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Muon Centrifuge question

  1. Oct 25, 2005 #1
    I'm speaking of the Muon decay experiment which showed no additional gravitationally related time effects in a centrifuge...
    Firstly does anyone have a real reference for the experiment? All I find are offhand mentions of it. The only identifying info I've gotten on it was that it was in 1966 and I'm unsure of the validity...

    My question is in trying to understand why a centrifuge has no general relativistic effects. What I've gotten to date is that it is only the gravitational potential that matters. Depth in the field.

    So does this mean that time effects are the same 30 miles from the surface of the sun as they are 30 miles from the surface of the moon for instance?
    IE: Intensity does not matter?

    Does this mean that the Sagnac effect is in no way a gravitational or GR effect? It is explicable in SR terms alone?
     
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  3. Oct 25, 2005 #2

    Astronuc

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    I have not heard of an experiment where muons are somehow developed in a centrifuge, where they decay.

    Muons are produced in particle interactions, usually from the decay of pions, but they are also produced in electron-electron/electron-positron collisions.

    Muons have very short half-lives ~2.2 microseconds, so it would be difficult to get them in a centrifuge, unless one places the centrifuge in beam of protons.

    Here is an example of more familiar muon experiments - http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html#c1
     
  4. Oct 25, 2005 #3

    JesseM

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    When you talk about gravitational effects here, do you mean due to earth's gravity, or just due to the G-forces experienced by something inside a centrifuge because it's being accelerated? As long as spacetime is flat you can analyze accelerated motion from the point of view of an inertial frame--a clock moving at velocity v in a frame will always be slowed down by a factor of [tex]\sqrt{1 - v^2/c^2}[/tex] in that frame, so just look at the object's velocity as a function of time v(t) in the frame you're using, and do the integral [tex]\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt[/tex] to find the total time elapsed according to the object's clock between two times [tex]t_0[/tex] and [tex]t_1[/tex] using the frame's time-coordinates.
     
    Last edited: Oct 25, 2005
  5. Oct 25, 2005 #4
    I've found mentions of the experiment all over the net....

    The idea is simply that there were no additional gravitational effects on time aside from the local lab-frame. All effects were exactly as predicted by SR.

    Hence a centrifuge does not mimic gravity in all ways. The explanation of the lack of GR effects is explained in terms of depth within the field from what I understand but I'm not sure I totaly grasp the explanation.


    I've found further hints toward identifying the experiment. Apparently it is mentioned in a book:
    "Gravitation" Misner, Thorne & Wheeler pg. 1055


    The best way to find it on the net is to look for clocks in a centrifuge... not muons because muons bring too much other stuff to the surface...
     
  6. Oct 25, 2005 #5

    JesseM

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    Sure, and since GR reduces to SR in flat spacetime, that also means all effects were exactly as predicted by GR as well. The two theories don't conflict, GR just deals with a broader range of cases than SR.
    I'm not sure that's true either...even though the problem can be analyzed using SR, I think it can also be analyzed using GR, and although I'm not sure how this analysis would work, it would probably transform time dilation due to velocity as seen in an inertial frame into time dilation due to a gravitational field in a non-inertial coordinate system where the particle in the centrifuge is at rest. My guess here is based on the GR explanation of the twin paradox, in which what looks like acceleration in an inertial frame is transformed into a uniform gravitational field in a coordinate system where the accelerating twin is at rest.
     
  7. Oct 25, 2005 #6

    Janus

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    You are deeper in a gravitational field 30 miles above the Sun then you are 30 miles above the moon, so there is a difference in gravitational time dilation.

    Local gravitational intensity does not matter however. Sitting on the surface of Uranus you are deeper in a gravity well than you are sitting on the surface of the Earth and the time dilation will be greater on the surface of Uranus. However, the surface gravity on Uranus is less than that of the Earth's.
     
  8. Oct 25, 2005 #7

    pervect

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    Let's see if we can nail the experiment down a bit better. On the page you mention, this seems like the most relevant quote

    in the bibliography this is further referenced
    "The anomolous magnetic moment of the negative muon", Nuovo Cimento 45, 281-286"

    There really isn't any great mystery of how GR gets the same result. You do a simple coordinate transformation to a new set of coordinates in eq 38.1

    t' = t
    x' = x*cos(wt)-y*sin(wt)
    y' = y*cos(wt)+x*sin(wt)
    z' = z

    dt^2 - dx^2 - dy^2 - dz^2 ->
    (1-(x'^2 +y'^2) w^2) dt^2 - dx^2 -dy^2 - dz^2 - 2w(y dx - x dy) dt

    and instead of a Lorentzian metric in dt, dx, dy, dz you get a non-Lorentzian metric. The metric coefficient that's responsible for time dilation is the one for dt^2, which goes from 1 to (1-w^2(x'^2 + y'^2))

    The clock at the origin of the coordinate system ticks the fastest, and the clock in the centrifuge ticks more slowly. The result is exactly the same.

    Note again that it is *not* accleration that causes time dilation. Metric coefficients of g_00 that not unity cause time dilation, and so does velocity (i.e. time is maximized when dx=dy=dz=0 in the Lorentzian metric).

    In the linear limit the metric coefficeint of g_00 can be identfied with gravitational potential energy, as other posters have mentioned.
     
    Last edited: Oct 25, 2005
  9. Oct 26, 2005 #8
    What determines depth in a gravitational field? I don't entirely follow. :blushing:


    Pervect:
    I don't actually have the book, thanks for some info on it. Do they give the reference for the publication of the experiment?

    It sounds like you are recapping that there is no time effects from the acceleration but there is, of cource, the expected kinematic shift.

    I'm just trying to fully grasp the difference between acceleration and depth in a gravitational field. How/why intensity does not matter...
     
  10. Oct 26, 2005 #9

    pervect

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    The reference was to Farley, et al, "The anomolous magnetic moment of the negative muon", Nuovo Cimento 45, 281-286"

    I'm not at all familiar with Nuovo Cimento, it's apparently an Italian physics journal. Google finds it online at http://paperseek.sif.it/, but not for the years in quesiton.

    To really understand how gravity causes time dilation in GR, you need to understand the GR concept of a metric.

    Hopefully you've heard of the Lorentz interval from special relativity? This is the quantity

    [itex] dL^2 = dt^2 - dx^2 - dy^2 - dz^2 [/itex]

    This quantity is the same for all observers in any inertial frame of reference, no matter what their velocity.

    In General Relativity, this is modified slightly. We compute the Lorentz interval via means of a metric, for instance it might be

    [itex] dL^2 = g_{00} dt^2 - g_{11} dx^2 - g_{22} dy^2 - g_{33} dz^2 [/itex]

    The metric coefficients in the above expression area the [itex]g_ij[/itex]. In general, they (the metric coefficients g_ij) are functions of the coordinates.

    The above equation (with some additional cross-terms which I've omitted such as g_01 dx dt, etc. ) is completely general and is valid for any coordinate system whatsoever, not just an inertial frame, via the proper choice of the coefficeints g_ij.

    You may or may not recognize the above expression as a "quadratic form" from linear algebra.

    Anyway, suppose that an object is not moving in space, and is in an inertial frame. This implies that we use the SR form of the expression for the Lorentz interval, and that dx,dy, and dz are all zero.

    We are left with dL^2 = dt^2. So we see that the Lorentz interval dL is just the time interval for an object that is in an inertial frame and not moving.

    Now, what happens when we switch to GR, and consider an object that is not moving in some general coordinate system? We find that dL^2 =g_00 dt^2. This means that g_00 plays the role of a time dilation factor. g_00 less than 1 imply "gravitational time dilation".

    What's really important is that the Lorentz interval is the same for all observers, this is the basis of relativity.

    The various forms of "time dialtion" are just a consequence of this fact. In the rest frame of an inertial particle, dL^2 = dt^2 and we have no time dilation.

    If we go to a moving frame of reference, dL^2 = dt^2 (1 - (dx/dt)^2 - (dy/dt)^2 - (dz/dt)^2), and we see that we have velocity induced time dilation.

    If you concentrate on the words, different coordinate systems all attribute time dilation to different phenomenon, and it can be confusing.

    If you focus on the Lorentz interval, all becomes clear. The Lorentz interval is the same for all observers, and the various explanations of "time-dilation" are all a consequence of this fact plus the exact defintion of the coordinate system being used.
     
    Last edited: Oct 26, 2005
  11. Oct 26, 2005 #10

    Janus

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    It is related to how much work is needed to move the same mass from that point in the field to a point an infinite distance away. It is also porportional to escape velocity: [itex]\sqrt{\frac{2GM}{R}}[/itex]. In fact,the gravitational time dilation equation:
    [tex]T = \frac{t_0}{\sqrt{1-\frac{2GM}{Rc^2}}}[/tex]
    can be gotten by simply replacing v in the SR time dilation equation with the equation for escape velocity.
     
  12. Oct 27, 2005 #11
    So, the sagnac effect and RLGs don't require the use of GR at all as I've been led to believe? It seems like SR alone is not only all that is needed but all that is appropriate.

    What is the reason why some believe the GR is the only theoretical treatment that will truly work?

    In this NATO paper, one of the authors says this very thing in section 2.1.3:
    http://www.rta.nato.int/Pubs/RDP.asp?RDP=RTO-AG-339

    It's obvious there is a piece I haven't fit into the puzzle.
     
  13. Oct 27, 2005 #12

    pervect

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    Everyone agrees that ring laser gyroscopes work, and that the Sagnac effect is real, if that's what you are after with the paper.

    Explaining the Sagnac effect within the context of SR has generated a surprisingly large amount of literature. For instance there is a $200 dollar book on this topic alone (which I don't own, way too pricy for me)

    http://www.amazon.com/exec/obidos/t...104-1896784-8131162?v=glance&s=books&n=507846

    However, my personal favorite paper on the best way to explain it (there are many approaches, all of which yield the same result, the Sagnac effect is real) is Tartaglia's paper

    http://arxiv.org/abs/gr-qc/9805089

    The abstract summarizes the approach nicely

     
    Last edited: Oct 27, 2005
  14. Nov 1, 2005 #13
    Though I've seen a least a few quotes in which Einstein himself postulated a universal reference frame that was the gravitational mean of the universe, it's not a very popular idea. :uhh:

    Though I didn't thoroughly examine the paper you mention, after perusing, it seems as though the author is proposing something similar to a universal reference frame...


    How widely respected is the reference he gives for this paradox in the common SR explanation of the Sagnac effect?
    [1] F. Selleri, Found. Phys. Lett., 10, 73 (1997)

    I need to find a copy to look at myself because between the Muon experiment and the Selleri paradox, it seems as though an adequate explanation has not yet been agreed upon by the scientific community... :confused:
     
  15. Nov 1, 2005 #14

    pervect

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    I don't understand why you think that, assuming you are referring to the Rizzi and Tartaglia paper.

    Semantic issues may be arising - everyone agrees that laser gyroscopes actually do function, so that measuring one's rotational velocity in an absolute sense is possible. However, it is not possible to use any similar instrument to measure one's linear velocity, thus there is no "universal reference frame".

    There is nothing in the Tartaglia paper to suggest that there is any way to measure the "absolute" linear velocity of an object, thus there is nothing in it to suggest an "absolute" reference frame according to the above definition.

    I don't know. I was able to find a paper by Foy and Selleri at http://arxiv.org/abs/gr-qc/9702055 which is by Selleri and talks about your muon experiment which is online.

    I wasn't terribly impressed (I think the Tartaglia paper is much better), but it is difficult to gain any objective measure of what the "scientific community" thinks. Citation counts might give an answer if enough papers have been published, but this could also give false indications - a paper which is widely criticized could wind up with a high citation count, a lot like a bad post on a bbs generates a lot of flames...

    NASA ADS citation service gave very low citation counts, I suspect that there database is simply not complete in this specific area (it's not oriented towards this topic area). Citebase seemed to be down, so I couldn't see if it fared any better. Without a large citation count, I don't think the question can be answered objectively.
     
  16. Nov 2, 2005 #15
    Last edited: Nov 2, 2005
  17. Nov 2, 2005 #16

    pervect

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    That appears mainly to be a list of papers published by Selleri, with only a few papers actually referencing his work.

    Still, one paper caught my eye:

    http://scholar.google.com/scholar?h...l.pdf+F.+Selleri,+Found.+Phys.+Lett.,+10,+73+

    For some reason the PDF version wouldn't display, but the google cached version did.

    It's really *amazing* what can get published in a peer reviewed journal. Even if it is a philosophy journal...

    Feh! Total nonsense.
     
  18. Nov 9, 2005 #17
    Well, as you might have guessed I'm still fond of reading nonsense becasue I like finding questions to which I currently do not know the answer. Then I can go about getting the answer!


    I lightly perused the paper you mentioned above and found an interesting quandry. Whyare clocks on satellites not affected by the gravitational potential of the sun more than that of earth?

    If the intesity is not important and the depth is then it stands to reason that the suns effects would not only be noticable in GPS satellites but it would by far be the greater effect by orders of magnitude.

    Edit: The change, not the overall effect I suppose...

    What am I missing?
     
  19. Nov 9, 2005 #18

    pervect

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    What's important is the total potential energy difference, the change in potential energy, as you put it.

    Using Google calculator and the formula E = -GM/r, for an object in a geostationary orbit in a circular orbit 42614 km from the center of the Earth we get for the energy change due to the Earth's gravity

    Ignoring the axial tilt of the Earth for ease of calculation, we can look at the change in potential energy due to the change in distance from the sun for the geostationary satellite

    (-G * mass of Earth) / (42 614 kilometer) = -9 355 103.16 m2 / s2

    ((-G * mass of the sun) / ((1 au) - (42 614 kilometer))) -
    ((-G * mass of the sun) / ((1 au) + (42 614 kilometer))) =

    -505 439.521 m2 / s2

    which is about 20 times less.
     
  20. Nov 10, 2005 #19
    I apologize for being so vague. I didn't mean the variation from one side of the planet to the other. I meant the variation from aphelion to perihelion.
     
  21. Nov 10, 2005 #20

    pervect

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    There are several time systems in use.

    The 5 million km difference between perihelion of the Earth's orbit and aphelion is one of the main reasons why Barycentric Dynamical time, (TDB) defined from the barycenter (colloquially the center of mass) of the solar system, is different from Terestrial Time (TT), the time of an earthbound clock at sea level. To an excellent approximation, all clocks on the geoid (at sea level) tick at the same rate, which is why TT doesn't need to specify a specific point on the geoid.

    TT is equivalent to the more common TAI time, except for a fixed offset. TAI time is different from the even more common UTC time (that you will hear on WWV and other time-broadcasting services) because of the buildup of "leap seconds" that keep UTC in synch with the rotation of the Earth.

    For some more detail, see

    http://www.hartrao.ac.za/nccsdoc/slalib/sun67.htx/node221.html#SECTION000515400000000000000

    [add]
    You may ask - why is TDB at the same rate as TT? Shouldn't it be at a different rate? The answer is that TDB was defined to include only the periodic terms. The sort of time which includes the rate difference from the barycenter to the Earth's orbit would be TCB (Barycentric Coordinate Time).

    http://tycho.usno.navy.mil/systime.html

    Hopefully I haven't messed up any details - anyway, you can see that there are several different time scales in use, which incorporate the relativistic effects.

    Because most of our clocks are located on Earth, TT or TAI (essentially equivalent but for an offset) are the most commonly used of the atomic times. UTC is used for civil time because it keeps the clock synchronized to sunrise and sunset within a second.
     
    Last edited: Nov 10, 2005
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