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Relativistic interpretation question?

  1. Mar 22, 2012 #1
    Two individuals A and B are traveling through space at a velocity in the "x" direction at and are next to each other so they can synchronize their clocks. Since there is no other available reference they believe they are not moving. Both have a duplicate shape charge which as they are in the same reference frame produce the same force on both. A's charge is facing the -x direction and B's in the +x direction.

    They also know the exact mass of the ship in their reference frame and the force of the charge so they know the velocity of separation and can correct for light communication time in comparing their clocks.

    They both simultaneously fire their shape charges and as there is no absolute reference frame and each are accelerated to a higher velocity relative to their initial reference frame would it be correct that:
    1) Both clocks would go slower.
    2) With respect to their compared measurement the clocks stay synchronized.

    If not why not.
     
    Last edited: Mar 22, 2012
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  3. Mar 22, 2012 #2

    HallsofIvy

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    No, because any statement about how fast a clock is going depends upon from which frame of reference it is observed. The moment they fire their charges, they are no longer stationary relative to each other. Each person would observe the other's clock as slower than their own (and so their own clock as faster).

    I don't know what "with respect to their compared measurement" means but, no, the clocks will not stay synchronized except to a person who remains stationary with respect to the first frame of reference (in which the two clocks were stationary before the acceleration.

    If not why not.[/QUOTE]
     
  4. Mar 22, 2012 #3

    ghwellsjr

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    In their initial common rest frame, both clocks will remain in sync and end up running slower than the coordinate time of that frame. However, they each will determine that the other one's clock is running slower than their own.
     
  5. Mar 22, 2012 #4
    When you say determine, how is the determination made?
     
  6. Mar 22, 2012 #5

    ghwellsjr

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    There are a couple ways. For example, if they know Special Relativity and since you said they know the speed they have accelerated to, they can just do a calculation.

    Or, without even knowing their relative speed, they can look at each others clock and compare the rate it's ticking at compared to their own which will allow them to calculate the Relativistic Doppler. From that, they can calculate their relative speed and from that they can determine the rate the other ones clock is ticking at according to their own frame of reference.
     
  7. Mar 22, 2012 #6

    pervect

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    Here's an example of how it could work:

    Suppose initially both clocks are sych'd to 0:00:00, before they're separated. And imagine each clock has a radar reflector or a radar repeater, that reflects or repeats radar signals, and appends a timestamp to it.

    1 second after separation, clock A sends a signal to clock B. Clock B recieves the signal at 2 seconds by its clock, and sends a signal back to A, along with the time of reception (2 seconds).

    A recieves the signal at 4 seconds. Because the speed of light is constant (which A can measure for himself and determine it to be the case), he knows that at the midpoint time, (1+4)/2 = 2.5 seconds, that B's clock read 2.0 seconds (because that's what the timestamp said).

    By symmetry, you can swap the roles of A and B - B determines that at 2.5 seconds, A's clock read 2.0 seconds.

    You can plot all this on a space-time diagram and determine that A's concept of simultaneous events comprises a different line on the space-time diagram than B's concept of simultaneious events, or the original frame's concept of simultaneous events.

    You can do other things - the radar method above also gives the distance at t=2.5 seconds, which is 1/2 the round travel trip time in light seconds, i.e. 1.5 light seconds. This gives the velocity of separation (1.5 / 2.5) * the speed of light.

    There's a fairly general principle that says that the ratio of reception to transmission is some constant. This is used in what's called the k-calculus approach to special relativity, but no calculus is involved, just high school algebra.

    In this example the ratio, k, is 2.0. So A's signal at time 1 is recived at some time k*1, and the refelcted signal is received at some time k*k*1. You can see in the example I worked previously k was 2.0, the value of k depends on the speed, and knowing k, you can compute the speed from the radar return data, simlar to the way it was computed in the example.
     
  8. Mar 22, 2012 #7
    Now if (I change the conditions) I have two sets of two two with two attached to one tether and two to another. Each pair at opposite ends have the same velocity so they now rotate uniformly in circle. Both of one tethered pair have the x+ charge and the other tethered pair have the x- charge. We also have a reference clocks similarly tethered at the same rotational velocity. All fire their shape charge at the same time.

    1) What happens to the x+ and x- clocks relative to the "reference" clocks
    2)What happens to the x+ relative to the x-clock.
    And why
     
  9. Mar 23, 2012 #8

    ghwellsjr

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    I'm sorry, I can't understand what you are describing. Why don't you print up your post and give it to one of your friends without explaining anything to them and ask them to tell you what they think it means. Then maybe your friend can give you some ideas on how to describe what you have in mind because it makes no sense to me.
     
  10. Mar 23, 2012 #9
    Hope this restatement helps.

    I have four individuals A,B,C,D that are traveling through space at the same velocity where A&B are traveling in one direction and C&D in the opposite direction and A&C are attached to one tether and B&D to another. Each pair at opposite ends have the same velocity so they now rotate uniformly in circle. We also have reference clocks similarly tethered at the same rotational velocity. Both A&C pair have the + direction charge and B&D have the - direction charge. Thus after firing the shape charges both have the same velocity change relative to the reference clocks, but due to the tether relative to each other travel in opposite directions around the circle. So when all four fire their shape charge simultaneously,

    1) What happens to the + charge and the -charge clocks relative to the "reference" clocks
    2) What happens to the +charge clocks relative to the - charge clocks.
    And why
     
  11. Mar 23, 2012 #10
    Have you tried calculating it yourself? What result did you get? The best way to learn any subject is to do it and you can always ask for assistance if you post your working.
     
  12. Mar 23, 2012 #11

    ghwellsjr

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    This is still very confusing. I would have to make a lot of assumptions about what you are describing to get to any sort of a sensible problem. Did you show this to one of your friends and ask him to tell you what it means without giving him any additional help?
     
  13. Mar 23, 2012 #12
    First this is not for class and I have no calculations to do. If I was in a class I would have the answers and would not be asking these questions here.

    Is this situation considered a non-inertial reference frame and how does that effect the results. I see references to it but no explanations or examples of it.
     
  14. Mar 24, 2012 #13
    Whether you are in a class or not, the best way to understand any subject is to try doing it yourself. It is also good practice not to expect everyone else to do the work for you without showing you are at least willing to put in as much effort yourself.

    Regarding the question, you appear to have four clocks, A to D, moving at equal speed in a circle. For reference, consider a fifth clock E at the centre. In the inertial frame of E, each of A to D is slowed by the usual time dilation factor due to its speed. If all the speeds are the same, they will appear slowed by the same amount.

    The frames of the individual clocks are of course not inertial so the comparisons are more complex. Overall, what you seem to be describing is similar to the situation of signals sent between communications satellites.
     
  15. Mar 24, 2012 #14

    ghwellsjr

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    Your interpretation is evidence of the impossibility of figuring out his scenario. One thing is clear, he had three tethers, you only have two. He didn't say anything about how you could put a fifth clock at the center. At the center of what??? It is also clear, from his question, that he is talking about six clocks, not four+one. Until he can present a clear scenario with clear questions, there is no point in trying to answer his questions.
     
  16. Mar 24, 2012 #15
    I agree, it's hard to discern the picture but if all you say is "I don't understand" then he has no way forward but to repeat his description. I've given my interpretation which is probably wrong but now he only has to correct my errors, and seeing in what way I misread the previous post should show him which bits need improving. It's not perfect but it's a way of moving forward.

    As stated, he has eight clocks and four tethers but to start with they move in pairs. The charges reverse the direction of four but leave the reference clocks unchanged. Hence A reverses direction to move the same way as C's reference partner. Because of that, I ignored the reference clocks as they add nothing. On the other hand, all four named clocks are moving in a circle so I added a fifth at the centre to provide an inertial reference.
     
  17. Mar 26, 2012 #16
    I have not seen any www info (e.g. wiki, etc) that actually show how to do a non-inertial example.

    Four labeled clocks (A,B,C,D) and two referece clocks.

    Six and three tethers. In the two referece clocks are also tethered.

    All move around in a circle with the same velocity and same centerpoint. One set of three (two lettered clocks and one reference clock) is on one side of the circle and the other three on the opposite side.

    The charges slow the velocity of one of the lettered tethered pair and increase the velocity of the other lettered tethered pair. The reference clock pair velocity is unchanged.

    It is with respect to the reference pair that I would like to understand. With respect to for example earth you can not see a clock in the center of the earth. The clocks you compare are mainly on or above the surface of the planet.

    Question 1) Does the "slowed rotation clock" since it moves away at a velocity relative to the reference clock show a time slowing while the "faster rotation clock" since it also is moving away at the same velocity from the reference clock also slow in time relative to the reference clock? Just as it did in an inertial frame in the first question I asked? If not why not?

    Later, as the "slower rotation clock" and "faster rotation clock" approched their original reference clock you spead up the "slower rotation clock" and slowed the "faster rotation clock" so that they were now sitting next to their original reference clock and you can looked at the time on all clocks.

    Question 2) Relative to the reference clock what would these two clocks read?

    With respect to question 2) if there is no prefered reference frame as often claimed of relativity then both clocks should slow. But I do not believe this is the correct answer.
    So there is ?only some times? no prefered reference frame?

    In real life you can only have a clock at some point on the surface away from the center so you can't actually "see" that clock as required for experimental verification. Any physical measurement could only compare the clocks that are not in the center.
     
  18. Mar 26, 2012 #17

    ghwellsjr

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    That's because non-inertial frames really are complex and there's no single standard way to do one and it's never necessary to use a non-inertial frame because any inertial frame will do.
    Thank you for providing this perfectly clear description.
    As I mentioned earlier, you can use any inertial frame you want and the easiest one to use is the same one you used to describe your scenario in.

    So the only thing you have to know to answer your questions is the faster a clock travels, the slower it ticks. So all six clocks start off ticking at the same rate. Now if we consider just three that are traveling together, one will start moving faster and ticking slower and one will start moving slower and ticking faster. After these two clocks make a trip around and rejoin the reference clock, the faster one will have accumulated less time and the slower one will have accumulated more time.

    Now the issue of what each clock actually sees of the other clocks, in general, whenever they are separating, they see the other one tickng slower and whenever they are approaching, they see the other one ticking faster, but you have to also take into account the light transit time.
    Hopefully, these questions have been answered. Like I said, since there is no preferred reference frame, any frame will work.
     
  19. Mar 28, 2012 #18
    There are basically two approaches but neither is easy. In fact I would have to sit and work it out carefully myself but I'll give you a rough guide which is hopefully nough to get you started.

    One is to choose an arbitrary inertial frame, the obvious one here being the centre of the circle and do all the calculations in that, then translate to an inertial frame which is momentarily comoving with the clock for which you want the answer. The other approach is to note that the centifugal forces in the accelerated frame of a moving clock are equivalent to gravity and apply some of the results of GR to what is seen (e.g. gravitational redshift). The two methods should give the same answers.

    So a bit like a dumbell with theree grouped on one side and three on the other, then the charges fire and each group starts separating. If that's roughly right, that's enough, my next answer probably applies even if that's slightly wrong.

    OK, the first thing to note is that you have two reference clocks (call them R1 and R2)which are moving around the common centre so you have to calculate for one at a time, they don't share a common frame. Of course by symmetry each will see the same results.

    True but you can use a hypothetical clock at that location and do the calculations in that frame then use the transforms to change to a real location. That's how GPS works for example.

    Relative to a hypothetical clock at the centre (which I called E before), each reference clock is moving with the same speed and at constant radius so they are slowed by the same amount. Relative to R1, clock E is moving in a circle so it would appear that E should be slowed relative to R1 but an accelerometer at R1 registers a non-zero value. Classically we would call that "centrifugal force" and class it as a pseudo-force but in this we can think of it as "artificial gravity". Clock E appears to be "uphill" from R1 so appears blue-shifted by an amount determined by the difference in "gravitational potential" and that exceeds the relativistic slowing due to its motion hence overall E ticks faster than R1

    That's right, tTo start with it is close to the reference so it is slowed only by the motion away but as that increases, it's angular velocity increases as does it's apparent gravitational potential so you have several effects to take into account and they all change with time. Even if the angular velocities around E are each constant, A for example will go a full circle and return to the reference clock where is started so it first moves away but later comes back, and is rotating as well in a complex spiral relative to say R1.

    I'm not going to spend my time working that out but hopefully I've explained enough for you to make a start. Others may help if you get stuck.

     
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