Consider an infinitesimal coordinate transformation, $$ x^{\mu} \rightarrow x'^{\mu} = x^{\mu} + \epsilon^{\mu}(x). $$ We can show that the metric tensor under such a transformation, up to first order in $\epsilon$, becomes $$ g'_{\mu \nu}(x') = g_{\mu \nu} (x) - \partial_{\mu} \epsilon_{\nu} - \partial_{\nu} \epsilon_{\mu}.$$ Similarly, the differential $dx^{\mu}$ transforms as $$ dx'^{\mu} = \frac{\partial x'^{\mu}}{\partial x^{\alpha}} dx^{\alpha} = dx^{\mu} + \partial_{\alpha} \epsilon^{\mu} dx^{\alpha}. $$ Then, the metric distance transforms as $$ds^2 = g_{\mu \nu } dx^{\mu} dx^{\nu} \rightarrow (g_{\mu \nu} (x) - \partial_{\mu} \epsilon_{\nu} - \partial_{\nu} \epsilon_{\mu} ) \; (dx^{\mu} + \partial_{\alpha} \epsilon^{\mu} dx^{\alpha})\; (dx^{\nu} + \partial_{\beta} \epsilon^{\nu} dx^{\beta}) \\ = g_{\mu \nu } dx^{\mu} dx^{\nu} + \partial_{\alpha} \epsilon_{\nu} dx^{\alpha} dx^{\nu} + \partial_{\beta} \epsilon_{\mu} dx^{\mu} dx^{\beta} - \partial_{\mu} \epsilon_{\nu} dx^{\mu} dx^{\nu} - \partial_{\nu} \epsilon_{\mu} dx^{\mu} dx^{\nu} \\ = g_{\mu \nu } dx^{\mu} dx^{\nu} = ds^2. $$ We see that it is invariant under an infinitesimal coordinate transformation.

Is this true for any coordinate transformation? If so, what makes, say for example, Lorentz transformations leaving the Minkowski metric invariant any special? It seems like any infinitesimal coordinate transformation would leave the metric distance invariant.


3 Answers 3


Is this true for any coordinate transformation?

Yes. Given a curve $\gamma$ through spacetime, the quantity $$\Delta s := \int_\gamma \sqrt{g_{\mu\nu} \mathrm dx^\mu \mathrm dx^\nu}$$

is a scalar, and is therefore invariant under arbitrary coordinate changes.

If so, what makes, say for example, Lorentz transformations leaving the Minkowski metric invariant any special?

When talking about the observations of a human observer, it is most convenient to work in Cartesian coordinates, in which the metric components take the form $$g_{\mu\nu} = \pmatrix{-1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0&0&1&0\\0&0&0&1}$$

In such a coordinate system, $x^0\propto t$ can be interpreted as the time on the observer's wristwatch, while $(x^1,x^2,x^3)$ can be interpreted as distances along axes which are orthogonal to the $t$-axis and to each other. Events which are quantified through such a coordinate system can be readily translated into the experiences of an human observer in a way familiar from elementary mechanics.

The fact that Lorentz transformations leave the Minkowski metric invariant mean that they correspond to the experiences of human observers moving relative to one another. Given one inertial frame corresponding to some observer (i.e. a Cartesian coordinate system in which the metric components take the form above), a Lorentz transformation maps you to another frame, corresponding to a different (but equally valid) observer. On the other hand, a generic coordinate transformation is always valid, but may not correspond directly to the observations of a human observer.


The metric and hence the distance is not invariant under arbitrary smooth transformations.

However, the metric and hence the distance, is invariant under smooth isometric transformations essentially as a tautology. Notice that iso-metric means 'same [distance] measurement'.

  • $\begingroup$ Good point! The isometric transformations are so ubiquitous to the physics discipline that we rarely worry to mention them explicitly. $\endgroup$ Jan 10 at 3:24
  • $\begingroup$ The OP is talking about passive coordinate transformations, in which the metric changes along with the coordinates in such a way to leave scalars such as $g_{\mu\nu} \mathrm dx^\mu \mathrm dx^\nu$ invariant. An isometric transformation $f$ is a transformation between pre-defined metric manifolds - those with pre-existing metrics $g$ and $g'$ - such that $g = f^* g'$. $\endgroup$
    – J. Murray
    Jan 10 at 12:32
  • $\begingroup$ @J. Murray Could we not understand the transformation from either of the active or passive views? However, your point is well taken in that the OP's transformations are mappings of the manifold to itself rather than mappings between different manifolds. $\endgroup$ Jan 10 at 16:45
  • 1
    $\begingroup$ @J.Murray: To my mind, a passive transformation is a coordinate change, it leaves the underlying manifold unchanged. An active transformation actually moves the points of a manifold around. The notion of a metric is irrelevant here. However, when there is a metric involved then we need to take account of it as you point out. $\endgroup$ Jan 10 at 17:00
  • $\begingroup$ @J. Murray It seems to me that a passive transformation of coordinates is iso-metric. Why do suggest that this is not so? $\endgroup$ Jan 10 at 17:07

The proof you have given for the invariance of $ds^2$ simply reflects the fact that scalar quantities are tensors of rank $0$ and thus have the same value in all coordinated systems (where the transformations are distance preserving), all scalars are invariant under such transformations, not just the metric distance between two points in space time. For example, the speed of light, another scalar, is the same in all coordinate systems. However, tensors of higher rank, like vectors and covectors, do not remain invariant under transformations of coordinates. These objects transform according to the very specific rules for the transformations of tensors. Familiar examples of tensors are the electromagnetic vector potential, $A^\mu$,(a tensor of contravariant rank 1) and the covector $\partial_\mu=\partial /\partial x^\mu$, (a tensor of covariant rank 1).

Now consider the wave equation:$$\partial_\mu\partial^\mu A^\nu={\partial^2A^\nu\over\partial (ct)^2}-{\partial^2A^\nu\over\partial {x^1}^2}=0.$$ Here we have a rank one tensor equation that is not necessarily invariant under any and all coordinate transformations, in fact, it is well known that if one tries to transform the coordinates according to the principle of Galilean relativity, that the equation will loose its familiar form. However, if the coordinate transformation is a Lorentz transformation, the equation is invariant, i.e. the equation in transformed coordinates can be found by simply replacing all the relevant quantities by their transformed counterparts:$${\partial^2{A^\nu}^\prime\over\partial(ct^\prime)^2}-{\partial^2{A^\nu}^\prime\over\partial{{x^1}^\prime}^2}=0.$$ Thus, the Lorentz transformation, the transformation that preserves the form of the Minkowski metric is also a symmetry of Maxwell's electromagnetic theory (cf. Jackson, 514-516). Albert Einstein's theory was in favor of the symmetry group of the Maxwell equations over that of the Newtonian laws (Galilean symmetry), hence the reason for stipulating the invariance of the Minkowski metric specifically.

  • $\begingroup$ "... thus have the same value in all coordinate systems" - not all co-ordinate systems. For example, what if we double the magnitude of all our basis vectors ? $ds^2$ is not invariant under this transformation. $\endgroup$
    – gandalf61
    Jan 10 at 16:28
  • $\begingroup$ @gandalf61 That is simply false. The spacetime interval is a coordinate-independent quantity always, by construction. $\endgroup$ Apr 18 at 22:31
  • $\begingroup$ @VincentThacker $ds^2$ is only an invariant if the co-ordinate transformation is an isometry. There are many co-ordinate transformations that are not isometries. And note that the phrase in brackets "where the transformations are distance preserving" was added to the answer after my comment was posted. $\endgroup$
    – gandalf61
    Apr 19 at 6:39
  • $\begingroup$ @gandalf61 No, this is completely false. You are misinformed. The length of a vector is always invariant. Albertus' original version is correct, as also stated in the first few lines of J.Murray's answer. $\endgroup$ Apr 19 at 7:33
  • $\begingroup$ This is unrelated to isometries because we are talking about changing coordinates of a single manifold. Diffeomorphisms and isometries are maps between manifold(s) and are much more general. $\endgroup$ Apr 19 at 7:43

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