Notices
Results 1 to 58 of 58
Like Tree6Likes
  • 1 Post By Farsight
  • 2 Post By btr
  • 1 Post By x0x
  • 1 Post By btr
  • 1 Post By btr

Thread: Non-inertial clock orbiting an inertial clock: "Do moving clocks always run slowly?"

  1. #1 Non-inertial clock orbiting an inertial clock: "Do moving clocks always run slowly?" 
    Member
    Join Date
    Jun 2014
    Posts
    32
    This is my first post. I am a total amateur when it comes to relativity. So hopefully someone can help.

    I've been scouring the internet to confirm a position stated by Don Koks at
    Do moving clocks always run slowly?

    It is not a position I have seen before. Namely-

    "Do moving clocks always run slowly?" According to the above source the answer is NO, when:

    1. An accelerating clock is orbiting a relatively stationary clock, and
    2. The distance between the two clocks is so minimal that any delay in signal exchanges is negligible.

    From the stationary clock's reference frame the accelerating clock appears to run SLOW
    From the accelerating clock's reference frame the stationary clock appears to run FAST

    IF this is correct then communications between the two reference frames will be conducted virtually simultaneously. One observer is time dilated while the other observer is time contracted......WOW!

    I think this scenario would be totally different if the orbiting clock was in uniform motion. I believe both frames would see each other's clock as running slow.

    Grateful for any confirmation. Are there any sources to support Don Kok's assessment?
    Reply With Quote  
     

  2. #2  
    Banned
    Join Date
    Jan 2013
    Posts
    652
    I've spoken to Don Koks, he's a pretty reasonable guy, and a better "relativist" than some. I read his article, and thought it was OK. However he spent a lot of time talking about the twins paradox, and didn't quite get to the bottom of things IMHO.

    Quote Originally Posted by Hal on Earth
    "Do moving clocks always run slowly?" According to the above source the answer is NO, when:

    1. An accelerating clock is orbiting a relatively stationary clock, and
    2. The distance between the two clocks is so minimal that any delay in signal exchanges is negligible.

    From the stationary clock's reference frame the accelerating clock appears to run SLOW
    From the accelerating clock's reference frame the stationary clock appears to run FAST

    IF this is correct then communications between the two reference frames will be conducted virtually simultaneously. One observer is time dilated while the other observer is time contracted......WOW!
    This is only like two observers situated at different elevations in a gravitational field. NIST have optical clocks which are so accurate that they can see a difference in clock rates when one is only 30cm above the other.

    Quote Originally Posted by Hal on Earth
    I think this scenario would be totally different if the orbiting clock was in uniform motion. I believe both frames would see each other's clock as running slow.
    That's right. SR time dilation is usually symmetrical, wherein each twin sees the other's clock running slower than his own. People say this is a paradox, but it's only like you and me being separated by distance. I see you looking smaller than me, and you see me looking smaller than you. But we don't shout paradox!

    Quote Originally Posted by Hal on Earth
    Grateful for any confirmation. Are there any sources to support Don Kok's assessment?
    Check out the Simple inference of time dilation due to relative velocity. Then replace Don's clocks with parallel-mirror light-clocks and draw the light paths. That really gets to the bottom of it.
    Hal on Earth likes this.
    Reply With Quote  
     

  3. #3  
    btr
    btr is offline
    Senior Member btr's Avatar
    Join Date
    May 2014
    Location
    Circumnavigating the photon sphere.
    Posts
    168
    Hello, Hal on Earth.

    In Special Relativity, the proper time an observer experiences is given by a kind of four-dimensional analogue of "length". Essentially, you use a modified version of Pythagoras's theorem in four dimensions to compute a kind of "length" for each tiny segment of their world-line (that is, the curve through spacetime representing their history), and then add up all of those little segments to get the total.

    If you are OK with calculus and line integrals, the actual formula Special Relativity gives for the proper time along a world-line looks like this:


    I've used units where the speed of light , to keep things from looking unnecessarily messy. If you compare that with the expression for the ordinary length of a curve in Euclidean 3-dimensional space, you can see a clear analogy:


    (I.e we just add up the little length segments along the curve, each segment contributing by Pythagoras's theorem.)

    Given two points (events) in spacetime, we can construct many curves joining them, and given the analogy above it should come as no surprise that the proper time along a curve depends on the actual shape of that curve. In Euclidean space, you find that the shortest curve (i.e. the curve with minimum above) between two points is a straight line. In spacetime (Minkowski space), you find that the longest curve (i.e. the curve with maximum above) between two points is a straight line. Why the difference? It comes down to the negative signs in the integral for above (I can explain further if you like).

    You can now see how the twin paradox described by Koks really works. Let the departure event in spacetime be D and the return-to-home event be R. The stay-at-home observer is unaccelerated, and so has a world-line which is a straight line between D and R. The travelling twin, on the other hand, has a curved world-line between D and R. An example of the sort of scenario described is depicted in the following diagram:


    By the above considerations, the proper time for the curved world-line must be less than the proper time for the straight one, and so the travelling twin is seen to have aged less than the stay-at-home one when they are reunited at R.

    Personally, I think that thinking about proper time along world-lines in the geometric manner above is less prone to cause confusion than the oversimplified "travelling clocks run slowly", which (as you've seen) muddies the waters in scenarios like the twin paradox. It also has the advantage of being applicable in General Relativity, where you can actually derive the (non-trivial!) equations of motion for a test particle in a gravitational field, just by demanding that it moves along a path of maximal proper time.
    Jilan and Hal on Earth like this.
    Reply With Quote  
     

  4. #4  
    Member
    Join Date
    Jun 2014
    Posts
    32
    Quote Originally Posted by Farsight View Post

    This is only like two observers situated at different elevations in a gravitational field. NIST have optical clocks which are so accurate that they can see a difference in clock rates when one is only 30cm above the other.
    Thanks for that reference. I looked up NIST's site and downloaded an article Optical Clocks and Relativity, which was useful.

    Quote Originally Posted by Farsight View Post
    That's right. SR time dilation is usually symmetrical, wherein each twin sees the other's clock running slower than his own. People say this is a paradox, but it's only like you and me being separated by distance. I see you looking smaller than me, and you see me looking smaller than you. But we don't shout paradox!
    A very nice analogy.

    Quote Originally Posted by Farsight View Post
    Check out the Simple inference of time dilation due to relative velocity. Then replace Don's clocks with parallel-mirror light-clocks and draw the light paths. That really gets to the bottom of it.
    Yes, I think that's the essence of it. After replacing the orbiting clock with a light-clock the orbit path becomes the zig zag perimeter of a 'circle' area. We replace the stationary clock with a light-clock and place it in the centre of this area. To mainatain close separation each mirror is virtually touching the perimeter of the circle; so there is a straight line path between the stationary mirrors. Clearly, the differential distances between orbiting path and 'stationary path' is what creates the the differential clock rates.
    Reply With Quote  
     

  5. #5  
    Member
    Join Date
    Jun 2014
    Posts
    32
    Hi BTR, thanks for your detailed reply. I am not however confident I understand one aspect.

    For a non-inertial clock 'A' closely orbiting an inertial clock 'B', would the proper time line in Minkowski space be longer for A than for B? So understood (if correctly so) then this seems consistent with the double diagonal path traced by the photon to be longer for A.

    Another interesting conclusion for the orbiting clock involves the frequency of light. Suppose both inertial and non-inertial clocks emitted light in all directions. The non-inertial clock would receive a HIGHER frequency of light in its frame of reference because, using the light-clock example, more photons would 'fit' between the mirrors that are further apart due to the photon's longer double diagonal path. Conversely, the inertial clock must receive a LOWER frequency. If this is correct, then here we have an example of 'red shifts' and 'blue shifts' occuring outside the example of the Doppler Effect, because the two clocks maintain a constant very close distance.
    Reply With Quote  
     

  6. #6  
    Member
    Join Date
    Jun 2014
    Posts
    32
    Thanks Farsight. I posted a reply which was auto-acknowledged by the admin function but it still hasn't been published. I'll just wait a while longer and re-post if it doesn't come up.
    Reply With Quote  
     

  7. #7  
    Banned
    Join Date
    Jan 2013
    Posts
    652
    Noted Hal.

    Quote Originally Posted by Hal on Earth
    Yes, I think that's the essence of it. After replacing the orbiting clock with a light-clock the orbit path becomes the zig zag perimeter of a 'circle' area. We replace the stationary clock with a light-clock and place it in the centre of this area. To maintain close separation each mirror is virtually touching the perimeter of the circle; so there is a straight line path between the stationary mirrors. Clearly, the differential distances between orbiting path and 'stationary path' is what creates the differential clock rates.
    Yep. Always draw the light paths in parallel-mirror light-clocks to get down to the fundamentals. Like btr said, what we're really dealing with is just Pythagoras' theorem. It's mentioned on Wiki. The light path for a moving clock is drawn like this: /\/\/\/\/\/\/\. Take half of this /\ and you've got yourself a right-angle triangle where the hypotenuse represents the speed of light c and the base represents your speed as a fraction of c. The height is the Lorentz factor 1/√(1-v²/c²), the reciprocal distinguishing time dilation from length contraction. Note that the thing that underlies invariant intervals is that the light-path lengths are the same. Proper time is merely the number of reflections. Pretty simple really.
    Reply With Quote  
     

  8. #8  
    Junior Member
    Join Date
    Jun 2014
    Posts
    1
    i am a newbie here , i wish to learn about time dilation ,,
    i simply cannot visualize it , please help ;
    sorry if i posted at wrongplace
    Last edited by KJW; 06-21-2014 at 05:13 PM. Reason: Changed "@" to "at".
    Reply With Quote  
     

  9. #9  
    btr
    btr is offline
    Senior Member btr's Avatar
    Join Date
    May 2014
    Location
    Circumnavigating the photon sphere.
    Posts
    168
    Quote Originally Posted by Hal on Earth View Post
    Hi BTR, thanks for your detailed reply. I am not however confident I understand one aspect.

    For a non-inertial clock 'A' closely orbiting an inertial clock 'B', would the proper time line in Minkowski space be longer for A than for B? So understood (if correctly so) then this seems consistent with the double diagonal path traced by the photon to be longer for A.

    Another interesting conclusion for the orbiting clock involves the frequency of light. Suppose both inertial and non-inertial clocks emitted light in all directions. The non-inertial clock would receive a HIGHER frequency of light in its frame of reference because, using the light-clock example, more photons would 'fit' between the mirrors that are further apart due to the photon's longer double diagonal path. Conversely, the inertial clock must receive a LOWER frequency.
    If A is orbiting the inertial observer B with uniform circular motion, then the world-line of A will be a helix through spacetime, and B's world-line will be the straight line on which the helix is centred.

    If we now consider one complete orbit of A from the point of view of B's inertial reference frame, we'll be looking at a segment of the helix joining two events with the same spatial coordinates but with different time coordinates. If you imagine joining those two events with a straight line (which would be parallel to B's world-line), the straight line looks like it would be shorter than the arc of the helix, but (recalling the negative signs in the integral for the proper time) the proper time for the straight line would actually be shorter.

    From the point of view of observer B, therefore, the clock of A runs slowly. Using the integral formula I gave before, you can show that from B's standpoint the frequency of A's clock is slowed by a factor


    where is the orbital speed of A as judged by B.

    It also follows that the frequency of B's clock is, from the standpoint of A, increased by the reciprocal of that same factor.

    If this is correct, then here we have an example of 'red shifts' and 'blue shifts' occuring outside the example of the Doppler Effect, because the two clocks maintain a constant very close distance.
    The effect exists, but it is still usually considered to be an example of the Doppler effect (specifically, the transverse Doppler effect), because it arises due to relative motion. Despite the fact that the distance between A and B is constant, the vector displacement varies.
    Reply With Quote  
     

  10. #10  
    Member
    Join Date
    Jun 2014
    Posts
    32
    Thank you BTR. Your explanation makes it easier to understand conceptually. Everything fits elegantly.

    I made a mistake earlier in this thread. Namely, the light clock example doesn’t work in ‘real conditions’ because the object which is in uniform circular motion is constantly changing direction so it is actually accelerating. Therefore the photon in the classic two mirror system escapes- and we can’t measure the double diagonal path.
    One solution by James West was to introduce a third mirror – a ‘floor mirror’- to keep the photon within the mirror system, and to therefore determine the path of the photon- but it works only for constant direction examples. (Citation West, Joseph. “A light clock satisfying the clock hypothesis of special relativity.” Eur. J. Phys. 28 (2007) 693-703. Or secondary article www.phys.org/pdf102850833.pdf )

    But how do we find the path in an orbiting mirror system? You need mirrors all round!

    I imagined an enclosed disc whose perimeter is made of a mirror surface. Therefore the photon cannot escape inside this light-clock, if we still call it that. I don’t know what shape this path will appear to an inertial observer, but you would suppose it must be regular. According to West’s FMEL design, for constant direction acceleration the path is kind of double wavy, but it still resembles a double diagonal path. So let's make the photon initially travel along the diameter. This suggests an orbiting photon trapped in a perimeter mirror disc will also take a wavy path between bounces. Or...does it bend around the circle in a WWWW shape instead? It is intriguing to consider.

    I’m sorry, but this topic became the subject of my new un-replied to thread (path of a trapped photon) which I might delete if the discussion picks up here.
    Reply With Quote  
     

  11. #11  
    Banned
    Join Date
    Jan 2013
    Posts
    652
    I'm think the parallel mirrors take the form of concentric cylinders. Then the light path is like this /\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\ but bent round into a circle.
    Reply With Quote  
     

  12. #12  
    btr
    btr is offline
    Senior Member btr's Avatar
    Join Date
    May 2014
    Location
    Circumnavigating the photon sphere.
    Posts
    168
    Quote Originally Posted by btr View Post
    If we now consider one complete orbit of A from the point of view of B's inertial reference frame, we'll be looking at a segment of the helix joining two events with the same spatial coordinates but with different time coordinates. If you imagine joining those two events with a straight line (which would be parallel to B's world-line), the straight line looks like it would be shorter than the arc of the helix, but (recalling the negative signs in the integral for the proper time) the proper time for the straight line would actually be shorter.
    My proof-reading skills were evidently not at full power when I wrote the above paragraph. The proper time for the straight line would actually be greater than that of the helical path, in line with what I said in my first post in this thread. Sorry about that.
    Reply With Quote  
     

  13. #13  
    x0x
    x0x is offline
    Senior Member
    Join Date
    Jun 2014
    Posts
    737
    You can get the ratio of clock rates relatively easily by knowing GR.
    Start with the Schwarzschild solution to EFE:



    Use the fact that



    Then:



    produces, by division with :



    Write the above for two locations , at radial distances and and you get:





    Divide the two:



    where i=1,2. If the orbit is circular, then but this is not the usual case.

    The above equation relates the proper times of two clocks , located at different radial distances and moving at different angular speeds. It is the foundation of the calculations for setting up the frequencies of the GPS clocks (such that they are synched up).



    Since the above (exact) formula is rather complicated, the GPS calculations use a power series approximation, making use of the fact that . The formula encapsulates the effects in:

    1. gravitational potential difference , via the term in
    2. radial speed difference (if it exists), via the term in
    3. angular speed difference, via the term in

    In general, the formula is expressed in terms of ratios of frequencies, not clock rates:

    Last edited by x0x; 06-26-2014 at 12:23 AM.
    Hal on Earth likes this.
    Reply With Quote  
     

  14. #14  
    Banned
    Join Date
    Oct 2013
    Posts
    157
    The above post reflects something I already complained about on TheScienceForum.

    Tex is too small.
    Reply With Quote  
     

  15. #15  
    btr
    btr is offline
    Senior Member btr's Avatar
    Join Date
    May 2014
    Location
    Circumnavigating the photon sphere.
    Posts
    168
    It's because the default, on this forum, is to display fractions, integrals and so on in the "text mode" style like so:

    Mark-up: [tex] \frac{1}{2}\int_{-\infty}^\infty e^{-\frac{1}{2}x^2} \text{d}x [/tex]

    Rendered output:

    That mode works OK-ish when the equations are in-line with text, but what you sometimes want is the "display style":

    Mark-up: [tex] \displaystyle \frac{1}{2}\int_{-\infty}^\infty e^{-\frac{1}{2}x^2} \text{d}x [/tex]

    Rendered output:
    Reply With Quote  
     

  16. #16  
    x0x
    x0x is offline
    Senior Member
    Join Date
    Jun 2014
    Posts
    737
    Thank you, it helped making the formulas more legible.
    Reply With Quote  
     

  17. #17  
    Banned
    Join Date
    Oct 2013
    Posts
    157


    This was just a copy and paste (busy.) This may be off topic, but meh

    Tex
    Reply With Quote  
     

  18. #18  
    btr
    btr is offline
    Senior Member btr's Avatar
    Join Date
    May 2014
    Location
    Circumnavigating the photon sphere.
    Posts
    168
    The \Large command (and its friends like \small) don't work in math mode, unfortunately, even in proper LaTeX.

    You can cheat, though, using an mbox like this:

    Mark-up: [tex] \mbox{\Large\( \displaystyle \frac{\alpha ^e \infty ^{1/\mu }}{\frac{\pi }{e}}\leq \int_{\psi }^{\sigma } \zeta \, d\gamma \)} [/tex]

    Rendered output:

    (I got that trick from here. Note that I also used \displaystyle to get nicer looking fractions and integrals.)
    Beer w/Straw likes this.
    Reply With Quote  
     

  19. #19  
    Member
    Join Date
    Jun 2014
    Posts
    32
    The classic light-clock in SR is a pure mathematical construct, not a physics construct, because it is assumed the mirrors are perfectly reflective and perfectly rigid. Nevertheless, we get the derivation of the Lorentz transform.

    Now, if it assumed a light-clock could be fashioned out of a hollow reflective sphere and a likewise neglect is paid to the fact there is no such thing as perfect reflection or perfect rigidity in physics, can we likewise derive the Lorentz transform?

    To simplify things, we can assume zero radial speed, but why not go further and neglect gravity too? In other words, let our orbiting clock have negligible mass, ditto an inertial clock which is identical in structure.

    So from


    we simply get



    Using units of c =1, let the sphere's diameter =1. Imagine the diameter is an axis so the sphere (or disc) rotates about an end point of that axis. (Picture a common circular fan with only one ‘axle’ and only one ‘blade’ which is the radial strut and diameter of the sphere). Actually, we need to allow a negligible gap to accommodate the inertial observer- a second light-clock of the same sphere design- which is in the centre of all this. So let’s neglect that gap too.

    The commencing photon in this sphere shaped light-clock is emitted along the empty space of the sphere's diameter. For each orbit (i.e. completion of a cycle or 1 radian) there must be a fixed number of reflections for a given angular speed. If, upon each reflection, the sphere clock also sent a pulse to the observer sphere clock, the observer would receive a frequency of pulses analagous with the frequency of reflections in a ‘standard’ light-clock travelling at the same speed. (However, a difference in this analogy needs to be stated. The ‘standard’ light-clock always travels at 'inertial speed' so if the frequencies were actually observed they would be symmetrical between 2 observers). If the inertial observer clock also sent a pulse to the orbiting clock, the pulses exchanged would be asymmetrically received.

    My point is: if we can derive the Lorentz transform using a ‘standard’ forward moving light-clock then why shouldn’t we be able to do it with an orbiting light-clock? I imagine the observed lengths travelled between each pulse received- if displayed on a computer screen- could be directly related to length contraction and length expansion respectively for the 2 observers.
    Reply With Quote  
     

  20. #20  
    x0x
    x0x is offline
    Senior Member
    Join Date
    Jun 2014
    Posts
    737
    Quote Originally Posted by Hal on Earth
    To simplify things, we can assume zero radial speed, but why not go further and neglect gravity too?
    You cannot "neglect gravity". The effect due to difference in gravitational potential is real and cannot be "neglected"


    In other words, let our orbiting clock have negligible mass, ditto an inertial clock which is identical in structure.
    The effect I explained has nothing to do with the "mass of the clock". Sorry, you do not seem to have been able to follow the explanation and you are trying to reduce it to your level of understanding. Physics doesn't work this way.
    Reply With Quote  
     

  21. #21  
    Member
    Join Date
    Jun 2014
    Posts
    32
    Well maybe I’m guilty of reducing too much.

    You gave a method from GR to compare proper time between clocks. Thank you, that was appreciated. But I’m not sure we need to consider gravity if the clocks are of negligible mass and are not close to any large body like a planet.

    I probably should have been clearer. I started this thread because I am interested in the asymmetry of exchanged signal frequencies between two small bodies: an inertial body and a non-inertial orbiting body. I was interested in how the Lorentz transform might be derived in this situation using a different kind of light-clock, such as a ‘trapped photon’ between mirrors either in a sphere, a disc or a cylinder.
    Reply With Quote  
     

  22. #22  
    x0x
    x0x is offline
    Senior Member
    Join Date
    Jun 2014
    Posts
    737
    To make gravitation negligible you need to make . The mass of the clocks does not enter in the equation.
    The answer is:



    If clock2 is in the center of rotation, then you can make in the above.
    Reply With Quote  
     

  23. #23  
    Member
    Join Date
    Jun 2014
    Posts
    32
    Quote Originally Posted by x0x View Post
    To make gravitation negligible you need to make . The mass of the clocks does not enter in the equation.
    The answer is:



    If clock2 is in the center of rotation, then you can make in the above.
    Are the frequency symbols correct? I made it

    Reply With Quote  
     

  24. #24  
    x0x
    x0x is offline
    Senior Member
    Join Date
    Jun 2014
    Posts
    737
    Quote Originally Posted by Hal on Earth
    Are the frequency symbols correct? I made it
    My formula is correct, yours is not. Why did you change it?
    Reply With Quote  
     

  25. #25  
    Banned
    Join Date
    Jan 2013
    Posts
    652
    Hal, I've drawn you a picture:



    There's a parallel-mirror light-clock in the middle, and another one with curved mirrors rotating around it. If you're standing in the centre and you deemed that the rotating clock was instantaneously moving from right to left at 0.5c, it would be going slower than the central clock by the usual Lorentz factor which works out to be 0.866. See wikipedia for the linear example and btr's post #9.
    Reply With Quote  
     

  26. #26  
    Member
    Join Date
    Jun 2014
    Posts
    32
    Quote Originally Posted by x0x View Post
    My formula is correct, yours is not. Why did you change it?
    My mistake. It was right the first time.

    I must be confusing the term 'frequency'.

    OK. The inertial clock is clock 2, and the non-inertial clock is clock 1. Clock 2 is at the centre of the rotation and the linear speed of clock 1 is, say 0.6c :
    The ratio of clock 2's frequency to clock 1's frequency is 0.8 (as the denominator is one). This implies = 1 and = 1.25. Then this means clock 2 receives clock 1's signals at 0.8 the rate compared to the rate clock 2 sends signals to clock 1.

    I think there is a jump here that is missing a piece in my logic.

    Quote Originally Posted by Farsight View Post
    There's a parallel-mirror light-clock in the middle, and another one with curved mirrors rotating around it. If you're standing in the centre and you deemed that the rotating clock was instantaneously moving from right to left at 0.5c, it would be going slower than the central clock by the usual Lorentz factor which works out to be 0.866. See wikipedia for the linear example and btr's post #9.
    That was the image I initially had in mind. But on further, er, reflection, it seemed to me this model relies on the 'instantaneous' argument. Which is fine if you want to picture it that way, but in practice the photon would not travel the double diagonal path you have shown because the mirrors are in acceleration (due to circular motion) so the photon would be left behind. (Like when you toss a ball vertically in a moving train that is accelerating the ball won't fall in your hand like it does for uniform motion).

    (I wondered if a spherical mirror system might lead to a different way of measuring relative signal frequencies exchanged between inertial and non-inertial observers but that's getting me nowhere)
    Reply With Quote  
     

  27. #27  
    x0x
    x0x is offline
    Senior Member
    Join Date
    Jun 2014
    Posts
    737


    Plug in any numbers you want.
    Reply With Quote  
     

  28. #28  
    Banned
    Join Date
    Jan 2013
    Posts
    652
    Quote Originally Posted by Hal on Earth View Post
    That was the image I initially had in mind. But on further, er, reflection, it seemed to me this model relies on the 'instantaneous' argument..
    Then just give yourself two concentric cylindrical mirrors. Methinks all this is simpler than you're thinking.
    Reply With Quote  
     

  29. #29  
    Member
    Join Date
    Jun 2014
    Posts
    32
    Quote Originally Posted by x0x View Post


    Plug in any numbers you want.
    OK. But let's put the numbers in an example. Here c=1. Linear speed of clock 1=0.6, and clock 2 is at the centre of the rotation, so =1, then



    This means clock 1 receives a signal at 1.25 times the frequency that clock 2 sent it- from clock 2's inertial frame of reference.

    The relationship between frequency, light and wavelength is given by



    A wavelength of 600 nm is orange in the visible spectrum. Let clock 2 send an orange light to clock 1. After doing the calculations we find clock 1 receives a wavelength of 480 nm which is blue in the visible spectrum.

    Conversely, we may say clock 1 sends a blue light to clock 2, which clock 2 receives as an orange light.
    Last edited by Hal on Earth; 07-05-2014 at 12:41 PM. Reason: [tex]f_1= \frac{1}{\sqrt 1- 0.6^2}=1.25[/tex] changed from [tex]f_1= \frac{1}{\sqrt 0.6^2}=1.25[/tex]
    Reply With Quote  
     

  30. #30  
    Member
    Join Date
    Jun 2014
    Posts
    32
    Quote Originally Posted by Farsight View Post
    Then just give yourself two concentric cylindrical mirrors. Methinks all this is simpler than you're thinking.
    ....then please show me how it works.
    Reply With Quote  
     

  31. #31  
    x0x
    x0x is offline
    Senior Member
    Join Date
    Jun 2014
    Posts
    737
    Quote Originally Posted by Hal on Earth
    A wavelength of 600 nm is orange in the visible spectrum. Let clock 2 send an orange light to clock 1. After doing the calculations we find clock 1 receives a wavelength of 480 nm which is blue in the visible spectrum.
    Correct.

    Conversely, we may say clock 1 sends a blue light to clock 2, which clock 2 receives as an orange light.
    No, the frame of the clock in the center of the circle is inertial, the frame of the orbiting clock is not, so, the two are not symmetric. You cannot simply invert the calculations.
    Reply With Quote  
     

  32. #32  
    Banned
    Join Date
    Jan 2013
    Posts
    652
    Quote Originally Posted by Hal on Earth View Post
    ....then please show me how it works.
    You just say one reflection is one tick. That's it. It's that simple. When the light has some horizontal or rotational motion as distinct from an up-and-down vertical motion, you've got time dilation which you work out based on Pythagoras' theorem. Make sure you read that wiki section on the simple inference of time dilation. It's almost too simple. I think you're struggling with this because you can't quite appreciate just how simple it is.
    Reply With Quote  
     

  33. #33  
    btr
    btr is offline
    Senior Member btr's Avatar
    Join Date
    May 2014
    Location
    Circumnavigating the photon sphere.
    Posts
    168
    Farsight is right, it's not really that complicated, it's just a set-up which creates the illusion of complexity.

    To begin with, note that when a photon is reflected by an ideal mirror the only momentum change for the photon happens in the direction normal to the mirror's surface. Therefore, in working out the photon's path in the central observer's rest frame we can pretty much ignore the fact that the mirrors are rotating about their common centre.

    The picture given in #25 illustrates a photon's path in the rest frame of the central observer. Let's keep track of the photon's position in plane polar coordinates as it bounces around. Now, consider one straight segment of the photon's journey, in which it travels from the inner mirror (of radius ) to the outer mirror (of radius ) and its coordinates change from to . The length of the line segment traced out is equal to the time between the bounces (measured by the central observer) multiplied by the speed of light, i.e. ; a little geometry reveals that


    The tangential distance covered by this segment of the zig-zag in the central observer's frame is , and so . Since the orbiting observer's velocity has no radial component, there is no length contraction in that direction and the radial distance covered in the orbiting observer's frame is just ; thus, (note the prime). Plugging all of this into the last formula,


    If the dimensions of the orbiting light clock are small in comparison with the orbital radius, the distance covered during one "tick" will be small in comparison with the orbital radius , i.e. the argument of the cosine function above will be small in comparison with unity. We can therefore expand the cosine in a power series and throw away terms of order , obtaining (after a little rearrangement of terms)


    You can see that when (the radius of curvature of the orbit) goes to infinity, the second bracketed term on the right hand side drops out (along with all the already-tiny higher order terms represented by the ellipsis ) and we recover the usual time dilation formula. We also recover that formula in the limit that the physical size of the clock goes to zero, since in that case and both go to zero, while their ratio becomes .
    Farsight likes this.
    Reply With Quote  
     

  34. #34  
    x0x
    x0x is offline
    Senior Member
    Join Date
    Jun 2014
    Posts
    737
    Actually, GR gives an exact answer to the question, not an approximate one.
    Reply With Quote  
     

  35. #35  
    btr
    btr is offline
    Senior Member btr's Avatar
    Join Date
    May 2014
    Location
    Circumnavigating the photon sphere.
    Posts
    168
    So does SR, since no gravitation is involved in the rest frame of the central observer, and I gave it in my previous post.
    Reply With Quote  
     

  36. #36  
    x0x
    x0x is offline
    Senior Member
    Join Date
    Jun 2014
    Posts
    737
    The difference is that I gave the exact solution. Yours is just an approximation. Actually, several approximations.
    Reply With Quote  
     

  37. #37  
    btr
    btr is offline
    Senior Member btr's Avatar
    Join Date
    May 2014
    Location
    Circumnavigating the photon sphere.
    Posts
    168
    I gave an exact solution to the actual problem given (concentric ideal mirrors separated by a finite amount), and then showed the limiting behaviour as the clock size tends to zero, in which limit the Lorentz time dilation formula is recovered.

    Your first attempt using the Schwarzschild metric was ludicrously wrong; your second attempt in #27 is only valid when the separation between the mirrors is negligible.
    Reply With Quote  
     

  38. #38  
    x0x
    x0x is offline
    Senior Member
    Join Date
    Jun 2014
    Posts
    737
    Quote Originally Posted by btr
    Your first attempt using the Schwarzschild metric was ludicrously wrong
    Nonsense.
    Reply With Quote  
     

  39. #39  
    Member
    Join Date
    Jun 2014
    Posts
    32
    Quote Originally Posted by x0x View Post

    No, the frame of the clock in the center of the circle is inertial, the frame of the orbiting clock is not, so, the two are not symmetric. You cannot simply invert the calculations.
    I meant separately the non-inertial body travelling at linear speed 0.6c might decide to send a blue light (480 nm) to the inertial body which would receive it as orange (600 nm). So the calculations are inverted. Or did you think I meant the orange light sent by the inertial clock ‘bounced back’ as blue? No I didn’t mean it that way. Btr's point below supports why the blue light sent by the rotating body will be received as orange by the inertial central body.

    Quote Originally Posted by btr View Post
    The effect exists, but it is still usually considered to be an example of the Doppler effect (specifically, the transverse Doppler effect), because it arises due to relative motion. Despite the fact that the distance between A and B is constant, the vector displacement varies.
    That was post 9. It is very interesting to consider the vector displacement as a spatial quantity that is expanding in the non-inertial orbiting frame. Or is that a contradiction in SR? Consider the scenario where the central clock sends orange light to a sensor in the orbiting clock that reacts by sending equivalent photons at the same frequency inside the orbiting mirror. These fresh emissions will find themselves being accommodated at a higher rate per reflection- because the 'distance' is longer. The orbiting observer therefore considers the higher frequency as blue light. Am I imagining a different kind of space? Or is it just time? Space is vector quantity, and time is supposed to be a scalar quantity.

    Quote Originally Posted by Farsight View Post
    It's almost too simple. I think you're struggling with this because you can't quite appreciate just how simple it is.
    I'm kicking myself now, because I already knew it. I was imagining only 2 mirrors in your picture in post 25 but when ALL of them rotate (in synchronised time) then of course the double diagonal path can be considered to move with the rotation. Simple.

    Btr, xOx thanks for the equations.
    Reply With Quote  
     

  40. #40  
    x0x
    x0x is offline
    Senior Member
    Join Date
    Jun 2014
    Posts
    737
    Quote Originally Posted by Hal on Earth
    I meant separately the non-inertial body travelling at linear speed 0.6c might decide to send a blue light (480 nm) to the inertial body which would receive it as orange (600 nm).
    Yes, I knew what you meant.

    So the calculations are inverted.
    No, it is not that simple. The situation is not symmetric, o you cannot simply "invert". What the inertial observer measures is (relatively) simple to compute. What the non-inertial observer measures is not so simple to compute unless you use the solution I provided you with, based on GR.

    Btr, xOx thanks for the equations.
    You are welcome!
    Reply With Quote  
     

  41. #41  
    btr
    btr is offline
    Senior Member btr's Avatar
    Join Date
    May 2014
    Location
    Circumnavigating the photon sphere.
    Posts
    168
    Quote Originally Posted by Hal on Earth View Post
    Consider the scenario where the central clock sends orange light to a sensor in the orbiting clock that reacts by sending equivalent photons at the same frequency inside the orbiting mirror. These fresh emissions will find themselves being accommodated at a higher rate per reflection- because the 'distance' is longer. The orbiting observer therefore considers the higher frequency as blue light. Am I imagining a different kind of space? Or is it just time? Space is vector quantity, and time is supposed to be a scalar quantity.
    Your conclusion is exactly right. If the central observer emits what he/she considers to be orange light, the orbiting observer will receive what he/she considers to be blue light. Conversely, if the orbiting observer emits what he/she considers to be blue light, the central observer will receive what he/she considers to be orange light. In this scenario the same equation


    can be used to determine from , or from , regardless of which observer is the emitter and which is the receiver (technically, assuming ideal frequency measurement apparatus that isn't affected significantly by the accelerations).

    Edited to add:

    Instead of picturing the photons as infinitesimal lumps with an abstract "frequency" property, imagine that the central clock is of identical construction to the orbiting one, and that the central observer emits a circular wave-front each time the central clock ticks. The number of wave-fronts detected by the orbiting observer between consecutive ticks of the orbiting clock will be greater than one, on average, as the orbiting photon has to travel further between reflections than the central one (as judged by the central observer). Conversely, if the orbiting observer emits a wave-front for each tick of its clock, there will be fewer than one of the arrive at the centre, on average, between consecutive ticks of the central clock, because the photon in the central clock has less distance to travel between reflections. The ratios in each case are obviously exact reciprocals of one another.
    Reply With Quote  
     

  42. #42  
    Member
    Join Date
    Jun 2014
    Posts
    32
    Quote Originally Posted by x0x View Post
    No, it is not that simple. The situation is not symmetric, so you cannot simply "invert". What the inertial observer measures is (relatively) simple to compute. What the non-inertial observer measures is not so simple to compute unless you use the solution I provided you with, based on GR.
    So using the GR method, could you show what frequency the orbiting clock (at 0.6c) sees when receiving orange light (600 nm) from the central clock? We assume identical mirror systems for both clocks and negligible separation.

    I must be missing a subtlety, because if and have identically separated mirrors- and the orbiting mirrors are almost touching the central mirrors- then doesn't the exactness of the solution fall short only by an infinitesimal? You and Btr seem to disagree on this point which is confusing.

    Quote Originally Posted by btr View Post

    Instead of picturing the photons as infinitesimal lumps with an abstract "frequency" property, imagine that the central clock is of identical construction to the orbiting one, and that the central observer emits a circular wave-front each time the central clock ticks. The number of wave-fronts detected by the orbiting observer between consecutive ticks of the orbiting clock will be greater than one, on average, as the orbiting photon has to travel further between reflections than the central one (as judged by the central observer). Conversely, if the orbiting observer emits a wave-front for each tick of its clock, there will be fewer than one of the arrive at the centre, on average, between consecutive ticks of the central clock, because the photon in the central clock has less distance to travel between reflections. The ratios in each case are obviously exact reciprocals of one another.
    It’s exactly how I visualised it. But is there a problem in considering each photon as a particle? Because more particles fit in between the mirrors you can consider the stream as having shorter wavelengths in relation to this 'transverse vector' which is the space between the mirrors. Conversely, the fewer the particles then the longer is each wavelength. This is just a mathematical relation. It leads to the suggestion wave particle duality is expressible in terms of "variable relativistic space", which is again just a mathematical relation that should be easy to show. (Maybe I'm thinking of an old fashioned relativistic mass concept).
    Reply With Quote  
     

  43. #43  
    btr
    btr is offline
    Senior Member btr's Avatar
    Join Date
    May 2014
    Location
    Circumnavigating the photon sphere.
    Posts
    168
    ...because if and have identically separated mirrors- and the orbiting mirrors are almost touching the central mirrors- then doesn't the exactness of the solution fall short only by an infinitesimal? You and Btr seem to disagree on this point which is confusing.
    You are exactly right, Hal on Earth. The same is true even if the orbit is very large, too. In fact, the larger the orbit in comparison with the mirror-clock dimensions, the better (you have to worry less about the corrections to the Lorentz time dilation formula which depend on the finite size of the orbiting clock's geometry).

    The General Relativity formulae posted earlier are just a distraction, so don't worry about them; there is no gravity to worry about in the scenario you constructed, so this is purely a a Special Relativity problem. I would also suggest that based on conversations in other threads, I would consider taking x0x's posts with a large pinch of salt; however, I recommend you look around these forums yourself and come to your own conclusion on that one.

    Quote Originally Posted by Hal on Earth View Post
    It’s exactly how I visualised it. But is there a problem in considering each photon as a particle?
    Technically we've made an idealisation, treating the photon as an infinitesimal point moving along null geodesics, but it is adequate for the problem at hand. As long as the size of the photon's wave packet is negligible compared to the mirror separation, you don't need to worry.

    Quote Originally Posted by Hal on Earth View Post
    Because more particles fit in between the mirrors you can consider the stream as having shorter wavelengths in relation to this 'transverse vector' which is the space between the mirrors. Conversely, the fewer the particles then the longer is each wavelength. This is just a mathematical relation. It leads to the suggestion wave particle duality is expressible in terms of "variable relativistic space", which is again just a mathematical relation that should be easy to show. (Maybe I'm thinking of an old fashioned relativistic mass concept).
    There is only a single photon "ticking" in each light-clock, so I'm not sure I follow you here. It's possible I'm being dense due to working late.
    Reply With Quote  
     

  44. #44  
    x0x
    x0x is offline
    Senior Member
    Join Date
    Jun 2014
    Posts
    737
    Quote Originally Posted by btr
    The General Relativity formulae posted earlier are just a distraction, so don't worry about them; there is no gravity to worry about in the scenario you constructed,
    If there is "no gravity", one makes (the Schwarzschild radius is null) in my formula, that's all. The General Relativity solution that I posted is the General solution to the problem, valid for ALL cases. I am still waiting for you to point out the "ludicrous errors in the derivation" . If you did not understand it, I could explain it to you. Which part did you not understand?
    Reply With Quote  
     

  45. #45  
    Senior Member
    Join Date
    Jan 2014
    Posts
    748
    cinci: Einstein calculates the time dilation for a clock circling a stationary clock. Starting with the Lorentz transformation for time,

    t' = (t - vx/c^2)/(1 - (v/c)^2)^.5

    Notice that x = vt for this form of the transformation, so

    t' = (t - v(vt)/c^2)/(1 - (v/c)^2)^.5

    Simplfying,

    t' = t(1 -(v/c)^2)/(1 - (v/c)^2)^.5 = t(1 - (v/c)^2)^.5

    which is the time dilation formula. Notice that this equation depends on the distance x being non-zero.

    We could write

    dt' = (dt - v(vdt)/c^2)/(1 - (v/c)^2)^.5

    and integrate it for small steps and determine the time dilation that occurs during the time Einstein's circling clock makes a circuit and the time dilation formula above would hold; the distance travelled would be 2pir.

    Now let's write the differential transformation for the circling clock:

    dt = (dt' - (-v)(vdt'))/(1 - (v/c)^2)^.5

    Here, the integration turns out differently. The integrated path the circling clock sees for the stationary clock is literally a circle whose radius approaches zero. So while the circling clock makes one circuit (or many), it perceives the stationary clock to travel zer distance. Therefore the integration of the vdt' term I zero. Now the integration produces

    t = (t' + 0)/(1 - (v/c)^2)^.5 = t'/(1 - (v/c)^2)^.5

    So in both cases,

    t'/t = (1 - (v/c)^2)^.5

    that is, observers on both clocks will agree that the circling clock runs slower.
    Reply With Quote  
     

  46. #46  
    Member
    Join Date
    Jun 2014
    Posts
    32
    Hi Cincirob

    I'm still trying to establish if this problem has been resolved in physics.

    You said we ‘integrate for small steps’. This is like integrating the nearly circular path for the small steps of a polygon. One author, J West, introduces co-moving inertial observers to arrive at a different conclusion than the one you cited, namely that ‘circling’ observers see the central observer’s clock to run fast, not slow.

    See free pdf : [0905.1859] The differential aging of inertial and non-inertial observers: The Eyewitness observations of a relativistic polygon traveler

    West states that to go from inertial observers to constantly accelerating observers is a large conceptual leap in special relativity.
    Moreover, he concludes the ‘circular’ moving observer judges the circumference and the radius to be adjusted by (expand?) by 1/γ, which means the circle, is (larger?) and Euclidian. This may seem surprising.

    Whatever the solution, I would like to ask you a question. By the reasoning you offered, if a rotating clock travelling at 0.6c receives light from the inertial central clock, and the rotation is very large or the rotating clock is very small, what is the factor that determines the circling body’s perceived frequency? (See post 29)

    I asked this question to xOx, who hasn’t replied. Presumably xOx would have said it was the gamma factor. However Btr thinks it is the reciprocal of the gamma factor (see posts regarding orange and blue light). For equal frequencies sent from their respective frames of reference, the rotating clock sees blue light while the central clock sees orange light.

    Does Btr's conclusion contradict that the non-inertial rotating clock sees the inertial central clock to run slow?
    Last edited by Hal on Earth; 07-03-2014 at 12:42 PM. Reason: distinguishing my interpretation from West's words re 1/γ
    Reply With Quote  
     

  47. #47  
    btr
    btr is offline
    Senior Member btr's Avatar
    Join Date
    May 2014
    Location
    Circumnavigating the photon sphere.
    Posts
    168
    Quote Originally Posted by Hal on Earth View Post
    ...I asked this question to xOx, who hasn’t replied. Presumably xOx would have said it was the gamma factor. However Btr thinks it is the reciprocal of the gamma factor (see posts regarding orange and blue light). For equal frequencies sent from their respective frames of reference, the rotating clock sees blue light while the central clock sees orange light.

    Does Btr's conclusion contradict that the non-inertial rotating clock sees the inertial central clock to run slow?
    The orbiting observer sees the central observer's clock run faster than his/her own, not slower, by the same argument I gave earlier (and in agreement with the last part of cincirob's post, "observers on both clocks will agree that the circling clock runs slower"). Otherwise there would indeed be a contradiction.

    On your point above regarding circumferences, note that for . Therefore, the orbiting observer thinks the circumference is less the rather than greater.
    Reply With Quote  
     

  48. #48  
    x0x
    x0x is offline
    Senior Member
    Join Date
    Jun 2014
    Posts
    737
    Quote Originally Posted by btr
    Your first attempt using the Schwarzschild metric was ludicrously wrong
    You have two options: prove your claim or retract it.
    Reply With Quote  
     

  49. #49  
    btr
    btr is offline
    Senior Member btr's Avatar
    Join Date
    May 2014
    Location
    Circumnavigating the photon sphere.
    Posts
    168
    Quote Originally Posted by x0x View Post
    You have two options: prove your claim or retract it.
    OK, since you specifically requested it.

    Note that I refuse to get into a debate about your error(s), as based on experiences in other threads it will turn into an unending and unproductive derail, so I will say my piece in this post and comment no further on the matter. If you want to have your say after this, please do, but this is my last post on the matter.

    With that out of the way, recall that the problem (see posts #1 and #10) was, effectively, to find the ratio of tick rates for two clocks with the following specifications:

    1. The inertial clock is a standard "light clock"; two parallel mirrors reflect a photon back and forth in perpetuum.

    2. The "orbiting" clock consists of two parallel mirrors undergoing uniform circular motion centered on the inertial clock, such that the mirror planes are always tangential to the orbit; it is implied that the separation between the mirrors is identical with that of the inertial clock, to permit meaningful comparison. This is to be modelled (see again post #10) by a pair of concentric circular mirrors which bound an annular region in which a photon is confined, executing a zig-zag path around the orbit. Farsight posted a diagram in post #25 illustrating this.

    The general SR formula for the case of "idealised" clocks (ones small enough that the details of the construction could be neglected) was given by me in post #3 (in integral form), and the simpler version for uniform circular motion was given in post #9. However, in post #10, Hal on Earth indicated that he/she wanted to explicitly find the path in the concentric mirror system, and see that the usual time dilation formula resulted. Therefore, the idealised clock approximation was of no use, and a more detailed geometric argument was required. This I gave in post #33, providing both an exact formula (the second main equation in the post) and an approximation valid for (the third main equation in the post). Thus, I showed that in the limit of idealised clocks, the usual time dilation formula was recovered, but that in the case of finite-dimensioned clocks of the type considered, correction terms arose.

    Rewinding now to post #13, you stated "you can get the ratio of the clock rates relatively easily by knowing GR," and quoted some formulas relating to a different problem, basically ignoring the interesting details of the one specified by Hal on Earth. For reference, I will quote your post in its form at the beginning of the month, in case it gets edited:

    Post #11:
    Spoiler Alert, click show to read: 
    Quote Originally Posted by x0x View Post
    You can get the ratio of clock rates relatively easily by knowing GR.
    Start with the Schwarzschild solution to EFE:



    Use the fact that



    Then:



    produces, by division with :



    Write the above for two locations , at radial distances and and you get:





    Divide the two:



    where i=1,2. If the orbit is circular, then but this is not the usual case.

    The above equation relates the proper times of two clocks , located at different radial distances and moving at different angular speeds. It is the foundation of the calculations for setting up the frequencies of the GPS clocks (such that they are synched up).



    Since the above (exact) formula is rather complicated, the GPS calculations use a power series approximation, making use of the fact that . The formula encapsulates the effects in:

    1. gravitational potential difference , via the term in
    2. radial speed difference (if it exists), via the term in
    3. angular speed difference, via the term in

    In general, the formula is expressed in terms of ratios of frequencies, not clock rates:


    Not only was that far more complicated than the SR derivation (and involved pulling the Schwarzschild solution out of thin air), but the formulae you gave where for two test particles orbiting a non-rotating gravitating body, adding spurious extra detail where not required (an entire gravitating body!) and leaving out the details that actually mattered (the finite dimensions of the orbiting light-clock). Anticipating that you might take the bizarre route of arguing that the central, inertial mirror constitutes the central source in the Schwarzschild metric, I will point out that the clocks were not posited as having any appreciably mass; furthermore, if you will argue that the mass of the central clock cannot be neglected, I will counter that the mass of the orbiting clock would then also be non-negligible, making the Schwarzschild solution (which assumes spherical symmetry) inapplicable.

    Hal on Earth, incidentally, politely indicated that you had modelled the wrong scenario in post #21 (after you insisted in post #20 that "you cannot neglect gravity" and that "the effect due to difference in gravitational potential is real").

    Post #20:
    Spoiler Alert, click show to read: 
    Quote Originally Posted by x0x View Post
    You cannot "neglect gravity". The effect due to difference in gravitational potential is real and cannot be "neglected"

    The effect I explained has nothing to do with the "mass of the clock". Sorry, you do not seem to have been able to follow the explanation and you are trying to reduce it to your level of understanding. Physics doesn't work this way.

    Post #21:
    Spoiler Alert, click show to read: 
    Quote Originally Posted by Hal on Earth View Post
    Well maybe I’m guilty of reducing too much.

    You gave a method from GR to compare proper time between clocks. Thank you, that was appreciated. But I’m not sure we need to consider gravity if the clocks are of negligible mass and are not close to any large body like a planet.

    I probably should have been clearer. I started this thread because I am interested in the asymmetry of exchanged signal frequencies between two small bodies: an inertial body and a non-inertial orbiting body. I was interested in how the Lorentz transform might be derived in this situation using a different kind of light-clock, such as a ‘trapped photon’ between mirrors either in a sphere, a disc or a cylinder.

    Anyway, you eventually conceded that Hal on Earth's scenario required you to set and = 0, eventually reproducing, via an incredibly obfuscated and physically unenlightening route which omitted all of the interesting geometric details of the problem actually specified, the approximate formula I gave way back in post #9 (the standard SR time dilation formula).

    -----------------------

    I will wrap up with an attempt to circle back round to a discussion of the actual topic, and point out that you will probably want to correct the mistake you made in posts #31 and #40:

    Post #31:
    Spoiler Alert, click show to read: 
    Quote Originally Posted by x0x View Post
    Quote Originally Posted by Hal on Earth
    A wavelength of 600 nm is orange in the visible spectrum. Let clock 2 send an orange light to clock 1. After doing the calculations we find clock 1 receives a wavelength of 480 nm which is blue in the visible spectrum.
    Correct.
    Quote Originally Posted by Hal on Earth
    Conversely, we may say clock 1 sends a blue light to clock 2, which clock 2 receives as an orange light.
    No, the frame of the clock in the center of the circle is inertial, the frame of the orbiting clock is not, so, the two are not symmetric. You cannot simply invert the calculations.

    Post #40:
    Spoiler Alert, click show to read: 
    Quote Originally Posted by x0x View Post
    Yes, I knew what you meant.

    No, it is not that simple. The situation is not symmetric, o you cannot simply "invert". What the inertial observer measures is (relatively) simple to compute. What the non-inertial observer measures is not so simple to compute unless you use the solution I provided you with, based on GR.

    You are welcome!

    Hal on Earth was quite right; if 2 sends orange light to 1, 1 sees it as blue, and if 1 sends blue light to 2, 2 sees it as orange. You agreed that if clock 2 (the inertial one) sends an orange light to clock 1, that clock 1 perceives it as blue. You then mistakenly denied, however, that if clock 1 sends a blue photon to clock 2 that clock 2 would detect it as orange. You firmly reinforced this erroneous position in post #40:

    (On a slightly separate note, you also wrongly asserted that General Relativity was the better tool to model the problem being considered. This is dead wrong, as this thread has illustrated, but also a common misconception. The facts are that not only is SR perfectly able to handle accelerated frames of reference, but in situations such as this, where no gravity is involved in the inertial observer's frame, GR actually reduces to SR, so why would you complicate things unnecessarily?)
    Reply With Quote  
     

  50. #50  
    x0x
    x0x is offline
    Senior Member
    Join Date
    Jun 2014
    Posts
    737
    Quote Originally Posted by btr
    Not only was that far more complicated than the SR derivation (and involved pulling the Schwarzschild solution out of thin air),
    Err, the question was (try reading the title of the thread again) Non-inertial clock orbiting an inertial clock: "Do moving clocks always run slowly?"
    NOTHING about any "light clock". The OP is about TWO clocks, one orbiting the other. You answered your buddy Farsight question, sharing his fascination with "light clocks" and NOT the question of the OP.

    The approach I presented is the standard approach in solving this type of problem as specified by the OP. For example, it produces the EXACT answer for the case of two clocks orbiting at different radii, as in the case of the GPS foundations. This is what the OP asks (check the title of the thread), but this is NOT what you answered. The solution in post 27 produces the EXACT answer for the ratio of frequencies of the GPS clocks. You are ignorant of this fact.

    but the formulae you gave where for two test particles orbiting a non-rotating gravitating body, adding spurious extra detail where not required (an entire gravitating body!)
    It is the GENERAL answer, valid for all cases. One can easily get the PARTICULAR solution that YOU produced, modulo the embarrasing approximations that you inserted. Tough.


    and leaving out the details that actually mattered (the finite dimensions of the orbiting light-clock).
    Err, the solution I provided is GENERAL, I understand you have a tough time comprehending this. You only need to make
    and out pops the SR solution, as it should.


    At this point I think that this is the closest you'll come to admitting that you were wrong and you don't know what you are doing.
    Reply With Quote  
     

  51. #51  
    x0x
    x0x is offline
    Senior Member
    Join Date
    Jun 2014
    Posts
    737
    Quote Originally Posted by btr
    in that direction and the radial distance covered in the orbiting observer's frame is just ; thus, (note the prime).
    The above WOULD have been true for a your "light clock" in uniform rectilinear motion.
    It is NOT true for the case of a rotating light clock. Hence, the error in your final answer.
    This (lesser) error comes in ADDITION to your bigger error of NOT having answered the OP question in the first place.

    By the above considerations, the proper time for the curved world-line must be less than the proper time for the straight one, and so the travelling twin is seen to have aged less than the stay-at-home one when they are reunited at R.
    Talk about non-sequitur, the answer (while correct) has NOTHING to do with what the OP is asking.
    Reply With Quote  
     

  52. #52  
    Member
    Join Date
    Jun 2014
    Posts
    32
    Quote Originally Posted by cincirob View Post
    that is, observers on both clocks will agree that the circling clock runs slower.
    Cincirob, I apologise. I carelessly jumbled your conclusion when I hand wrote your equations, and read off my notes rather than the screen. Stupid me.
    Reply With Quote  
     

  53. #53  
    Senior Member
    Join Date
    Jun 2014
    Posts
    551
    I read this part of
    Do moving clocks always run slowly?
    A commonly heard phrase in the realm of special relativity is "Moving clocks run slowly". But—even in the context of special relativity—is it always true? The answer is no. It's only true when a clock's ageing is measured in an inertial frame. This assumption of inertiality might not always be stated explicitly in textbooks, but it's always there.
    But not the rest of it. Perhaps I’ll read it some other time. But there’s one thing I’d like to point out and that’s the fact that the term special relativity is defined as relativity in flat spacetime in an inertial frame of reference. Most people don’t know this last part but it’s true. Don changed the meaning in his web page. If you look on page 3 of A first Course in General Relativity – Second Edition by Bernard F. Schutz you’ll see that the author writes
    Although the concept of relativity is old, it is customary to refer to Einstein’s theory simply as ‘relativity’. The adjective ‘special’ is applied in order to distinguish it from Einstein’s theory of gravitation which acquired its name ‘general relativity’ because it permits us to describe physics from both the accelerated and inertial observers and is in that respect a more general form of relativity.
    I’ve seen this definition in many textbooks. After all, the second postulate does refer to what happens in inertial frames. So you can’t apply that principle to a non-inertial frame. Therefore it you can’t apply all of a theories principles then you can’t apply the principle itself.
    Reply With Quote  
     

  54. #54  
    x0x
    x0x is offline
    Senior Member
    Join Date
    Jun 2014
    Posts
    737
    Quote Originally Posted by physicist
    I’ve seen this definition in many textbooks. After all, the second postulate does refer to what happens in inertial frames. So you can’t apply that principle to a non-inertial frame. Therefore it you can’t apply all of a theories principles then you can’t apply the principle itself.
    So, what do you do when you have to solve the OP problem? How do you deal with rotating frames?
    Reply With Quote  
     

  55. #55  
    btr
    btr is offline
    Senior Member btr's Avatar
    Join Date
    May 2014
    Location
    Circumnavigating the photon sphere.
    Posts
    168
    Quote Originally Posted by Physicist View Post
    But there’s one thing I’d like to point out and that’s the fact that the term special relativity is defined as relativity in flat spacetime in an inertial frame of reference. Most people don’t know this last part but it’s true. Don changed the meaning in his web page. If you look on page 3 of A first Course in General Relativity – Second Edition by Bernard F. Schutz you’ll see that the author writes

    I’ve seen this definition in many textbooks. After all, the second postulate does refer to what happens in inertial frames. So you can’t apply that principle to a non-inertial frame. Therefore it you can’t apply all of a theories principles then you can’t apply the principle itself.
    Caution: quibbling about SR/GR demarcation ahead, so feel free to skip.

    My experience has been diametrically different, for whatever reason, and as a result I wouldn't agree with Schutz at all in that passage. To me, the domain of SR is flat, infinite spacetime, whatever coordinates are used. Presumably you, Schutz and I would agree that if we decided to perform a coordinate transformation from inertial rectangular coordinates to inertial spherical polar coordinates, that we had not left the domain of the special theory. I don't really see any persuasive reason that fully general coordinate transformations in Minkowski space should be classified any differently.

    I have not carried out an exhaustive review, but the texts I can think of off the top of my head are either silent on the matter or seem to share that viewpoint. Some notable examples: Chapter 6 of MTW's weighty tome Gravitation is devoted to the topic of accelerated frames of reference in the special theory, and uses them to introduce Fermi-Walker transport; they also treat the Thomas precession using accelerated frames in SR. Chapter 1 of Hobson's General Relativity: An Introduction for Physicists briefly discusses accelerated observers and Rindler horizons within the context of SR. Chapter 3 of Rindler's own Relativity: Special, General and Cosmological and chapter 7 of Koks's Explorations in Mathematical Physics: The Concepts Behind an Elegant Language both explain how to handle accelerated observers in SR.

    As for what the SR postulates say, there is some fair agreement between authors (in my experience, whatever that is worth), and I'd say that D'Inverno's formulation in chapter 2 of Introducing Einstein's Relativity is reasonably representative:

    Quote Originally Posted by Ray D'Inverno
    Postulate I. Principle of special relativity:
    All inertial observers are equivalent.
    ...
    Postulate II. Constancy of velocity of light:
    The velocity of light is the same in all inertial systems.
    I see these postulates (and the various equivalent formulations) as constraining the laws of physics by saying how they must behave under Lorentz transformations, rather than limiting the application of SR to inertial systems. To give an analogy, it is as though I were to describe a linear transformation by its action on a basis of a vector space rather than on every vector; I do not mean to limit the domain, I am just giving the minimum information to completely define the transformation.

    None of it makes any difference, anyhow, since in any real problem we'd apply the same methods and arrive and the same answers; we'd just disagree over what to call the thing we were doing.

    Quibble over.
    Reply With Quote  
     

  56. #56  
    Member
    Join Date
    Jun 2014
    Posts
    32
    From all your inputs and some consequent thinking, here’s what I figure about rotating clocks. However, there are no direct references to Minkowski space here; I'm going for an intuitive approach that may well be enhanced by an understanding of 4 vectors and so on.

    We introduce an ordinary old fashioned circular clock(with cogs etc) and an observer. For the moment, the details of their respective positions in space and time are not called for.

    This circular idealised mechanical clock has one hand. The clock emits a signal (a light pulse) for each completed revolution of the dial by the hand. An observer receives each signal and notes the interval between them. The observer then converts the interval pictorially into the circumference of a circle. The method should be obvious:

    When the interval=2, the radius =2, the circumference =12.5663706
    When the interval =2.5, the radius =2.5, the circumference=15.70796325
    When the interval =3.33333333, the radius =3.33333333, the circumference= 20.94394891
    The area of this pictorial circle is given by . When the circumference is divided by the area for each radius we get the following ratios:

    When the radius=2, the ratio=12.5663706/12.5663706 =1
    When the radius=2.5, the ratio=15.70796325/19.63495406 =0.8
    When the radius=3.3333333, the ratio =20.94394891/34.90657802 =0.6
    Now, concurrent to this exercise, consider multiple pairs of flat (do we need concentric?) mirrors that face each other. Let the separation between the mirrors=1, and c=1, to keep it simple. For idealised mirror systems (plural) in a rotation, the path of a photon being reflected between the moving mirrors- in relation to each mirror struck- is a double diagonal path. Note that an inertial observer at the centre of the rotation sees a straight path. (Contrast this with the familiar single direction light-clock where the inertial observer sees double diagonal paths).

    The double diagonal path for inertial motion, where the distance L between the mirrors=1, is given by

    = . But because we are dealing with angular velocity which expresses constant acceleration this expression is analogous to .


    We compare the double diagonal path with each (pictorial) radius, which is the same:

    When the double diagonal path=2, the radius =2
    When the double diagonal path=2.5, the radius =2.5
    When the double diagonal path=3.33333333, the radius=3.33333333.
    Next, we compare our rotating mirror data with the ratio of the circumference to the area of the pictorial circle. The ratio was given previously, so

    When linear speed=0, the ratio=1
    When linear speed=0.6, the ratio=0.8
    When linear speed=0.8, the ratio=0.6
    These relations are pleasingly familiar. They belong to the Lorentz factor. Hence the ratio of the circumference to the circle’s area is derivable from where linear speed is analogous to speed, v. Indeed, had we performed the reciprocal operation to derive each ratio (by dividing the area by the circumference) we would have gotten ratios of 1, 1.25 and 1.66667 respectively, which corresponds to


    Recalling, the ratio was defined as the circumference divided by the area. The circumference is the edge of a curve which we may assign a thickness to. When linear speed=0, thickness =1, when linear speed=0.6, thickness =0.8 and when linear speed =0.8, thickness =0.6. Therefore the thickness of the circle’s circumference (edge) contracts parallel to the direction of the circle’s expansion by the usual Lorentz factor. The reciprocal expression is that the circle's area expands by the reciprocal of the Lorentz factor.

    Now, suppose the clock is represented as a sphere rather than a disk. The surface area = and volume=. We use similar logic for finding the previously used ratio, except we use the ratio of the surface area to the volume instead. A little extra work was needed here to find the same result. For each linear speed, find the ratio, then divide that found ratio by the ratio corresponding to linear speed=0 (which is always 1.5). The results will be the same - you get the length contraction factor. The sphere’s surface (or “edge”) therefore contracts parallel to the direction of expansion if that additional step is permitted.

    Alternatively, we may perform the reciprocal operation by dividing the volume by the surface area and then dividing that ratio by the ratio corresponding to linear speed=0 (which is always 0.666667).

    This leads to an intuitive way of conceptualising asymmetric perception of frequencies. But first, we assume all the mathematical complications of separation between bodies, size of each body, length of rotation etc have been ironed out to oblivion by the usual limit methods. It's all flat geometry without gravity.

    If you draw a single constant rectilinear wavy line (light or other e/m radiation)and let it pass through the centre of circles of various sizes you can count the waves inside each circle. That number tells you the different frequencies perceived in each circle for each measured unit of respective proper time, even though your wavy line remains constant. (Example: 5 (x 10^17) waves per second will fit in a circle with a radius of 2, and 6.25 (x10^17) waves per second will fit in a circle with a radius of 2.5. These waves will be perceived as orange and blue respectively)

    We may also say each circle represents the relativistic area (or volume) of a rotating clock. More on this:

    We can imagine each circle as a sphere, and its volume as a “relativistic volume”. It is relativistic in the sense the volume pictorially increases but the clock’s hand moves with unchanged energy. (Energy is conserved, only the energy/mass mix differs). Simply, the hand moves more slowly relative to the pictorially greater distance. In fact time is slowed by 1/γ because the energy of the clock’s hand is the same as the energy for a corresponding “smaller” clock. Therefore the hand has more "dimension" to cover for the same energy.

    Are we justified in substituting the term “relativistic mass” for “relativistic volume”? Sorry, but that was a long winded introduction to my question. I think the answer may be a qualified “yes” because "relativistic mass" is arguably ill-defined anyway. In this present conception of mass, the volume (or area) is time and the surface area (or circumference edge) is distance. Pretty weird. We got to this point by using a novel application of the mirror system for accelerated frames!

    Perhaps time is the “inner” property while space is the “outer” property of our spherical universe. But that’s just a personal view.
    Last edited by Hal on Earth; 07-07-2014 at 02:04 PM.
    Reply With Quote  
     

  57. #57  
    Senior Member
    Join Date
    Jun 2014
    Posts
    551
    Hal on Earth – When a clock is in a gravitational field the rate at which it ticks is governed not only by its speed but also by the potential where the clock is located. Consider how the term special relativity (SR) is defined in a text such as A First Course in General Relativity by Bernard F. Schutz. In that text he defines SR as that which pertains only to non-accelerated observers. Schutz explains that the term general permits us to describe physics from the point of view of both accelerated and inertial observers.

    Please keep in mind that while SR can’t address accelerated observers by definition, it can easily address accelerated particles.

    This is rooted in how Einstein viewed gravity. This is all clearly outlined in his little book The Meaning of Relativity.

    I’ve seen how people love to nick pick about definitions and I find that to be a waste of time. Therefore this will most likely be all I have to say on the subject.
    Reply With Quote  
     

  58. #58  
    Member
    Join Date
    Jun 2014
    Posts
    32
    Quote Originally Posted by Physicist View Post
    Hal on Earth – When a clock is in a gravitational field the rate at which it ticks is governed not only by its speed but also by the potential where the clock is located. Consider how the term special relativity (SR) is defined in a text such as A First Course in General Relativity by Bernard F. Schutz.
    I don’t disagree. In this example the rotating clock and the inertial central observer may be two points in space but not near a gravitational field like a planet.

    Quote Originally Posted by Physicist View Post
    Please keep in mind that while SR can’t address accelerated observers by definition, it can easily address accelerated particles.
    It's early days, but I have shown a way that SR does address accelerated observers, which does not seem to be textbook early 20th Century SR.

    Quote Originally Posted by Physicist View Post
    This is rooted in how Einstein viewed gravity. This is all clearly outlined in his little book The Meaning of Relativity.
    Thanks for reminding me I have a Kindle version! In the chapter on Special Relativity Einstein states.
    If we place two similar clocks (rotating with K’ [the rotating clock’s frame], one upon the periphery, and the other at the centre of the circle, then, judged from K [the central clock], the clock on the periphery will go slower than the clock at the centre. The same thing must take place, judged from K’ , if we do not define time with respect to K’ in a wholly unnatural way (that is, in such a way that the laws of with respect to K’ depend explicitly upon the time).
    I'm not sure what Einstein meant by “unnatural”. But his main point is the rotating clock will agree that the central clock will judge the rotating clock as going slower. This means the rotating clock must judge the central clock as going fast. And this is exactly the point Don Koks made, referenced in my opening post.

    Then directly after the quote Einstein says
    Space and time, therefore, cannot be defined with respect to K’ [the rotating clock’s frame] as they were in the special theory of relativity with respect to inertial systems.
    So Einstein, I think, is making the simple point that for inertial frames two clocks would always see each other’s clock as running slow. But when a non-inertial frame is introduced this does not hold, as shown when the previously stated conclusion was reached.

    Einstein’s next sentence was
    But according to the principle of equivalence, K’ may also be considered a body at rest, with respect to which there is a gravitational field (field of centrifugal force, and force of Coriolis). We therefore arrive at the result: the gravitational field influences and even determines the metrical laws of the space-time continuum.
    Now that’s interesting. Here Einstein seems to suggest that, by the equivalence principle, the degree of time dilation for the rotating clock must be equal to the time dilation for a clock in an equivalent gravitational field. I'm not sure exactly how this equivalence is determined, but someone might have a formula.

    The EP in my mind is part of the framework of SR- and not separate to it-as was shown by the novel derivation of the Lorentz factor using “light-clock mirrors” for a rotating frame:

    If you consider a large number of (ideal) mirror pairs that face each other and rotate about the central observer, then the photon will strike a mirror belonging to a different mirror pair. But from the perspective of each mirror struck the path taken by the photon is a double diagonal one. Well, not exactly I should say. The paths start to bend the faster the mirrors rotate, simply due to the shape of the circle. So you need to translate the rotational motion into a straight line to see how the theorem of Pythagoras can be used. (I had to refine my idea here).

    Now, when I considered a sphere (of spacetime?) the length of radius was based on the signal interval received.

    In a 2D depiction, when the radius is taken as the combined lengths of the double diagonal path then time expands by the factor “area/circumference”. This means distance contracts by the factor “circumference/area”. In a 3D depiction, we consider the changing surface to volume ratio of the sphere (RL) and the “at rest” surface to volume ratio (R). You find RL/R is the length contraction factor. To find the time dilation factor you just perform a reciprocal division, i.e. the changing volume to surface area ratio, so that RTR is the time dilation factor.
    Reply With Quote  
     

Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •