# [SR] - Photonic Reference Frame

1. Oct 30, 2012

### RagingCain

Recently learned about Special Relativity, so my brain is running perpetual circles around itself asking questions.

So, I figured I would come here and ask where could I find out more information about the Lorentz Factor being undefined at a luminal velocity?

Or more basically, where do I find the current most correct/accepted theory and mathematics for describing the motions of Photons?

2. Oct 30, 2012

### Staff: Mentor

This forum has a FAQ with a brief discussion:

The Usenet Physics FAQ also has a useful page on this topic:

Mathematically, it is obvious that the Lorentz Factor is undefined when v = c; just plug v = c into

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

and observe that $\gamma$ is undefined because its denominator is zero.

"Photons" in SR are objects that move at speed c in all reference frames. Put another way, photons are objects that move on lightlike (or null) worldlines; in a standard spacetime diagram, these are 45-degree lines (because we use units in which c = 1), as shown here:

http://en.wikipedia.org/wiki/Minkowski_diagram

3. Oct 30, 2012

### RagingCain

That pretty much covers what I have already observed in my head. I am also familiar with Minkowski diagram. While it helps the observation of sub-luminal up to the luminal, it still leaves me completely lost at understanding the reference frame of an individual Photon.

Would you be willing to advise me on how I should adjust my perception/clean up/fix logical mistakes? Other than the thought experiment doesn't make sense (like in the link.) I have gotten that twice, and to me it seems very reasonable to let it go, but at the same time if one is willing to do things correctly, by acceptable notions of physics one should explore this thought experiment like any other. I find it somewhat frustrating that the people I have talked to don't want to entertain these ideas while applying science and logic. I think there is more to Special Relativity than meets the eye. So given that, please understand, I have been told its "almost useless" thinking about this type of situation.

To better understand the individual properties of light, I have taken to creating a reference frame on an individual photon, traveling in a wave like motion at velocity C.

The wave nature of the photon is assumed sinusoidal, and I base my functions off of Maxwell's Wave Function for Electromag.

Y(X) = A Sin (kx + ωt + ∅)

Such that, A = Amp, K = Wave Number, ω = Angular Freq, t = time, ∅ = Phase.

Now, assuming constant c, and that this Photon is coming from a non-variable / single direction device such as a perfect laser, I managed to conclude a few basic concepts. But first frame relativity would require another observational point. I decided to look at another Photon along this same wave at the same time, and I would call it colinear - or cowave to be more precise. The problem with this is that there is no other perceivable photon in front or behind to an individual photon. Logically, you traveling at the speed of light, the guy behind you is traveling at the speed of the light in the same direction, and the guy in front of you is traveling, again, at the speed of light in the same direction. It is impossible for you to "perceive" anyone else forward or aft. I quickly broke down some math because I need to know if this scenario is logically feasible, otherwise I am spinning my wheels going insane. Thus I looked at the predictability of a Photon with a wave function, and wondered if there was a way of Mathematically showing a Photon could exist in a position of our choosing along this wave.

Y(X) = A Sin (kx + ωt + ∅) is transformed to thus,

Photon Zero: Y(X1) = A1 Sin (k1x1 + ω1t1 + ∅1)
Photon One (Chose by Observer): Y(X2) = A2 Sin (k2x2 + t2 + ∅2)

Considering the approach is for a cowave photon, this reduces the functions complexity.
|A1| and |A2| should be identical at a given time.
K1 = K2, same wave number.
ω1 = ω2, same ang. freq.
t1 = t2, results observed (if only in theory) at identical times.
∅1 = ∅2 should be equal or zero, zeroth for this observation.

Now given that it is wave like motion, I have determine that both Photons are in a constant state of X -> Y -> X vector motion.

This lead me to believe that a Photon is only at velocity C in the X-Direction, when at peak amplitude of the wave motion, either A * 1, or A * -1. So I began contemplating the Photons on sequential peaks.

I broke down the equations further for analysis of Delta X, since X1 cannot equal X2. I.e. observation of two distinct photons.

Knowing how Sine waves function, we know the maximum value determined by this equation is A or -A. Also since the Sine wave intersects the X-Axis, we can also choose a Photon capable of meeting at the x-Axis, at the same time, which correlates to the minimum distance between the two Photons.

So Photon 1, in relation to Photon 0, has a Y(X) = - A, when P0 has A, and a minimum distance of ΔX when A = 0 for both.

Maximum Distance => Opposite Peaks = D = √[(ΔY)^2 + (ΔX)^2]
Minimum Distance => A = 0 = d = ΔX

Maximum Distance => D = √[(2A)^2 + (ΔX)^2] => √[4(A)^2 + (ΔX)^2]

Solving for ΔX requires some basic algebra of the equations above:
Y(X1) = A1 Sin (k1x1 + ω1t1 + ∅1)
Y(X2) = A2 Sin (k2x2 + t2 + ∅2)

Transformed into:

Y(X1) = A Sin (kx1 + ωt)
Y(X2) = A Sin (kx2 + ωt)

Assuming Maximum Distance occurs when A * 1, -1 is reached, we solve for the corresponding Sine function.

1, -1 occurs at Pi/2, 3Pi/2

kx1 + ωt = Pi/2
kx2 + ωt = 3Pi/2

x1 = Pi/2k - ωt/k
x2 = 3Pi/2k - ωt/k

ΔX = X2 - X1 = Pi/K

So we choose a Photon 1, such that ΔX = Pi / k.

Maximum Distance => D = √[4(A)^2 + (Pi / k)^2]

Now the reason why I did the above, was just to show mathematically to myself, that if two Photons are capable of being treated as particles, that I should come up with a logically derived formula even though neither Photon can observe the existence of another.

Given Photon 0, I should be able to find another Photon that exists in this manner.

Now at this given snapshot in time, both of these Photons are travelling at velocity C in the X direction. According to SR formula, the Lorentz Factor is undefined, or more specifically, the relative velocities become undefined with one another. However, Newtonian Physics seems to be more understandable, and perhaps more applicable strangely, as the Newtonian relative velocity is simply zero, which is something I consider to be correct. There is also the oscillation between the Y-Axis in opposing directions in which two Photons are peaking at +/- C in which Special Relativity should would have to be used again. I.e. These two photons travelling at +/-C upon reaching the X-Axis or when ΔX = Pi/K, the velocity of P0 should be in the -Y direction @ C, and P1 should be in the Y-Direction @ C. There is oscillations, and changes, in the Photon's position with respect to one another, its just not in the X-Axis. Two objects will have a relatively zero velocity with respect to one another traveling in identical directions at identical velocities. Co-wave Photons, technically, are the only thing in the universe that would represent a Newtonian Relative Velocity because there is simply no change in reference frame nor velocity (without outside interference.) However, it doesn't seem to work well with a wave function, just a co-linear function.

I have also considered what a single coherent reference frame would probably mean to the Photons and began to think about the concept of time ever so slightly in this reference frame. If there is one unique reference frame for this cowave, it would stand to suggest that the entire wave, every photon would be in the exact same reference frame from start to finish. It could explain why light only deviates in relation to outside factors, there is no internal change to this system. I.e. Light from another Star travelling thousands of years to us is the "same" light emitted from the Star with some external factors influencing it. Unfortunately this idea and many others are just rattling around in my head. I have more "raw" ideas in my head but are almost fantastical without any solid groundwork to begin breaking it down logically what I mean.

I would love some help it determining what I should strive to better understand or fix errors in my logic early on, as I have somewhat become obsessed with this frame of reference. It is almost like I see the entire Universe in my head as a manipulated wave with the basic unit relative to a "Photonic" reference frame. In other words, all observable space is basically Photonic and the smallest possible frame is the Photonic reference frame itself. Explaining why the velocity of light is constant to us, is that its part of the underlying fabric of perceivable space. Again just "fantasy" :D

Last edited: Oct 30, 2012
4. Oct 30, 2012

### Fredrik

Staff Emeritus
There's no such thing. The FAQ post that Peter linked to tells you all you need to know to understand why there can't be an inertial coordinate system in which a massless particle is at rest. And this old post of mine explains why the usual procedure for how to associate an inertial coordinate system with the motion of a particle won't work then the particle moves at the invariant speed.

5. Oct 30, 2012

### Staff: Mentor

There is no such thing. Did you read the links I gave?

There are coordinates called "null coordinates" or "light cone coordinates" that have the property that one of the coordinate axes is the worldline of a photon. See here:

http://en.wikipedia.org/wiki/Light_cone_coordinates

However, these coordinates can't serve as "reference frames" in the ordinary sense; they don't have all of the required properties.

You could start by asking small, focused questions about specific things. For example, this...

...doesn't really help, because I don't know what, specifically, doesn't make sense to you. It all makes sense to me. (I assume you mean the thought experiment about pursuing a beam of light with velocity c.)

Sure, no problem. The thought experiment shows that, if Maxwell's Equations are correct, light can't work the way ordinary material objects work; light has to travel at the same speed, c, regardless of the motion of the observer, and no observer (i.e., no material object) can travel at c. At what point are you having trouble getting to that conclusion?

I can't help what other people have told you or what attitude they have taken. I can only go by what you have said here, and as I said above, what you have said here doesn't give me enough to go on in trying to help you. So however frustrating it may seem to you, you'll have to, as I said, start with small, focused questions about specific things that don't make sense to you, and we can go from there.

I think it's way too soon in this discussion to go into this in detail; if the simple thought experiment doesn't make sense to you, it will be extremely difficult for you to do this without going off the rails almost immediately. But I'll try, since it's possible that it will help speed things up.

Ok so far.

You haven't even got one "observational point" yet; the photon itself doesn't qualify, since nothing that moves at the speed of light can be an "observational point". That's what "the photon doesn't have a reference frame" means. You need a material object moving slower than light. The laser source would do; its worldline provides an "observational point", and another material object, such as a person with a detector at some spatially separated point, would provide a second.

Huh? I don't see why this would be the case. There can certainly be multiple photons released from the laser at different times and all moving in the same direction.

Not possible; nothing that is traveling at the speed of light can serve as an "observational point". See above.

Multiple photons released at different times and all moving in the same direction can certainly exist, as I said above. Is that what you mean? From the rest of your post, I'm not sure, and I can't really give any further useful comments since it looks to me like you've gone off the rails at this point. That's why I suggested taking things in smaller steps to start with.

It looks like the key thing you are struggling with is why a photon, or any object moving at c, can't serve as an "observational point". And that seems to go back to the thought experiment. So again, I think it would be a good idea to focus on exactly what doesn't make sense to you in the thought experiment.

6. Oct 30, 2012

### Fredrik

Staff Emeritus
I think that the OP was thinking that "from a photon's point of view" any distance along the line of motion in space is "Lorentz contracted" to 0. You and I know that there's no such things as "a photon's point of view" or a Lorentz contraction to 0, but the OP is still struggling with these things.

7. Oct 30, 2012

### RagingCain

I did read both the links and simply got rushed in concluding my earlier post.

Sorry to clarify, I have been told these things by others:
Due to the undefined nature of the Lorentz factor and that Photons can't be treated as an observational point, this thought experiment is invalid.

I guess I simply wanted to explore the motion of Photons given the tools at hand, which I admit are not a lot :).

I just have a very strong gut feeling that there is something very special about the relationship formed between photons inside an electromagnetic wave. I was also bothered by this one undefined value (divide by zero) and just simply started exploring from there.

Stating that a photon isn't able to observe, is fully understood, but there isn't anything like a photon. I just figured that the best way to understand the system would be looking at multiple parts of the system simultaneously.

I also don't believe the photon is at relative rest to one another even traveling at the same speed, due to vector based velocities along the wave function, the velocities are constantly changing both magnitude and direction.

So perhaps a better, more focused, question would be, how does one begin to analyze an individual Photon?

8. Oct 30, 2012

### Staff: Mentor

I can expand on this a little bit. Einstein asked himself what it would be like to move at the same speed as a light beam. The answer he came up with is that it would be like seeing the light beam "frozen", i.e., an oscillation that varies in space but does not change with time. The problem was that he knew there was no such solution of Maxwell's Equations; Maxwell's Equations only have solutions that describe waves moving at c, i.e., oscillations that change with time.

Einstein's conclusion was that if Maxwell's Equations are correct, then light can't work the same way as a material object does, because you can adjust your speed to be at rest relative to a material object, but you can't adjust your speed to be at rest relative to a light beam; it will always be moving at c. And since in order to have an "observational point", you need to be able to adjust your speed to be at rest relative to that "point", a light beam can't serve as an "observational point".

It's an object that moves at c in any inertial reference frame. Or, if we use units in which c = 1, which is standard in relativity, it's an object that moves at speed 1. If you draw its worldline on a spacetime diagram, it will be a 45-degree line in such units. For example, see here:

http://en.wikipedia.org/wiki/Minkowski_diagram

The key fact about such a line is that a Lorentz transformation takes the line into itself, whereas it does not take other lines (such as the ct and x axes) into themselves. (Note that by "Lorentz transformation" here I mean an ordinary one with v < c.) Basically this is just another way of saying that the relativitistic velocity addition law,

$$w' = \frac{w + v}{1 + vw/c^2}$$

where $w$ is some object's speed in one frame and $w'$ is its speed in a second frame moving at v relative to the first, gives $w' = c$ if $w = c$. whereas it gives $w' \neq w$ for $w < c$.

I should note that a photon is viewed as a "particle" in this model, or more precisely a very short pulse of light (like a laser fired for a very short time). The wave nature of light requires more complex models, but for many relativity problems you don't really need to deal with the wave nature of light.

9. Oct 30, 2012