Question About a light beam .

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Question! "About a light beam..."

The question is can a light beam cover more than 300,000km in 1 sec? (I do realize that there's a reason why "c" is called a constant...)

Or maybe I'm asking the wrong question.. if the light beam could leave marks on the surface it's hitting while the light beam is traveling, if the light beam starts covering more than 300,000km in a second, how would the marks on the surface look?

To better understand the setting I'm talking about, see pic:
[PLAIN]http://img693.imageshack.us/img693/9719/gc0001.jpg
Confused TC is confused:confused:
 
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Superluminal velocities are not uncommon, but they're not really meaningful.

The light "spot" only has meaning in our interpretation of it. No photon has moved faster than c.

Here's a simpler scenario:
I can shine a flashlight on the moon's left limb, then flick it super fast to its right limb. The spot of light will cross the moon's face in excess of c. But the spot is only conceptual. The photons that hit the left limb have no relation to the ones that hit the right limb. And they are all only traveling at c.

Another one:
I can fire a machine gun at a target 1/2 mile to the north, then swing it around in a half second till I'm firing it at a target half mile to the south. My "spot" of bullets will have moved from north to south at 30,000fps, even though no bullet is traveling faster than the muzzle velocity of 2,200fps.


how would the marks on the surface look?
Fainter.

Just like the machine would spread its bullets out over the arc, so the laser would spread its photons. Fewer photons per unit distance = fainter light.
 
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Variations on your question come up rather frequently here. In fact, someone posted about it in the General Physics forum just yesterday!

https://www.physicsforums.com/showthread.php?t=425424

The spot on the screen can move faster than c because it isn't a physical object. The individual photons in the beam travel radially outwards at c, in different directions.

Note that there is no way for a person sitting on the screen at one location to use this moving spot to send a message to someone at another location on the screen, without communicating somehow to the person at the center of the arc who is holding the laser.
 


DaveC426913 said:
I can fire a machine gun at a target 1/2 mile to the north, then swing it around in a half second till I'm firing it at a target half mile to the south. My "spot" of bullets will have moved from north to south at 30,000fps, even though no bullet is traveling faster than the muzzle velocity of 2,200fps.

Yes, this analogy cleared up things for me, however does this mean that a light beam also has a "rate of fire"?

For example could we determine how many photons traveled through space in 1ms? given that we know the girth of the beam?
 


All you need to know is the power output of the laser, and the wavelength of the light that it emits.

Suppose the power output is one milliwatt. That means it emits 0.001 joule of energy per second. Suppose further that the wavelength \lambda is 600 nm. Now you can calculate the energy per photon using

E = hf = \frac{hc}{\lambda}

and use that to convert "joules per second" to "photons per second."
 
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From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...

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