How do length measurements with measuring sticks depend on the metric in general relativity?

Question

Suppose I have a metric $\mathrm{d}s^2 = (c\mathrm{d}t)^2 - \mathrm{d}x^2 - (A\mathrm{d}y)^2 -\mathrm{d}z^2$, with A a constant (larger than zero and independent of $x,y,z,t$). Suppose I have a collection of identical measuring sticks. Suppose that I put such sticks along the x-axis ($y=z=0$), one after the other, until the point $x=1$ is reached. Suppose I need $N$ of them to do that. Now if I do the same along the y-axis ($x=z=0$), how many would I need to reach $y=1$? Also $N$? Or $N/A$? $N*A$, perhaps?

Now suppose there is also a cross-term $B\,\mathrm{d}y\mathrm{d}t$ in the metric ($B$ a constant unequal to zero). Does this change the number of measuring sticks needed to reach $y=1$?

Answer 1

This is a good opportunity to discuss and illustrate how coordinate representations of tensorial objects can really obscure their true nature.

Suppose I have a metric $\mathrm ds^2=(c\mathrm dt)^2−\mathrm dx^2−(A\mathrm dy)^2−\mathrm dz^2$, with $A$ a constant (larger than zero and independent of $x,y,z,t$).

At a glance, it appears that this metric encodes some kind of spatial anisotropy - that distances in the $y$ direction are somehow different from distances in the $x$ and $z$ directions. But this is not the case. Note that if we start with ordinary Minkowski spacetime with $\mathrm ds^2 = c^2 \mathrm dt^2 - \mathrm dx^2 - \mathrm dy^2 - \mathrm dz^2$ and then perform a coordinate transformation $(ct,x,y,z)\mapsto (ct',x',y',z') = (ct,x,y/A,z)$, we would obtain your expression; in that sense, the presence of the constant $A$ is a coordinate artifact.

We can define the proper length $L_p$ of a measuring rod to be $\int \sqrt{-\mathrm ds^2}$ where the integral is taken along the unique$^\ddagger$ geodesic which connects the endpoints of the rod at fixed $t$, in an inertial frame in which the rod is at rest. This is a coordinate-independent definition which gives

$$L_p = \int \sqrt{\mathrm dx^2 + A^2\mathrm dy^2 + \mathrm dz^2}$$

If one end of the rod is placed at the origin and the other end is placed at some position $x_0$ on the $x$-axis, then we would have $L_p = \int_0^{x_0} \mathrm dx =x_0$. This is just what you'd expect. However, if we place the rod along the $y$-axis, then we find that $L_p = \int_0^{y_0} A \mathrm dy = Ay_0 \implies y_0 = L_p/A$.

The interpretation is simple - the $y$-coordinate is just stretched by a factor of $A$ relative to the $x$- and $z$-coordinates. There's no interesting physics at work here, we've just picked a weird way to label points in spacetime. The proper distance (again, taken along the unique geodesic at fixed $t$) between the origin and the point $y=1$ on the $y$-axis is $A$ times larger than the proper distance between the origin and the point $x=1$ on the $x$-axis; if it takes $N$ measuring rods to span the latter distance, it will take $N\cdot A$ rods to span the former.

---

Now suppose there is also a cross-term $B\mathrm dy \mathrm dt$ in the metric ($B$ a constant unequal to zero). Does this change the number of measuring sticks needed to reach $y=1$?

No. Note that once again, this is a weird coordinate artifact. If we start in Minkowski spacetime with a metric $\mathrm ds^2 = c^2 \mathrm dt^2 - \mathrm dx^2 - \mathrm dy^2 - \mathrm dz^2$ and then perform the coordinate transformation $$(ct,x,y,z)\mapsto (ct',x',y',z') = \left( \sqrt{1-\left(\frac{B}{2A}\right)^2} ct, x, \frac{y + \frac{B}{2A}\sqrt{1-\left(\frac{B}{2A}\right)^2}ct}{A}, z\right)$$ $$\implies (ct,x,y,z) = \left(\frac{ct'}{\sqrt{1-\left(\frac{B}{2A}\right)^2}},x',Ay'-\frac{B}{2A} ct', z'\right)$$

then your new metric will be (dropping the primes)

$$\mathrm ds^2 = c^2 \mathrm dt^2 - \mathrm dx^2 - A^2 \mathrm dy^2 - Bc\mathrm dt \mathrm dy - \mathrm dz^2$$

The oddness of this metric once again reflects the fact that we are using odd coordinates; this time, we are scaling both $y$ and $t$ and choosing our coordinate origin to be moving with velocity $\frac{B}{2A}\sqrt{1-\left(\frac{B}{2A}\right)^2}c$ in the $y$-direction (I've implicitly assumed that $B/2A < 1$ for this transformation).

In any case, since this can be obtained from the Minkowski metric via a simple coordinate transformation, physical predictions must be the same. If you go through the argument with the measuring rods again, we will be taking measurements at constant $t$ which means that the $\mathrm dt\mathrm dy$ term will drop out and everything else will be the same.

---

$^\ddagger$In general, if you fix your timelike coordinate to some constant value and then choose two points on the resulting time slice, there will not be a unique geodesic joining the two. Even if there is, for curved spacetimes there are generally no global inertial frames, so this construction is generally ambiguous.

Answer 2

Let's employ a stick that has a natural length equal to $L$ in its own rest frame. For events with no temporal separation (so $dt = 0$) such a stick will extend between events having $\Delta s = L$.

$N$ such sticks laid end to end will cover a proper length $N L$ at any instant of time (i.e. on the hypersurface $dt = 0$).

For events along the $x$ direction we have $dy=dz=0$ and $dx = ds$ therefore the sticks will reach from $x=0$ to $x=\Delta s = N L$. So if our units are such that this location is at $x=1$ then we have that the length of a stick is $L = 1/N$.

Now let's put these sticks along the $y$ direction. They extend between the origin and the event at $\Delta s = N L = 1$ along the $y$ direction. The change in $y$ coordinate between these events is $$ \Delta y = \int_0^1 \left| \frac{dy}{ds} \right| ds = \int_0^1 \frac{1}{A} ds = \frac{1}{A} $$ Thus $N$ sticks extend to $y = 1/A$. It follows that to reach all the way to $y=1$ you will need $NA$ sticks.

Looking back now at the original metric, you can see that if $A > 1$ then a movement along the $y$ direction by $\delta y = 1$ corresponds to a larger proper distance $\delta s$ than a movement along the $x$ direction by $\delta x = 1$. So no wonder more sticks are needed.

Answer 3

For your calculations, we'd assume you are looking at a fixed time slice. Of course, putting the sticks along the axes would take some time, but your metric is static(it's components are independent of time) thus it won't matter. So, $dt=0$. Thus the cross terms ($dtdy$) won't contribute. This answers the second part of your question.

Now for constant time, the metric takes the form $ds^2 = dx^2 +A^2 dy^2$ (sorry I changed the signature of the metric. I'm a strict (-,+,+,+) guy).

So, when you're going along the $x$ axis, $ds = dx$. For moving along $y$ direction you have $ds = A dy$. Hence you may write

$$dy =\frac{1}{A} dx \implies y = \frac{x}{A}$$

Thus going 1 unit distance in $y$ direction is same as going $A$ distance along $x$. So, you'll need $A*N$ sticks.

To see, how the cross terms contribute try calculating the velocity of light along $x$ direction and $y$ direction separately.