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Minkowski spacetime

  1. Sep 13, 2008 #1
    Hello,
    I've had my first couple of lectures on general relativity. Actually, we started by talking about special relativity. We were taught the SR uses Minkowski spacetime and that the displacement (squared) between any two events is given as follows:
    ds^2 = (c^2)(dt^2) - dx^2 - dy^2 - dz^2

    Now, please correct wherever you guys spot I got something wrong in my understanding:
    1) The spatial coordinates have a minus sign because we want light to have a displacement of 0, and only light can have (c^2)(dt^2) equal to the sum of the spatial coordinates, and thus achieving a slope of 1 in a space vs time diagram.
    2) What does the displacement ds actually mean? If light has a displacement of 0, then what does it mean? Does it not move through spacetime :S ?
    3) The professor said that ds^2 is invariant under any coordinate change... can anyone explain why this is so?
    4) The professor also rapidly wrote this equation towards the end of the lecture but I really didn't understand where it came from and what it means... could someone tell me more or less what it is? The equation is: ds^2 = g(ab) dx(a) dx(b)
    a and b are bottom indices of g, and then they're upper indices for both dx's, so that there is a summation over both (Einstein convention).
    I believe g(ab) is something which he called the metric tensor... but he didn't really explain it that much, maybe next lecture. However I want to understand how this equation connects with the other one.

    I would appreciate any help you guys can give me. Take care.
     
  2. jcsd
  3. Sep 13, 2008 #2

    JesseM

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    ds is what is called the line element. In GR you are allowed to use absolutely any smooth coordinate system, but you have to specify the metric in that coordinate system, which gives you the line element at every point; then if you integrate the line element between pairs of events on any spacetime curve expressed in that coordinate system, it gives you c * the proper time along that curve between the pair of events you chose (i.e. if the proper time was 5 seconds, the integral of ds would be 5 light-seconds). For curves representing the paths of slower-than-light objects (what are called 'timelike curves'), the proper time represents the time elapsed on a physical clock that has that curve as its worldline, which of course is a physical quantity that should always be the same along a given curve regardless of what coordinate system you use (but in different coordinate systems on the same spacetime, the metric will be expressed differently, and thus the equations for the line element at each point will be different). For a curve representing a light beam the integral of ds should always be zero, and for a curve that would require an object to move FTL to have it as its worldline (what is called a 'spacelike curve'), ds^2 is negative, so the integral of ds is an imaginary number. But in all three cases the integral of ds along a given curve is a physical quantity that doesn't depend on your choice of coordinate system.

    For an inertial coordinate system in flat spacetime, the line element has the simple form that you wrote, ds^2 = c^2*dt^2 - dx^2 - dy^2 - dz^2. For a non-inertial coordinate system (whether in flat spacetime or curved spacetime) the line element will generally be more complicated. For example, if we're looking at the curved spacetime around a black hole, and we use Schwarzschild coordinates on this spacetime, then p. 116 of this book gives the line element as:

    [tex]ds^2 = (1 - \frac{2m}{r}) c^2 dt^2 - \frac{1}{(1 - \frac{2m}{r})} dr^2 - r^2 ( d\theta^2 + r^2 sin^2 \theta d\phi^2 )[/tex]
     
  4. Sep 13, 2008 #3

    Hurkyl

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    As is often the case, looking at special values of a new mathematical gadget can help build intuition!


    Suppose you are measuring some curve (or just a line segment) for which t is a constant. Then, on this curve, dt=0 and things simplify to
    ds² = -dx² - dy² - dz² = -(dx² + dy² + dz²)​

    And if we are measuring a line segment upon which (x,y,z) is a constant, things simplify to
    ds² = c² dt²​

    Do these give you any idea about how we might want to intuit ds?



    The value of ds² at a point on a curve is just a number; coordinate changes do not change the values of numbers. The major point is that the formula
    ds² = c² dt² -dx² - dy² - dz²​
    in certain coordinate charts. (specifically, those called 'inertial' in special relativity) In other coordinate charts, this formula won't hold.


    (Changing things slightly, so that we can do the arithmetic with matrix algebra) Basically, what he's saying that you can write this as a matrix equation:

    [tex]ds^2 = d\mathbf{x}^T G d\mathbf{x}[/tex]

    where, relative to our chosen coordinate basis, [itex]d\mathbf{x}[/itex] is the vector
    [tex]\left[
    \begin{array}{c} c dt \\ dx \\ dy \\ dz \end{array}
    \right] [/tex]

    If we further assume G is a symmetric matrix, can you solve that equation for G?



    I said I was changing things slightly: in reality, the coordinate representation of G should look more like 1x4 matrix whose entries are 1x4 matrices, and the products involved are a little more subtle.
     
    Last edited: Sep 13, 2008
  5. Sep 13, 2008 #4
    Thanks for your help guys (keep 'em coming! :P)
    So how can one understand that the line element of light is 0... I mean what does that mean? Light clearly changes position in spacetime... or does it not?
     
  6. Sep 13, 2008 #5
    But the Minkowski metric is not positive definite. In Euclidean space ds^2=0 if and only if distance between two points are zero and thus the points coincide. But certainly this is not so in Minkowski space, spacetime interval can be zero without having two events in spacetime coincide.
     
  7. Sep 13, 2008 #6

    Hurkyl

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    Isn't (1) an answer to (2)?
     
  8. Sep 13, 2008 #7
    what is ds? ds can be written this way: ds^2=dx^2+dy^2+dz^2-cdt^2.

    does that look familiar? how about this? c^2=a^2+b^2.


    if x y and z are nonzero and your time is zero then your ds will be real. thats spacelike. likewise any real ds is spacelike.

    if your x y and z are zero and your time is nonzero then your ds will be imaginary. thats timelike. likewise any imaginary ds is timelike.

    ds will always be either entirely real, entirely imaginary, or zero.
     
    Last edited: Sep 13, 2008
  9. Sep 13, 2008 #8
    I'm not sure this formula is meant to be applied to light. its applied to moving massive objects. massive objects can approach the speed of light but can never reach it.

    isnt ds whats called the proper distance (for spacelike events) and proper time (for timelike events)?

    http://en.wikipedia.org/wiki/Proper_distance
    http://en.wikipedia.org/wiki/Proper_time


    a ds of zero between 2 events means that they occur at the same place at the same time. but the formula breaks down for 2 events that require an object to be moving at the speed of light to got from one to the other.
     
    Last edited: Sep 13, 2008
  10. Sep 13, 2008 #9

    JesseM

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    As an analogy, student85 may want to consider the distance between points in the complex plane, using the normal pythagorean formula. For instance, take the distance between the origin and 2 + 2i. According to the pythagorean formula, the distance would be [tex]\sqrt{(2 - 0)^2 + (2i - 0)^2}[/tex], which of course is equal to [tex]\sqrt{4 - 4} = 0[/tex]
     
  11. Sep 13, 2008 #10
    I thought that in the complex plane the distance from 0 to 2 + 2i was [tex]\sqrt{(2 - 0)^2 + (2 - 0)^2}[/tex] which is [tex]\sqrt{8}[/tex]... because you do not actually take the i into account when calculating distances in the complex plane.
     
  12. Sep 13, 2008 #11
    I guess that would just depend on your metric. the point is that applying the ordinary Pythagorean theorem to a space in which one dimension is imaginary gives one a minkowski space. I think thats called hyperbolic space too.
     
  13. Sep 13, 2008 #12
    Not quite. I am still a little confused as to the meaning of ds. In euclidean space it means the distance between two points.
     
  14. Sep 13, 2008 #13
    So, the time dimension is imaginary?
     
  15. Sep 13, 2008 #14
    well if a particle could travel at the speed of light between 2 points then from the particles point of view (the particle considers itself to be stationary) the starting point and ending point are the same and zero time elapses. of course thats impossible but it does fit the equation. so the proper time and proper distance between the 2 events would, in fact, be zero.
     
    Last edited: Sep 13, 2008
  16. Sep 13, 2008 #15
    in this equation it is. in some equations inductance and mass are imaginary (both of which, interestingly enough, involve the concept of time through rate of change).
     
  17. Sep 13, 2008 #16

    JesseM

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    It's true that the modulus of a complex number is just equal to the distance from the origin under the regular Euclidean metric. But I don't think there's any requirement that you use any particular metric on the complex plane, I meant that if you define distance in terms of the sum of squares of a real-valued leg and an imaginary leg, the total distance between separate points can be zero.
     
  18. Sep 15, 2008 #17

    DrGreg

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    Given

    [tex]ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2[/tex]​

    1. If dt = 0 then [itex]\sqrt{-ds^2}[/itex] is just Euclidean distance (relative to the observer).

    2. If dx = dy = dz = 0 then [itex]\sqrt{+ds^2}/c[/itex] is just time measured by the observer.

    3. If ds = 0, you have the equation of a photon.

    4. If you perform a Lorentz transform, the answer you get for ds in the new coordinates is identical to the answer you got in the old coordinates. (ds is invariant.) This should be a simple exercise in algebra.

    5. Therefore, the answer to your question, given a curve in spacetime, is:

    5a. If [itex]ds^2 > 0[/itex] at every event along the curve, the curve is defined to be timelike, it is the worldline of a massive particle and [itex]\int \sqrt{+ds^2}/c[/itex] is the proper time of the particle. (The reasoning behind this is that at each event along the curve, you could perform a change of coordinates to get dx = dy = dz = 0 at that event, without changing ds.)

    5b. If [itex]ds^2 = 0[/itex] at every event along the curve, the curve is defined to be null (or lightlike), and it is the worldline of a photon.

    5c. If [itex]ds^2 < 0[/itex] at every event along the curve, the curve is defined to be spacelike and [itex]\int \sqrt{-ds^2}[/itex] is the proper length of the curve. (The reasoning behind this is that at each event along the curve, you could perform a change of coordinates to get dt = 0 at that event, without changing ds.)
     
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