In almost general case, the space-time metrics looks like: \begin{equation} ds^2 = g_{00}(dx^0)^2 + 2g_{0i}dx^0dx^i + g_{ik}dx^idx^k, \end{equation} where $i,k = 1 \ldots 3$ - are spatial indeces.

The spatial distance between points (as determined, for example, by the stationary observer): \begin{equation} dl^2 = \left( -g_{ik} + \frac{g_{0i}g_{0j}}{g_{00}}\right)dx^idx^j = \gamma_{ik}dx^idx^j, \end{equation} where \begin{equation} \gamma_{ik} = -g_{ik} + \frac{g_{0i}g_{0j}}{g_{00}} , \end{equation}

And we can rewrite merics in a form: \begin{equation} ds^2 = g_{00}(dx^0 - g_idx^i)^2 - dl^2, \end{equation} in the last expression $g_i$ is: \begin{equation}\label{vecg} g_i = -\frac{g_{0i}}{g_{00}}. \end{equation}

My question in general souns like, how do the velocity of a certain particle are determined by different types of observers? Or special case: which type of the observer can determine velocity $v^2 = \frac{dl^2}{g_{00}(dx^0 - g_idx^i)^2}$?

I am confused by the fact that time $\sqrt{g_{00}}(dx^0 - g_idx^i)$ is not the proper time of any observer. What is this time?

I had draw the space-time diagram to clarify my question. Here $x^0$ - coordinate time, $t$ - "orthogonal" time defined as $dt = dx^0 - g_idx^i$. Is it correct? enter image description here

  • $\begingroup$ Hi. It would depend on the type of matter distribution if you have curved spacetime. The most common case is in cosmology for example, where observers move along "flow lines" along the fluid, so an observer's 4-velocity is $(1,0,0,0)$. $\endgroup$ Commented Oct 17, 2017 at 19:27
  • $\begingroup$ The formula is valid also in flat spacetime dealing with non-orthogonal coordinates. You are computing just the velocity using the proper time evaluated along a $x^0$-curve... Tomorrow I will write an explicit answer. $\endgroup$ Commented Oct 17, 2017 at 21:11
  • $\begingroup$ @Valter Moretti "proper time evaluated along..." which type of an observer belong this proper time? $\endgroup$
    – Sergio
    Commented Oct 18, 2017 at 6:05
  • $\begingroup$ @Sergio, I am too busy. I hope later today I will able to answer your question. It is related to the so-called Born metric.... $\endgroup$ Commented Oct 18, 2017 at 8:04
  • $\begingroup$ Which book/paper do you get these formulas? Landau vol 2, 2nd ed §89 motivates your formula that it is the time in point B (dx going from A to B) at the time x0 at point A. (it is in the chapter of time independent gravitational field, and I dont know if it applies generally) $\endgroup$
    – lalala
    Commented Dec 27, 2023 at 14:37

2 Answers 2


I assume that, referring to local coordinates $x^0,x^1,x^2,x^3$, the vector $\partial_{x^0}$ is timelike and $\partial_{x^i}$ are spacelike for $i=1,2,3$.

I henceforth use the signature $-,+,+,+$.

Notions like velocity can be defined in general coordinate systems but some precautions are necessary.

  • First of all $x^0$ is not the physical time measured along timelike curves $x^i=$ constant (i=1,2,3) representing observers at rest with the coordinates and thus parametrized by the coordinate $x^0$. The physical time, measured with physical clocks, is the proper time along these curves $d\tau = \sqrt{-g_{00}} dx^0$. In practical terms, a "small" displacement $$\Delta a \partial_{x^0}$$ along this temporal axis corresponds to a physical interval of time $$\Delta\tau = \sqrt{-g(\Delta a \partial_{x^0},\Delta a \partial_{x^0})}= \Delta a \sqrt{-g_{00}}$$

  • A spacelike $3$-surface $\Sigma_{x^0}$ defined by fixing $x^0=$ constant can be interpreted as the rest space of the coordinate system provided a suitable (Euclidean) metric is defined on it. This is not the usual metric $h$ induced by $g$ in the standard way $h(X,Y):= g(X,Y)$ for $X,Y$ tangent to $\Sigma_{x^0}$. The definition of the physically correct metric arises form the constraint that light-like paths must have constant velocity $1$. It is not difficult to prove that (e.g., see Landau-Lifsits' book on Field Theory sect. 84 ch.10 for a nice physical "proof") the appropriate metric is just that you wrote. If you consider a pair of vectors tangent to the rest space $\Sigma_{x^0}$, i.e., $$X= \sum_{i=1}^3 X^i\partial_{x^i}\quad \mbox{and} \quad Y= \sum_{i=1}^3 Y^i\partial_{x^i}$$ then the physical scalar product is $$\gamma(X,Y) := \sum_{i,j=1}^3 \left(g_{ij} - \frac{g_{i0}g_{j0}}{g_{00}}\right) X^iY^j\:.$$ Notice that $\gamma$ is defined on all vectors in spacetime not only those tangent to $\Sigma_{x^0}$: If $$X= X^\mu\partial_{x^\mu}\quad \mbox{and} \quad Y= Y^\nu\partial_{x^\nu}$$ $$\gamma(X,Y) := \left(g_{\mu\nu} - \frac{g_{\mu 0}g_{\nu 0}}{g_{00}}\right) X^\mu Y^\nu = \sum_{i,j=1}^3 \left(g_{ij} - \frac{g_{i0}g_{j0}}{g_{00}}\right) X^iY^j$$ and it automatically extracts the spatial part of them, since $\gamma(X, \partial_{x^0})=0$.

Now we pass to the definition of velocity of a particle with respect to the said reference frame. The story of the particle is a curve $x^\mu = x^\mu(s)$, the nature of $s$ does not matter. The tangent vector to this curve is $$X= \frac{dx^\mu}{ds} \partial_{x^\mu}$$
We can write, if the curve is sufficiently smooth, $$x^\mu(s+ \Delta s) = x^\mu(s) + \Delta s X^\mu(s) + O((\Delta s^2)) \:.$$ During the interval of parameter $\Delta s$, the particle runs in $\Sigma_{x^{0}(s)}$ an amount of physical space
$$\Delta l = \sqrt{\gamma\left( \Delta s X,\Delta s X\right)} = \Delta s \sqrt{\gamma(X,X)}$$ up to second order $\Delta s$ infinitesimals.
The corresponding amount of physical time is extracted from the orthogonal projection of $\Delta s X$ along the time axis $\partial_{x^0}$ taking its normalization into account: $$T = g\left( \Delta s X, \frac{\partial_{x^0}}{\sqrt{-g_{00}}}\right) \frac{\partial_{x^0}}{\sqrt{-g_{00}}} = \Delta s \frac{X^0 g_{00} + \sum_{i=1}^3 g_{0i}X^i}{-g_{00}}\partial_{x^0}\:.$$

According to my first comment above, the length (with respect to $g$) of this vector is just the amount of physical time spent by the particle in the interval $\Delta s$ of parameter. This time is measured by the proper time of a clock moving along the $x^0$ axis. $$\Delta \tau = \sqrt{-g(T,T)} = \Delta s \frac{X^0 g_{00} + \sum_{i=1}^3 g_{0i}X^i}{\sqrt{-g_{00}}}$$ up to second order $\Delta s$ infinitesimals.

In summary, the velocity of the particle referred to the coordinates $x^0,x^1,x^2,x^3$ is $$v = \lim_{\Delta s \to 0}\frac{\Delta l}{\Delta \tau} = \sqrt{-g_{00}}\frac{\sqrt{\gamma(X,X)}}{X^0 g_{00} + \sum_{i=1}^3 g_{0i}X^i}$$ so that $$v^2 = -g_{00}\frac{\gamma(X,X)}{(X^0 g_{00} + \sum_{i=1}^3 g_{0i}X^i)^2} $$ This is the rigorous expression of the formula you wrote. The vector $X$ in your case has components $X^\mu = dx^\mu$. The minus sign in front of the right-hand side is just due to the different choice of the signature of the metric.

COMMENT. These notions are not related with the choice of a curved spacetime. Everything is valid also in Minkowski spacetime referring to non-Minkowskian coordinates with $g_{0k}\neq 0$ (and possibly $g_{00} \neq -1$). A standard example are coordinates at rest with a rotating platform with respect to an inertial system in Minkowki spacetime.

  • $\begingroup$ Can you, please, clarify me about your answer with space-time diagramm. I had draw it and added to may question, but I do not completely sure of its correctness. $\endgroup$
    – Sergio
    Commented Oct 20, 2017 at 12:34
  • $\begingroup$ @Sergio Yes, barring problems with the different signature, it seems to me that your picture is correct. $\endgroup$ Commented Oct 20, 2017 at 16:03
  • $\begingroup$ The crucial observation is the "line of simultaneity" you correctly indicated. To be precise, that line should be orthogonal to the temporal lines. The $x^0$ constant 3-surfaces are not orthogonal to $x^k=$ constant ($k=1,2,3$) lines just because $g_{0k} \neq 0$. $\endgroup$ Commented Oct 20, 2017 at 16:06
  • $\begingroup$ I have changed the diagram. Now "line of simultaneity" orthogonal to the temporal lines. Is it correct? $\endgroup$
    – Sergio
    Commented Oct 20, 2017 at 16:46
  • $\begingroup$ Yes, it seems correct! $\endgroup$ Commented Oct 20, 2017 at 16:48

You write

The spatial distance between points (as determined, for example, by the stationary observer)...

But don't forget spatial distance is relative to the observer; recall "length-contraction" as taught in introductory special relativity. It is only those observers which are comoving with your given coordinate system $x^\mu$ which determine your spatial metric $\gamma_{ik}$. These observers have 4-velocity $(1/\sqrt{g_{00}},0,0,0)$, under your +--- metric signature. The spatial metric you give was derived by Landau & Lifshitz in their fields textbook (see $\S84$ in the 1994 edition). Landau & Lifshitz do state this assumption of comoving observers, and also give conditions for when such observers exist. Different observers will determine a different spatial 3-metric. I am writing a paper on such topics, and will link it when completed.

The discussed properties should be understood as only local, meaning in the immediate vicinity of the observer; this is particularly true in curved spacetime.

As for relative velocity in curved spacetime, suppose $\mathbf u$ and $\mathbf v$ are two observers (4-velocities) at the same event. Then their relative 3-velocity has Lorentz factor $$\gamma=\mathbf u\cdot\mathbf v$$ assuming metric signature +---. See Carroll's textbook.


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