Rotational speed and time dilation

Discussion in 'Pseudoscience' started by BdS, Sep 6, 2013.

  1. BdS Registered Senior Member

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    Hey

    Does radial location rotation speed have anything to do with time dilation?

    Example. the Earth has an equatorial circumference of 40030.17Km and diameter of 12742Km, in 24 hours you will travel 40030.17Km or 40030.17Km / 24 = 1667.92Km/h in space from the earths spin rotation. If you move 1 Km up in a tall building you are now on a circumference of 40036.45Km, diameter of 12744Km and travel 40036.45Km in 24hrs or 40036.45Km / 24 = 1668.18Km/h

    Does traveling at different speeds at different circumferences have an effect on time dilation?
     
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  3. origin Heading towards oblivion Valued Senior Member

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    Yes. You can use satellites in geosynchronous orbit for GPS as an example time dilation effects of rotation due to a difference in the radius.

    A question like this should be in the science section - unless you plan on taking this to the 'dark side'.

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  5. MarkM125 Registered Senior Member

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    This question shouldn't be in psuedo, it should be in Physics & Math. It's a question about actual physics.

    The answer is naturally yes. The relativistic kinematics of rotational motion can be described by invoking a set of three basis four vectors

    \( e_{r} = (cos \ \omega t, \ sin \ \omega t, \ 0, \ 0) \\ e_{\phi} = (-sin \ \omega t, \ cos \ \omega t, \ 0, \ 0) \\ e_{t} = (0, \ 0, \ 0, \ 1) \)

    The derivatives of the above can then be written as

    \( \frac{De_{r}}{D \tau} = \frac{dt}{d \tau} \omega e_{\phi} = \gamma \omega e_{\phi}\\ \\ \frac{De_{\phi}}{D \tau} = -\frac{dt}{d \tau} \omega e_{r} =- \gamma \omega e_{\phi}\\ \)

    The position, velocity, and acceleration can then be computed.

    \( x^{\mu} = re_{r} + te_{t} \\ v^{\mu} = r \gamma \omega e_{\phi} + \gamma e_{t} \\ a^{\mu} = -r \gamma ^{2} \omega ^{2} e_{r} \\ \)

    Since the magnitude of four velocity must always be equal to -1, we can find the magnitude of \(v^{\mu}\) and set it equal to -1.

    \( r^{2}\gamma^{2}\omega^{2} - \gamma ^{2} = -1 \\ \gamma^{2}(1 \ - \ r^{2}\omega^{2}) = 1 \gamma = {\frac{1}{\sqrt{1 \ - \ r^{2}\omega^{2}}}} \)

    So, orbits of greater angular velocity result in larger amounts of time dilation.

    However, although higher clocks would run slightly faster because of this fact, the effect of gravitational time dilation strongly outweighs it. So, clocks at higher altitudes always run slower on earth

    EDIT: The above statement was a mistake. Gravitational time dilation, of course, increases as one approaches the center of the mass in question. See rpenner's post.
     
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  7. rpenner Fully Wired Valued Senior Member

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    Almost -- clocks at higher altitudes run faster since effect of gravitational time dilation which slows clocks closer to the Earth outweighs the special relativistic tendency to make moving clocks slow.

    (Just as real mathematicians can be bad at arithmetic, sometimes physicists assured of the precision of the models under discussion miscount the number of minus signs. And Obama is doing nothing to help.

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  8. MarkM125 Registered Senior Member

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    Wow, that was a bad mistake. I have absolutely no idea why I wrote that. I can't even think of a reason that I did that as a mistake. Anyways, thanks for the correction, I'll point out the error in an edit.
     
  9. Tach Banned Banned

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    Not necessarily, the difference in clock rate is given by the expression:

    \(d\tau_E-d\tau_S \approx \frac{dt}{2}[r_{sch}(\frac{1}{R_S}-\frac{1}{R_E})+\frac{v_S^2-v_E^2}{c^2}]\)

    where:

    \(d\tau_E\)=period of ground based clock
    \(d\tau_S\)=period of satellite based clock
    \(R_E\)=Earth radius
    \(R_S\)=satellite orbit radius (assume circular orbit)
    \(v_E\)=Earth tangential speed
    \(v_S\)=satellite tangential speed
    \(r_{sch}\)=Earth Schwarzschild radius (approx 9mm)
    \(dt\)=coordinate time interval

    If the satellite is geosynchronous, then:

    \(d\tau_E-d\tau_S \approx \frac{dt}{2}(R_S-R_E)[(R_S+R_E)\frac{\omega^2}{c^2}-\frac{r_{sch}}{R_SR_E}]\)

    The expression \(r_{sch}(\frac{1}{R_S}-\frac{1}{R_E})+\frac{v_S^2-v_E^2}{c^2}\) can be positive, zero or negative, it all depends on the satellite orbital radius and its speed compared to the speed of the Earth.
    For example, if \(v_S=\sqrt{v_E^2+c^2r_{sch}(1/R_E-1/R_S)}\), the ground clock and the satellite clock run at the same exact rate. Granted, this is a very large speed due to the contribution of the \(c^2r_{sch}(1/R_E-1/R_S)\). Generally, satellites move around 8km/s, the term \(c^2r_{sch}(1/R_E-1/R_S)\) is of the order of 30-40km/s, so, it is not impossible.
    SR should not be invoked in solving this, this is a pure GR application.
     
    Last edited: Sep 7, 2013
  10. origin Heading towards oblivion Valued Senior Member

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    Please, no politics except in the politics section.
     
  11. rpenner Fully Wired Valued Senior Member

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    Doing the math for high floor in a building of human scale like the OP, the radial term from GR dominates.
    Doing the math for high Earth orbit with closed trajectories and altitude above about twice the radius of the Earth, the the radial term from GR dominates.
    Doing the math for high circular Earth orbits with altitude above about half the radius of the Earth, the radial term from GR dominates.

    In these cases higher = faster clocks compared to terrestrial clocks.

    But for a near-parabolic orbit during its closest approach to Earth when \(R_0 \lt r \lt 3 R_0\) then there is (I think) the possibility that Earth based clocks will see it tick slower when it close.
    And for circular orbits when \(R_0 \lt r \lt \frac{3}{2} R_0\) then I am fairly sure the Earth based clocks will see it tick slower.

    This should be intuitive in that if the Earth was airless, one could be in orbit at sea-level -- at the same height as everyone else but moving faster than a rifle bullet.

    But I don't think that was the question that BdS asked or that MarkM was trying to answer.
     
  12. Tach Banned Banned

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    The math that I showed covers all the cases (both the "BdS" building case and the "origin" GPS case). While the case of the building is straightforward, the case of the GPS depends not only on radius but also on the speed, as I showed, so you can get:

    -Earth clock faster than satellite clock
    -Earth clock slower than satellite clock
    -Both clocks ticking at the same rate
     
  13. BdS Registered Senior Member

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    497
    How does acceleration fit into time dilation then? An object being accelerated at the same acceleration the gravitational field accelerates at would give the same amount of time dilation? the object being accelerated would experience exactly the same effects as gravity? part of the equivalence principle?

    eg. On Earth at sea level grav acceleration is 9.8m/s. If an object was acclerating at 9.8m/s then it would experience the same time dilation as the object at sea level under the effects of gravity.
     
  14. BdS Registered Senior Member

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    497
    Acceleration effect + current speed effect + gravity effect = Time dilation
     
  15. Janus58 Valued Senior Member

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    Acceleration, of itself has no effect on time dilation. This is known as the Clock Hypothesis. It has been tested to very high g values using high speed centrifuges.

    Essentially, here's what they did: You place radioactive samples on a centrifuge and spin them up at various speeds and radii. By varying both speed and radius, you can get different combinations of speed and acceleration. For instance, you can have the sample traveling at the same speed but under different accelerations, or you could have the same acceleration at different speeds. All these tests show that is was only the speed of the sample at the end of the arm that had any effect on the decay rate of the sample.
     
  16. BdS Registered Senior Member

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    Found this FAQ page.

    http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html

    Counter intuitive, but the clock hypothesis has much evidence supporting it, even in particle accelerators.

    The Equivalence principle is not equivalent in the time dilation aspect. Interesting but confusing

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    ACCELERATION HAS NO EFFECT ON TIME DILATION! only the extra speed caused by the acceleration is added.

    How does gravity cause TD then? I thought it was from the induced acceleration, any links or guidance appreciated.

    At what speed will traveling at sea level on earth balance the gravity TD to speed TD to 0 on the object moving? if I've worked it out correctly, theoretically it will involve slowing down compared to the earths rotation speed.
     
  17. Janus58 Valued Senior Member

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    Gravitational time dilation is due to the difference in gravitational potential, not due to a difference in gravitational acceleration.

    For example, if you were to imagine a uniform gravitational field that did not change in strength with altitude, and placed two clocks at different heights in it, the higher clock would run slower, even though both clocks experience the same force of gravity[i/].

    A real example is the fact that the surface gravity of Uranus is actually less than that of the Earth's, yet a clock on Uranus would run slower than one on the Earth because it is at a lower gravitational potential.

    Also, there is an equivalence between gravity and acceleration when it comes to time dilation. If you had an accelerating rocket, with a clock in the nose and one in the tail, the clock in the tail will run slower, even though both clocks are under the same acceleration.

    In terms of a rotating frame, when working from the rotating frame, time dilation works just as it would if you were in a non-rotating frame with a gravity field which increases with distance from an axis. IOW, time dilation is related to the work needed to move a mass inward towards the axis.

    Here's the thing, you can either treat things from the non-rotating frame, where time dilation is due to the tangential velocity at a given point, or you can treat it from the rotating frame, in which case it is due to the potential difference between points of the frame. In either case you get the same answer What you cannot do is do both. You don't apply both the potential difference and the tangential velocity.
     
  18. BdS Registered Senior Member

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    Thank you Janus

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  19. Gerasimov Registered Member

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    The main advantage of pseudoscience is that you can come up with all that is possible.
    The main drawback is that it's not working.
     
  20. BdS Registered Senior Member

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    Yeah and make things up like, take it to the dark side.

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    I've been trying to figure out a way how physical time dilation exists.

    For speed TD Im at:
    The faster you travel the more momentum/mass you have. The object traveling now has extra mass = to the extra speed/momentum. Because of the extra mass the objects atomic forces/fields are affected, strengthened proportional to the extra mass.

    From the experiment I will try explain, "You place radioactive samples on a centrifuge and spin them up at various speeds and radii." From the extra momentum you give the radioactive samples their mass increases and the extra mass affects their atomic forces/fields. The radioactive samples bound lattice changes (probably a bound lattice tension) and causes the detectors to measure a frequency change. We call this change time dilation, but the physical effect of this frequency shift would be the radioactive samples atomic forces/fields changing and causing the emitter to emit a different frequency.


    For gravitational TD Im at:
    Cosmic scaled forces/fields induce strength to the objects atomic forces/fields that are located in the cosmic scaled forces/fields. The radioactive samples frequency is x at sea level with the cosmic induce contribution. If we move the sample up 1Km into a weaker cosmic scaled force/field the induced contribution is reduced and the radioactive samples lattice is affected (probably a bound lattice tension) to weaker induction and the emitted frequency changes. With the weakened grav/magnetic induction on the radioactive sample the frequency increases and we say time is faster. Again the cause of the frequency change would be strengthening/weakening atomic forces/fields in the radioactive sample.


    You'll never know if you dont try... still working/learning on it to try pin down which forces/fields will be responsible and the maths involved to scrap or prove the ideas. Any feedback negative or positive is always appreciated...

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  21. BdS Registered Senior Member

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    To expand on this, not only has the higher altitudes clock accelerated compared to the lower altitudes clock on earth. The higher altitudes clock has also changed its average orbital distance from the sun. What would the effect be considering that? the higher altitudes clock would have also accelerated in orbital speed around the sun too? and its average orbital distance would have also changed in the milky way

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