The Eternal Perspective of Photons: A Scientific Exploration

[Jesus Rodriguez]

I'm trying to ask a simple question, which is probably a fatal mistake but...

According to accepted Einsteinian relativity, say I'm traveling at the speed of light. I understand I can't get to that speed. Suppose I was born at that speed, I'm a photon, whatever. From my "photonic perspective" does time pass by? Is it all one big "now", so that from my perspective I'm eternal, or does time merely slow to some finite crawl?

If time does change, even if very slowly, I can understand and have no need of the next question.

If time from the photon's perspective does not change, then how is the perspective of the photon not eternal? In other words, photons are created and destroyed (changed) all the time. Yet if from their perspective time does not change then something does not make sense to me. A particle for which time does not change should have no beginning nor any end.

I'm confused somewhere...

JR

 

[pervect]

I'm trying to ask a simple question, which is probably a fatal mistake but...

According to accepted Einsteinian relativity, say I'm traveling at the speed of light. I understand I can't get to that speed. Suppose I was born at that speed, I'm a photon, whatever.

Here is your mistake. Your are supposing impossible things, and then asking us to give you an answer.

This leads to mathematical nonsense.

Lewis Caroll (who was a mathemetician in real life) has a very amusing and prefectly valid mathematical proof that illustrates the perils of impossible assumptions.

Given 2+2=5, prove that I am the Queen of England.

proof:

2+2=5, but 2+2=4. Therefore 5=4. Therfore 2=1.

Now, me and the Queen are 2. But 2 = 1. Therefore me and the Queen are 1. Therfore I am the Queen of England.

(Or was it the King? It doesn't matter, the proof works either way :-)).

Anyway, the moral of the story is:

Please don't assume that you can move at the speed of light. You can't.

 

[robphy]

Here's a related discussion
https://www.physicsforums.com/showthread.php?t=105484Time Dilation Question

(see the posts near #6)

 

[DaveC426913]

Please don't assume that you can move at the speed of light. You can't.

The issue here is not one of whether or not he can move at the speed of light. The issue is one of whether or not a photon can "experience".

From my "photonic perspective" does time pass by? Is it all one big "now",

No. Yes.

If time from the photon's perspective does not change, then how is the perspective of the photon not eternal? In other words, photons are created and destroyed (changed) all the time. Yet if from their perspective time does not change then something does not make sense to me. A particle for which time does not change should have no beginning nor any end.

I believe I see your point. A given photon exists only from the point where it is emitted from one atom until it is absorbed by another. That means that, even though a photon does not experience time, it does nonetheless, experience a "lifetime". While every point in its lifetime is experienced simultaneously, it is a discreet, and possibly quite small, set of points.

 

[pervect]

The issue here is not one of whether or not he can move at the speed of light. The issue is one of whether or not a photon can "experience".
No. Yes.

It is definitely a mistake to import the baggage of frames of reference applicable to a massive observer to a photon - 3 space + 1 time does not work.

This is a frequently asked question, and the frequently given answer is basically that it's not a good question.

For more info, there's always the sci.physics.faq entry

http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/headlights.html

 

[Soul Surfer]

The best way of thinking about it is to say that a photon does not experience time from the instant of its creation in some interaction to the instant of its destruction in another

 

[Jesus Rodriguez]

The best way of thinking about it is to say that a photon does not experience time from the instant of its creation in some interaction to the instant of its destruction in another

Thank you for the responses, (the links were useful). This makes better sense to me now, and as FAQ link says, I don't at all feel stupid for asking the question.

Typically, a good set of answers generate further questions. The link explains how when the equation for relative velocities is used about an object traveling at the speed of light a "meaningless" answer is obtained, signified by an attempt to divide by zero. Some here take that to mean the question itself is somehow meaningless.

However there is an underlying assumption: is mathematics sufficient to contain philosophy? Though no expert I believe that question received a resounding no some time back. That is to say, mathematics does not wholly encompass methods of thinking or valid logical expressions. I believe that while Godel's theorems set a clear limit in any system that involves counting (mathematics), the much older form of logic (sometimes called "term logic") is not so afflicted.

Thus, it may be that some questions while meaningless in mathematical terms, may in fact have a sensible construction nonetheless in other systems, such as natural languages. I find the quote above and that about a photon having no life, but nonethelss a measurable lifespan satisfying answers, and again thank the respondents.

Put in other ways: there are valid questions, and answers, that can not be set in a mathematical framework.

Or, abusing a cliche: "There is more in Heaven and Earth than CAN be dreamt of in mathematical philosophy."

Again admitting I'm a total neophyte, I await comments.

JR

 

[ZapperZ]

However there is an underlying assumption: is mathematics sufficient to contain philosophy? Though no expert I believe that question received a resounding no some time back. That is to say, mathematics does not wholly encompass methods of thinking or valid logical expressions. I believe that while Godel's theorems set a clear limit in any system that involves counting (mathematics), the much older form of logic (sometimes called "term logic") is not so afflicted.

All you need to do is find something in physics that is devoid of an underlying mathematical description. You'll discover that there's none. So here, it is the assumption that there's something that can't be described mathematically that has no validity.

http://nedwww.ipac.caltech.edu/level5/March02/Wigner/Wigner.html

Zz.

 

[TheAntiRelative]

Well heck, if we're going to bring philosophy and reasoning into it then things can get a lot more interesting.

Mathematically describe categorizations.

It's a matter of opinion whether or not concepts that Math cannot describe are valuable or not. Zapper ascribes to the "not valuable" when it comes to physics. However I can argue that physics is a categorization which is somewhat subjective.

Zero is a concept that I'm sure Zapper would find valuable but it is still a concept and does not exist in reality. That's why you can't divide by it.

Hand me 100 nothings...

So non-physical concepts obviously have some small place in physics as well and the usefullness of a concept is subjective. So I assert that it is not an incontrovertible fact that the question is meaningless. It is instead one of the places that philosophy and reasoning touch physics and the answer is subjective.

I also postulate that reasoning that requires concepts that cannot be described mathematically is more valuable than math alone. I'll use the difference between computers and humans as my proof. I believe humans are better overall problem solvers than computers even though computers far exceed humans in speed of calculation and errorless math.

 

[ZapperZ]

I also postulate that reasoning that requires concepts that cannot be described mathematically is more valuable than math alone. I'll use the difference between computers and humans as my proof. I believe humans are better overall problem solvers than computers even though computers far exceed humans in speed of calculation and errorless math.

If you think mathematics is analogous to number crunching, then you have mathematics all wrong.

Again, just show me a physical phenomenon that defies mathematical description. If not, all of this are idle speculation without basis. Now THAT certainly defies mathematical description!

Zz.

 

[Jesus Rodriguez]

All you need to do is find something in physics that is devoid of an underlying mathematical description. You'll discover that there's none. So here, it is the assumption that there's something that can't be described mathematically that has no validity.

http://nedwww.ipac.caltech.edu/level5/March02/Wigner/Wigner.html

Zz.

Elegance, a concept that Einstein claimed was an underlying assumption to his theoretical musings. Please translate elegance into a mathematical formula.

JR

 

[ZapperZ]

Elegance, a concept that Einstein claimed was an underlying assumption to his theoretical musings. Please translate elegance into a mathematical formula.

JR

Elegance is not a physical concept. It's a quality that human gives, like "beauty", something Einstein also described as what a physical equation should be.

Furthermore, ask 10 people the meaning of elegance, and you get 10 different description. Ask 10 people if an object is "elegant", and you get 10 very subjective answer.

Is this what you would consider as a valid quality to describe a physical world, that it simply depends on TASTES? How would you like it that your electronics work simply based on someone's mood and perceived human qualities?

Try again.

Zz.

 

[Hurkyl]

Zero is a concept that I'm sure Zapper would find valuable but it is still a concept and does not exist in reality.

Zero is no less "real" than one. Or, if you prefer, one is no more real than zero.

That's why you can't divide by it.

No, that's not why you can't divide by it.

Hand me 100 nothings...

Already have, and I tossed in 250 extra for free.

 

[Longstreet]

Also the photon always has zero energy in it's reference. So I guess there isn't anything there *to* pass with time.

 

[jtbell]

Also the photon always has zero energy in it's reference. So I guess there isn't anything there *to* pass with time.

A photon doesn't have a reference frame in which it is stationary, at least not one that you can meaningfully apply the equations of relativity to. At best, "reference frame of a photon" is purely a figurative or metaphorical term, with no physical content (in the context of relativity theory).

 

[Longstreet]

You people are so pedantic. You know what I mean. In the limit of accelerating in the direction of a photon the photon will lose all energy. In other words the reference of the photon is where it's proper energy is zero (since it's proper mass is zero). Of course I'm not worring about the quantum problem here of actually measuring the photon to make sure you're chasing it as it would be impossible for any real observer to even know where it is going.

Since energy is the rate of change of a system, there is not rate of anything in the reference of a photon. So if the absense of a passage of time concerns you any, it kind of makes sense since there would be no change anyway.

 

[robphy]

Just a few comments to illuminate some issues that seem to be often glossed over... (for which I've begun to realize for myself).

First, all terms (e.g. "reference frame", "photonic perspective", "simultaneously", etc...) need to be defined precisely. This is one important role of mathematics (namely, a mathematical model of the physics one wishes to describe).

Here are some "reasonable" properties of a "reference frame" of a massive particle in SR (whose worldline has an everywhere timelike tangent vector). [I am going to emphasize the geometric structures to avoid dealing with and trying to interpret certain algebraic formulas that break down when [probably inappropriately] applied to a massless particle.]

• If A and B are distinct events on that worldline, either A is in the causal past of B (so that A can influence B), or vice versa.
• Its Minkowski-arc-length along the worldline is nonzero and can be associated with a clock carried by the particle... this clock marks the particle's "proper time".
• The hyperplane Minkowski-orthogonal to that tangent vector does not contain that tangent vector... and can be called the particle's "space at an instant" (a "moment of time").
• The events on this hyperplane are regarded to be "simultaneous" according to this particle since:
  • these events are assigned the same time coordinate as read off by the particle's wristwatch/proper-time (e.g., by a radar method [at least for nearby events]: send off a light signal at wristwatch time t1, receive its echo off the distant event at wristwatch time t2, assign to that distant event the time-coordinate (t1+t2)/2)
  • these events are spacelike-related (and therefore not causally-related) to each other
• For an inertial massive particle in SR, the entire Minkowski spacetime is foliated by these hyperplanes... meaning that the entire spacetime is sliced into nonintersecting hyperplanes, so that every event in spacetime is assigned a (but certainly not all the same) time-coordinate.

I think the list above seems reasonable.

What are the analogous statements for a photon (a massless particle) in SR (whose worldline has an everywhere lightlike [a.k.a. null] tangent vector)?

• (Does "time stop" for a photon?)
  If A and B are distinct events on that worldline, either A is in the causal past of B (so that A can influence B), or vice versa.
  [still TRUE... so (to me) it DOES NOT make sense to say that "time stops" or that all of its events occur "simultaneously" since there is certainly a sense of causal-sequence of the photon's events.]
• (Does it make sense to define "proper time" for the photon?)
  Its Minkowski-arc-length along the worldline is ZERO. So, there may be a problem here. Maybe one can define it...but is it useful? Don't ignore the previous point.
• (Does it make sense to call this hyperplane "space" for the photon?)
  The hyperplane Minkowski-orthogonal to that tangent vector DOES contain that tangent vector [joining two causally-related events]... this is a feature of the Minkowskian geometry of SR.
• The events on this hyperplane are NOT all spacelike-related to each other... There are events on this plane (namely along the tangent vector) that are causally-related to each other.
• For an inertial massless particle in SR, the entire Minkowski spacetime is NOT foliated by these hyperplanes. So, there are many events that are not assigned a full set of coordinates. [One might argue here that one really needs a congrunce of worldlines.]

So, it seems to me that there are some problems trying to define a reference frame for a photon.

My $0.03

 

[ZapperZ]

I believe that what robphy just did it EXTREMELY valuable. This is because there are many people here who are using these terms without understanding that there are precise definitions and properties as defined in physics. And since we ARE here discussing physics, when you use such terms in ridiculous ways, and the rest of us starts to scratch our heads, do NOT simply assume that we are being "pedantic".

Look at the term you are using. If you do not have a clue on the underlying mathematical descriptions of it, chances are, you don't know what it is. So leave open the possibility that MAYBE, you used it in an inconsistent fashion, or maybe even in ways that it wasn't meant to be used.

Zz.

 

[quantumdude]

In other words the reference of the photon is where it's proper energy is zero (since it's proper mass is zero).

Longstreet, the people here are not being pedantic, you're just being sloppy. The reference you refer to here simply does not exist, and it has nothing to do with anything in quantum theory. And your statement that energy is the 'rate of change of a system' is completely devoid of meaning. If you talk rubbish around these parts, you can fully expect to be called out on the carpet for it by the memebership here.

 

[crazycalhoun]

I'm trying to ask a simple question, which is probably a fatal mistake but...

According to accepted Einsteinian relativity, say I'm traveling at the speed of light. I understand I can't get to that speed. Suppose I was born at that speed, I'm a photon, whatever. From my "photonic perspective" does time pass by? Is it all one big "now", so that from my perspective I'm eternal, or does time merely slow to some finite crawl?

It doesn't make sense to talk about a photon's perspective/frame of reference/whatever. But let's pretend that a photon did have a reference frame. Obviously it'd be at rest relative to itself, so $d\tau = 0$.

If we try to take the norm of the four-velocity we end up with $\frac{dx^{\alpha}}{d\tau} \frac{dx^{\beta}}{d\tau} = \frac{0}{0}$, which as we all know is undefined.

 

[bernhard.rothenstein]

I'm trying to ask a simple question, which is probably a fatal mistake but...

According to accepted Einsteinian relativity, say I'm traveling at the speed of light. I understand I can't get to that speed. Suppose I was born at that speed, I'm a photon, whatever. From my "photonic perspective" does time pass by? Is it all one big "now", so that from my perspective I'm eternal, or does time merely slow to some finite crawl?

If time does change, even if very slowly, I can understand and have no need of the next question.

If time from the photon's perspective does not change, then how is the perspective of the photon not eternal? In other words, photons are created and destroyed (changed) all the time. Yet if from their perspective time does not change then something does not make sense to me. A particle for which time does not change should have no beginning nor any end.

I'm confused somewhere...

JR

Have you ever tried to synchronize clocks comoving with photons?

 

[Mortimer]

For whoever is interested I've put this post that is related to this discussion in the "Independent Research" forum.

 

[Longstreet]

I'm not trying to define some rediculus thing like 1+1=3. My argument is that you can approach a frame of reference (in a limit) where the kinetic energy and the momentum of the photon is zero. If you don't want to call this the photon's reference it's fine by me. I'll start calling it the limiting frame. Just don't talk about how it's not physically meaningfull. I think the problem is meaningfull because if there actually was any energy in my example then there would be infinite energy in every other frame. You may say "duh", but that's a good thing in my perspective.

As per robphy, I do think it is useful to know some definitions. But I think they should be used as long as you don't lose the origional subject, and I'm not saying you're doing that. I'm saying that's why my posts might apear "sloppy". However, they are not because they are consistant even if I don't use pedantically correct terminology to describe the physics to someone who is asking about them. I believe my example is still valid. If you have a primary reference frame with a photon, and then set increasing faster secondary reference frames in the direction of the photon so that the secondary reference is at c, then both the secondary reference frame and the photon has similar properties from the point of view of the primary reference frame. If the secondary reference frame is not the reference of the photon, then you cannot send a third reference to see the secondary reference frame either, and so on. You end up with having an infinite number of reference frames going in exactly the same direction with exactly the same velocity, and still not being equal to each other.

As for time not passing, I think we can safely use the reference of our normal reality to say that if you put a clock on a ship that speeds around at c and then comes back, that the clock will not have ticked one iota. If it hasn't ticked I don't see how time has passed for it.

Btw, pedantic isn't wholey a bad word. (<-edit: well, it is a little bit bad

 

[Dense]

So if a photon does not experience time, even though it was emitted four billion years before it was absorbed by the Hubble's lens, does that mean you can say that light travels at the speed of time?

 

[pervect]

I'm not quite sure where this is leading, but I do have a few comments:

If we take a null vector and a Minkowski metric, the "natural" way to span the hyperspace perpendicular to that vector is via another null vector and two space-like vectors.

I.e suppose we take (t,x,y,z) as our coordinate system, and $\eta_{\mu\nu}$ = diag(-1,1,1,1) as our metric.

Then p = (1,1,0,0) is a null vector, because $$\eta_{\mu\nu} p^{\mu} p^{\nu} = 0$$

If a vector q is orthogonal to p, we must have $$\eta_{\mu\nu} p^{\mu} q^{\nu} = 0$$

If we let q = (a,b,c,d) we find (-a+b=0)

Thus, a "natural" basis for the set of orthogonal vectors (which forms a 3-d subspace) is

(1,-1,0,0) [another null vector]
(0,0,1,0) [a spacelike vector]
(0,0,0,1) [a spacelike vector]

However, it's possible to span the space in other ways, i.e

(1,-1,1,0) (spacelike)
(1,-1,0,1) (spacelike)
(0,0,1,1) (spacelike)

Other more exotic possibilities exist as well, but Kruskal coordiantes for instance are widely used to describe black holes without any annoying singularities at the horizon, and these coordiantes involve a pair of null coordinates usually called u and v.

I'm not particularly used to working with null coordinates, however.

None of this means or should be taken to mean that photons have a "perspective".

 

[pervect]

I'm not trying to define some rediculus thing like 1+1=3.

People rarely deliberately define 1+1=3. Unfortunately, it is quite possible to do so accidentally, by mistake.

The consequences of this are just as severe when it is done by accident as when it is done on purpose - given a false assumption,one can prove anything is true.

 

[Longstreet]

What I mean is I'm not trying define the velocity of light as zero in any reference frame: c = 0.

So if a photon does not experience time, even though it was emitted four billion years before it was absorbed by the Hubble's lens, does that mean you can say that light travels at the speed of time?

Speed, in my mind, is the rate of change of something with respect to time. So the speed of time might be 1sec/sec, which is really unitless but anyway...

A wave both has a group velocity, which I might think of as the reference frame since I'm alone on that one, and the phase velocity. The phase velocity is funny because the phase is not something that physically travels through space. It changes only with respect to time at points in space, but each "peak" or "creast" of the wave, which represents the phase, falls to successive space values giving you something that looks like a velocity.

If I'm remembering correctly, since my notes are a bit scattered, if the group velocity is u, then the phase velocity is c^2/u. Notice that any wave less than the speed of light has phase velocity greater than c. That's fine because remember the phase velocity is just an illusion anyway. But you'll also notice that at c, the phase velocity is also c. This is what we are familiar with in electromagnetic waves. So there is a kind of 1:1 relationship between the speed of light and the rate of change of the EM wave itself if you want to think of it that way. However, I'm sure this is slopy and you can ignore my rediculous accusations as flights of fancy.

 

[robphy]

I'm not quite sure where this is leading, but I do have a few comments:

If we take a null vector and a Minkowski metric, the "natural" way to span the hyperspace perpendicular to that vector is via another null vector and two space-like vectors.

I.e suppose we take (t,x,y,z) as our coordinate system, and $\eta_{\mu\nu}$ = diag(-1,1,1,1) as our metric.

Then p = (1,1,0,0) is a null vector, because $$\eta_{\mu\nu} p^{\mu} p^{\nu} = 0$$

If a vector q is orthogonal to p, we must have $$\eta_{\mu\nu} p^{\mu} q^{\nu} = 0$$


If we let q = (a,b,c,d) we find (-a+b=0)

Thus, a "natural" basis for the set of orthogonal vectors (which forms a 3-d subspace) is

(1,-1,0,0) [another null vector]
(0,0,1,0) [a spacelike vector]
(0,0,0,1) [a spacelike vector]

However, it's possible to span the space in other ways, i.e

(1,-1,1,0) (spacelike)
(1,-1,0,1) (spacelike)
(0,0,1,1) (spacelike)

Other more exotic possibilities exist as well, but Kruskal coordiantes for instance are widely used to describe black holes without any annoying singularities at the horizon, and these coordiantes involve a pair of null coordinates usually called u and v.

I'm not particularly used to working with null coordinates, however.

None of this means or should be taken to mean that photons have a "perspective".

Are you saying that:
given p=(1,1,0,0) [a null vector],
one has
q=(1,-1,0,0) [another null vector]
with $$\eta_{\mu\nu} p^{\mu} q^{\nu} = 0$$ ?

$$\eta_{\mu\nu} p^{\mu} p^{\nu}=p_t p_t - p_x p_x - p_y p_y -p_z p_z = (1)(1)-(1)(1)-(0)(0)-(0)(0)=0$$... so p is null.
Likewise
$$\eta_{\mu\nu} q^{\mu} q^{\nu}=q_t q_t - q_x q_x - q_y q_y -q_z q_z = (1)(1)-(-1)(-1)-(0)(0)-(0)(0)=0$$... so q is null.
However,
$$\eta_{\mu\nu} p^{\mu} q^{\nu}=p_t q_t - p_x q_x - p_y q_y -p_z q_z = (1)(1)-(1)(-1)-(0)(0)-(0)(0)=2$$... so p is not Minkowski-orthogonal to q. [In your required equation, (-a+b=0) means of course a=b ].

Note on a spacetime diagram, p and q look like they are orthogonal... but that is Euclidean-orthogonal, of course. p and q form (up to conventional factors) the light cone coordinates.

There is a little theorem that goes: (from Geroch, Mathematical Physics)
Let n be a nonzero null vector, and v some vector with g(n,v)=0. Then either v is spacelike or v is a numerical multiple of n (hence null).

Certainly p,q, and the two spacelike vectors span Minkowski spacetime... but they don't form a mutually orthogonal basis.

 

[JesseM]

I'm not trying to define some rediculus thing like 1+1=3. My argument is that you can approach a frame of reference (in a limit) where the kinetic energy and the momentum of the photon is zero. If you don't want to call this the photon's reference it's fine by me. I'll start calling it the limiting frame. Just don't talk about how it's not physically meaningfull.

One problem is that while some quantities might be well-defined in this limit, others would not, because by taking the limit in different ways you could get different answers. For example, what is the velocity of another photon as seen in a photon's limiting frame? Well, you could take the limit as both photon's velocities approach c at the same rate, in which case you'd conclude the other photon should be at rest in the limit. Or you could take the limit by assuming the other photon's velocity is c and letting the first photon's velocity approach c, in which case you'd conclude the other photon's velocity is still exactly c as seen in the first photon's limiting frame.

 

[Longstreet]

Let's see if I can get latex to work:

$$\nu' = \nu \sqrt{ \frac{1-u/c}{1+u/c}}$$

That's your run of the mill doppler shift. We have our base frame which represents normal reality. Our test frame is u relative to our frame. We have a laser mounted on our base frame shooting light in the direction of our test frame. If you take the limit of u to c, the frequency of the light in the laser beam goes to zero. The constant c is still c in the limit. However, my point is there is nothing too go at c. The phase is still technically changing at 3x10^8m/s in the limiting frame, but the wavelength is infinitly long.

 

[pervect]

$$\eta_{\mu\nu} p^{\mu} q^{\nu}=p_t q_t - p_x q_x - p_y q_y -p_z q_z = (1)(1)-(1)(-1)-(0)(0)-(0)(0)=2$$... so p is not Minkowski-orthogonal to q. [In your required equation, (-a+b=0) means of course a=b ].

You're right, I was quite confused.

I was ultimately interested in defining a coordinate system. It looks like the right way to do it is to say that given a plane defined by two space-like vectors x and y (and their linear combinations) that there are two unique (except for their magnitude) null vectors perpendicular to that plane (for example z+t and z-t for the x-y plane), and that these 4 vectors (two space-like and two null) span the space-time and can thus serve as a coordinate system.

Hopefully I got it right this time around (I've seen null coordinates used, but I haven't used them much personally).

 

[JesseM]

Let's see if I can get latex to work:

$$\nu' = \nu \sqrt{ \frac{1-u/c}{1+u/c}}$$

That's your run of the mill doppler shift. We have our base frame which represents normal reality. Our test frame is u relative to our frame. We have a laser mounted on our base frame shooting light in the direction of our test frame. If you take the limit of u to c, the frequency of the light in the laser beam goes to zero. The constant c is still c in the limit. However, my point is there is nothing too go at c. The phase is still technically changing at 3x10^8m/s in the limiting frame, but the wavelength is infinitly long.

This is if you take the limit where the velocity of the frame relative to the emitter is approaching c but the light wave is moving at c in every frame. But getting back to the point I made in my last post, why couldn't you also take the limit as the velocity of the frame relative to the emitter is approaching c, and the velocity of the wave being emitted is also approaching c (while its frequency remains constant)? This is a different way of trying to decide what the wavelength of an electromagnetic wave moving at c relative to the emitter should be in a "limiting frame" which is itself moving at c relative to the emitter. Since you get different answers depending on how you calculate the limit, this shows that there is no well-defined limit here.

edit: sorry, I think I messed up there...regardless of how fast the wave is moving, if the relative velocity between you and the emitter is approaching c, then the time between ticks of the emitter's clock should be approaching zero and therefore the frequency should be approaching zero as well. So frequency is not one of the quantities that will fail to have a well-defined-limit in the "limiting frame".

 

[caroso]

Good you're discussing this and making hypothesis, who knows one day man can be "teletransported" or faxed (I doubt cos is too much information and nobody would want to be exterminated in the process or be duplicated like Jet Li -in that science fiction movie-) but if that is improbable and not impossible, in that future nobody would have to bear radical statements like "It is definitely a mistake to import the baggage of frames of reference applicable to a massive observer to a photon - 3 space + 1 time does not work." You'll see, if Eliah ever traveled in a charriot of fire, HE HAD TO BE TRANSFORMED INTO LIGHT, can you imagine the phosphorus-chrononaut? Of course, I'm just using the Greek word for light as an example to trigger the imagination. If some people already think about impossibilities, it will be impossible for them, that's a fact.