# A photon travels in a vacuum from A to B to C. From the point of view of the photon, are A, B, and C at the same location in space and time?

As other answers have pointed out, there is no point of view or frame of reference that keeps up with a photon. Never the less, the idea that such a frame of reference exists as the limit of infinite boosts is a very natural one that comes up over and over. Here is why there are problems with that idea.

Suppose start at rest in a certain frame of reference, and accelerate at 1 g for 1 sec. This give you a new speed. Do this again and again.

As you travel faster and faster an observer in your starting frame sees you traveling closer and closer to the speed of light, and your clock running slower and slower, and your ruler getting shorter and shorter. The limit of these measurements is you traveling at the speed of light, your clock stopped, and your ruler contracted to $0$ length.

It is natural, but wrong, to suppose that at this point your frame of reference is the same as a photon. Therefore photons experience no time, and see the entire universe as contracted to a point.

You don't get closer to catching up to a photon.

After each boost, you can measure the speed of a photon. Each time, it is still passing you at the speed of light. In this sense, you are no closer to catching up to its speed.

Mathematically, this is like advancing from 1 to 2 to 3, etc. At each step you are no closer to infinity. But the limit of this sequence is infinity. There are logical problems with infinity, and the limit of this sequence is not a state where you are sitting on a point named infinity. Mathematically, this sequence diverges and there is no limit. The definition of an infinite limit is that given any value, after enough steps you will pass that value.

It is the same with catching up to the speed of light. Given any speed slower than light, after enough boosts you will be going faster than that speed. But there is never a state where you are going the speed of light.

The limit state doesn't match what we see when we observe photons.

Photons travel at a finite speed. As they advance, they change phase. So the idea that they are in a frame where no time passes and all points along their path have been compressed into the same point is wrong.

The Minkowskian geometry of spacetime has a notion of observer-independent “distance” between points that is very different from the Euclidean notion of distance. In Euclidean geometry, two points with zero distance between them are the *same* point. In Minkowskian geometry, two points with zero spacetime distance between them can be *different* points in spacetime.

Mathematically, this is because the spacetime metric

$$ds^2=dx^2+dy^2+dz^2-c^2dt^2$$

is not a positive-definite metric like the Euclidean metric

$$ds^2=dx^2+dy^2+dz^2$$

is.

All of the spacetime points along a photon’s path (A, B, C, etc.) are different points. But these points have zero spacetime distance between them. Furthermore, this distance is zero in all Lorentz frames.

The photon doesn't have a point-of-view. There are no inertial frames that are light like, and there is really nothing to be gained by saying "what-if?".

If you put the origin of your coordinates at $B$, and line the $z$-axis with the photon's momentum, then the three events are (with c=1):

$$ [(-A,0,0,-A), (0,0,0,0), (C,0,0,C)]$$

where $A>0$ and $C>0$.

The best you can do is boost along the $z$-axis with speed $\beta$. In this frame, the three events are:

$$ [(-fA,0,0,-fA), (0,0,0,0), (fC,0,0,fC)]$$

where:

$$ f = \gamma(1-\beta) = \frac{1-\beta}{\sqrt{1-\beta^2}} = \sqrt{\frac{1-\beta}{1+\beta}}$$

turns out to be the relativistic Doppler shift factor.

This can be rewritten as:

$$ [(-A',0,0,-A'), (0,0,0,0), (C',0,0,C')]$$

Though you can make $f\rightarrow 0$ as $\beta \rightarrow 1$, nothing has really changed (since $A$ and $C$ are arbitrary).