How to Prove Minkowski Metric Invariance Under Lorentz Transformations

The components of the Minkowski metric are not unchanged under every change of coordinates. They retain the same Minkowski form under the homogeneous Lorentz transformations that relate inertial frames; translations also preserve intervals, as discussed below.

The central result is

$$\Lambda^{\mathsf T}\eta\Lambda=\eta,$$

where $\Lambda$ is the Lorentz-transformation matrix and $\eta$ is the Minkowski metric matrix. This identity is exactly what preserves the spacetime interval.

The Minkowski metric and spacetime interval

Use coordinates

$$x^\mu=(x^0,x^1,x^2,x^3)=(ct,x,y,z),$$

and choose the mostly-plus sign convention

$$\eta_{\mu\nu}=\operatorname{diag}(-1,1,1,1).$$

For an infinitesimal displacement $dx^\mu$, the squared spacetime interval is

$$ds^2=\eta_{\mu\nu}\,dx^\mu dx^\nu =-(c\,dt)^2+dx^2+dy^2+dz^2.$$

Some texts instead use the mostly-minus convention $\operatorname{diag}(1,-1,-1,-1)$. All the arguments are identical; only the overall signs change.

Matrix proof of invariance

Let a Lorentz transformation act on the coordinates as

$${x'}^\mu=\Lambda^\mu{}_{\nu}x^\nu.$$

For displacements, this gives

$$d{x'}^\mu=\Lambda^\mu{}_{\nu}\,dx^\nu.$$

The interval measured in the primed frame is therefore

$$\begin{aligned} (ds')^2 &=(dx')^{\mathsf T}\eta\,dx'\\ &=(\Lambda dx)^{\mathsf T}\eta(\Lambda dx)\\ &=dx^{\mathsf T}\Lambda^{\mathsf T}\eta\Lambda dx. \end{aligned}$$

Thus, if

$$\Lambda^{\mathsf T}\eta\Lambda=\eta,$$

then

$$(ds')^2=dx^{\mathsf T}\eta dx=ds^2.$$

This is the proof. The equation $\Lambda^{\mathsf T}\eta\Lambda=\eta$ is also the defining condition for a Lorentz matrix.

In index notation, the same condition is

$$\eta_{\rho\sigma}\Lambda^\rho{}_{\mu}\Lambda^\sigma{}_{\nu}=\eta_{\mu\nu}.$$

It says that the metric pairing of any two four-vectors is preserved:

$$(\Lambda u)\cdot(\Lambda v)=u\cdot v.$$

Direct verification for a boost

Consider a boost in the $x$ direction with $|v|<c$. Define

$$\beta=\frac{v}{c}, \qquad \gamma=\frac{1}{\sqrt{1-\beta^2}}.$$

With $x^0=ct$, the transformation is

$$\begin{pmatrix} ct'\\x'\\y'\\z' \end{pmatrix} = \begin{pmatrix} \gamma&-\beta\gamma&0&0\\ -\beta\gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix} \begin{pmatrix} ct\\x\\y\\z \end{pmatrix}.$$

Call this matrix $\Lambda$. Its nontrivial $2\times2$ block is

$$B=\begin{pmatrix} \gamma&-\beta\gamma\\ -\beta\gamma&\gamma \end{pmatrix}, \qquad \eta_2=\begin{pmatrix}-1&0\\0&1\end{pmatrix}.$$

Compute the transformed metric block:

$$\begin{aligned} B^{\mathsf T}\eta_2B &= \begin{pmatrix} \gamma&-\beta\gamma\\ -\beta\gamma&\gamma \end{pmatrix} \begin{pmatrix}-1&0\\0&1\end{pmatrix} \begin{pmatrix} \gamma&-\beta\gamma\\ -\beta\gamma&\gamma \end{pmatrix}\\[4pt] &= \begin{pmatrix} -\gamma^2+\beta^2\gamma^2&0\\ 0&\gamma^2-\beta^2\gamma^2 \end{pmatrix}\\[4pt] &= \begin{pmatrix}-1&0\\0&1\end{pmatrix} =\eta_2, \end{aligned}$$

because $\gamma^2(1-\beta^2)=1$.

The $y$ and $z$ coordinates are unchanged, so the complete matrix satisfies

$$\Lambda^{\mathsf T}\eta\Lambda=\eta.$$

Equivalently, substituting the Lorentz formulas directly gives

$$-(c\,dt')^2+dx'^2+dy'^2+dz'^2 =-(c\,dt)^2+dx^2+dy^2+dz^2.$$

Tensor-component interpretation

There are two closely related statements that are often mixed together.

First, the metric is a geometric tensor. Its components transform under a coordinate change according to

$$g'_{\alpha\beta} =\frac{\partial x^\mu}{\partial x'^\alpha} \frac{\partial x^\nu}{\partial x'^\beta} g_{\mu\nu}.$$

If $x'=\Lambda x$, then $x=\Lambda^{-1}x'$, so in matrix notation

$$\eta'= (\Lambda^{-1})^{\mathsf T}\eta\Lambda^{-1}.$$

For a Lorentz transformation, this is again $\eta$. Therefore every inertial Cartesian coordinate system has the same metric components.

Second, one may actively transform a four-vector while keeping the coordinate basis fixed. In that case the preserved inner product is expressed as

$$(\Lambda u)^{\mathsf T}\eta(\Lambda v)=u^{\mathsf T}\eta v,$$

which directly gives $\Lambda^{\mathsf T}\eta\Lambda=\eta$.

The coordinate and active viewpoints use inverse matrices in different places, but they express the same physical invariance.

What is and is not invariant?

The individual coordinates $t$, $x$, $y$, and $z$ are not invariant. Different inertial observers generally assign different times and spatial positions to the same events. What is invariant is the metric interval constructed from them:

$$\Delta s^2=-c^2\Delta t^2+\Delta x^2+\Delta y^2+\Delta z^2.$$

For finite event separations, replace $dx^\mu$ by $\Delta x^\mu$; the same matrix proof applies.

Translations such as $x'^\mu=\Lambda^\mu{}_{\nu}x^\nu+a^\mu$ also preserve intervals, because the constant translation cancels when taking differences. Lorentz transformations together with translations form the Poincaré group.

Common mistakes

• Using an ordinary orthogonal condition. In Euclidean space the condition is $R^{\mathsf T}R=I$. In Minkowski spacetime it is $\Lambda^{\mathsf T}\eta\Lambda=\eta$.
• Forgetting the metric signature. The formulas depend on whether $\eta=\operatorname{diag}(-1,1,1,1)$ or $\eta=\operatorname{diag}(1,-1,-1,-1)$ is used.
• Calling the coordinates invariant. Lorentz transformations change coordinates; they preserve the interval and the metric pairing.
• Mixing active and passive transformations. Decide whether vectors or coordinate components are being transformed, and use the corresponding matrix or inverse matrix consistently.
• Treating arbitrary coordinate changes as Lorentz transformations. Curvilinear coordinates, accelerated coordinates, and general coordinate transformations need not leave the metric components equal to $\eta$ everywhere.