Consider a star 100 light years away. We say the light left the star 100 years ago. But, for the beam of light, time has slowed down, so it actually left the star more than 100 years ago, technically, to an observer on earth, right?
- Jinx
Um, for a photon in the beam of light, no time at all is passing, so it has always been between the fusion event which produced it and the absorbsion by your retina. Travel is instantaneous to the photon. The time dilation expereinced by the photon does not change the measurement in our frame of reference one iota.
No, it actually left the star 100 years ago. A light year, as I’m sure you’re aware, is a measure of distance equalling the distance a photon travels in a year. This works out to 186,000 miles/sec * 3600 sec/hour * 24hr/day * 365.25 day/year ~= 5869713600000 miles. This is true regardless of whether time is slowed down from the perspect of the photon or not. Your confusion comes in when you try to think of how one would perceive time traveling the speed of light. This question is meaningless because one cannot travel the speed of light.
OK, then please explain: As I recall being once taught, Einstein’s theories concerning time were proven correct by scientists finding cosmic rays (IIRC) passing through earth’s atmosphere and decaying much later than expected. Hence, they were found to have travelled a greater distance through the atmosphere than expected. The reason, as I understand it, is that classical equations were used…and not Einstein’s equations which adjust the relativity of time.
Unless I have the story wrong, and/or I am misunderstanding what a cosmic ray is (and what it decays into, if anything at all), isn’t this an example where the distance travelled by a particle is different to the observer? And if you do agree with this logic…would the same be true for a star’s light waves per my OP?
- Jinx
Others have already given good answers, but I’ll add my own, probably redundant one anyway.
For anything traveling at sub-light speed from your example star to Earth, the distance will be shorter than 100 light years. The faster the speed, the more the distance is contracted, and the shorter is the perceived travel time. This contraction is negligible for the speeds of ordinary human experience, but will shrink dramatically as the speed approaches that of light.
It reaches its culmination for objects that actually travel at light speed, like light. For such objects, the travel distance vanishes to zero, and the travel time is also zero.
So basically, photons have no age, at least in their own reference frames.
It’s true that Einstein’s theory of Special Relativity is supported by the detection of particles, muons, arriving “miraculously” at the Earth’s surface, when classical physics predicts they wouldn’t live long enough. These particles are the product of cosmic rays though, generated in the Earth’s upper atmosphere, and not cosmic rays themselves. In any case, none of these particles are photons. They have mass, and therefore do not travel at light speed. (Cosmic ray particles are radioactively stable, and can cross the galaxy, intact.)
Yep, it is an example of time dilation — and nope, it doesn’t apply to light waves.
I should have said, “predicts more of them wouldn’t live long enough.” Some muons would reach ground level, even pretending time dilation didn’t exist. Much fewer of them though.
What if you were not observing this by “Retina”,but by another form of light observered by a human,(or non) observer in a different observers spectrum of colors,a totally different outlook.
Its possible,YES!
As others have stated, the particles of interest are muons. They can be produced when cosmic rays struck the upper atmosphere and travel at very high (but sublight) speeds and are quite different from photons.
Just to confuse the issue more (or maybe not), there the star has three positions.
When it emitted the photons we see, it was in one place. But in the 100 years (in this example) that the light has been travelling, the star has moved. So it has the position where we ‘see it’, and it has the position to which it moved through space in the intervening time. But the universe is expanding, so it has that position as well.
Would you remind rephrasing that? For some reason I can’t seem to parse your post well enough to get any clear meaning from it.
It seems like you’re implying that light in the non-visible spectrum behaves differently than visible light (with respect to the theory of relativity), which is not the case.
I might be wrong in this assumption, but it works for me. Due to the fact that photons have no mass, light itself is not subject to the time/space/mass distortion effects of traveling at light speed.
I don’t think that’s right. As people have said, it is really correct to say that from a photon’s perspective no time passes as it travels. From our perspective, a hundred years pass as it travels. And of course it’s our perspective that we care about.
I realize it seems a little odd to talk about a photon’s “perspective.” Statements of the form “If it were possible for you to experience this thing which it is physically impossible to experience, then it would look like this” seem kind of screwy. But it is meaningful to make statements about time intervals in the reference frame of the photon. All I’m saying is that in the photon’s reference frame, the time interval between the start of its journey and the end of its journey is zero.
If you want to get somewhat technical about it, this is because photons travel along null-geodesics of space-time – i.e. paths along which the space-time interval is zero. However, in the photon’s reference frame, any two events that occured at the photon (such as its departure from a distant star and its arrival at earth) have no spacial separation. Thus, for the space-time interval to be zero, the time difference between the events must be zero in the photon’s reference frame.
Can this be attributed to the shortening observed along the axis of travel of objects approaching C? What I mean is that, from the photon’s point of view, the universe is traveling at precisely C (no preferred reference frame and all that), so that, as far as the photon is concerned, relativistic shortening has caused the universe to collapse into a two dimensional plane.
Is this a reasonable explanation of what’s going on, or am I grossly over simplifying?
Nope, there’s no such thing as “the photon’s reference frame”. Since you know about null geodesics, you must know the transformation equations, too. Try plugging in v=c and see what happens.
The whole idea of “time as seen by the photon” is meaningless.
Both. We can’t talk meaningfully about a reference frame moving at c. But we can talk about a set of reference frames where the speed approaches c. In a reference frame moving at very high speed (very close to c), distances would be compressed a great deal. As the speed approaches c, the distance travelled in that frame approaches 0. As oversimplifications go, then, it would not be unreasonable to say that “in the photon’s frame, the distance travelled is 0”. Strictly speaking, that statement is meaningless, but it can be taken as shorthand for the correct statement “to the extent that a reference frame’s speed is close to c, the distance travelled is close to 0”.
Yeah, I shouldn’t have said “the photon’s reference frame”. You can’t have v = c, but you can talk about the limit as v goes to c, which is really what I should have done.
Actually, I just noticed that Chronos already said this, so I guess this post is kind of redundant. But since I have it all typed anyway . . .
Ah, I think I get it. It’s similiar to a function that is undefined at x=A, but that has a finite value for limit(x–>A).