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I appreciate if this question has been posited before and easily findable by Google searching, but as of yet I haven't found anything to answer this. I'm sure I'm making an incorrect assumption in the fact that if the distance formula in Euclidean space is merely finding the Pythagorean distance by considering the hypotenuse made by each extension of an object in $x,y,z$ in space, that I'd expect, if the spacetime interval had indeed this analog to invariant "distance" in special relativity, it would be Pythagorean too, but it is (assuming the events in question are only in the spatial dimensions $x$ and $t$):

$$(\Delta s)^2 = (c\Delta t)^2 - (\Delta x)^2$$

Now, I'm sure that this is a much more nuanced thing that may not have much to do with the Pythagorean theorem, but it's cannily close to it, and since I've been told the spacetime interval is fairly close to an analog for Pythagorean distance in that of Minkowski space, I'd think there would be some key difference, as it is eerily close.

I suppose my question is: what is the intuition behind this formula, and why is it not Pythagorean? I was looking through some similar questions notably this one, but I'm having trouble still getting my head around why the signature of the metric is negative.

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    $\begingroup$ I'm sorry but how is this not a duplicate of the question you link yourself? $\endgroup$ Commented Nov 28, 2017 at 17:58
  • $\begingroup$ Not exactly, I think. His question has to do with what the geometric interpretation of the equation is, and mine is more acutely to do with the signature of the metric itself. It may be asking for a geometric interpretation to guide it, but any answers I've looked through haven't helped with my confusion so far, partially due to the fact that my knowledge of the subject so far is very new and understanding the answers are proving difficult. $\endgroup$
    – sangstar
    Commented Nov 28, 2017 at 18:01

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The quantity "$s^2$" (the squared-interval) is the analog of the square of the radius of a circle. For fixed s, $$s^2=(ct)^2-x^2$$ is the set of events that are "equidistant from the origin"... in the sense that a wristwatch on an inertial observer will elapse $s$ ticks from origin to this set of events. This curve is a hyperbola, and it plays the role of the circle (as a curve of constant separation from the origin).

Although there is a minus sign there, it's pythagorean in the sense that it's referencing two vectors that are "perpendicular" in that geometry. (Their dot-product in that geometry is zero.) The best way to appreciate "perpendicular" is that: when the "radius vector" meets the "circle", the "tangent to the circle" is "perpendicular to the radius".

Play around with my simulation (where time runs to the right)
https://www.desmos.com/calculator/awgqxtkqcc

In the way I think about things, I prefer to change units so that I have, effectively, $$s^2=t^2-(y/c)^2,$$ where $c$ is a convenient conversion constant.

Try adjusting the E-slider: E=1 is Minkowski, E=0 is Galilean, and E=-1 is Euclidean in $$s^2=t^2-E(y/c)^2.$$

enter image description here

By the way, in the Minkowski case there are still Pythagorean triples... but not in the places a Euclidean would expect. Here are some Pythagorean triples for time-dilation on a spacetime diagram drawn on rotated graph paper. (Update: You can find some more Pythagorean triples in this energy-momentum diagram... You'll have to draw in the legs yourself.)

enter image description here

enter image description here

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The spacetime metric shows a negative sign either on the time cooordinate or the spatial ones (depending on the convention on the signature) as a direct consequence of the principles of special relativity, that is:
1) The laws of physics are the same in any IRF (inertial reference frame)
2) The speed of light is a constant in any IRF
The outcome is the Lorentz transformation.
The metric tensor is built consequently.
Let us consider two IRF's, S and S'. The distance is defined as:
$ds^2 = \eta_{\mu\nu} dx^{\mu} dx^{\nu}$ in S
$ds^2 = \eta'_{\mu\nu} dx'^{\mu} dx'^{\nu}$ in S'
The principles of special relativity require that the metric tensor is the same in the two IRF's: $\eta_{\mu\nu} = \eta'_{\mu\nu}$
To determine the components of the metric we may consider a particular Lorentz transformation, for instance S and S' with aligned spatial coordinates and S' moving in the +x direction relative to S with speed $\beta$ and define t = 0 in both frames when the spatial origines coincide:
$t' = \gamma t -\gamma \beta x$
$x' = -\gamma \beta t + \gamma x$
y' = y
z' = z
What you have to do now is to express the primed differentials vs. the unprimed ones and equate the coefficients of the products $dx^{\mu} dx^{\nu}$. The components of the metric may be shown vs. $\eta_{tt}$, for instance. You will find a negative sign which distinguishes the time coordinate from the spatial ones. Then to fix the parameter you require that in a t = const section of spacetime the distance complies to the Euclidean geometry.
Usually the metric is written as:
$\eta_{\mu\nu}$ = diag(-1,1,1,1)

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