The Photon's Perspective Taboo

In summary, the "Photon's Perspective" taboo is based on the assumption that photons exist and that they have a particle/wave duality nature. However, once you attempt to take on the reference frame defined by light cone coordinates, the standard formulas for relativistic physics break down.
  • #36


OB 50 said:
However, if at any point someone suggests looking at the universe from the perspective of the photon, the discussion is instantly over.

OB50, sometimes the hard part is formulating the right question. Or approch the problem from first formulating some useful questions. Dale seems to know the right question.

It could go something like this. What is the transformation from coordinates X to coordinates X', where v-->c? v is the velocity of the orign of X' in coordinates X.

Next map a photon--a wave or a particle, from one coordinate system to the another. Does this help?
 
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  • #37


A question should be answerable in at least same terms as it is put in. So if OP has asked a mathematically non rigorous question, the answer should be that much at the minimum.

Physics will have the last word, but that also means someone needs to have the first word. Thats where OP comes in.
 
  • #38


BTW the inexistence of a photon rest frame has some direct physical consequeces, for example, e+ + e- -> gamma (electron and positron annihilate forming only one photon) is impossible because there is no frame where gamma is at rest (while there is such frame for the center of mass of e+ and e-)
 
  • #39


atyy said:
There is something called "light cone coordinates". However, the reference frame defined by these coordinates do not form an inertial reference frame, which is the sort of reference frame in which the "standard formulas" hold.

Nevertheless, a coordinate system can be constructed.

Fredrik said:
It's the limit for the set of inertial frames, but we are allowed to define and use coordinate systems that aren't inertial frames. A coordinate system is just a function [itex]x:U\rightarrow\mathbb R^4[/itex] where U is an open subset of spacetime. (There are some technical conditions as well. The most important one is that if x and y are coordinate systems, the function [itex]x\circ y^{-1}[/itex] which represents a change of coordinates, must be differentiable infinitely many times).

I can see no a priori reason to discount coordinate maps that are not infinitely differentiable nor, for that matter, noninvertible. To derive some properties of a coordinate system whos origin has a relative velocity, v→c to an inertial coordinate system, it is sufficient, and apparently necessary to consider those that are neither.

By maps, I am referring to the composition, [itex]z = x\circ y^{-1}[/itex], and it inverse, z-1,

Fredrik said:
Some of it overlaps with some of what I said, but at the very least I said a lot more than that. And I really wouldn't talk about "the time dilation effect at velocity c from the reference frame moving at c". You can't just set v=c and expect things to remain well-defined. You have to talk about the limit v→c instead.

OK by me.
 
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  • #40


Phrak said:
Nevertheless, a coordinate system can be constructed.

Yes, I agree, so it's not "nevertheless". :smile:
 
  • #41


atyy said:
Yes, I agree, so it's not "nevertheless". :smile:

:rolleyes: Everthemore? :rolleyes:
 
  • #42


Phrak said:
:rolleyes: Everthemore? :rolleyes:
:smile::smile:
 
  • #43


http://img14.imageshack.us/img14/8451/perspectiveofaphoton.jpg
 
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  • #44


Translation? The photons maps to anywhere/everywhere in the universe?

The problem I have with provding a mathematical solution is that it is wide open to being interpreted semantically. i.e. contributors are now free to describe the photon's "experience" by metaphorizing the math as they see fit.
 
  • #45


DaveC426913 said:
Translation? The photons maps to anywhere/everywhere in the universe?

Thanks for responding, Dave. In the (a',b',x',y') coordinate system, the orign of (a,b,x,y), and therefore (ct,x,y,z) as well, would be smeared over the b' axis.

The temptation is to construct a primed Minkowski space (ct',x',y',z'), with the usual metric. But this doesn't follow immediately, that I can see. The points in (a',b') are a topological points space; a metric would be an addition premise, which would be overly speculative, etc, etc.

The problem I have with provding a mathematical solution is that it is wide open to being interpreted semantically. i.e. contributors are now free to describe the photon's "experience" by metaphorizing the math as they see fit.

I agree. Though, in the OP's defense, until you know the question you need to ask, how do you ask the question? The title I used was the filename I chose give by the name of this thread.

I am answering the question, "what is the coordinate system where vx --> c". To be strict, it appears not a coordinate system but a topology.

You seem to be concerned about infringing on the PF rules, but there is no new theory here, just special relativity, unless I've made an error.

A photon is an extended object isn't it? One needs a family of coordinate transforms.

btw, have I taken any liberties with the meaning of limits?
 
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  • #46


DaveC426913 said:
Translation? The photons maps to anywhere/everywhere in the universe?

Maybe you meant a beam of light traveling in the x direction. The beam lies along the a axis. This maps to a'=0. b is spread all over finite b', and y'=y=0, z'=z=0. Nothing can be said about x' and t' without premising a metric (I think), if it's even meaningful to do so. Is that what you meant? The x and t coordinates remain what they were, the line segment x=ct, lying on the light cone.
 
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