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How to explain time dilation easily to others, maybe like this ?

  1. Nov 6, 2011 #1
    My idea of explaining time dilation to others (average people) was : see a light wave as the time, person A standing still sees the time (light wave) passing, person B moving sees lesser time passing when going in the direction of the light wave (while "looking" to the light wave) because he is moving.

    Ok, you can tell this but you have to proove if you are right.

    So, I started my topic, https://www.physicsforums.com/showthread.php?t=536987" , consider only answer #11 with picture, answer #18 (ignore formula Lorenz, was wrong), and last answers from #39 (rest is nonsense because of mistakes).

    So I show person A was predicting a time dilation without Lorenz (by measure the total length of the passing light wave in his rest frame, ΔXa_light, for himself and thinking for person B, ΔXb_light).

    Person A calculated the next formulas :

    ΔXb_light = ΔXa_light . (1 - V/C).

    ΔTb_light = ΔTa_light . (1 - V/C).

    Let's see how formula's are after Lorenz, by making an equation for the lightspeed in the rest frame of B (only the line c can be converted from restframe A to B, no other speedlines, than Lorenz goes wrong).

    I found : (no mistakes this time)

    ΔXb_light = γ . (ΔXa_light - V/C . ΔXb_seen_from_a).

    ΔTb_light = γ . (ΔTa_light - V/C2 . ΔXb_seen_from_a) ..... or ... ΔTb_light = γ . ΔTa_light . (1 - V2/C2)

    In both formula's is clear to see that the length of the passing light wave (and time) is depended from ΔXb_seen_from_a (movement person B) in a way liking on person A it saw (so the prediction by A was right).

    Question: have I now the facts to explain time dilation in this way ?
     
    Last edited by a moderator: Apr 26, 2017
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  3. Nov 6, 2011 #2

    phinds

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    Uh, you seem to be making it awfully hard for as an explanation for a beginner. Try this:

    http://www.phinds.com/time%20dilation/ [Broken]

    EDIT: I say you've made it hard, 'cause you've used equations w/o a graphic, but graphics are what makes it easier for the average person to understand (see my link)
     
    Last edited by a moderator: May 5, 2017
  4. Nov 7, 2011 #3
    Hi Phinds,

    I explain it only with this sentence : "see a light wave as the time, person A standing still sees the time (light wave) passing, person B moving sees lesser time passing when going in the direction of the light wave (while "looking" to the light wave) because he is moving".

    The rest of my story is only to proove my sentence for physicists, otherwise you don't tell the truth. Graphics are also difficult to understand for avarage people, they look to it and they believe it and say I understand maybe to please you and forget it, one sentence can do more I believe and I tell it just like something in real life. E.g. A standing still looks to a passing train and sees more wagons passing than B moving in the direction of the train. Time dilation is never easy to explain/understand but now I can compare it with just something from daily life. That's my thought ...
     
    Last edited by a moderator: May 5, 2017
  5. Nov 7, 2011 #4

    ghwellsjr

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    Then why doesn't it work the same way for trains?

    If you can say that you "see a light wave as the time", why can't I say I "see the train as the time"? If you can say that person B (moving) sees fewer waves go by him than person A (stationary) and that illustrates time dilation, then why can't I say that person B (moving) sees fewer wagons go by him than person A (stationary) and that illustrates time dilation?

    I've told you over and over again in your other threads that you are demonstrating Dopper shift, not time dilation.

    I've told you over and over again in your other threads to think about your same illustration with person B going in the opposite direction. Now he will see more waves (or wagons) going by him than person A. Does that prove or illustrate time contraction? Again, both of these examples illustrate Doppler and unless you can show that the Doppler is different for trains than it is for light, you haven't addressed the issue of time dilation.
     
  6. Nov 7, 2011 #5
    Hi Gwh,

    If you can say that you "see a light wave as the time", why can't I say I "see the train as the time"? If you can say that person B (moving) sees fewer waves go by him than person A (stationary) and that illustrates time dilation, then why can't I say that person B (moving) sees fewer wagons go by him than person A (stationary) and that illustrates time dilation?[/QUOTE]

    No because a train is not responceable for time dilation, light is. The times for trains are corrected by Lorenz via the "timetables" for light. And in the beginning of my topic I was speaking about more or lesser waves but that's on the end of my topic not the case anymore (that's why I say skip all that answers between, that gives confusion) . My formulas are just Einstein and if you calculate further it's just ΔTb_light = 1/γ . ΔTa_light and ΔXb_light = 1/γ . ΔXa_light. As well distances and times are lower/smaller compared to A.

    That's possible what A thought, but I compare the formula's before and after Lorenz and see in both that the movement of B (ΔXb_seen_from_a) is subtracted (or partly) from the total length of the light wave ΔXa_light. The waves are the same. On the end you say, everything is smaller also time and distance (length of light waves, but same waves), but the formula let's see what's happening between that result and before Lorenz.

    I know that all is symmetric in the SRT (see end of my topic) , but that's another story/step to explain. Here I talk only over time dilation for B compared to A not the other way. This step of symmetric can't be used with trains, unless you change the direction of the train. With light is the direction not important (gives time dilation in both directions, must be, compared to A or B, because of the constant light speed and limit of speed).
     
  7. Nov 7, 2011 #6

    ghwellsjr

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    I'm not talking about the symmetric relationship between A and B with respect to time dilation. I'm talking about B traveling toward A and seeing more waves than A.

    Let me try it this way:

    Suppose A and B are at the same place at the same time. As soon as A turns on his light, B starts moving away from A. They are both counting the waves. B will count fewer waves than A, as you stated. But after awhile, B reverses direction and starts moving toward A, all the while continuing to count the waves. Now he will be counting them at a faster rate than A. When he gets back to A and stops, both A and B will have counted exactly the same number of waves, won't they?

    What happened to the time dilation?
     
  8. Nov 7, 2011 #7
    Gwh, I agree both count the same waves, you think that I don't agree as in the beginning of my old topic and you repeat that all the time, not me. That confuses now for others. THat's just not anymore what I said.

    I just say it looks that your own movement (B) is subtracted from the original light wave as A thought.

    The Lorenz formula says, it's partly subtracted but that's too much and corrected by γ (mostly > 1). Maybe it has no meaning it is subtracted in the formula because the end result is only all is smaller (resized) but you are not aware, also the clock is ticking slower.

    The end result is only that all is smaller (same waves but smaller, compared to A).

    You could think in a moment, the result that you move and the light wave is going slower for you, results in a effect that you see the same light wave but smaller (same waves, just resized, but there is a time dilation now).

    Maybe you have or know an animation that shows all the moments of moving with as end result you see all smaller (compared to A).

    Maybe is this the confusion now, I speak about the length of all waves together (so the light signal), maybe you think I speak over the length of 1 wave.
     
  9. Nov 7, 2011 #8

    ghwellsjr

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    Well if you agree they both count the same waves, then counting waves does not illustrate time dilation because if both A and B had clocks that started with the same time on them, then B's clock would have less time on it when they got back together.

    And the reason why counting waves does not illustrate time dilation is because your statement, "see a light wave as the time", is not correct. Light is not time and it doesn't represent time. You need a clock to represent time. You can make a clock using light bouncing between a couple mirrors, but without the mirrors, there is nothing to count. Counting light waves as they progress past you is not counting time.

    I have already shown you an animation that demonstrates time dilation, length contraction, and relativity of simultaniety but you didn't like it.
     
  10. Nov 7, 2011 #9
    I always found the light clock on a train to be the best way of convincing people of time dilation.
     
  11. Nov 7, 2011 #10

    dlgoff

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    Last edited by a moderator: May 5, 2017
  12. Nov 8, 2011 #11

    A.T.

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    This one?

    https://www.youtube.com/watch?v=C2VMO7pcWhg
     
  13. Nov 8, 2011 #12

    A.T.

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    Light is always moving at the same speed for you, no matter how you move.
     
  14. Nov 8, 2011 #13

    Dale

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    This doesn't work because it is refering to the longitudinal Doppler effect, not the transverse Doppler effect.
     
  15. Nov 8, 2011 #14

    ghwellsjr

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    That's one of the best animations but this the one I made:

    https://www.youtube.com/watch?v=dEhvU31YaCw

    I showed it to digi99 on his thread https://www.physicsforums.com/showthread.php?t=547461"
     
    Last edited by a moderator: Apr 26, 2017
  16. Nov 9, 2011 #15
    I look to all animations and answer all later (spent too much time to my subject and lesser to my own profession for the moment). Maxwell is also very interesting but it all needs time and I get that in time.

    Using a mirror is only the equipment you need to measure the passing light wave, I go not so far in that for explaining, this makes it more complex than needed to explain time dilation.

    I don't understand why you see light not as a time signal in your own reference system, in Lorenz you need it to correct our times.

    But all is relative and also you need a reference point to express time otherwise it has no meaning (in this case person A is our reference point, generally our planet is a reference point for us).

    When I say the light wave is going slower for you, I mean time goes slower for you, lightspeed c is the same (distances are going smaller too).

    That's all going smaller is good to understand. Every moment the light waves (time) are going slower for you, this will be immediately corrected by smaller distances, so immediately everything (compared to person A) will be going smaller. Just like my formula (I mean Lorenz) says.

    Maybe is all designed in our space we can't detect the source, it does not mean that the same laws are valid in other spaces. Physicists are speaking more and more about parallel worlds, I have to read about all that stuff but I can feel it already somehow.
     
  17. Nov 9, 2011 #16

    ghwellsjr

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    Do you know what Doppler is? Do you know what time dilation is? Are you aware that if we consider a Frame of Reference where A and the source of the light waves are at rest, then it doesn't matter which direction B is traveling, he will have the exact same time dilation, as long as his speed is constant? Are you aware that B will detect a different Doppler frequency of the light waves than what A is detecting depending on his direction? If he is traveling toward A, he will detect a higher frequency for the light. If he is traveling away from A, he will detect a lower frequency for the light. But the time dilation will remain the same and can be determined from the Doppler but not in the way you are doing it because you are trying to say that because B counts fewer waves while traveling away from A, that demonstrates time dilation. And so I ask you, if that is true, what does it demonstrate when B travels toward A and counts more waves?
     
  18. Nov 9, 2011 #17

    ghwellsjr

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    When I said that counting light waves is not the same as a clock, I was talking about person B who is moving relative to the distant light source located with person A and counting those light waves. Person B can use a light source as a clock, even without mirrors, as long as he is carrying the light source with him. I explained this all in your first thread on this subject:
    And as long as we know that person A (observer 1) remains stationary and person B (observer 2) continues at a constant speed, then each of them can use the ratio of the waves counted from the other one divided by the waves from counted from their own light source and this will be the correct Relativistic Doppler from which they can each calculate the relative time dilation of the other one. They will both get the same ratio and they will both calculate the same time dilation. This calculation of the time dilation from the Relativistic Doppler continues to work correctly even when person B turns around and comes back to person A.

    However, it is not obvious that while person B is traveling with respect to person A's light source that his own light source will be time dilated and emitting a frequency that is lower than person A's and that's why I keep saying that your oversimplified teaching demonstration doesn't work. If you don't arbitrarily say to your student, "Oh, by the way, person B is experiencing time dilation so his light source is emitting waves at a lower frequency than person A, trust me", then the illustrations will be just like normal Doppler for waves traveling in a fixed medium like water or air.

    If you look up Doppler Effect in wikipedia, you will see that the formula for the detected frequency by an observer moving away from a stationary source is:

    f = [(c-v)/c]f0 = f0(1-v/c)

    This is exactly the equation you came up with in your first post and in your other threads.

    Now if you look up Relativistic Doppler Effect, you will see that the formula for the detected frequency by an observer moving away from a stationary source is:

    f = f0√[(1-v/c)/(1+v/c)]

    Do you see the difference? There is a squareroot operation and a division by (1+v/c) which you don't have in your equation.

    I sure hope this clears up this matter for you.
     
  19. Nov 17, 2011 #18
    Hi Gwh, I had a computercrash, so I could not answer sooner.

    If I use this formula ΔXb_light_seen_from_a = ΔXa_light . (1 - V/C) in this formula ΔXb_light = γ . (ΔXa_light - V/C . ΔXb_seen_from_a) I get :

    ΔXb_light = SQRT(1 - v2/c2) / (1 - V/C) . ΔXb_light_seen_from_a

    This looks indeed the Relativistic Doppler effect (if ΔX can be replaced by frequencies too).

    But this was not meant in my topic, it's a coincidence that both formulas are the same. But in my topic I was expressing time ΔTb_seen_from_a and came to my formula.

    You are still talking about counting waves (I don't), and indeed you get the Doppler effect while holding an object (device) in a light wave. So maybe we have to find something in the future that it can be in another way, if it was necessary for me, but it is not.

    But I have already confirmed that you never will see my effect, a slower going light wave, because by nature distances are immediately corrected, so the light speed will be the
    same. Finally you will see only smaller light waves (if could be showed to you).

    But still can't be proofed that I may not explain that on my website and to other interesting people. In the Lorenz formula ΔXb_light = γ . (ΔXa_light - V/C . ΔXb_seen_from_a) you still see that ΔXb_seen_from_a is subtracted (partially) from ΔXa_light, maybe it has no meaning.

    Yes, because V + C = C is only possible with a slower going time on the left side of the equation (forced by light, the time signal).
     
  20. Nov 17, 2011 #19
    But I have some questions about time dilation I not really understand to come further (and I find it all very interesting).

    1) I understand that the situation between A and B is SRT symmetric, so if they compare clocks they both see the same time. In fact there is no time dilation in this situation to messure (you can not suppose if you are standing still, that your clock is going slower because somebody else is going to move compared with you, let's say an astronaut in space) ? Does this mean that a time is measured in this case included a time dilation (and same of course) ?

    2) Only if forces are used, a time dilation is to measure ?

    3) But why is a Muon lifetime to measure ? (let's say his time is always 4 seconds if standing still (A), if moving (B) his 4 seconds takes 6 seconds for (A), so we see on our clock 6 seconds, if the Muon could show a clock it should show 4 seconds, where is now the symmetry)

    4) Can you express time with light waves between two points in space e.g. point 1 is our reference point Earth and point 2 has a constant speed going into space ?
     
  21. Nov 19, 2011 #20
    This is interesting Gwh, so you say you can calculate time dilation from the Relativistic Doppler. Does this mean you can calculate the time dilation between een star and Earth if you could isolate it's light ?
     
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