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Why does Einstein say that a clock slows when it is moving?

  1. Dec 27, 2009 #1
    In http://www.bartleby.com/173/12.html" [Broken] Einstein says the following:
    But what is he saying here?

    That 1 proper second in the reference frame K is equal to gamma proper seconds in the moving frame K0, leading to the conclusion that the seconds in K are larger than the seconds in K0 and that therefore the moving clock slows?

    No it isn't that, for the proper seconds in K0 have been transformed by the application of the Lorentz Transformation equations into coordinate time (as we term it), so he is saying that 1 proper second in the reference frame K is equal to gamma coordinate seconds in the moving frame K0.

    So Einstein is actually describing a conversion from one set of units to another; that conversion being performed using the Lorentz Factor.

    And does the moving clock slow? Well, no, for what we see is that 1 unit of time in the reference frame K is equal to gamma coordinate units, so if anything one would say that the moving clock goes faster, albeit in smaller seconds.

    Grimble:uhh::confused::uhh:
     
    Last edited by a moderator: May 4, 2017
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  3. Dec 27, 2009 #2

    HallsofIvy

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    What he is saying is that a person in frame K sees t'= 1 second tick off a clock in frame K0 while he sees [tex]t = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex] tick off his own. t and t' do NOT measure "how long a second is", they measure what time is shown on a clock. Since t is larger than t', a clock in K0 shows time changing more slowly than in K.
     
  4. Dec 27, 2009 #3
    Yes but since the formula he is using is [tex]t = \frac{t^'}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex] with [itex]t^'[/itex] set to 1, we have an equation where the dimensions are not the same on each side and, as I understand it, this is just plain wrong.
     
  5. Dec 27, 2009 #4

    Pengwuino

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    t' = 1 second, the units remain
     
  6. Dec 27, 2009 #5

    HallsofIvy

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    Why would you say the dimensions are wrong? v and c are both in "distance per time" units so [itex]v^2/c^2[/itex] is dimensionless and so is
    [tex]\frac{1}{\sqrt{1- \frac{v^2}{c^2}}}[/tex]
    Both sides have dimensions of time.

    (Surely you are not thinking that "setting t= 1" removes the time units on the right? t is a time measure. Setting t= 1 means setting t equal to "1 second" or "1 minute" or "1 hour".)
     
  7. Dec 27, 2009 #6
    What sort of second is that? A proper second, or a coordinate second, for they are surely separate units, are they not? After all the Lorentz factor is that which is used to convert between the two.



    Wait a minute there, you cannot be suggesting that an equation that has units of days on one side and milliseconds on the other has the same units, just because they both are units of time, can you? For that would be the equivalent of saying we could have feet on one side of an equation and centimetres on the other and that would be valid, as they are both dimensions of distance.

    No, what the formula is saying is that 1 unit of proper time is equal to gamma seconds of coordinate time, that is coordinate time has units that are smaller than proper units by a factor of gamma.

    Grimble
     
  8. Dec 27, 2009 #7
    I don't know what you mean by "proper second" and "coordinate second", and what is the difference between them. As far as physics is concerned, there is only one type of "second" - the second shown by a well-designed clock. Einstein predicted that the "second" shown by a moving clock lasts longer in comparison to the "second" shown by the clock at rest. This is the relativistic time dilation.

    Eugene.
     
  9. Dec 27, 2009 #8
    likewise by symmetry a person in frame K0 sees t=1 seconds tick off a clock in frame K while he sees [tex]t' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex] tick off his own. Since t' is larger than t, a clock in K shows time changing more slowly than in K0

    Have I stated this correctly? Velocity is relative yes?
     
  10. Dec 27, 2009 #9
    If you and I are moving with respect to each other, then I say that your second (the second shown by your co-moving clock) is longer than mine. On the other hand, you say that my second is longer than your second. Both of us are correct. That's relativity.
     
  11. Dec 27, 2009 #10
    Hello Grimble

    All this has already been addressed at length in a thread started by you entitled, as far as I can remember, Time Dilation Formula.
    Matheinste
     
  12. Dec 27, 2009 #11
    Observation versus clock count. I see how both parties see the frequency of light flashes emitted by the other as slower due to their relative velocity away from each other. But suppose both observers agree to fire retro rockets after 1 million flashes (one flash per second) locally in their own frame and come to a mutual relative velocity of zero. And at that point they both agree to turn off the flasher. How many total flashes will each see before they see the flashing stop? I say both will count 1 million flashes. So what do you mean when you say a different amount of time passes in one frame versus the other?
     
  13. Dec 31, 2009 #12
    Let us say we have two identical clocks, one in system A, and one in system B.

    Clock A records 10 seconds while clock B records only 8 seconds.

    So for clock B more time is passing between successive 'ticks' (as Einstein put it): clock B is running slow.

    But is it? Or is time running faster? If more time is passing between two ticks of an identical clock does that not imply that it is time passing at a faster rate?
     
  14. Dec 31, 2009 #13

    HallsofIvy

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    In this formula, a "second" is "a second as measured on the observer's clock". No, that does not convert between two different "kinds" of seconds.





    No, I am not suggesting nor did I say such a thing. There is only one "kind" of second here, one second as measured on the observers clock.

    No, that is NOT what the formula is saying.

     
  15. Dec 31, 2009 #14

    HallsofIvy

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    Yes, of course, both will count 1 million flashes. But each will see the others flashes as coming slower than their own and will believe that he started his retro rockets before the other.
     
  16. Dec 31, 2009 #15

    HallsofIvy

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    No, there is NOT "more time passing between two ticks" because it is those ticks that are measuring time. An observer in system A sees B's clock record 10 seconds only 2 seconds after his. He concludes that B's clock is running slower than his and that therefore time is running more slowly in B's system.
     
  17. Dec 31, 2009 #16
    The rate of ticking of any clock IS the rate at which time is passing. There is no "real" rate of time passing that a clock's rate of ticking can be compared with. The rate of ticking is THE time for any clock. Time does not pass at some absolute rate, some clocks recording more of it and some less depending on their relative motion. All clocks are on an equal footing.

    Matheinste.
     
  18. Dec 31, 2009 #17
    OK, so if two clocks measure the same second, and it is longer for one clock than the other, but it is still the same second?

    A measure, of any kind, is surely based upon some standard, and that standard is what measurements are relative to?

    Then if the length of a second is different for two observers, yet is still equal to one second in each case, then the standard second for each observer has to be different?

    And if the standard is different for each observer, then, surely, they must be measuring time on separate scales that are appliccable to each observer?

    Or are you saying that there is, somehow a single "universal" second, or somekind of "Absolute" second that is measured differently by observer's moving at different relative velocities?

    If a second has different durations, depending on the conditions under which it is measured the does this not imply defferent scales depending on those conditions?

    And if all seconds are the same then where do the concepts of proper time and coordinate time come from?

    Grimble
     
  19. Jan 1, 2010 #18
    I would say there is only the local time in the observers inertia frame.

    Let's say we made ten clocks all the same and tested them while they were all in the same inertia frame. We find they all run at the same rate. We then send all ten into unique inertia frames. In each frame we do various low speed physics experiments we will find that physics is working as expected when we use our local copy of the clock. Each local clock, each local time is as expected in terms of physical/chemical/biological processes.

    If we want to translate between frames we can. The amount of local clock time in frame A (call it Ta) will be different than the amount of local clock time in frame B (Tb).

    All our experience is in a slow Newtonian world we have no intuitive understanding of the relativistic mix of time and place. We have to use the math. We can always translate time and place in frame X into time and place in frame Y. It just feels wrong.
     
  20. Jan 1, 2010 #19
    Or to paraphrase "Give me a place to stand and a lever and I will move the world" we could say "Give me a place to stand and I will give you time (a clock to use)".
     
  21. Jan 2, 2010 #20
    Which is referred to as Proper time?

    And, as these clocks will all be subject to identical physical laws then all the clocks will be keeping identical time?

    I don't understand what you mean here. Is it that the local time in different inertial frames will be different?

    Are you saying that, if we are in frame C, moving at the same relative velocity to frame A and to frame B, and we translate their local clock times, we will get different results???

    So, are you saying that your identical clocks keep the same time, but that the time is different in each inertial frame?

    Let me ask that another way; if we have identical clocks that keep identical time in one inertial frame, and we then distribute them among other inertial frames where they are subject to identical conditions (identical laws of physics), can you please explain to me how they can possibly experience different local times?

    But it is only the difference between what is experienced/measured locally and what is experienced/measured when the subject and object are in relative motion.

    "no intuitive understanding" (sometimes expressed as 'counter intuitive') - Is this not the one idea that is used to try and make relativity seem mysterious, as if it is some sort of magic that follows the laws of physics - yet doesn't? That enable statements to be made that otherwise would be seen as illogical?

    e.g. 'seconds' seem to be some sort of magical unit, that is the same everywhere, yet can be different sizes in different places. It is like having two marks on a rod that define a metre and saying that that is a metre what ever the temperature of the rod.

    One MUST define the conditions under which one is measuring.

    Something is wrong somewhere, Grimble
     
  22. Jan 2, 2010 #21
    Because a different amount of time elapsed for each clock.

    If the laws of physics are the same for two clocks, and they both keep proper time, then if more time elapses for one clock than the other, they will show different times.

    In SR, the reason two clocks show different readings isn't because of anything physical about the clocks. The clocks are just measuring devices. It's the actual time elapsed being different that is the point.
     
  23. Jan 2, 2010 #22
    But that is just what is not happening is it?

    Just step back a moment and consider what Einstein was doing.

    He was trying to resolve the apparent dilemma of his first postulate and his second being mutually exclusive.

    For if the first were true, and physical laws were the same in 'Galilean' ssytems (inertial frames) then the speed of light cannot be universal and vice versa.

    And Einstein went on to show how that could be true by shewing how space and time varied when viewed by a moving observer.

    Saying that the times in separate inertial frames are different contradicts the first postulate.

    If that were true, there would be no need for special relativity! For the problem could be resolved by merely saying the time scales were different!

    The difficulty that Einstein resolved was how the speed of light could be constant, while the clocks in different inertial frames did keep the same time.

    The answer surely lies in the observing frame seeing the extra time required pass in the moving frame, whilst still taking the same (shorter) time to pass in the observer's own frame.
     
  24. Jan 2, 2010 #23
    Yes, that's exactly what is happening.
    Saying that the elapsed time between given events are different in different inertial frames certainly doesn't violate the first postulate. The first postulate says that the laws of physics are the same in each frame. There is no law of physics that says the time elapsed between given events is constant, or that the rate of a clock is constant. Time is a variable, and frame dependent.

    The speed of light being a constant, not a variable, is a law of physics in SR, so must have the same value in each frame.

    The only way that the laws of physics would require the clocks to read the same elapsed time between events is if the events were the same relative to each frame, ie a symmetrical situation.
    They keep the same time in their own respective frame, not relative to each other. And if each clock keeps good time in its own rest frame, and the actual elapsed time between events is different in different frames, the clocks must show different elapsed time between those events if the laws of physics are the same in each frame.
    I don't know what this means.

    The bottom line is that a clock is a measuring device that measures (not determines) elapsed time. If a good clock measures an elapsed time of 10 hours, it's because 10 hours of proper time elapsed. Period. If not, the clock isn't a valid clock in SR/GR.
     
  25. Jan 2, 2010 #24
    Time is not a law of physics.

    We measure time with things we call clocks which are mechanisms or systems which have a cyclic nature and whose workings are governed by the laws of physics. At the end of every cycle the mechanism returns to exactly the same state it was in at the end of the previous cycle. As the mechanisms of these clocks, whatever they may be, are governed by the laws of physics and as the first postulate says that these laws are the same in all inertial systems, the behaviour of clocks must be the same in all inertial systems. Therefore, if two identical clocks, moving inertially relative to each other, record different elapsed times it is because the elapsed times ARE different, whatever time thay are measuring, even if your reasoning or intuition says that these times should be the same. It is not because the clocks are behaving differently.

    Think of a clock as some device that generatres ticks. The interval between these ticks is the unit of time. All identical clocks, by definition, whatever there inertial motion, generate the same unit interval between ticks. The observers role is to count these ticks, or by means of some sort of readout to read the number of ticks. The counted number of ticks is a measure of the time elapsed. It is a measure of the total number of EQUAL unit time intervals. If the count of ticks of the clocks is different it is because they have measured different numbers of equal unit time intervals.

    Matheinste.
     
  26. Jan 2, 2010 #25
    In order to get to other inertial frames each clock has to experience a force and accelerate. In the process the mass of the clock and each of its parts increases as [tex] \gamma m_0 [/tex]. So how can you say that a clock that "now" (sorry) masses twice as much as another clock should tick tock in the same way?

    They were only identical clocks before we started to boost (accelerate) them.

    The problem is that in every frame except the local frame time and place are not orthogonal. So not only do observers in different frame have different times they measure different distances. For example one observer on earth on observer in spaceship traveling with a gamma of 50 to a star 50 light-years away and stopping. At the end of the trip the earth person says the ship traveled 50 light-years distance in a time of 50 years (approximately not going to do the math). The ship observer says I traveled one light-year distance in one years time. They do not agree on time nor place! Of course the traveler can say "now" that I am back in the initial inertia frame I know the distance in this frame was 50 light-years but the great thing is I was not doing my traveling in this frame! So my age is only one year older than when the trip started. So it would be fair for him or her to say in the English language effectively I was traveling at 50 times the speed of light or to say effectively time was going 50 times slower for me.

    I think it has to do with the translation between an English language representation and a mathematical representation that gets people so angry. Please note I used the English words "effectively I was" I did not use the words "I was".
     
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