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I can't understand inertial reference frames!

  1. May 22, 2006 #1
    Hi,

    Please help me, I can't make head or tail of the concept of an inertial reference frame. What is an inertial reference frame? In what fundamental way does it differ from a noninertial reference frame if all motion is relative? Thanks for your help.

    Molu
     
  2. jcsd
  3. May 22, 2006 #2

    vanesch

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    Very good question: the definition of an inertial frame is a subtle concept ; moreover it is dependent on the paradigm in which you work (Newtonian, Special relativity, general relativity...).

    There is some circularity in the definition of an inertial frame: you define particles to be free of interaction if they are in uniform motion wrt the frame, and you define an inertial frame as one in which free particles have uniform motion :bugeye:.

    Now, in practice, you can solve this by saying that you can be pretty sure that certain observable objects are essentially free of interaction, and that in certain frames their motion does indeed look like it should. Then you can say that this frame is probably close to an inertial frame for the purposes of your inquiry, and the object is a free object for the purposes of your inquiry.


    The best thing to do, in Newtonian physics, for instance, is to say that it is a POSTULATE of the theory that there is an inertial frame, and to postulate that there are things which are particles free of interaction.

    Now, things become clearer in general relativity: there, indeed, there is no such thing as an "inertial frame" and all frames are on equal footing. But now, the concept what a coordinate system means has become more involved (for instance, it's not up to you to say what is the "time" coordinate, and what is a "space" coordinate, and sometimes, these can switch roles !).
     
    Last edited: May 22, 2006
  4. May 22, 2006 #3
    So am I to understand that the concept of inertial reference frames in Newtonian mechanics is an imprecise concept better un--understood?
     
  5. May 22, 2006 #4

    rcgldr

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    Last edited: May 22, 2006
  6. May 22, 2006 #5

    vanesch

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    I wonder how you determine that your inertial frame has "constant velocity".
    Velocity with respect to what ? An inertial frame ? :rolleyes:
     
  7. May 22, 2006 #6

    vanesch

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    No, it is well understood. But as you saw by the "attempt at definition" by the link given in Jeff Reid's post, it's difficult not to be circular if you want to stay totally general in its definition.

    As I said, in *practice* there's no difficulty. For many things, a coordinate frame attached to the surface of the earth is "good enough". In fact, we don't find any "free objects" at the surface of the earth (apples falling from the tree do NOT describe a uniform motion), but we don't seem to mind, and say that it is due to a force working on them, the force of gravity.
    If a coordinate system attached to the surface of the earth is not good enough, we do with a coordinate system through the center of the earth, and with the axes oriented to the "fixed starry sky". That's a much better inertial frame. Things that were incomprehensible in the former one, are now understood, such as, say, the Coriolis force. We recon that it is "an effect of the rotation of the earth".
    An even better inertial frame would be one that is centered on the sun, and with its axes pointing to the "fixed starry sky".
    That's about as good as you can do in Newtonian mechanics. This is the frame in which many objects which can be considered somehow "free" are in uniform motion... but there aren't any, of these "free" objects ! They all are in orbits around the sun, or around a planet, or... because they ALL are subjected to the force of gravity... And that was Einstein's idea: let's include this unavoidable "force of gravity" into the definition of "inertial frame".
    General relativity is the result of a deep reflexion of what it really means, an "inertial frame".
     
  8. May 23, 2006 #7

    rcgldr

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    See if the laws of inertia apply from the inertial frame reference as mentioned by the web site I linked to and many others. The point was that with an accelerating frame of refernence the laws of ineritia wouldn't apply. Again I defer to the web site and several others that define inertial frame of reference in the same manner.
     
  9. May 23, 2006 #8
    But to use this definition one must first define the concept of force externally and then observe if in a particular frame particles have 0 acceleration when no force is being applied.

    But I believe force itself is defined from newtons second law as the time derivative of momentum. You define that force is the rate of change of momentum, but then in some particular frames say that even though the momentum-derivative is non-stationery there is no force. The definition looks like a FAPP (For All Practical Purposes) one to me rather than a theoretically sound physical one.
     
  10. May 24, 2006 #9

    rcgldr

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    True, but again, deferring to the web site I linked to (and many others), you can simply imagine a human in the frame of reference, where forces can be felt. You could also use an accelerometer.
     
  11. May 25, 2006 #10
    So we somehow decide what forces are acting and then see if they produce the right acceleration to check if the frame is inertial? How is this decision taken?
     
  12. May 25, 2006 #11
    Well with respect to the uniform background that you need to define. I mean, without an Euclidean background, Newtonian mechanics (which is described in terms of vectors) has no meaning.

    In this context, an inertial frame of reference is that frame onto which no net forces is acting. Thus, a frame with constant velocity with respect to the Euclidean background-space.


    marlon
     
  13. May 25, 2006 #12
    But without the presence of ether, you can't determine velocity against background-space! Galilean relativity forbids it.
     
  14. May 25, 2006 #13
    But we are talking about classical physics here, not general relativity. The concept of inertial frame does not mean anything in general relativity. That's the entire point of Einstein's work.


    marlon
     
  15. May 25, 2006 #14

    vanesch

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    :confused:

    Newtonian physics is not done on E^3 x E^1 but on a fibre bundle with base E^1 (time) and fibre (E^3). A fixed "background" Euclidean space is more "Aristotelian". In Newtonian physics, there is no natural identification between the different fibres E^3. Such an identification is exactly what is an observer frame. There needs to be enough structure on the fibre bundle to allow for the differentiation between inertial and non-inertial observers, but there should not be enough structure as to single out a preferred natural identification between the bundles (as is the case in E3 x E1).

    Penrose has a great exposition about that in his book "The road to reality", in chapter 17, where he shows that Newton - who didn't know about fibre bundles - struggled a long way, and finally resolved for a "background space" in order to formulate his laws. It's a great read, and one which provides the insight of what's the difference between the Aristotelian view (an absolute space, and an absolute time), and the Newtonian (Galilean) view, with the inherent impossibility to have an absolute space.
     
  16. May 25, 2006 #15

    vanesch

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    Yes, exactly. That's the difference between an "Aristotelian" and a "Galilean" spacetime. In a galilean spacetime, there's no natural identification between the E3 spaces "at different times", while in the "Aristotelian" view, an absolute point is an absolute point.

    That's what I said in my previous post.
     
  17. May 25, 2006 #16
    :rolleyes: This really is a bit of a cheap way out, no ?

    We do not need to work with fibers when talking about Newtonian Physics. Hell, we should not be talking about fibers because of historical relevance. So, once again, one can define an ordinary Euclidean base which is used as background in the "Newtonian world". It is with respect to this base that all the necessary vectors are defined and how our intro physics books are written. So i can only repeat what i have stated before as an answer to your "question".

    Actually, this exactly why i like Newtonian physics. Once we need to start working with differential geometry and related tensor formalisms, things get more complicated. Eventhough the latter formalisms provide the correct physical description i think their value should not be overestimated (indeed, i am referring to general relativity now). Just look at the amount of problems we are able to solve by just using the old fashioned Newtonian physics. What problems are solved by GR ? (er, this is a rhaetorical question). All industry that we know off is based upon classical physics and QM, but where's GR ? That is also why i say that Newton's work is far more valuable than Einstein's.

    regards
    marlon
     
  18. May 26, 2006 #17
    I said Galilean relativity, not general relativity.
     
  19. May 26, 2006 #18
    But irrespective of whether we CAN take the background as E^3xE (which I believe we can not once Galilean relativity, which was not used by Newton but became a part of classical mechanics because Newton was wrong and Galileo was right), there is no method whatsoever for measuring a velocity against that background spacetime. It is theoretically impossible to determine a reference frame to be fixed with respect to the E^3, even if it existed. That's the whole point about the ether fiasco. There exists no absolutely stationery reference frame.

    Molu
     
  20. May 26, 2006 #19

    vanesch

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    Of course you can do that, just as you can do general relativity in an Euclidean space. But then you miss the essential concept. You can even do Euclidean geometry in a vector space, with an origin. The problem is that you've introduced arbitrary, unphysical concepts, and afterwards you have then to postulate that for one or other strange reason, the "detection" of these arbitrary concepts is "impossible". It is much clearer (though more sophisticated) to start directly with the right vision.

    So, yes, you can say that there is "an absolute space and an absolute time" (which is essentially an Aristotelian viewpoint), and start working from there. And then you miss the entire content of the principle of Galilean relativity, which was EXACTLY the point, that there is no natural identification between "space points" at two different times, but that this is an entirely observer-dependent notion.

    This is usually solved (correctly) in courses on Newtonian physics, by saying that *by postulate* there exists (at least one) inertial observer. It's this observer which can then "fix" absolute space for all times.
    But then there's nothing that sets _this_ inertial observer apart from another one, and we'd be confronted with an operational difficulty: HOW do you find, operationally, this SPECIAL inertial observer for which absolute space is fixed ? You can't. When you start working from this "special" inertial observer, you arrive at a point where you find out that it is, in all respects, impossible to discern from "another" inertial observer which is in uniform motion wrt the original, "special" one. So this "absolute" observer was a totally arbitrary choice. But that means that there's no physical meaning to be attached to this absolute space in the first place, because it is totally arbitrary.
    This viewpoint is EXACTLY the same as the "Lorentz ether" view in special relativity. You can specify indeed, an absolute luminoferous ether, and then postulate that things contract and times dilate wrt to this Lorentz ether. But no serious course on special relativity takes on this view, because it would make you totally MISS the MEANING of special relativity (although all calculations would be correct). In the same way as it makes you miss the meaning of special relativity by introducing an ether, an absolute euclidean space in Newtonian physics makes you miss entirely the content of the principle of Galilean relativity, of which the content is exactly, that there is NO natural way to identify "space points" at two different instances of time, and that this identification is a totally observer-dependent concept.
    It is the same as doing Euclidean geometry in R^3. You miss the entire idea behind the concept of an Euclidean space when you *identify* an Euclidean space with R^3 - although of course all results will be correct.
     
  21. May 26, 2006 #20

    vanesch

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    Yes, indeed. The mathematical structure should not introduce fundamental concepts which are - in principle - operationally undefinable. This is indeed, what goes conceptually wrong with things such as "absolute space" or "ether".
     
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