Does Aether Exist?

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Consider two particle accelerators. You will not achieve the same results by having a stationary reference and a loop at v=X, or two loops each going in opposite direction at 1/2X. All four reference points will see the same velocity, but the collision output would be different for each accelerator. The output on the target will allow us to tell which is which, which may not be obvious in any of the references. From that we can then calculate the order of the references.

The illusion can work with two reference but not with three or more.
 
The device that I promised:

A ten meter long support. Length only counts and be rigid enough not to bend.
Lay horizontally with the length. In the middle (center of gravity) a joint which allows full rotation of 360 degrees horizontally.
At one end is put a laser with a slim beam.
At the other end is put a receiver that is a webcam with a minimum resolution of 5M, or something like that and that is connected to a monitor.
A laser beam displacement on the receiver of value 0.1 mm shall be seised on the monitor. A displacement of 0.1 mm (reported to 10m) indicates a speed of 3km / s. I am convinced that such precision is sufficient.

In a certain position, set the receiver laser system so that on the monitor appears the bright point exactly in the middle. Also here intersect the coordinates X and Y which are graded.

Now rotate the device. The bright point on the monitor can go to minus or plus. Where to find the maximum minus must calibrate again the device, bringing the bright point to the origin, on the monitor.

Now rotate the device 90 degrees and the value indicated is the X component of the speed vector (Y component is zero).
To find the component z the device must be positioned with the rotation plane perpendicular to the axis X. Now we are able to find component Z.

The earth speed vector is done by vectorial sum of component X and Z (Y = 0).
 
The device that I promised:

A ten meter long support. Length only counts and be rigid enough not to bend.
Lay horizontally with the length. In the middle (center of gravity) a joint which allows full rotation of 360 degrees horizontally.
At one end is put a laser with a slim beam.
At the other end is put a receiver that is a webcam with a minimum resolution of 5M, or something like that and that is connected to a monitor.
A laser beam displacement on the receiver of value 0.1 mm shall be seised on the monitor. A displacement of 0.1 mm (reported to 10m) indicates a speed of 3km / s. I am convinced that such precision is sufficient.

In a certain position, set the receiver laser system so that on the monitor appears the bright point exactly in the middle. Also here intersect the coordinates X and Y which are graded.

Now rotate the device. The bright point on the monitor can go to minus or plus. Where to find the maximum minus must calibrate again the device, bringing the bright point to the origin, on the monitor.

Now rotate the device 90 degrees and the value indicated is the X component of the speed vector (Y component is zero).
To find the component z the device must be positioned with the rotation plane perpendicular to the axis X. Now we are able to find component Z.

The earth speed vector is done by vectorial sum of component X and Z (Y = 0).

You don't think that LIGO would have picked it up by now?

LIGO is an interferometer with arms 4km in length, designed for looking for gravity waves. If an aether existed, because even in a single run it examines every possible rotation in a 2d plane (it's fixed on the earths surface, which means it rotates with the earths surface about its axis, which has the same net effect as spinning it about its own) and the experiment has effectively been repeated at multiple... Let's call it phase angles (Earth-Sun-Motion Vector) because there have been multiple runs at different times of the year. The November 2005 run had an accuracy of one part in 10[sup]21[/sup].

LIGO has, to my knowledge, produced no data to support the existence of an Aether.
 
Trippy #65:
Your last -->You don't think that LIGO would have picked it up by now?

, , , Not if the aether is too 'fine' (< Planck lengths) to be detected by LIGO, or any other detector constructed of matter!
 
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. . . by the way, Trippy . . . has LIGO ever detected any gravity waves . . . there appears to be a lot of that stuff (gravity) around, you know . . .
 
I don't know the working principle of LIGO.

What connection is between the light and the gravitational field (waves?)?

LIGO basically works on the same principle as the michelson-morley experiment.

LASER gets shone at a beam splitter.
Beams travel down two arms, each 4km long.
Beams enter a Fabry–Pérot cavity, bounce around 75 times, then leave it and travel back down the arms.
Under 'normal' circumstances, the beams return to the splitter half a wavelength of out phase, and destructively interfer, resulting in no light reaching the detector.
When a gravity wave passes through, it changes the length of one or both arms, which changes the phase at the splitter, and results in light reaching the detector.

The Aether, as you appear to be describing it, should be expected to deflect the beam in each arm, but to different degrees, which should be expected to result in a signal with a 24hr periodicity emerging in the data.

Now, you might argue that they would discard the signal, and you might be right, however, they operate two detectors seperated by 3002km, and the expectation WRT gravity waves is that the same signal would be expected to emerge at both detectors seperated by about 10ms. In the case of the Aether behaving in the way you appear to be describing it, one might expect that the same signal would be present at both sites, with a phase difference of something like 1.8 minutes
 
LASER gets shone at a beam splitter.
Beams travel down two arms, each 4km long.
Beams enter a Fabry–Pérot cavity, bounce around 75 times, then leave it and travel back down the arms.
Under 'normal' circumstances, the beams return to the splitter half a wavelength of out phase, and destructively interfer, resulting in no light reaching the detector.
When a gravity wave passes through, it changes the length of one or both arms, which changes the phase at the splitter, and results in light reaching the detector.
This experiment has nothing to do with what I have affirmed.
The Aether, as you appear to be describing it, should be expected to deflect the beam in each arm, but to different degrees, which should be expected to result in a signal with a 24hr periodicity emerging in the data.
I never said that.
I said "Aether" in which the light propagates, and it is an absolute reference system for the speed, and nothing else.
 
This experiment has nothing to do with what I have affirmed.

I never said that.
I said "Aether" in which the light propagates, and it is an absolute reference system for the speed, and nothing else.

Then what, precisely are you trying to measure with this setup:

The device that I promised:

A ten meter long support. Length only counts and be rigid enough not to bend.
Lay horizontally with the length. In the middle (center of gravity) a joint which allows full rotation of 360 degrees horizontally.
At one end is put a laser with a slim beam.
At the other end is put a receiver that is a webcam with a minimum resolution of 5M, or something like that and that is connected to a monitor.
A laser beam displacement on the receiver of value 0.1 mm shall be seised on the monitor. A displacement of 0.1 mm (reported to 10m) indicates a speed of 3km / s. I am convinced that such precision is sufficient.

In a certain position, set the receiver laser system so that on the monitor appears the bright point exactly in the middle. Also here intersect the coordinates X and Y which are graded.

Now rotate the device. The bright point on the monitor can go to minus or plus. Where to find the maximum minus must calibrate again the device, bringing the bright point to the origin, on the monitor.

Now rotate the device 90 degrees and the value indicated is the X component of the speed vector (Y component is zero).
To find the component z the device must be positioned with the rotation plane perpendicular to the axis X. Now we are able to find component Z.

The earth speed vector is done by vectorial sum of component X and Z (Y = 0).

If it's not the amount of deflection of the LASER beam being caused by the Aether? (emphasis mine)
 
Step 1.
What do you think a short laser beam will hit an object that rotates the point where you "target the" or
will hit a point taking into account the distance to the object, speed of light and rotation speed of the object?
Step 2.
I affirm that the second version is correct, namely: "it will hit a point taking into account the distance to the object, speed of light and rotation speed of the object".

The same will happen if the target has a velocity perpendicular to the axis of the laser-target.
A short laser beam will not hit the target in the point spotted, it will hit a point taking into account the distance to the target, speed of light and the speed of the target.
Step 3.
The same scenario as in the previous example except that the source has the same speed as the target.
The short laser beam will hit the target in the same place as in the previous example.
Why? Because the light is not influenced by the speed of source .
The source and target are at rest relative to each other.
Then what speed could be measured?
Is the speed relative to "Aether" which is a medium of propagation of light.

So, the "Aether" is an absolute reference system.
What don't you agree?
 
No one has YET discussed the hypothesis that aether does not interact with any material detector . . . .thus it is not detected . . . .
 
No one has YET discussed the hypothesis that aether does not interact with any material detector . . . .thus it is not detected . . . .

If that were the case its existence would have no effect on anything, so it would not matter that we could not detect it.

However, even empty space interacts with matter. From that interaction we get curved space in the form of both geodetic and frame dragging effects. Both experimentally tested and verified.
 
So you will hit the sun in the point spotted and not the point that will be on the trajectory after 8min?

If you aim a laser light at the sun it will intersect where the sun was some 16 minutes earlier. That could even be miss.

First pointing at the sun you are pointing at where is "was". Second during the time it takes the light to get to the sun it will have moved another 8 minutes further. To make things even worse it is the Earth's spin that is changing the sun's position in the sky not the sun's own motion.

But the sun does move faster around the galaxy than the earth does around the sun, so things get more complicated the more deeper you go.
 
If you aim a laser light at the sun it will intersect where the sun was some 16 minutes earlier. That could even be miss.

First pointing at the sun you are pointing at where is "was". Second during the time it takes the light to get to the sun it will have moved another 8 minutes further. To make things even worse it is the Earth's spin that is changing the sun's position in the sky not the sun's own motion.

But the sun does move faster around the galaxy than the earth does around the sun, so things get more complicated the more deeper you go.
Yes you are right but I did not want to complicate it because neither is understood by many.

But I must make a correction:
"If you aim a laser light at the sun it will intersect where the sun was some 16 minutes earlier." -where the sun will be after 16 minutes.
"First pointing at the sun you are pointing at where is "was"." -where will be.
"Second during the time it takes the light to get to the sun it will have moved another 8 minutes further."-right.
 
Yes you are right but I did not want to complicate it because neither is understood by many.

But I must make a correction:
"If you aim a laser light at the sun it will intersect where the sun was some 16 minutes earlier." -where the sun will be after 16 minutes.
"First pointing at the sun you are pointing at where is "was"." -where will be.
"Second during the time it takes the light to get to the sun it will have moved another 8 minutes further."-right.

When you look at the sun you see the light that left it 8 minutes earlier, it is 8 minutes further into the day at that instant. So.., If you point at the sun you are pointing at where it was.

If you shine a laser at where the sun now appears to be (where it was 8 minutes ago) the light will take 8 minutes to reach where you thought the sun was and by that time the sun will have moved another 8 minutes for a total of 16 minutes into the day, so you will have missed it by 16 minutes.

It only takes 8 minutes for the laser light to reach the sun. If you want to actually hit it aim 16 minutes ahead of where it appears to be, now.

But this does not account for the sun's orbital motion around the Milky Way or the Earth's orbital motion around the sun.

And none of this accounts for the curvature of space, as per GR.
 
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