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When writing the metric in Minkowski space, how can we distinguish between the past and the future? I understand the answer after drawing the light cone but I want to know how we get that by just considering the quadratic form of the metric: $$ ds^2= -dx_{1}^2+ dx_{2}^2+ dx_{3}^2 + dx_{4}^2 $$ without visualizing the light cone.

Thank you.

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Although the question is about Minkowski space, I think it may be helpful to consider the more general case first. It's not possible to distinguish past and future simply based on the metric. In fact, there are spacetimes that are not even time-orientable. This is similar to the way in which a Mobius strip is not an orientable surface.

So the same considerations apply for Minkowski as for any other spacetime. Its metric doesn't automatically endow it with an arrow of time. The arrow of time has to be put in by hand, in addition to the metric. This happens to be possible in Minkowski space because it's time-orientable. In Minkowski space you typically express your arrow of time by choosing coordinates such that the positive $t$ axis points in that direction. But the metric is invariant under a change of coordinates $t\rightarrow -t$, so there is no way to get that from the metric itself.

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  • $\begingroup$ Thanks for the answer. Do you mean that the metric signature doesn't play any role? $\endgroup$ – Dory Sep 20 '13 at 15:55
  • $\begingroup$ @Dory: Right. Flipping the signature of the metric from $+---$ to $-+++$ is purely a matter of notational convention, with no physics involved. $\endgroup$ – user4552 Sep 21 '13 at 17:11
  • $\begingroup$ Another question.What about flipping the metric signature from +--- to ++--? What's wrong with choosing the metric to be like the second type? $\endgroup$ – Dory Sep 22 '13 at 7:03
  • $\begingroup$ @Dory: General relativity doesn't have any requirements on the signature. We just don't happen to live in a universe with two time dimensions. $\endgroup$ – user4552 Sep 22 '13 at 23:08
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You may consider having a "relative" arrow of time, in the case where you consider $2$ events $x$ and $y$ such as $\Delta s^2 = (x^0 - y^0)^2 - (\vec x - \vec y)^2 > 0$, in some inertial frame $F$.

Suppose that $x_0 > y_0$. Then you can do a Lorentz tranformation to go from the initial inertial frame $F$ to an other inertial frame $F'$. By Lorentz invariance, we have $\Delta s'^2 = (x'^0 - y'^0)^2 - (\vec x' - \vec y')^2 = \Delta s^2$

But you have too $x'_0 > y'_0$. So, in all inertial frames $F$, you have a relative arrow of time between the events $x$ and $y$, meaning that a physical process at $y$ could be the cause of a physical process at $x$, but not the inverse.

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    $\begingroup$ The sign of $\Delta s^2$ is the same for the past and future light cones. $\endgroup$ – user4552 Sep 21 '13 at 17:15

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