Invariant spacetime and time dilation

Discussion in 'Physics & Math' started by Confused2, Jun 11, 2020.

  1. Confused2 Registered Senior Member

    Messages:
    607
    We often see 'time dilation is caused by something'. The 'something' may be true and unhelpful or true and misleading or true and unfathomable or downright wrong.

    Here I'm saying time dilation is caused by the geometry of spacetime. Could be any of the above.

    The assumptions are flat spacetime, no accelerations, no tricks and the simplest possible notation.

    The general idea is that the spacetime interval (I'm calling it 's') is the same in every frame - you might call it the invariant spacetime interval if you wanted to. I'll give it in full (simplest possible notation) and then tone it down to show how nice it really is. A # means 'comment'.

    In full, where x is the distance travelled in the x direction, y and z in y and z directions (all at right angles) and t is the measured time.
    s²=x²+y²+z²-c²t² # it's not as bad as it looks.
    This is all distances measured between events.
    Let's have some events.
    Alice starts her stopwatch (event A) and after time t she burps (event B).
    Two events, A and B which both take place in the same place in her frame so x=0,y=0,z=0
    so we get the invariant spacetime interval
    s²=-c²t² # not so bad is it?
    Let's say some guys are driving past at speed v in the x direction. They get their own frame and I'm going to put a dash (') after everything relating to their frame. In the car frame Alice is moving past at velocity v' in the x direction, the time interval between Alice starting her stopwatch (event A) and burping (event B) is t'.
    so x'=v't' # distance*time (consult Newton for clarification if needed).
    Substituting v't for x' in the spacetime interval equation
    s²=v'²t'²-c²t'² # this is the guys spacetime interval
    We said (agreed?) the spacetime interval between events is the same for both the car guys and Alice - in her frame it was -c²t²
    so
    -c²t²=v'²t'²-c²t'² # Alice's spacetime interval = Guys spacetime interval

    change sign, divide by c²,extract t' and square root both sides
    c²t²=c²t'²-v'²t'²
    t²=t'²-v'²t'²/c²
    t²=t'²(1-v'²/c²)
    t=t'√(1-v'²/c²)
    as √(1-v'²/c²) is less than 1 it follows that the time in Alice's frame is less than the time in the Guys frame.
    So time dilation is a caused by the geometry of spacetime.
     
    Last edited: Jun 11, 2020
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  3. Halc Registered Senior Member

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    226
    The only comment I have is that if the car is moving at v in Alice's frame, then Alice is moving at -v in the car frame. Doesn't effect the interval calculation since v gets squared, but just saying.

    It is definitely one of the ways to explain it, yes.
     
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  5. paddoboy Valued Senior Member

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    25,607
    And spacetime geometry depends on the mass/energy within that region.
    Time dilation as well as length contraction are really easy to understand and fathom, when one considers that the speed of light is a finite constant.
     
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  7. Confused2 Registered Senior Member

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    607
    Many thanks for your correction and comment. In my (long) struggle to grasp SR I found calculating (without thinking) was the most helpful way to make progress.
     
  8. Confused2 Registered Senior Member

    Messages:
    607
    The OP was intended as (one of) the simplest ways to look at SR so no tricks and no mass/energy in the region.

    The spacetime interval encapsulates the speed of light being finite and constant in every frame.

    I have a light clock on my desk which 'works' using the invariant spacetime interval - the constant speed of light appears rather nicely. I might post it if I can work out how to do subscripts in BBcode (or possibly Latex).
     
  9. Confused2 Registered Senior Member

    Messages:
    607
    Oops, missed out a dash in the second line above - should be
    "Substituting v't' for x' in the spacetime interval equation"
     
  10. Halc Registered Senior Member

    Messages:
    226
    Well, XML doesn't work here: X[sub]a[/sub]

    Latex also fails: ##X_a## or [latex]X_a[/latex]

    3rd try using tex: \(X_a\)

    All three of these fail in this post, but I definitely see tex being used successfully in the tutorial. Has something broken? The delimiters are obviously being intercepted, but I'm not getting what is advertised.
     
    Last edited: Jun 11, 2020
  11. Neddy Bate Valued Senior Member

    Messages:
    2,108
    Your 3rd try seems to work in the post, just not in the "preview" pane before hitting post. There are many bugs in this forum software.
     
  12. Confused2 Registered Senior Member

    Messages:
    607
    Probably still some mistakes ..

    Staying with the invariant spacetime interval..
    s²=x²+y²+z²-c²t²

    For a light clock
    Basically a pulse of light bouncing between two mirrors.
    Code:
    
    The path of the pulse in the clock frame
    ______
       |
    ^  |
    y__|___
    x->
    
    The path of the pulse when the clock is moving at velocity Vx in the x direction with respect to the observer
    ______________
        /\  /
    ^    /  \  /
    y__/____\/_____
    x->
    
    I give up trying to edit this.
    Should be straight up and down in the clock frame and /\/\/\ for the observer.
    
    The invariant spacetime interval is (still)
    s²=x²+y²+z²-c²t²
    Currently nothing happens in the z direction so I'll go with
    s²=x²+y²-c²t²


    The separation of the mirrors is y
    The observer frame gets a dash after everything.

    In the clock frame
    y=ct (Newton, velocity*time)
    so
    y²=c²t²
    substituting for y in the spacetime equation

    s²= 0 + c²t²-c²t² = 0 # !
    In the observer frame

    \(x'=v'_{x}t'\)
    \(y'=v'_{y}t'\)
    squaring
    \(x'^{2}={v'_{x}}^{2}t'^{2}\)
    \(y'^{2}={v'_{y}}^{2}t'^{2}\)
    Substituting x² and y²
    \(s^{2}={v'_{x}}^{2}t'^{2}+ {v'_{y}}^{2}t'^{2} -c^{2}t'^{2}\)
    In the clock frame we found s²=0
    so
    \({v'_{x}}^{2}t'^{2}+ {v'_{y}}^{2}t'^{2} -c^{2}t'^{2}=0\)

    ----------
    In passing the vector addition of the velocities..
    \({v'_{x}}^{2}t'+ {v'_{y}}^{2}t' = c^{2}t'^{2}\)
    Paddoboy will be pleased to see the speed of light is c in both frames.
    There may be a chicken and egg relationship where the invariant spacetime interval is actually the egg not the chicken.
    ----------

    We found y²=c²t² for the clock and y is the same in both frames so we can substitute this into our equation for the observer frame with both impunity and advantage to get
    \({v'_{x}}^{2}t'^{2}+ c^{2}t^{2} -c^{2}t'^{2}=0\)
    Change signs
    \(-{v'_{x}}^{2}t'^{2}- c^{2}t^{2} +c^{2}t'^{2}=0\)
    Shuffle
    \(c^{2}t^{2}= c^{2}t'^{2}-{v'_{x}}^{2}t'^{2}\)
    So, as in the burping example
    \(t^{2}= t'^{2}(1-{v'_{x}}^{2}/c^{2})\)
    \(t= t'\sqrt{1-{v'_{x}}^{2}/c^{2}}\)

    Hopefully that was a fun.
    Tex ... never again! 20 minutes to write and over 2 hours to do the Tex.
     
    Last edited: Jun 11, 2020
  13. Neddy Bate Valued Senior Member

    Messages:
    2,108
    I think I can do it...

    Code:
    
    The path of the pulse in the clock frame
    ______
       |
    ^  |
    y__|___
    x->
    
    The path of the pulse when the clock is moving at velocity Vx in the x direction with respect to the observer
    ______________
         /\    /
    ^   /  \  /
    y__/____\/_____
    x->
    
    
    It was giving me a difficult time as well, but somehow it worked out.
     
  14. Neddy Bate Valued Senior Member

    Messages:
    2,108
    Yeah, I decided not to use Tex a long time ago, for that same reason.
     
  15. Confused2 Registered Senior Member

    Messages:
    607
    Muons? Twin Paradox?
     
  16. DaveC426913 Valued Senior Member

    Messages:
    14,619
    I think you're missing the point.

    With TEX, your readers spend 20 minutes understanding what you wrote.
    Without TEX, your readers spend (the 2 hours you saved) by trying to understand what you wrote - and giving up.

    TEX benefits the communicatee directly.
    TEX benefits the communicator indirectly ... by directly making the communicatee's life easier.


    It would have taken you less time to draw this in GIMP (free graphic editor), and copy/paste it straight into your post.

    A labelled diagram is worth a thousand words...
     
    Last edited: Jun 11, 2020
  17. Confused2 Registered Senior Member

    Messages:
    607
    Muons.

    I had five things in mind when I started this series of posts.

    1/The spacetime interval inherently gives time its own dimension which may (or may not) be clearer than flattening time onto space and being puzzled by the result.

    2/The constant speed of light dropped out nicely in the light clock post.

    3/It clarifies (hopefully) the relationship between a clock moving through (say) the Earth frame and a clock that is stationary in the Earth frame.

    4/The 'Muon experiment' is one half of the twin paradox <- this has been contraversial in the past and may be again.

    5/To dismiss this (following) resolution of the twin paradox the reader has to (belatedly) argue against the spacetime interval as a valid method.

    Muons.

    Bob is piloting the Muon, Alice stays at home on the ground. The count stations are 10km apart (vertically) and the Muon velocity (v') is 0.98c in the Earth frame. Bob doesn't move in the Muon frame except to start and stop his stopwatch.

    In the Muon frame Bob sees the first count station go past and starts his stopwatch, he stops his stopwatch as the second count station goes past.
    The stopwatch shows an elapsed time t.
    Plugging this into the spacetime interval:
    s²=-c²t²

    In the Earth/Alice frame we know (Newton again)
    y'=v't' # y' is the distance between count stations
    y'²=v'²t'²
    Plugging these values into the spacetime interval:
    s²=v'²t'²-c²t'²
    The spacetime interval is the same in both frames so
    -c²t²=v'²t'²-c²t'²
    Change signs, divide by c and extract t'
    t²=t'²(1-v'²/c²)
    t=t'√(1-v'²/c²)
    v'=0.98c so √(1-v'²/c²)=0.2
    So Bob's elapsed time (t) is 1/5 of the time Alice measures.

    Which is why Muons reach the ground (and why the twin paradox isn't a paradox?)

    The Twin Paradox.
    ---------------------------

    The spacetime interval is the same going up as down.
    The elapsed time (t) in Bob (the traveller) frame is up + down
    2t=2t'√(1-v'²/c²) where t' is the elapsed time in the Earth/Alice/stay at home frame.
    If v'=0.98c
    t=0.2t'

    That's it! Done.
     
    Last edited: Jun 12, 2020
  18. Neddy Bate Valued Senior Member

    Messages:
    2,108
    So, in the earth frame, the tick rate of the muon clock is 1/5 the rate of the earth clocks. So if the muon clock elapses one tick, the earth clocks elapse 5 ticks.

    However, there is still the slight problem (but not a paradox), that in the muon frame, the tick rate of both of the earth clocks (upper and lower) is 1/5 the rate of the muon's own clock. The resolution to this slight problem (not a paradox) is simply that the two earth clocks are out of synch in the muon frame, (even though they are in synch in the earth frame), such that the lower earth clock is already ahead by 4.8 ticks when the muon passes the upper earth clock which displays 0.0 ticks, and thus when the muon reaches the lower earth clock it displays 5.0 ticks, having only elapsed 0.2 ticks during the muon's journey. This is because 0.2 ticks is 1/5 the single tick of the muon's own clock.
     
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  19. phyti Registered Senior Member

    Messages:
    458
    Confused2;
    True in terms of calculation, but not in terms of physical processes.

    Fig.a, a light clock in a rest frame, consisting of emitter/detector ED and a mirror M.
    Light (blue) oscillates in any direction p, perpendicular to direction of motion x.
    One clock cycle = 2.

    Fig.b, the clock moves at .5c. The light pulse has a constant speed, but can change direction. The pulse now consists of two components, vt which compensates for clock motion, and ut the active part of the clock. After 1/2 cycle their is a gap between the pulse and M. The reason for this is, 'the propagation speed of light in space is independent of its source'. If a material object had been projected in the p direction at speed w, it would have simultaneously moved at v acquired from the clock and w, thus intercepting M. The clock is therefore generating cycles at 87% capacity. The observer moving with the clock is regulated by biological processes which are also operating at 87%, and is thus not aware of any changes in the clock process.

    Fig.c, the vt component is masked by the observer motion, so he is only aware of the vertical ut component, which he interprets as ct'. If u is interpreted as c then t' becomes local/proper time, and constant for all speeds.

    The expression for the invariant interval can be extracted from the light clock.

    Please Register or Log in to view the hidden image!

     
  20. Confused2 Registered Senior Member

    Messages:
    607
    I've never actually tried to do it (probably should have). How? Just x and t would establish the principle.
     
  21. phyti Registered Senior Member

    Messages:
    458
    Confused2;

    It would be nice if there was a standard posting method for all forums.
    I use an equation editor which is almost as easy as writing on paper.
    Then paste them into a doc. format, then translate to a pdf format.
    I would never waste my time with tex or any variation.
     

    Attached Files:

  22. Confused2 Registered Senior Member

    Messages:
    607
    I tried to follow your light clock analysis and eventually looked at another of your light clock diagrams.
    http://www.sciforums.com/threads/ho...ion-by-experiment.162037/page-17#post-3626412
    This seems to be an analysis based on the fact that light expands spherically in every frame regardless of the velocity of the source - not a light clock.
    A 'light clock' is a pulse of light bouncing between two mirrors - that really doesn't seem to be what you have drawn.
    From a light clock (perhaps rashly) I would expect to see | and /\/\ not O. I may not be the only one to have been puzzled by your light clock diagrams over the years.

    I'll look at it again as a section through an expanding sphere of light.
     
  23. phyti Registered Senior Member

    Messages:
    458
    Confused2;

    My version of light clock uses 1 mirror M and an emitter/detector ED sending single pulses (blue) instead of a single photon oscillating between 2 mirrors. The light gray lines are measurements. The arc shows all possible positions for a photon after 1/2 cycle. The speed of the clock determines the angle between the vertical and light path.
    The speed of light is fixed but the direction can vary. In fact, it must vary, else there would have to be a preferred direction of oscillation.

    In view-a, you are at rest with the clock, and see the pulse move up and down vertically for 1 clock cycle.

    In view-b, you watch the same clock move past you at .5c. The light (blue) has started its zig-zag path. We only need to show its progress compared to the rest clock 1/2 cycle. In both the a and b views, the light has moved the same distance in space of 1 unit.
    The right angled triangle shows the progress is only .87 that of the rest clock.

    In view-c you are moving with the clock at .5c. The clock indicates .87. You don't question it as being slow, since your biological clock is also running at .87 of its rate at rest. You agreed by definition that 1 clock cycle requires 1 vertical oscillation, so that becomes your standard while moving with the clock.
    The cvt triangle always shows the local/proper time of the moving clock, as the ut component. The calculations with the graphics show the gamma factor is contained within the geometry of the light clock.
     

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