectors for an offset to be detected, or what would be the widest sphere for the detectors to be configured in to register an offset, or both (assuming there actually ever would be an offset registered).

    The following are the results I'm imagining:

    a) The front surface of the light hits all detectors simultaneously. Shortly after so does the back surface.
    b) The front surface of the light hits all detectors simultaneously. Shortly after the back surface hits the detectors (or stops hitting the detectors, rather) in an offset sort of way.
    c) Both the front and back surface spheres of the light wall register delays of the same 'shape' and timing.
    d) Both the front and back surface spheres of the light wall register a delay of the same shape, the back surface registering an even larger offset but in the same direction.
    e) The front surface hits the detectors in an offset way and the back surface stops hitting the detectors but is offset in a different direction(!)

    I am not sure if this is an imaginary repeat of the Micheson-Morley experiment, which answer I should expect or what they ought to mean.
     
    Last edited: Jun 10, 2010
  2. jcsd
  3. Jun 10, 2010 #2

    russ_watters

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    That's not what SR predicts nor what the MMx shows. They predict/show that the speed of light between any two observers that are stationary wrt each other is always constant. Therefore, there will never be any delay.

    Think about it this way: the MMx apparatus could be mounted inside your sphere and provide the results for your experiment.
     
  4. Jun 11, 2010 #3
    Thx.
    Forgive me if this is obvious: This sounds like it implies then that if observer or object A is moving towards observer (or object) B at speed x and object A turns it light on, the light will exceed the constant c and travel towards object B at speed x + c(?)
     
    Last edited: Jun 11, 2010
  5. Jun 11, 2010 #4
    That is not the implication, and a brief perusal of Einstein's 1915 paper(s) will explain that nicely.
     
  6. Jun 11, 2010 #5

    russ_watters

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    Though you haven't explicitly said so, you're introducing an external frame of reference from which to measure the speed of light. That's not how you measure speed. The speed is measured by taking the distance between A and B and dividing by the time for the the light to traverse it. That's it. You're taking that speed and adding the speed that the whole device is moving relative to an external reference frame. That's not what speed is.

    That the speed isn't measured against an external frame of reference is the point of the first posulate of SR and what the MMx shows. The first postulate says that the laws of the universe are the same for any inertial reference frame. That means s=d/t regardless of what inertial frame of reference you are in.
     
  7. Jun 12, 2010 #6
    I am thinking a clittle clearer now.

    I have two sticks in space with light bulbs on the ends, say bulbs A & B on one stick, bulbs C & D on the other. Both sticks are straight and are the same length.

    I hold both sticks upright, next to each other lining up bulbs A & C at the top and bulbs B, D at the bottom. I switch on bulbs A & C together and simultaneously drop the first stick so it is falling (or it was already moving and just flying past the other stick but the bulbs were level when the lights were switched on).

    Light from A should reach B at the same time that light from C reaches D because both the top bulbs are stationary wrt to their bottom bulbs, both are level when it started and both sticks are the same length.

    But stick one is falling and these are really long sticks, so by the time light reaches bulb B from A the sticks should no longer be level. To my thinking, at some point light from bulb A on stick one ought to pass bulb D on stick two before light from C has.

    But how can this be if everything was equal and level at the start? Does SR make this possible?
     
    Last edited: Jun 12, 2010
  8. Jun 12, 2010 #7
    It is the Relativity of Simultaneity. http://www.pitt.edu/~jdnorton/Goodies/rel_of_sim/index.html
     
  9. Jun 12, 2010 #8
    so light from bulbs A and C WILL both reach D at the same time, and will both reach bulb B at the same time but slightly later? and if I measured the distance between A (or C) when the light started to separately B and D when the light reaches each resp., d/t will show the same speed for light for A --> B, A --> D, C --> B, C --> D.
     
  10. Jun 13, 2010 #9
    I suppore as both sticks can't occupy exactly the same space there would always be a slight diagonal for A --> D and C --> B meaning still the same speed for light throughout(?) but slightly longer times & distances than A --> B and C --> D.
     
  11. Jun 13, 2010 #10

    russ_watters

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    That one sentence of mine had a slightly imprecise wording: the requirement is for an inertial (non-accelerating) frame of reference, not merely a stationary one. When you drop the stick, you are adding an acceleration, essentially changing the experimental setup in the middle of the experiment. This can be dealt with, but it isn't the simple test case of SR. SR is for inertial reference frames.
     
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