Help on Standing Waves, related to the Silvertooth experiment

[lidal]

Imagine a laser light source pointing towards a mirror at 90 degrees , so that the light will reflect back on itself. A standing wave will be created. Assume that a standing wave detector (SWD) is placed in front of the mirror, between the laser and the mirror. The thickness of the SWD is much smaller than the wavelength of the light, for good resolution. It is obvious that the DC voltage detected by the SWD will vary as the SWD is moved along the path of light, relative to the mirror, because the SWD would move relative to the standing wave. But assume that the mirror and the standing wave sensor (SWD) are mounted on a common linear stage that can be moved relative to the laser source, so that the distance between the SWD and the mirror is always constant as the stage is moved relative to the laser. The question is: will the voltage at the SWD vary as the stage is moved relative to the laser, along the path of the light beam? The obvious answer is: NO. This is because the standing wave is locked to the mirror and moves with it, so that the SWD is always stationary relative to the standing wave. Hence there will be no variation of voltage at the SWD. The distance between the mirror ( or SWD) and the laser is irrelevant.

 

However, I saw an article in which the author reported an experiment in which he replicated the Silvertooth experiment. ‘ A Replication of the Silvertooth Experiment ‘ . The link to the article is:

 

http://www.conspiracyoflight.com/Silvertooth/Silvertooth.html

 

The experimenter (the author) concluded that the Silvertooth’s effect is due to temperature. My problem here is not whether Silvertooth’s experiment is valid or not.

 

Actually, this experiment is basically different from Silvertooth’s experiment because the standing wave is locked to the mirror. Yet the author reported that the voltage at the SWD varied as the linear stage was moved. This experiment even violates fundamental accepted knowledge of elementary physics. It implied that the voltage at the SWD will vary as the linear stage is moved relative to the laser, even if there is no relative motion between the mirror and the SWD.

 

Actually, in the experiment, the mirror is dithered and the change in phase of the AC voltage at the SWD is observed. Since the mirror and the SWD are co-moving, the phase of the SWD voltage should not vary as the stage is moved relative to the laser.

 

Am I right or where did I go wrong? Please help in pointing out the problem. Thank you for your time.

[Strange]

Can you provide a reference to this "Silvertooth Experiment"? All I could find was mentions on various crank science and conspiracy sites.

[lidal]

Can you provide a reference to this "Silvertooth Experiment"? All I could find was mentions on various crank science and conspiracy sites.

 

It is hard to find the kind of reference you mean, i.e. reference to it in mainstream journals or sites.

 

http://arxiv.org/pdf/1109.2681.pdf

http://adsabs.harvard.edu/abs/1990PhyEs...3..161W

http://arxiv.org/PS_cache/physics/pdf/0612/0612201v2.pdf

[swansont]

Imagine a laser light source pointing towards a mirror at 90 degrees , so that the light will reflect back on itself. A standing wave will be created. Assume that a standing wave detector (SWD) is placed in front of the mirror, between the laser and the mirror. The thickness of the SWD is much smaller than the wavelength of the light, for good resolution. It is obvious that the DC voltage detected by the SWD will vary as the SWD is moved along the path of light, relative to the mirror, because the SWD would move relative to the standing wave. But assume that the mirror and the standing wave sensor (SWD) are mounted on a common linear stage that can be moved relative to the laser source, so that the distance between the SWD and the mirror is always constant as the stage is moved relative to the laser. The question is: will the voltage at the SWD vary as the stage is moved relative to the laser, along the path of the light beam? The obvious answer is: NO. This is because the standing wave is locked to the mirror and moves with it, so that the SWD is always stationary relative to the standing wave. Hence there will be no variation of voltage at the SWD. The distance between the mirror ( or SWD) and the laser is irrelevant.

 

 

The obvious answer is yes, because if you move the source relative to the mirror, you lose the standing wave. It is not "locked" to the mirror. Move the source a half-wavelength away and the signal drops to zero.

[studiot]

You can see for yourself what happens with a standard construction laser.

 

Point the laser in some safe but convenient direction.

Move a target along in the beam.

The laser line will impact the target as a small disk of light.

This diameter of this disk waxes and wanes as you move the target along in the beam.

 

When using such as laser for levelling or alignment purposes you need to take this phenomenon into account.

 

[lidal]

 

The obvious answer is yes, because if you move the source relative to the mirror, you lose the standing wave. It is not "locked" to the mirror. Move the source a half-wavelength away and the signal drops to zero.

Thank you for the response.

But suppose that the distance of the SWD from the mirror is adjusted so that the incident and reflected lights interfere at the SWD with complete constructive interference, hence a maximum voltage detected by the SWD. This will happen if the distance between the SWD and the mirror is an integral multiple of the half wavelength?, say N half wavelengths. Next the stage is moved away from the laser by an arbitrary distance. In this position,again there will be complete constructive interference because the distance between the SWD and the mirror is still N half wavelengths.

 

The standing wave is locked to the mirror because its amplitude is always zero at the mirror surface.

Can you provide a reference to this "Silvertooth Experiment"? All I could find was mentions on various crank science and conspiracy sites.

http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html

http://www.infinite-energy.com/iemagazine/issue59/adissidentview.html

E. W. Silvertooth, special relativity, Nature 322 (August 14, 1986)

[swansont]

Thank you for the response.

But suppose that the distance of the SWD from the mirror is adjusted so that the incident and reflected lights interfere at the SWD with complete constructive interference, hence a maximum voltage detected by the SWD. This will happen if the distance between the SWD and the mirror is an integral multiple of the half wavelength?, say N half wavelengths. Next the stage is moved away from the laser by an arbitrary distance. In this position,again there will be complete constructive interference because the distance between the SWD and the mirror is still N half wavelengths.

 

The standing wave is locked to the mirror because its amplitude is always zero at the mirror surface.

The incoming wave's phase is dependent on the distance of the source from the mirror. If it's not an integral number of half-wavelengths, you will get some destructive interference, and that will reduce the signal.

[Strange]

This will happen if the distance between the SWD and the mirror is an integral multiple of the half wavelength?,

 

Surely, it is the distance between the source and the mirror that needs to meet this condition?

[lidal]

The incoming wave's phase is dependent on the distance of the source from the mirror. If it's not an integral number of half-wavelengths, you will get some destructive interference, and that will reduce the signal.

 

The ampltude of the voltage detected at the SWD depends only on the phase relationship between the incident and reflected waves. The incoming wave's phase is dependent on the distance of the SWD from the laser but the phase of the reflected wave also changes as the laser is moved, so that the phase relationship between the incident and reflected waves will stay the same.

[swansont]

 

The ampltude of the voltage detected at the SWD depends only on the phase relationship between the incident and reflected waves. The incoming wave's phase is dependent on the distance of the SWD from the laser but the phase of the reflected wave also changes as the laser is moved, so that the phase relationship between the incident and reflected waves will stay the same.

 

But if the incoming phase changes, having the same relationship means the amplitude changes. "Same phase as incident" ≠ "constant phase" If you move completely out of phase, there is no signal at all.

 

And reflection can give a phase change, so saying they have the same phase relationship is wrong. (180º phase change of reflected from a medium with a higher index, as in this case)

http://hyperphysics.phy-astr.gsu.edu/hbase/waves/standw.html

 

That's why you get destructive interference if the incident wave doesn't strike at a node.

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/interf.html

[lidal]

 

Surely, it is the distance between the source and the mirror that needs to meet this condition?

It is the distance between the detector (SWD) and the the mirror that should be an integral multiple of half wavelengths, for maximum voltage at the detector. Changing the distance of the laser from the stage ( co-moving detector and mirror) will not affect whatever voltage detected at the detector.

[swansont]

It is the distance between the detector (SWD) and the the mirror that should be an integral multiple of half wavelengths, for maximum voltage at the detector. Changing the distance of the laser from the stage ( co-moving detector and mirror) will not affect whatever voltage detected at the detector.

 

Sure it will, if you are sending coherent signal in and detecting the amplitude. If you have the detector at a node, there is no signal from the incident wave. If the signal hits the mirror at an antinode, there is no reflected signal. You have destructive interference. That means a reduced signal.

 

This is how resonant cavities work. There are no supported signals unless the separation between the two ends are an integral number of half-wavelengths.

[lidal]

 

Sure it will, if you are sending coherent signal in and detecting the amplitude. If you have the detector at a node, there is no signal from the incident wave. If the signal hits the mirror at an antinode, there is no reflected signal. You have destructive interference. That means a reduced signal.

 

This is how resonant cavities work. There are no supported signals unless the separation between the two ends are an integral number of half-wavelengths.

 

I am gateful for the help. I will take some time to understand what you mean by this.

 

But let me also present the problem mathematically:

 

An ideally coherent laser source emits light wave according to :

A sin ωt

The incident wave at the detector ( SWD) will be :

A sinω( t - t_i) , where t_i is the time delay from laser source to detector

t_i = D/c , where D is the distance between the source and the detector.

 

The reflected wave at the detector will be:

- A sinω( t - t_i - 2L/c ) , where L is the distance between the detector and the mirror

 

The negative sign above is because the wave will reverse phase by 180 at reflection.

 

Sum of incident and reflected waves at the SWD will be:

 

A sinω( t - t_i) + - A sinω( t - t_i - 2L/c ) = A sin ( ω t - ωt_i) + - A sin( ωt - ωt_i - 2ωL/c )

 

We can see that t_i , hence distance D from laser, does not affect the voltage at the detector because the phase difference between the incident and reflected waves at the detector is always 2ω L/c , as we can see from the equation above.

[Sensei]

You can see for yourself what happens with a standard construction laser.

 

Point the laser in some safe but convenient direction.

Move a target along in the beam.

The laser line will impact the target as a small disk of light.

This diameter of this disk waxes and wanes as you move the target along in the beam.

 

When using such as laser for levelling or alignment purposes you need to take this phenomenon into account.

It's normal inverse square law.

https://en.wikipedia.org/wiki/Inverse-square_law

 

Actually, this experiment is basically different from Silvertooths experiment because the standing wave is locked to the mirror. Yet the author reported that the voltage at the SWD varied as the linear stage was moved. This experiment even violates fundamental accepted knowledge of elementary physics. It implied that the voltage at the SWD will vary as the linear stage is moved relative to the laser, even if there is no relative motion between the mirror and the SWD.

 

Actually, in the experiment, the mirror is dithered and the change in phase of the AC voltage at the SWD is observed. Since the mirror and the SWD are co-moving, the phase of the SWD voltage should not vary as the stage is moved relative to the laser.

 

Am I right or where did I go wrong? Please help in pointing out the problem. Thank you for your time.

I draw you image, and take photo.

See attachment.

Laser has initial energy E₀, or intensity I₀,

after traveling distance d₀, it's reflected from mirror, and reverse direction back to source,

and travel distance d₁,

and hit detector. Traveling total d=d₀+d₁ distance.

 

In second example light is traveling d=d₂+d₃ distance.

 

Distance between laser and detector remain the same x.

 

Yet another version of it, is when distance between detector and mirror remain the same (move together), and only distance from laser and mirror changes.

Edited September 4, 2015 by Sensei

[studiot]

 

Sensei:

Posted Today, 01:58 PM

studiot, on 04 Sept 2015 - 11:06 AM, said:

You can see for yourself what happens with a standard construction laser.

 

Point the laser in some safe but convenient direction.

Move a target along in the beam.

The laser line will impact the target as a small disk of light.

This diameter of this disk waxes and wanes as you move the target along in the beam.

 

When using such as laser for levelling or alignment purposes you need to take this phenomenon into account.

It's normal inverse square law.

https://en.wikipedia...erse-square_law

 

 

Definitely not so.

 

FYI optically focussed beams are not subject to the inverse square law.

 

Read carefully what I said again.

I am not sure if the periodic variation is due to the modulation in the beam rather than the red laser light frequency.

[Sensei]

Definitely not so.

FYI optically focussed beams are not subject to the inverse square law.

 

Definitely, they are.

If you have 1mm laser diameter 1m from laser source, what will be diameter of this laser if you point at f.e. Moon, 384,000 km or so .. ?

Do you think so it'll be also 1mm diameter.. ?

[swansont]

 

 

Sum of incident and reflected waves at the SWD will be:

 

A sinω( t - t_i) + - A sinω( t - t_i - 2L/c ) = A sin ( ω t - ωt_i) + - A sin( ωt - ωt_i - 2ωL/c )

 

We can see that t_i , hence distance D from laser, does not affect the voltage at the detector because the phase difference between the incident and reflected waves at the detector is always 2ω L/c , as we can see from the equation above.

 

It's a standing wave, so you really want the phases in terms of the position rather than time, but the concept is what's important:

 

What happens when t_i changes, because of a change in D? If that makes the first term zero, then the whole equation goes to zero. (2ω L/c being a multiple of π) If it makes the first term just smaller, then the second term is smaller as well (though your - sign implies they just cancel)

 

IOW, phase difference ≠ phase. The phase tells you the size of the incident wave at a point. The phase difference tells you the relation between the size of the incident and reflected wave. So the phase difference says "they are the same size" — that's all. It does not say what that size is — that's dictated by the phase of the incident wave.

 

Definitely, they are.

If you have 1mm laser diameter 1m from laser source, what will be diameter of this laser if you point at f.e. Moon, 384,000 km or so .. ?

Do you think so it'll be also 1mm diameter.. ?

 

It will probably follow Gaussian optics

 

http://www.newport.com/Gaussian-Beam-Optics/144899/1033/content.aspx

 

In the far field the waist varies linearly with x, but not in the near field (equation below figure 2) and I think the near field is what's under discussion

[lidal]

It's normal inverse square law.

https://en.wikipedia.org/wiki/Inverse-square_law

 

I draw you image, and take photo.

See attachment.

Inverse square law.png

Laser has initial energy E₀, or intensity I₀,

after traveling distance d₀, it's reflected from mirror, and reverse direction back to source,

and travel distance d₁,

and hit detector. Traveling total d=d₀+d₁ distance.

 

In second example light is traveling d=d₂+d₃ distance.

 

Distance between laser and detector remain the same x.

 

Yet another version of it, is when distance between detector and mirror remain the same (move together), and only distance from laser and mirror changes.

 

This may be helpful. It also means that as the distance between the laser and the stage is changed, the amplitude of the voltage detected at the SWD will also change because, even if the phase relationship between the incident and reflected waves does not change, the relative amplitudes of the incident and reflected waves will change so that the detector voltage will also change, if the distance from source to detector is not too large, which is the case of the experiment I mentioned. But let me try to figure out how this can help me explain the experiment I mentioned.

[Sensei]

It will probably follow Gaussian optics

 

[article]

 

Isn't about how light is propagating through transparent/semi-transparent medium, such as water, glass, etc.. ?

"we would find that a Gaussian source distribution remains Gaussian at every point along its path of propagation through the optical system."

Vacuum in cosmos is not quite "optical system".

 

Obviously photons going though medium will be reflected, or absorbed. That's why we can see laser beam in water. Reflected by medium photons stop being part of original laser beam.

In perfectly ideal transparent medium, laser beam would remain invisible.

Edited September 4, 2015 by Sensei

[swansont]

 

Isn't about how light is propagating through transparent/semi-transparent medium, such as water, glass, etc.. ?

"we would find that a Gaussian source distribution remains Gaussian at every point along its path of propagation through the optical system."

Vacuum in cosmos is not quite "optical system".

 

Obviously photons going though medium will be reflected, or absorbed. That's why we can see laser beam in water. Reflected by medium photons stop being part of original laser beam.

In perfectly ideal transparent medium, laser beam would remain invisible.

 

"Optical system" also includes the free space in between optical components. In between two lenses comprising a telescope, for example. The beam never gets down to a point, for example — there is always some finite size to it — and the beam is described by the equations in the link. The beam does not follow a simple 1/r^2 expansion during the part when it expands, and certainly doesn't follow it when it's being focused down, which is the specific example studiot gave.

[lidal]

Consider a VHF signal source conncected to a cable that is shorted ( zero resistance) at its far end.

From electromagnetic theory, the first voltage maxima of the resulting standing wave always occurs at quarter wavelength from the shorted end.

Even if the distance between source and load ( short circuit) is changed by changing the length of the cable, the position of the first voltage maxima relative to the shorted end will not change.

[swansont]

Consider a VHF signal source conncected to a cable that is shorted ( zero resistance) at its far end.

From electromagnetic theory, the first voltage maxima of the resulting standing wave always occurs at quarter wavelength from the shorted end.

Even if the distance between source and load ( short circuit) is changed by changing the length of the cable, the position of the first voltage maxima relative to the shorted end will not change.

 

Does the value of the voltage stays the same?

[studiot]

I am suprised at your responses to my post, Sensei.

You are normally the one who comes up with simple homebru experiments to illustrate points of physics.

 

I noticed this effect when testing a construction laser prior to use.

The laser is designed to focus the beam in as parallel a manner as practicable and is used by producing a spot on a special target.

Perhaps the english expression waxing and waning caught you out.

What I meant was that the spot diameter varies periodically with distance from the laser, from a few mm to perhaps one cm and back.

[John Cuthber]

I am suprised at your responses to my post, Sensei.

You are normally the one who comes up with simple homebru experiments to illustrate points of physics.

 

I noticed this effect when testing a construction laser prior to use.

The laser is designed to focus the beam in as parallel a manner as practicable and is used by producing a spot on a special target.

Perhaps the english expression waxing and waning caught you out.

What I meant was that the spot diameter varies periodically with distance from the laser, from a few mm to perhaps one cm and back.

Laser beams don't generally do that.

[studiot]

 

Laser beams don't generally do that.

 

 

 

So explain the observation.