The Lorentz Transformation for a time interval has the term (1-v2/c2). This goes to zero for the coordinate system moving along with a photon. If a photon does not experience time, yet we can measure the time it takes for a photon to transverse an interval, can a photon actually be a valid model for light?
2006-07-11 05:58:45 · 13 answers · asked by dhwrenchmsn 1 in Science & Mathematics ➔ Physics
In Special Relativity, the velocity of one reference frame is only determined in reference to another reference frame. If a photon experiences even the smallest unit of time, that would transform to an infinity amount of time in any other reference system (the rest of the universe). If we take the photon as the reference system at rest, then no one else can measure an interval of time, since the photon would then have to experience infinite time. That is why I think a is photon not real (does not exist in spacetime) and therefore is not a acceptable model for light.
2006-07-11 10:56:28 · update #1
Einstein asked this question to himself, and the answer he came up with helped him develop the theory of relativity. This was one of his famous thought-experiments. He tried to imagine riding on a beam of light, and concluded that time would seem to stand still.
So, good question, the first part anyway. As for the second part, well, it just depends on your context. In some senses a photon is not real. In others, it is quite real. You would need to be more specific. It's like asking, is your personality real? Is your distant ancestor real? Is color real? Etc.
2006-07-11 09:27:51 · answer #1 · answered by stanheidrich 2 · 0⤊ 0⤋
True, the Lorentz equation has that term, but it would apply to a frame of reference that is moving relative to the observer's. Relative to the photon (and any reference frame moving at the same speed as the photon), it would still experience time flow, but everything else would be at a standstill.
Furthermore, the photon model is the only valid model for certain experiments, just like the wave model is valid only for the others.
2006-07-11 13:22:59 · answer #2 · answered by dennis_d_wurm 4 · 0⤊ 0⤋
The Lorentz transformation is used to convert observations in one reference frame to another. The other being the moving one. If we use the photon as the other reference frame, the equation does explode since we can't divide by 0. Therefore, photons don't experience time. Proof: ever heard of the cosmic background radiation. It is the light emitted from the big bang at the surface of last scattering. A very long time has elapsed since then (13-14 billion years) so photons don't' experience time in the same sense as we do. Like mentioned in a previous answer, it's not as though everything is instantaneous (in our ref. frame). However in the photon reference frame, there is no such thing as time so it does appear to be instantaneous (in photon ref. frame) because no time has gone by in photon ref. frame since there is no such thing as time. It's usually confusing because of the reference frames.
2006-07-11 13:51:17 · answer #3 · answered by jerryjon02 2 · 0⤊ 0⤋
Lorentz Transformation does not imply that a photon doesn't experience time but merely provides a means to relate the measurements of an event from one reference to another. Photons do experience time in the sense that it takes an interval for them to get from one point in space-time to another. Its not like they get from point A to point B, say from the Milky way to Andromeda, instantaneously.
2006-07-11 13:28:26 · answer #4 · answered by Romeo 3 · 0⤊ 0⤋
If a photon moves at the velocity of light then the answer is NO, it does not experience time.
and i believe there are many things in the Universe that does not experience time other than light, photons, or even inside the black holes.
and I think God (Allah), created such things for the human mind to believe in the eternal life after life mentioned in the Holy Qur'an.
I know I wont gain 10 points for mentioning Islam, but thanks for the 2 points :p
2006-07-11 13:08:43 · answer #5 · answered by Anonymous · 0⤊ 0⤋
We measure the time it takes for a photon to transverse an interval in our coordinate system. The time the photon experiences internally doesn't influence this measurement.
2006-07-11 13:03:05 · answer #6 · answered by Anonymous · 0⤊ 0⤋
photon is an amount of energy so it s not an element and the light consist of many photons with a different wave length so if the photon is not real we cant see any thing
the photon velocity makes the realty of time which the time to move a photon from one point to another and that s why the time revolving the speed of light according to Einstein theory.
2006-07-11 13:47:24 · answer #7 · answered by ziad 2 · 0⤊ 0⤋
Time is relative.
An observer standing on a photon would not experience time because he is moving at light speed.
However, we detect the motion of the photon because we can experience time.
2006-07-11 13:04:29 · answer #8 · answered by MeteoMike 2 · 0⤊ 0⤋
the photon model of light is a model
it works to predict many aspects of the behavior of light
it is a useful model
the wave model of light is also useful
the whole essence of relativity is that as observer of the light, our time experience is different than that of the light itself, or of an observer traveling with the light
2006-07-11 13:04:29 · answer #9 · answered by enginerd 6 · 0⤊ 0⤋
When particles that decay travel at high speeds (like 90% the speed of light) their decay takes longer than it should, meaning internally less time has passed for the particle than the time we view. This is the case with muons created in the upper atmosphere and moving towards the ground: They should decay before reaching the ground, but they don't. They take longer to decay. At 100% the speed of light internally no time at all would pass.
2006-07-11 14:49:24 · answer #10 · answered by jeffcogs 3 · 0⤊ 0⤋