Rotating a laser beam faster than light?

[gonegahgah]

Thanks mgb.

So if you are weightless in deep space and someone is accelerating away from you they will have a component of animating slower than you added due to GR rules and anyone near a planet will have a component of animating slower than you added due to GR rules; ignoring the SR component for the moment.

That gives us an added GR component of:

in very deep space & steady (none) | very deep space & accelerating (slower)
----------------------------------------------------------------------------
near planet not accelerating (slower) |

I've left the fourth quadrant empty.
Let's say someone is in a car and they accelerate along the road will this add a GR component of slow down to how fast we see them animating on top of that due to the gravity?
Or are they mutually exclusive and only the greater of the two prevails?
Or does gravity prevail always?

 

[madmike159]

If you shon a laser on to the inside wall of a circle 1 light second away, it would have a circumference of about 60 light years.

Sorry no idea where 60 LY came from.

 

[gonegahgah]

mgb, if you don't know the answer are there any people you know that you find out for me.

There seems to be another thread on relativity and acceleration at https://www.physicsforums.com/showthread.php?t=273652".
Does this invalidate anything that we have here so far?

 

[JesseM]

So if you are weightless in deep space and someone is accelerating away from you they will have a component of animating slower than you added due to GR rules and anyone near a planet will have a component of animating slower than you added due to GR rules; ignoring the SR component for the moment.

GR is not needed to deal with acceleration in deep space, it's only needed for situations involving gravity. SR can deal just fine with an accelerating object of negligible mass in deep space.

 

[gonegahgah]

Thanks Jesse. That was what you said in the other thread. So are you saying that mgb was incorrect about acceleration being equivalent to gravity; specifically with respect to the effect upon time dilation. Are they non-equivalent in this specific matter?

 

[Jonathan Scott]

Thanks Jesse. That was what you said in the other thread. So are you saying that mgb was incorrect about acceleration being equivalent to gravity; specifically with respect to the effect upon time dilation. Are they non-equivalent in this specific matter?

I haven't been following this thread, but locally the effect of acceleration and the equivalent gravitational potential gradient on clocks is identical, so for example a clock at the front of an accelerating spaceship appears to run slightly faster than one at the back. This can be calculated by Special Relativity.

Also, if you consider something being accelerated in a rotating ring (like an Arthur C Clarke space-station) I think (if I remember correctly) that you will find that if you analyze the difference in time dilation between two different levels by Special Relativity given the velocity difference, you get the same result as if you analyze the difference based on the illusion of gravity seen in the frame of an observer in the ring.

 

[Fredrik]

Thanks for jumping ahead on a question I was wondering Fredrik comparing relative speed time dilation (SR) and accelerated or gravitational time dilation (GR). So they are completely unrelated and can be treated separately?

I wouldn't say that they are unrelated. The standard SR time dilation describes the difference between clocks on straight (but not parallel) world lines. The time dilation due to acceleration describes the difference between clocks on different hyperbolic world lines. The different shapes of the hyperbolic world lines are often caused by the clocks being attached to opposite ends of the same solid object.

Also, I consider the time dilation due to acceleration to be a part of SR. (It's of course relevant in GR too, but not just in GR).

The scenarios you're describing are comparing a clock on a straight world line with a clock on a hyperbolic world line. This is just a small complication added to the standard scenario, because all we have to do is to compare the straight world line to the tangents of the hyperbolic world line, and those are just a bunch of straight lines. So we can still use the standard time dilation formula.

SR time dilation produces a slowing of time as does GR time dilation. That is:
- if something is in greater gravity it will animate slower by our time.
- if something is under greater acceleration it will animate slower by our time.
- if something is going faster then it will animate slower by our time.
And this is accumulative isn't it?
ie. if you combine gravity with different relative velocity you sum their time dilation amounts.

This looks correct to me, but it's possible that I have overlooked something.

So if you are weightless in deep space and someone is accelerating away from you they will have a component of animating slower than you added due to GR rules

GR has nothing to do with it, and if you're only interested in how much slower the other clock ticks relative to yours "right now", you can completely ignore the acceleration. Just compare your world line to the tangent of the other clock's world line by using the standard time dilation formula.

and anyone near a planet will have a component of animating slower than you added due to GR rules;

If you both use rockets to stay at a constant distance from the planet, then yes.

Let's say someone is in a car and they accelerate along the road will this add a GR component of slow down to how fast we see them animating on top of that due to the gravity?

Only if you're in the car too but behind the driver (and I still consider this to be SR, not GR). If you're standing on the side of the road, you can compare your world line to the tangents of the car's world line, and that means that you're just going to be using the standard formula for time dilation.

 

[gonegahgah]

So in other words, if you are steady in deep space and something accelerates relative to you it will appear to animate slower; but if you are steady in gravity and something accelerates relative to you - on the same world line - it will not animate slower than you?

So in deep space we apply the accelerating rule but in gravity we discard the accelerating rule completely and instead substitute it with the world line rule?

 

[gonegahgah]

Also, if you consider something being accelerated in a rotating ring (like an Arthur C Clarke space-station) I think (if I remember correctly) that you will find that if you analyze the difference in time dilation between two different levels by Special Relativity given the velocity difference, you get the same result as if you analyze the difference based on the illusion of gravity seen in the frame of an observer in the ring.

That's quite interesting Jonathan. Will have to look at that idea.

 

[Fredrik]

So in other words, if you are steady in deep space and something accelerates relative to you it will appear to animate slower; but if you are steady in gravity and something accelerates relative to you - on the same world line - it will not animate slower than you?

If you're "steady in gravity" (which I interpret as being held at a constant altitude above a fixed point on the surface of a non-rotating spherical mass) and something accelerates relative to you, then it's not on the same world line. Don't forget that a world line is a curve in spacetime, not in space, so if two objects move on the same paths in space, but with different velocities, their world lines are different.

If the other clock is accelerating relative to you but staying at your altitude the whole time, it should speed up, not slow down, because its world line is closer to a geodesic (free fall) than yours. (A higher "sideways" velocity will bring it closer to being in orbit. An orbit is a free fall that misses the Earth but fails to escape it. The world line of an object in free fall is a geodesic. A geodesic is a curve of maximum proper time. So a clock with a higher "sideways" velocity should tick faster).

For example, suppose that you're standing on the ground of a non-rotating planet right next to a train track that's been built along a great circle around the planet, and that you have two synchronized clocks. You put one of them on the train and send it off to go around the planet. When it comes back it will be ahead of yours, unless of course its speed around the planet was much faster than the speed needed for a low altitude orbit. If it goes fast enough, it will behind yours when it gets back.

So in deep space we apply the accelerating rule but in gravity we discard the accelerating rule completely and instead substitute it with the world line rule?

Actually there's only one rule: What a clock measures is the integral of $\sqrt{-g_{\mu\nu}dx^\mu dx^\nu}$ along the curve in spacetime that represents its motion. In other words, it measures the proper time of its world line. (In an inertial frame in 1+1-dimensional SR, the expression above can be simplified to $\sqrt{dt^2-dx^2}$).

The "standard time dilation in SR", "time dilation due to acceleration in flat spacetime" and "time dilation due to curvature of spacetime" are all special cases. The standard time dilation formula compares the proper times of two straight world lines. What I've been calling "time dilation due to acceleration" compares the proper times of two hyperbolic world lines associated with opposite ends of a solid object. Gravitational time dilation is also usually about comparing the world lines on opposite ends of a solid object (e.g. two different floors of the same building), but in curved spacetime.

In complicated situations, it's pointless to try to use the rules that apply to special cases separately. You're just going to have to calculate the proper time along each clock's world line.

 

[MeJennifer]

For example, suppose that you're standing on the ground of a non-rotating planet right next to a train track that's been built along a great circle around the planet, and that you have two synchronized clocks. You put one of them on the train and send it off to go around the planet. When it comes back it will be ahead of yours, unless of course its speed around the planet was much faster than the speed needed for a low altitude orbit. If it goes fast enough, it will behind yours when it gets back.

I question the correctness of the phrase I marked in red.

 

[Austin0]

I haven't been following this thread, but locally the effect of acceleration and the equivalent gravitational potential gradient on clocks is identical, so for example a clock at the front of an accelerating spaceship appears to run slightly faster than one at the back. This can be calculated by Special Relativity.
___________________________________________________________________-

Hi I am a little confused. I had thought that the desynchronization of clocks was due to clocks running slightly faster in the BACK due to acceleration?. Is this incorrect?

Thanks

 

[JesseM]

Thanks Jesse. That was what you said in the other thread. So are you saying that mgb was incorrect about acceleration being equivalent to gravity; specifically with respect to the effect upon time dilation. Are they non-equivalent in this specific matter?

According to the equivalence principle, if you're in a room in a very small region of spacetime that's sitting in a gravitational field, physics will look exactly the same as it does in an accelerating room in empty space. If both the floor and the ceiling of the room are at a fixed height in the gravitational field, I believe this would be equivalent to the floor and the ceiling in empty space undergoing Born rigid acceleration, in which the distance from one to the other in each one's instantaneous inertial rest frame at any given moment remains constant from one moment to another. This means that in a given inertial frame, it will appear that the floor and ceiling are accelerating at different rates (each is experiencing a constant G-force, but the G-force experienced by the floor is slightly larger)--their worldlines will look like the hyperbolae shown in the second diagram on this page (and note the paragraph immediately below that diagram which says 'We can imagine a flotilla of spaceships, each remaining at a fixed value of s by accelerating at 1/s. In principle, these ships could be physically connected together by ladders, allowing passengers to move between them. Although each ship would have a different proper acceleration, the spacing between them would remain constant as far as each of them was concerned.'). Because the velocity of the floor and ceiling are different at any given moment in this inertial frame, the time dilation factor of clocks at the floor and ceiling will be different, and this will cause the one on the floor to run slower by the same amount as the clock on the floor in the room that's at a fixed height in a gravitational field (this is my understanding anyway, I haven't actually done the math, but see pervect's post #75 on this thread).

 

[Alfi]

To be more precise - a line joining the position of all the photons at a point in time would form a spiral. It's important to remember that individual photons travel in a straight line (as robbie said)

does this imply that there is a limit to the rotational speed of the laser shooting it's photons one at a time based on the frequency. Even with our fastest ejection of individual photons,
If the laser was spun fast enough, would there be blank spots on the screen light years away?
( photons would hit further than one wavelength apart)

 

[Fredrik]

Hi I am a little confused. I had thought that the desynchronization of clocks was due to clocks running slightly faster in the BACK due to acceleration?. Is this incorrect?

Thanks

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.

 

[mgb_phys]

does this imply that there is a limit to the rotational speed of the laser shooting it's photons one at a time based on the frequency. Even with our fastest ejection of individual photons,
If the laser was spun fast enough, would there be blank spots on the screen light years away?
( photons would hit further than one wavelength apart)

There are always gaps since the photons come out individually, it's just that there are so many of them that it appears as a constant beam. As you sweep the laser over larger distances the gaps between them become larger
You can easily work out how many photons a laser emits per second from the power (the number of J/s ) and the energy of a photon of that wavelenght ( E = hc/wavelength)

 

[gonegahgah]

mgb, I wasn't aware there was a relationship between wavelength and the number of photons. Don't get me wrong; I think this whole sweeping laser notion in no way represents faster than speed motion; but I was just curious about the inference I got from what you posted about a relationship. Are you saying one wavelength is equivalent to one photon? Could you explain this further for me, please?

 

[JesseM]

mgb, I wasn't aware there was a relationship between wavelength and the number of photons.

A single photon can have any wavelength, but there is a relationship between a photon's energy and its wavelength (for an individual photon, relation[/url] says E=hf where f is the frequency and h is Planck's constant, and f is related to the wavelength $$\lambda$$ by f = c/$$\lambda$$ so E = hc/$$\lambda$$), so if you know the power of a light wave hitting a surface (energy/time) and you know its wavelength, that tells you the number of photons hitting the surface per unit time.

 

[gonegahgah]

So ergo one frequency (& wavelength) is implied by that to correspond to a single photon.

Someone is supposed to have taken a photo of light.
The photo of a 'photon' that I saw comprised multiple wavelengths in a single 'photon' not just a single frequency (or wavelength).

 

[JesseM]

So ergo one frequency (& wavelength) is implied by that to correspond to a single photon.

I don't understand what you mean by this, or how it's supposed to follow from what I said. A single photon can travel a distance that corresponds to multiple wavelengths, do you disagree? Likewise, if multiple photons hit a surface, the distance between the positions where they hit can be much larger or smaller than each photon's wavelength.

 

[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.
___________________________________________________________________________
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.