How can we physically determine the metric in SR? How can we physically determine a coordinate system where $g$ is diagonalized?

Question

Here I'll use theoretical units with $c=1$. Suppose we are given two coordinate systems $(t, x)$ and $(\tilde{t}, \tilde{x})$, and suppose they are related by $t = \tilde{t} - \tilde{x}/2$ and $x = \tilde{x}$. How can we tell which one has its time axis perpendicular to its spatial axes?

Equivalently, how can I tell that the metric $g$ in the $(t, x)$-coordinates is of the form $g = dt^{2}-dx^{2}$ and not $g = dt^{2} - (3/4)dx^{2} + dt\; dx$?

Is there any sequence of physical experiments I could do to select the "correct coordinate system" where the metric is diagonalized? Equivalently how can I physically deduce the metric in a given coordinate system? Is it even possible to fully determine the metric?

Answer 1

I think your questions aren't well-posed since you aren't giving definitions of your coordinates (or are making implicit assumptions).

I will show how radar experiments (in the spirit of the Bondi k-calculus) will define a metric $ds^2=dt^2-dx^2$.

(This is based on my answer https://physics.stackexchange.com/a/508251/148184 to Minkowski Metric Signature .)

(This is based on a diagram from Bondi's "E=mc2: An Introduction to Relativity" (http://www.worldcat.org/title/emc2-an-introduction-to-relativity/oclc/156217827). I corrected a typo in the original.)

After a time $T$ after separation-event-O, Alice sends a signal to Bob. That signal is received by Bob at an event-Q when his watch reads $kT$, where $k$ is a proportionality constant that doesn't depend on $T$, but only on the relative-velocity between Alice and Bob.

Alice receives the echo when her watch reads $R=k(kT)$, where we have used the same proportionality constant [invoking the principle of relativity... with the usual assumptions of homogeneity and isotropy (as done in Euclidean geometry and Galilean relativity)].

Alice assigns space and time components of the displacement $\vec{OQ}$ as $$\Delta t_{OQ}= (R+T)/2 \qquad \mbox{half of the sum [the mid-time]}$$ and $$\Delta x_{OQ}= (R-T)/2 \qquad \mbox{half of the difference [half of the round-trip]}$$

Thus, $$\Delta t_{OQ}=(k^2+1)T/2\qquad\mbox{and}\qquad \Delta x_{OQ}=(k^2-1)T/2.$$ ​ This can be used to determine the relation between $k$ and the relativity velocity $v_{BwrtA}$.

$$v_{BwrtA}=\frac{\Delta x_{OQ}}{\Delta t_{OQ}}=\frac{k^2-1}{k^2+1}.$$ With a little algebra, $$k=\sqrt{\frac{1+v_{BwrtA}}{1-v_{BwrtA}}}\qquad\mbox{Doppler factor}.$$

Note that $RT=k^2T\ T=(kT)^2=\left(\Delta t_{OQ, Bob}\right)^2$, the elapsed time of $\vec{OQ}$ according to Bob's watch. This suggests that product of radar times $RT$ is an invariant.. it depends only on $\Delta t_{OQ, Bob}$ but not $k$. For Bob’s radar measurement of $\vec{OQ}$, his radar times are equal (since his reception coincides with his transmission).

One could introduce a third observer

to derive this (and the Lorentz transformations). See my linked answer above for details.

The bottom line is that $RT$ is an invaraint.
In terms of $t$ and $x$ (as defined above), we have $$RT=(\Delta t+\Delta x)(\Delta t-\Delta x)=(\Delta t)^2-(\Delta x)^2.$$ Thus, we generalize the result as $$ds^2=dt^2-dx^2.$$

Answer 2

Assume we are working in special relativity with $D+1$ dimensional spacetime. Suppose we have an inertial coordinate system $(x^{\mu})$, so objects that don't interact with anything move in uniform motion according to these coordinates. If we can determine another coordinate system where the metric $g$ is known, and if we know how to transform between the old and new coordinate systems, then we can determine $g$ in the original coordinates.

Since the metric is symmetric (i.e. $g_{\mu\nu} = g_{\nu\mu}$), there are $$ \binom{D+1}{2}+(D+1) = \frac{(D+1)(D+2)}{2} $$ many degrees of freedom for the metric. We can consider the following parts of $g$:

• $g_{00}$
• $g_{ii}$ and $g_{i0}$
• $g_{ij}$ where $i < j$.

The first bullet describes only one component, the second bullet describes $2D$ many components, and the last bullet describes $\binom{D}{2} = D(D-1)/2$ many components. In total, we describe $$ 1 + 2D + \frac{D(D-1)}{2} = \frac{(D+1)(D+2)}{2} $$ many components, which is exactly the number of degrees of freedom of $g$. For each component, we will either transform coordinates so that the component is eliminated, or we will set up a system of equations that will allow us to determine the values of components. We will proceed in three steps.

Part 1. Let $g_{S} = g_{ij}dx^{i}dx^{j}$ be the spatial part of $g$. We will find a new coordinate system in which $g_{S}$ is diagonalized. To do this we will exploit the Lorentz force law as follows. Send a stream of electrons in any direction. By applying an (approximately) uniform magnetic field to the electron beam in various ways, we are able to find/construct spatial axes that are orthogonal to one another according to $g_{S}$. This gives us a new system $(y^{i})$ of coordinates in which $$ g_{S} = g'_{11}dy^{1}dy^{1} + \cdots + g'_{DD}dy^{D}dy^{D} $$ for some reals $g'_{11}, \ldots, g'_{DD}$.

By transforming from $(x^{\mu}) = (x^{0}, x^{i})$ to $(y^{\mu}) = (x^{0}, y^{i})$, we obtain $g = g'_{\mu\nu}dy^{\mu}dy^{\nu}$ where $g'_{ij} = 0$ for all $i\ne j$. The components $g'_{ij}$ with $i < j$ are determined to be zero. This leaves us with the remaining $2D+1$ components to be determined.

Part 2. To determine the rest of the coefficients, we proceed as follows. We assume that the two-way speed of light is always $c=1$, but for the sake of generality we won't assume a fixed one-way speed of light in coordinates $(y^{\mu})$.

The one-way speed of light in a direction can be found by sending a light signal from the origin of our coordinates to a clock at a distance $L$ from the origin (according to our coordinates). Then we take $$ \frac{L}{(\text{receiver clock at reception}) - (\text{transmitter clock at transmission})}, $$ and this will be the one-way speed of light according to the $(y^{\mu})$-coordinates. Hence the one-way speed of light in any direction is a known quantity.

For each index $i$ where $1\le i\le D$, we let $c_{+}^{i} > 0$ be the one-way speed of light in direction $+y^{i}$ and let $c_{-}^{i} < 0$ be the (negative of the) one-way speed of light in direction $-y^{i}$. Let $1\le k\le D$ and consider shooting two light beams in the $+y^{k}$ and the $-y^{k}$ directions. For light beams going to the $\pm y^{k}$ directions, $$ dy^{k}/c_{\pm}^{k} = dy^{0} \qquad\text{ and }\qquad dy^{i} = 0 \;\text{ for }\; i\ne k. $$ If $c_{+}^{k}$ is infinite, then any fraction with an infinite denominator is interpreted to equal zero. Likewise for $c_{-}^{k}$. Knowing that $ds^{2} = 0$ for null geodesics, we find \begin{align*} 0 &= g'_{\mu\nu}dy^{\mu}dy^{\nu} \\[1.0ex] &= g'_{00}dy^{0}dy^{0} + g'_{k0}dy^{0}dy^{k} + g'_{0k}dy^{k}dy^{0} + g'_{kk}dy^{k}dy^{k} \\[1.0ex] &= \left[ g'_{00}/(c_{\pm}^{k})^{2} + 2g'_{k0}/c_{\pm}^{k} + g'_{kk} \right] dy^{k}dy^{k}. \end{align*} Since $dy^{k}$ is nonzero for light beams moving in the $\pm y^{k}$ directions, we obtain \begin{align*} 0 &= g'_{00}/(c_{+}^{k})^{2} + 2g'_{k0}/c_{+}^{k} + g'_{kk}, \\ 0 &= g'_{00}/(c_{-}^{k})^{2} + 2g'_{k0}/c_{-}^{k} + g'_{kk} \end{align*} for $1\le k\le D$. This provides $2D$ equations that are linear in the metric components.

Part 3. Lastly, consider the time interval from $y^{0} = 0$ to $y^{0} = 1$, and set $g'_{00}$ to be the negative square of the time $T$ elapsed there, so that $g'_{00} = -T^{2}$.

From parts 2 and 3, we obtain $2D+1$ many equations with $2D+1$ components to solve for. Without proof, I claim this is a solvable system of equations, and so all components $g'_{\mu\nu}$ are known in the $(y^{\mu})$-coordinates.

By transforming back to the $(x^{\mu})$-coordinates, we find $g_{\mu\nu}$'s in the original coordinate system, as desired.