The Hyperbolic Geometry of Einstein's Relativity

How changing your velocity means traversing a hyperbolic space,

Maria Nicolae, 30 March 2026

In a previous post, I teased that there is an intimate connection between Einsteinian relativity and hyperbolic (Lobachevskian) geometry, but that it was "a story for another time". Well, that time has come; in this post, I'll elaborate on that remark and illustrate this connection.

In Einsteinian relativity, velocities do not trivially add. Rather, it gets harder to accelerate as you get closer to the speed of light. If you associate with the space of velocities a geometry in which the "distance" between two velocities is how hard you have to accelerate to change from one to the other, this geometry turns out to be hyperbolic. In this post, I'll show that this is the case. Starting from Einstein's postulates, I derive how velocities transform, and use this to derive a geometry of velocity-space expressed in the language of differential geometry as a Riemannian manifold. If you need an overview of differential geometry, I recently wrote about it here. Finally, I show that this geometry is hyperbolic.

How Velocities Transform

To find out how velocities transform, I will start with how spacetime itself transforms, and then derive from that the transformation rule for velocities by differentiating paths through spacetime.

Lorentz Transformations of Spacetime

Einsteinian relativity is axiomatised by Einstein's postulates:

the laws of physics are the same in all inertial frames of reference (inertia), and
the speed of light is the same for all observers.

From these, we can figure out which transformations of spacetime are valid, i.e. keep the laws of physics the same. These are called the Lorentz transformations.

Let the spacetime coordinates of a point before transformation be ${\bf x} = [t, {\bf r}]^T$ , where $t$ is the time coordinate and $\bf r$ are the spatial coordinates, and let ${\bf x}' = [t', {\bf r}']^T$ be the coordinates after a Lorentz transformation. The first postulate tells us that inertial (constant-velocity) paths through spacetime remain inertial when they transform. Such inertial paths are represented by straight lines in $\bf x$ spacetime coordinates, so Lorentz transformations map straight lines to straight lines. Furthermore, these transformations should keep parallel lines parallel, since whether or not two objects collide should be something that all observers agree upon. Thus, Lorentz transformations are linear transformations, and can be represented by a matrix multiplication

{\bf x}' = \Lambda{\bf x}

(1)

where $\Lambda$ is a Lorentz transformation matrix. In block matrix form, this is

\begin{bmatrix} t'\\ {\bf r}' \end{bmatrix} = \begin{bmatrix} \Lambda_{tt} & {\bf\Lambda}_{t{\bf r}}^T\\ {\bf\Lambda}_{{\bf r}t} & \Lambda_{\bf rr} \end{bmatrix} \begin{bmatrix} t\\ {\bf r} \end{bmatrix}.

(2)

The second postulate tells us that Lorentz transformations preserve the light cone $\left|\left|{\bf r}\right|\right| = ct$ . Adding this condition to the previous condition of linearity constrains the Lorentz transformations up to an overall scaling. Obviously, scaling is not a valid Lorentz transformation, since the laws of physics are not scale-invariant. Thus, the last constraint we need is that, for all Lorentz matrices $\Lambda$ , $\det{\Lambda} = 1$ .

Examples of Lorentz transformations include spatial rotations

\Lambda_\text{rotation} = \begin{bmatrix} 1 & {\bf 0}^T \\ {\bf 0} & R \end{bmatrix}

(3)

where $R$ is a spatial rotation matrix, as well as velocity changes (boosts)

\begin{align*} \Lambda_\text{boost}({\bf v}) &= \left[\begin{array} \gamma & \frac{\gamma}{c^2}{\bf v}^T\\ \gamma{\bf v} & I + \frac{\gamma-1}{v^2}{\bf v}{\bf v}^T \end{array}\right]\\ \gamma &= \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \end{align*}

(4)

where $v = \left|\left|{\bf v}\right|\right|$ . All Lorentz transformations can be expressed as compositions of these.

Lorentz Transformations of Velocity

To figure out how Lorentz transformations affect velocity, I'll consider a path through spacetime ${\bf x}(\tau) = [t(\tau), {\bf r}(\tau)]^T$ and relate the velocity to the derivative of that path. The Lorentz transformation of the spacetime path is

{\bf x'}(\tau) = \Lambda{\bf x}(\tau).

(5)

Because differentiation and matrix multiplication are both linear, the derivatives of the original and transformed paths are also related by the Lorentz transformation:

\frac{d{\bf x}'}{d\tau} = \Lambda\frac{d{\bf x}}{d\tau}.

(6)

Expanding this into a block matrix, as per Equation 2, we obtain

\begin{bmatrix} \frac{dt'}{d\tau}\\ \frac{d{\bf r}'}{d\tau} \end{bmatrix} = \begin{bmatrix} \Lambda_{tt} & {\bf\Lambda}_{t{\bf r}}^T\\ {\bf\Lambda}_{{\bf r}t} & \Lambda_{\bf rr} \end{bmatrix} \begin{bmatrix} \frac{dt}{d\tau}\\ \frac{d{\bf r}}{d\tau} \end{bmatrix}.

(7)

With the original and transformed velocities being

\begin{align*} {\bf v} &= \frac{d\bf r}{dt}\\ {\bf v}' &= \frac{d{\bf r}'}{dt'}, \end{align*}

(8)

and dividing both sides of Equation 7 by $dt/d\tau$ ,

\begin{bmatrix} \frac{dt'}{dt}\\ \frac{dt'}{dt} {\bf v}'\\ \end{bmatrix} = \begin{bmatrix} \Lambda_{tt} & {\bf\Lambda}_{t{\bf r}}^T\\ {\bf\Lambda}_{{\bf r}t} & \Lambda_{\bf rr} \end{bmatrix} \begin{bmatrix} 1\\ {\bf v} \end{bmatrix}.

(9)

Thus,

{\bf v}' = \frac{1}{\Lambda_{tt} + {\bf\Lambda}_{t\bf r}\cdot\bf v} \left({\bf\Lambda}_{{\bf r}t} + \Lambda_{\bf rr} {\bf v}\right);

(10)

the "recipe" for transforming a velocity vector ${\bf v}$ is:

prepend $1$ to the vector ${\bf v}$ to get a spacetime vector,
multiply this vector by the Lorentz transformation matrix $\Lambda$ , and
turn the result into a velocity vector by dividing the spatial components (last ones) by the time component (the first).

This is a projective transformation of velocity space, named as such because the last step projects the spacetime vector onto the spatial subspace.

The Geometry of Velocity-Space

I now associate with the set of velocities a geometry, in which the "distance" between two velocities is how much you have to accelerate to change from one velocity to the other:

s({\bf v}_a, {\bf v}_b) = \int_{{\bf v}_a}^{{\bf v}_b} a\:d\tau

(11)

where $a$ is the proper acceleration and $\tau$ is the proper time, which are the acceleration and time measured in the moving reference frame. We want to express this in terms of a Riemannian metric

s({\bf v}_a, {\bf v}_b) = \int_{{\bf v}_a}^{{\bf v}_b} ds = \int_{{\bf v}_a}^{{\bf v}_b} \sqrt{d{\bf v}\cdot g({\bf v}) d{\bf v}}.

(12)

To do this, there are two properties of the velocity distance metric that we can use. First is the nonrelativistic limit, for which

s({\bf 0}, d{\bf v}) = \left|\left|d{\bf v}\right|\right|

(13)

and therefore

g({\bf 0}) = I.

(14)

The second property is that this distance is preserved by Lorentz transformations, which are the isometries of this geometry. Namely, given a Lorentz transformation of the start and end velocities $({\bf v}'_a, {\bf v}'_b)$ ,

s\left({\bf v}'_a, {\bf v}'_b\right) = s({\bf v}_a, {\bf v}_b).

(15)

To compute $d s$ in a way that will reveal the form of the metric tensor $g({\bf v})$ , I use this second property for the Lorentz transformation $\Lambda = \Lambda_\text{boost}(-{\bf v})$ , in terms of which ${\bf v}' = 0$ , which then lets me apply the first property:

\begin{align*} ds &= s({\bf v}, {\bf v}+d{\bf v})\\ &= s({\bf 0}, d{\bf v}')\\ &= \left|\left|d{\bf v}'\right|\right|. \end{align*}

(16)

The transformed differential velocity $d{\bf v}'$ in this expression can be evaluated as a differential of Equation 10:

\begin{align*} d{\bf v}' &= \frac{1}{\Lambda_{tt} + {\bf\Lambda}_{t\bf r}\cdot({\bf v}+d{\bf v})} \left[{\bf\Lambda}_{{\bf r}t} + \Lambda_{\bf rr} ({\bf v} + d{\bf v})\right]\\ &= \frac{1}{\Lambda_{tt} + {\bf\Lambda}_{t\bf r}\cdot{\bf v} + {\bf\Lambda_{t\bf r}}\cdot d{\bf v}} \left(\underbrace{{\bf\Lambda}_{{\bf r}t} + \Lambda_{\bf rr} {\bf v}}_{\bf 0} + \Lambda_{\bf rr} d{\bf v}\right)\\ &= \frac{1}{\Lambda_{tt} + {\bf\Lambda}_{t\bf r}\cdot{\bf v}} \Lambda_{\bf rr} d{\bf v}. \end{align*}

(17)

Substituting the form of the boost matrix from Equation 4, this becomes

\begin{align*} d{\bf v}' &= \frac{1}{\gamma - \frac{\gamma v^2}{c^2}} \left(I + \frac{\gamma-1}{v^2}{\bf v}{\bf v}^T\right) d{\bf v}\\ &= \frac{1}{\gamma\left(1-\frac{v^2}{c^2}\right)} \left(I + \frac{\gamma-1}{v^2}{\bf v}{\bf v}^T\right) d{\bf v}\\ &= \gamma \left(I + \frac{\gamma-1}{v^2}{\bf v}{\bf v}^T\right) d{\bf v}\\ \end{align*}

(18)

Substituting this back into Equation 16,

ds = \left|\left|\gamma\left(I + \frac{\gamma-1}{v^2}{\bf v}{\bf v}^T\right) d{\bf v}\right|\right|.

(19)

Then, by squaring both sides and replacing the vector norm with an explicit dot product

\begin{align*} ds^2 &= \left|\left| \gamma \left(I + \frac{\gamma-1}{v^2}{\bf v}{\bf v}^T\right) d{\bf v}\right|\right|^2\\ &= \gamma^2 \left(I + \frac{\gamma-1}{v^2}{\bf v}{\bf v}^T\right)d{\bf v} \cdot \left(I + \frac{\gamma-1}{v^2}{\bf v}{\bf v}^T\right)d{\bf v}\\ &= \gamma^2 d{\bf v} \cdot \left(I + \frac{\gamma-1}{v^2}{\bf v}{\bf v}^T\right)^T \left(I + \frac{\gamma-1}{v^2}{\bf v}{\bf v}^T\right)d{\bf v}\\ &= d{\bf v} \cdot g({\bf v}) d{\bf v}, \end{align*}

(20)

we finally find the metric tensor

\begin{align*} g({\bf v}) &= \gamma^2 \left(I + \frac{\gamma-1}{v^2}{\bf v}{\bf v}^T\right)^T \left(I + \frac{\gamma-1}{v^2}{\bf v}{\bf v}^T\right)\\ &= \gamma^2 \left(I + \frac{\gamma-1}{v^2}{\bf v}{\bf v}^T\right)^2\\ &= \gamma^2 \left(I^2 + \frac{2\gamma-2}{v^2}{\bf v}{\bf v}^T + \frac{\gamma^2-2\gamma+1}{v^4}{\bf v}{\bf v}^T{\bf v}{\bf v}^T\right)\\ &= \gamma^2 \left(I^2 + \frac{2\gamma-2}{v^2}{\bf v}{\bf v}^T + \frac{\gamma^2-2\gamma+1}{v^2}{\bf v}{\bf v}^T\right)\\ &= \gamma^2 \left(I^2 + \frac{\gamma^2-1}{v^2}{\bf v}{\bf v}^T\right)\\ g({\bf v}) &= \gamma^2 \left(I^2 + \frac{\gamma^2}{c^2}{\bf v}{\bf v}^T\right). \end{align*}

(21)

Velocity-Space is Hyperbolic

By construction, the metric in Equation 21 is spherically symmetric, so from this point on, I work with two-dimensional velocities ${\bf v} = [v_x, v_y]^T$ for simplicity. The differential line element for this metric is

\begin{align*} ds^2 &= \gamma^2 \left[\left(1+\frac{\gamma^2}{c^2}v_x^2\right)dv_x^2 + \left(1+\frac{\gamma^2}{c^2}v_y^2\right)dv_y^2 + \frac{2\gamma^2}{c^2}v_xv_ydv_xdv_y\right]\\ &= \gamma^2 \left[dv_x^2 + dv_y^2 + \frac{\gamma^2}{c^2}\left(v_xdv_x + v_ydv_y\right)^2\right]. \end{align*}

(22)

To analyse this, I first convert it to polar coordinates $(v, \theta)$ for which

\begin{align*} v_x &= v\cos\theta\\ v_y &= v\sin\theta. \end{align*}

(23)

The differentials of these are

\begin{align*} dv_x &= \cos(\theta)dv - v\sin(\theta)d\theta\\ dv_y &= \sin(\theta)dv + v\cos(\theta)d\theta, \end{align*}

(24)

and substituting these into Equation 22 gives us

\begin{align*} ds^2 &= \gamma^2 \left[\left(\cos(\theta)dv - v\sin(\theta)d\theta\right)^2 + \left(\sin(\theta)dv + v\cos(\theta)d\theta\right)^2 + \frac{\gamma^2}{c^2}\left((v\cos\theta)(\cos(\theta)dv-v\sin(\theta)d\theta) + (v\sin\theta)(\sin(\theta)dv+v\cos(\theta)d\theta)\right)^2\right]\\ &= \gamma^2 \left[\cos^2(\theta)dv^2 + v^2\sin^2(\theta)d\theta^2 - v\sin(\theta)\cos(\theta)dvd\theta + \sin^2(\theta)dv^2 + v^2\cos^2(\theta)d\theta^2 + v\sin(\theta)\cos(\theta)dvd\theta + \frac{\gamma^2}{c^2}\left(v\cos^2(\theta)dv - v^2\sin(\theta)\cos(\theta)d\theta + v\sin^2(\theta)dv + v^2\sin(\theta)\cos(\theta)d\theta\right)^2\right]\\ &= \gamma^2 \left[dv^2 + v^2d\theta^2 + \frac{\gamma^2}{c^2}(v dv)^2\right]\\ &= \gamma^2 \left[\left(1+\frac{\gamma^2v^2}{c^2}\right)dv^2 + v^2d\theta^2\right]\\ ds^2 &= \gamma^4 dv^2 + \gamma^2 v^2 d\theta^2. \end{align*}

(25)

Finally, to make the geometry clear, I switch to "azimuthal equidistant" coordinates $(\rho, \theta)$ , where I define $\rho$ as the distance from the origin

\begin{align*} \rho &= \int_0^v ds\\ &= \int_0^v \gamma(v')^2 dv'\\ &= \int_0^v \frac{1}{1-\frac{v'^2}{c^2}} dv'\\ \rho &= c \tanh^{-1} \frac{v}{c}. \end{align*}

(26)

This is the function of velocity for which the one-dimensional relativistic addition of velocities is linear

\rho(v_\text{sum}) = \rho(v_1) + \rho(v_2);

(27)

the nondimensionalised version of this quantity, $\zeta = \rho/c$ , is known as the "rapidity" in the parlance of Einsteinian relativity. From this, we obtain

\begin{align*} v &= c \tanh \frac{\rho}{c}\\ dv &= \frac{1}{\cosh^2 \frac{\rho}{c}} d\rho\\ \gamma &= \cosh \frac{\rho}{c}. \end{align*}

(28)

Substituting these into Equation 25, we obtain

\begin{align*} ds^2 &= \cosh^4\left(\frac{\rho}{c}\right) \frac{1}{\cosh^4 \frac{\rho}{c}} d\rho^2 + c^2\tanh^2\left(\frac{\rho}{c}\right)\cosh^2\left(\frac{\rho}{c}\right)d\theta^2\\ ds^2 &= d\rho^2 + c^2\sinh^2\left(\frac{\rho}{c}\right) d\theta^2. \end{align*}

(29)

We can see from this line element that this is the geometry of the hyperbolic plane, which you may recognise from the previous post about differential geometry. Circles around the origin of radius $\rho$ have circumferences that increase exponentially with $\rho$ , rather than linearly (as in the Euclidean plane) or sinusoidally (as on the surface of a sphere). We also see from this that $c$ is the characteristic scale of the curvature of this hyperbolic plane; for $\rho \ll c$ , the geometry is approximately Euclidean.

What This Tells Us

The fact that there is this connection between Einsteinian relativity and hyperbolic geometry tells us, first and foremost, that mathematical formalisms for one apply to the other. The Cartesian velocity coordinates $\bf v$ in the ball of radius $c$ , together with the metric in Equation 21, is a representation of hyperbolic geometry. Specifically, it is what is called the Beltrami-Klein model of hyperbolic geometry, which has the unique property that it represents geodesics of the hyperbolic geometry as straight lines in the model's Cartesian coordinates. Its isometries, then, must be those projective transformations (mapping straight lines to straight lines) which preserve the $c$ ball; as we have seen, these are the Lorentz transformations.

As an example of what hyperbolic geometry can teach us about Einsteinian relativity, there is the fact that hyperbolic geometry, like all curved spaces, exhibits holonomy. This is the phenomenon where, if you move around in a curved space without turning, you can nonetheless end up facing a different direction to what you started in. Consider, for example, moving on the curved surface of our spherical Earth: if you start out facing north at 0°N 0°E, walk forward to the North Pole, then walk right to the equator (0°N 90°E), and finally walk backwards to 0°N 0°E, you're back where you started, and are now facing east instead of north, despite having never turned your body. A similar phenomenon happens in hyperbolic geometry, and therefore in Einsteinian relativity: the combined effect of two boosts is not necessary a pure boost, so it's possible to rotate simply by accelerating in different directions, though the rotation is miniscule if your velocity boosts are much smaller than $c$ . Physicists call this the Wigner rotation.