What is a reference frame?

Discussion in 'Physics & Math' started by James R, Oct 13, 2004.

  1. James R Just this guy, you know? Staff Member

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    Since this question is causing all kinds of problems in a number of threads discussing relativity, I thought I would post an explanation of the concept, here.

    You can think of a reference frame as follows:

    Take an infinite number of rods of equal length, and an infinite number of identical clocks. Connect these up in a cubical lattice, so that each cube is constructed from 12 of the rods. At each corner of each cube, place one of the clocks.

    The resulting (imaginary) structure stretches over the whole universe.

    Now, suppose that each rod is L metres long. Follow the procedure below to synchronise the clocks.

    1. Designate 1 particular clock as the coordinate (x,y,z) = (0,0,0)

    The "nearest neighbour clocks" of this clock exist at the following coordinates:

    (L,0,0) , (0,L,0), (0,0,L), (-L,0,0), (0,-L,0), (0,0,-L).

    2. At time T on the (0,0,0) clock, send a light signal from that clock outwards in all directions.

    3. When the light signal is received by a clock at distance nL from the (0,0,0) clock, set the clock receiving the signal to read T + nL/c, where c is the speed of light.

    For example, when the light signal reaches the clock at (L,0,0), the time on that clock should be set to T + L/c. This is the same time as is displayed on the (0,0,0) clock at the time the (L,0,0) clock receives the signal, since the time taken for the light to travel distance L is L/c.

    4. Once the synchronisation procedure has been carried out, we have an infinite set of clocks spread throughout space, all of which display the same time.

    This system of synchronised clocks and rods is a reference frame.

    Suppose an explosion occurs somewhere and at some time in space. To specify the location of the explosion, we count the number of rods from the (0,0,0) clock to the position where the explosion occurred, in each of the three coordinate directions. That gives the spatial coordinates of the explosion (x,y,z). To get the time coordinate, t, of the explosion, we simply read off the time displayed on the clock sitting at the point where the explosion occurred.

    Thus, every event in spacetime has a unique set of spacetime coordinates (x,y,z,t).

    Some points to note:

    • A reference frame covers the whole of space and time.
    • All objects in spacetime exist in all reference frames.
    • The spacetime coordinates of a single event are unique for a particular reference frame.

    There are an infinite number of possible reference frames. For example, set up another lattice of clocks and rods, and have it move in the x direction at some speed v. This new set of clocks and rods needs a new (0,0,0) clock, which can be any chosen clock in the set. It also needs its own synchronisation. Coordinates in this new frame must be given new labels, such as (x',y',z',t').

    It is obvious that events which occur in one reference frame will, in general, have different coordinates (x,y,z,t) than the coordinates in some other reference frame (x',y',z',t').

    Often, we are concerned with how to translate coordinates from one reference frame to another.

    3 examples are given. Coordinates in frame 1 are given without the prime (') symbol. Coordinates in frame 2 have the prime (') symbol.

    Example 1 - rotated coordinate system at rest

    Reference frame 1 is set up.
    Reference frame 2 is chosen such that the (0,0,0)' clock is at exactly the same location as the (0,0,0) clock. The z' direction is chosen to be the same as the z direction. The x' direction is chosen to be rotated by an angle of q from the x axis.

    The coordinate tranformation is:

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

    Example 2 - Galilean tranformation

    Reference frame 1 is set up.
    Reference frame 2 is chosen so that, in frame 1, the (0,0,0)' clock is located at position (vt,0,0) at any time t. In other words, frame 2 moves in the positive x direction at speed v relative to frame 1, and is not rotated in any way relative to frame 1. Universal time is assumed.

    This results in Galilean transformation:

    x' = x - vt
    y' = y
    z' = z
    t ' = t

    Example 3 - Lorentz transformation

    Reference frame 1 is set up.
    Reference frame 2 is chosen so that, in frame 1, the (0,0,0)' clock is located at position (vt,0,0) at any time t. In other words, frame 2 moves in the positive x direction at speed v relative to frame 1, and is not rotated in any way relative to frame 1. The postulates of Special Relativity are assumed.

    This results in Lorentz transformation:

    x' = g(x - vt)
    y' = y
    z' = z
    t ' = g(t - xv/c<sup>2</sup>)

    where g is the Lorentz factor 1/sqrt(1-(v/c)<sup>2</sup>) and c is the speed of light.
     
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  3. Pete It's not rocket surgery Moderator

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    Thank you James, that's helped firm up a couple of things I was shaky on.
     
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  5. Pete It's not rocket surgery Moderator

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    Example 4 - translated coordinate system at rest

    Reference frame 1 is set up.
    A clock (a, b, c) is chosen in Reference frame 1.
    A time offset (d) is chosen.
    Reference frame 2 is chosen such that the (0,0,0)' clock is at exactly the same location as the (a,b,c) clock.
    The time of the Reference frame 2 clock is set to the time of the Reference frame 1 (a,b,c) clock plus the time offset d.

    The coordinate tranformation is:

    x' = x + a
    y' = y + b
    z' = z + c
    t' = t + d
     
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  7. Pete It's not rocket surgery Moderator

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    Example 5 - reflected coordinate system at rest

    Reference frame 1 is set up.
    Reference frame 2 is chosen such that the (0,0,0)' clock is at exactly the same location as the (0,0,0) clock, and is synchronized with it.
    The y' direction is chosen to be the same as the y direction.
    The z' direction is chosen to be the same as the z direction.
    The x' direction is chosen to be in the opposite direction to the x direction.
    Reference frame 2 is chosen such that the (0,0,0)' clock is at exactly the same location as the (a,b,c) clock.

    The coordinate tranformation is:

    x' = -x
    y' = y
    z' = z
    t' = t
     
  8. Pete It's not rocket surgery Moderator

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    How about a shear transform - is that at all useful in this context?
     
  9. geistkiesel Valued Senior Member

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    2,471

    James R; I understand your scheme ,but I observe no practical structure in your model, no physiocal reality and as an application of a pure mathematical-geometrical abstraction. Why not define real inertial frames in real situations, where the clocks are real and doing real "clocking" and where concepts lik eequivalence of frames can be defined, where "inertial' has a physical meaning? such as:

    Physical attributes* of an inertial frame of reference, the embankment, Ve:
    1. Mass (grams): 5 x 10^27.
    2. Volume (km3) = 1.09 x 10^12
    3. Surface area (km^2) = 5.11 x 10^8.
    4. circumference (km) = 40047.
    5. radius km 6378
    6. escape velocity km/sec = 11.2.
    7. Gravity acceleration (cm/sec2) = 980
    8. earth direction vector rate of change = 10^-8 degrees/sec.
    9. Velocity components (km/sec):
      • rotation. = .496
      • orbital = 29.8
      • solar (galactic) 208
      • earth/sun radiusk =m 1.49 x 10^8
      • Sun distance moved in one year km 208x31472262 = 6,546 X 10 ^9
      • angle subtended by earth/sun radius and sun distance covered /year = tan^-1 1.49x10 ^8/6.546x10^9 = 1.303 degrees.
    10. There are no measured affects on Ve due to the intrinsic motion of Ve.
    11. The prediction of position for past, presdent and future surface and volume coordinates is as exact as conditions demand.
    12. The Ve inertial mass overwhelmingly exceeds that of any other frame(s) , combined, in inertial mass, this alone being sufficient to grant Ve official “preferred frame of reference status”, whether liked or not.
    13. All Ve surface borne frames of reference inherit all the attributes of Ve motion and,
    14. From the foregoing all Ve – Vn relative measurements necessarily require Vn to accelerate to acquire Vn > 0 and hence any assumption that Vn = 0 with respect to Ve thereafter is a proved physical impossibility, so why give it mathematical equivalence?
    15. all Ve based references frames Vn, for all n, all necessarily suffer acceleration in order to create a state of relative velocity > 0.
    16. Vn can only claim Vn = 0 under the conditions that Ve - Vn = 0.
    17. Ve never acquires a velocity other than the known motion discussed above, which mdeasurable a straight-line trajectory.
    18. The Ve trajectory is moving measurably in a staight line and is affectivley inclusive in Newton's law of motion that an object at rest remains so; an object moving in a straight line continues to do so, eternallly. This includes, of course, photons of light
    19. the equivalence of inertial frames is mostly a myth.

    *Handbook of Astrionautical Engineering Volume 1 Chapter 2
     
  10. James R Just this guy, you know? Staff Member

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    geistkiesel:

    That's right.

    The word "inertial" has a very specific meaning already.

    The reason we don't tie the concept of a reference frame to particular physical systems such as the Earth is that doing so adds nothing useful to the concept, and in fact detracts from its usefulness. It distracts unnecessarily from what is important in particular problems.

    William of Occam said, famously, "Entities should not be multiplied unnecessarily."

    If we need to use characteristics of the Earth or the solar system, we can use them. But for most problems involving reference frames, these characteristics are irrelevant. And in fact, introducing them goes against the philosophy of relativity, in particular, which holds that no particular reference frame has preferential status.
     
  11. geistkiesel Valued Senior Member

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    You, of course, repeating other statements then that SR is purely mathematical and imaginary and has absolutelyh , has nothing to do with iron, aluminium, water, wood, smashing pumpkins and people counting crows.

    Then you are comitting an atrocious act of misleading the public with your assertion that SR tells it like it is, reality, I mean. You speak with forked tongue, white man.
     
  12. guthrie paradox generator Registered Senior Member

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    What benefit does defining "real" places and times give you?
     
  13. geistkiesel Valued Senior Member

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    Somethings, some real things, that aren't imaginary.

    What do you get ouit of rejecting reality for the imaginary things? Too much for you to handle, the real stuff, I mean?
     
  14. James R Just this guy, you know? Staff Member

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    Are you jumping on the MacM band wagon, geistkiesel? That only your own prejudices are real, and everything else is imaginary?

    I've previously explained to you the definition of velocity as being the time derivative of displacement, and acceleration as being the second derivative. A person on a train sees the train tracks move relative to the train, which in turn means the tracks have a velocity, and possible acceleration. That's not imaginary. The tracks actually do change their position relative to the train. To deny that is the real denial of reality.
     
  15. MacM Registered Senior Member

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    As in your twins arguement however, do not the tracks lack the history of GR due to acceleration forces? So acceleration is not part and parcel of such exchangability as you claim as a second derivative of time and distance.
     
  16. James R Just this guy, you know? Staff Member

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    Sorry, MacM. I don't understand the point you are trying to make.
     
  17. MacM Registered Senior Member

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    You have referred to velocity being the 1st derivative and acceleration being the 2nd derivative of time and distance change. You also referred to the fact that there is reciprocity in the velocity between the tracks and the moving train observer.

    My point was to clarify the fact the the train observer has gone through acceleration and the tracks didn't. Even though the tracks see a relativity velocity change it is not physically reciprocal to the train observer in that there is no force of accelertion.

    It is not true then that the observed track velocity by the train observer is directly applicabable to the velocity observed by the observer along side the tracks of the moving train.

    It is an identical circumstance to the arguement made in the twins paradox case. The two observers do not have a totally symmetrical history.
     
  18. James R Just this guy, you know? Staff Member

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    MacM:

    It is not correct to say that only the train accelerates. In the train's frame, the track accelerates.

    What I assume you are saying, to put it in technical terms, is that only the track's frame is inertial. The train frame is non-inertial.

    In that case, we appear to be in agreement. I am still not sure why you raised the issue, though.
     
  19. geistkiesel Valued Senior Member

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    The train OBSERBVER that considers the earth moving is an idiot. You knOw and everybody knows, the ground doesn't accelerate. Only Vn accelerates and it only this acceleration that we observe relative motion between Ve and al other Vn.
    to the train observefr the ground accelerates>

    Did you ever try dropping a golf ball from a train that is considered at rest with respect to the Ve? Which way will the ball bounce, assuming sufficient fricitonal contact with a relatively smooth/even pavement below the moving train? Whic way will the ball bouince?
     
  20. MacM Registered Senior Member

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    This long effort to expose Relativity is really starting to pay off by golly. Now we can see how you pick and choose when to say certain things.

    Historically you have used the force of acceleration against me when I stated that B would see A accelerate etc. You made it clear that A never actually accelerates in that it lacks any non-inertial forces.

    Now you want to blow off my using that same arguement. Make up your mind. Is your memory failing or do you just deliberately say whatever sounds professional at the moment when you need to redirect a discussion?.

    Always with trying to be the guy making corrections or clarifications. Well none were needed here. Thank you.

    Well, that was a long way around to being in agreement. It is really that tough for you to just say "You are right MacM?"

    I raised the issue since you tried to blow off QQ by saying the track acccelerates. It doesn't. Don't repeat what I have already said and that is that the train observer sees a changing velocity but that the track isn't really accelerating because it is inertial and experiences no non-inertial forces.
     
  21. Persol I am the great and mighty Zo. Registered Senior Member

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    Oh come on now. At least try.

    The ball bouncing in the direction the ground is moving has nothing to do with the ground being at absolute zero and the train having velocity.

    If you throw a golf ball at the train it will bounce in the direction the train is moving. This has nothing at all to do with absolute zero.

    And most of all, the ground IS moving. Even if you feel like believing the Earth is the center of the universe, it is still revolving.
     
  22. James R Just this guy, you know? Staff Member

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    MacM,

    You seem to be confused.

    I have no idea what this "A" and "B" are that you are talking about, since there have been no references to that in the thread, I think. I assume that by "B" you mean the train, and "A" the tracks.

    You have also contradicted yourself.

    First you say:

    You're making unsupported assertions about what I have said in the past, here, but never mind. The crux of this is that you claim that the train sees the tracks accelerate, which is true.

    But then you flip-flop and say:

    You really need to make up your mind about what you believe. Either you believe the track accelerates in the train's reference frame, or you don't.

    If you don't, you're just wrong. If you do, we have no argument.

    I don't really care either way. Why don't you go away and think about it for yourself?
     
  23. MacM Registered Senior Member

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    Afraid not. It is you that seem confused and forgetful.

    No. I said "Historically". Historically I had made reference to the fact that the moving observer in one frame when considered at rest see the other observer moving and also that each saw the other accelerate.

    You correctly commented that the other didn't actually accelerte that one remained inertial whie the orther actually accelerted which was accompanied by the inherent forces of acceleration.

    No I didn't. I said each sees a change in velocity as acceleration but clarifeid that only one actuallyaccelerates since one lacks the forces of accelertion.

    False. I said more technically correct than you have implied to QQ, that while the train sees the tracks undergo a change in velocity, that it isn't actually acceleration because it remained inertial.

    My mind hasn't changed. Your hasn't changed I'm sure but your presentation has changed claims from being acceleration to (historically, not in this thread) not being acceleration because it remained interial.

    It sounds like we have an arguement. I believe you have just said the tracks accelerate. Is that correct? If so where is the accelerating force on the tracks.?

    I'm sure you could use the rest. You seem to be getting wishy washy a sign of burn out, saying so many things to try and make a point that you are forgetting what you have said before.
     

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