Special Relativity, Time, and the speed of light ...?

You got. You just don't see it. From the Frame of Reference of a motionless observer, photons don't move relative to each other. That is why time must stop for the photons.

Thought Experiment:

At c time stops.

Thus:

velocity = distance/time

Distance = velocity * time

(We know that light has a non-zero velocity =⇒ If photons are not moving with respect to the ship (distance = 0) then time must = 0)

The photons would be traveling at the same velocity as the ship, so they are motionless with respect to the ship as observed by an outside observer.

An observer on the ship sees the photon moving at c but since time has stopped, the above is true.

Imagine a "photon clock"

The person traveling at c is in a transparent space ship exactly 1 light second long. There is a perfect mirror at each end of the ship with a photon bouncing back and forth between the 2 mirrors. Each time the photon hits a mirror the clock ticks one second. Now take the ship to c.

The observer can see the photon just as it bounces off the rear mirror. Since the ship and the photon are traveling at the same speed (c) the photon appears to be motionless with respect to the ship (and going c with respect to the observer’s frame of reference. For an observer on the ship it takes forever for the photon to travel the length of the ship and for the clock to tick 1 second. If the ship was going a little less than the speed of light it might take years as observed from outside for the clock to tick 1 second. The observer on the ship would still say the ship was 1 light second long and that the photon traveled at the speed of light.

That clock is dead accurate for the frame of reference it is in.

Whether or not time is passing depends on the frame of reference. For the Photon on the ship time has stopped and it is not moving with respect to the ship. For the Photon as observed from outside the ship, time is still passing and it is moving in that frame of reference.

"If time stands still at the speed of light"

The Lorentz transforms in their derivation *cannot* be applied to light itself. So these formulae *cannot* be used to understand light.

"how can light photons have any relatively motion with respect to one another?"

They do, so clearly no problem. Not all photons move in the same direction...

"Or is it that at the speed of light, photon motion (constant speed) is completely independent of time?"

Not a question Science can answer, unfortunately.

"But then, if space and time continuously adjust themselves in order to maintain a constant speed of light, space and time must need a certain finite amount of time in order to make the necessary corrections for light speed."

Well, the silliness of applying the Lorentz transforms to light is the very reason such *cannot* be done.

"What does it mean, time stands still? Does that mean, t = 0 or t = 1?"

Well, it is your fantasy, what makes sense to you? I like t_emission = t_absorption, except that I do not think photons are "timeless" in the sense you mean.

[EDIT:

"Neb", photons do not have a "frame of reference", and "ds^2" is invariant in Special Relativity, not General. So you should really not just say "relativity [et al]", because this is incorrect.

]

The separation between two events in space-time (ds^2) is invariant in relativity. This means that all frames of reference agree on the value of ds^2. Different reference frames will disagree on how much of the ds^2 is space and how much of that is time, but will still agree on ds^2. Something that is moving thru space-time is always at rest in their own frame of reference so they would measure their motion thru space-time solely by the value of a clock in their reference frame. So your question depends on what a clock would read if it could be in the reference frame of light.

Space-time is naturally separated into three regions depending on the value of ds^2. It can be positive, negative, or 0. The reason why it can be negative or 0 is because the time contribution to ds^2 has the opposite sign of the space contribution to ds^2. So,

If ds^ >0, this represents the region of space-time we 'live' in - it's everything that goes slower than light

If ds^2 < 0, this represents the faster than light region which is not attainable

If ds^2 = 0, this is the region where light travels. As I said before. ds^2 always measures what a clock would read in the rest reference frame (which is the speed of light in this case) so a clock would read 0 elapsed time if it could ride along with the light beam. Weird, huh?

*Edit OzoneGuy -

Everything has a frame of reference, including photons. In the reference frame of a photon, ds^2 is 0 and it's proper time (as read by a clock) is 0

ds^2 is invariant in both special and general relativity. Don't forget that ds^2 represents an infinitesimal line element at a point in space-time. In both special and general relativity ds^2 is fully defined by the metric tensor - special relativity simply restricts the reference frames to inertial frames whereas there is no such restriction in general relativity. Don't confuse the concept of general covariances (which allows any reference frame and preserves the invariance of ds^2 at a point in space-time) with the idea that the metric can change as you move from one location to another location in space-time due to space-time curvature.

*Edit in response to Bob D1

Sorry for the confusion there - a point in space-time is often referred to as an event even though nothing actually has to happen there. So, a less confusing way of putting it is that ds^2 is the separation between two points in space-time.

An easy way of looking at is just using the Pythagorean theorem in a simple x-y coordinate system. The separation between any point in the x-y space and the origin of the coordinate system can be thought of as forming the hypotenuse (ds) of a right triangle with the other two legs being the distance along the x and y axis (dx and dy) so that just using the Pythagorean theorem then ds^2 = dx^2 + dy^2. Now overlay a new coordinate system x'-y' (a new reference frame) that is just rotated with respect to x-y. We now measure the legs of the right triangle with our new coordinate system we will get different values for the legs (dx' and dy') than we had before. But did we change the length of the hypotenuse by just adding a rotated coordinate system? No. The length of the hypotenuse hasn't changed so ds^2 remains invariant : ds^2 = dx^2 + dy^2 = dx'^2 + dy'^2. This idea remains true when you add a time coordinate although the time length dt^2 is negative.

Strictly speaking, ds^2 = dx^2 + dy^2 is a differential. You have to picture the right triangle shrunk down to an infinitesimal size to get to Einstein's General relativity.