Rotating a laser beam faster than light?

[gonegahgah]

My apologies. I should have worded that better. My question isn't about the intensity, nor arrival timing of photons, nor how far photons can travel.

You said that for an "individual photon ... E=h/f"
I assumed you meant E(individual photon) = h / f. Is that correct?
If that is correct it would imply that one cycle of light through a point (or a wavelength) corresponds to a self-contained single photon.

 

[jtbell]

I assumed you meant E(individual photon) = h / f. Is that correct?

That is correct.

If that is correct it would imply that one cycle of light through a point (or a wavelength) corresponds to a self-contained single photon.

No, it doesn't. Why would you think it does?

 

[Austin0]

Yes, it's incorrect. In the original rest frame, the rocket keeps getting shorter due to Lorentz contraction. This means that the front and rear can't have the same acceleration profile. The rear must accelerate faster than the front for the rocket to get shorter. What a clock measures is the integral of \sqrt{dt^2-dx^2} along its world line, and since the rear is accelerating faster, the integral along its world line gets a bigger contribution from the dx² term. Because of the minus sign in front of it, that makes the integral smalller. That's why the clock in the rear will be behind the clock in the front.

Fredrik --thanks for your responce but I am still confused.
As I understand it, the desynchronuzation of clocks is responcible for the invariance of the measured speed of light? The only way this seems to make sense is if the clocks at the rear are running ahead of the clocks in front. Ie: showing a later time.
This seems to mean that if you have distributed , synchronized clocks in a rest frame and then accelerate the frame , the clock in the rear would end up running ahead of one in the front. Does this make sense or am I seriously missing something??

 

[JesseM]

You said that for an "individual photon ... E=h/f"
I assumed you meant E(individual photon) = h / f. Is that correct?
If that is correct it would imply that one cycle of light through a point (or a wavelength) corresponds to a self-contained single photon.

Why would it imply that? Can you explain your reasoning?

 

[Fredrik]

As I understand it, the desynchronuzation of clocks is responcible for the invariance of the measured speed of light?

I don't know if it works that way, i.e. if you can prove the invariance using only what we know about desynchronization of clocks attached to the same solid object. Maybe you can, but it seems to be an unnecessarily complicated way to think about it. I think of SR as a physical theory that consists of a mathematical model (Minkowski space) and a set of postulates that identifies things in the model with things in the real world. The metric can be used to derive anything about the model, so I'd rather say that the Minkowski metric is the cause of both desynchronization and invariance.

Anyway, Einstein's second "postulate" says that the speed of light is the same in all inertial frames. It's not the same in arbitrary coordinate systems.

The only way this seems to make sense is if the clocks at the rear are running ahead of the clocks in front. Ie: showing a later time.

I don't see why.

This seems to mean that if you have distributed , synchronized clocks in a rest frame and then accelerate the frame , the clock in the rear would end up running ahead of one in the front. Does this make sense or am I seriously missing something??

If the red "this" refers to what I said, then the red "ahead" should be "behind". Anyway, if you take a grid of rulers and synchronized clocks and accelerate them by starting a tiny rocket attached to each clock when they all show the same time, and then shut them off when they have reached velocity v in the original rest frame, then they will be doing inertial motion again, but they won't be synchronized, so the numbers you can read off the grid won't agree with the coordinates assigned by an inertial frame.

 

[gonegahgah]

Why would it imply that? Can you explain your reasoning?

Well, you said that E=hc/λ.
I assume you were transfering that through (by the progression of the paragraph) as meaning E(individual photon) = hc/λ. So constant divided by wavelength.

I read your words and this to say that one photon = one wavelength. If the photon were multiple wavelengths joined end to end then the total energy of an individual photon would be (number of wavelengths) * hc / λ as the equation implies that it is per wavelength.

Maybe you could clarify it for me by showing how you would work out how many photons arrived given the energy and time.

 

[Austin0]

Expansion: Again, think of the rod as a spring. If you accelerate the rod, it will shorten the way a spring does until the force of the coiled spring equals the force of the acceleration and the pressure wave of the start of the acceleration propagates through it. Then it will continue to accelerate as a shortened rod.
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Hi One question.
Given that the rate of propagation [speed of sound] is very slow compared to intermolecular elastic reactions , wouldn't both the force and the compression move as a reciprocal occilation with no net compaction or actual motion at the point of acceleration until the whole system attained sufficient energy to begin moving as a body?

In the spring context; Wouldn't continuous acceleration result in continued expansion /compression of the spring with the force of acceleration never reaching equilibrium with the intermolecular tensile force in a state of maximum compression?

This is assuming that the magnitude of the force does not exceed the systems ability to distribute it fast enough.

This is a question I have mulled over before in other contexts so would appreciate any clarification.