https://www.forbes.com/sites/startswitha...-universe/

EXCERPT: . . . All of the equations of special relativity break down at the speed of light. Time doesn't pass for your surroundings. All distances along your direction-of-motion contract down to zero. Redshifts and blueshifts occur in infinite amounts.

It might be very tempting to intuit, based on this, that since the distances along your direction-of-motion contract down to zero, the Universe becomes two-dimensional to you. That time doesn't pass — so it's timeless — and it would appear just as a plane: with infinite length contraction. And therefore, that a photon, seeing you replace an apple with a banana on your desk, would experience the presence of both at once.

But what happens in reality is, perhaps, even more surprising.

A photon cannot see or experience anything, as it turns out. It's true that time doesn't pass for a photon: in relativity, it represents what we call a null geodesic. It travels from its point-of-origin to its point-of-termination: from where an interaction creates (or emits) it to where another interaction destroys (or absorbs) it. This is exactly what happens whether it's emission/absorption, emission/reflection, a scattering interaction, or any type of interplay with another particle.

When you ask what a photon would "see," you are assuming that it's possible for something to interact with a photon and for the photon to experience that interaction somehow. Yet all it experiences are two "things" during its existence: the interaction that creates it and the interaction that destroys it. Whether there is a photon that persists after the destruction, such as via scattering or reflection, is immaterial. All that a photon experiences are those two events at the endpoints of the photon's journey.

This is why we demand that we do our relativity calculations in an inertial reference frame. We can calculate how a photon redshifts or blueshifts if we use a reference frame moving slower than the speed of light, but not from the photon's reference frame. From an inertial frame of reference, we can calculate the distance between its emission and absorption point, but not from the photon's reference frame. We can calculate its light-travel time, from any inertial reference frame, but not from the photon's reference frame.

The problem is that the photon's reference frame isn't an inertial reference frame: In an inertial reference frame, there are physical laws which don't depend on the motion of anything external to the system. Yet for a photon, the physical rules it obeys depend exclusively on everything going on external to it. You cannot calculate anything meaningful for it from the photon's reference frame alone....

MORE (details): https://www.forbes.com/sites/startswitha...-universe/

EXCERPT: . . . All of the equations of special relativity break down at the speed of light. Time doesn't pass for your surroundings. All distances along your direction-of-motion contract down to zero. Redshifts and blueshifts occur in infinite amounts.

It might be very tempting to intuit, based on this, that since the distances along your direction-of-motion contract down to zero, the Universe becomes two-dimensional to you. That time doesn't pass — so it's timeless — and it would appear just as a plane: with infinite length contraction. And therefore, that a photon, seeing you replace an apple with a banana on your desk, would experience the presence of both at once.

But what happens in reality is, perhaps, even more surprising.

A photon cannot see or experience anything, as it turns out. It's true that time doesn't pass for a photon: in relativity, it represents what we call a null geodesic. It travels from its point-of-origin to its point-of-termination: from where an interaction creates (or emits) it to where another interaction destroys (or absorbs) it. This is exactly what happens whether it's emission/absorption, emission/reflection, a scattering interaction, or any type of interplay with another particle.

When you ask what a photon would "see," you are assuming that it's possible for something to interact with a photon and for the photon to experience that interaction somehow. Yet all it experiences are two "things" during its existence: the interaction that creates it and the interaction that destroys it. Whether there is a photon that persists after the destruction, such as via scattering or reflection, is immaterial. All that a photon experiences are those two events at the endpoints of the photon's journey.

This is why we demand that we do our relativity calculations in an inertial reference frame. We can calculate how a photon redshifts or blueshifts if we use a reference frame moving slower than the speed of light, but not from the photon's reference frame. From an inertial frame of reference, we can calculate the distance between its emission and absorption point, but not from the photon's reference frame. We can calculate its light-travel time, from any inertial reference frame, but not from the photon's reference frame.

The problem is that the photon's reference frame isn't an inertial reference frame: In an inertial reference frame, there are physical laws which don't depend on the motion of anything external to the system. Yet for a photon, the physical rules it obeys depend exclusively on everything going on external to it. You cannot calculate anything meaningful for it from the photon's reference frame alone....

MORE (details): https://www.forbes.com/sites/startswitha...-universe/