# In Einstein's General Relativity, do the space-time dimensions curve?

In Einstein's General Relativity, do the space-time dimensions curve according to the positions of stars, planets, and masses?

• According to GR, space-time does curve because of the massive objects (and because of energy and momenta, in general). Not sure what you mean by space-time 'dimensions' getting curved. – Dvij Mankad Apr 30 '17 at 0:07
• We have three spatial dimensions x1, x2, x3 and a fourth dimension x4. Can all these dimensions curve in GR? – girlphysicsmajor Apr 30 '17 at 0:12
• The curvature is an effect that is ascribed to the spacetime as a whole - not to any one particular dimension of it. But yes, generically speaking, all the dimensions behave differently as compared to how they would behave in flat spacetime. – Dvij Mankad Apr 30 '17 at 5:00
• In the presence of mass, can the x dimension curve differently from the y or z dimension? Or do all space time dimensions always curve in the exact same dimension? – girlphysicsmajor Apr 30 '17 at 17:28
• As I mentioned in my previous comment, there is nothing like the curvature in one particular dimension. We only talk about the curvature of the whole of spacetime at a particular point. So, it seems meaningless to ask whether all the dimensions curve in the same way or not. I am not an expert at differential geometry but at least in General Relativity, that is not how the things work. Maybe looking up something on Reimann Curvature might help you. That is the concept of curvature that we use in General Relativity. – Dvij Mankad Apr 30 '17 at 17:32

No, the dimensions do not curve. It's the metric that changes, and induces curves and bends in the spacetime geometry, and consequently the geodesics as well.

The Einstein's equation is $G_{\mu \nu} = 8 \pi G \ T_{\mu \nu}$. When you add mass/energy in an otherwise empty space, you are basically adding a nonzero energy momentum tensor $T_{\mu \nu}$. This nonzero $T_{\mu \nu}$ then dictates what $G_{\mu \nu}$ will be so that it solves the Einstein's equations. $G_{\mu \nu}$ is made up entirely of the metric tensor $g_{\mu \nu}$, so you get a solution for the metric tensor $g_{\mu \nu}$ in the presence of some $T_{\mu \nu}$.

$g_{\mu \nu}$ encodes the information on how distances are calculated in a general spacetime (flat or curved). In the presence of some nonzero $T_{\mu \nu}$, the resultant $g_{\mu \nu}$ will be different from the metric tensor in flat spacetime $\eta_{\mu \nu}$. As a result, the spacetime geometry gets distorted into some (non flat) shape.

• So are you saying that while the spacetime geometry curves, the spacetime dimensions do not? Are you saying that the x axis, representing the x axis, is always the exact same entity in GR, and that the coordinates never stretch nor curve? – girlphysicsmajor Apr 30 '17 at 16:26
• You say that the dimensions do not curve? Does spacetime curve? Do the spacetime dimensions curve? Does only space time curve, while the spacetime dimensions remain flat? – girlphysicsmajor Apr 30 '17 at 16:30
• Are you saying that the spacetime dimensions provide a fixed background which can never bend nor curve? – girlphysicsmajor Apr 30 '17 at 16:38
• @girlphysicsmajor Since the metric tensor $g_{\mu \nu}$ describes spacetime geometry, yes, spacetime 'curves'. The dimensions do not. What is your definition of spacetime dimensions 'curving'? Dimensions, quantified by coordinates, are just labels in a classical field theory like general relativity that pinpoint the location of events in spacetime. I don't understand what it means for dimensions to bend. – Avantgarde Apr 30 '17 at 17:31
• The Guardian is a newspaper. Newspapers and other similar forms of media typically 'explain' physics in very crude, misleading and probably even incorrect ways. If you want to learn physics, learn from a textbook. Misner T W's book is a classic and I suppose it's very good, though I've never studied from it. And no, coordinates don't change in the presence of mass. Why would they? – Avantgarde May 1 '17 at 3:34

Space is curved if you don't come back to your starting point when you walk around a square. Or equivalently, you wind up at different points if you walk east-then-north vs north-then-east.

The surface of the Earth is curved in this sense. It doesn't show for a small square. But try a really large square. Start on the equator.

• Walk 1/4 of the way around the world to the east. Turn left and walk 1/4 of the way around the world to the north. You are at the north pole.
• Walk 1/4 of the way around the world to the north. Turn right, and walk 1/4 of the way around the world. (OK, it isn't east because coordinates are weird at the north pole.) But you are on the equator.

In GR, a mass causes distortions of distance and time. If you are in orbit, the distance to the center of a star is deeper than you would expect from dividing the circumference traced out by the orbit by $2\pi$. Time runs slower at the surface than in orbit.

Space-time is 4 dimensional, so you get an extra direction you can walk around the block. You can also wait a while.

So trace out this "square" where one side is distance, and the other time. Start at a point above the star.

• Have a person at the top wait a bit, then find the point/time a distance X below him at that time.
• Find the point/time a distance X below the top person right now. Have someone at that bottom point wait a bit.

Time is slower at the bottom. In his travel through time, the bottom person passes through the the point/time the top person picks out. But when he does, he isn't done waiting yet.

• So do the 4 space-time dimensions actually move when they curve? – girlphysicsmajor Apr 30 '17 at 3:33
• Consider that each point in 4-D spacetime has its own coordinate system. And, to make use of this coordinate system in terms of multiple points separated by some distance you need to compute how coordinates at one point change to become useful at another point. This is a coordinate transformation and the key idea of working with general relativity. – K7PEH May 7 '17 at 18:27

Your question, strictly speaking, doesn't make much sense, but I think I understand your confusion, so I'll try to make an according answer.

What is dimension?

Since your account name is "girlphysicsmajor", I assume you are a physics major, so you probably know some linear algebra. You know, then, that given a vector space $V$, the dimension of $V$ is the natural number that gives the cardinality of a maximal set of linearly independent vectors.

So the dimension of a space = the amount of independent directions you can go in.

There are more sophisticated notions of dimension (Lebesgue covering dimension, Hausdorff dimension for example), but these are useful mainly when the space in question is very unstructured. The curved spacetime of general relativity is what mathematicians call a differentiable manifold. This structure is more general than a vector space structure, but the vector space notion of dimension can be used here as well (I don't want to detail how).

So even for GR, the number of dimensions = the number of linearly independent directions. Curvature has nothing to do with this.

What you mistake for dimension (I think):

Coordinate systems. In special relativity you describe physics using pseudo-Cartesian coordinates, a quartett of numbers $(ct,x,y,z)$, for which the $(x,y,z)$ are cartesian coordinates, and the $t$ time coordinate is also "cartesian" related to the other three in the sense that the line element can be written as $$ds^2=-c^2dt^2+dx^2+dy^2+dz^2.$$ If the pesky minus sign wasn't there, then this would be a cartesian coordinate system. With the minus sign involved, we often call this a "pseudo-cartesian" coordinate system.

There are 4 coordinates, because spacetime is 4 dimensional, but the coordinates themselves are not "dimensions".

In particular, nothing stops you from using noncartesian coordinate systems even in special relativity. For example, you can transform the spatial coordinates to spherical polar coordinates, then you have the line element $$ds^2=-dt^2+dr^2+r^2(d\vartheta^2+\sin^2\vartheta d\varphi^2),$$ but this STILL describes the same spacetime.

What I think your question is:

I think you are asking that if spacetime is curved does that mean that the $t,x,y,z$ "coordinate lines" are curved lines then?

This question is not well posed, because in a general curved spacetime, there is no analogue of cartesian coordinates.

It is true, that curved spacetimes will always be described by coordinate systems whose coordinate lines are curved lines, but a general curved spacetime will never ever distinguish any particular coordinate system from any other.

What is curvature, then?

Consider flat spacetime with cartesian coordinates, let $(ct,x,y,z)=(x^0,x^1,x^2,x^3)=(x^\mu)_{\mu=0}^3$. The line element is $$ds^2=-c^2dt^2+dx^2+dy^2+dz^2=\eta_{\mu\nu}dx^\mu dx^\nu,$$ where summation on the repeated indices is assumed, and $\eta_{\mu\nu}$ is 0 if $\mu\neq\nu$, 1 if $\mu=\nu=1,2,3$ and -1 if $\mu=\nu=0$.

Now, we make a general coordinate transformation $$x^\mu=x^\mu(u^0,u^1,u^2,u^3).$$

The differentials are then $$dx^\mu=\frac{\partial x^\mu}{\partial u^\nu}du^\nu,$$ (repeated indices summed!), so $$ds^2=\eta_{\mu\nu}dx^\mu dx^\nu=\eta_{\mu\nu}\frac{\partial x^\mu}{\partial u^\sigma}\frac{\partial x^\nu}{\partial u^\kappa}du^\sigma du^\kappa=g_{\sigma\kappa}(u)du^\sigma du^\kappa,$$ where $$g_{\sigma\kappa}(u)=\eta_{\mu\nu}\frac{\partial x^\mu}{\partial u^\sigma}\frac{\partial x^\nu}{\partial u^\kappa},$$ and I have indicated that these "$g$" coefficients now depend on the new coordinates $u=(u^0,u^1,u^2,u^3)$.

This new line element describes the same flat spacetime too, though. But you can ask the question that if you are given some totally unknown coordinate system $(u^0,u^1,u^2,u^3)$, and a line element as $$ds^2=g_{\mu\nu}(u)du^\mu du^\nu,$$ is it possible to find a coordinate system $(x^0,x^1,x^2,x^3)$ such that in that coordinate system, the line element is in standard form $ds^2=\eta_{\mu\nu}dx^\mu dx^\nu$?

In general, this is not possible. When it is not possible, we say that spacetime is curved. How not possible this is, is measured by an object called the Riemann curvature tensor. If the curvature tensor is zero, spacetime is flat, and can be described with cartesian coordinates. If the curvature tensor is nonzero, then spacetime is curved and there is simply no such coordinate transform that would bring the line element to its standard form! This is what we call curvature.

Is it possible for a coordinate to be curved?

The answer is no, not really. For the purposes of curvature, you cannot decompose a manifold as some sort of "direct product" $\mathbb{R}\times \mathbb{R}\times \mathbb{R}\times \mathbb{R}$ and "measure the curvature of each coordinate" separately or anything like that.

In fact, one dimensional manifolds have no curvature and the curvature of two dimensional manifolds can be described by a single function. By contrast, the curvature tensor in four dimensions has 20 independent components. From this alone it is clear that you cannot simply decompose a manifold to a "sum" or "product" of lower dimensional manifolds and calculate the curvature separately (after all $2+2=4$ for dimensions but $1+1\neq 20$ for curvature components).

In some cases, it is possible to have metrics, for which "some parts of it are flat". For example, the "spatially flat FLRW metric" used often in cosmology, is given by $$ds^2=-d\tau^2+a^2(\tau)(dx^2+dy^2+dz^2)$$. As you can see, the "constant time" spatial slices have a spatial metric of $ds^2_{\text{space}}=a^2(\tau_0)(dx^2+dy^2+dz^2)$. Performing the coordinate transform $x\mapsto x/(a^2(\tau_0))$, $y\mapsto y/(a^2(\tau_0))$ etc. will give $ds^2_{\text{space}}=dx^2+dy^2+dz^2$ and from this we can see that the spatial slices are flat three dimensional manifolds.

But you cannot perform this procedure for the entirety of spacetime to obtain a "flat spacetime", despite the fact that all spatial slices are flat. And the reason you cannot perform this procedure is because for a single $\tau=\tau_0$ constant time slice, the coefficient $a^2(\tau_0)$ is a constant. But if you consider all of spacetime, $a(\tau)$ is a function of $\tau$, so if you defined $x'=x/a^2(\tau)$, you'd have $dx'=dx/a^2(\tau)-(2x/a^3(\tau))d\tau$, and you wouldn't get a flat metric anymore.

• You write: "Is it possible for a coordinate to be curved? The answer is no, not really." So is the answer no? No not, not really, but kind of? Is it possible for the geometry of the spacetime dimensions to be curved? Can it be curved differently in the x, y, z, and x4 dimensions? – girlphysicsmajor May 1 '17 at 18:31

A (somewhat crude) analogy may help- Consider you have 2 blank papers. On one you draw straight margins along the edges(call this sheet A). On the other, you draw curved( in any way whatsoever) margins(call this sheet B). Then, roll up A( or crumple it if you like). Now look at both the sheets. The margins of both look similar, of course. But which one is curved? Certainly not both, even though the margins look alike. Only A is actually curved!

The point is; the "margins" are akin to "dimensions". And (what should now be obvious) this suggests that the dimensions themselves are not guaranteed signposts to check for curvature.

As a side note, you may like to know that just by changing your coordinate systems you could fool yourself that space is 'curved' even though it is not.( the margins may fool you when you compare the crumpled and flat sheet; a 'good' crumple would mean they may look exactly similar!). A subtle interplay of cartesian and curvilinear coordinates is involved, which I will skip. You may want to read up on Christoffel symbols though.

Note- General relativity is much more complicated. You don't have one sheet; you have a sheet at every point of the body you are looking at. This is only a 2 dimensional analogy to try to explain a 4 dimensional phenomenon. Please keep that in mind.

Hope this helps.