Does Light Experience Time?

[Dennisg]

Does light experience time? If anything could travel at the speed of light would it experience time?

 

It would seem that photons now arriving at earth from the edge of the universe experience the big bang and then find our earth as it is now with nothing in between. To them it would appear that our world was created in the big bang.

[iNow]

From the perspective of a photon, no. If you could somehow experience time from a photon's perspective, you would be everywhere all at once. This has to do with the speed of light and how objects moving at slower speeds are experienced (still) relative to light speed.

 

The mind boggles, I know. I had a really tough time the first time I read about this. Unfortunately, I really don't have the training or knowledge to explain this much more, so I'm sure one of the other much more qualified people who post here at SFN will come in after me to give you a much clearer answer.

 

Welcome to SFN, Dennisg.

Edited August 27, 2008 by iNow

[booker]

From the perspective of a photon, no. If you could somehow experience time from a photon's perspective, you would be everywhere all at once.

 

This is nonsense.

 

Does light experience time? If anything could travel at the speed of light would it experience time?

 

It would seem that photons now arriving at earth from the edge of the universe experience the big bang and then find our earth as it is now with nothing in between. To them it would appear that our world was created in the big bang.

 

That's pretty much it. Let's say you were to somehow travel unaccelerated, along with the peak of a lightwave. From source to destination no time will have elapsed. No distance will have been covered.

 

Since you are interested in this sort of thing, you might want to consider what two observers would have to say about spacetime from the perspective of two different wave crests. Do they perceive that they are cojacent to each other, a finite distance away, over the horizon?

 

As I understand it, P.A.M. Dirac had something to say about this sort of inertial frame.

 

If you follow this language, start with a Lorentz bost in the x direction. Take the eigenvalue of the matrix. This will result in a change of coordinates to (x+ct, x-ct, y, z). In this coordinate basis a Lorentz boost is a contraction along one dimension and a dialation along the other. It seems to be the coordinate system useful in examining the conditions where v-->c.

Edited August 27, 2008 by booker
multiple post merged

[iNow]

This is nonsense.

Prove it (also, you basically said the same exact thing I did just with different words, so whatever, dude).

 

The real challenge is that there is no such thing as a valid reference frame for a photon since it's never at rest, which is perhaps the ONLY reason you could validly assert that my answer to this thought experiment is "nonsense."

 

 

I'm pretty sure I read about it for the first time in a book called "In Search of Schrodinger's Cat" by John Gribbin.

[Pete]

From the perspective of a photon, no. If you could somehow experience time from a photon's perspective, you would be everywhere all at once.

That is a misconception. For that to be true one would have to have a clock moving with the photon so as to confirm that all the events occured at the same time. In any case a photon moves on a geodesic and there is no way to construct a geodesic which could fill the entire universe. The idea of a photon having a perspective is nonsense.

 

Pete

[Klaynos]

Photons are timeless.

 

This is quite easy to show:

 

[math]\Delta t_0 = \gamma \Delta t_s[/math]

 

Where:

[math]\Delta t_0[/math] is the time interval in external observer's frame.

[math]\Delta t_s[/math] is the time interval in co-moving frame.

 

But v=c so:

 

[math]\gamma = \infty [/math]

 

Thus any finite interval [math]\Delta t_s[/math] in the reference frame of the massless particle will be perceived to be infinite.

 

So when massless particles (which must be travelling at c, I can show that as well if you like) are not interacting with massive particles time is effectively stopped, we describe this as being timeless. Because of this photons do not change in flight, and another result is that as neutrinos do change in flight they must not be travelling at c, and therefore can't be massless...

[Pete]

Photons are timeless.

 

This is quite easy to show:

 

[math]\Delta t_0 = \gamma \Delta t_s[/math]

 

Where:

[math]\Delta t_0[/math] is the time interval in external observer's frame.

[math]\Delta t_s[/math] is the time interval in co-moving frame.

 

But v=c so:

 

[math]\gamma = \infty [/math]

 

Thus any finite interval [math]\Delta t_s[/math] in the reference frame of the massless particle will be perceived to be infinite.

 

So when massless particles (which must be travelling at c, I can show that as well if you like) are not interacting with massive particles time is effectively stopped, we describe this as being timeless. Because of this photons do not change in flight, and another result is that as neutrinos do change in flight they must not be travelling at c, and therefore can't be massless...

For your derivation to have any physical meaning it requires that a clock be moving at the speed of light and that's impossible. What you've shown is a limit and a limit is not an actually number, its a concept.

 

Pete

[Kyrisch]

A limit can have a value. [math]\lim_{x \to 8} x = 8[/math]

[swansont]

On the one hand you have this: The Lorentz transformation will get you into the "timeless" frame, if you take that limit, but how does one transform back? (I have a vague recollection of a treatment of this, but don't recall any details about the treatment or the quality of the site) The photon's frame is not inertial. That gives me pause when someone says the photon experiences no time, or the photon is timeless. I think it's more apt to say that time has no meaning in the photon's frame, which doesn't quite mean the same thing.

 

OTOH, you have the notion that if you do a Feynman diagram analysis (path integral) for a photon, you integrate over all paths. Of course, the effect of most of these cancel out, but IIRC there are circumstances where an interaction takes place (optical Aharanov-Bohm effect) away from the classical path of the light. Of course this is QM, and "classical path" is a problem — photons are described by wave functions, with a spatial extent. But that's all in how you interpret the physics, and not so much in the physics itself.

[Pete]

A limit can have a value. [math]\lim_{x \to 8} x = 8[/math]

 

Sorry. I meant to say that infinity is not a number.

[booker]

The idea of a photon having a perspective is nonsense.

 

Pete

 

Then do the math yourself taking v-->c, or read about it. Singular solutions in one coordinate chart are not singular in another. If you want more nonsense, in your words, consider an observer over your local event horizon.

 

Surely we can take limits, or we can limit our discussion to physics without calculus

Edited August 28, 2008 by booker

[ajb]

swansont is absolutely right here. As the photon has no rest frame it makes no sense to ask about the "photon's point of view".

 

When people say stuff like "photons experience no time" or "photons are timeless" they are describing the calculation which Klaynos did. This calculation shows that time has no meaning at v=c. It should not really be thought of as the "time in the photons frame" other than very loosely.

[Pete]

Then do the math yourself taking v-->c, or read about it.

I'm very well versed in relativity, thank you. take on the value which it approaches. The notation v-->c means "Let v become closer and closer and closer to the value of c." In this case the ratio of the times recorded on two clocks becomes infinite where one clock is moving with speed c while the other is at rest. For any finite value of c this is possible. But not for the limit since one would have to have a clock moving at the speed of light for this to be true and that is impossible.

 

Note: There is a mathematical distinction between lim(x->a) f(x) and f(a). I.e. it is not always true that lim(x->a) f(x) = f(a). It is also not always true that what is true for all finite x is also true for lim(x->a) f(x)

 

 

In the mean time I recommend that you "read about" what limits mean in more detail. For example: Suppose S_n is a partial sum which, for any finite value of n, is an integer. This does not mean that lim(n->infinity) S_n is also an integer. If could be a real number, like pi for example. What is true for arbitrary values of a quantity does not mean that its true for the limit of that quantity.

Surely we can take limits, or we can limit our discussion to physics without calculus

The answer is yes, I know how to take a limit. However I do question you're understanding of the limit process. What reason do you have to believe that I'm not well educated in mathematics and/or relativity? Is it because I disagreed with your interpretation of the facts?

 

booker - What's with the attitude?

 

Pete

Edited August 28, 2008 by Pete

[FayeKane]

Klaynos was correct, as were a couple of others. Here's what's going on, in a nutshell:

 

Time is a "direction", like any of the other three. But because of the signature of the four-dimensional metric equation (the interval), all lengths in that direction are negative numbers when described in inches or meters, etc..

 

Photons are moving through space at c.

Mass is energy moving through time at c.

 

The two directions are orthogonal.

 

Antimatter is energy moving through time at c in the other direction.

 

Acceleration in three dimensions is rotation in four dimensions. If you accelerate a mass to to c, you have rotated 90 degrees in 4D. You then have extent in space, but not in time, that is, you have converted your mass to energy.

 

No, photons do not experience time. They can't. Their momentum vector is rotated 90 degrees away from the "direction" of time.

 

-- faye kane, homeless brain

http:/blog.myspace.com/fayekane

Edited September 5, 2008 by FayeKane
minor addition

[ajb]

Klaynos was correct, as were a couple of others. Here's what's going on, in a nutshell:

 

Time is a "direction", like any of the other three. But because of the signature of the four-dimensional metric equation (the interval), all lengths in that direction are negative numbers when described in inches or meters, etc..

 

Ok, I know what you are trying to say here; time-like, null-like and space-like intervals (vectors).

 

 

Photons are moving through space at c.

Mass is energy moving through time at c.

 

The two directions are orthogonal.

 

Photons moving at c through space, all good. Mass is energy moving through time at c? Can you clarify what you mean? From the point of view of the mass he travels at speed 0 through space and speed 1 (one second per second) through time.

 

Antimatter is energy moving through time at c in the other direction.

 

This is (almost) the old Feynman interpretation of antimatter. I don't think it is thought about much these days.

 

Acceleration in three dimensions is rotation in four dimensions. If you accelerate a mass to to c, you have rotated 90 degrees in 4D. You then have extent in space, but not in time, that is, you have converted your mass to energy.

 

No idea what you mean here. 4-acceleration of a (massive) particle is easy to describe in special relativity.

 

 

No, photons do not experience time. They can't. Their momentum vector is rotated 90 degrees away from the "direction" of time.

 

The idea of photons experiencing time is not well founded. No rest frame. End of story.

 

Generally in relativity theory you have to be very careful cutting space-time into space and time. The Lorentz transformations tell you that you have transformations that mix the two and so any cutting (generally) violates relativity. And for sure is not in the spirit of the theory.

[Pete]

Mass is energy moving through time at c? Can you clarify what you mean? From the point of view of the mass he travels at speed 0 through space and speed 1 (one second per second) through time.

When I see people make this statement they are usually referring to the magnitude of the 4-velocity, which, of course, is c.

 

Pete

[ajb]

Ok, we have

 

[math]X^{\mu} = \frac{dx^{\mu}(\tau)}{d\tau}[/math]

 

for "some" path.

 

For massive particles we can always reparametrise so that [math]\tau[/math] is the proper time and we call X the 4-velocity. And we can show that

 

[math]\eta_{\mu \nu}X^{\nu}X^{\mu} = -c^{2}[/math],

 

ok.

 

But is that what FayeKane said?

 

In the rest frame I guess it is ok to say that matter moves through time a speed c.

Edited September 5, 2008 by ajb

[Pete]

Ok, we have

 

[math]X^{\mu} = \frac{dx^{\mu}(\tau)}{d\tau}[/math]

 

for "some" path.

Hmmm! Interesting notation. I've never seen anyone use [math]X^{\mu}[/math] for four-velocity

 

For massive particles we can always reparametrise so that [math]\tau[/math] is the proper time. And we can show that

 

[math]\eta_{\mu \nu}X^{\nu}X^{\mu} = -c^{2}[/math],

 

ok.

 

But is that what FayeKane said?

No. If she did then we wouldn't have to guess.

In the rest frame I guess it is ok to say that matter moves through time a speed c.

I hate it when people say that!

 

Pete

[ajb]

I just used X as it is a vector, no other assumption at the first stage. Only later do we insist on it being timelike and then we can use the proper time. I think usually people use U.

 

I too don't like that saying!

 

Also, trying to do the same for null vectors (photon paths) wont work.

[swansont]

Discussion of light and predestination has been moved

 

http://www.scienceforums.net/forum/showthread.php?t=35085

[Tsadi]

From the perspective of a photon, no. If you could somehow experience time from a photon's perspective, you would be everywhere all at once. This has to do with the speed of light and how objects moving at slower speeds are experienced (still) relative to light speed.

 

The mind boggles, I know. I had a really tough time the first time I read about this. Unfortunately, I really don't have the training or knowledge to explain this much more, so I'm sure one of the other much more qualified people who post here at SFN will come in after me to give you a much clearer answer.

 

Welcome to SFN, Dennisg.

 

No.

 

From the perspective of a photon, you would have no perspective at all. A photon moves a duration of zero time, and moving through absolutely no space. That does not mean it experiences all periods all at once, but implies strongly that it experiences no periods at all.

 

If a photon could be alive, its life would be simultaneously its death.

[iNow]

I think I've already clarified that the photon does not have a valid reference frame, so where exactly am I mistaken, again? Anything else is just interpretation, analogies to try to help our limited minds understand.

 

Nobody is arguing that the analogy is the reality.

[Tsadi]

I think I've already clarified that the photon does not have a valid reference frame, so where exactly am I mistaken, again? Anything else is just interpretation, analogies to try to help our limited minds understand.

 

Nobody is arguing that the analogy is the reality.

 

 

This part:

 

If you could somehow experience time from a photon's perspective, you would be everywhere all at once.

 

This is not true my friend. It does not experience all places of its ''travels'' all at once.

 

The bottom line i pointed out was, it experiences zero travel. There are no places it experiences, not in time, and not in space.

[Norman Albers]

Lightpaths are null geodesics.

[swansont]

Lightpaths are null geodesics.

 

In our frame. Not a photon's frame.