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Special Relativity for the visual thinker - testing my understanding

  1. Jun 15, 2012 #1

    For me it all starts with what I agree is counter intuitive:

    The Speed of Light

    The speed of light. An innocuous phrase. Stating a seemingly obvious fact that light moves at a particular speed, in the same way that anything that moves has a speed.

    Yet what does it mean?

    The speed of light in a vacuum is constant; but relative to what?

    The speed of light is constant and the same relative to any observer.

    If we measure the speed of light between two points in space, without regard to whether those points are moving, then we are measuring how long it takes to travel a fixed distance without regard to any movement of the source nor of the recipient.

    Take, for example the light from a star (that lies in the solar plane). As it orbits the sun the Earth is travelling at around 70,000 mph, sometimes towards that start and sometimes away from it. Yet, contrary to all common sense and expectations, the speed of the light from that star is always 'c', the speed of light.

    How can that be?

    Einstein gave us the answer when he said that time is not absolute. This new way of regarding time supported Einstein's Special Relativity, which showed that all could be resolved once one accepted that the measurement of time could depend on where it was measured from.

    Einstein's Light Clock, which he defines as part of his 'thought experiments' serves to show how this works.

    The light clock comprises no more than pulses of light sent to a distant mirror and thence reflected back to the light source. The mirror is a fixed distance so that the light returns in a set time, 1 second is commonly used.

    http://img651.imageshack.us/img651/4208/fig1lightclock.jpg [Broken] Uploaded with ImageShack.us[/PLAIN]

    Fig 1 light clock

    But if the clock, in its Frame of Reference, is moving at velocity v, relative to an observer; that observer will see the clock travel the distance vt in the time t that the light takes to reach the mirror.

    So, during the time, t, that the light travels to reach the mirror, the clock has travelled an additional distance,vt.

    The Greek letter τ (tau) is commonly used to denote Proper Time, that is time measured by an observer on a standard clock that he is travelling with. A clock that he is stationary to.

    But is the time τ the same as time, t, for the moving clock?

    That is, is the time in the clock's Frame of Reference the same when measured by an observer, relative to whom, the clock is moving?

    Fig 1 above, gives us the answer in no uncertain terms, for by the application of Pythagoras it tells us

    τ² = (ct)² - (vt)²
    τ = ct√ 1 - v²/ c² or
    t = γτ

    where c = 1
    and γ = 1
    √ 1 - v²/ c²

    γ is also known as the Lorentz factor which is much used in Special (and indeed in General) Relativity

    But that is just formulae, which are difficult to picture, so let us put some figures into the scene:

    http://img29.imageshack.us/img29/5997/fig2rotations.jpg [Broken] Uploaded with ImageShack.us[/PLAIN]
    Fig 2 rotations

    Part 1 shows the traditional (Galileian) view of how the light pulse would appear for a clock moving at 0.6c.

    In the second that the light would travel to the mirror, the clock will have moved 0.6 light seconds along the x axis, combining to create a diagonal path, that the Pythagoras theorem would give a length of: √1 + (0.6)² = √1.36 = 1.166 light seconds. Which is further than the light could travel in 1 second.

    However, in Part 2 we can see the Relativistic view and how far the moving light will have travelled, along the diagonal path in that second; and we see it will have reached the point 0.8,0.6 in the coordinates, or Frame of Reference, of the observer for whom the clock is moving.

    Yet an observer travelling with that clock would see the light reach the mirror, 1 light second away, in that time, so we know that in that second the light has reached the mirror.

    This seeming paradox is what took the genius of Einstein to resolve.

    Indeed we can see this in Part 2 of Fig 2. For the y axis which is measuring the distance travelled by light in light seconds can also represent the time in seconds of the stationary observer.

    And we see, on that time scale, that, while the light in the moving clock has travelled for 1 second, only only 0.8 seconds have passed for the stationary observer.

    This means that, when the Moving clock has travelled for one second, it will have travelled 0.6 light seconds relative to the stationary observer; one light second along the diagonal path, in the moving frame of reference of the clock; yet only 0.8 light seconds in 0.8 seconds in the stationary observer's frame of reference.

    And this leads us to the inevitable conclusion that the time it takes for the pulse of light to reach the mirror, is measured differently in the two frames of reference. An observer moving with the clock would measure 1 second, while an observer for whom the clock was moving would measure only 0.8 seconds.

    So how well does this agree with the formula we deduced earlier?

    We said that: t = γτ

    and that when: v = 0.6c, γ = √1 – (0.6)² = √1 – 0.36 = √ 0.64 = 0.8

    Therefore: t = 0.8 τ
    or τ = 1.25 t

    or 1 second in the moving clock's Frame of Reference, measured by the stationary observer, has the same duration as 0.8 seconds, measured in that observer's own Frame of reference.

    And, observing the clock's rotated Frame, that same stationary observer will record, that having taken the light 1.25 sec to reach the mirror, the clock would, by then, have travelled 0.75 light seconds.
    http://img809.imageshack.us/img809/9615/fig2arotationslc.jpg [Broken] Uploaded with ImageShack.us[/PLAIN]
    Fig 2a Length Contraction.

    In this diagram I have added, in green, the distance traveled by the Moving clock, as measured by the stationary observer, at the time when the light hits the mirror. This is 1.25 seconds as measured by the stationary observer, meaning that the mirror will have traveled 0.75 light seconds at v = 0.6c in that 1.25 seconds. Compared with the 0.6 light seconds in the 1 second that has passed for that same stationary observer in his own frame.*
    So 0.75 seconds in the moving frame (in the direction of travel) has contracted to 0.6 seconds in the stationary frame.

    Having said all that, how do we put it together so we can see how it all relates?

    Well, let's refine Fig 2, part 2 to see just what is happening there and to bring out the essential facets of the rotation.

    In Fig 3 we have three diagrams, the first two showing how the observers' coordinate systems are identical (as each is seen by the observer within the frame), the observers in each frame seeing and measuring in Proper units. However, in the third diagram, we can see that the rotation of the moving Frame has effectively contracted the coordinate scales from the perspective of the Stationary Observer.

    http://img403.imageshack.us/img403/6982/fig3hyperbolicrotation.jpg [Broken] Uploaded with ImageShack.us[/PLAIN]

    Fig 3 hyperbolic rotation

    It is worth noting that the time it takes for the light to reach the mirror in the moving clock is 1.25 seconds. Comparing this with the 1 second it takes, measured by an observer travelling with the clock (for whom the clock is stationary) we can say that the moving clock has slowed; that the time taken is dilated.

    We can also see that, from the perspective of the Stationary observer, the clock has moved 0.75 light seconds in the 1.25 seconds that the Stationary observer measures to have passed in the clock's Frame; and that that same Stationary observer measures as 0.6 light seconds in only 1 second in their own Frame; that that length has contracted

    So how does that work?

    If we draw the rotation of each axis separately, where the moving Frame is rotated by the angle β (where Sin β = v/c), we can extract representations of the individual dimensions; including those two often misunderstood old Gentlemen, Time Dilation and Length Contraction, to show how they fit into the bigger picture :

    http://img207.imageshack.us/img207/8650/fig4tdlcrotation.jpg [Broken] Uploaded with ImageShack.us[/PLAIN]

    Fig 4 TD & LC rotations

    The 1 second duration in the rotated moving Frame has the same duration, measured by the stationary observer, as 0.8 seconds measured on his own frame's coordinates. This is described as the time in the moving frame being 'dilated'.

    In the same way the 0.75 light second length in the moving Frame has the same magnitude, measured by the stationary observer, as 0.6 light seconds measured on his own frame's coordinates. This is described as the length in the local frame being 'contracted'.

    This leads us to the inevitable conclusion that Time dilation and Length Contraction are the same process, the same phenomenon, described from opposite points of view.

    Indeed we can also see that it is only the stationary observer's measures of the rotated moving Frame where, in order to avoid exceeding the speed of light, the elements comprising those measurements have been changed. The unit sizes have been 'contracted', by the factor, γ (the Lorentz factor), while the number of units has been 'dilated', also by the factor, γ.

    As can be seen in Fig 3, the coordinate scales in both frames are identical when measured locally, it is only when an observer measures a moving frame that 'dilation' and 'contraction' are encountered, and it is an intrinsic part of those phenomena that the 'moving' measurement has the same magnitude as the stationary measurement; it being the scales of the coordinates that have changed rather than their overall sizes.

    If we now juxtapose the two diagrams in Fig 4, we can build a visual representation of the relationship between the stationary observer and how they see the moving clock's Frame of Reference.

    http://img94.imageshack.us/img94/5895/sfor.jpg [Broken] Uploaded with ImageShack.us[/PLAIN]

    Fig 5 rotating the clock's FoR

    Now, taking into account that both axes are rotated to the same degree, it makes sense to me to represent this visually as a rotation of the whole frame. I know that this is not an accepted procedure, but neither is there anything saying that it is incorrect, as far as I am aware, and bear in mind that this is just an attempt to visualise how the elements of SR relate.

    And remember, all we are doing here is visualising how this will look to the stationary observer, observing the moving clock's Frame of Reference. Nothing actually rotates; rotation is purely a mathematical representation of what is being observed.
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 15, 2012 #2


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    If you want to visualize how velocity, time dilatation and length contraction relate in a single diagram, then put proper time instead of coordinate time on the axis. Coordinate time is then represented by the length of the worldline:

  4. Jun 15, 2012 #3
    Wow, great link, thanks !
  5. Jun 15, 2012 #4
    Yes exactly. Note that as you are discussing the making sense of it, it may be useful to write more precisely "the measured speed" of light. In SR the expression "the speed of light" is defined as the measured speed of light, but in conceptual discussions this is often confounded with metaphysical speed of light, in the sense of what supposedly "really happens".

    Directly related to that, you added "contrary to all common sense and expectations". Well, certainly at first contrary to all expectations, but not contrary to common sense. There are different ways (depending on one's philosophy) to make perfect sense of it and there's nothing new about that.

    And if I correctly understand the continuation of your post, that's precisely what you try to achieve for yourself. Correct?

    Maybe more later.
  6. Jun 15, 2012 #5
    That is how it is drawn; ct is a standard convention for a time axis - c, the speed of light x time.
  7. Jun 15, 2012 #6


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    I know. But if you want to show those relationships geometrically it is better to use proper time.
  8. Jun 15, 2012 #7
    Nice post Grimble,

    A very concise explanation of a "difficult to explain" concept.
  9. Jun 15, 2012 #8
    Awesome link!
  10. Jun 15, 2012 #9
    You appear to have made a mistake here. If the primed variables are measurements made in the moving frame, then it should be:

    " Where t = γt', x = x'/γ " and not " Where t' = γt, x = x'/γ " as you stated.

    The ct and ct' labels on the left graph should be reversed.

    This means your next diagram is not technically correct:

    because ct' should be less than ct, not greater.

    I think you are trying to do something like this:


    but not quite got it right.
    Last edited by a moderator: May 6, 2017
  11. Jun 15, 2012 #10
    That is exactly what I am doing!
    in the link you refer to proper time = 0.87 when coordinate time = 1 (v = 0.5c)
    in my diagram proper time = 0.8 when coordinate time = 1 (v = 0.6c)
    Our diagrams are both of the same thing.
  12. Jun 15, 2012 #11
    I'm sorry but proper time, t = 0.8 seconds, which is dilated to coordinate time t' = 1 second. t = γt'

    How can proper time be less than coordinate time?

    Einsrtein gave us t = t' in chapter XII point (5) http://www.bartleby.com/173/12.html

    That is exactly the way round they are in Figs 1, 2, 2a, 3
  13. Jun 15, 2012 #12
    OK, I was assuming that the unprimed variables were coordinate values. In that case your diagram implies proper length x = 0.8 metres, transforms to coordinate length x' = 1 meter, which is length expansion, not length contraction. As far as I can tell the blue grid in diagram 5 is proper time versus coordinate distance and the red grid is coordinate time versus proper distance and neither grid shows the measurements of a single observer using his own stationary clocks and rulers.

    For some reason you have the expression (t=γt') in the above quate agreeing with what I said and contradicting the (t'=γt) in diagram 3. Was that just a typo?

    Easy. In fact you demonstrated it above. t=0.8 seconds, t'=1 second. Therefore t<t' and proper time is less than coordinate time.

    I don't think he gave us t=t', but it is hard to tell what he did give us from the smudged blurry equations in that text.
    Last edited: Jun 15, 2012
  14. Jun 15, 2012 #13

    Easy, see here.

  15. Jun 16, 2012 #14
    No, it implies that 1 meter coordinate length transforms to 0.8 metres proper length which is length contraction.
    The red grid, representing the moving frame measured by the stationary observer, with the primed axes, has coordinate units.
    The blue grid, representing the stationary frame measured by the stationary observer, with the unprimed axes, has proper units.
    yes, sorry:redface:

    Absolutely right - a temporary mind blip!:redface:

    Last edited: Jun 16, 2012
  16. Jun 16, 2012 #15
    By convention, ct is proper time while ct' is coordinate time
  17. Jun 16, 2012 #16
    Agreed! Proper time is less than coordinate time; Proper time is dilated to give coordinate time.

    :redface: Sorry! :redface:

    :redface: Getting tired :redface:
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