Four-Velocity Vector

In simple terms, the four-velocity represents the velocity measured with respect to an object’s proper time $ \tau $.

$$ U^\mu = \frac{dx^\mu}{d\tau} $$

The scalar product between the contravariant and covariant four-velocity is constant in all inertial frames, making it a relativistic invariant:

$$ U^\mu U_\mu = c^2 $$

This value remains the same for every observer, regardless of the chosen inertial reference frame.

Explanation and Derivation

The four-velocity is defined as the derivative of the position four-vector $x^\mu = ( ct, x, y, z )$ with respect to the object’s proper time $\tau$:

$$ U^\mu = \frac{dx^\mu}{d\tau} $$

This describes how a particle’s spacetime coordinates change with respect to its own proper time - that is, the time measured in the frame where the particle is at rest.

Expanding the four components gives:

$$ U^\mu = \left( \frac{d(ct)}{d\tau}, \frac{dx}{d\tau}, \frac{dy}{d\tau}, \frac{dz}{d\tau} \right) $$

The first term is the temporal component ($ct$), while the remaining three are spatial ($x, y, z$).

The derivative with respect to $\tau$ tells us how much each coordinate changes per unit of proper time.

Since proper time and coordinate time are related by the Lorentz factor $ \gamma $,

$$ d\tau = \frac{dt}{\gamma}, \qquad \text{where } \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $$

we can rewrite the derivative as:

$$ U^\mu = \frac{dx^\mu}{d\tau} = \frac{dx^\mu}{dt} \cdot \frac{dt}{d\tau} $$

Using $ d\tau = \frac{dt}{\gamma} $, we find $ \gamma = \frac{dt}{d\tau} $, and thus:

$$ U^\mu = \gamma \frac{dx^\mu}{dt} $$

This step converts the reference from the object’s proper time to the coordinate time $t$ measured by an external observer.

Expanding the derivatives yields:

$$ U^\mu = \gamma \frac{d}{dt}(ct, x, y, z) $$

$$ U^\mu = \gamma \left( \frac{d(ct)}{dt}, \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \right) $$

In essence, we multiply the ordinary components of velocity by $\gamma$.

$$ \frac{d(ct)}{dt} = c $$

$$ \frac{dx}{dt} = v_x $$

$$ \frac{dy}{dt} = v_y $$

$$ \frac{dz}{dt} = v_z $$

Hence, the explicit components of the contravariant four-velocity are:

$$ U^\mu = (\gamma c, \gamma v_x, \gamma v_y, \gamma v_z) $$

or, in compact notation:

$$ U^\mu = (\gamma c, \gamma \vec{v}) $$

The temporal component is $\gamma c$, while the spatial part is $\gamma$ times the ordinary velocity $\vec{v}$.

To obtain the covariant four-velocity, we lower the index using the Minkowski metric with signature $(+ - - -)$:

$$ U_\mu = g_{\mu\nu} U^\nu $$

Here, $ g_{\mu\nu} $ is the metric tensor that reverses the signs of the spatial components:

$$ g_{\mu\nu} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} $$

Multiplying out gives the explicit components of the covariant four-velocity:

$$ U_\mu = (U_0, U_1, U_2, U_3) = (\gamma c, -\gamma v_x, -\gamma v_y, -\gamma v_z) $$

or more compactly:

$$ U_\mu = (\gamma c, -\gamma \vec{v}) $$

As expected, the temporal component remains positive, while the spatial ones change sign - reflecting the pseudo-Euclidean structure of spacetime.

Now, let’s compute the scalar product between the contravariant and covariant forms:

$$ U^\mu U_\mu = U^0 U_0 + U^1 U_1 + U^2 U_2 + U^3 U_3 $$

Substituting the explicit components:

$$ U^\mu U_\mu = (\gamma c)(\gamma c) + (\gamma v_x)(-\gamma v_x) + (\gamma v_y)(-\gamma v_y) + (\gamma v_z)(-\gamma v_z) $$

$$ U^\mu U_\mu = \gamma^2 (c^2 - v_x^2 - v_y^2 - v_z^2) $$

$$ U^\mu U_\mu = \gamma^2 (c^2 - v^2) $$

where $v^2 = v_x^2 + v_y^2 + v_z^2$.

Since $\displaystyle \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$, substituting gives:

$$ U^\mu U_\mu = \left( \frac{1}{\sqrt{1 - v^2/c^2}} \right)^2 (c^2 - v^2) $$

$$ U^\mu U_\mu = \frac{c^2 - v^2}{1 - v^2/c^2} $$

$$ U^\mu U_\mu = \frac{c^2 - v^2}{ \frac{c^2 - v^2}{c^2}} $$

$$ U^\mu U_\mu = (c^2 - v^2) \cdot \frac{c^2}{c^2 - v^2} $$

$$ U^\mu U_\mu = c^2 $$

All terms involving $v$ cancel out, leaving a final result of $c^2$, independent of the object’s speed.

In other words, $U^\mu U_\mu = c^2$ is a true invariant: it remains the same for every inertial observer, even though $\gamma$ and $\vec{v}$ vary from frame to frame.

While the individual components of $U^\mu$ transform between reference frames, its “length” in spacetime - the norm of the four-velocity - remains constant and equal to $c^2$.

This proves that the scalar product between $U^\mu$ and $U_\mu$ is a relativistic invariant, identical for all observers.

A Practical Example

Consider a spaceship traveling at $0.6c$ along the $x$-axis:

$$ v = 0.6c $$

Compute the Lorentz factor:

$$ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{\sqrt{1 - 0.6^2}} = \frac{1}{\sqrt{0.64}} = 1.25 $$

The factor $\gamma$ expresses how much the ship’s proper time runs slower than coordinate time.

The contravariant four-velocity is therefore:

$$ U^\mu = (\gamma c, \gamma v_x, \gamma v_y, \gamma v_z) $$

Since motion occurs only along $x$:

$$ U^\mu = (\gamma c, \gamma v, 0, 0) $$

Substituting the numerical values:

$$ U^\mu = (1.25c, 1.25 \times 0.6c, 0, 0) $$

$$ U^\mu = (1.25c, 0.75c, 0, 0) $$

The first component represents the “temporal velocity,” while the others form the spatial part scaled by $\gamma$.

The covariant form, using the Minkowski metric, is:

$$ U_\mu = (1.25c, -0.75c, 0, 0) $$

The spatial components change sign due to the metric.

Now compute the scalar product:

$$ U^\mu U_\mu = U^0 U_0 + U^1 U_1 + U^2 U_2 + U^3 U_3 $$

$$ U^\mu U_\mu = (1.25c)(1.25c) + (0.75c)(-0.75c) + (0 \cdot 0) + (0 \cdot 0) $$

$$ U^\mu U_\mu = 1.5625c^2 - 0.5625c^2 = 1.0c^2 $$

$$ U^\mu U_\mu = c^2 $$

This concrete example confirms that the scalar product between the two forms of $U$ is independent of velocity and always equals $c^2$ - a fundamental relativistic invariant.

And that’s the key idea.