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Help understanding special relativity concepts please.

  1. Aug 13, 2009 #1
    I should probably start by saying that I'm not too interested in the mathematics yet, I'm after conceptual understanding. I'll get the maths easy when i've got the concepts.

    ok, this is what i've got so far, stop me when i get it wrong...


    the black lines represent the inertial coordinate system (one time and one spatial axis) of an observer, who is passed at t=0 by another inertial observer (red line) travelling at velocity V wrt the coordinates. at the same instance, two photons are emmitted from the origin. these are the blue lines (c = 1).

    since both observers were present at the event, both observers see the photons an equal distance from their own location, hence the dotted lines which represent constant time for each respective (colour coded) observer. (obviously the red line is parralell to the x' axis not shown) this is simultineity fallen out of the window.

    its also pretty easy to get length contraction, as any distance along the x' axis will be shorter when projected onto the x axis, though i'm not sure that this is the correct understanding as it puts a limit on the contraction (1/root2 of the original length)

    time dilation is however just confusing me. i can see that the "stationary" observer measures time along the vertical t axis, but is it right to say that the primed observer measures time along the t' axis? that means between 0 and 2 on the diagram the primed observer (according to the non primed observer) has experienced more time than the non primed. it looks to me like (again according to the non primed observer) the red line is longer, and a clock following this path will have ticked more often than the black line clock.
    this doesn't fit what i've read so far about time dilation, so where am i going wrong? can I even use this diagram to explain time dilation?

    thanks very much.
  2. jcsd
  3. Aug 13, 2009 #2
    Hello earlofwessex,

    The dashed t and x axes have to be calibrated to get the actual values. Many text books on SR will show you how to do this using parabolas. Unfortunately I do mnot have the drawing skills to show you how. But, very roughly, to give the idea, if you draw the upper half of a parabola, cutting the t axis perpendicularly at a distance 1 along the t axis, then the point at which the t dashed axis cuts this parabola represents a distance of 1 along the t dashed axis.

  4. Aug 13, 2009 #3


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    You can't get time dilation just by looking at the geometric distance along the t' axis between two events in the diagram, no. Likewise you can't get length contraction by looking at the geometric length of an interval on the x' axis when projected on the x axis. If you take two events along the t' axis, and the distance between them in the unprimed frame is [tex]\Delta x[/tex] while the time between them in the unprimed frame is [tex]\Delta t[/tex], then obviously the geometric length along the t' axis between them would be given by the Pythagorean formula, so it'd be [tex]\sqrt{\Delta t^2 + \Delta x^2}[/tex]. However, to find the actual time between these events in SR, you have to use something similar spacetime interval which differs from the Pythagorean formula by a minus sign--it would be [tex]\sqrt{\Delta t^2 - (1/c^2)*\Delta x^2}[/tex] (normally the spacetime interval is designed to give something with units of distance rather than time, so it would normally be written as [tex]\sqrt{\Delta x^2 - c^2 \Delta t^2}[/tex], but this will differ from the formula above only by a constant imaginary factor). If you actually draw in the ticks of time and distance along the x' and t' axis, they appear stretched by the same amount when drawn in the context of the x and t coordinate system.

    There's a further complication that the two lengths being compared in the length contraction equation are not directly analogous to the two times being compared in the time dilation equation--in the time dilation equation you're looking at the time interval between a single pair of events in both frames, while in the length contraction equation you're looking at the distance between two parallel worldlines along a surface of constant time in both frames (i.e. you're measuring the distance between the front and back end of an object at a single moment). Below you can see a diagram I drew up for another thread you might find helpful (click it to enlarge if it appears small)--you can ignore the stuff about the "temporal analogue of length contraction" and the "spatial analogue of time dilation" if you find it confusing, they are just meant to show that you could come up with a pair of spatial quantities more directly analogous to time dilation (i.e finding the distance in both frames between a single pair of events) and vice versa.

  5. Aug 14, 2009 #4


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    At least not in a Minkowski diagram (coordinate time as the vertical axis). In a diagram with proper time as the vertical axis, you can see both time dilation & length contraction directly by looking at the geometric length of an interval projected on the axes. The coordinate time is still visible as well, as the length of every world line:
  6. Aug 14, 2009 #5
    ok cool i get that, it explains why the axis are stretched. but is there any reason why that must be the case? oh and this might be a stupid question, but is there only one parabola shape? (ie, only one possible line which cuts through t=1 and converges on the blue line) my intuition says yes. actually isn't there a function? sinh or cosh or something.

    when you say actual time, i'm assuming you mean time measured by the red t' line. I also get that the minus sign in LaTeX Code: \\sqrt{\\Delta t^2 - (1/c^2)*\\Delta x^2} stretches the axis, which corresponds to the parabola mentioned by matheinste, but where does the minus sign come from?

    ha ok, simultiniety prevents you from measuring the distance between two points at the same "now" for both points according to a moving observer.

    doesn't that mean there are two contributing factors to the observed contraction? one from the contraction itself, and one which is a kind of "time-like" illusion due to the surfaces of constant time not being parallel. I would have thought that would have been more apparent from the equations.

    ... temporal just means "time" like spacial means "distance" right?

    thanks to both of you for pointing out what i've been missing out on.
  7. Aug 14, 2009 #6
    so what i've drawn is a minowski diagram? co-ordinate time is the time measured by an object at rest in its own inertial co-ordinate frame? whats proper time?

  8. Aug 14, 2009 #7


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    yes, or: coordinate time is the proper time of a clock at rest
    proper time of a clock is what the clock measures. if the clock moves, it's proper time is less than coordinate frame.
  9. Aug 14, 2009 #8


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    Time measured by a physical clock which has the red t' line for its worldline, yes.
    I suppose it comes from the Lorentz transformation which relates the coordinates of an arbitrary event in the primed frame to the coordinates of the same event in the unprimed frame (with the assumption that each frame assigns coordinates physically using local measurements on a system of rulers and clocks at rest in that frame, with the clocks of each frame synchronized according to the Einstein simultaneity convention). The Lorentz transformation itself comes from the two fundamental postulates of relativity, the first being that the laws of physics should work the same in every inertial frame, the second being that each inertial frame should measure light to move at c in all directions. You can also use these two assumptions to derive the time dilation and length contraction directly without first deriving the Lorentz transformation--are you familiar with the light clock thought-experiment? Normally it's only used to derive time dilation, but once you've done that you can consider a different orientation where the axis between the two mirrors is parallel to the direction of motion rather than perpendicular to it, and from this derive the length contraction equation.
    What does "the contraction itself" mean? If length is defined in terms of a simultaneous measurement of the position of the front end and the position of the back end, there's no way to measure the contraction independent of a simultaneity convention.
  10. Aug 14, 2009 #9


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    On a technicality, the "parabolas" referred to in this thread are really hyperbolas not parabolas.
  11. Aug 14, 2009 #10


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    Some of the things you're concerned about are answered in this post, from a thread about "the photon's point of view".
    I prefer the above definition of proper time over the one JesseM gave you. Actually what he said should be thought of as an axiom of special relativity rather than as a definition of proper time. I would state the axiom like this: "A clock measures the proper time of the curve in spacetime that represents its motion".
  12. Aug 14, 2009 #11
    wow thanks for all your replies.

    so, time measured by a clock when viewed from its own rest frame is called co-ordinate time, while that same measurement when viewed from another rest frame is called proper time. - its a way of identifying whos perspective is being talked about.
    as far as the definition goes, I apprecieate it but its way over my head for the moment. i don't know what a metric is, let alone how to integrate one along a curve. but anyway. i'll get there eventually. whats a curve in spacetime if all objects involved are travelling in a straight line? (isn't that a condition of SR?)

    actually, deriving the transformations is part of my goal here.
    I am familiar with the light clock thought experiment, and know how to get time dilation from it, I'll have a go at getting length contraction in the way you suggest.

    I don't even know what my next question is yet...

    I suppose that "the contraction itself" means the spacial analogue of time dilation, which looking at your drawing looks more like an expansion. but thats cleared something up for me, i always wondered why time and space appeared to stretch in the opposite direction.

  13. Aug 14, 2009 #12


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    Last edited: Aug 14, 2009
  14. Aug 14, 2009 #13


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    Here is another Lorentz transform image.
    Notice that you can see time dilation for both observers here. Specifically, if you start at the origin and you go along the x'=0 line you see that it crosses the t=2 line before it crosses the t'=2 line (moving clocks go slow in the unprimed frame). Similarly if you start at the origin and you go along the x=0 line you see that it crosses the t'=2 line before it crosses the t=2 line (moving clocks go slow in the primed frame).

    I will leave it to you to see how length contraction works.
  15. Aug 14, 2009 #14


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    You've got it backwards here--proper time is the time along a worldline (between two events on that worldline) as measured by a clock that moves along that worldline, so if the clock is moving inertially, proper time is equal to the coordinate time in the clock's rest frame (but proper time can be generalized to clocks that aren't moving inertially too). The time in the observer's frame is that observer's coordinate time.
    That's not a condition actually, in SR you can calculate proper time along non-straight paths if you want (see the discussion here). SR just has the condition that spacetime itself has not been curved by gravity. As an analogy, Euclidean geometry requires that you are dealing with a flat 2D plane rather than a curved surface like the surface of a sphere, but it can deal with curved lines in this plane like circles.
    Yeah, the spatial analogue of time dilation is where you pick two events that are simultaneous in one frame, then figure out the distance between the same two events in a different frame--the distance will be larger, not smaller as with length contraction.
  16. Aug 15, 2009 #15
    ok thanks, i get the diagram, including the length contraction. is it significant that the areas of the black squares and the white "squares" are the same? if it is, whats the argument which says they have to be? (if they do that is)

    interesting link. it reminds me of a silly question i have: what would the wheel of a relativistic train look like? since a point on its rim would go from being at rest in the rails frame, and then be moving at... well, i think theres something about relativistic velocity addition to stop it surpassing c, so if the train were travelling very close to c, the top of the wheel would be moving hardly any faster than the train. i suppose it would be egg shaped, and most of the wheel would be collected at the top. :confused:
  17. Aug 17, 2009 #16


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    The wheel is circular in the train's rest frame, so in the ground's rest frame it's going to look like any other Lorentz transformed circle, i.e. flattened (Lorentz contracted) in the direction of motion.
  18. Aug 17, 2009 #17


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    I don't know about the areas, but the scale is fixed by the first postulate. Specifically:
    Let A = the distance from the origin to the intersection of x=0 and t=2
    Let A' = the distance from the origin to the intersection of x'=0 and t'=2
    Let B = the distance from the origin to the intersection of x=0 and t'=2
    Let B' = the distance from the origin to the intersection of x'=0 and t=2

    Then by the first postulate of relativity A/B = A'/B'.
  19. Aug 18, 2009 #18
    I certainly dont claim any knowledge in this area, but are you sure?
    imagine an object fixed to a point on the rim of a wheel which is rolling (without slip).
    (as a visual aid, let the object spin freely on its pivot, so that it remains the right way up whilst the wheel rotates)
    when the point on the rim that the object is fixed to is at the top, the object will be moving faster than the centre of the wheel (galileo said twice as fast), whilst when the point on the rim is in contact with the ground, the object will be verging on stationary.

    now let this wheel travel at relativistic speeds. since lorentz contraction depends on speed, the object will be more contracted at the top of its path than at the bottom.

    thanks, it seems silly i missed something so obvious.
    and of course, as the speed of the prime frame varies, a fixed point in the prime frame traces the parabola (hyperbola) in the unprimed frame mentioned in the first few posts. and also explains the - sign in the spacetime interval.
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