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Does Light have an inertial reference frame?
So, this is something that I have been puzzling about for a bit. We know from Einstein that at the speed of light time goes to 0. We also know that when the speed of light is measured in any reference frame it is a constant, and as long as the medium it is propogating through is the same, then the same constant.
So, lets assume for a moment that we are in the reference frame of light itself. Time doesn't propogate. So without delta T there can be no movement. So, that means as far as light goes, all things and all places have to exist simultaneously, regardless of when they happen in a non - light intertial reference frame.
Or to say it a different way. With some cool equipment I can measure light going from point A to point B. There is a distance and a time that it takes and I come up with the speed of light. However, if I am the light (no pun intented) then I have to be at points A and B simultaneously. Regardless of how far apart or how long it takes when measured in a different reference frame.
I don't really see how to reconcile that other than to think that light doesn't have an intertial reference frame, but that seems like a cheat. I've read a number of books about light and relativity and have never seen this addressed.
Any takers?
Thinkingblade
So, my term was not precise. I am aware that light is massless therefore no inertia. That isn't really my question. I also know that time does not pass for a photon and that C is a constant in all intertial reference frames.
So, to Jose - yes, this I already understood, it wasn't my question, though I appreciate your response.
To eye - you are closer to my question, but your answer has problems with it. First, on board the photon time cannot "seem to pass normally" delta T is zero. As far as I understand Einstein's writings on the matter that is as fixed as the value of C itself relative to inertial reference frames. Therefore "velocity" as we think of it is either undefined or infinity - take your pick. So nothing can come and go.
Irv, interesting. It is perfectly legal to divide by 1-i, it just means in this case that T=0. A number of "real" electrical phenomena come from "imaginary" solutions. Is there some "mathematically" wrong with the solution?
Thanks all.
5 Answers
- Irv SLv 71 decade agoFavorite Answer
Yes, that question is a bit deeper than usual.
At C the reference frame involves an imaginary number.
[A bit from the Lorenz transform: ( .../1-( 1- V^2/C^2))^1/2]
When V = C, you're dealing with dividing by '1 - i '.
It's not a 'real' frame.
(Sorry if you haven't encountered imaginary numbers yet,
but they become necessary here.)
You could regard it as 'outside of time'.
That photon is the result of an electron jump through a space where it's existence is 'forbidden' .
It might be said to fly through space, frozen in time, until it can undo the forbidden reaction.
But don't take that as any more than a flight of fancy on my part.
- oldprofLv 71 decade ago
You are asking if time passes for a photon. It does not, not on board the photon. But of course it does for outside observers (and who in their right mind would ride a photon). So we onlookers do have a delta t and there is that delta S (change in spatial location) we see the photon do.
As there is a delta S you are not at A and B concurrently, as far as the outside observer is concerned. On board the photon, time seems to pass normally, but whizzes by outside (off photon) at delta t ~ infinity. The universe will come and go instantaneously as you peer off photon to the cosmos around you.
And why would a photon have an "inertial" reference frame...it has no inertia. It is massless. Now you, on the other hand, would be in deep yogurt because you have mass and you can't possibly keep up with that photon. Poor Thinkingblade...he was such nice guy too. : - (
PS: But time does seem to pass normally on the photon. That results because, on the photon, I am at rest with that reference frame. The fact that time has stopped on the photon relative to the outside does not change the observeration that time is normal to me on the photon.
- Anonymous1 decade ago
Postulate 2 of Special Relativity: Light travels at c in all inertial reference frames.
Therefore: there is no inertial reference frame in which light is at rest.
QED.
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