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Space Time

  1. Dec 27, 2011 #1
    Why in deriving invariant distance in space time we use Pythagorean Theorem with a negative sign?
  2. jcsd
  3. Dec 28, 2011 #2
    c^2 = a^2 + b^2 + t^2

    if we think of time as being imaginary we get

    c^2 = a^2 + b^2 - t^2
  4. Dec 28, 2011 #3


    Staff: Mentor

    Because it matches the results of physics experiments.

    Basically, if we take a good clock, and we calculate the "length" of the path that the clock takes through spacetime using the above formula then we find experimentally that the clock measures that "length" as the elapsed time on the clock.
  5. Dec 28, 2011 #4
    Like many things in physics, when one first come across it one thinks "How did he/she ever figure out THAT?" I don't think it's a simple intuition for most of us. Is the Pythagorean Theorem itself intuitive?? It didn't seem to be for several thousand years.

    My impression is that some really smart dudes played with various mathematical ideas and interpretations in the late 1800's and early 1900's trying to figure out descriptions(models) of light, ether, and Maxwell's equations for electromagnetic radiation...they all seemed mixed up together somehow.

    We don't usually hear about what are the many likely formulations that didn't work. Einstein sure didn't "invent" that formulation; he took what was some available mathematics and it was his physical insights that gave him the ability to use that math to describe observational results. He did the same thing in general relativity using Riemann [curvature] which a mathematician friend found for him)as a basis for explaining his physical insights.

    [Now that I think of it, that was the basis for an early formulation of the strong nuclear force....by Gabriele Veneziano....he noticed the Euler beta-function had some characteristics which matched physical observations. It worked but nobody knew why until
    string theory.]

    Here is how Sean Carroll explains the space time interval:


    For some more insights, check here:
    Last edited: Dec 28, 2011
  6. Dec 28, 2011 #5
    Try calculating [itex](\Delta x')^2+(c\Delta t')^2[/itex] in the unprimed frame with the Lorentz Transformation. Does it give you [itex](\Delta x)^2+(c\Delta t)^2[/itex]? No, it gives you something long and hairy. Now try doing the same thing with [itex](\Delta x')^2-(c\Delta t')^2[/itex]. This time you'll find that [itex](\Delta x')^2-(c\Delta t')^2=(\Delta x)^2-(c\Delta t)^2[/itex].
  7. Dec 28, 2011 #6
    By pure chance I came across a related historical note that might be of interest:


  8. Dec 28, 2011 #7
    The -c2 appended onto that term in that equation acts as a correction factor. The negative is there because time is imaginary. The c2 is there to act as a correction factor for the way we measure time. If we were to measure things very fundamentally, time and space would be measured in the same way. But we don't, so we must add the c2.

    See http://www.quora.com/Physics/How-di...arated-by-time-are-also-seperated-by-distance and read the comments on the answer.
    Last edited: Dec 28, 2011
  9. Dec 28, 2011 #8
    Here is another perspective:


  10. Dec 28, 2011 #9

    I thought it is negative because time has an inverse relationship with length.

    What does "time is imaginary" mean?

    It's not the way "we" measure it. It's the way it is measured. The physics of measuring time includes distance doesn't it?

    In any case we do measure time and space the same with intervals.
  11. Dec 28, 2011 #10
    Since time is unlike any other dimension, that equation calls it imaginary to make it clear that the time dimension is different than space dimensions. There are probably deeper reasons than that, though.

    Time is part of the distance between two events in space-time (the other parts being physical distance). We traditionally measure this distance with seconds, but it needs to be converted to meters for the space-time distance equation. The c*t in (ct)2 can be viewed sort of like velocity*time = distance.
  12. Dec 29, 2011 #11
    I see another poster used the term "imaginary". Is this used in math/physics? It seems worse then reffering to gravity as a "fictitious force". (seems like an oxymoron)

    Is there a less vague definition of "imaginary" then the context above?

    An which equation are you reffering too with the -c2?
  13. Dec 29, 2011 #12


    Staff: Mentor

    Imaginary just means [itex]\sqrt{-1}[/itex]. The imaginary time concept has been out of use for several decades now, similar to the concept of relativistic mass. Imaginary time works in inertial frames in SR, but doesn't generalize to non inertial frames or curved space times.
  14. Dec 29, 2011 #13
    Ha, is it the transition to understanding what equivalent means in the context of time = length, energy = mass.

    Why not just, "the other side of the coin" eh Dalespam :smile:

    "Imaginary time works in inertial frames in SR, but doesn't generalize to non inertial frames or curved space times."

    I don't know GR at all, does that statement mean equivalence isn't so simple in GR? Are there no inertial frames in GR (i interprut curved spacetime as acceleration of somesort)? Said differently does GR ruin this awsome symmetry thing? (specifically complicates it)

    How neat & tidy with; the "space" is time=length, the "stuff" is energy=mass. So there are two coins, space and stuff :smile:

    lol I actual played around with the square root of -1 to see what is meant. If I consider squared and square root litteraly (as in lengths of space). Speaking loosely, I can see the signs distinguish between (have the same relationship as) length and time. I haven't thought much of it yet, but a long shot from here it looks like there is no quanta of time, in the same sense as length. Perhaps in that sense it was termed "imaginary".
    Last edited: Dec 29, 2011
  15. Dec 29, 2011 #14
    Does the squared in c2 come into play because in velocity*time = distance, time is being used twice? Once in velocity and again in duration of velocity.

    I guess Im asking the why is c squared there.
    Last edited: Dec 29, 2011
  16. Dec 29, 2011 #15
    Instead of thinking of it like -c^2t^2 think of it as -(ct)^2. Here, you can see the c*t is referring to velocity*time, so we get a distance. So, just like the other terms in that equation, we need to square it.
  17. Dec 29, 2011 #16
    Maybe I need to ask it differently,

    Why does c have to be squared (why are the terms squared in the first place)?

    I'm pretty sure I've heard it has to do with the units (physical units I assume). Im just trying to piece that together.
  18. Dec 29, 2011 #17
    Nope, it's just the pythagorean theorem. The distance between any two events in a two-dimensional world is √(x2 + y2). In a three dimensional world with an "imaginary" time dimension being measured in seconds, it's √(x2 + y2 + z2 - (ct)2)
  19. Dec 29, 2011 #18
    Cool stuff, thanks guss.

    I still can't see it though.

    lets say the value for cordinate x is 3 and y is 5. What is the "function" of this part; (3+3+3)+(5+5+5+5+5).

    Asked differently,
    Why does the value of length along a dimension have to be squared in order to calculate the distance to an other value of length along a different dimension. (oohhh triangulating :redface:)

    Maybe if I play with right angle triangles looking for this it'd make it more clear.
    Last edited: Dec 29, 2011
  20. Dec 29, 2011 #19


    Staff: Mentor

    In introductory SR you use only inertial coordinate systems and generally simple algebra. Once you are a little more advanced you combine space and time into a single geometric structure, spacetime. There are several ways to do this.

    One is to use a Euclidean line element [itex]ds^2 = dx_0^2 + dx_1^2 + dx_2^2 + dx_3^2[/itex] with coordinates [itex](x_0,x_1,x_2,x_3)=(i c t, x, y, z)[/itex] where i is the imaginary number [itex]i=\sqrt{-1}[/itex]. This is the "imaginary time approach". It is nice because the coordinates all have units of length and it allows you to use the familiar Euclidean line element to measure distances.

    If you expand the line element then you can see immediately that it is equal to [itex]ds^2= -c^2 dt^2 + dx^2 + dy^2 + dz^2[/itex]. With this approach the coordinates can be stripped of their units and the units can be collected in an object called the metric.
    -c^2 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1
    So now the line element can be written in terms of the metric [itex]ds^2=g \cdot dx \cdot dx[/itex] where [itex]dx=(dx_0,dx_1,dx_2,dx_3)=(dt,dx,dy,dz)[/itex]. This seems a little more cumbersome, but it is clearly equivalent. It also gets rid of the imaginary time thing.

    However, at this point, there is no solid reason to prefer one approach over another, it is only once you go to non-inertial coordinate systems that the metric approach becomes truly useful. For instance, suppose that you wanted to use a rotating coordinate system. Now, the line element is: [itex]ds^2=-(1-r^2\omega^2)dt^2+2\omega r^2 dt d\theta + dr^2 + r^2 d\theta^2 + dz^2[/itex]. This fits easily into the metric approach with a metric of
    r^2 \omega ^2-1 & 0 & r^2 \omega & 0 \\
    0 & 1 & 0 & 0 \\
    r^2 \omega & 0 & r^2 & 0 \\
    0 & 0 & 0 & 1
    and coordinates of [itex](x_0,x_1,x_2,x_3)=(t,r,\theta,z)[/itex], but it cannot fit into the imaginary time approach without converting back to inertial coordinates.

    Once you add gravity it becomes impossible to find a set of coordinates which cover the entire space and are everywhere inertial, so in that case you are forced to abandon the imaginary time approach.

    That is what I meant about the imaginary time concept not generalizing.
    Last edited: Dec 29, 2011
  21. Dec 29, 2011 #20
    if you draw a spacetime diagram, with time (ct) with c being the speed of light "calibrating speed" on the y axis, and space (x) on the x axis then you can display this. Pythagorean theorem would lead to s2=x2+(ct)2. This gives a setup which gives (in the way of the axes) x2+y2=s2. This, when drawn on a spacetime diagram, gives a circle, which would mean that the same event could happen in both the past and present of another. This would obviously violate causality. To explain this, imagine an event 0, which is situated at the origin, and an event A which is s distance in spacetime away, this could be anywhere on the circle, so it could be, as i say both in past and present, depending on how you are relative to it. So the only other option is the minus sign in the pythagorean theorem, as you had a triangle, god knows why that works, it just does. It allows hyperbolic lines on the diagram instead of a circle, if you draw 4 lines which go through the origin and bisect the right angle between the axes, then you find that if you drew the hyperbolas they would track these 45 degree lines. If you drew the 4 hyperbola in between the lines, you would see that 1 is in the past only, one is in the present only, and 2 cross the space axis, so still create the problem of viloation of causality; however, simple y=mx+c shows use that the gradient of the 45 degree lines are 1, and therefore represent the speed c in space. If any line is shallower than this, then it is travelling faster than c. So, if nothing can travel faster than c, the 2 hyperbola that cross the spaces axes are invalid, as no event on them can effect the event at the origin. This proves that hyperbolic spacetime (minkowski space time) with the amended pythagorean formula just works.
    At the end of the day, if something works, it just does. This is just a theory remember, it could be that spacetime is some very odd geometry that we have no clue about, but for the moment this hyperbolic way seems viable and so we stick with it.
    Hope this gives some insight as to why the minus sign is used. :)
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