Hi Guys, I just wanted to post a very simple question that I want to be absolutely clear about.

It was prompted by activities in another thread and involves how light behaves when projected at 90 degrees to the vector of the source.
The first diagram shows our light source at rest with the laser light beams drawing a straight line.

http://www.paygency.com/Diagrams/AV1.jpg

The second diagram shows our light source at velocity, thus the beams of laser light touch the outer ring behind the source.

http://www.paygency.com/Diagrams/AV11.jpg

I do realise that this is not a very mathematical approach but what I am trying to clarify is that light gains no momentum by the vector velocity of the source, when projected 90 degrees to that vector.
I think it is fairly self explanitory but if the question needs to be clarified let me know....

Does light gain momentum from the velocity of it's source when projected 90 degrees to that sources vector?

Edit: the diagrams are showing an exageration of the effect for the purposes of clarity.

Your second image is wrong. The laser strikes the same point of the circle regardless of its velocity.

Your second image is wrong. The laser strikes the same point of the circle regardless of its velocity.

Unfortunately, not according to E. Skyler that has made such tests for the past (6) years. This must (and should be) investigated to either verify or falsify his results. It is a damn important issue but as yet nobody seems interested.

Wonder why. :bugeye:

*********************** Extract ***********************
http://www.physicsnews1.com/article_12.html

(11) Ryan Skyler and I performed our first Earth Motion Detector test in the month of May near Seattle, Washington, USA. The laser was directed due north over a horizontal distance of 150 feet (45.72 m) to a paper target. During a 72 hour period, the red laser dot scribed three counter-clockwise rotations about the target's center. Each such rotation was of an elliptical shape 3/8" in width by 1/4" in height. A time-lapse film was made of two of the rotations before the camera expired. Since the test was performed outdoors, I think the early-morning dew proved too much for the camera's electronics. Over a distance of 150', a movement of 3/16" to one side of the target's center is not a lot of movement in a 12 hour period. Such a shift in position would be easy to miss if one was not specifically expecting it to appear. If the distance were extended to a mile, the ellipse would be drawn of a more obvious size at 13.2" wide by 8.8" high. But power and focusing difficulties along with laser jitters due to refraction of the energy beam in the intervening thermal currents present their own problems. A clear improvement would be to run the test the entire distance through a vacuum.


(12) What did I make of these results? To find the speed Earth was traveling at right angle to the horizontal direction of our test, I considering that the target was successful in moving 3/16" to one side in the time it took for the laser energy-wave beam to travel the 150' distance. In order to find Earth's speed in this direction, I first needed to determine the time it took for the non-sideways moving laser beam to travel the 150' distance to the sideways-moving target.

Problem 1: Determine the time light takes to travel 150' through air.

Convert light-speed
in air to ft/sec = 185,944 mi/sec * 5280 ft/mi
= 981,784,320 ft/sec

Time for light to travel 150 ft = 150 ft / 981,784,320 ft/sec
= 0.00,000,015,3278,3 sec (1.527830470953131539114415679403e-7 sec)
(Windows calculator - scientific mode)

Problem 2: Convert 3/16 in to decimal ft.

Decimal ft = 3 / 16 in / 12 in/ft
= 0.1875 in / 12 in/ft
= 0.01,562,5 ft

(13) Now that I have the distance in feet that Earth traveled sideways in the time it took for the laser beam to reach the target, let us determine Earth's sideways speed in both ft/sec and miles/sec to see overall if the resulting answer makes sense.

Problem 3: Determine Earth's speed as measured in the direction 90 degrees sideways to the horizontal direction of the laser beam.

Earth speed = distance / time
= 0.01,562,5 ft / 0.00,000,015,3278,3 sec
= 102,269.2 ft/sec

Problem 4: Convert speed in ft/sec to mi/sec to mi/hr.

Earth speed = ft/sec / ft/mi
= 102,269.2 ft/sec / 5280 ft/mi
= 19.36,917 mi/sec (SI units = 31.17 km/sec)

Earth speed = mi/sec * sec/hr
= 19.36,917 mi/sec * 3600 sec/hr
= 69,729 mi/hr (SI units = 112,218.17 km/hr)

(14) How does 69,729 mi/hr or 19.37 mi/sec compare with Earth's orbital speed about the Sun? Let's see...

Problem 5: Calculate Earth's orbital speed about the Sun

Earth orbital distance per year = orbital radius * 2 * Pi
= 92,955,820 mi * 2 * 3.14159
= 584,058,642.44 mi/yr (SI units = 939,953,152 km/yr)

Hours per year = day/yr * hr/day
= 365.25 day/yr * 24 hr/day
= 8,766 hr/yr

Seconds per year = hr/yr * sec/hr
= 8,766 hr/yr * 3600 sec/hr
= 31,557,600 sec/yr

Earth's orbital speed = orbital distance (mi/yr) / orbital time (hr/yr)
= 584,058,642.44 mi/yr / 8,766 hr/yr
= 66,627.73 mi/hr / 3600 sec/hr
= 18.5 mi/sec

(15) As you can see, this Earth-Motion Detector has indicated a velocity within 5% of Earth's orbital velocity about the Sun. While 69,729 mi/hr is an interesting value, there are many factors that need to taken into consideration before such a figure has any chance of being accepted as Earth's absolute speed through space. First consider that since this test is horizontal on a rotating Earth, the possibility exists that never at any time is the laser beam traveling at right angle or directly toward or directly away from Earth's absolute motion through space. If these directions are not reached then the maximum range of indication will not be reached resulting in a somewhat lower speed indication than actually exists.

(16) Also worth considering is Earth's involvement in its orbit of the Sun plus Earth's involvement in its orbit of the galaxy known as the "Milky Way". Since the Sun has a certain speed of absolute motion through space, this speed will add to or subtraction from Earth's speed provided the Sun's direction of absolute motion is not in line with Earth's axis of orbit of the Sun. The same situation is in effect as this solar system orbits the galactic core. Any core speed will add to or subtraction from the Sun speed again provided that the galactic core's direction of absolute motion is not in line with the Sun's axis of orbit of that core.

(17) There are a lot of motions out there to consider. It is quite possible for the galactic core to have one direction of motion with the Sun in possession of a motion in the opposite direction and finally Earth with yet a third direction of motion. My point is that initially the results of this test may not seem sensible. But over time the validity of the results produced by this device will become clear.

(18) At this point it is regretful that the fixed test is so slow, taking 24 hours to give a complete reading. Things can be speeded up if a large rotateable object is handy. I imagine a laser Earth-Motion Detector performing on the deck of a ferry boat that is being rotated by a tug, or on an aircraft carrier for a longer range. Another way to get more immediate results from a fixed-to-Earth test is to amplify the laser's motion as it draws an oval path on the target. In place of a terminating target, a concave mirror can be used to effectively amplify the red dot's oval motion. After reflection, the red laser light-wave beam will pass back through the concave mirror's focal point and on to a large target with a hole at its center. The incoming laser light-wave beam passes through the hole in the target, reflects on the concave mirror and then returns to illuminate the target's back side. Here the laser's motion is so great you can actually watch it move along the target's surface like watching the movement of the minute hand on a large outdoor clock. Of course it still takes 24 hours to complete the elliptical path. But at least there is no delay in viewing its movement on the target.
************************************************** *****

Your second image is wrong. The laser strikes the same point of the circle regardless of its velocity.

Thanks Janus,
So I can state quite clearly that emitted light does gain orthogonal momentum from the velocity of the light source. according to conventional scientific thought?

Your second image is wrong. The laser strikes the same point of the circle regardless of its velocity.
Janus58,
Let us take this one step at a time.

When the frame is at rest, Fig 1 of theQQ drawing the emitted photons behave as shown.

Now ask yourself he question: When at rest the system emits photns as shown. Befoe the photons strike the "circle" the ship moves in the direction indicated. Just so we remain on the same page, the frame moves after the photons are in flight.

Were will the photons and walls of the circle meet?

I say the frame will pull away from the photons as the motion of the photons, at this point is certainly independent ofhe motion of the frame.

Your response is ambiguous in the sense that you have not distinguished the result stemming from:
1. SRT or
2. The frame imparts a momentum component to the emited photons.

If we acept your answer to QQ then the frame must drag the photons along and the mirror will have no affect on the light. If the photons have an imposed forward momentum component then the velocity vector for the light will increase. The velocity component moving to the right will be C. The velocity component of the light in the direction of motion of the frame will be v. The total photon light vector then is V = (C^2 + v^2)^1/2 > C.

Geistkiesel

(16) Also worth considering is Earth's involvement in its orbit of the Sun plus Earth's involvement in its orbit of the galaxy known as the "Milky Way". Since the Sun has a certain speed of absolute motion through space, this speed will add to or subtract from Earth's speed provided the Sun's direction of absolute motion is not in line with Earth's axis of orbit of the Sun. The same situation is in effect as this solar system orbits the galactic core. Any core speed will add to or subtraction from the Sun speed again provided that the galactic core's direction of absolute motion is not in line with the Sun's axis of orbit of that core.


Dayton Miller found that the sun is dragging the solar system along in the direction perpindicular to the plane of he solar system. The earh axis then points (generally) in the direction of motion of the sun motion. This would impose a 'constant" velocity component parallel to the laser beam motion of your experiment. Miller found a definte "ether drag" compnent along the earth spin axis. Hs final conclusion was the earth's motion is due south.

Here is link to Miller. (http://www.orgonelab.org/miller.htm)

If Miller is correct the earth's orbit motion around the sun is always nearly orthogonal to he sun motion.

An interesting result occurs if one considers the hree velocity components of the earth motion. The spin and earth-sun orbit velociy components are nearly paallel. The spin velocity of the earth at the equater is approx. .5km/sec, the earth-sun orbit velociy is appox. 30 km/sec, the sun motion (in direction of earth axis) is approximately 208 km/sec. Using these numbers and 1.5 x 10^8 km as the radius of the earth-sun orbit circle and we construct a triangle using the earth-sun distance and the distance of the sun motion in one year, (208 km/sec) x 31557600 sec = 6.56 x 10^9 km. The earth (planets) motion then is helical whee one complete helical segment is spread out over 6.56 x 10^9 km.

Consider earth spin and a recurring day and night a point on the earth surface. The pointdescribes a helix that stretches (208 km/sec)x (24x3600 sec) = 1.79 x 10^6 km with a (earth) radius 6378 km. I calculate that one days motion results in the helical segment due to spin that has a .2 degree angle from start to finish.

Does this ring any bells?
Geistkiesel

Thanks Janus,
So I can state quite clearly that emitted light does gain orthogonal momentum from the velocity of the light source. according to conventional scientific thought?
QQ, You conclude Janus58 correctly. When we calculate the total light speed due to all known velocity components we see:

(c^2 + v^2)^1/2 > C.

I suppose this gives new meaning to the term, "the speed of light is constant".

Geistkiesel

so when you fire a pulse of light at the moon reflector do you give yourself a lead time or do you wait until the mirror is sited as a straight line from the laser.

so when you fire a pulse of light at the moon reflector do you give yourself a lead time or do you wait until the mirror is sited as a straight line from the laser.
I would "Kentucky Windage" and calculate the simultaneous arrival of the target and bullet.
Geistkiesel

so when you fire a pulse of light at the moon reflector do you give yourself a lead time or do you wait until the mirror is sited as a straight line from the laser.
I made a hasty answer above which I am stuck with, I suppose. It would be technically difficult to determine all the parameters. The mean moon-earth distance being 384,000 km and the speed of light being 300000km means I have one second of flight time to correct for. The earth is in motion but lateral motion will not affect the trajectory, so it appears I can consider my platform stable and aim in the same way I aim at the moving ducks when the county fair opens up the shootng galleries.

the moon moves 1km/sec wrt the earth, so accuracy of sighting will be critical. I am not sure what the area of the laser beam would be after 300000km but I will assume the diameter of the laser projected on the moon surface at 30 meters (this turns out to be a very tight specification, see below, 7km dispersion of beam on the moon - I retract the 30 meter number.).

Now when see the moon it has moved one unit light distance one second before. If I aim at the center of the moon I would miss by a km. S I determine the direction of the moon motion and aim 1 km ahead.

I think it was MacM that reported a similar sitution. If I remember correctly MacM told about a situation that was contra to an "absolute motion" thread I was working on. If the position of the laser on the earth were such to be able to determine an instantanteous earth-moon velocity correlation the lateral motion of light question which is currently being scrutinized in his forum can be quantfied.

Provide the laser, when and where, QQ and I will give it a shot. I say if I get less than 1/2 km error that I "hit the target". You will have to provide the target also, I have already provie some of he speculation.

[Here is a description of MacDonald Obsevator (Univ. of Texas) (http://www.as.utexas.edu/astronomy/research/research.html)

I do recall that laser reflections off the moon surface are a common activity. Smewhere there is an apparatus for the shot already in place.
Some laser background I pared is quoted below. (http://www.brookscole.com/physics_d/templates/student_resources/003026961X_serway/optional/lasers.html)"We shall describe several other applications that should serve to illustrate the wide variety of laser uses. First, lasers are used in precision long-range distance measurement (range-finding). It has become important, for astronomical and geophysical purposes, to measure as precisely as possible the distance from various points on the surface of the earth to a point on the moon's surface. To facilitate this, the Apollo astronauts set up a compact array, a 0.5 m square of reflector prisms on the moon, which allows laser pulses directed from an earth station to be retro-reflected to the same station. (See Section 35.7 on prism reflectors.) Using the known speed of light and the measured round-trip travel time of a 1 ns pulse, one can determine the earth-moon distance, 380 000 km, to a precision of better than 10 cm. Such information would be useful, for example, in making more reliable earthquake predictions and for learning more about the motions of the earth-moon system. This technique requires a high-power pulsed laser for its success, since a sufficient burst of photons must return to a collecting telescope on earth and be detected. Variations of this method are also used to measure the distance to inaccessible points on the earth. "

Here is Set Up informationfor Experiments and some technical data. (http://www.lpi.usra.edu/expmoon/Apollo11/A11_Experiments_LRRR.html)
"Laser beams are used because they remain tightly focused for large distances. Nevertheless, there is enough dispersion of the beam that it is about 7 kilometers in diameter when it reaches the Moon and 20 kilometers in diameter when it returns to Earth. Because of this very weak signal, observations are made for several hours at a time. By averaging the signal for this period, the distance to the Moon can be measured to an accuracy of about 3 centimeters (the average distance from the Earth to the Moon is about 385,000 kilometers).

The Laser Ranging Retroreflector experiment has produced many important measurements. These include an improved knowledge of the Moon's orbit and the rate at which the Moon is receding from Earth (currently 3.8 centimeters per year) and of variations in the rotation of the Moon. These variations in rotation are related to the distribution of mass inside the Moon and imply the existence of a small core, with a radius of less than 350 kilometers, somewhat smaller than the limits imposed by the passive seismic and magnetometer experiments. These measurements have also improved our knowledge of changes of the Earth's rotation rate and the precession of its spin axis and have been used to test Einstein's theory of relativity.


--------------------------------------------------------------------------------
What else do we have to do?

Geistkiesel

MacM,
See my post below to QQ. MacDonald Observatory in West Texas has all the hardware and technical talent for a righteously pure a laser moon shot. QQ asked how would one aim the to hit a spot on the moon.
How do we get a few seconds of observatory time? Which I also asked of QQ.
Geistkiesel

Thanks Janus,
So I can state quite clearly that emitted light does gain orthogonal momentum from the velocity of the light source. according to conventional scientific thought?

Yes, with respect to the frame that the source is moving relative to. But this gain of momentum is not reflected as any change in the speed of the light but as a change of frequency.

MacM,
See my post below to QQ. MacDonald Observatory in West Texas has all the hardware and technical talent for a righteously pure a laser moon shot. QQ asked how would one aim the to hit a spot on the moon.
How do we get a few seconds of observatory time? Which I also asked of QQ.
Geistkiesel

Since I see no post below, I think you meant above. I looked at it and while I found nothing clearly spelled out I did note the following and was going to post this reply in any case:

Introduction
The McDonald Laser Ranging Station (MLRS) is a dedicated laser ranging station capable of measuring round trip light travel times to a constellation of artificial earth satellites and lunar retro-reflectors to a precision of about 1 centimeter and time of laser firing to about 35 picoseconds. Data from this station as well as 30-40 similar satellite-capable systems and one other regularly contributing lunar-capable system around the world are used for a variety of scientific pursuits including study of the earth's gravitational field, plate tectonics, earth's orientation in space, high precision time transfer, relativity, lunar and solar system dynamics, and providing high precision orbits for GPS and ocean top mapping missions.

MLRS History

1 - If the target remained in the line of sight one would not need any ranging.

2 - The ranging specified of 35 pico-seconds calculates to a positional change of 0.0105m over the distance to the moon at the speed of light.

3 - I would not expect to see rotational velocity transferred to alight beam. That would not makes sense but rotation should (if current physics is correct) induce a lateral motion of the beam along the tangency to the rotation.

This correlates to the specified precision of hitting the target. That however
doesn't really make a lot of sense since it neither seems to consider angular velocity of rotation relative between earth and the moon nor any affect due to earth's orbit velocity.

I think this subject merits further exploration. We may find our answer in this area of current technology but we will need to do some digging since I would not expect them to openly point to any conflict with relativity in their work, just as they improperly credit absolute gamma calculations with supporting SRT in GPS.

As a staarting point I see earth's rotation at a surface velocity of 463.8m/sec and light travel time to the moon at 385,000km as having the tangent induced motion of 595.21 m after 1.28 seconds of travel time.

That is if lateral light beam motion were infact to be induced by rotation.

I don't see such correction clearly mentioned? There is also the added complexiety here that the moon's orbit velocity is about 1,000m/s.

Since it co-rotates with the earth I get a net 684m shift of the target during the light beam transit. That is based generally on some rounded numbers and not precise.

If there is no lateral motion transfer due to earth's rotation and the laser beam moves as in line of sight then the target should move almost 1,300m.

Without lateral motion from earth's orbit the target moves 37,000m.!

Unfortunately this seems to run counter to E. Skyler claims. I do not see any such large adjustments in lunar targeting. But then that would be a function of where in the cycle the moon is targeted. It may not be targeted during its orthogonal position to the earth's orbit.

Light gains no momentum because it has no mass, its am atter of change in frequency of the wave.

Light gains no momentum because it has no mass, its am atter of change in frequency of the wave.
Can you describe how light is carried along with the emitting frame motion by a change in frequency and not velocity, or momentum?

Your claim that momentum cannot be gained because light is massless does not explain away experimental evidence that light shining on objects delivers momenum change in the target object. Are you saying that light has no momentum to exhange with the universe, but when light strikes an object the frequency change in the light is interpreted by the tarhget mass as momentum transfer?

Are there any experimental results that support the "frequency change" of light when striking a mirror oriented parallel with the motion of the light source induces a change in direction of the light? Does each reflection of the light frm the mirror surface induce a frequency change in the light? For each reflection?

Does the light frequency increase or decrease and how does the frequency variation induce a change of direction in the light?

Your post was published with no supporting materials or argument. Are you a universal source of information and knowledge, or do you wear the cloak of scientist? You, a confessed atheist, whatever that is supposed to nmean as a personal description ahve 127 posts in Sciforums, yet I have not seen any post other than thios one that are addressed to the Physics and Math group. Where do you publish your materials?

Geistkiesel

Light gains no momentum because it has no mass, its am atter of change in frequency of the wave.

Addendum to previous post replying to Odin'ism:

Is the Homepage you claimed when you registered here at Sciforums your homepage or one you adopted? Your homepage is a silly collection if usless junk and more enlightening it claims to be no more than this.

Fnally, Odin, described in the literature as a God is your chosen handle for Forum posting is contradixtory to a claim of atheism is it not?

So which is your religion: adoration of a preferred god or atheism, or ambiguity?

Geistkiesel

Geistkiesel:
Go easy on the guy! He made a silly statement with out researching it first.

Odin'ism:
Research "light sails" and how they work (solar particles are one source of momentum transfer(not really necessary), and the other is the momentum transferred by light itself)

Janus,
Yes, with respect to the frame that the source is moving relative to. But this gain of momentum is not reflected as any change in the speed of the light but as a change of frequency.
I am a litle confuse Janus.
The speed of light has never been in question and I wonder the relevance of this post.
I am attempting to define the othogonal changes imparted by the velocity of the source which I already assume do not effect the speed of light but only where a laser beam may go, it's direction.

The question behind this is obviously a very simple one:

"if orthagonal momentum creates change in the emmitted light in any manner what so ever then an ability to find an absolute rest velocity is possible." It is also possible that if the changes can be measured then absolute velocity relative to absolute rest can be measured. In other words the actual velocity of the craft can be determined by using the changes to teh beam of light due to velocity.

Now also I keep in mind that this is not a new question and I am confident that it has been already answered and dismissed by conventional thought.

So I ask again to seek clarity:

What effects occur on a beam of laser light due to the orthogonal velocity of it's source and can these effects be used to determine when the source is at absolute rest, or zero velocity relative to the universe in general?

Janus,

I am a litle confuse Janus.
The speed of light has never been in question and I wonder the relevance of this post.
I am attempting to define the othogonal changes imparted by the velocity of the source which I already assume do not effect the speed of light but only where a laser beam may go, it's direction.

The question behind this is obviously a very simple one:

"if orthagonal momentum creates change in the emmitted light in any manner what so ever then an ability to find an absolute rest velocity is possible." It is also possible that if the changes can be measured then absolute velocity relative to absolute rest can be measured. In other words the actual velocity of the craft can be determined by using the changes to teh beam of light due to velocity.



You've got this exactly backwards. If the light behaved as you showed it in your diagrams, then you would have a way of testing for absolute rest. Because then the momemtum of the light would change in respect to the source. I could be alongside the source and check to see where the laser hit. if it hit anywhere except where it hits as shown in the first diagram, I would know my absolute velocity by measuring where it hit. On the other hand, a person who remains in that state of absolute rest would measure no change in the the momentum of the light, regardless of whether the source was moving or not.

In the case where the light always strikes the same point, and I, alongside the source, could never determine my absolute velocity because the point at which the light hits does not change, nor does the light's momentum change with respect to me.

To an outside observer the lights momentum would change depending on whether the source was moving with respect to him or not, but he cannot use this as evidence of a state of absolute rest. The light would show the same change of momentum with respect to this observer even if you considered the light source as stationary and the observer as moving. Therefore, he can not determine whether it is him, the source, or both that are moving. In fact, the distinction becomes meaningless. All he can say is that he and the source have a relative velocity with respect to each other.

QQ wrote:

and can these effects be used to determine when the source is at absolute rest, or zero velocity relative to the universe in general?

No. Many experiments (MM et al) have demonstrated that there is no priveleged frame of reference for the universe.

janus and super L thanks again for your posts. I was aware of what you were probably going to say. And i understand exactly that according to SRT lights direction can not determine any velocity other than relative velocity.

I just posted the two diagrams to be absolutely clear on the SRT position by posting what I thought would be totally wrong or backwards as Janus has described.
I am confident that the underlying question was fairly obvious. And I might add a farely common one at that.

the answers yo have given lead me to ask the next question which i know has no easy answer and that is if what you say is correct how can light be deemed as anything but relatively invariant to the observer.

Meaning that light is invariant to all observeres but relatively so.

The reasoning being that if the same light behaves differently for each observer then how can the claim of invariance with out the qualification of relative unvariance be maintained?
Now I know there is an answer out there that still supports SRT and also that it aint that simple to provide it.

QQ:
Now I know there is an answer out there that still supports SRT and also that it aint that simple to provide it.

Amen brother. I know you can show it with the math (not to me! :confused: ) but I still have no way to visualize it from a physical point of view.

janus and super L thanks again for your posts. I was aware of what you were probably going to say. And i understand exactly that according to SRT lights direction can not determine any velocity other than relative velocity.

I just posted the two diagrams to be absolutely clear on the SRT position by posting what I thought would be totally wrong or backwards as Janus has described.
I am confident that the underlying question was fairly obvious. And I might add a farely common one at that.

the answers yo have given lead me to ask the next question which i know has no easy answer and that is if what you say is correct how can light be deemed as anything but relatively invariant to the observer.

Meaning that light is invariant to all observeres but relatively so.

The reasoning being that if the same light behaves differently for each observer then how can the claim of invariance with out the qualification of relative unvariance be maintained?
Now I know there is an answer out there that still supports SRT and also that it aint that simple to provide it.
Right you are QQ, but have you seen anything except the "photon dragging" explanation of how the light gets carried along with the frame? Without dragging the SRT has nothing. Without photon dragging SRT gets discarded in the trash bin of interesing but useful physical theories.
]
Geistkiesel

Since I see no post below, I think you meant above. I looked at it and while I found nothing clearly spelled out I did note the following and was going to post this reply in any case:



1 - If the target remained in the line of sight one would not need any ranging.

2 - The ranging specified of 35 pico-seconds calculates to a positional change of 0.0105m over the distance to the moon at the speed of light.

3 - I would not expect to see rotational velocity transferred to alight beam. That would not makes sense but rotation should (if current physics is correct) induce a lateral motion of the beam along the tangency to the rotation.

This correlates to the specified precision of hitting the target. That however
doesn't really make a lot of sense since it neither seems to consider angular velocity of rotation relative between earth and the moon nor any affect due to earth's orbit velocity.

I think this subject merits further exploration. We may find our answer in this area of current technology but we will need to do some digging since I would not expect them to openly point to any conflict with relativity in their work, just as they improperly credit absolute gamma calculations with supporting SRT in GPS.

As a staarting point I see earth's rotation at a surface velocity of 463.8m/sec and light travel time to the moon at 385,000km as having the tangent induced motion of 595.21 m after 1.28 seconds of travel time.

That is if lateral light beam motion were infact to be induced by rotation.

I don't see such correction clearly mentioned? There is also the added complexiety here that the moon's orbit velocity is about 1,000m/s.

Since it co-rotates with the earth I get a net 684m shift of the target during the light beam transit. That is based generally on some rounded numbers and not precise.

If there is no lateral motion transfer due to earth's rotation and the laser beam moves as in line of sight then the target should move almost 1,300m.

Without lateral motion from earth's orbit the target moves 37,000m.!

Unfortunately this seems to run counter to E. Skyler claims. I do not see any such large adjustments in lunar targeting. But then that would be a function of where in the cycle the moon is targeted. It may not be targeted during its orthogonal position to the earth's orbit.
How is the laser beam spread to 5 - 7 km diameter on the moon surafce handled in regard to accuracy that you described? The .0105meters accuracy you quoted isn't the lateral position of the moon-earth system is it?

With a 1 second flight time, the laser beam spreading to 5 km diameter, the moon moving approximately 1 km during beam flight time all add up to me as needing an explanation. You said,

"2 - The ranging specified of 35 pico-seconds calculates to a positional change of 0.0105m over the distance to the moon at the speed of light. "

How were these numbers determined?

What have I missed?

Geistkiesel

Geistkiesel:
Go easy on the guy! He made a silly statement with out researching it first.

Odin'ism:
Research "light sails" and how they work (solar particles are one source of momentum transfer(not really necessary), and the other is the momentum transferred by light itself)

Careful here SL. The latest seems to be some question if light sails will actually work based on how radiometers work.

MacM,

I know how radiometers work. Are you saying that a light sail might not work in a perfect vacuum, with light pressure alone?

How is the laser beam spread to 5 - 7 km diameter on the moon surafce handled in regard to accuracy that you described? The .0105meters accuracy you quoted isn't the lateral position of the moon-earth system is it?

With a 1 second flight time, the laser beam spreading to 5 km diameter, the moon moving approximately 1 km during beam flight time all add up to me as needing an explanation. You said,

"2 - The ranging specified of 35 pico-seconds calculates to a positional change of 0.0105m over the distance to the moon at the speed of light. "

How were these numbers determined?

What have I missed?

Geistkiesel

Not sure yet. Haven't had time to do more research but something doesn't add up either in the ranging information and/or E. Skylers test claims.

MacM,

I know how radiometers work. Are you saying that a light sail might not work in a perfect vacuum, with light pressure alone?

That is what some scientist are claiming. Remember the radiometer rotates opposite the direction momentum would have it rotate. :confused:

MacM,

Please clarify your last post if you could.

MacM,

Please clarify your last post if you could.

I may have to back peddle on the radiometer example a bit. The following link seems to suggest the reverse operation of these devices is a function of insufficient vacuum.

I have never seen it claimed before that they will rotate in the momentum direction. I have always seen it claimed and showing it operating from the black surface and not propelled by the reflective surface.

Not sure how accurate this piece is but I'll look further.

Here is a little better description:

********** Extract ***************
http://math.ucr.edu/home/baez/physics/General/LightMill/light-mill.html

But there is a problem with this explanation. Light falling on the black side should be absorbed, while light falling on the silver side of the vanes should be reflected. The net result is that there is twice as much radiation pressure on the metal side as on the black. In that case the mill is turning the wrong way.
******************************************



In any case I will look for the information on solar sails.

MacM:

A simple vane radiometer works this way:

There is a good but not perfect vacuum in the bulb. The black side absorbs more light radiation and therefore heats more than the white side. Molecules of rarefied air that come into contact with the black surface are heated more than ones that contact the white side and therefore rebound from the black surface with more force, pushing the black side forward.

In a perfect vacuum, enough light pressure could cause the vanes to rotate, the direction depending on which side you shone the light. I have seen references to this somewhere.

MacM:

A simple vane radiometer works this way:

There is a good but not perfect vacuum in the bulb. The black side absorbs more light radiation and therefore heats more than the white side. Molecules of rarefied air that come into contact with the black surface are heated more than ones that contact the white side and therefore rebound from the black surface with more force, pushing the black side forward.

In a perfect vacuum, enough light pressure could cause the vanes to rotate, the direction depending on which side you shone the light. I have seen references to this somewhere.


I'm inclined agree. As I just posted the links state how they turn backwards but then give the less than perfect vacuum resolution and say it should turn the other way. I think that is where the sceintific debate is. Some are not convienced it will happen, even in the depths of space.

Time will tell.