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SR question

  1. Jan 6, 2005 #1

    quasar987

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    Here's postulate (13.1) now...

    First of all what does he mean by "distinguish"? And why would that be in contradiction with the postulate of SR?

    Thanks.
     
  2. jcsd
  3. Jan 7, 2005 #2
    I don't understand your question. Please explaine it better.
     
  4. Jan 7, 2005 #3

    quasar987

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    Well, I transcripted a passage from the book "Mechanics" by Keith Symon, in which he uses the word "distinguish", of which I don't know what meaning he gives to it. I would appreciate if someone could explain in what sense it is used here. Maybe a way to do that would be to restate the sentence using other words. Or maybe to state what "distuinguish" means in relativity vocabulary.

    I also ask why would the fact that there is an experimental way of "distinguishing" between two inertial coordinate systems be in conflict with the postulate of relativity, postulate which I have also transcripted for convenience.
     
  5. Jan 7, 2005 #4

    jcsd

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    We have two inertial frames A and B, in frame A frame B is moving with speed v and in frame B frame A is moving wityh speed w. It follows that w = f(v) and v = g(w). So far we haven't mentioned the principle of relativty, but you should see that it imposes the conditon that the functions f and g are infact the same function, so we now have the conditon f(v) = v, i.e. Therefore f is just the identiy function (due to the arbitary nature of v) and it follows that v = w.
     
    Last edited: Jan 7, 2005
  6. Jan 7, 2005 #5

    quasar987

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    I don't see what the function f and g represent. The only thing I could see is if they are the function "measurement of speed". But for exemple, the speed w of frame A as measured in frame B is the product of the measurement using meter sticks and clocks at rest in B. It never make use of the speed v. So what is f(v) ?
     
  7. Jan 7, 2005 #6

    JesseM

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    To "distinguish" two frames means to observe that the laws of physics don't obey the same equations in each frame's coordinate system. For example, in an accelerating reference frame the ordinary laws of Newtonian mechanics don't give correct predictions--you have to introduce an additional "fictitious force" whose strength is proportional to the acceleration, like the "centrifugal force" felt by an object moving in a circle.
     
  8. Jan 7, 2005 #7

    jcsd

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    As I said f and g both turn out to be the identiy fucntion, but they are merely the speed of one frame as measured in another.

    You need to get away to from how they are actually measured: ask yourself this can f be a constant function? No, otherwise this suggests that in any given frame all objects represnting frames move through it at the same speed. Can it be a function of anyhting other than v? No other wise it suggets that two objects at rest in the same rest frame will move at different speeds to each other in other frames.
     
  9. Jan 7, 2005 #8

    quasar987

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    Answer to both jcsd and JesseM's last post: Excuse my beign slow of understanding, but what law of physics exactly would have a different form in A and B as a consequence of that two objects' speed, that are at rest in B, could be measured to give different values in A?
     
  10. Jan 7, 2005 #9

    JesseM

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    The "postulate of special relativity" says that, if you have two inertial reference frames, the laws of physics must have exactly the same form in both coordinate systems. The argument of your textbook is that if two inertial reference frames did not get the same value for the other frame's speed in their own coordinate system, then this postulate of SR would be violated. So it's sort of a "proof by contradiction" to show that the two frames must each get the same value for the other frame's speed.

    But, as I said, if you have an accelerating (non-inertial) reference frame, the laws of physics do not have the same form in this frame's coordinate system as they would in an inertial frame's coordinate system, at least not in special relativity (in general relativity they might, I'm not sure).
     
    Last edited: Jan 7, 2005
  11. Jan 7, 2005 #10

    Phobos

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    If you measured a difference, then you may conclude that one reference frame is better (truer) than the other, which contradicts SR. For example, you may think that the motion in one frame is the proper/true motion and the other frame's motion is relative to that. But the laws of physics must be the same in both inertial reference frames.
     
  12. Jan 7, 2005 #11

    quasar987

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    That, I understand now. I know this is what Symon is saying. I just don't see how it is true that if two inertial reference frames did not get the same value for the other frame's speed in their own coordinate system, then the laws of physics would not have exactly the same form in both coordinate systems.

    For instance, there is no change in the law of inertia. If a body is at rest in A, it will be seen as moving with constant velocity w by B, and if (to complicate a little), an object moves with constant velocity u in B, then it will be seen in A as moving with constant velocity v + u. (following the notations introduced by jcsd in post #4)

    Then for what laws will we be able to compare the two coordinate system and say 'ah!, the laws governing this particular phenomenon are different' ?
     
  13. Jan 7, 2005 #12

    jcsd

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    Well as I said, when transforming properties between inertial frames the function must be dependt on velocity only.

    If you like the the only property that can be assigned to a frame (as we know alraedy we have set of laws that are invariant under rotation and translation) is it's relative velocity.
     
  14. Jan 7, 2005 #13

    JesseM

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    Hmm, this is actually kind of tricky. Let's consider a universe with one space and one time dimension, with one frame S that uses coordinates x and t, and another frame S' which uses coordinates x' and t'. If S sees S' moving at velocity v1 in its own frame, then to transform S-coordinates to S'-coordinates you need a tranformation like this:

    x' = f(x,t,v1)
    t' = g(x,t,v1)

    If S' sees S moving at velocity v2 in its own frame, then in order for the postulate of relativity to be true, the same functions should allow you to transform back from S'-coordinates to S-coordinates:

    x = f(x',t',v2)
    t = g(x',t',v2)

    In both Newtonian mechanics and relativity, it's true that if v1 = v, then v2 = -v. For example, here is the Galilei transformation:

    x' = x - v*t
    t' = t

    and

    x = x' - (-v)*t
    t = t'

    Here is the Lorentz transformation:

    x'=gamma(x - v*t)
    t'=gamma(t - v*x/c^2)
    where gamma = 1/squareroot(1 - v^2/c^2)

    and

    x=gamma(x' - (-v)*t')
    t=gamma(t - (-v)*x'/c^2)
    where gamma = 1/squareroot(1 - (-v)^2/c2)

    But suppose it was no longer true that v1 always equalled -v2. Suppose we had two frames S and S' such that the velocity of S' in the S-frame was v while the velocity of S in the S'-frame was -2v...would it be impossible to find two functions f and g that satisfied these relations?

    x' = f(x,t,v)
    t' = g(x,t,v)

    and

    x = f(x',t',-2v)
    t = g(x',t',-2v)

    Of course, you'd also need them to satisfy the additional restraint that the origin of each coordinate system, x=0 and x'=0, actually is moving at the correct velocity in the other coordinate system.

    It may be that there is some simple trick to show that it is impossible to find a pair of functions f and g that satisfy these requirements, but I don't see it offhand.
     
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