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Help - Speed of Light

  1. Apr 18, 2015 #1
    I am a new user of this site, and have not studied physics in forty years. However, I was reading Janus' Special Relativity Primer at http://www.thephysicsforum.com/special-general-relativity/25-special-relativity-primer.html [Broken], with its excellent animated diagrams, but am having considerable trouble getting my head around his statements concerning the 'participants' determination of what has happened. Can someone please give me some assistance?

    I am having problems with the concept that light emitted in one reference frame will still have a speed of c (approx. 186000 mps) when measured in another reference frame. I am aware that this is a fundamental postulate of special relativity, but does not seem to stand up to scrutiny.

    I have created some diagrams to try to help myself see what is happening, but have been unable to include them with this plea for assistance.

    I have tried to imagine a conveyor belt – it seems like a good way to create a second reference frame – as long as can be imagined, certainly much longer than could be created. On the belt will be placed a carriage, likewise rather longer than could be built easily, in fact one light second (186000 miles) in length. An observer, called the moving observer (MO) will be sitting in the carriage at the mid-point, with access to two clocks, one at each end and synchronised with each other both by actual time displayed and the ticking rate. There will also be two light detectors, one at each end of the carriage, facing the rear.

    So far, this is like a fairly normal thought experiment in relativity. The two clocks are also synchronised to a ‘static’ clock which is being monitored by a static observer (SO), though the synchronisation only needs to be by the ticks. The conveyor belt will be moving at a speed of c/2 relative to the static observer’s frame. The speed is only to make the maths easy.

    When the carriage is placed on the belt, it will be carried at c/2 as shown in the diagrams. When the rear of the carriage reaches a point one light second to the left of the static observer, it will trigger a very short, very intense burst of light from an overhead source.

    The rear detector in the carriage will almost instantly detect the light and trigger the clock to record its time. Later, the light will arrive at the front of the carriage, where the front detector will likewise trigger its clock to record the time.

    Now, if all necessary wiring etc. at each end of the carriage is identical in length, conductivity etc., the time difference should give the actual time for the light to traverse the carriage. My question is what will be the time difference between the clocks?

    I have performed my calculations (below) based on the observations of the static observer, as the light source is in his reference frame, and the light can be envisaged as reaching the distances indicated by the blue arc (if I could have included my diagrams).

    Obviously, the carriage does not need to be closed, the conceptual experiment could be performed on the conveyor belt, in a completely open situation.

    Can someone please tell me the flaw in my maths (quite likely, maths was never really my strongest subject), or in my reasoning (less likely, I have always been rather good at logic)?

    The following summarises my logic and calculations, the belt is moving from left to right as the static observer sees it.

    A moving observer (MO) sits on a carriage which has been taken a long enough distance from a static observer (SO)
    Carriage has open rear, allowing light to enter unhindered
    Carriage will be placed on conveyor belt which can move at c/2 (mps)
    Carriage has detectors at front and rear, facing backwards
    Detectors connected to clocks with identical mechanisms (cables exactly same length etc.)
    Moving observer equidistant from clocks, prior to moving, clocks set to be in sync by time interval and time displayed
    Static observer has clock which is in sync with moving observer's clocks by time interval only
    Bright light source will open for short time period (compared to time light takes to travel length of carriage
    Trigger at rear of carriage will open light source, light will be detected almost instantly by detector at rear of carriage, causing clock to record that time
    Detector at front of carriage will detect light when it reaches front of carriage causing front clock to record time
    As all wiring, cabling are identical, any time delay in the recording will be identical at front and rear of carriage

    To static observer, if there were no Lorentz-Fitzgerald transformation, light will reach front of carriage after covering 2 light seconds of distance (carriage is 1 light second long)
    But, because carriage is moving, need to get actual distance from MO's point of view, by applying LFT, this will be D(MO) = D(SO) * (√(1 - v2/c2))
    D(due to movement) = 2 * (√(1 - (c2/4)/c2))
    D(due to movement) = 2 * (√(1 - 1/4)
    D(due to movement) = 2 * √(3/4)
    D(due to movement) = √3 light seconds

    To static observer, this will take √3 seconds to occur (light moves at c for him), but for time from MO's point of view, we need to apply a LFT
    T(MO) = √3 * (√(1 - (c2/4)/c2))
    T(MO) = √3 * √(3/4) = √3 * √3/2
    T(MO) = 1.5 seconds

    But distance is only 1 light second from MO's point of view, so light travels at 2c/3 when created in 'static' reference frame and observed in 'moving' reference frame

    It is worth noting that, if the light reflects from the front wall, applying the same reasoning as above, it will take 0.5 seconds (from MO’s point of view) to reach the rear of the carriage, thus its average speed will be c.

    I apologise if the solution to this puzzle is really simple, and I am merely being thick, but it might help me get my head around the most basic bit of relativity, and start learning again.
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Apr 18, 2015 #2


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    One light-second in length according to which observer? The moving observer (the one sitting at the midpoint of the carriage and moving with it)? Or the "static" observer? It makes a difference because of length contraction.

    I don't understand. If the clocks on the carriage are moving with the belt, they can't be synchronized both with each other (as moving clocks) and also with static clocks. That's not possible. You have to pick one synchronization convention or the other.

    Again, one light-second according to which observer? The static one, or the one moving with the carriage? They will disagree on this distance because of length contraction.

    You can. Use the "upload" button at the lower right of the post editor to upload the file, then insert a link to it (the upload widget should tell you what the link address is) in your post.

    This is not possible; see above.

    This makes it seem as if the carriage is supposed to be one light-second long according to the static observer. Is that the case? If so, it is inconsistent with what you say further on:

    It can't be 1 light-second for the static observer and also for the moving one; that's not possible.

    I think you need to re-work your analysis in the light of the above. A general hint: the best way to construct such scenarios is to figure out what are all the events of interest, and assign them all coordinates in one inertial frame--for example, the frame of the "static" observer, in which the carriage is moving. Then Lorentz transform all of those coordinates to the moving frame. If you try to mix frames while you are assigning coordinates, it's very easy to get mixed up.
  4. Apr 18, 2015 #3
    Thank for replying. As you can see, I am struggling somewhat with this concept, and thanks for your assistance. I have had a short period of thinking about it, and your direction has helped a bit. I shall do some deeper thinking, and see if I can get it straight in my mind.

    Last edited by a moderator: Apr 18, 2015
  5. Apr 19, 2015 #4


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    It's also worth getting into the habit of adding "...in the belt's rest frame", or whatever, any time you say "length", "speed", "time" or "simultaneous".
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