Does Light Experience Length Contraction?

Let's think clearly about length contraction.

In the frame of reference in which an object is at rest, the measured length of an object is $L_0$.

In a frame of reference in which the object is moving with velocity $v$ parallel to the length of the object, the measured length of the object is

$$L'(v) = L_0\sqrt{1 - \frac{v^2}{c^2}} $$

Now, in your question, you wrote:

However, in massless particles $v=c$, so the Lorentz factor becomes $\infty$, meaning that an object traveling at $c$ will have $0$ length.

The problem here is that, for a massless particle, there is no frame of reference in which the object is at rest; a massless particle has speed c in all frames of reference.

Thus, the proper length, $L_0$, doesn't exist so the formula above is not valid for massless particles.

This shouldn't be too surprising since the Lorentz transformations, from which the length contraction formula above is derived, doesn't exist for $v = c$ since the Lorentz factor

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

is undefined for $v=c$.

This brings us back to the question of whether or not a massless object can have extent along the direction of motion since we cannot use length contraction, as you have, to reason that it can't.

I had intended to address the open question above as an update after spending some time thinking about it while mowing. In the meantime, another answer has been given that approaches this question in essentially the same manner. For what it's worth, here's the addendum.

Consider the question: Given the equation for the world line of a uniformly moving particle in some frame of reference, what is the equation in another relatively moving frame of reference?

Working in 1-D and with standard configuration, assume a particle's world line in the unprimed frame of reference is given by

$$x(t) = ut + x_0$$

where $u$ is the velocity of the particle and $x_0$ is the position when $t=0$.

In the primed frame of reference, which has velocity $v$ in the unprimed frame, the particle's world line is given by

$$x'(t') = u't' + \frac{x_0}{\gamma_v(1 - \frac{uv}{c^2})} $$


$$u' = \frac{u - v}{1 - \frac{uv}{c^2}}$$

Now, assume we have two particles with world lines given by

$$x_1(t) = ut$$

$$x_2(t) = ut + d$$

Clearly, we have

$$x_2(t) - x_1(t) = d$$

$$x'_2(t') - x'_1(t') = \frac{d}{\gamma_v(1 - \frac{uv}{c^2})}$$

It's straightforward to show and, indeed, intuitive that, except for one special case, there is a maximum value when $v = u$

$$x'_2(t') - x'_1(t') = \frac{d}{\sqrt{1 - \frac{u^2}{c^2}}} = d_0 $$

The value $d_0$ has physical significance in that it is the separation of the world lines in the frame of reference in which the particles are at rest.

This is physically significant since, in this frame of reference only, the separation can be measured without requiring synchronized, spatially separated clocks. This separation, $d_0$ is thus an invariant, an objective quantity of physical significance.

In terms of $d_0$ and the speed $u$, we can write

$$d(u) = d_0 \sqrt{1 - \frac{u^2}{c^2}}$$

which is the familiar length contraction formula.

However, for the case that $|u| = c$ (the world lines are light-like), we cannot set $v = u$; the Lorentz factor is not defined for $v = c$.

For $u = c$, we have

$$x'_2(t') - x'_1(t') = \frac{d}{\gamma_v(1 - \frac{cv}{c^2})} = d\sqrt{\frac{1+\frac{v}{c}}{1- \frac{v}{c}}}$$

Thus, for the case that the particles are light-like, there is no maximum value of $d$; no $d_0$ that we can attach physical significance to.

Lorentz' law only applies to physical objects, i.e. those with rest mass. Wavelength is a parameter -- as you probably know, particles with mass have a wavelength related to their momentum -- so it's not sensible to discuss wavelengths changing with speed for photons whose speed is always $c$ .

As an aside, I find it hard to conceive of a way to assign a "length" to a massless particle travelling at $c$ in the first place :-) .