What distinguishes time from space in Quantum Field Theory?

Question

Consider the following expression for a general QFT action:

$$ S ~=~ \int_0^t\mathrm dt~L ~=~\int_0^t\mathrm dt\int_\mathbb {R^3}\mathrm d^3x~\mathcal L ~=~\int\mathrm d^4x~\mathcal L.$$

Here we have clearly singled out time as the "worldline parameter" for calculating $S$, because we want to calculate how a system evolves in time (in some frame). But what's special about time (in particular this choice of time coordinate)? Of course time is manifestly distinguished from space by the metric $\eta_{\mu\nu}$, but a-priori is there anything special about time (it's not clear that we would have any entropy considerations, for example)? In particular we can consider Lorentz transformations that take this time $t$ to some other coordinate. It feels like in principle we should be able to chose any 3-D subspace of $\mathbb R^4$ and flow that along the spacetime manifold.

I'd really appreciate someone with a detailed understanding of the physics and/or geometry of the situation giving some clarifying remarks (I understand there may not be a clear-cut answer).

(Btw. I am aware of the aware of various arrows of time and have read some relevant discussion, e.g. What grounds the difference between space and time?)

Answer 1

Indeed, you are free to choose any time-like direction as a direction in which the system evolves. For example the ADM formalism of General Relativity is based on this idea. In this formalism you take a 3-dimensional space-like surface as our world. Then the time will be related to the normal to this surface.

Very good detailed description of the ADM construction can be found in "A relativist's toolkit" by Poisson.

I guess that the demand that evolution should go in a time-like direction is related to some causality features. What is true is that you cannot do a Lorentz rotation from a space-like to a time-like direction.

Answer 2

Here two physically relevant facts distinguishing space form time in Minkowski spacetime (however everything follows can be generalized to every spacetime with suitable requirements).

1) This is general. Consider, in Minkowski spacetime, two events $p$ and $q$. If $p$ and $q$ stay in the $3$-space of a reference frame, then among the curves with spacelike tangent vector joining these points there is not the shortest curve, but you can always find a curve with length smaller than any fixed $\epsilon>0$. Obviously, if you consider only the curves lying to the rest space, there is the shortest one: the segment joining $p$ and $q$.

If $p$ and $q$ stay along a curve with timelike vector, for instance on the temporal axis of a reference frame, there is the longest curve joining them (instead of the shortest one).

2) This is properly referring to Quantum Field Theory. More precisely to fields solving $\delta \int_{\Omega} {\cal L}(\phi, \partial \phi) d^4x =0$ with $\phi$ vanishing on the boundary of $\Omega$, also before being quantized.

Assuming that the Lagrangian is quadratic in the first derivatives of $\phi$ with the metric $\eta$ as quadratic form, you obtain equations for $\phi$.

These equations, to be solved, need initial conditions. That is, you have to assign $\phi$ and $\partial_n \phi$ on a $3$-surface, whose $n$ is the normal unit vector.

Well, the problem is well-posed, i.e. there exists a unique solution and it continuously depends on initial data (with a suitable choice of topology and in a certain class or sufficiently regular initial data) if the surface can be described by spatial coordinates, for example the rest $3$-space of an inertial frame (but the result is general with some requirements I do not describe here). Otherwise, if just one coordinate on the $3$-surface is timelike the problem gives rise to some pathology (lack of existence, lack of uniqueness or discontinuous dependence from initial data). Without well-posedness of initial data problem to quantize fields (with some undefined meaning for "to quantize" since the standard procedure is not feasible) turns out to be very difficult.

Answer 3

In (classical or quantum) field theory, time is distinguished by a (classical) observer. The observer moves along a world line $x(s)$ parameterized by its eigentime $s$, with timelike derivative $\dot x(s)$. This distinguishes a direction of time, and the observer's space is orthogonal to it.

Different observers will generally have different directions of time, and any two of them can be turned into each other by a proper Loretz transformation.

Note that proper Loretz transformation always transform timelike vectors into timelike vectors, never into lightlike, spacelike, antilightlike, or antitimelike vectors. (For example, there is no Lorentz transformation that transforms the $t$-axis (i.e., $x_0$-axis) into the $x_1$-axis.) Thus one is not completely free to select a 3D subspace, but only those 3D affine subspaces that are orthogonal to a timelike vector (and hence consist of mutually spacelike points). These are indeed fully equivalent, for each one there is an observer for which it defines its space.

See also Chapter A7: Time and space in my theoretical physics FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html

Answer 4

An important area of QFT where time is particularly special is the fact that the inner product of two fields is done over all space at a particular time (or over a spacelike hypersurface, if you're in curved spacetime).

In relativity itself, just the fact that the metric is (-+++), or (+---), depending, already has a lot of consequences separating time from space in many ways, especially the causal structure of spacetime.

It feels like in principle we should be able to chose any 3-D subspace of R4 and flow that along the spacetime manifold.

This in particular is a problem because while it has been proven that if you have a nicely behaved spacetime, you can take a spacelike hypersurface and "flow" it along a timelike curve, a generic spacetime with an arbitrary hypersurface may not work. This is related to the Cauchy problem of general relativity (the ability to extend the local solution of a PDE to the entire spacetime).

Answer 5

As the other answers have pointed out, there are geometric differences between space and time. However, if you want to know "how a field evolves" you have to specify what you mean. For example, we usually assume we can measure an entire field $\phi(x, y, z)$ at every point in space at a specific time. There are, however, applications where you might be measuring the field at a specific plane in space over time -- in this case you might make the longitudinal direction normal to the plane your independent variable and time as an independent variable. This is particularly useful in boundary value problems, which are not a big deal in QFT from what I understand but they could be elsewhere.

This appears in accelerator physics applications, where you might want to specify your coordinates in terms of a closed ideal trajectory and deviations from it, to which end your "time" is interpreted as the time of flight it took for a particle to get to that particular azimuth in a storage ring.

I think in practice, for most field theoretical applications, time is just the more convenient independent variable since you are not usually riding on a Galilean transformation of the coordinates and you are not dealing with boundary value problems.