Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Time scales in Inertial Frames of Reference

  1. Mar 21, 2010 #1

    Grimble

    User Avatar

    Hello, I am trying to understand the relationships of the time scales that obtain within different Inertial Frame of Reference. Not when viewing one frame from another, I am quite happy with the Lorentz factor for that. No it is how the local time scale, that measured by a stationary, local obsever, situated at the origin of that frame's coordinates compares with the local time scale of another IFoR

    The coordinates of any Inertial Reference Frame, employing stationary clocks, will be measured in proper units and have the same magnitude in any such frame.

    A stationary clock situated at the origin of the frame's coordinates will, by definition, measure proper time.

    The distance from that clock to any other stationary clock showing the same time, in that same frame of reference, will, by definition, measure proper distance.

    It has to be the same in any IFoR for the following reasons:

    i.If two IFoRs are at rest with one another they are both effectively in the same IFoR and share the same proper time.

    ii.If they are moving relative to one another, i.e. have a relative velocity, each can still be considered to be at rest and must, therefore, still measure the same proper time.

    iii.Every IFoR obeys the same simple physical laws, therefore identical, synchronised, clocks situated in such frames must keep identical time.

    iv.If the proper time in IFoRs COULD be anything other that identical, then the differences could negate the need for Special Relativity! As any conflicts between Einstein's first and second postulates could, possibly, be explained by the differences in the measurements of time.

    v.The proper times of two IFoRs can be calculated from a third IFoR by means of the Lorentz Transformation equations; and, if that third IFoR was permanently positioned at the midpoint between the two IFoRs in question, those calculated proper times would have to be identical. Otherwise we would be contravening the Special Principal of Relativity, Einstein's 1st Postulate, - The laws of physics are the same in all inertial frames of reference, in other words, there are no privileged inertial frames of reference.

    vi.If two IFoRs are moving with a constant relative velocity with respect to one another, then the movement of one, being a combination of time and distance relative to the second, must be the reciprocal of the movement of that second one with respect to the first. Therefore they must be using the same proper units.

    I don't know whether this will be considered to be ATM or not, but no doubt some will claim so, but to me it is what Einstein described.

    Grimble
     
    Grimble, Mar 21, 2010
  2. jcsd
    Phys.org - latest science and technology news stories on Phys.org
  3. Mar 21, 2010 #2

    JesseM

    User Avatar
    Science Advisor

    You're using "proper time" in a way that seems a bit confusing. Proper time is a property of particular worldlines, like the worldline of a clock; If you pick two events on that worldline, the proper time between them is just the time that would be measured between the events by a clock moving along that worldline. I don't think it's meaningful to talk about proper time unless you have specified the worldline you want to calculate the proper time along, and the two events on that worldline that you want to calculate the proper time between. Thus sentences like "The proper times of two IFoRs can be calculated from a third IFoR" are ambiguous--are you just saying that the proper times between events on the worldlines of clocks which are at rest in two different frames can be calculated from the perspective of a third frame?
     
    JesseM, Mar 21, 2010
  4. Mar 22, 2010 #3

    Grimble

    User Avatar

    Hello Jesse, No, well, maybe, let me explain what I am thinking:

    If an observer with a clock is stationary at the origin of the coordinates of an IFoR, then being stationary beside a clock he is measuring proper time. So that stationary clock at the origin of an IFoR is the worldline that is being measured.

    And so for any IFoR, I am referring to the worldlines of stationary clocks at their origins.

    Such a clock couuld also be considered to be setting the time scale for that IFoRs coordinates.

    So we take two IFoRs with a velocity v between them. (A & B)

    Any measurement on such a clock, in either of our two Frames, could be transformed, by LT, to a coordinate measurement in another IFoR, (C), that was permanently midway between A & B.

    Then, because the relative velocity between A & C, and between B & C would both be v/2, the units of measurement in A and the unit measurements in B, transformed identically by LT, in C (using v/2) would give the same result.

    Therefore the unit size of proper time scales would be the same in any IFoR. (And similarly those of proper length too)
     
    Grimble, Mar 22, 2010
  5. Mar 22, 2010 #4

    JesseM

    User Avatar
    Science Advisor

    What does "measuring proper time" mean? Do you think proper time has something to do with frames of reference? It doesn't, it's just a characteristic of worldlines like I said. Every correctly-functioning clock is measuring proper time along its own worldline (or anything traveling alongside it), regardless of whether the clock is stationary in an inertial frame or accelerating.
    Sure, but it would be incorrect vocabulary to call the time scale in an inertial frame "proper time".
    Coordinate measurement in C of what, exactly? A single clock in frame A can really only measure the time between two events on its worldline (unless you know the time light from an event struck the clock and you know the distance D between the event and the clock in this frame, then you can figure out the time of the event itself by subtracting D/c from the clock's reading when the light struck it). For events that occurred on the worldline of a clock at rest in A, you can just use the time dilation equation rather than the full Lorentz transform to figure out the time between the events in C. On the other hand, for an arbitrary events, you have to know both the time between them in A and the distance between them in A in order to use the Lorentz transformation equation to find the time between them in C.

    Also, is there any significance to the fact that C's velocity is between A and B? What I say above would be true regardless of C's velocity relative to A.
    Don't understand what it would mean to transform "units of measurement"--the LT transforms time and distance coordinates of particular events (or time and distance intervals between particular pairs of events).
    Again, coordinate time in a given frame should not be referred to as the frame's "proper time", nor should coordinate distance be referred to as "proper length" (which refers to an object's length in its own rest frame, which may be different than its length in whatever frame you're using to describe it). And even if you're talking about coordinate time scales, I don't understand what it means to say the "unit size" is "the same". Do you just mean that if the same experiment is performed in both frames, like both frames looking at caesium 133 atoms at rest in their own frame and timing how long it takes for them to oscillate 9,192,631,770 times, they will each get the same answer for the coordinate time (in this case exactly one second)? If you're not talking about coordinate time in different frames having the same physical definition involving a physical apparatus at rest in that frame, then in what sense are you using the phrase "the same"?
     
    JesseM, Mar 22, 2010
  6. Mar 25, 2010 #5

    Grimble

    User Avatar



    I'm sorry, for any misunderstanding, but you are misinterpreting what I am saying. Take the passage above, that I have emboldened. Under your definition there, which fits the scenario I specified, my observer would see the clock adjacent to him, clock measuring proper time. I was not trying to define proper time, I was trying to select a situation in which it would be true to say that the clock specified would be measuring proper time, in the context of an IFoR


    So if a clock is situated such that it could be measuring the time scale of an IFoR and it also happens to be measuring proper time, we can't say so? because that would be using vocabulary incorrectly?

    But could we say that the time scale for that frame's time coordinates was, in-fact, measuring time using the same time scale as proper time, as measured by that clock?


    OK, let us say two ticks, one second apart on the clock at rest in frame A and that we can use time dilation to figure out the time between those two 'events' in frame C.

    Yes the significance I am seeing is that if C is permanently midway between A & B then the relative velocity between A & C will be v/2 and that between B & C will also be v/2. And, therefore, that the relationship between measurements in A and measurements in C (due to time dilation) would be the same as the relationship between measurements in B and measurements in C (due to exactly the same time dilation),

    And these are somehow measured on a scale that has no unit of measurement?

    When something is measured its dimensions are compared with a standard 'scale' to determine how far its dimensions extend along that scale.

    If a clock is measuring proper time, then on what scale is that proper time being measured? If it is not measured against (compared to) a well defined scale, then what is it measuring and how can it be compared to or calculated from any other measurement? It cannot just define itself or where would we be?

    If you are referring to coordinate time as the time measured against that frame's coordinates then I can understand what you are saying.
    Although if that frame's time coordinate is taken from a stationary clock at that frame's origin (not, I think an unreasonable idea) then as "Every correctly-functioning clock is measuring proper time along its own worldline (or anything travelling alongside it)" that time scale is also that clock's proper time and then that 'proper time' must be measured on the same scale as that frame's coordinate time scale.

    Another point to consider: two clocks, each stationary in its own IFoR.
    Those two IFoRs at rest with one another.
    Those two clocks will keep identical proper time.
    They will effectively be synchronised.
    If their relative velocity changed and they became, once more, two IFoRs, would the proper time 'ticks' of those clocks still be 'synchronised?
    Is their any reason why they shouldn't be, for either clock could be considered to have remained stationary. The movement of each was only relative to the other.
    (I don't pretend to know the answer to this...)



    OK, consider this:
    Two IFoRs, relative velocity v; a stationary clock in each 'ticking' once per second, proper time; a third IFoR moving with equal velocity v/2 with respect to each of the first two.
    The Lorentz factors between the third frame and each of the first two will be identical.
    The one second 'ticks' from each of the clocks, time dilated, will give values in the third frame that are equal in duration.

    These measurements are linked by mathematical formulae, they have a mathematical relationship, the same is true of all bodies, clocks, frames of reference. Those measurements can be converted from one to the other. They can be compared.
    There must therefore be some sort of hypothetical absolute scale?

    Can you follow what I am trying to say?

    Grimble
     
    Grimble, Mar 25, 2010
  7. Mar 25, 2010 #6

    JesseM

    User Avatar
    Science Advisor

    OK, if you specify that you're talking about the proper time between two specific events on a specific clock's worldline, that's fine. But do you agree that every clock is "measuring proper time" along its own worldline, regardless of whether it is inertial or accelerating? If so, perhaps you can see why I thought there was some misunderstanding when you said If an observer with a clock is stationary at the origin of the coordinates of an IFoR, then being stationary beside a clock he is measuring proper time, since the clock is "measuring proper time" regardless of whether the first part of the sentence about being stationary in an inertial frame happens to be true, it sounded like you were saying it was a required condition for the clock to be measuring proper time.
    You can say that the same clock is measuring both of those, but you still should distinguish them from one another because they aren't the same in general.
    Sure, in that case the time between these events will be greater by a factor of gamma in C's frame.
    Yes, the relationship would be the same. I don't see where you're going with this though.
    Sure it'll have some units of measurements, like light-seconds and seconds perhaps. But you aren't transforming units of measurements between frames, you're transforming coordinates. For example, you could have two events E1 and E2 that happen 10 light-seconds apart in frame A, and two other events E3 and E4 that also happen 10 light-seconds apart in frame A. If we thought we were "transforming units" that would suggest a fixed ratio between light-seconds in frame A and light-seconds in frame B, meaning if E1 and E2 happen 5 light-seconds apart in frame B, then E3 and E4 should also happen 5 light-seconds apart in frame B. But of course it needn't work that way, because the time between E1 and E2 might be different than the time between E3 and E4 in frame A, which allows the distance between E1 and E2 to be different in frame B than the distance between E3 and E4.
    Space and time coordinates (and intervals) in an observer's inertial rest frame are ideally defined relative to a grid of rulers and synchronized clocks (synchronized using the Einstein synchronization convention) at rest relative to the observer--that's how Einstein defined coordinate measurements in his original 1905 paper.
    There are a variety of ways to construct a standard clock using some physical process that seems to occur at a regular rate (regularity defined in terms of consistency with other physical processes that all seem to have fairly constant ratios of tick rates relative to one another when they happen at rest relative to one another). For example, you can define "1 second" as 9,192,631,770 photon emissions of a caesium 133 atom (see here). Then for any physical object, 1 second of proper time could be defined as 9,192,631,770 emissions on an atomic clock using caesium 133 atoms which travels along the same path through spacetime as that object.
    Yup, just the time coordinate that frame assigns to events.
    It's taken from synchronized clocks at rest in that frame stationed throughout the grid of rulers that define position coordinates, that way every event can be assigned position and time coordinates in a local manner. For example, if you see a firecracker go off in your telescope, and you see the explosion happened right next to the 12 light-second mark on the ruler defining your x-axis, and you also see that the clock attached to the 12 light-second mark read a time of 3 seconds as the firecracker went off right next to it, then you can assign the event of the explosion coordinates x=12 light-seconds, t=3 seconds in your rest frame.
    But proper time only applies to events that happen along the worldline of the clock whose proper time you're talking about. If one event happens at x=11 light-seconds when the clock there reads t=1 second, and another event happens at x=12 light-seconds when the clock there reads t=3 seconds, then there's a coordinate time of 2 seconds between these events, but those 2 seconds are not "proper time" along the worldline of any of the clocks at rest in your frame which define your coordinate time, since none of these clocks have both events on your worldline (and if both events happened on the worldline of an object moving at 0.5c, then the proper time between the events for that object would not be 2 seconds but rather 2*sqrt(1 - 0.5^2) = 1.732 seconds).
    I think "keep identical proper time" isn't really meaningful, proper time is always defined solely between events on that clock's worldline, so you can't talk about how much proper time passed on clock B between two events on clock A's worldline, and ask whether it's "identical" to the amount of proper time that passed on clock A between those events. Perhaps you mean something like this: if we pick an event A1 on clock A's worldline and an event B1 on clock B's worldline that's simultaneous with A1 according to both frame's definition of simultaneity, and then pick an event A2 on clock A's worldline and an event B2 on clock B's worldline that's simultaneous with A2 in both frames, then the proper time for clock A between A1 and A2 is the same as the proper time for clock B between B1 and B2. This is true, but it makes use of frame-dependent notions of simultaneity, whereas proper time is always defined in frame-independent terms.
    "Synchronized" has no frame-independent meaning (it just means that they show the same reading at the same time-coordinate in a given frame), if they are synchronized in coordinate systems where they are at rest they will be out-of-sync in other coordinate systems, and vice versa.
    Since "synchronized" is frame-dependent you have to specify what frame you're talking about. The could still remain synchronized in a frame where they both had equal speeds (like your frame C which was midway between A and B), but in other frames they wouldn't be.
    Don't know what you mean by "absolute time scale". Of course we can compare the time between arbitrary events in different frames, but if you have two pairs of events E1&E2 and E3&E4, different frames can disagree about whether more time passed between E1&E2 or more time passed between E3&E4, and there can be no frame-independent physical truth about the matter. That would conflict with what most people mean by "absolute time", if you mean something different you'll need to define it.
     
    JesseM, Mar 25, 2010
  8. Mar 27, 2010 #7

    Grimble

    User Avatar

    Yes, I am sorry, it was badly worded on my part.

    OK, point taken.

    From this point on I understand what you are saying but there are some points that bother me:
    (I am only concerned with SR here, so let us assume that all bodies are inertial and have their own IFoR and therefore no gravitational effects)

    1. Time Dilation and Length contraction.
    1.1.Any body in space has its own scales of measurement, independent of any other body in space.
    1.2.It doesn't matter how many other bodies there are moving at different relative velocities each bodies coordinates will be unaffected by those other bodies.
    1.3.It is only when one body observes another's coordinates that TD/LC occurs.
    1.4.An observer in one body does not measure another, moving, body directly it takes the measurements made by the observed body and transforms them (LT) to determine what those measurements would have been if it could have measured them directly.
    1.5.If a body does this to derive measurements for more than one body, moving at different relative speeds it will have used different Lorentz factors for those transformations. So what coordinate scale is it using? 1 light-second for example would have different magnitudes depending upon whether it were measured within the bodies own frame of reference or within one of the observed IFoRs.

    2.Proper time.
    2.1.For stationary bodies, within IFoRs, their worldlines, being geodesics in flat spacetime, would be parallel.
    2.2.The scales for their individual worldlines (e.g. an atomic clock using caesium 133 atoms which travels along the same path through spacetime as that object.) would be identical, as the laws of physics operate the same in any IFoR (1st postulate)
    2.3.I can see no way that, with identical physical laws, their dimensions could differ.
    2.4.There seems to be a reluctance to apply any sort of definition to how proper times are related. They have to be related or we would be unable to transform between proper time and coordinate time.

    3.Coordinate time
    3.1.If the scales for the coordinates of a reference frame are set by standard clocks and standard rulers, in the native, local, time frames of an IFoR, existing as an IFoR, their can be no suggestion of movement as movement has to be relative to some other body and between IFoRs it has to be reciprocal.
    3.2.So all IFoRs must have the same time scales as they all obey the same laws.
    3.3.It just seems as if their has to be a lot more structure than modern relativists seem to allow.

    Help! Grimble
     
    Grimble, Mar 27, 2010
  9. Mar 27, 2010 #8

    JM

    User Avatar

    Grimble, I think you are addressing an important point, namely the units of measure used in two different IFOR's. A fundamental principle of physics says that the units of measure must be the same on both sides of an equation. Thus when writing T = t/m, where T and t are times and m is Einsteins 'gamma', the units of T and t must be the same. Both traceable to a common standard, such as noted by JesseM. The time interval between one-second ticks of the T clock must be the same as the interval for t. And so the common statement that 'moving clocks run slow' must be taken as a 'short cut' for a more exact statement.
     
    JM, Mar 27, 2010
  10. Mar 27, 2010 #9

    DrGreg

    User Avatar
    Science Advisor
    Gold Member

    An analogy might help. Is "1 km North" the same as "1 km Northwest"? How would you convert "1 km N" to a NW length? How would you convert "1 km NW" to a N length?

    From the point of view of someone facing North, you might say 1 km NW is equivalent to 707 m N. From the point of view of someone facing Northwest, you might say 1 km N is equivalent to 707 m NW. Neither view is actually wrong. And there is no "absolute scale" that we can convert both distances to, which everyone would agree to.

    What is missing from the above is a second dimension. The truth is that 1 km NW is equivalent to 707 m N and 707 m W. 1 km N is equivalent to 707 m NW and 707 m NE. When expressed in this way, this is analogous to using the full Lorentz transform, which deals with both time and space, instead of thinking of time dilation and length contraction separately.

    Also, in the phrase "1 km NW is equivalent to 707 m N and 707 m W", we could say that "1 km NW" is a "proper" measurement, because that is measured directly by travelling the distance, but "707 m N" and "707 m W" are both "coordinate" measurements within a north-v-west coordinate system.
     
    DrGreg, Mar 27, 2010
  11. Mar 28, 2010 #10

    Grimble

    User Avatar

    Hello DrGreg, that is a very interesting analogy, which I have tried to picture.

    http://img706.imageshack.us/img706/9236/rotatedframe1.th.jpg [Broken]

    This is an example of one of the diagrams that I had drawn whilst pondering Lorentz transformations, Minkowski rotations, Time Dilation, Length Contraction and the Twin Paradox. As it goes a long way in my mind to linking them all.

    What do you think? Is it on the right lines?
    Grimble.
     
    Last edited by a moderator: May 4, 2017
    Grimble, Mar 28, 2010
  12. Mar 28, 2010 #11

    DrGreg

    User Avatar
    Science Advisor
    Gold Member

    If we are talking about just one dimension of space (x), we need only a two dimensional (t,x) spacetime diagram; there's no need to bring in a third dimension which seems to be what your diagram is attempting?

    I attach a standard spacetime diagram for v = 3c/5, [itex]\gamma = 5/4[/itex], [itex]\gamma v = 3c/4[/itex]. The left hand diagram has the blue axes horizontal and vertical. The right hand diagram is exactly the same except that the red axes have been realigned to be horizontal and vertical. These are diagrams in two dimensions only; you can measure the position of an event either in red coordinates or blue coordinates.

    I can't work out how you managed to transform (10, 6) into (8, 4.8). (10,6) should transform into (8,0). (6,10) should transform into (0,8). The Lorentz transform in this case is

    [tex]ct' = \frac{5ct - 3x}{4}[/tex]
    [tex]x' = \frac{5x - 3ct}{4}[/tex]​

    and the inverse is

    [tex]ct = \frac{5ct' + 3x'}{4}[/tex]
    [tex]x = \frac{5x' + 3ct'}{4}[/tex]​
     

    Attached Files:

    DrGreg, Mar 28, 2010
  13. Apr 8, 2010 #12

    Grimble

    User Avatar

    I must apologise for my delay in replying to you Dr. Greg, but I needed to be confident that I am answering your points correctly and the formatting can be a little tricky until one grows accustomed to it.

    Not at all, I am considering two, two dimensional frames, where one is rotated out of the plane of the other.

    Yeeees, but you are using different axes, so it is to be expected that you have different coordinates. I would have thought that a few moments taken to understand my diagram, to comprehend the axes I have used, and the transformations will be obvious.

    No arguments there, and if one uses them correctly, the transformations work out. Let me shew you:

    we have the following coordinate values, first yours, then mine:
    ct' = 8, ct = 10, x' = 0, x = 6
    ct' = 10, ct = 8, x' = 6, x = 4.8

    then using the above transform we have, using your coordinates:

    [tex]8 = \frac{50 - 18}{4}[/tex]

    [tex]0 = \frac{30 - 30}{4}[/tex]​

    which obviously works; and secondly using my coordinates:

    [tex]10 = \frac{40 - 0}{4}[/tex]

    [tex]6 = \frac{24 - 0}{4}[/tex]​

    The difference, that my transformations are only 'time dilation' and 'length contraction'

    [tex]ct' = \frac{5ct}{4}[/tex]

    [tex]x' = \frac{5x}{4}[/tex]​

    without the need to add the second terms of the transformations that allow for the movement of the origins as the effects of the relative velocity. For in my diagram that is shewn by the paths in the two frames.

    The problem here is that with your flat diagram of rotation the two axes of your moving frame are rotated in contra directions, whereas my rotation, of the ct',x', frame out of the plane of the ct,x frame, drawn in three dimensions is much more representative of what is actually occurring.

    Compare with this diagram of the relationship of a single dimension (ct or x) from two IFoRs with a relative velocity of 0.6c
    http://img41.imageshack.us/img41/287/specialrelativitydiagrar.jpg [Broken]

    Imagine a Minkowski diagram of a single dimension (x or ct) of an IFoR.
    Then add the same dimension from another IFoR, moving at constant velocity v with respect to the first.
    We label these IFoRs A and B, colour coded red and green.
    The separation of the two horizontal lines represents the relative velocity v.

    Now the coloured diagonals represent the rotation of the moving frame relative to the one at rest, so, as either can be taken as moving, we see this from both views, colour coded.

    (Now, a word concerning my representation of proper units as drawn in this diagram; the coordinate units, being a transformation according to the Lorentz factor, from each side, using the same relative velocity will be identical and for ease of representation I have drawn them to a common scale. But if this offends ones sense sensibilities, each could be drawn to a separate scale without affecting the overall sense of the diagram.)

    The important concept here is that the rotated axis of the observed (moving) IFoR retains its same scale for the proper units, but their perpendicular projection onto the rest axis of the 'Observing' IFoR results in contracted (coordinate) units.

    While the elongation of the rotated axis, to meet the vertical projection of its end point, gives the dilation (increase in the number of units).

    And if one uses this diagram to calculate; let us say the length contraction; taking the axes to be the x axes of the two IFoRs we find that the figures from the diagram agree with the x' = x/γ formula (viz: compare the red and green figures on the diagram)
     
    Last edited by a moderator: May 4, 2017
    Grimble, Apr 8, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook