Are space and time interchangeable?

Question

There was a previous question asked that I don't understand (nor did many other people apparently). My question is hopefully different.

On a Wikipedia article about speed of gravity it states:

Formally, $c$ is a conversion factor for changing the unit of time to the unit of space.

Does this really mean $c$ provides a conversion from space to time and visa versa? I mean does it technically, or even practically observably, allow for you to turn space into time and visa versa in an equivalent way $E=mc^2$ allows conversion of matter into energy and visa versa (that we see in nuclear power plants and the synthesis of matter in CERN etc)? Are there any well known / hypothesised examples of this? I mean you can't take a N meter cubed volume of space and turn it into $seconds^3$ using $\dfrac{volume}{c^3}$ or visa versa right? (I'm not even sure how to get my head around seconds cubed, energy and mass converting between each other is strange enough!)

Answer 1

There is a sense in which space and time are interchangable, but there is also a very important sense in which they are distinct.

In relativity (both flavours) we treat spacetime as a four dimensional manifold with three spatial dimensions and one time dimension - I'll explain the difference in a moment. Any observer can choose a set of coordinates with themselves at the origin, then they can measure out their three spatial axes with their ruler and measure their time axis with their clock.

But different observers won't agree on their axes. For example if I'm moving relative to you then your time axis will appear to me like a mixture of my time and space axes. So time and space are interchangeable in the sense that what looks like time to you looks like a mixture of time and space to me. If you're interested I talk about this some more in What is time, does it flow, and if so what defines its direction?. This mixing up of the time and space dimensions is behind phenomena like time dilation and Lorentz contraction.

However, while different observers won't agree on what constitutes space and what constitutes time, they will always agree that there are three spatial axes and one time axis. The difference is how these dimensions appear in the metic tensor.

Suppose we take good old Euclidean space and you move a distance $dx$ along the $x$ direction, $dy$ along the $y$ direction and $dz$ along the $z$ direction. The total distance moved, $ds$, is given by Pythagoras' theorem:

$$ ds^2 = dx^2 + dy^2 + dz^2 $$

In special relativity we have a similar total distance moved in 4D spacetime, and it's given by the metric:

$$ ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 $$

But note that the time term $dt^2$ appears in the equation with a negative value. This is the key difference between space and time. The spatial terms have a different sign to the time term. In sense space and time are quite distinct.

Finally, that factor of $c$ is needed because when you add quantities they have to have the same units. So in the equation for the metric we cannot just add $dt^2$ and $dx^2$ because that would be adding seconds squared to metres squared. Multiplying by $c$ converts the units of time to be the same as the units of distance.

To see why $c$ is the speed of light have a look at What is so special about speed of light in vacuum?.

Answer 2

In a way, yes, in general relativity.

As mentioned in other answer, infinitesimal square of the distance in flat 4D spacetime (aka Minkowski space), appearing in special relativity, is given by \begin{equation} ds^2=-c^2 dt^2 + dx^2 + dy^2 + dz^2. \end{equation}

Things become a bit more complicated in general relativity, because the spacetime can be curved. In general, it becomes \begin{equation} ds^2=g_{\mu\nu}dx^\mu dx^\nu, \end{equation} where $g_{\mu\nu}$ is something called a metric tensor and we use Einstein summation convention in which summation over repeated indices is implied, i.e. RHS is summed over $\mu$ and $\nu$ ranging from 0 to 3 in 4D spacetime. In spherical coordinates, this looks like \begin{equation} ds^2=-c^2 dt^2 + dr^2 + r^2(d\theta^2+\sin^2 \theta d\phi^2). \end{equation}

In special relativity in Cartesian coordinates, $g_{\mu\nu}$ is just a diagonal matrix with $g_{00}=-1$, $g_{11}=1$, $g_{22}=1$, $g_{33}=1$ and other entries zero, but in general relativity there are many options.

In classical general relativity, only fundamental difference between space and time is the difference in sign of eigenvalues of the metric tensor.

Now, if we have a spacetime with a point mass, then we have Schwarzschild metric, given by: \begin{equation} ds^2=-\left(1-\frac{2GM}{c^2r}\right)c^2 dt^2 + \left(1-\frac{2GM}{c^2r}\right)^{-1}dr^2 + r^2(d\theta^2+\sin^2 \theta d\phi^2), \end{equation} where $M$ is mass.

The distance $r_s=2GM/c^2$ is called Schwarzschild radius. If mass is contained inside its Schwarzschild radius, it is a black hole with event horizon (point of no return) at distance $r_s$.

Now, notice that, in metric, coefficients multiplying $dt^2$ and $dr^2$ switch signs when $r$ becomes smaller than $r_s$. This means that, in a way, radial and temporal coordinates have switched their roles as space and time coordinates!

This also means that $r$ will inevitably grow smaller for unfortunate space traveler who crossed the event horizon, just like e.g. $t$ inevitably grows bigger for anybody in flat spacetime. The center of black hole stops being something "over there" once you crossed the horizon, it literally becomes your future.