Does the space collapse to a singularity, for a photon?

Question

Using some elementary concepts, I've arrived at the apparent conclusion that all the space seen from a particle moving at the speed of light always collapses to a singularity at distance 0 from the particle itself. Looking for contrary proofs (or errors).

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Knowing Lorentz factor is

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

• $v\quad$ velocity of a moving object in reference to the observer
• $c\quad$ speed of light

Knowing that the velocity of a photon is always $c$ in every reference system

$$\Rightarrow \gamma(c) = \frac{1}{\sqrt{1 - \frac{c^2}{c^2}}} \rightarrow \infty$$

Using the Lorentz contraction formula

$$\Delta L' = \frac{\Delta L_0}{\gamma}$$

with

• $\Delta L'$ the distance from the inertila reference frame seen by the moving reference frame

• $\Delta L_0$ the distance from the moving reference system seen by the inertial reference system

• $\gamma$ the Lorentz factor

$\Delta L' \rightarrow 0$ always

All the space$^*$ collapses to a singularity point overlying the reference frame moving at speed of light.

$^*$seen by the very reference frame moving at $c$

The computation is valid for an inertial reference system put in any point in space, for that I used the generalization "all space".

Answer 1

Your calculation is OK, but your conclusion is too short-sighted:

Assume that there was a reference frame of a photon. Than the photon could take measures of its observations and by these measures, it could draw the picture of the universe.

However, as you saw, all measures of the photon would be zero. That means that the photon is not able to distinguish any interval of spacetime, all spacetime intervals are zero according to its measurements. And that means nothing else than that there is no reference frame of the photon, because a reference frame which gives always zero is useless, and it is even no reference frame.

As a result, your calculation confirms (and proves) simply the fact that a photon has no reference frame.

Answer 2

It is usually a misunderstanding when questions are asked about the inertial frame of a photon. Photons do not have a rest frame. No particle without a rest mass has a rest frame. When you look at the math, you will end up dividing by zero, and it will lead to infinities.

Now what you could really ask, and what there is in your math, that is how something (traveling near speed c) would look like when viewed from an observer at rest. Now that object that travels near speed c would seem to be contracting as it nears speed c (when viewed from an observer at rest).

To see this, you need to look at the thing very close to the speed of photons. It is very good to look at the frame of a neutrino. The neutrino is as close you can get to the speed of light and as little mass as you can get. When you try to look at a macro object traveling at that speed (of a neutrino) you will figure out from the math that an observer at rest will see the object contract (its length/spatial extension) to close to zero (at a certain axle parallel to the travel direction). You are correct that length contraction will affect this object when viewed from an observer at rest.

Now you are talking about how this object would view the rest of the world from its own reference frame. Now along the axis of travel, you are correct that length contraction would make distances seem close to zero from the reference frame of the object (along the axis of travel).

Now let's look at the time dilation. An object traveling near speed c (if it had a clock with it) would see its own clock tick normal. But let's check what this object would see on its own clock when traveling from the Sun to Earth. Would it see 8 minutes on its own clock? No. It would see an amount of time much less. Close to zero. For an object traveling near speed c, it would seem as the travel from the Sun to Earth took almost no time.

Now when an observer at rest would look at the object travel from the Sun to Earth, the observer would see 8 minutes elapse on its own clock. That is why we say that it takes 8 minutes for a neutrino to reach Earth. That is how much time elapses on the clocks on Earth. But on the neutrino's frame, the clock elapses almost no time.

Now you cannot tell what it would look like from a photon's frame, since it has no rest frame. But you could tell that it should be 8 minutes for the photon to reach Earth. But for the photon, no time passes, it is not moving in the time dimension. For the photon, the spacetime distance from the emission at the Sun to the absorption on Earth is 0. We call this a lightlike worldline. That is why for the photon we could say that it does not experience time as we do (who have rest mass).

Now you could say that for the photon the spacetime distance is 0, we call that lightlike worldline, and that is why we say that spacetime distances (along the axis of travel) for the photon are contracted to 0.

Answer 3

Since you are looking for contrary proofs or errors there are a couple of clear errors with the derivation.

The first is that division by zero is undefined. The Lorentz factor at c is undefined. In the limit as v approaches c it goes to infinity, but at c it is undefined.

The second is that the length contraction formula does not apply. The formula relates the length in a frame where it is at rest to the length in a frame where it is moving. There is no frame where a photon is at rest. So that length does not exist and therefore cannot be compared to any other length.

Answer 4

Space collapses for photos with our constructive reference not with other reference .. "our" means the observer perspective(who's velocity is considerably less than of light's) . By logic, there's a possibility that space may not collapse for photons with other constructive reference like a observer who is travelling at speed more than that of light.