Curved Light Beams

No. I think not curved. The light beam consists of a specific number of photons. Each photon will travel in a straight line from its point of origin. Since the origin rotates, the "stream" of photons will be emmited from a sequentially changing position much like water from a hose that is swept back and forth.

Also, if you throw a rock from a rotating platform the rock takes a straight line trajectory (except for the effects of gravity) from the moment it leaves your fingers.

I used the word "beam" whereas you have used the word "stream". Are we not describing the same thing? If a stream of photons is curved much as a stream of water from a hose then why not a beam of photons.

What is a beam? It is a number of photons in motion. The individual photons are all moving at the same velocity, but have a different trajectory relative to their origin in the case of the rotating emitter.

So, beam and stream are being used as the same thing in my explanation. The important thing is that each photon moves in a straight line trajectory. It doesn't mean every photon has the same trajectory in your experiment.

exactly that but light does curve because with a mass say of a black hole will pull the photons due to gravitational pull acurve the light from its original path.

Here is another analogy. Imagine a rifle bolted to a large disk that rotates. When the trigger is pulled the bullet begins to move outward. If the rifle was in a static position the bullet simply spins out of the barrel and goes on its merry way. However, this rifle is moving in a radial direction about an axis. At the point where the bullet is fired at the breach of the barrel, there is a radial force imparted on the bullet by the barrel. Without the barrel the bullet would simply move in a perfectly straight line (ignoring gravity and wind). However, since the barrel is moving it will impart an acceleration laterally on the bullet. While the angular velocity of the barrel is constant, the radial velocity is not. The velocity of the breach of the barrel is less than the muzzle. This imparts acceleration laterally on the bullet all the way down the barrel.

When the bullet leaves the barrel that acceleration is zero and the bullet resumes a straight line trajectory from that point without a radial component. This is different then if the same rifle was mounted to a train rolling down a straight and level track at constant velocity. In the latter example the bullet does have two vector components to its trajectory.

In both the radial and linear examples the free flight trajectory (after the bullet leaves the barrel) is always a straight line. The same applies to light. In order for a photon to have a circular trajectory there must be some force applied to the photon (i.e., gravitational acceleration). Once the photon is emitted from the rotating disk there is no angular acceleration to alter the photon's course, which is a straight line from the point of origin at the time of emission.

Yet another analogy is if you have a large rotating disk and roll a marble coated with paint outward from the disk's center. The marble will roll in a straight path relative to the fixed ground, but will paint an arc on the disk (ignoring friction).

The new trajectory at the moment the bullet leaves the barrel will be the vectorial resultant of the 2 velocities, the linear and the angular. and it will form an arc which is the resultant of the ratio between the 2 velocities.

Look at a rotary water sprinkler.

I was doing some math,: when the angular velocity equal the linear velocity, the bullet will not leave the barrel. Something ain't right here. I'll try some more and let you know.

Wangito

In ordinary physics, I think of the term radial velocity to mean a velocity with its direction along a radius of the circle in question. (In astronomy, however, the term is often used to mean "line of sight".) I think of the velocity of a particle at the periphery of a disc as an instantaneous tangential velocity or instantaneous circumferential velocity.

I also think of incident velocity as the velocity of something "coming in" rather than something going out.

So between the question, and Hero's responses, I'm a little confused. When Hero says, "there is a radial force imparted on the bullet by the barrel" I think he means there is a tangential force imparted. A radial force, to me, would mean a force along the length of the barrel – the force imparted by the BANG.

I think the rifle analogy is pretty good. Suppose the muzzle velocity is 100 m/s. Suppose the disc on which the rifle is mounted is 2 m radius, and the rifle muzzle is located at the circumference of that disc. We adjust the disc speed to give a surface speed at the circumference of 100 m/s. As the bullet leaves the barrel, its velocity in the tangential direction will have been accelerated to 100 m/s. Its velocity in the radial direction will be 100 m/s. Thus, its velocity over the ground will be the vector sum of those velocities: 141.4 m/s at 45 degrees to the rifle barrel at the instant of barrel exit.

Now replace the rifle with the pointer. Speed up the disc to a surface speed 99% of c. Ignore dilation. Therefore, a photon will be moving at roughly 1.4 times the speed of light, in a straight line. (Of course we can't realistically ignore dilation.)

However, the rifle analogy would be closer if the rifle were a machine gun that could fire, let's say, 100 bullets as the disc rotates through 30 degrees. Then, if we snap a picture from far above, (using a strobe light) the bullets would be arrayed in a curve (despite the fact that, individually, they are traveling in 100 separate straight lines). Is the apparent curve of our bullets (photons) the equivalent of a curved "light beam"? Suppose the pointer equipped disc were shrouded in a slight fog. Would you see the beam apparently curving?

Recalculate the situation with the pointer, but with the disc surface speed at .5 c. Include the effects of dilation.

"So between the question, and Hero's responses, I'm a little confused. When Hero says, "there is a radial force imparted on the bullet by the barrel" I think he means there is a tangential force imparted. A radial force, to me, would mean a force along the length of the barrel – the force imparted by the BANG."

Yes, you are correct. The radial component I spoke of is that of the disk's rotation about its axis which is tangental to the bore of the barrel. Good catch.

"Millisecond Pulsars" are graphic examples of this effect. Here you have a "hot spot" emitting xrays from the surface of a rapidly-spinning neutron star. Each xray photon travels outward from the star in a straight line, but when viewed from a point on the spin axis high above the pulsar (high >> 1 ly) the overall shape of the aggregate is a tightly-wound spiral. But when looking toward the pulsar in the plane of the spiral, all one sees is a blinking xray "beacon" whose photons have travelled in a straight line (geodesic to purists) from the star to the observer.

-e

The beam is straight in four-dimensional spacetime and curved in three dimensional space.

Not necessarily. The beam can span a region of locally-flat spacetime in which the beam's geodesic is straight in both cases.

-e

Those are some interesting threads and analogies but did the original question get answered?

It seems to me the bullet, having mass plus a radial vector imparted by the barrel, would behave differently than a massless photon emitted from a crystal.

I agree, at least part way. This is subject where I feel like I am on a see-saw, tilting between readily observed stuff that makes sense, and other stuff I take mainly on faith.

For example, photons are massless, in the sense that they have no invariant mass, but act as if they have mass, given that they have momentum and can push things(like solar sails) around.

Consider this from Wikipedia:

To illustrate the significance of these formulae, the annihilation of a particle with its antiparticle must result in the creation of at least two photons for the following reason. In the center of mass frame, the colliding antiparticles have no net momentum, whereas a single photon always has momentum. Hence, conservation of momentum requires that at least two photons are created, with zero net momentum. The energy of the two photons — or, equivalently, their frequency — may be determined from conservation of four-momentum. The reverse process, pair production, is the dominant mechanism by which high-energy photons such as gamma rays lose energy while passing through matter.

I think that Europium's explanation of the pulsar is pretty close to the answer to the question. But the original poster has not been back, so it's hard to know what he had in mind. If he is thinking in more "easily observable world" terms, then the answer is that light travels in straight lines, especially if the source is moving at ordinary motor speeds. But that is only my opinion -- and this is a field in which I really only parrot the words of others.

In this scenario, a particle's mass plays no part in its trajectory. So long as no other force acts on the particle once it leaves the source, the particle's trajectory will be a straight line.

For the sake of simplicity and convenience, let's just assume all particles emitted by the source travel through a region of flat spacetime that is completely devoid of fields, errant electrons, protons, μ-mesons (all "-ons" for that matter), gas, dust, aliens, and curved-spacetime-invoking obfuscators visiting our Universe from the "O" (not "Q") dimension, and so forth.

Unlike with the bullets shot from a spinning gun, the vector sum of the radial and tangential velocities of the emitted photons cannot exceed c. However, the vector sum does determine the apparent angle at which the photons are emitted from the source. For an observer standing on the surface of neutron star, photons emitted vertically from the hot spot will not appear to have been emitted vertically to an outside observer (Earth), but at an angle to the surface normal. Consequently the "vertically-emitted" photons will appear to have been emitted before the hot spot was directly in line between the observer and the center of the star. However, once the photon leaves the emitter its trajectory outward is a straight line.

I mention this because some millisecond pulsars spin so fast that a point on the star's equator is moving at appreciable fraction of the speed of light. The current record is held by an approximately 20-mile-diameter pulsar spinning at an incredible 716 revolutions per second. This means that its equator is moving at about 24% of the speed of light.

By comparison, the fastest kitchen blender I know of is the Vita-Mix. Its blades spin at a paltry 37,000 RPM. Close but no cigar.

-e

So... if we power a merry-go-round of just 155 kilometers diameter with a Vita-Mix motor, the kids at the edge get to ride at light speed!

Ken wrote: "So... if we power a merry-go-round of just 155 kilometers diameter with a Vita-Mix motor, the kids at the edge get to ride at light speed!"

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What kids?

-e

I believe the beam will travel at a straight line, but it will look as if it's cured because the observer is stationary. Think of it this way, you have a child in a carousel, and you are looking at her. At the point she is traveling towards and away from you, her would be seen as cured. But as the point where she is perpendicular to you, you will be looking straight at her. Now, suppose you are in the carousel with her and looking straight at her, and she will be look straight at you because you and her are in the same traveling velocity.

MidniteFighter