# Why is time special?

In Special Relativity, the spacetime interval between two events is $s^2 = -(c{\Delta}t)^2+({\Delta}x)^2+({\Delta}y)^2+({\Delta}z)^2$ giving the Minkowski metric $\eta_{\mu\nu}=\text{diag}(-1, 1, 1, 1)$. What is the justification for making time have a negative coefficient, and how closely is that related to the 2nd law of thermodynamics? Sure, by letting $\eta = \text{diag}(1, 1, 1, 1)$, we get a pretty boring spacetime, and the boosts in the Poincaré group become trig instead of hyperbolic functions, but what's the physical reasoning behind this?

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Calling anthropic principle? lol –  Alan Rominger Aug 12 '11 at 16:07
well time needs to be something different than space (otherwise it wouldn't have acquired a different name, now would it?). so $\text{diag}(1,1,1,1)$ would be just 4D space. –  luksen Aug 12 '11 at 16:19
Time is special because it's not spatial :) But the negative coefficient in time is a convention, the metric signature can be (+, −, −, −) or (−, +, +, +), so for some people the space is special, not the time ;) –  Andyk Aug 12 '11 at 16:51
@Zassounotsukishi Well, of course it's anthropic! I should have emphasized the part of my question relating $\eta_{00}$ to the 2nd law of thermodynamics. What's the (is there a?) relationship there? –  David Souther Aug 12 '11 at 16:54
@ANKU so what you're saying is that we live in a 1-dimensional world, but the direction of time has 3 degrees of freedom. I guess I'm curious as to how entropy increases from left to right :-P ... or does it? –  Alan Rominger Aug 12 '11 at 17:32

I think this is a case of the mathematics being designed to model reality. As you say, making the time component of the metric positive would give a space that doesn't match what we observe. In particular, the negative component for time allows us to disconnect regions of space that aren't causally linked. In other words, the fact that the speed of light is finite and a maximum means that we must describe space-time with a shape that keeps causally disconnected regions separate. The necessary shape is reflected in the choice of the sign of $\eta_{00}$.

That's how I understand anyway...

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What if $\eta_{\mu\nu}=\text{diag}(-1, -1, 1, 1)$? –  Alan Rominger Aug 12 '11 at 16:33
What becomes important for many GR type applications is the trace of the metric (it comes up, for instance, in Stokes Theorem for Curved Spacetime). You want a non-zero trace, so -1,-1,1,1 wouldn't be very good! :P –  Benjamin Horowitz Aug 13 '11 at 5:19
Stokes Theorem in curved spacetime is $\int_M d^n x \sqrt{|g| } \nabla_\mu V^\mu=\int_{\partial M}d^{n-1}y \sqrt{| \gamma |} m_\mu V^\mu$, ($\gamma$ is the induced metric on $\partial M$) basically integration doesn't work with a trace = zero metric. :( –  Benjamin Horowitz Aug 13 '11 at 5:38
@Warrick your answer feels the most correct at this point, and I can see the different signs as clearly highlighting (and requiring) that causality separation. As I tried to clarify in my comments on the Q itself, what is the relationship between causality and the 2nd law of Thermo? –  David Souther Aug 14 '11 at 4:14
I've never seen a fully relativistic formulation of thermodynamics so I'm not aware of any link between the "specialness" of time and the Second Law. –  Warrick Aug 15 '11 at 7:10

There is a direct link between the minus sign in the metric and thermodynamics. Because the sign is negative, positive energies cannot be rotated to negative energies, and it makes sense to say that the energy of a physical system is always positive. This gives rise to thermodynamic partitioning.

Unlike energy, spatial momentum randomizes with signs, so there is little point in considering the partitioning of momentum. The momentum in a thermal environment will partition around the mean CM velocity of the environment, which can be taken to be zero. But the energy partitions with an extra parameter, the temperature, controlling equilibrium, and there is no change of coordinates which zeroes out the equilibrium energy.

Of course, the same holds in the Galilean space-time, the $c\rightarrow\infty$ version of Minkowski spacetime, so it does not force the issue in any way.

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