# Do photons experience time?

1. May 12, 2015

### Physt

I've often heard the speed of light and time dilation related from the standpoint of a particle as being in constant motion at the speed of light with the vector rotated slightly off the temporal dimension and toward any combination of the three spatial dimensions.

Does this mean photons experience no time?

How does this relate to redshift seen over large distances? Does the photon still experience a slight amount of time to be redshifted?

If photons experience some small quantity of time, is there a way to calculate the actual constant speed through spacetime?

2. May 12, 2015

### Staff: Mentor

This model often appears in pop science presentations, but it has serious limitations and can easily lead to misunderstandings, so I don't recommend using it. One interesting indicator is that, AFAIK, none of the scientists who push this model in pop science presentations actually use it for anything in their scientific work; they all use the standard spacetime model of relativity.

No, it means the concept of "experienced time" does not apply to photons.

The redshift is a change in the observed frequency of the photon; it is best viewed as a change in the relationship of the photon and the observer, not as a change in the photon itself.

3. May 21, 2015

### EM_Guy

My understanding is that in the frame of reference of any photon, the emission and absorption of the photon are simultaneous events.

4. May 21, 2015

### Ibix

There is no such thing as the frame of reference of a photon. Brian Greene notwithstanding.

5. May 21, 2015

### robphy

If displacement vector from the emission event to the absorption event along a photon's spacetime trajectory is not the zero vector, then the absorption event is in the causal future of the emission event---and thus these events cannot be simultaneous events... and absorption occurs after emission.

6. May 21, 2015

### EM_Guy

Not even in the limit?

I think that my point is that as an inertial frame of reference gets closer and closer to c, time gets really dilated, and length gets really contracted. So, in the limit (if we can speak of the limit), an inertial frame of reference that is moving at 0.999999999999 (repeating) c, the time between the emission and absorption of the photon become zero, and the distance between the location of emission and absorption shrinks to zero. From our frame of reference, the events are clearly not simultaneous. From our frame of reference, a photon that was emitted from a star in some distant galaxy takes millions of years to travel millions of light-years across the universe to the place where the photon eventually is absorbed. But to the photon, the emission and absorption are simultaneous. The two events took place at the same time and in the same place.

I'm not a physicist, so if this thinking is wrong, please explain.

7. May 21, 2015

### EM_Guy

By the way, I know that a frame of reference that is getting closer to closer to c is an accelerating frame of reference, and thus not an inertial frame of reference. I'm asking us to consider the fastest possible inertial frame of reference that there is, which in the limit, sounds to me like it would be the frame of reference of the photon itself. You can say that such an inertial frame doesn't exist, for if it did, then light would have to be going at the speed of light relative to the "photon-observer"! Ack! Nevertheless, in the limit, the time between the two events is approaching zero, right?

8. May 21, 2015

### Staff: Mentor

No. Objects with zero rest mass, like photons, are fundamentally different, physically, from objects with nonzero rest mass, like us. The concepts of "elapsed time" and "simultaneous events" only make sense for the latter types of objects; they simply do not have any meaning for the former types of objects. The fact that you can do a mathematical process that looks like taking a limit as $v$ approaches $c$ does not mean that mathematical process necessarily tells you anything meaningful about physics.

9. May 21, 2015

### Ibix

Robphy has already given one argument. Here are two more.

The speed of light is constant in all inertial reference frames. In the rest frame of a photon the speed of light would have to be both zero and 3x108m/s.

Alternatively you can substitute v=c into the Lorentz transforms and see what the time between emission and absorption is by that method. The answer is not zero, athough it is if one assumes the the interval equals the proper time for light-like paths. Contradictory answers from maths is generally a hint that you're doing something illegitimate.

Your argument holds good up to 0.9...9c, but there is a qualitative difference between c and any other speed, not just a quantitative one.