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

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 $$y=1$$?

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.

• Thanks! I think I understand it better now. Maybe I should post it as a different question, but if, for this case, one measures the spatial distance from the time it takes a light signal to travel between two points, then the result is the same as one would obtain with rods? – anoniem Mar 3 at 18:09
• @anoniem I would suggest asking that as a separate question - I'd be interested in answering it, if I see it. The first case with no cross term works out just fine, and you can see that without much work. The second case has an additional subtlety, which is that by choosing a weird coordinate $y$ which is not orthogonal to $t$, a stationary object still has a changing $y$-coordinate. This is a (relatively simple) example of a frame dragging effect. – J. Murray Mar 3 at 21:45

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.

• Thanks! I'm understanding a bit more now about the meaning of the metric. – anoniem Mar 3 at 17:20

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.

• Maybe I am mistunderstanding something, but if the distance between $y=0$ and $y=1$ is $\Delta s= A \Delta y = A$, wouldn't we need $NA$ sticks to reach the point $y=1$? Isn't your equation $dy = \frac{dx}{A}$ really saying that if you use $N$ sticks in the $y$-direction, you will only reach the point $y = 1/A$ (rather than $y=1$)? – Jakob KS Mar 3 at 12:35
• Thanks for your detailed answer, I'm understanding a bit more now about the meaning of the metric. Is Jakob KS' remark correct that your reasoning seems to say that one would need $NA$ sticks? This is what Andrew Steane also argues. Btw, is it wrong to say that one and the same stick has a different length, depending on its orientation? – anoniem Mar 3 at 14:14
• @JakobKS Thanks a lot! Big mistake on my part. Fixed it! – Ari Mar 3 at 14:29
• @anoniem Yes I did a calculational mistake! Now I've edited it. To answer your question, No. A stick's length is same independent of the orientation. It's just the rulers in one direction is scaled (see first part of Andrew Steane's answer). – Ari Mar 3 at 14:34