Do photons travel instantaneously?

[Stevexyz]

TL;DR

Does Special Relativity give us an insight to the Nature of Time?

Special Relativity tells us - the faster things travel their time is slower relative to a stationary observer.
Do massless particles, like photons traveling at the speed of light, experience zero time and in their frame of reference travel "instantaneously".

 

[phinds]

Do massless particles, like photons traveling at the speed of light, experience zero time and in their frame of reference travel "instantaneously".

No. We debunk this misunderstanding so often here on PF that we have a FAQ on it. I forget where it is but I'm sure someone here will post a link for you.

Here it is:
https://www.physicsforums.com/threads/rest-frame-of-a-photon.511170/

 

[PeroK]

Special Relativity tells us - there is no concept of absolute motion nor of a stationary observer. Motion is relative.

 

[Stevexyz]

I am new to this site. Please let me clarify my question.
Clocks on a spacecraft traveling at 10% light speed would slow down by 0.5%. At 99.9999% light speed the clocks will slow down by 99.86% compared to clocks on earth. Taking this towards the limit - the photons of light 'experience' zero time (instantaneous travel time) in their journey to earth.
Is there a flaw in this logic? Thanks

 

[phinds]

I am new to this site. Please let me clarify my question.
Clocks on a spacecraft traveling at 10% light speed would slow down by 0.5%. At 99.9999% light speed the clocks will slow down by 99.86% compared to clocks on earth. Taking this towards the limit - the photons of light 'experience' zero time (instantaneous travel time) in their journey to earth.
Is there a flaw in this logic? Thanks

Yes. The flaw is that "experience" in the sense you mean it requires an inertial (rest) frame. Photons don't HAVE a rest frame.

 

[PeroK]

Is there a flaw in this logic? Thanks

Yes, you cannot always take a limit and get a physically meaningful result.

 

[PeterDonis]

Is there a flaw in this logic?

Yes. @phinds linked you to our brief FAQ that describes the flaw.

To expand on the flaw a bit:

Taking this towards the limit

You can't. A photon moving at $c$, i.e., a lightlike object, is not a "limit" of timelike objects moving closer and closer to $c$ relative to some fixed inertial frame. A Lorentz transformation cannot change a timelike vector into a null vector, or vice versa. They are two fundamentally different kinds of things.

the photons of light 'experience' zero time

Because photon worldlines are null, not timelike, the concept of "experienced time" is meaningless for them. That is not the same as "experience zero time", which would require the concept of "experienced time" to be meaningful for photons, which it is not.

 

[Nugatory]

Clocks on a spacecraft traveling at 10% light speed would slow down by 0.5%. At 99.9999% light speed the clocks will slow down by 99.86% compared to clocks on earth. Taking this towards the limit - the photons of light 'experience' zero time (instantaneous travel time) in their journey to earth.
Is there a flaw in this logic?

We could just as correctly consider the Earth to be traveling at 10% or 99.9999% of lightspeed while the spacecraft is at rest - and then it would be the Earth clocks that are slowed down. Now if we take this towards the limit (which is not in fact valid) we would conclude that we on Earth are the ones experiencing zero time. That makes little enough sense that something has to be wrong with the logic somewhere.

There are two things going on here. First is that it’s not right (not exactly wrong either, but terribly misleading) to say that time slows down for a moving clock. To understand what is really going on with time dilation you have to understand the relativity of simultaneity, an essential concept that is left out in most casual articles about relativity. We have a bunch of older threads explaining how RoS is related to time dilation; if I have time I’ll post up some links.

Second, the time dilation formula you used in your calculations to get the .5% and 99.86% numbers is derived using assumptions that are equivalent to $v<c$, so doesn’t apply in the $v=c$ case. The infinity that appears when you set $v=c$ is just a routine division by zero error, the way the math tells you that you can’t use it that way.

 

[anuttarasammyak]

@Stevexys
The function
\gamma(v)=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}
is defind for v<c. Surely
\lim_{v\rightarrow c-0}\gamma(v)=+\infty
\lim_{v\rightarrow c-0}\gamma(v)^{-1}=+0
But
\gamma(c),\frac{1}{\gamma(c)}
are not defined.

 

[A.T.]

Taking this towards the limit - the photons of light 'experience' zero time (instantaneous travel time) in their journey to earth.

What journey? You have to apply the limit for all aspects, including length contraction. If the particle's speed approaches c relative to Earth, the distance to Earth in the particle's frame approaches 0.

 

[Stevexyz]

Does this mean the photon traveling from a distance star, in its frame, experiences zero distance and zero time?

 

[phinds]

Does this mean the photon traveling from a distance star, in its frame, experiences zero distance and zero time?

NO. Photons do not "experience" ANYTHING. They have no rest frame. Please read the posts in this thread carefully.

 

[Nugatory]

Does this mean the photon traveling from a distance star, in its frame, experiences zero distance and zero time?

No. The phrase "in its frame" is completely meaningless because there is no such thing. If a photon had a frame it would be at rest using that frame - but there is no frame in which a photon is not moving at speed $c$ and therefore no frame that we can call "the photon's frame".

This is the answer in the very first reply to your post, and the subsequent posts are just different ways of saying the same thing. The answer is not going to change no matter how often you ask.

 

[Ibix]

Does this mean the photon traveling from a distance star, in its frame, experiences zero distance and zero time?

"Time" in everyday speech can correspond to at least two different concepts in relativity. Neither can be defined for things traveling at lightspeed. Nor can you really define distance "in the frame of a photon", largely because "the frame of a photon" is a contradiction in terms. Attempting to define one leads to you requiring two parallel vectors that are also perpendicular.

 

[PeroK]

Do massless particles, like photons traveling at the speed of light, experience zero time and in their frame of reference travel "instantaneously".

No.

Please let me clarify my question.
- the photons of light 'experience' zero time (instantaneous travel time) in their journey to earth.

No.

Does this mean the photon traveling from a distance star, in its frame, experiences zero distance and zero time?

No.

 

[Stevexyz]

We could just as correctly consider the Earth to be traveling at 10% or 99.9999% of lightspeed while the spacecraft is at rest - and then it would be the Earth clocks that are slowed down. Now if we take this towards the limit (which is not in fact valid) we would conclude that we on Earth are the ones experiencing zero time. That makes little enough sense that something has to be wrong with the logic somewhere.

There are two things going on here. First is that it’s not right (not exactly wrong either, but terribly misleading) to say that time slows down for a moving clock. To understand what is really going on with time dilation you have to understand the relativity of simultaneity, an essential concept that is left out in most casual articles about relativity. We have a bunch of older threads explaining how RoS is related to time dilation; if I have time I’ll post up some links.

Second, the time dilation formula you used in your calculations to get the .5% and 99.86% numbers is derived using assumptions that are equivalent to $v<c$, so doesn’t apply in the $v=c$ case. The infinity that appears when you set $v=c$ is just a routine division by zero error, the way the math tells you that you can’t use it that way.

Thanks for your reply. Putting it in a thought experiment. Imagine traveling to a distant star. On reaching the star, several decades have past as shown on the spaceship's clocks and on our aging bodies. Now consider doing the same journey but incredibly close to the speed of light (not possible in practice but not violating any physical laws) we would see the clocks have changes only by a few minutes and our bodies are unchanged. Wouldn't it be reasonable to extrapolate and predict that if we were to travel at a smidgen less than speed of light, it would appear (to us astronauts) to have taken close to zero time?

 

[phinds]

Wouldn't it be reasonable to extrapolate and predict that if we were to travel at a smidgen less than speed of light, it would appear (to us astronauts) to have taken close to zero time?

Yep but what does that have to do with your original question? Travelers are not light. Do you want to talk about light or massive objects?

 

[A.T.]

Wouldn't it be reasonable to extrapolate and predict that if we were to travel at a smidgen less than speed of light, it would appear (to us astronauts) to have taken close to zero time?

Yes, in their frame very little time would have passed, and the target would have moved very little. This is looking at inertial movement between two points only and doesn't deal with the acceleration phases.

 

[PeroK]

Thanks for your reply. Putting it in a thought experiment. Imagine traveling to a distant star. On reaching the star, several decades have past as shown on the spaceship's clocks and on our aging bodies. Now consider doing the same journey but incredibly close to the speed of light (not possible in practice but not violating any physical laws) we would see the clocks have changes only by a few minutes and our bodies are unchanged. Wouldn't it be reasonable to extrapolate and predict that if we were to travel at a smidgen less than speed of light, it would appear (to us astronauts) to have taken close to zero time?

A better experiment is what happens at CERN, where particles are accelerated to close to the speed of light (relative to us). But, no matter how much energy we give them, they never reach the speed of light.

Moreover, the particles themselves do not change as they are accelerated: there is no sense in which the particles have an absolute speed - they're just the same as they were when they were at rest (relative to us).

In that sense a particle traveling at near the speed of light (relative to us) is still no nearer the state of a photon than we are. In fact, in the particle's reference frame it is we who are traveling close to the speed of light and - by your implication - must be feeling a little light-like.

This is why there is no sense that a particle accelerated to close to the speed of light is getting any closer to a photon. And why extrapolating to the limit is, in this case, physically meaningless.

 

[Stevexyz]

Yep but what does that have to do with your original question? Travelers are not light. Do you want to talk about light or massive objects?

Yes travellers are not light. They are heavy like me. And that's why they can't remove that smidgen to actually travel at the speed of light. If their interstellar time has reduced from many decades to minutes, would it be unreasonable for them to conjecture that if they lost all their weight and became massless, like photons, the travel time would be zero. I am sorry the special relativity equations blow up when we try to enter these values. The photons are also described by quantum mechanics which does not have time. In QM, time is not something we can measure. Time is not “observable” in QM.

 

[PeterDonis]

If their interstellar time has reduced from many decades to minutes, would it be unreasonable for them to conjecture that if they lost all their weight and became massless, like photons, the travel time would be zero.

No. As has already been pointed out, this answer will remain the same no matter how many times you ask the question.

You are also continuing to miss the point that light, moving at $c$, is not a "limit" of objects moving closer and closer to $c$ relative to you. Relative to particles inside the accelerator at CERN, you are moving very close to $c$. In other words, "how close you are to $c$" is frame-dependent, and has no physical meaning. So arguments based on "limits as you approach $c$" are frame-dependent and have no physical meaning either.

The photons are also described by quantum mechanics which does not have time. In QM, time is not something we can measure. Time is not “observable” in QM.

First, to describe photons using QM, strictly speaking, you have to use QFT, which combines QM with special relativity. So all the things we have said about light in SR also apply to photons in QM/QFT.

Second, while "time" as it appears in non-relativistic QM is indeed not an observable (it's a parameter), that does not mean there is no way to "measure time" in QM. It is easy to find observables that can serve as "clocks" in QM to a good enough approximation for all practical purposes. Indeed, current atomic clocks have a precision of something like 15 decimal places, and rely on QM for their function.

 

[Nugatory]

I am sorry the special relativity equations blow up when we try to enter these values. The photons are also described by quantum mechanics which does not have time. In QM, time is not something we can measure. Time is not “observable” in QM.

Whatever you've read about quantum mechanics, you've misunderstood it - saying something is not an "observable" doesn't mean that we can't measure it.

 

[A.T.]

... like photons, the travel time would be zero...

There would be no "travel" because the traveled distance collapses to zero, so it doesn't make sense to talk about "travel time".

 

[PeterDonis]

There would be no "travel" because the traveled distance collapses to zero, so it doesn't make sense to talk about "travel time".

No, this is not correct. The Lorentz transformation is simply not valid mathematically for $v = c$. And, as has already been pointed out, there is no valid "limit" in which we can let $v \rightarrow c$ in the Lorentz transformation formula and get a sensible result. Lorentz transformations simply act in fundamentally different ways on timelike vectors vs. null vectors. So the correct statement is that there is no valid "frame" at all for a photon, and the concept of "distance traveled" makes no sense along a null worldline any more than the concept of "elapsed time" does.

 

[haushofer]

-"What do photons experience?"

-"Well, let's transform to their rest frame to find out!"

-"But isn't that impossible according to special relativity ?"

-"Gosh, you're right, I guess your question was ill-defined then."

THE END

 

[phinds]

THE END

Ah, would that it were and hope that it is but fear it is not.

 

[Herman Trivilino]

Imagine traveling to a distant star. On reaching the star, several decades have past as shown on the spaceship's clocks and on our aging bodies. Now consider doing the same journey but incredibly close to the speed of light (not possible in practice but not violating any physical laws) we would see the clocks have changes only by a few minutes and our bodies are unchanged.

Presumably you mean a speed close to the speed of light relative to Earth.

Wouldn't it be reasonable to extrapolate and predict that if we were to travel at a smidgen less than speed of light, it would appear (to us astronauts) to have taken close to zero time?

Sure, but that would again be a speed measured relative to Earth. People on board the spaceship would not agree that light is traveling just a smidgen faster than the ship, relative to the ship.

 

[Stevexyz]

-"What do photons experience?"

-"Well, let's transform to their rest frame to find out!"

-"But isn't that impossible according to special relativity ?"

-"Gosh, you're right, I guess your question was ill-defined then."

THE END

So we can't expect special relativity to provide answers/insights about photons and time (and other particles that travel at the speed of light)?

 

[Stevexyz]

Presumably you mean a speed close to the speed of light relative to Earth.
Sure, but that would again be a speed measured relative to Earth. People on board the spaceship would not agree that light is traveling just a smidgen faster than the ship, relative to the ship.

Yes, the speed of spacecraft would be measured relative to Earth. The occupants would experience/record different transfer times depending their speed. Why would it be wrong to them to infer that traveling at the speed of light (as photons/ghosts/souls) would be instantaneous? In all cases the spacecraft or light-beam would take a finite time when seen from Earth.

 

[phinds]

Why would it be wrong to them to infer that traveling at the speed of light would be instantaneous?

Because for things that traveling at the speed of light the concept of time is meaningless. How many times and in how many different ways are you going to want us to keep answering the same question? As we have told you over and over, the answer isn't going to change.