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Why is the Time component in the Space Time Interval negative?

  1. Jul 13, 2011 #1
    Why is the Time component in the Space Time Interval negative????

    The space time interval is defined to be:

    ds^2=dt^2-dx^2-dy^2-dz^2

    or depending on the convention used it may also be:

    ds^2=-dt+dx^2+dy^2+dz^2

    The equation is defining distance using the pythagorean theorem. This results in the flat spacetime metric being defined as:

    -1000
    0100
    0010
    0001

    If the distance between two points in cartesian coordinates is given by:

    d^2=(x-x2)^2+(y-y2)^2+(z-z2)^2

    Then why is the spacetime interval not defined to be:

    d^2=(t-t2)^2+(x-x2)^2+(y-y2)^2+(z-z2)^2

    and the metric would then be:

    1000
    0100
    0010
    0001

    By the same logic???

    Any help is appreciated.

    Thanks
     
  2. jcsd
  3. Jul 13, 2011 #2
    Re: Why is the Time component in the Space Time Interval negative????

    A. Einstein took a great interest in Maxwell`s equations. He (apparently) was puzzled as to why the speed of light in vacuum was fixed, regardless of the speed of the source of the light.

    In particular, if the source moves with speed `v` then (according to Galileo) the light should move with speed c±v.

    So instead of using the motion of material bodies as central, he thought of using the motion of light (in vacuum, as in empty space) as central.

    Now whether it was he, or one (or more) of his mathematician friends, the characteristic surfaces of the 3D wave operator were used to define a line element/metric. And that`s why the metric appears as it is, and not how you suggest.

    Addendum:

    See for example, chapter VIII of "Introduction to partial differential equations with applications" By E. C. Zachmanoglou, Dale W. Thoe (@Google books).
     
    Last edited: Jul 13, 2011
  4. Jul 13, 2011 #3

    WannabeNewton

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    Re: Why is the Time component in the Space Time Interval negative????

    Well I don't know if this is the ONLY reason but consider a photon: a photon moves on null geodesics such that ds^2 = -c^2dt^2 + dx^2 = 0 so dx^2 / dt^2 = c^2 and in this way all observers agree on the speed of light.
     
  5. Jul 13, 2011 #4

    atyy

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    Science Advisor

    Re: Why is the Time component in the Space Time Interval negative????

    It isn't based on logic. It's based on experiment. This property makes the speed of light the same in every inertial frame. This is why the metric is called "pseudo-Riemannian".
     
  6. Jul 13, 2011 #5
    Re: Why is the Time component in the Space Time Interval negative????

    You must mean a worldline (?)
     
  7. Jul 13, 2011 #6
    Re: Why is the Time component in the Space Time Interval negative????

     
  8. Jul 13, 2011 #7

    WannabeNewton

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    Science Advisor

    Re: Why is the Time component in the Space Time Interval negative????

    Pardon? Don't quite understand your question. A photon moves on a null geodesic such that the tangent vector to the geodesic has zero norm, in accordance with events being null - like separated along a photon's trajectory.
     
  9. Jul 13, 2011 #8
    Re: Why is the Time component in the Space Time Interval negative????

    Would all observers agree on the speed of light because of the invariance of spacetime distance between events as observed in different reference frames? (with the null geodesic as you suggest)
     
  10. Jul 13, 2011 #9
    Re: Why is the Time component in the Space Time Interval negative????

    Isn`t that GR (?)
     
  11. Jul 13, 2011 #10
    Re: Why is the Time component in the Space Time Interval negative????

    Look at like this. The spacetime interval using the the +--- convention is ds^2= (cdt)^2-dx^2-dy^2-dz^2. ds^2 is actually c^2 dtau^2 where tau is the proper time. By dividing both sides by c^2 you get dtau^2 = t^2- (dx^2-dy^2-dz^2)/c^2. For a particle travelling at the speed of light, the proper time tau is always zero so if dx^2, dy^2 or dz^2 are non zero positive real numbers, dt has to be equal and opposite in order for dtau to be zero. If you had a ++++ metric any non zero spatial displacement (squared) would require dt to be an imaginary number in order for dtau to be zero. The +--- or -+++ formalisms are not just conventions but dictated by nature. You could also write c^2 = ((cdt)^2-dx^2-dy^2-dz^2)/dtau^2 and note that c is a constant of nature. If we had a +++ metric any change in the spatial displacement would would require imaginary time in order for c to be constant. It is nature that decides c is a constant and that is just the way things are.

    Of course you could get around all this by writing the metric as ds^2 = (ic dt)^2+dx^2+dy^2+dz^2 where i is the imaginary number to force a ++++ metric and indeed in the past some people have done this, but it is not generally used these days probably because it would be confusing.
     
    Last edited: Jul 14, 2011
  12. Jul 14, 2011 #11
    Re: Why is the Time component in the Space Time Interval negative????

    A final thought. If the spacetime interval had signature (++++) then what would ds²=0 represent?
     
  13. Jul 14, 2011 #12

    JDoolin

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    Gold Member

    Re: Why is the Time component in the Space Time Interval negative????

    It has something to do with the difference between the trigonometric functions and the hyperbolic trig functions.

    Solutions of dy/dx=-x/y lead to x^2 + y^2 = Constant. These are circles.

    Solutions of dx/dt = t/x lead to t^2 - x^2 = Constant. These are hyperbolas.

    One big difference is that if you have something in front of you, you can turn around and make it to your left, then behind you, then to your right, then in front of you again. It all forms one continuos circle.

    On the other hand, you can't take an event that's in your future, and and turn until it is to your left, then in your past, then to your right, then in your future again. They are four separate hyperbolas.
     
  14. Jul 14, 2011 #13
    Re: Why is the Time component in the Space Time Interval negative????

    With the ++++ signature I suppose ds^2=0 could only represent no change in space or time coordinates. Would this be the metric without considering special relativity?

    I know that we work in units where c=1, so the null interval c^2*dt^2=dx^2+dy^2+dz^2 becomes dt^2=dx^2+dy^2+dz^2 This is the distance travelled for a light particle because it is effectively saying ct=x. So a spacetime interval could be >0 ("timelike") if the particle is slower than light. If the interval is "spacelike" it is <0. Would this imply faster than light travel for the particle? (Because -c^2dt^2+dx^2+dy^2+dz^2)? If so can an interval be less than zero? Perhaps it can but the event must be casually unrelated. It seems the metric signature is this way (-+++) in order to uphold the celestial speed limit of C.
     
  15. Jul 14, 2011 #14

    Dale

    Staff: Mentor

    Re: Why is the Time component in the Space Time Interval negative????

    One of the very important things about these metrics is the class of transformations that leave them unchanged. E.g. using the Cartesian coordinates rotations and translations do not change the distance between two points.

    Now, from experiment we know that the speed of light is unchanged by boosts, so we want to find a metric which preserves the speed of light. The equation for a spherical pulse of light is:
    [tex]c^2 dt^2 = dx^2 + dy^2 + dx^2[/tex]
    [tex]0= -c^2 dt^2 + dx^2 + dy^2 + dx^2[/tex]

    So any transformation which preserves the Minkowski metric also preserves the speed of light but the same is not true for the Euclidean metric.
     
  16. Jul 14, 2011 #15
    Re: Why is the Time component in the Space Time Interval negative????

    The diagram at the left represents a guy in a red rocket moving to the left (with respect to the black rest system) at relativistic speed and another blue guy moving to the right with the same speed. Special relativity tells us that the two will experience different cross-section views of the 4-dimensional universe. Accordingly, when the red guy is at station 9 along his X4 world line his 3-D cross-section view of the 4D universe shows the blue guy to be at station 8. But, when the blue guy is at station 9 his 3-D cross-section view of the 4D universe shows the red guy to be at station 8. Each sees the other guys clock running slow.

    Now, using the triangular relation diagrammed on the right, you can derive the metric beginning with the actual Pythagorean Theorem as your intuition seems to prefer. We simply recognize distances layed out in the 4-dimensional space. Look at the right triangle formed by the X4 coordinate of the red guy's system and the X1 and X4 coordinates of the blue guy's system. Just write down the Pythagorean equation directly for the hypotenuse and sides of the triangle. The blue X4 coordinate is the hypotenuse in our example. But, you can now solve the equation for just the red guy's X4 coordinate if you wish. And this is the form for the metric. Nevertheless, the Pythagorean Theorem still works for the spatial distances manifest in the diagram.

    SR_Coordinates_PythagoreanTheorem.jpg
     
    Last edited: Jul 14, 2011
  17. Jul 14, 2011 #16

    JDoolin

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    Gold Member

    Re: Why is the Time component in the Space Time Interval negative????

    If you know
    (1) how to multiply matrices
    (2) the Maclaurin series expansion for e^x, sin(x), cos(x), sinh(x), and cosh(x)
    then you can verify the following:


    [tex]{K1,K2,K3}= \left( \begin{array}{cccc} 0 & \beta_x & 0 & 0 \\ \beta_x & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \left( \begin{array}{cccc} 0 & 0 & \beta_y & 0 \\ 0 & 0 & 0 & 0 \\ \beta_y & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \left( \begin{array}{cccc} 0 & 0 & 0 & \beta_z \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \beta_z & 0 & 0 & 0 \end{array} \right)[/tex]

    [tex]{S1,S2,S3}= \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -\theta_1 \\ 0 & 0 & \theta_1 & 0 \end{array} \right) \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \theta_2 \\ 0 & 0 & 0 & 0 \\ 0 & -\theta_2 & 0 & 0 \end{array} \right) \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & -\theta_3 & 0 \\ 0 & \theta_3 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right)[/tex]


    Then [itex]e^{K1}, e^{K2}, e^{K3}[/itex] evaluate to the Lorentz Tranformations; i.e. accelerations in the x, y, z direction. and [itex]e^{S1}, e^{S2}, e^{S3}[/itex] evaluate to rotations in the x, y, z planes.

    You notice there is one negative sign in the S1, S2, S3 definitions, but there is no negative sign in the Boost1, Boost2, Boost3 definitions.

    As an example note that
    [tex]
    \left (\begin{matrix} 0 & \theta \\ -\theta & 0 \end{matrix} \right ) ^2
    =
    \left (\begin{matrix} 0 & \theta \\ -\theta & 0 \end{matrix} \right ) \left (\begin{matrix} 0 & \theta \\ -\theta & 0 \end{matrix} \right )=

    \begin{pmatrix} -\theta^2 & 0\\ 0 & -\theta^2 \end{pmatrix}
    [/tex]

    and
    [tex]\left (\begin{matrix} 0 & \theta \\ -\theta & 0 \end{matrix} \right ) ^3 = \begin{pmatrix} -\theta^2 & 0\\ 0 & -\theta^2 \end{pmatrix} \left (\begin{matrix} 0 & \theta \\ -\theta & 0 \end{matrix} \right ) = \begin{pmatrix} 0 & -\theta ^3\\ \theta^3 & 0 \end{pmatrix}[/tex]



    Taking e to the power of the matrix

    [tex]e^\left (\begin{matrix} 0 & \theta \\ -\theta & 0 \end{matrix} \right )[/tex]

    The first seven terms of the Maclaurin series expansion yields:

    [tex]
    \begin{pmatrix}
    (1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!} )
    &
    (\theta - \frac{\theta^3}{3!}+\frac{\theta^5}{5!} )
    \\
    ( -\theta + \frac{\theta^3}{3!}-\frac{\theta^5}{5!} )
    & (1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!} )
    \end{pmatrix}[/tex]

    and you can see this is leading to:


    [tex]
    \begin{pmatrix}
    \cos(\theta) & sin(\theta) \\
    -\sin(\theta) & cos(\theta)
    \end{pmatrix}
    [/tex]

    This final matrix is
    \begin{pmatrix}
    \frac{\Delta \vec u_{x'}}{\Delta \vec u_{x}} &
    \frac{\Delta \vec u_{x'}}{\Delta \vec u_{y}} \\
    \frac{\Delta \vec u_{y'}}{\Delta \vec u_{x}} &
    \frac{\Delta \vec u_{y'}}{\Delta \vec u_{y}}
    \end{pmatrix}

    Which relates the change in the unit vector in the pre-rotation coordinates, to the unit vectors in the post-rotation coordinates.

    For instance, if

    [tex]\vec u_x \overset {def} =
    \begin{pmatrix}
    1\\ 0
    \end{pmatrix}
    [/tex]

    and the rotation maps that point

    [tex]\begin{pmatrix}
    1\\ 0
    \end{pmatrix}\rightarrow \begin{pmatrix}
    cos(\theta)\\-sin(\theta)

    \end{pmatrix}[/tex]

    then

    [tex]\frac{\Delta \vec u_{x'}}{\Delta \vec u_{x}}=cos(\theta)[/tex]

    and

    [tex]\frac{\Delta \vec u_{y'}}{\Delta \vec u_{x}}=-sin(\theta)[/tex]

    There are several bits and pieces of this that I haven't worked out yet. For instance, why is it that

    [tex]\begin{pmatrix}
    \frac{\Delta \vec u_{x'}}{\Delta \vec u_{x}} &
    \frac{\Delta \vec u_{x'}}{\Delta \vec u_{y}} \\
    \frac{\Delta \vec u_{y'}}{\Delta \vec u_{x}} &
    \frac{\Delta \vec u_{y'}}{\Delta \vec u_{y}}
    \end{pmatrix} = e^{S1} [/tex]

    etc. I mean I can see that it is true, but I can't see why it is true. (Edit: More specifically, what would make someone think to take e to the power of a matrix. I guess it wouldn't have to be any more complicated than somebody saying "hey, I noticed these two things were equal." )

    (This discussion extends from https://www.physicsforums.com/showthread.php?t=430956 and specifically Jackson's Classical Electrodynamics.)
     
    Last edited: Jul 14, 2011
  18. Jul 14, 2011 #17
    Re: Why is the Time component in the Space Time Interval negative????

    YES. The spacelike interval implies c<v. SR does not allow the possibility of a material particle being accelerated from v<c to v>c. However, it does allow the possibility of particles moving at FIXED v≥c.

    "It seems the metric signature is this way (-+++) in order to uphold the celestial speed limit of C."

    Also, (+---).
     
  19. Jul 15, 2011 #18
    Re: Why is the Time component in the Space Time Interval negative????

    (ct)2 = x2+y2+z2

    Let's ignore the z-axis to make it simpler, work in 2d+time ...

    (ct)2 = x2+y2

    Since no length contractions exist wrt axes orthoganal wrt the axis of motion (motion is along +x here), then y=Y where Y is of the other system. But Y=cTau, and so y=cTau. So ...

    (ct)2 = x2+(cTau)2

    and so ...

    (cTau)2 = (ct)2-x2

    or equivalently, after multiplying by -1 ...

    -(cTau)2 = -(ct)2+x2

    Note that this equation has Tau of one system left of the equal sign, and x & t of the other system right of the equal sign. So it defines Tau as a function of the other system x,t.

    In imaginary systems, i2 = -1. So the above may be rewritten as ...

    (icTau)2 = -(ct)2+x2

    and since the length of the spacetime interval per Minkowski is s = icTau, then since (icTau)2=s2, the above becomes ...

    s2 = -(ct)2+x2

    More generically ...

    s2 = -(ct)2+x2+y2+z2

    So s is a length of the X,Y,Z,Tau system (numerically equal to time Tau), but is derived here in terms of the other system x,y,z,t. This is why the metric can vary from the traditional d = ++++ and instead become s = -+++ .

    GrayGhost
     
  20. Jul 15, 2011 #19
    Re: Why is the Time component in the Space Time Interval negative????

     
  21. Jul 15, 2011 #20

    JDoolin

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    Gold Member

    Re: Why is the Time component in the Space Time Interval negative????

    Another way of looking at it is that when you are talking [itex](x,y)[/itex] coordinate distance, you can imagine concentric circles around either particle and ask, what concentric circle the other particle is on. The answer involves a [itex]r=\sqrt{x^2 + y^2}[/itex]

    With [itex](x,t)[/itex] event distances, you can imagine "concentric" hyperbolas around one of the events and ask which hyperbola the other event is on. The answer involves a [itex]\tau = \sqrt{t^2-x^2}[/itex] or [itex]s=\sqrt{x^2-t^2}[/itex] (depending on which quadrant you're in.)
     

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