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I Query on basic mechanics of Special Theory of Relativity

  1. Feb 27, 2017 #1
    The Special Theory of Relativity is based on the observations of an outside observer on a system travelling at the speed of light.

    There is no single point of reference to determine speed. So speed is determined relative to another object.
    This being the case an Apple travelling at the speed of light relative to an Orange is only saying the relative speed of these two objects is the speed of light (SOL).

    1 The Apple is stationary and the Orange is travelling at SOL.
    2 the Orange is stationary and the Apple is travelling at SOL.
    3 The Orange is travelling at ½ SOL in relation to the mid point between the two objects and the Apple is travelling in the opposite direction at ½ SOL in relation to the mid point between the two objects.

    All three of these statements have the same result when considering only the Apple and the Orange.
    Therefore observations from the Apple to the Orange are the same as from the Orange to the Apple.

    My questiona are, how can there be any bias toward time dilatation from one to the other? How can time slow down for one when the observations from the Apple to the Orange are the same as from the Orange to the Apple?
    If the theory holds true making the observations from the Apple, then it also holds true making the observations from the Orange. Each would cancel the other out.
     
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  3. Feb 27, 2017 #2

    Orodruin

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    No it isn't.

    Apples and oranges are massive objects and therefore cannot travel at the speed of light.

    In order to understand this (taking away the "travelling at the speed of light part" and replacing it for "moving very fast relative to") you need to understand the relativity of simultaneity. Events that are simultaneous for the apple are not (necessarily) simultaneous for the orange. Not taking relativity of simultaneity into account is the main source of many of the apparent paradoxes that you will find in relativity (which really are not paradoxes once you understand the theory).
     
  4. Feb 27, 2017 #3

    Dale

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    As @Orodruin mentioned above, apples and oranges have mass, and therefore their relative velocity is some v<c.

    The observations do not cancel out. In addition to the relativity of simultaneity it is important to understand what the time dilation formula means. The time dilation formula says that a moving clock's proper time is slower than the coordinate time of a set of synchronized clocks. So the orange clock's proper time is slow compared to the apple system of synchronized clock's coordinate time and the apple clock's proper time is slow compared to the orange system of synchronized clock's coordinate time.
     
  5. Feb 27, 2017 #4
    Or slower. And if the system has a nonzero mass then the speed must be less than that of light.

    Correct. The situation is symmetrical.

    One observation can't cancel out another. They are separate observations made by separate observers. Each will observe the other's clock to be running slow compared to theirs. This seems like a paradox with no solution, but the solution lies in an understanding of simultaneity. If the apple is to measure the rate of the orange's clock, then the orange will need two clocks. This is because you need two events so you can measure the amount of time that elapses between them. Orange will need one clock next to the apple for the first event, and since the orange is moving it'll need another clock that will be next to the apple for the second event. Orange will have to synchronize these clocks, but the apple will not agree that he has done so correctly because events that are simultaneous for the orange are not simultaneous for the apple. The difference explains why apple observes orange's clocks to run slow when orange knows that apple's clocks are running slow.

    By symmetry you can do the same with the apple and consider it in motion so that it will need two clocks.
     
  6. Feb 28, 2017 #5
    OK thanks all. You have shown me that my issue is with simultaneity.
     
  7. Feb 28, 2017 #6
    Proper time is time in observer’s reference frame or rest frame, it is invariant. It is better to say that time interval measured by single moving in reference frame ##K## clock will be shorter than time interval measured by two (at least) Einstein – synchronized clocks of reference frame ##K##. Sure, in every point ##x## of reference frame ##K## moving clock will show less time than any clock of reference frame

    Dilation_1.jpg pic upload
    Dilation_2.jpg host images

    Also good picture is there (chapter time dilation) http://www.pstcc.edu/departments/natural_behavioral_sciences/Web Physics/Chapter039.htm

    because it shows two clocks in the point of departure and point of arrival of light clock. This is small but IMHO very important detail. Quite often popular books omit this detail.

    Moving clock C starts at clock A and arrives to clock B. If readings of clock C and A were the same (for example 12) , readings of clock B and clock C (when the clocks are in immediate vicinity) will be different (for example C will show 3PM and B will show 6PM).
    If observer C wants to measure time dilation of clock A, he must hire an assistant and add another clock C2 at another spatial position in his reference frame and synchronize that clock. He can use Einstein convention (i.e. velocity of light to that clock will be c). Then time interval measured by clock A will be shorter than time interval measured by clocks C and C2.

    Animation is here: https://en.wikipedia.org/wiki/Time_dilation#/media/File:Time_dilation02.gif

    This set of clock is often named as a REST FRAME of observer. Velocity of light in the rest frame is the same in all directions.
    Every observer in Special Relativity conducts measurements by introducing his own rest frame, by means of hiring assistants and filling space with Einstein - synchronized clocks.

    Thus, time dilation can be measured by a TEAM of observers.

    We can imagine a row of synchronized clocks of reference system K. each denoted by letter - A, B, C, D ….. Z. Then a person with single clock on his wrist (A’ for example) moves in this reference system K and compares readings of his clock with these clock A-Z successively. When he comes to the clock Z, his clock A’ shows gamma times less time, than clock Z. Thus, clock Z shows gamma times more time, than his own.
    At this point Z this clock A’ immediately turns back and starts travelling in reverse direction, passing by clock Z, Y, X ….. C, B and finally arrives into point A of reference frame K. Clock A’ compares readings with clocks Z-A successively again and sees, that it dilates itself gamma times, i.e. every clock on the way shows gamma times more time. When clock A’ arrives into point A, clock A’ shows gamma times less time than clock A, and clock A shows gamma times more time than clock A’.

    This discrepancy of clock readings when they meet again is often called clock paradox or Twin paradox.



    Vice versa. If apple will do as you proposed, apple will see that orange clocks run faster, as I mentioned above. If apple wants to measure dilation of orange’s clock, apple should do that from its own rest frame, i.e. by means of putting two synchronized clocks at points of departure and arrival. It is the source of confusion very, very often. We say that object moves if he changes spatial position in our frame.

    All that goes straight from the Lorentz transformations.

    ## T = \frac {t'_{x'}+ \frac {v'} {c^2} x'} {\sqrt {1-( \frac {v} c)^2}} ## (1)

    ##T## is clock readings that belongs to reference frame ##K##, taken in point ##x'## at moment of time ##t'_{x'}## of reference frame ##K'##, and ##t'_{x'}## reading of clocks that belongs to reference frame ##K'## in the point ##x'## of reference frame ##K'##

    How to interpret Lorentz transform for time?

    Transformation demonstrates, that time ##T## of reference frame ##K## (in which it does not depend of ##x## coordinate or any other coordinate) is universal in reference frame ##K## and each point of this frame.

    Now let's fix point ##x'##, for example ##x'=0##. In this case this transformation will look like that:

    ## T= \frac {t'_{0'}} {\sqrt {1-( \frac v c)^2}}## (2)

    ##T## is clock reading of reference frame ##K## taken in point ##x'=0## (in the origin ##O'## of reference frame ##K'##), and ##t'_{o'}## is time in the reference frame ##K'##, in particular in the origin ##O'##.

    We can take ## \frac {dT} {dt'}## when ##x'## is fixed and will get ##{dT}/{dt'} = \frac 1 {\sqrt {1- \frac {v^2} {c^2}}} ##

    According to (2) it is not time ##t'_{o'}## which is showed by single clock in the point ##O'## runs slower, but time ##T## , which is "distributed" through all reference frame ##K## and taken in the origin ##O'## of reference frame ##K'## runs faster (relatively to time ##t'_{o'}## that is in the origin ##O'## of frame ##K'##). Time dilation comes by means of transfromation of (2) into:

    ##t'_{o'}=T \sqrt {1- {\frac {V^2} {c^2}}} ##

    It is correct that ##T>t'## and ##t'<T##. It is also true that ##T'>t## and ## t<T'##. But that ##t<t'## and ##'t<t## from different points of view is nonsense.

    But, you can measure ticking rate of clock which is very, very far away from you if that clock emits radiation (light in visible spectrum, for example). You have just to look at it (or direct your measuring device, or tube) when this source of radiation and you are at points of closest approach.

    You have to do nothing special, just to direct you tube (or gaze) transversely to direction of motion of the source and wait. For example, you stay at point Y of y axis and look into origin.

    Soon or later source passes the origin and you see a brief flash in your tube. This flash will be of different color than proper color of the source, it will be reddish due to dilation of source’s clock, i.e. moving clock oscillates slower. This is so called transverse Doppler Effect.

    Of course, due to light – time correction apparent position of the source is different from its actual position and when you see the light the source has already moved into another place.

    This measurement can be (and had been) actually done and doesn’t need consideration of spatially separated and synchronized clocks.

    Interesting to note that time dilation had been measured with high precision by means of Mossbauer Effect based rotor experiments (Champeney and Moon, Hay et all, Kundig, Kholmetskii and Yarman, Fridman et all). Clock hypothesis says that clock dilates purely due to relative motion and not acceleration.

    In these experiments they place source of radiation in the center, and absorber on the rim of rotating disc (or vice versa) and measure transverse Doppler Effect. Results show, that if absorber rotates it measures blueshift of radiation, if absorber is in the center it measures redshift of radiation. That means, that rotating clock (absorber or source) ACTUALLY dilates.

    Distance between source and absorber doesn’t change. Relative to what absorber moves and dilates?

    Good to note, that if source and absorber are placed on opposite sides of the disc, they will not measure any frequency shift (any time dilation), since both dilate at the same magnitude. That was confirmed by Champeney and Moon in 1961. Though it is 100% clear. Source and absorber have equal linear velocities, and energy of photon cannot vanish to nowhere.

    The same effect (absence of Transverse Doppler effect) will be in case of rectilinear motion, if source and observer think that they move at equal relative velocities. In this case source has launch a flash when the both source and observer will be at equal distances from the normal between their patches of motion. In this case source launches a photon at oblique angle to direction of its motion backward. An observer tilts detector forward at the same angle. They have to do that because of aberration of light. These angles are connected by aberration of light formula,

    Moving observer will see, that source appears in the front and radiation frequency becomes higher (more blue). The faster he moves, the more oblique angle at which source appears to him will be. Thus, observable clock runs faster and faster because observer's own clock dilates. This effect is often called relativistic beaming.
     
    Last edited: Feb 28, 2017
  8. Feb 28, 2017 #7
    May I ask where you read or heard this? Because it seems to me to be quite the opposite. (The notion of relativity was first fleshed out by Galileo in which he argued uniform speed is undetectable in the bottom deck of a boat, so if special relativity is the extension of that to include electromagnetism, the presumably it would include speeds less than that of light.)
     
  9. Feb 28, 2017 #8
    Yes. And more. It's the time that elapses between two events that occur at the same place in that frame.

    No. Let us suppose that apple's clock reads noon when a worm enters the apple. When apple's clock reads 3:00 pm, the worm exits the apple. These two events (worm enters, worm exits) are separated by a proper time of 3 hours. The two events occur at the same place in apple's rest frame.

    Now, suppose orange comes zipping past apple when apple's clock reads noon. Orange sets his clock to noon. Orange and apple share the same location at noon, when the worm enters the apple.

    Let's suppose orange's speed is ##\frac{\sqrt{3}}{2}c \approx 0.866 \ c##, relative to apple. When the worm exits the apple, orange's clock is now located far from apple, but suppose orange has two clocks separated by the appropriate distance such that the second clock is next to apple when the worm exits. Having properly synchronized his clocks, orange will note that both his clocks read 6:00 pm when the worm exits. He will thus reckon that the worm spent 6 hours inside the apple and will thus conclude that the only way apple's clock could show an elapsed time of 3 hours is if apple's clock is running slow.

    But apple, having watched orange synchronize his clocks, will note that orange made a mistake. Had he properly synchronized them his second clock would have read 1:30 pm when it passed by apple just as the worm was exiting. Thus apple will note that orange's clocks are running slow compared to his own, showing that only 1.5 hours elapsed instead of 3.
     
  10. Feb 28, 2017 #9

    Dale

    Staff: Mentor

    I am sure that you know what you mean, but I really don't like this statement. In the spacetime interval ##dt## and ## d\tau## are different quantities. They are called coordinate time, and the proper time respectively. Only the proper time is invariant, the coordinate time is not. And "time in observer's reference frame" is a little ambiguous, but seems to refer to coordinate time.
     
  11. Mar 9, 2017 #10
    Ok so I have done some more research into simultaneity, and explanations of this so often seem to come back to the fast moving train example.

    Thanks for the links Bartolomeo, they are very useful and I have extracted the following example from the first one.

    train.JPG
    Can I explain my issues with this description of simultaneity and invite comments/corrections?

    I am presuming the two events at A & B occur simultaneously within frame S

    There is a time delay from the events occurring to when observer O registers them (the time it takes for light to travel from A and B to O).

    I assume observer O’ is making his observation at the exact same instant that O is. This is at the point in time when light from the events reaches O.

    Also I am assuming that O’ is exactly at the same point as O at the time of the observations.

    My conclusions from the above are that light from the events reaches O at a single moment in time. Therefore, reaching O’ at this single moment in time. The thing is, a single moment in time is an instant and does not have a change in time. So there cannot be any distance travelled within the instant and so velocity is irrelevant. O and O’ are at a single position at this instant. Just as the light from the events are at this single position at this instant. Both observers will register the same events at the same time.

    If my above assumptions are wrong and the train explanation is about O & O’ being at the same point when the events occur and their observations being made after this when light reaches them individually, then why not just discuss the difference between O and an observer standing on the platform between O and B within the same frame S? They would both register the events occurring differently.

    What I’m asking is, in what circumstance does the difference in observations occur?
     
  12. Mar 9, 2017 #11

    Ibix

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    That isn't quite what's being described in the text you provided, but OK.

    In the version of the experiment you are describing, there's no difference in observation [1]. Both observers receive the flashes together. The difference is in how they interpret that.

    Under what circumstances would you receive two flashes at the same time? Either the explosions happened at the same time at the same distance from you, or they happened at different times at different distances. You said the explosions were simultaneous in S. So in that frame, they must have happened at the equal distances from O. But they must have happened before O and O' were co-located, so they didn't happen at equal distances from O'. If they weren't at equal distances from O' then they must have happened at different times in S'.

    [1] That's not quite true - the Doppler effect says that the two frames will see different light frequencies.
     
  13. Mar 9, 2017 #12
    That's not a presumption. It's a conclusion. There's a single event that occurs at O, namely the arrival of the two flashes. But the point concerns two other events, the explosion at A and the explosion at B. They are simultaneous in frame S not because of some presumption. It's a conclusion based on the fact that O is equidistant from both A and B and the fact that the signals that travel to O do so at the same speed.

    Yes, and since the signals travel the same distance at the same speed, those two time delays are equal. Hence the conclusion that the explosions are simultaneous.

    No, he makes two observations, one before and one after. The one before is the arrival of the signal from B and the one after is the arrival of the signal from A.

    But the thing is, those two signals also traveled equal distances at equal speeds to reach O'. Thus, the explosion at B must have occurred before the explosion at A. This last point doesn't seem to be made explicitly in the excerpt you posted.
     
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