Relativity equation - calculating the algebraic limit

[skynelson]

Greetings,
A colleague questioned an assumption of mine in a recent paper. The assumption was that in the limit as v->c, the result is that t = L/c. This leads to the commonly discussed idea that light experiences zero proper time, and zero proper distance. (by the way, is this a commonly held perspective?) My paper is the first in a series, reasoning that light experiences no passage of time (or space) as it travels, yet due to the "Simultaneity principle", it always arrives at a point L/c into the future from when it was emitted. This value, L/c, is what I'm trying to verify here.

I would like to present my reasoning here as to why that mathematical result is true, and not simply a hand-waving assumption. Your feedback would be helpful. I will start by calculating the value for t' in the limit as v->c. I will then invert the idea to show that t = L/c.

(By the way, I understand and sympathize with many people's feeling that the question of what life is like from the perspective of a photon is a ridiculous question. I agree, mostly, in that it involves undefined quantities and is something that cannot be experienced, or even described, in human terms. This is fundamental. In my current work, however, I find it very instructive in illuminating the nature of light to ask the question "how does light 'experience' the universe?", because the undefined quantities involved lead to some very interesting possible conclusions. So there is my apology...)

Setup: Imagine event A is the emission of a photon from the surface of the sun. Event B is the absorption of that photon by a solar panel on a satellite orbiting Earth.

1) t' = gamma ( t - xv/c^2 )
the unprimed frame represents the sun/satellite system. The prime frame represents time as experienced by a photon (I know...just bear with me, and confirm that my mathematics is correct).
2) in the mathematical limit as v->c, gamma increases without bounds, but we also find that the term in parentheses ( t - xv/c^2 ) goes to zero. this is because t = x/v is the time taken for the light beam to pass between event A and event B (as measured by a satellite-bound observer). So this can be written as ( x/v - xv/c^2 )
3) in the limit as v->c, we can write it as:
x/c * ( 1/beta - beta )
as mentioned, this goes to zero in the limit as beta->1.

4) we can now rewrite the original equation as:
t' = t * (x/c) * ( 1/beta - beta) / ( sqrt( 1-beta^2 ))

5) we can therefore graph this quantity to understand its behavior as v->c. I have graphed t' as a function of beta. (see the attachment). Indeed, t' not only approaches zero as v->c, it HITS zero as v=c, and the point seems to be a well-defined limit. As far as I can see, there is nothing in this mathematical result which prohibits the interpretation that light experiences zero proper time.



That concludes the first step, showing that the value for proper time in the limit as v->c is zero.


If there are no problems with that, I then want to invert the problem.
1) if t'->0 as v->c, then we can use our original equation
t' = gamma ( t - xv/c^2 )
to write:
0/(infinity) = ( t - xv/c^2 ) as v->c
(?right?)

2) obviously, if this is correct, in this case:
t = xv/c^2 has to be true, to make the equation work.

3) finally, plugging in v->c (which is our requirement), this equation approaches:
t = xc/c^2
t = x/c
(or t = L/c)

This result is important, because in the first of my papers I present the perspective that if light indeed experiences zero proper time and zero proper space, the simultaneity effect essentially dictates that each particle of light will always arrive t = L/c into the future from any inertial systems perspective. We cannot truthfully say that light moves from its own perspective (since it has no 'experience' of Time and space), but from an inertial system's perspective it does move, as a result of the simultaneity effect, much like the strings of yarn in the elevator from a previous post. good example!

The point is that light moves through time and space, not because it's experiencing motion, but because reality is refreshed, like a TV screen, according to the simultaneity principle, at exactly the ratio t = L/c. the result is the apparent traveling of a photon at that rate.

This is a very shortened description of the paper. this simple new perspective on light has many possible interesting and profound consequences, which I point out in the papers.

(you can contact me if you want to read the original papers...www.SkyNelson.com)

 

[Mentz114]

4) we can now rewrite the original equation as:
t' = t * (x/c) * ( 1/beta - beta) / ( sqrt( 1-beta^2 ))

It seems very dubious that this has any meaning when beta = 1. If you allow 0/0 to have a value, it is possible to prove that 2=3 and all kinds of other nonsense.

The point is that light moves through time and space, not because it's experiencing motion, but because reality is refreshed, like a TV screen, according to the simultaneity principle, at exactly the ratio t = L/c. the result is the apparent traveling of a photon at that rate.

See what I mean ?

 

[JesseM]

If two clocks are at rest relative to one another and synchronized in their own frame, and the distance between them in their rest frame is L, then if I am moving inertially at speed v along the axis between them, in my rest frame the two clocks will be out-of-sync by exactly vL/c^2. So, I suppose you could say that in the limit as my speed relative to the clocks approaches c, the clocks approach being out-of-sync by exactly cL/c^2 = L/c (and of course the distance between the clocks also approaches 0 in my frame, so the time to move between them approaches 0, while the clocks also approach being completely frozen in their rate of ticking). But all of this only really makes sense as a statement about limits, and there are reasons the notion of a "photon's own frame" doesn't really make sense even in terms of limits--for example, what would you say the speed of one photon is in the "frame" of another photon? You could take the limit of a slower-than-light observer watching a photon in the limit as the observer's speed approaches c, in which case the answer would be that the photon is still moving at c in the limit, or you could take the limit of two slower-than-light observers moving alongside each other as both their velocities approach c, in which case the answer would be that each one sees the other at rest.

 

[atyy]

I did not check your calculation, but posts #28,29 may be relevant:https://www.physicsforums.com/showthread.php?t=252523&page=2

This guy has great notes:
http://people.maths.ox.ac.uk/~nwoodh/

 

[atyy]

I still haven't looked at your calculation.

But your assertion that the proper time for a photon is zero is true. The technical terms for these are "light cone" or "null cone" and "causal structure". A text that develops relativity from the light cone structure is "General Relativity" by Malcolm Ludvigsen (http://books.google.com/books?id=YA8rxOn9H1sC).

The refreshment of spacetime is not true within classical special relativity or relativistic quantum field theory, both of which are generally agreed to be accurate to the best current experimental precision. It is postulated (in some sense) in many attempts to formulate a quantum theory of gravity. All current attempts may eventually fail but I find these interesting:
http://arxiv.org/abs/gr-qc/9210011
http://arxiv.org/abs/gr-qc/0309009
http://arxiv.org/abs/0801.0861

One thought concerning your refreshment of spacetime: have you taken into account that light is an oscillation, and in principle can any wavelength, no matter how short, as long as it is not zero; it can also have any frequency, no matter how high, as long as it is not infinite?

 

[atyy]

light is an oscillation, and in principle can any wavelength, no matter how short, as long as it is not zero; it can also have any frequency, no matter how high, as long as it is not infinite?

I should of course qualify that these statements are only known to be true up to the highest energy currently available experimentally, or inferred to exist from observations of space.

 

[skynelson]

Mentz: i am not dividing 0/0. I'm dividing 0/infinity. or we could rewrite it as multiplying 0* sqrt(1-beta^2), which also goes to zero at v->c. this is standard practice, not totally unlike calculating the value of the electric field as zero, when r->infinity. nothing unreasonable.

 

[skynelson]

atyy,
thank you for the feedback. light being an oscillation in time does necessarily not pose a problem for the fact that light experiences zero proper time. if light is described by the wave function psi = exp(i {kx-wt}), it can exist as a wave function, a simple mathematical probability. in the process of the creation of space and time, through the simultaneity effect, at the ratio 'c', the wave function is expressed as an oscillation in the coordinates x and t.
I do recognize this as sort of a meaningless perspective game, on one hand. yet I have found some good reasons for doing this.
by the way, I'm sorry that I directed you to my website without having the paper there. I have not published it yet onto my web site, but you can e-mail me through the website and I can send it.

 

[JesseM]

Setup: Imagine event A is the emission of a photon from the surface of the sun. Event B is the absorption of that photon by a solar panel on a satellite orbiting Earth.

1) t' = gamma ( t - xv/c^2 )
the unprimed frame represents the sun/satellite system. The prime frame represents time as experienced by a photon (I know...just bear with me, and confirm that my mathematics is correct).

This is the equation for relating the time coordinate of a single event in two frames, but if you want to compare time intervals in two frames between A and B, what you're really looking at is (time coordinate of event B in frame 1) - (time coordinate of event A in frame 1) and comparing it with (time coordinate of event B in frame 2) - (time coordinate of event A in frame 2). So, even if the primed frame was the frame of some object moving slower than light, this would only make sense as an equation for the time the object experiences between A and B if we assume event A happened at x=0, t=0 in the unprimed frame and x'=0, t'=0 in the primed frame...is it safe to say you are making this assumption? If so, then of course if the object is moving at speed v in the unprimed frame, its position at time t will always be vt, so the equation would be:

t' = gamma * (t - tv^2/c^2) = gamma * t * (1 - v^2/c^2)

and since gamma = (1 - v^2/c^2)^(-1/2), this is equal to t * (1 - v^2/c^2)^(1/2), or t / gamma. And of course, delta-t' = delta-t / gamma is the standard time dilation equation, where delta-t' is the time between two events on a clock's worldline in the clock's own rest frame, and delta-t is the time between these same two events in the frame of an observer moving at speed v relative to it.

2) in the mathematical limit as v->c, gamma increases without bounds, but we also find that the term in parentheses ( t - xv/c^2 ) goes to zero. this is because t = x/v is the time taken for the light beam to pass between event A and event B (as measured by a satellite-bound observer). So this can be written as ( x/v - xv/c^2 )
3) in the limit as v->c, we can write it as:
x/c * ( 1/beta - beta )
as mentioned, this goes to zero in the limit as beta->1.

4) we can now rewrite the original equation as:
t' = t * (x/c) * ( 1/beta - beta) / ( sqrt( 1-beta^2 ))

5) we can therefore graph this quantity to understand its behavior as v->c. I have graphed t' as a function of beta. (see the attachment). Indeed, t' not only approaches zero as v->c, it HITS zero as v=c, and the point seems to be a well-defined limit. As far as I can see, there is nothing in this mathematical result which prohibits the interpretation that light experiences zero proper time.

This is correct, but it's a lot easier to see if you just use the time dilation equation delta-t' = delta-t / gamma. Again, delta-t is the time between two events on an object's worldline in the frame of an observer who sees the object moving at speed v, and delta-t' is the time between these events in the object's own frame. You can see that if delta-t is some fixed finite number, then as v approaches c and gamma approaches infinity, delta-t' approaches zero, meaning that the proper time for the object between A and B approaches zero as its velocity approaches c.

If there are no problems with that, I then want to invert the problem.
1) if t'->0 as v->c, then we can use our original equation
t' = gamma ( t - xv/c^2 )
to write:
0/(infinity) = ( t - xv/c^2 ) as v->c
(?right?)

2) obviously, if this is correct, in this case:
t = xv/c^2 has to be true, to make the equation work.

As I said above, your equation only makes sense in terms of time intervals if we assume event A happens at 0,0 in both coordinate systems, in which case it must be true that x = vt, or t = x/v. So in the limit as v approaches c, t approaches x/c, and of course in the limit as v approaches c, xv/c^2 will also approach x/c.

3) finally, plugging in v->c (which is our requirement), this equation approaches:
t = xc/c^2
t = x/c
(or t = L/c)

Again, your equation only makes sense if we assume that A happened at x=0, t=0 in the observer's frame, so if B is a second event on the worldline of a photon that started at A, then naturally it must be true that x = ct, because it's moving at the speed of light. So, this equation is telling us something pretty trivial.

This result is important, because in the first of my papers I present the perspective that if light indeed experiences zero proper time and zero proper space, the simultaneity effect essentially dictates that each particle of light will always arrive t = L/c into the future from any inertial systems perspective.

What do you mean by "the simultaneity effect"? Naturally it is obvious that if something is moving at velocity v, the time for it to travel a distance L will be L/c...this is just as true in Newtonian physics as it is in relativity.

The point is that light moves through time and space, not because it's experiencing motion, but because reality is refreshed, like a TV screen, according to the simultaneity principle, at exactly the ratio t = L/c. the result is the apparent traveling of a photon at that rate.

? What does it mean to say "reality is refreshed" at a certain ratio? The equation is just telling you the trivial fact that speed is (change in position)/(change in time), which can be rearranged as (change in time) = (change in position)/speed, so if the change in position is denoted as L and the speed is c (both in terms of the observer's frame), then the change in time must be L/c.

 

[atyy]

atyy,
thank you for the feedback. light being an oscillation in time does necessarily not pose a problem for the fact that light experiences zero proper time. if light is described by the wave function psi = exp(i {kx-wt}), it can exist as a wave function, a simple mathematical probability. in the process of the creation of space and time, through the simultaneity effect, at the ratio 'c', the wave function is expressed as an oscillation in the coordinates x and t.

I wasn't thinking that light being an oscillation is a problem for it having zero proper time. Light does have zero proper time.

I was thinking that light being an oscillation is a problem for the idea that reality is refreshed. The rate at which reality is refreshed must be higher than the highest oscillation frequency currently known. Special relativity imagines that there is no highest frequency, and is (I hope) wrong for a high enough frequency, but we haven't experimentally reached that frequency yet.

The idea that light does not "experience" time because it has zero proper time is meaningful in some limited ways. For example, light moves relative to us, so we can use the successive peaks and troughs of a light wave moving past us to measure time. In this sense, we "experience" time. However, the speed of light is the same for light of all frequencies, so light does not move relative to light. If two light waves of different frequencies set out in the same direction, any particular peak in one of the waves will not move relative to the nearest peak or trough in the other wave. It will never experience the peaks and troughs of the other wave moving past it. In this sense, a light wave does not "experience" time.

 

[skynelson]

Atyy, I would actually think that the "refresh rate"(as you so kindly obliged the term) is c. the frequency is unrelated, or rather unlimited, by c. this concept puts no limit on the WIDTH of the "refresh" wave. the speed of the wave defines time and space (by the ratio c). The "width" I would consider to be what defines energy and momentum... the width would be measured by delta-t, which would correspond to a photon of energy delta-E > hbar/delta-t. there is no upper (or lower) limit to this. the same would be true for space, delta-x.

I know much of this seems like a trivial look at stuff that is already well understood, but sometimes this is a useful process.

 

[skynelson]

? What does it mean to say "reality is refreshed" at a certain ratio? The equation is just telling you the trivial fact that speed is (change in position)/(change in time), which can be rearranged as (change in time) = (change in position)/speed, so if the change in position is denoted as L and the speed is c (both in terms of the observer's frame), then the change in time must be L/c.

Thank you, Jesse for your time and careful run through. yes, I am performing a trivial calculation here to prove to myself that the Lorentz transforms are equally valid to light in the limit that v = c as they are to objects moving at lesser speeds. I wonder if a simple reanalysis of these trivial facts can open some new doors; namely, for one, that light's holographic properties arise from the fact that it experiences zero proper time and space. actually, I don't just wonder, I've worked on it. I worked hard to complete this first paper, and all of the comments here have helped me finish. Perhaps you'd like to read it, pick through it with your fine toothed comb. :-)

if you e-mail me at theskyband (at) gmail.com I would be happy to forward it to you.

cheers,
Sky