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Practical Use For Length Contraction?

  1. Aug 11, 2004 #1
    I'm looking through One Two Three...Infinity by George Gamow. After having explained length contraction with the example of two ships passing in space who each view the other as contracted he comments in a footnote:

    "Of course this is all a theoretical picture. Actually if two rocket ships passed each other traveling at such speeds as we are here considering, the passengers on either ship would not be able to see the other at all - any more than you can see a bullet fired from a rifle at a fraction of this speed."

    One two Three...Infinity, by George Gamow
    -p.101, Chapter V: Relativity of Space and Time

    This brings up an important question about the general usefullness of length contraction. Since it could never be seen with the naked eye, anyway, I wonder in what regard it ever has any practical importance.

    How is length contraction "seen" and taken into consideration in the areas of physics that actually use relativity?
     
    Last edited: Aug 11, 2004
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  3. Aug 11, 2004 #2

    selfAdjoint

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    When a cloud of particles approaches a target at high speed in an accelerator experiment, its shape, which is spherical in its own rest frame, becomes a flattened pancake in the frame of the target. Typically the analysis is done in the center of mass frame in which both the cloud and the target are dilated. Then you have to calculate the probabilities of interaction and the expeected tracks of the products in this frame using Lorentz transformations.
     
  4. Aug 11, 2004 #3

    Janus

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    Well, if I were traveling to Alpha Centauri at .9726c, I would get there in a little over a year rather than in over 4 years because the distance between the Earth and Alpha C. would have contracted to 1 lightyear instead of 4.3 lightyears. That seems of pretty practical importance.
     
  5. Aug 11, 2004 #4
    Can they actually "see" this in some way, say by reflecting an EM signal off the cloud and getting a "flattened" reflection?
    I'm not familiar with the "center of mass frame" concept at all. You say, in this frame, both the other frames dilate?

    I thought the contraction might have a bearing on particle physics, but was confused about how they could possibly sense the contraction of an individual particle, it's so small and fast. Applying it to a "cloud" makes more sense.
     
  6. Aug 11, 2004 #5
    Janus is right. Length contraction has a big importance in space travel and cosmoligy. If you want to travel at high speeds, you musttake into acount the length contracion or you will overshoot your target.
     
  7. Aug 11, 2004 #6

    plover

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    If you found the center of mass of one of the clouds, it would identify a point in the midst of the cloud that traveled along with the cloud. But, there is no reason why you can't treat both clouds as a single system, and find the center of mass of that system. This point (at any given moment) would simply be:
    (c1m1 + c2m2) / (m1 + m2)​
    where cn is the current position of the center of mass of cloud n, and mn is the total mass of cloud n.

    This point will move relative to the lab in a way that depends on the relative velocities and masses of the two clouds. E.g. for clouds with similar mass and velocity, it would be almost stationary relative to the lab.

    However, this overall center of mass, csys, can be taken as the reference for an inertial frame. The 'center of mass frame' is thus frame that treats that overall center of mass as being at rest. As both clouds are moving relative to csys, they are both subject to the effects derived from the Lorentz transform (i.e. both would appear length contracted in this frame).
     
  8. Aug 11, 2004 #7
    Thanks Plover, I sorta, almost got that. You are saying that the "target" is another cloud?

    I'm also still wondering if length contraction can ever actually be "seen" in any way, such as by reflecting an EM signal off of something that is contracted and detecting a "flattened" signal.

    Other applications I wondered about are the GPS. Since the satellites are orbiting at high speed relative to the earth's surface, would they actually ever be going fast enough that some correction has to be made for contraction in order that the measurement of location not be off somehow?

    When we went to the moon, did the speeds get fast enough that they had to correct for contraction of the distance?
     
  9. Aug 11, 2004 #8

    selfAdjoint

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    An orbiting satellite is going about 5 miles a second, give a little take a little. so 5/186,000 is its beta, or .000026882, and the length contraction factor is given by [tex] \sqrt{1 - \beta^2} = \sqrt {.99999999} = 1 [/tex] (on my calculator), or much too small to measure.
     
    Last edited: Aug 11, 2004
  10. Aug 11, 2004 #9
    I see what you mean.

    People frequently say relativity is used in the GPS and, speculating off the top of my head, I thought it might have to do with length contraction throwing off the "pinpointing" of a location. I guess it's something else.
     
  11. Aug 11, 2004 #10

    plover

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    The way I set it up there are two colliding clouds of particles. I think in some experiments it might make sense to consider one of the clouds a "target", but other experiments would just have two clouds of the same type of particle moving through the accelerator in opposite directions. However, the principle would work the same way for a single cloud of particles and a target that was stationary in the lab frame.
    I suspect so. I'm not coming up with anything obvious, but there are certainly people here who would know better than I do.
    I remember hearing that the GPS calculations require relativistic adjustments in order to be accurate, but I don't know the details.
    Not even close... :biggrin:

    I have no idea what the actual speed of the spacecraft was but here's a hypothetical calculation:

    The moon is a bit under 420,000 km away. Let's double that since the path is an arc not a straight line (note: this is overkill). Let's say it takes them 12 hours to get there (this is a severe underestimate). Then:

    d = 2 * 420,000 km
    t = 12 * 3600 sec

    and the average velocity would be d/t which comes out to about 20 km/sec which is less than one hundreth of one percent of lightspeed.

    Unless I'm being a bonehead with the calculations, the length contraction observable by our hypothetical Earth-Moon craft would be less than a meter for the entire Earth-Moon distance. While I imagine there are instruments which could (in theory anyway) detect that small a deviation (which could be the answer to your other question above), it's certainly not within the margin of error for course calculations... :wink:
     
  12. Aug 11, 2004 #11
    I see, now. I couldn't figure out where selfAdjoint's target went, and where the second cloud came from. (You're supposed to apply the zoobenz transformation when going from the frame of another poster into your own for the benefit of a zoobie)
    This notion comes to me from something vague I heard about having to apply relativity to radar, but I may have misrememmered what I heard.
    Yeah, people mention GPS in the same breath with relativity alot, but I've gotten the impression you have to know tons about the GPS before you can understand why it needs to be applied.
    If I recall correctly, it took something like four or five days for them to get to the moon, so, yeah, I guess there's no way this would come up as something to be dealt with at the speeds we are capable of.

    So far, then, particle physics is in the lead with the most practical use of length contraction.
     
  13. Aug 12, 2004 #12

    selfAdjoint

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    The thing about the GPS clocks is that there are TWO relativistic effects bearing on them:
    1) The clocks should run faster than those on the ground because they are higher in the earth's gravity field than the ground, from GR.

    2) The clocks should run slower because they are moving fast relative to the clocks on the ground, from SR.

    So these countervailing effects have both to be applied to get the right adjustment for the GPS clocks. Turns out the GR factor is bigger, so the GPS time runs faster than the ground time flow.
     
  14. Aug 12, 2004 #13

    pervect

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    The main effect on GPS is a time dilation effect and not a Lorentz contraction. The gravitational effect from the altitude of the satellite turns out to be of opposite sign from the velocity induced time dilation and much larger.

    See for instance

    http://www-astronomy.mps.ohio-state.edu/~pogge/Ast162/Unit5/gps.html

    velocity effect : clock is slow by 7 microseconds/day
    gravity effect: clock is fast by 45 microseconds/day

    required accuracy - 20-30 ns (note that c = 1 ft/ns)

    So in the course of a day, without relativistic corrections, the error would be 50x larger than the required accuracy.

    The current state of the art of atomic clocks is accuracy to one part in 10^15 (for ground based clocks) which is .1 ns/day.

    http://www.bldrdoc.gov/timefreq/cesium/fountain.htm

    A space based atomic clock is apparently planned in the future

    http://www.bldrdoc.gov/timefreq/cesium/parcs.htm
     
  15. Aug 12, 2004 #14
    Thanks selfAdjoint and Pervect for explaining how relativity ties in to the GPS. That wasn't nearly as difficult to understand as I'd anticipated.

    Thanks for the links, Pervect. (It will take some time for me to digest the one about the cesium clock and how it works, though.)

    -Zooby
     
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