When Does an Observer Become an Inertial Observer?

Discussion in 'Physics & Math' started by Mike_Fontenot, Aug 2, 2019.

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  1. James R

    James R Just this guy, you know? Staff Member

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    arfa brane:

    It's not the time interval that matters so much as the relative sizes of the relevant force(s).

    For example, in a laboratory on the Earth's surface, the acceleration of the lab due to the Earth's rotation is $0.03 ms^-2$. If the system you're examining in the lab involves measurements where that acceleration will have significant effects, then you'll have to take the Earth's rotation into account. What is "significant" will, of course, be determined by the level of accuracy of the experimental quantity you're trying to measure. If you're making a rough estimate of the acceleration due to the Earth's gravity in the lab, for instance, then the value is about $9.80 ms^-2$, so an error of 3 parts in 1000 might well be acceptable.

    More to the point, in that example the weight and centrifugal force on the pucks both act vertically in the lab, while you're probably interested mainly in the horizontal motion of the pucks. The curvature of the Earth is small in the lab, so the deviation of the vertical direction across the air table will also be small.

    Depends on what you're trying to measure/calculate.

    i.e. if the observer is in an inertial frame, he is inertial.

    Only to the extent that any frame moving at constant velocity relative to an inertial frame is also an inertial frame. You can't turn a non-inertial frame into an inertial one. The best you can do is to choose to ignore the non-inertial effects - i.e. decide you're willing to put up with the non-inertial effects.
     
    James R, Aug 27, 2019
    #101
    exchemist likes this.

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  3. QuarkHead

    QuarkHead Remedial Math Student Valued Senior Member

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    Then either you didn't read or didn't understand my post #77.

    It is to some extent a matter of semantics - is gravitation a force or a form of geometry? I incline to the latter (so did Einstein as it appears).

    Look......given (almost) any gravitational field whatever, in a sufficiently small region of spacetime, the spacetime metric is constant. This implies this region is of curvature zero (flat). I would call this zero gravity.

    This is entirely consistent with the view that spacetime is a 4-manifold, which by definition, on scales as small as you want (almost) is indistinguishable from the "plane" \(R^4\) with, in this case, the Minkowski metric.

    This is GR geometry in a nutshell
     
    QuarkHead, Aug 27, 2019
    #102

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  5. arfa brane

    arfa brane call me arf Valued Senior Member

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    An inertial frame of reference is really just a choice of a velocity of zero. Since velocities are relative, you can also choose to be an inertial observer by choosing some material object (perhaps one with a centre of mass) to be at zero velocity.

    If you want to launch rockets and guide them around the solar system, a natural choice is the sun as the zero-velocity object; it isn't going to orbit the centre of the galaxy by much, relative to the months or years of spaceflight. GR isn't just about local geometry but also about how much it can change, given some interval of time.
     
    arfa brane, Aug 28, 2019
    #103

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  7. James R

    James R Just this guy, you know? Staff Member

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    No, I don't think so. The curvature depends on various derivatives of the space and time coordinates.

    If you're saying, essentially, that all free-falling frames are inertial, then I agree with you. Is that what you're saying?
     
    James R, Aug 29, 2019
    #104
  8. James R

    James R Just this guy, you know? Staff Member

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    No. Whether a frame is inertial or not depends on acceleration, not velocity.

    No, that is not sufficient to make a frame inertial. You need your material object to also have no acceleration. And not only in the sense of no coordinate acceleration, but also in the sense of flat spacetime.

    What you're really saying there is that forces on your rockets from things that are external to the solar system are negligible for your purposes of guiding the rocket around the solar system, or something similar to that. Choosing the sun as a reference point for a coordinate system doesn't make the sun's motion inertial all of a sudden.
     
    James R, Aug 29, 2019
    #105
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  9. QuarkHead

    QuarkHead Remedial Math Student Valued Senior Member

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    Various derivatives? Space AND time coordinates?

    Look, the connection is defined as the first derivative, with respect to coordinates, of a symmetric tensor field, which in the case of Riemann geometry, is the metric field. The curvature is in turn defined as the first derivative (again with respect to coordinates) of the connection.

    Ergo, the curvature field is the second derivative of the metric field.

    Of course the connection might require some explaining - I am willing to do that, but I doubt it would be popular on this forum as it currently is.
     
    QuarkHead, Aug 29, 2019
    #106
  10. phyti

    phyti Registered Senior Member

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    If an inertial frame is defined as 'a system of objects with no relative motion
    between any of those objects, for a given time interval', then that same system
    in a uniform gravitational field is also inertial.
    What do you think?
     
    phyti, Sep 5, 2019
    #107
  11. QuarkHead

    QuarkHead Remedial Math Student Valued Senior Member

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    I think I don't quite agree with your definition of an inertial frame.
    For me, an inertial frame refers to a set of global coordinates relative to which an object can be considered to be at rest which can be transformed to another, equally valid, set of global coordinates relative to which it can be considered to be in any state of uniform motion (other that light speed).
     
    QuarkHead, Sep 5, 2019
    #108
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