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It seems as though symmetry is a driving force behind theoretical physics. With symmetry in mind, should we expect that the number of time dimensions should be the same as the number of spatial dimensions?

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  • $\begingroup$ Related: physics.stackexchange.com/q/10651/2451 . Related (since there are at least 3 spatial dimensions): physics.stackexchange.com/q/43322/2451 and links therein. $\endgroup$ – Qmechanic Aug 7 '14 at 1:06
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    $\begingroup$ And which symmetry would it be that lets you suspect multiple time dimensions? $\endgroup$ – ACuriousMind Aug 7 '14 at 1:07
  • $\begingroup$ Time and space are not different in the framework of special and general relativity (and therefore should not be different in quantum gravity). You could measure time in metres if you want to, just multiply by $c$. There is no symmetry here. $\endgroup$ – SuperCiocia Aug 7 '14 at 1:27
  • $\begingroup$ @Harold So since they are the same thing, symmetry doesn't apply. This answers my question. $\endgroup$ – zeta Aug 7 '14 at 1:46
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    $\begingroup$ Even in the absence of symmetry arguments people have definitely looked at this kind of thing. The resulting space-time with more than one temporal dimension is called 'ultrahyperbolic'. A google or arXiv search on ultrahyperbolic space-time should be a good starting point $\endgroup$ – Arthur Suvorov Aug 7 '14 at 2:32
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Relativity treats spacetime as a four dimensional manifold equipped with a metric. We can choose any system of coordinates we want to measure out the spacetime. It's natural for us humans to choose something like $(t, x, y, z)$, but this is not the only choice. Even in special relativity the Lorentz transformations mix up the time and spatial coordinates, so what looks like a purely time displacement to us may look like a mixed time and space displacement to a moving observer. In general relativity we may choose coordinate systems like Kruskal-Szekeres coordinates where there is no time coordinate in the sense we usually understand the term.

So in this sense there is a symmetry between space and time coordinates because there is no unique separation between space and time. This is the point Harold makes in the comments.

But this does not answer your question. Regardless of what coordinate system we use, locally the metric will always look like:

$$ ds^2 = -da^2 + db^2 + dc^2 + dd^2 \tag{1} $$

where I've deliberately used arbitrary coordinates $(a, b, c, d)$ to avoid selecting one of them as time. If you look at equation (1) it should immediately strike you that there are three + signs and one - sign, so there is a fundamental asymmetry. It's the dimension with the - sign that is the timelike dimension and the ones with the + sign are spacelike. Whatever coordinates we choose we always find that there is are three spacelike dimensions and one timelike. We call this the signature of the spacetime and write it as $(-+++)$ or if you prefer $(+---)$.

Which brings us back to your question why are there three spacelike dimensions and one timelike dimension? And there is no answer to this because General Relativity contains no symmetry principle to specify what the signature is. If you decided you wanted two or three timelike dimensions you could still plug these into Einstein's equation and do the maths. The problem is that with more than one timelike dimension the equations become ultrahyperbolic and cannot describe a universe like the one we see around us.

So the only answer I can give to your question is that there is only one time dimension because if there were more time dimensions we wouldn't be here to see them.

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Tegmark has a nice discussion of this. He essentially argues that more than 1 timelike dimension would remove our ability to predict the future from the present. In math terms, more than one timelike dimension would render our differential equations ultrahyperbolic, and ultrahyperbolic PDEs are not well posed for initial data given along spacelike hypersurfaces.

See this article for a more thorough discussion.

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