Laymen question about relativity

Discussion in 'Physics & Math' started by Doctor Dread, Oct 6, 2017.

  1. Maxila Registered Senior Member

    Messages:
    156
    Hi Doctor,

    I know it been a while since you posted this but no one answered specifically to some problems in the context of your scenario. It appears you know the man could not travel at the speed of light but rather a speed close to it instead. That will be the assumption, a relative speed that results in a gamma of 876000 (about 99.99999999993483C pretty fast!). I’ll use your light-like calculations as approximations for the man traveling “near the speed of light”.

    You made a mistake in the reconciliation for the man in orbit relative to the observer on Earth; saying the number of orbits witnessed on Earth (25,200), which was 1 hour for them was 100 years for the man in orbit. That is incorrect; the correct observation for the person on Earth after 1 hour of Earth time would be to see the man age 1/876000 of an hour or .000411 seconds for the equal number of orbits they both witnessed during those relative time periods.

    When 1 hour has passed for the orbiting man they will still both agree on the number of orbits seen 45 million (assuming your math was correct); however they will not only disagree on the time that passed (1hr vs 100 years), what you didn’t recognize is they will also disagree on the distance traveled . The distance traveled around the Earth for the orbiting man would have been 1/876,000 less than the distance seen for the observer on Earth. This might help you re-evaluate you conclusions.

    As for the comment “The faster you go the "slower" YOU age, not slower you go through time”; in your relativistic example, ageing more slowly is a direct result of the relative difference in the space-time, meaning time was passing “slower” for the orbiting man.
     
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  3. Neddy Bate Valued Senior Member

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    1,430
    Yes, the orbiting man would record the earth as being extremely length-contracted along the axis of relative motion. This would lead the orbiting man to conclude that the distance traveled is smaller by a factor of 1/gamma, just as you say. Since the orbiting man's distance traveled and his recorded time are both affected by the same factor, this results in both men agreeing on the relative speed between them.
     
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  5. Schmelzer Valued Senior Member

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    3,703
    You are not crazy at all, this is simply the classical, original interpretation of special relativity supported by Lorentz and Poincare, which is today known as the Lorentz ether.

    It is simply another interpretation of special relativity, thus, gives the same predictions about all physical experiments. It is easier to understand for laymen, and Bell has argued in a paper named "how to teach special relativity" that it makes sense to teach it because it gives better intuitions.

    A similar interpretation is possible also for the equations of general relativity. See http://ilja-schmelzer.de/ether
     
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  7. Confused2 Registered Senior Member

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    443
    Trying a simpler example...
    There are two stations 10km apart.
    Alice is standing on one station - the stations are obviously 'stationary' in Alice's frame.
    Bob travels on a train between the two stations.
    Now the subtle bit...
    Bob travelled 10km between the stations which were stationary in Alice's frame so clearly he isn't stationary in Alice's frame.
    Now the killer... Bob and his clock are stationary in his frame - he just sees the stations fly past.
    Major difference here...
    Bob's clock hasn't moved in Bobs frame
    Bob's clock has moved 10km in Alice's frame.
    Alice's clock is everywhere the same in Alice's frame (another subtle point).
    First event is Bob at first station
    Second event is Bob at second station
    1/ In Alice's frame the events are separated by time and distance.
    2/ In Bob's frame the events are only separated by time.

    Until/unless you understand 1/ and 2/ then I don't think you stand any chance with relativity.
     
  8. Confused2 Registered Senior Member

    Messages:
    443
    Since my earlier attempt to derive time dilation from first principles didn't seem to go too well I'm going to start with an assumption - maybe that will go better.
    The assumption is that the spacetime interval (s) is the same for all observers.
    s²=x²-c²t²
    Following on from my last post...
    Using lower case for Alice's frame and upper case for Bob's..
    s²=x²-c²t²
    and
    s²=X²-c²T²
    where (this is all High School algebra from now on)...
    x²-c²t²=X²-c²T²
    In Alices frame the distance between stations is the same as the velocity multiplied by the time travelled so we have
    (vt)²-c²t²=X²-c²T²
    and in Bobs frame the distance between the events is 0 so
    (vt)²-c²t²=-c²T²
    or
    v²t²-c²t²=-c²T²
    extract t²
    t²(v²-c²)=-c²T²
    divide by c²
    t²(v²/c²-1)=-T²
    multiply both sides by -1
    t²(1-v²/c²)=T²
    or
    t²=T²/(1-v²/c²)
    or
    t=T/√(1-v²/c²)
    since √(1-v²/c²) is always less than 1 it follows (since we're dividing the elapsed time in Bob's frame by it) it follows that the elapsed time in Alice's frame is always greater than the elapsed time in Bob's frame because we are looking at events separated by distance and time in Alice's frame and the same events separated only by time in Bob's frame. It also follows naturally that Bob is your time-dilated uncle.
     
  9. Neddy Bate Valued Senior Member

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    1,430
    Nice derivation.

    Just to add a little more information, (and possibly some confusion): As long as the train is in uniform translational motion, the situation is perfectly reciprocal, because you could have considered two train cars and only one station, instead of two stations and one train car. In that case you could have derived the reciprocal time dilation equation:
    T=t/√(1-v²/c²)
     
  10. Confused2 Registered Senior Member

    Messages:
    443
    Thanks for that - it's a party piece of mine - I think you may well be the first person to have actually read it (equations and all that). I personally dislike starting with s²=x²-c²t². If we were looking at (say) a Pythagorean triangle with distance on one side and time on another we'd have to find the conversion constant between time and distance that makes the triangle a 'right-angle'. By a quirk of nature the constant k and the speed of light happen to be the same so we plug in the speed of light as though that defines the constant whereas I think it makes more sense (as geometry) to go with k and later comment that c 'happens' to have the same value as k. With some knowledge of the Standard Model it might be reasonable to say "Of course that k and c are the same." - but I don't have that knowledge.

    This is (of course) correct and difficult (or impossible) to read from a graph (or equation) without first knowing what the graph (or equation) is actually a graph (or equation) of - usually one clock moving between (synchronised) clocks located in another frame.
     
  11. Schmelzer Valued Senior Member

    Messages:
    3,703
    I prefer another way to introduce Lorentz transformations.

    Start with a wave equation. $\square u(x,t) = (\partial_t^2 - c^2 \nabla^2) u(x,t) = 0$. Assume you have found some solution $u_0(x,t)$. You may want to construct other solution. Then, there is a simple method: Make a Lorentz transformation. Namely, if $t' = t'(x,t), x'=x'(x,t)$ is some Lorentz transformation, then $u_v(x,t) = u_o(x'(x,t),t'(x,t))$ is another, different solution of the same wave equation. This method works for every such wave equation, also for water waves or sound waves, if you use the appropriate wave speed as c. The resulting solution is named the Doppler-shifted solution.

    The next step is understanding that there are even more equations where the same trick works. So, one can add some "mass term" $(\square + (mc^2)^2) u(x,t) = 0$, so that the resulting waves obtain a speed smaller than c. You can also do the same trick with the Maxwell equations, and with the Dirac equations. You can combine different equations with the same c, and add a quite general reaction term $r(u_1(x,t), u_2(x,t),\ldots)$, where essentially the only restriction is that it has to be computed at the same point $(x,t)$. In other words, you can include the wave equations of all particles and fields described by the Standard Model of particle physics.

    The result is that if you have some solution of the equations of the SM, you can apply the Lorentz transformation and obtain another one, a Doppler-shifted one.

    The next step is to see what happens now if you construct a clock out of solutions of fields described by the SM. Of course, once you have one solution which describes this clock, you can apply a Lorentz transformation and have another solution of the same equations. And one can easily find out some properties of this clock. Namely, there have to be events when the clock shows the time $\tau_0$ or $\tau_1$, and these events will have some location in space, $x_0$ resp. $x_1$. Then you can take a look at similar events in the Doppler-shifted solution. What are these "similar events"? Very simple, given that we have applied a particular Lorentz transformation to construct the Doppler-shifted solution, one can see that the similar events are simply obtained by applying the Lorentz transformation to the events $(x_0,\tau_0)$ resp. $(x_1,\tau_1)$. So, we can simply apply the same transformation $\tau' = \tau'(x,\tau), x'=x'(x,\tau)$ to the events to find out the corresponding events. By construction above, the Doppler-shifted solution has the same value as the original solution at the original events.

    And, essentially by construction, one cannot distinguish the two solutions simply by observing the clocks themselves. But, if the original solution was with a clock at rest, $x_0=x_1$, the new solution is a moving one, $x'_0\neq x'_1$. And it is measuring as "the same time" 1 for different time intervals, $\tau_1-\tau_0 \neq \tau'_1=\tau'_0$. So, moving clocks are time-dilated.

    http://ilja-schmelzer.de/ether/introduction.php for more details.
     
  12. Neddy Bate Valued Senior Member

    Messages:
    1,430
    Yes, I agree. I had not seen that approach before.

    Based on your posts #16, #18, and #20, I would have thought you would have taken the usual approach of using a vertical light-clock. This results in a right-triangle which has a hypotenuse with a length of ct, one leg with a length of vt, and the other leg with a length that is simply the height of the vertical light-clock itself.

    That is what phyti tried to do in post #14, although I don't fully understand the wording of that post, or the significance of the "circular arc" shown in the diagram.

    No, I don't see it that way. The units on each side of the right-triangle should simply be lengths ct, vt, etc. as explained above. The only underlying assumption needed is that the speed of light is the same in both reference frames. This lets you design a simple clock made out of a photon traveling a known distance. The vertical light-clock is the easiest to use, because you don't have to worry about length-contraction in that case.
     
    Last edited: Nov 25, 2017
  13. Confused2 Registered Senior Member

    Messages:
    443
    Your maths is beyond me... given time I'll try to use what I can do to attempt to gain some insight into what you suggest. For the present I'm staying with the basic SR I think I understand to find out whether or not I do actually understand it.
     
  14. Neddy Bate Valued Senior Member

    Messages:
    1,430
    Using your convention of lower case for Alice's frame and upper case for Bob's, with a standard vertical light clock on the train with Bob, you could derive the time dilation equation from a right-triangle with a hypotenuse of length ct, a horizontal leg of length vt, and a vertical leg of length cT, as follows:

    (ct)² = (cT)² + (vt)²
    c²t² = c²T² + v²t²
    c²t² - v²t² = c²T²
    t²(c² - v²) = c²T²
    t²(c²/c² - v²/c²) = c²T² / c²
    t²(1 - v²/c²) = T²
    t² = T² / (1 - v²/c²)
    t = T / √(1 - v²/c²)
     
  15. Confused2 Registered Senior Member

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    443
    Last edited: Nov 26, 2017
  16. phyti Registered Senior Member

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    259

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    If you plot time dilation, might as well add length contraction.

    On the left, A is the rest frame, with a mirror M attached at the far end of a stick d units long. B has a copy of the stick with a mirror M'. At the origin, A and B send a signal (blue) to their mirror.
    A calculates the B-clock as running slow by 1/g, and the B-stick lc by 1/g.
    On the right, B calculates the A-clock as running slow by 1/g, and the A-stick lc by 1/g.
    (g=gamma)
     
  17. Write4U Valued Senior Member

    Messages:
    5,085
    Isn't it the exact opposite? If you are passing the earth @ "c", to you the earth would not appear as a blur, but would appear to stand still, and everything on it. Is that not what Einstein described by "riding a beam of light"?

    Am I understanding this correctly?
     
  18. Write4U Valued Senior Member

    Messages:
    5,085
    Roger Antonsen demonstrated this: if we have 4 columns of squares stacked 3 squares high, a diagonal line from corner to corner always equals 5. This would assume the balloon rises 30 ft. as we walk 40 ft. Start watching a 7:50
    https://www.ted.com/talks/roger_antonsen_math_is_the_hidden_secret_to_understanding_the_world
     
    Last edited: Nov 29, 2017
  19. Confused2 Registered Senior Member

    Messages:
    443
    I have to admit I can't make any sense of this. I think it was phyti who first drew my attention to light clocks (thank you) - some years and some light clocks later it all started to make sense. My own experience is that no amount of construction lines helped - keep it simple and write out the equations in both frames with the speed of light the same in both and time dilation will emerge.
    Pythagoras is famous for working out (for a right angle triangle and only for for a right angle triangle)
    The square on the hypotrenuse is equal to the sum of the squares on the opposing two sides.
    If one side is 3 feet then the square on that side is 9
    If another side is 4 feet then the square on that side is 16
    Adding the squares we get 25
    25 is the square of 5
    so
    If the opposing sides are 3 units and 4 units and the hypotenuse is 5 units then the triangle will be a right angled triangle.
    See (for example) https://en.wikipedia.org/wiki/Pythagorean_theorem
     
  20. Write4U Valued Senior Member

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    5,085
    Is it not remarkable that mathematics which started with the number 2 and evolved with the concepts of more numbers, has enabled us to make sense of natural patterns which emerge from the relationship between those numbers.
     
  21. NotEinstein Registered Senior Member

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    545
    That's the first time I'm hearing mathematics started with the number 2; fascinating! Do you have a link to some more information about this?

    Oh, and natural patterns don't emerge from the relationship between numbers.
     
  22. Write4U Valued Senior Member

    Messages:
    5,085
    Yes, Mario Livio mentioned it as natural fundamental recognition of pairs associated with human sensory organs and appendages, i.e. 2 eyes, 2 ears, 2 nostrils, 2 breasts, 2 arms, 2 legs, 2 feet.

    IOW, the mirrored symmetry, which is a pattern of the right side and the left side of the human body and most bipedal animals.
    That's true, as you stated it, but patterns existed before we invented numbers to symbolize them and the most fundamental pattern is the mirror image of something.

    Interestingly, Lobster DNA produces a complete right side and a complete left side of the body. This was proven by the fact that some lobsters have 2 different colors which are divided exactly through the middle of the entire length of the lobster body.

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    It has been shown that many animals can make a distinction between more and less. Experiments with Lemurs (a distantly related species) clearly showed they can recognize the size of quantities just as fast as humans. Of course they don't count, but it seems a hardwired cognitive part of the brain.

    As hominid intelligence evolved it seems natural that the number 2 would be the first number to be given a symbolic value of plurality. The number 1 is not plural, it is singular.


     
    Last edited: Dec 3, 2017
  23. NotEinstein Registered Senior Member

    Messages:
    545
    Do you have a link, please?

    But isn't the number one even more prevalent?

    Not sure if that's the most fundamental pattern, but I mostly agree, yes.

    (No comment)

    Yes, but doesn't the fact you have to explicitly exclude the number one strongly suggest it is the first number to be invented?
     

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