adityakiran18 said:
hi ghwellsjr,
Thanx 4 the reply...
Actually I am afraid its not about that...This is the situation I was talking of:
I am standing at (0,-2) and facing X-axis...
A light ray is moving along positive X-axis direction and is at the origin(0,0) at present
(i mean its nearby to me)...
...There is no object anywhere...Its just the single ray of light...
Can I still see that..?
Assume linear propogation as I said single light ray(like a LASER source,,etc)...Dont consider huygen's wave or imaginary spheres...
Because i do not know about all the special cases how light behaves when it is not within a vacuum away of gravitational fields i will answer the question only for that special situation in which we assume you are in a perfect vacuum and there is no gravity field present.
So what we have here is the first postulate (some like to call it axiom) of special relativity.
Light will ALWAYS travel at C in a vacuum absent of any gravity.
Therefore, if you hold a flashlight towards any direction. The photons it emits will travel at C away of you. It would be impossible to see those photons unless they hit an object and bounce back. At least that is what one has to assume when he wants to arrive at the formulas Einstein arrived at for special relativity.
My knowledge about QM is too minimal to even attempt a guess on which scale it would change the formulas if photons would not travel as Einstein has imagined them to travel or as he deliberately postulated to "keep things simple enough" for his first shot.
Anyway, if we want to arrive at the formulas Einstein arrived at we HAVE TO consider those idealistic photons all traveling at C or at least "lightbeams" traveling at C within a vacuum away of gravity.
l*gamma = l' with
gamma = sqrt(1 - (v^2/c^2))
as an example of one of the formulas you will arrive at if you follow the two postulates for special relativity strictly.
(There are actually more postulates required but then it gets almost philosophical if we dive that deep)
And here is the clue. YOU yourself, seen from your own reference system DO NOT MOVE at all. You have to consider yourself standing still, and everything else is moving around you.
Your eyeballs move at exactly zero m/s within your reference frame. Remember it is called relativity for a reason. Everything moves relative to each other.
You could see someone ELSE's eyeballs move at CLOSE TO but never reach C, following a photon closely and getting distance to it very very slow.
The mindbending part here is that the eyeballs of the other guy you see traveling close to C, just a tiny bit slower than the photon they follow, will, when seen from that other guy's perspective(reference system) see that photon traveling at C (and so will everyone else in any inertial reference system - i repeat ALL WILL SEE THAT PHOTON TRAVEL AT C no matter which inertial reference system they are observing it from), and this guy or his eyeballs will have to consider himself NOT moving.
If your mind did not blow up yet, continue reading...
This is not easy at all to get your mind around. It took me months. But after drawing this
http://img204.imageshack.us/img204/2425/89825284.jpg
representing two observers traveling at v= 0.5c RELATIVE to each other (Both consider themselves not moving, but see each other passing at 0.5c) i finally managed to arrive myself at the formulas, using only the two postulates(or axioms if you prefer).
I now KNOW that Einstein was right first hand, FINALLY! (As long as the two postulates are valid. It would be "easy" to adjust the formulas if QM told us that photons don't behave as idealistic as assumed in the postulates but nevertheless we should be very close to same formulas still).Those are the two postulates required to build the formulas for special relativity. This, maths and maybe some geometry will get you the formulas.
1. Light(photons? lightbeams?) in a vacuum away of any gravity field will ALWAYS travel at C in all inertial frames of reference.
2. The laws of physics are the same in all inertial reference systems.
Or as i would say.. if two reference systems A and B are equal in quality, then there is no reason to consider any of the two special.
Whatever an observer a within a ref system A sees in ref system B under certain conditions, an observer b in system B will see the same if same conditions in A are met.
If for example observer a sees a ruler which is 1 meter within b's reference system (not moving relative to b) shrink by 80% to 0.8m, then b will see a different ruler which is 1 meter within a's reference system (not moving relative to a) also shrink by 80% to 0.8m.
Why? Because neither a's reference system nor b's reference system are special.
The laws of physics are the same in both reference systems(2nd postulate).
a sees b moving and considers himself motionless. b sees a moving and considers himself motionless. They move relative to each other at a certain speed. Their relative speed to each other is a common factor which cannot be used to assume either reference system A or B being special in any way.
Start drawing two diagrams representing two reference systems, and follow strictly the postulates like i did. The yellow lines are lightbeams. As you can see, the diagrams are drawn in a way, so llightbeams moving at C are at 45° always (see Minkowski diagrams)
After you understood this, your next stop will be the twin paradoxon most likely. The key to this is that acceleration changes the reference system the accelerating twin is in which creates an asymmetrical situation. One twin being in the same reference system at all times, while the other twin is passing through several reference systems by accelerating at times.
