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Real examples of the space-time interval

  1. Mar 9, 2014 #1
    If the space-time interval is calculated according to:

    ds2=dx2+dy2+dz2-c2*dt2 what might be an example of 2 events in space-time separated by this interval?

    Suppose for example ds2 =1 (or a constant) and the first event was at the origin where would the second events lie?

    Am I looking at the (edge of ) the spacetime cone that I have sometimes heard talk of?

    What if the first event is not at the origin?

    Has any one a link where I could read about it in detail?

    I hope I have the very basic understanding of the situation more or less correct.
     
  2. jcsd
  3. Mar 9, 2014 #2

    tiny-tim

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    no, the surface of that cone is ds2 = 0

    ds2 = 1 is entirely outside the cone, in the "space-like" region

    it is a balloon which at time zero has radius r = 1 where (r2 = x2 + y2 + z2),

    and at general time t has radius √(1 + t2) :wink:
    ??:confused: just move the origin
     
  4. Mar 9, 2014 #3

    stevendaryl

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    An "event" is a point in spacetime. It is specified by giving a location and a time. Any two events have an associated interval separating them.

    So if the first event has coordinates [itex](x_1, y_1, z_1, t_1)[/itex] and the second event has coordinates [itex](x_2, y_2, z_2, t_2)[/itex], then we define:

    [itex]\delta x = x_2 - x_1[/itex]
    [itex]\delta y = y_2 - y_1[/itex]
    etc.

    Then the interval is given by:
    [itex]\delta s^2 = \delta x^2 + \delta y^2 + \delta z^2 - c^2 \delta t^2[/itex]

    A "light cone" is what you get in the special case in which you pick [itex](x_1, y_1, z_1, t_1)[/itex] and consider all possible events [itex](x_2, y_2, z_2, t_2)[/itex] such that [itex]\delta s^2 = 0[/itex]. These are the events that are reachable from [itex](x_1, y_1, z_1, t_1)[/itex] by a light signal traveling at constant speed c.
     
  5. Mar 9, 2014 #4

    Simon Bridge

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    I think you have become confused.

    You may have event 1 occurring at P1=(x1,y1,z1) at time t1, and event 2 happens at position P2=(x2,y2,z2) at time t2 ... these two events are separated in space by R=|P2-P1| and time by T=|t2-t1|. This is how you are used to doing things. It does not matter if you put the origin at P1 or P2 or somewhere else.

    If you specify P1 with T and R, then you know that P2 must appear somewhere on the sphere radius R in 3-space, and it must be at time t1+T or t1-T.

    This description treats time as some sort of backdrop that everything else happens against.

    The space-time interval is the 4D equivalent of the distance between two points in 3D.
    i.e. http://en.wikipedia.org/wiki/Spacetime#Spacetime_intervals

    So if you specify E1 = (0,0,0,0) and a space-time interval of r, then you know that E2 must appear on a 4D surface described by the space-time interval equation, centered on the origin of the 4D coordinate system. You can see how it works from an example...

    to simplify things, restrict space to the x axis so all coordinates are (x,ct)
    specify E1 at the origin, give a space-time interval of -1
    then E2 must be somewhere on the curve given by ##x^2-(ct)^2=1##light-second
    plot the curve that results.

    Try it again for an interval of -1 light-seconds (and 0 light-seconds).

    I'm guessing you have a pop-science understanding - which seems to have got garbled.
    I'll suggest the very accessible space-time FAQ: Relativity and FTL for a more firm grounding on the core concepts.

    It covers most of your questions.
     
  6. Mar 9, 2014 #5
    thanks for all your answers (and the link). It will take me a good while to digest them (I hope).

    Is there a concrete example of an actual space-time interval that has been measured ? (distances between the sun and the earth ?) Or do they only differ measurably from classical measurements when the 2 events are connected by relativistic transmissions (if that makes linguistic sense) or long periods of time?

    Do these measurements come into their own when used to measure distances in particle physics?
     
  7. Mar 9, 2014 #6

    Simon Bridge

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    It takes everyone a while to get use to it.

    Every separation is a space-time interval - our intrinsic sense of space and of time just give us different experiences for these things. Our senses are "mired in time"... something to do with the way memory works.

    It is only when we examine these experiences very carefully that the need to treat space and time as different aspects of a whole becomes apparent.

    The application of relativity to particle physics is an example of when careful examination shows us the need to treat space and time on an equal footing.

    But space-time intervals are not things you measure (directly) - there is no space-time "meter" to lay out between events. Instead you measure space (using rulers) and time (using clocks) and you can put these together to work out the space-time interval.

    The space-time interval is nice to use because it does not depend on the observer.
     
  8. Mar 9, 2014 #7
    I am just musing probably but if the 2 events were connected by a beam of light that took the longest possible route between the 2 events (in the absence of any other external gravitational influence) could that be considered to be some kind of a meter? (I realize I am parading my ignorance but wouldn't two particles bend /attract light in the absence of any other nearby massive bodies?)

    Would that theoretical (or wholly false ?) trajectory bear any relation to the spacetime interval?
     
  9. Mar 9, 2014 #8

    stevendaryl

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    Very true, but there is a physical interpretation: With [itex]\delta s^2[/itex] defined by

    [itex]\delta s^2 = \delta x^2 + \delta y^2 + \delta z^2 - c^2 \delta t^2[/itex]

    • If [itex]\delta s^2 > 0[/itex], then [itex]|\delta s|[/itex] is the spatial distance between the two events, as measured in a rest frame in which the two events are simultaneous.
    • If [itex]\delta s^2 < 0[/itex], then [itex]\dfrac{|\delta s|}{c}[/itex] is the time between the two events, as measured in a rest frame in which the two events take place at the same location (that is, the rest frame of an observer who travels at constant velocity between the events).
     
  10. Mar 9, 2014 #9

    Simon Bridge

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    Light always takes the shortest possible route.

    How?

    Note: as stevendaryl points out - if the two events take place in the same place, then the time between them is the space-time interval ... similarly, if they take place at the same time, then the separation between them is the space-time interval. In these special cases you can technically measure the interval ... however, in general, different observers will disagree about when and where events take place.

    In general relativity, space-time is curved by local energy densities - mass is very dense energy. Light travels in "straight lines" (called "null geodesics") along this curved space-time. Since we see space and time separately, it looks to us like the light sometimes follows a curve.

    Note:
    The space-time interval you showed above is for flat space-time - works for "locally flat" though.
    The deflection on the scale of a fundamental particle is too small to bother with anyway.

    Sure - the space-time interval would be zero for any events connected by a pulse of light.
     
    Last edited: Mar 9, 2014
  11. Mar 9, 2014 #10
    I know it seems silly (thanks for answering anyway) but my idea was that this beam (or trajectory) of light would resemble a very long , bendy piece of string -which is a bit like a physical measuring rod.

    If the light took the shortest route rather than the longest I don't think it would affect my "argument".

    Why would the deflection caused by a fundamental particle on a photon be any different (in the long run) to that caused by a black hole?

    If the 2 particles were the only influence wouldn't one of them eventually attract the photon (in theory) -or would the photon achieve escape velocity in short order (there are only 2 particles /events and 1 photon in my universe)?

    I realize this seems silly - I wish I could dream up more realistic unrealistic scenarios.
     
  12. Mar 9, 2014 #11

    Simon Bridge

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    The thing about the string is that it has a well defined 3D length that you can compare other distances to. What would you use for the length of, say, a laser beam? Remembering that in order to measure the space-time interval - this length must be invariant.

    You can use a laser beam as a 3D ruler, or as a clock for that matter. What you are trying to do is use it as a 4D ruler for space-time.

    There's no "if" about it. But you have yet to make an argument. I'm trying to get you to think about what a space-time interval means.

    i.e. any two events connected by the laser-beam would have a space-time interval of zero.
    That could be treated as a measurement of the space-time interval but I don't think that's what you wanted.

    Magnitude.
    Some effects are too small to be worth calculating.

    I don't understand why you need two particles fro this discussion - a laser beam aimed tangent to the surface of the Earth shows no measurable deflection due to the Earth so what chance does a fundamental particle have?
    The effect of electromagnetism is much stronger on that sort of scale.

    If the particles could gravitationally capture the photon, then their escape velocity would be higher than the speed of light - so the particles would be black holes.

    In a finite Universe, with an infinite amount of time, then I figure the photon would eventually get absorbed ... all else remaining equal.

    As diverting as this is, you will find yourself better equipped if you revise the fundamental concepts before engaging in more thought experiments along these lines.
    Right now I believe your original question has been answered.
    Enjoy.
     
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