Here's an explicit way to construct a light worldline in special relativity (assuming flat spacetime), using spherical coordinates.
Consider the following world line, parameterized by the affine parameter $\lambda$:
\begin{align*}
x &= \int r\cos\theta\cos\phi\ \mathrm{d} \lambda\\
y &= \int r\cos\theta\sin\phi\ \mathrm{d} \lambda\\
z &= \int r\sin\theta\ \mathrm{d} \lambda\\
ct &= \int r\ \mathrm{d}\lambda
\end{align*}
Note that for now we assume $r,\theta,\phi$ may depend in any way on $\lambda$. As we will presently demonstrate, this world line still satisfies the condition of being a null curve.
Proof that this is a null curve:
A null curve with coordinates $x^{\mu}$ is defined in special relativity as one in which
$$ \eta_{\mu\nu}\frac{\mathrm{d} x^{\mu}}{\mathrm{d}\lambda} \frac{\mathrm{d} x^{\nu}}{\mathrm{d}\lambda} = 0 $$
Which in our case translates to:
$$ \left(\frac{\mathrm{d} x}{\mathrm{d}\lambda}\right)^2+
\left(\frac{\mathrm{d} y}{\mathrm{d}\lambda}\right)^2+
\left(\frac{\mathrm{d} z}{\mathrm{d}\lambda}\right)^2-
c^2\left(\frac{\mathrm{d} t}{\mathrm{d}\lambda}\right)^2 = 0 $$
Calculating the derivatives squared is straightforward:
\begin{align*}
\left(\frac{\mathrm{d} x}{\mathrm{d}\lambda}\right)^2 &= r^2\cos^2\theta\cos^2\phi & \left(\frac{\mathrm{d} y}{\mathrm{d}\lambda}\right)^2 &= r^2\cos^2\theta\sin^2\phi \\
\left(\frac{\mathrm{d} z}{\mathrm{d}\lambda}\right)^2 &= r^2\sin^2\theta & c^2\left(\frac{\mathrm{d} t}{\mathrm{d}\lambda}\right)^2 &= r^2
\end{align*}
So we get:
\begin{align*}
\eta_{\mu\nu}\frac{\mathrm{d} x^{\mu}}{\mathrm{d}\lambda} \frac{\mathrm{d} x^{\nu}}{\mathrm{d}\lambda}&=
r^2\cos^2\theta\cos^2\phi + r^2\cos^2\theta\sin^2\phi +
r^2\sin^2\theta - r^2 \\
&= r^2\cos^2\theta+r^2\sin^2\theta - r^2\\
&= r^2 - r^2 \\
&= 0
\end{align*}
Which proves that this is a null curve.
Now, null curves for light in flat spacetime must correspond to straight lines, because there's no curvature that can bend the light by assumption. This means that this curve must also satisfy the geodesic equation, which in SR is:
$$ \frac{\mathrm{d}^2 x^\mu}{\mathrm{d}\lambda ^2} = 0 $$
Looking at the derivatives we computed $\large \frac{\mathrm{d}x^\mu}{\mathrm{d}\lambda}$, we see that this condition implies all five functions $\sin\theta$, $\sin\phi$,$\cos\theta$, $\cos\phi$, $r$, cannot depend on $\lambda$. For if any one of them depended on $\lambda$, their derivative in $\lambda$ would not vanish, contradicting the above condition for being a geodesic.
This means that we can integrate the derivatives $\large \frac{\mathrm{d}x^\mu}{\mathrm{d}\lambda}$ with respect to $\lambda$ to recover the worldline. Doing that, we see that all four coordinates of the curve take the form:
$$ x^\mu = x^\mu|_{\lambda=0} + \lambda\cdot\frac{\mathrm{d} x^\mu}{\mathrm{d} \lambda} $$
Which we can write more explicitly, as:
\begin{align*}
x &= x_0 + \lambda r\cos\theta\cos\phi\ \\
y &= y_0 + \lambda r\cos\theta\sin\phi\ \\
z &= z_0 + \lambda r\sin\theta\ \\
ct &= ct_0 + \lambda r
\end{align*}
Which as expected, simply correspond to the motion of a light ray along a straight line, with some constant projection angles $\theta$ and $\phi$.
References: this answer is based on my solution of exercise 1.13. from the book "Problem Book In Relativity and Gravitation", Alan P. Lightman, William H. Press, Richard H. Price, Saul A. Teukolsky
