Michelson–Morley experiment: Did it disprove the existence of ether?

[John232]

This is a pretty useless statement. The modern concept of curved spacetime has nothing to do with the early 20th century concept of the luminiferous aether. Spacetime has only geometric properties, it has no material properties. Specifically, it does not have a velocity.

Technically it does have a velocity when dealing with expansion. But still useless, as I have not gained anything by trying to visualize it thinking of it this way besides a better mental picture of it.

 

[John232]

You talk about a "spacetime" ether - that's Minkowski's ether, and you say that it is warped. Fine. [Edit: I would say that in it different observers see events from a different angle, swapping time for space or vice versa].
Anyway, it's a literal interpretation of the equations, and in that sense it's simple. But as you said, it's also strange and "warped". In that sense it's not really simple but difficult.

I said that one may just as well stick with Lorentz's ether, which is similar to Newton's "space". That one is not warped nor does it permit to swap time for space; instead objects Lorentz contract and processes slow down. And there is nothing mystical about it, it's straightforward and quite easy to understand. However, the equations describe appearance only, so that an aspect of reality remains hidden - some people dislike that idea.

The same relativistic equations apply to both physical models.
So, take your pick, but don't mix them up!

The more I tried to understand why spacetime is dialated due to velocity the more mystical it seemed to become to me. If you look at the light triangle to derive SR using pythagreons, two sides of the triangle have the same length (ct). It is only by assigning a different time to each observer that the triangle itself becomes solvable.

The hypotunus is longer than the other side that is also ct. So then the added forward velocity as seen from one observer that sends the beam at this angle would have to create a force that makes the observer see that photon travel at c even though another observer sees it take a different trajectory that allows it to travel at a shorter distance at the same speed c.

So then there is this type of mystical force that is connected between spacetime and photons that makes you observe them to always travel at c that warps spacetime to accomplish this goal and it all comes from haveing an object that always travels at the same speed no matter what velocity or trajectory you are observing it from. I would say that was really strange. It is the only constant that is an objects velocity and it is though it gives it the power to warp reality just so that everything agrees with its constant bidding.

SR really describes how light has to warp reality so that it is always measured to travel at the same speed. Otherwise there would be no need for sepereate times for observers and any other object would just have to travel at some other speed that is not a constant. But the constant itself only allows for two other variables to be altered from its speed and that is space and time that is used to measure its velocity.

 

[Dale]

I would say that was really strange.

"Strange", sure, but hardly "mystical". All of SR can be summarized by the Minkowski metric:
ds² = -c²dt² + dx² + dy² + dz²

The minus sign makes the first term makes it a little bit strange, but since we are all used to the Euclidean metric:
ds² = dx² + dy² + dz²

it is not too big of a leap.

 

[harrylin]

The more I tried to understand why spacetime is dialated due to velocity the more mystical it seemed to become to me. If you look at the light triangle to derive SR using pythagreons, two sides of the triangle have the same length (ct). It is only by assigning a different time to each observer that the triangle itself becomes solvable.
[..]
So then there is this type of mystical force that is connected between spacetime and photons that makes you observe them to always travel at c that warps spacetime to accomplish this goal and it all comes from haveing an object that always travels at the same speed no matter what velocity or trajectory you are observing it from. I would say that was really strange. It is the only constant that is an objects velocity and it is though it gives it the power to warp reality just so that everything agrees with its constant bidding.

SR really describes how light has to warp reality so that it is always measured to travel at the same speed. Otherwise there would be no need for sepereate times for observers and any other object would just have to travel at some other speed that is not a constant. But the constant itself only allows for two other variables to be altered from its speed and that is space and time that is used to measure its velocity.

I don't think so!
Look at my post #131:
- I assumed no mystical force between spacetime and photons
- my calculation has zero warping of reality by light.

 

[roineust]

For some people, who see physics also by using math, this must look like the very same question, repeating itself again and again, each time in different wording, and if this is the situation, I am truly sorry for that, but I have to ask:

What is the relation, and is there a relation, between Newton's first law (uniform motion) and a photon's behavior? Is it correct to say that Newton first law just plainly doesn't apply to a photon?

Thanks,
Roi.

 

[Dale]

What is the relation, and is there a relation, between Newton's first law (uniform motion) and a photon's behavior?

Yes, one geometrical way of thinking of the postulates essentially says that in reference frames where the worldlines of inertial objects are mapped to straight lines the worldlines of light pulses are also mapped to straight lines with slope=c.

 

[John232]

I don't think so!
Look at my post #131:
- I assumed no mystical force between spacetime and photons
- my calculation has zero warping of reality by light.

I wasn't talking about you. I was using a derivation based on (ct)^2+(vt)^2=(ct)^2.
The triangle has no dimensions unless you replace one of the time variables with t'.
This is because c is a constant and cannot be changed so the time variable on one side and the hypotunus have to be different(time for an observer has to change to maintain the constant c). So then you know that the person traveling measures c to be the same but his time would have to be different to measure his photon going straight down a shorter distance from him so then the first side ct would have to be ct'. This will give a smaller value of time for the observer traveling since that side of the triangle is a shorter distance...

 

[roineust]

Please take a look at the here attached diagram:

In other words: Physics tells us, that we know exactly what the velocity of the rocket between 3 and 4 is, even without measuring, only by knowing that there is no gravitation and no power pushing the rocket anymore, but we can not know what the speed of a photon is, on its way between 1 and 2 (or to be even more clear, between 5 and 6), without measuring its velocity?

Thanks,
Roi.

 

[Dale]

Physics tells us, that we know exactly what the velocity of the rocket between 3 and 4 is, even without measuring, only by knowing that there is no gravitation and no power pushing the rocket anymore, but we can not know what the speed of a photon is, on its way between 1 and 2 (or to be even more clear, between 5 and 6), without measuring its velocity?

Huh? What are you talking about? What is the velocity between 3 and 4 and how do we know it?

 

[harrylin]

I wasn't talking about you.

Of course not. I try to relate to the topic, which is MMX. And that it does not require warping anything by light. [edit: remember, you wrote: "SR really describes how light has to warp reality"]

I was using a derivation based on (ct)^2+(vt)^2=(ct)^2.
The triangle has no dimensions unless you replace one of the time variables with t'. [..]

Ehm... the first equation implies vt=0. And the dimension is length. Sorry, here I give up!

 

[harrylin]

For some people, who see physics also by using math, this must look like the very same question, repeating itself again and again, each time in different wording, and if this is the situation, I am truly sorry for that, but I have to ask:

What is the relation, and is there a relation, between Newton's first law (uniform motion) and a photon's behavior? Is it correct to say that Newton first law just plainly doesn't apply to a photon?

Thanks,
Roi.

Newton: "Every body perseveres in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed"

In relativity theory a photon is a wave packet and not a "body" that can be "in rest". And a force can not really be "impressed" on a wave packet.

Still, in special relativity a photon perseveres in its state of uniform motion at speed c in a right line, unless it is compelled to change that state of motion by interaction with matter. Does that help?

Harald

 

[harrylin]

Please take a look at the here attached diagram:

In other words: Physics tells us, that we know exactly what the velocity of the rocket between 3 and 4 is, even without measuring, only by knowing that there is no gravitation and no power pushing the rocket anymore, but we can not know what the speed of a photon is, on its way between 1 and 2 (or to be even more clear, between 5 and 6), without measuring its velocity?

Thanks,
Roi.

No why would you think that??

 

[roineust]

What I am trying to do, and if it is simply wrong, please try to explain to me why, is something like this:

We know that 1 and 2 measure in their frames both 300000kps. But SR says that it is at once the same velocity and is not the same velocity.

What I am trying to say is:

(it=light velocity)

There are only 2 possibilities: it is the same (between 1 and 2), or it is not the same.
If it is not the same, then it must change somewhere on the way.

Now, in this second case, I want to take a frame, say between 5 and 6, and reduce proximity both between 5 and 6 until it tend to 0, and as well between 5 and 6 and 2, until the distance between light and detector 2 also tend to 0.

Wouldn't that prove that the only location that light can change it's velocity, is when its actually 'one' with the detector at 2?

I know,
This is very problematic, the way I say all that.
But, I could never say it in equations.

So, I try, and excuse me if it is considered wrong.

Roi.

 

[Dale]

1 and 2 and 5 and 6 appear to be locations, not frames. In any frame the speed of light will be c at locations 1 and 2 and c at locations 5 and 6 and c at any location inbetween. Why would you think that it is changing speed anywhere?

 

[harrylin]

What I am trying to do, and if it is simply wrong, please try to explain to me why, is something like this:

We know that 1 and 2 measure in their frames both 300000kps. But SR says that it is at once the same velocity and is not the same velocity.

What I am trying to say is:

(it=light velocity)

There are only 2 possibilities: it is the same (between 1 and 2), or it is not the same.
If it is not the same, then it must change somewhere on the way.

Now, in this second case, I want to take a frame, say between 5 and 6, and reduce proximity both between 5 and 6 until it tend to 0, and as well between 5 and 6 and 2, until the distance between light and detector 2 also tend to 0.

Wouldn't that prove that the only location that light can change it's velocity, is when its actually 'one' with the detector at 2?

I know,
This is very problematic, the way I say all that.
But, I could never say it in equations.

So, I try, and excuse me if it is considered wrong.

Roi.

Dear Roi,

I already explained that once or twice to you in this thread (and I think also ghwellsjr).

What did you not understand (or what did you understand) of my post #93, after reading the further explanation in post #128?

And my post #131, did you understand the meanings (c-v) and (c+v) in the equations?
From your last post, I think that that is essential. And if you are allergic to formula's, did you try to plug in numbers? v=0.8c is a convenient choice for number examples.

Here's such a number example:
You are floating on a river and you see a motorboat pass that has a fixed speed of 5 km/h in the water. You also measure that the embankment passes by at 2 km/h. How do you calculate the time that the motorboat needs to go from one place to another? and back?

Harald

 

[John232]

Of course not. I try to relate to the topic, which is MMX. And that it does not require warping anything by light. [edit: remember, you wrote: "SR really describes how light has to warp reality"]

Ehm... the first equation implies vt=0. And the dimension is length. Sorry, here I give up!

vt=d The distance the photon has traveled. The only reason why vt would seem to equal zero is because the same equation is put on two sides of a right triangle. But instead of like a normal object the velocity isn't seen to be different even though it has a different trajectory that is longer. So in order for us to perceive it to have the same speed with a different longer trajectory then we have to observe the object emmitting the photon to have a smaller measurement of time since they also agree on its speed but observe it to travel a shorter straight trajectory perpendicular to its direction of motion. So then we could watch them measure the photons speed and get the same answer as we did even though it took a different trajectory.

So then t has decreased inorder for them to measure a shorter d instead of c becasue c does not change.

 

[John232]

I got mixed up vt is the distance the object traveled. But ct is the distance the photon traveled. vt would also turn out to be zero but assigning them independant times allows it to be solved so that the photon and the object in question traveled a real distance.

 

[harrylin]

Dear Roi,

I already explained that once or twice to you in this thread (and I think also ghwellsjr).

What did you not understand (or what did you understand) of my post #93, after reading the further explanation in post #128?

And my post #131, did you understand the meanings (c-v) and (c+v) in the equations?
From your last post, I think that that is essential. And if you are allergic to formula's, did you try to plug in numbers? v=0.8c is a convenient choice for number examples.

Here's such a number example:
You are floating on a river and you see a motorboat pass that has a fixed speed of 5 km/h in the water. You also measure that the embankment passes by at 2 km/h. How do you calculate the time that the motorboat needs to go from one place to another? and back?

Harald

For a real number example I forgot to give a distance between the two places. Let's say 21 km on the embankment for round numbers.

 

[roineust]

harrylin,

If I understand correctly, and I think you gave me numbers that let me know I understand what you mean, then that would be 21/(2+5)=21/7=3 hours.

Can I meanwhile ask a question?
Here is an improved diagram, why doesn't it prove that the speed of light in vacuum, is an absolute number and not a relative number?

Roi.

 

[harrylin]

harrylin,

If I understand correctly, and I think you gave me numbers that let me know I understand what you mean, then that would be 21/(2+5)=21/7=3 hours.

Yes that's right - for a boat sailing along with the river. And how long does the return trip take?
Slowly but surely I am getting you on track to be able to calculate MMX yourself. ;-)

Can I meanwhile ask a question?
Here is an improved diagram, why doesn't it prove that the speed of light in vacuum, is an absolute number and not a relative number?
Roi.

Roi, I never heard of an "absolute number" versus a "relative number"... What do you mean with that?
When I said that it is the same speed c in a "relative" sense, I meant with "the same" that the number is the same!

I also tried to clarify with examples and equations what I meant with that, and now I started to explain it to you with numbers.

 

[roineust]

What I am trying to ask, if and why, is it, or is it not correct, to say:

Time dilation phenomenon can not be applied on the velocity of light in vacuum - Because this 'absoluteness' of the velocity of light, presumably proved in the previous diagram.

In case this last statement is correct, then my next question is:

Therefore, the phenomenon of time dilation can not be applied to the whole of the space within a given frame, since light will occupy some of this space.

This last statement is probably where I am wrong, but I would like very much to understand why.

 

[harrylin]

What I am trying to ask, if and why, is it, or is it not correct, to say:

Time dilation phenomenon can not be applied on the velocity of light in vacuum - Because this 'absoluteness' of the velocity of light, presumably proved in the previous diagram.

In case this last statement is correct, then my next question is:

Therefore, the phenomenon of time dilation can not be applied to the whole of the space within a given frame, since light will occupy some of this space.

This last statement is probably where I am wrong, but I would like very much to understand why.

Actually, I already gave you the answer to the first question in post #131.
There I showed how a two-way signal in a moving apparatus is time delayed by the same factor as a crystal.
As a matter of fact, it's much easier to explain the time dilation of a light clock than that of a crystal!

Therefore I am explaining you how to calculate such things by means of the classical number example with a river. But you did not answer my question about the return trip of the motorboat...

Then it will be easy for you to understand the c+v and the c-v in post #131. And next we can move on to time dilation and length contraction. And not to forget relativity of simultaneity (and only then will you able able to understand these answers here!).

So, it's a bit the inverse of what you expected:

- the answer to your first question is in a certain way, no: a moving light clock has time dilation as I showed, because light speed doesn't change.

- but the answer to your second question is yes - time dilation is about objects and is a function of the speed of those objects. Within a standard measurement system ("frame"), clocks are synchronized in such a way that the system appears to be in rest in space. By definition is the time dilation of your measurement system zero.

Cheers,
Harald

 

[Dale]

What I am trying to ask, if and why, is it, or is it not correct, to say:

Time dilation phenomenon can not be applied on the velocity of light in vacuum

Time dilation applies to times, not velocities. Velocities transform via the velocity addition formula, not the time dilation formula.

Therefore, the phenomenon of time dilation can not be applied to the whole of the space within a given frame, since light will occupy some of this space.

Time dilation will apply to the measurement of time throughout the whole space, regardless of whether or not there is light present.

 

[roineust]

harrylin,

I don't understand exactly what you mean by cross trip, but if you mean from one bank of the river, to the other bank, and the current speed is 2kph, and the boat speed is 5kph, and the river width is, say 1km, then I think the time to cross to the other bank would be:

1/(root((2^2)+(5^2)))=1/(root(29))= ~0.185 hours= ~11 minutes.

Would that be correct?

Roi.

 

[harrylin]

[..]
- but the answer to your second question is yes - time dilation is about objects and is a function of the speed of those objects. Within a standard measurement system ("frame"), clocks are synchronized in such a way that the system appears to be in rest in space. By definition is the time dilation of your measurement system zero.
Harald

I now realize that DaleSpam understood what Roi meant while I misunderstood it - thus the answer to Roi's second question was in fact a big NO, for the laws of physics apply everywhere!

 

[harrylin]

harrylin,

I don't understand exactly what you mean by cross trip, but if you mean from one bank of the river, to the other bank, and the current speed is 2kph, and the boat speed is 5kph, and the river width is, say 1km, then I think the time to cross to the other bank would be:

1/(root((2^2)+(5^2)))=1/(root(29))= ~0.185 hours= ~11 minutes.

Would that be correct?

Roi.

I don't think that I wrote "cross trip" - see below!

But you already answered another question that is relevant for MMX.

And yes your answer is almost right - it's just trigonometry.
Funny enough, it's a question that Michelson also had wrong the first time that he calculated it:

The boat has a constant speed of 5 km/h, let's call that velocity c.
Now replace the boat by light.
Also, replace the river bank by the MMX apparatus (moving at velocity v).
Then, in 1881 Michelson calculated 0.20 hours for your example! But when he did his famous experiment in 1887, he calculated it correctly and he found more than 0.20 hours. Do you understand why it will take longer when there is a strong river current?

But I wrote "How do you calculate the time that the motorboat needs to go from one place to another? and back?" and "Yes that's right - for a boat sailing along with the river. And how long does the return trip take?".
The end question is: how long will it take to go from A to B downstream and immediately back upstream to A?

If you have that, and you solved the little error in your calculation here above, then you can do Michelson's calculations. And after that, you may have the basic knowledge to follow the things we discussed with you.

 

[Buckleymanor]

I don't think that I wrote "cross trip" - see below!

But you already answered another question that is relevant for MMX.

And yes your answer is almost right - it's just trigonometry.
Funny enough, it's a question that Michelson also had wrong the first time that he calculated it:

The boat has a constant speed of 5 km/h, let's call that velocity c.
Now replace the boat by light.
Also, replace the river bank by the MMX apparatus (moving at velocity v).
Then, in 1881 Michelson calculated 0.20 hours for your example! But when he did his famous experiment in 1887, he calculated it correctly and he found more than 0.20 hours. Do you understand why it will take longer when there is a strong river current?

But I wrote "How do you calculate the time that the motorboat needs to go from one place to another? and back?" and "Yes that's right - for a boat sailing along with the river. And how long does the return trip take?".
The end question is: how long will it take to go from A to B downstream and immediately back upstream to A?

If you have that, and you solved the little error in your calculation here above, then you can do Michelson's calculations. And after that, you may have the basic knowledge to follow the things we discussed with you.

Now replace the boat with light.
Is this not a supposition that is unsupported.
How can you .

 

[roineust]

harrylin,

I think that the answer to your original question, which now I probably understand, is 5kph-2kph=3kph, and 21km/3kph=7hours. So the round trip time would be: 3hours+7hours=10hours.

Roi.

 

[harrylin]

harrylin,

I think that the answer to your original question, which now I probably understand, is 5kph-2kph=3kph, and 21km/3kph=7hours. So the round trip time would be: 3hours+7hours=10hours.

Roi.

Yes indeed!

Now the same question, which you already started, if one wants to cross a 1 km wide river straight over and then return back home.
If you will just aim your boat straight to the other side, then you will arrive on the other side after 0.2 hours, but the stream will have drifted you from course, like this (you want to get to from A B but you end up somewhere else):

Sketch by a swimmer floating in the water:

A ->v
----------
¦
¦c
¦ river
----------
B ->v

Sketch by an observer on the wall:

A
----------
¦
.¦ river ->
..¦
----------
B

So, you should aim your boat slightly upstream in order to get straight to the other side. How long will it take to get straight to the other river bank, like this:

A
----------
/
/ river ->
/
----------
B Harald

 

[harrylin]

Now replace the boat with light.
Is this not a supposition that is unsupported.
How can you .

The calculation is the same! The boat in this example has a constant speed in the water, just as light is supposed to have a constant speed in space.

 

[Buckleymanor]

The calculation is the same! The boat in this example has a constant speed in the water, just as light is supposed to have a constant speed in space.

That is the point.The boat might have a constant speed in this example in water but it does not consistently behave this way.However light is supposed to have a constant speed in space allways which in part is derived from the boats behavioure in water.
Is it not comparing different mediums with the same calculations and arriveing at the same conclusion because the numbers fit.

 

[harrylin]

Oops I put some arrows in the wrong direction.
Roi, here it is corrected with a little elaboration:

Yes indeed, you got the first calculation right!

Now the same question, which you already started, if one wants to cross a 1 km wide river straight over and then return back home.
If you will just aim your boat straight to the other side, then you will arrive on the other side after 0.2 hours, but the stream will have drifted you from course, like this (you want to get to from A B but you end up somewhere else):

Sketch by a swimmer floating in the water (the river banks moving to the left):

A <-- v
----------
¦
¦c
¦ river
----------
B <--v

Sketch by an observer on the wall (the river moving to the right):

A
----------
¦
.¦ river -->
..¦
----------
B

So, you should aim your boat slightly upstream in order to get straight to the other side. How long will it take to get straight to the other river bank, like this:

A
----------
/
/ river -->
/
----------
B Edit: I'll sketch how to calculate it with trigonometry.
The boat goes diagonally to the other side at speed c in the water:

<--- speed of A
|.../
|../c
|./
<--- speed of BHarald

 

[harrylin]

That is the point.The boat might have a constant speed in this example in water but it does not consistently behave this way.However light is supposed to have a constant speed in space allways which in part is derived from the boats behavioure in water.
Is it not comparing different mediums with the same calculations and arriveing at the same conclusion because the numbers fit.

Dear Buckleymanor,

An illustration is rarely perfect, and obviously a boat is not a wave. It's just an easy to understand illustration of how to calculate the propagation of a light ray between moving points.
If Roi understands how a boat is delayed due to relative motion, he can also calculate and understand how light propagates in a moving MMX interferometer; then he may be able to follow the discussions in this thread including the answers to his questions.

 

[harrylin]

It does not show that there is no ether, it shows that the Earth is not rotating through an ether. There are other possibilities such as an ether that is dragged along with Earth that would show the same result, but these are mostly considered superfluous.

Actually, that second part of that answer is completely wrong!
MMX was by design insensitive to rotation. For detecting rotation the Sagnac device was used and Michelson made a larger version to detect the rotation of the earth.
- http://en.wikipedia.org/wiki/Michelson–Gale–Pearson_experiment

 

[roineust]

I think that would be:

root((5kph^2)-(2kph^2))=root(21)kph=~4.6kph
1km/4.6kph=~13minutes, one way straight line.
13minutes*2=~26minutes, round trip from bank to bank straight lines.

Is that correct?

 

[harrylin]

I think that would be:

root((5kph^2)-(2kph^2))=root(21)kph=~4.6kph
1km/4.6kph=~13minutes, one way straight line.
13minutes*2=~26minutes, round trip from bank to bank straight lines.

Is that correct?

Yes that is correct!
And if the river were 21 km wide, the round trip would take about 2x4.6= 9.2 hours.

Now compare that with your earlier calculation for the round trip over the same distance downstream and upstream; you will see that that differs. For that trip you calculated 10 hours.

That is the basis of Michelson's experiment.
He assumed that the Earth is flowing through a light ether, and that it would be possible to detect that motion by comparing the return trip of light in two different directions. In his time it was impossible to directly measure the round trip time over such a short distance, but a change in the roundtrip time could be made visible with his interferometer.
However, he did not find a significant difference. So, he had made a mistake, but where?

Lorentz and Fitzgerald proposed that perhaps matter contracts because it is held together by electromagnetic fields. Heaviside had calculated for charges in motion that those fields contract in the direction of motion, by what is now know as the Lorentz factor.

For the calculation above, the Lorentz factor is (I write SQRT where you write root):
1 / SQRT(1 - 4/25) = 1/0.92.

So (here the boat example falls flat on its face but just consider the calculation!), if the distance of the embankment would shrink like that due to the motion of the river, then the return time as you first calculated would not be 10 hours but 9.2 hours. That is exactly the same as what you get for a trip to the other side of the river and back!
As a matter of fact, it would be the same in all directions.

If you now look again at my post #131, you may be able to understand it this time.
There I only discuss the one-dimensional problem. You may recognize c+v as the light going in the opposite direction as the apparatus, and c-v as the apparatus running ahead of the light, so that the light must catch up - and that takes longer.

Here follow the same formula's with the numbers of the boat example plugged in.

Michelson thought:
t1 = L/(c+v) t1 = 21/(5+2)
t2 = L/(c-v) t2 = 21/(5-2)
t1+t2 = T = [L(c-v)+L(c+v)] / [(c+v)(c-v)]
T = 2L * c /(c² - v²)
T = 2L/c * 1/(1 -v²/c²) T = 2x(21/5) x 1/(1-4/25) = 10

Cheers,
Harald

 

[Buckleymanor]

Lorentz and Fitzgerald proposed that perhaps matter contracts because it is held together by electromagnetic fields. Heaviside had calculated for charges in motion that those fields contract in the direction of motion, by what is now know as the Lorentz factor.

I would like Roi to understand light propagation and clear up some of my own misunderstandings on the subject as well.
What's unclear is that the Lorenze Fitzgerald contraction appears to be a propasition.
Light propagation is a form of electromagnetism which does not contract because it's speed is constant, where as electromagnetic fields do.
So why do some fields contract but not others.

 

[harrylin]

I would like Roi to understand light propagation and clear up some of my own misunderstandings on the subject as well.
What's unclear is that the Lorenze Fitzgerald contraction appears to be a propasition.
Light propagation is a form of electromagnetism which does not contract because it's speed is constant, where as electromagnetic fields do.
So why do some fields contract but not others.

How could a propagation speed "contract"? Instead, fields are maintained at the speed of light.

This proposition necessarily followed from special relativity, which was the outcome of the above considerations: in order to explain such null effects as of MMX, all material objects and fields have to contract in the direction of motion if we assume that the speed of light is constant in all directions.

 

[roineust]

Please see attached diagram,

I still don't understand, why there is no way to explain, in very general terms, or words, and without having light going both ways, but only one way, why this arrangement will not work as I think it will, while in the moving frame.

I know you are trying to explain to me many times, why both beams will always arrive together, no matter what the velocity is...But I still don't understand why.

Is there no middle way between understanding in words, and understanding in math?

 

[Dale]

I have tried to make some calculations, and got to the conclusion, that such an arrangement, will become possible, only when atomic clocks will reach the 10^-21 seconds accuracy level

That sounds like a numerical precision error. Would you like to post your work, and I can run it on Mathematica using arbitrary-precision numbers.

 

[roineust]

Thanks DaleSpam,

The idea is that the light in this arrangement travels a certain way, and that the arrangement is accelerated to a certain velocity. The presumable difference in indication of clock light detectors (at the end), arises supposedly from the time dilation caused by velocity, within the middle way clock (the rhombus shape), so there is a need to be able to detect such a time dilation difference between the two clock light detectors in the moving apparatus.

Therefore, the size of such a device, to be accelerate to a certain velocity, is dependent on the way that is chosen in order to accelerate it. I think there could be two different ways. Either using a Jet, Rocket, Satellite etc...which is in the speed range of 0.0001C-0.000001C, or, if it is possible to make this device really really small, say micro to nano size (I saw that there already exists an atomic clock on a chip), then it might be possible to accelerate the device to say, 0.05C?

For the type of device on a jet, the size could be no more than, say 10 meters long?
It takes light to cross 10 meters ~0.00000003 seconds? at 0.00001C, time dilation factor is ~1.00000000004. Multiplication of these numbers gives ~10^-9.

For the type of device in an accelerator, the size would be, say 0.0001 meter? It takes light to cross this length, ~0.0000000000003 second, and at 0.05C the time dilation factor is ~1.0018. Multiplication of these numbers gives ~10^-13.

Now I see that I might be wrong with the estimation of 10^-21.

It seems to be 10^-9 to 10^-13.

Is this calculation correct?

If it is, then it seems that technologically wise, it is very feasible, and even not astronomically costly to build such an experiment (the jet version).

So maybe such an experiment has already been done in the past, since such accuracy was already achievable some decades ago?

If such an experiment was not done, then of course it all comes back, to my effort to understand ,why is it so clear, that there is no need to build such an experiment?

Why is it so clear, that this device will always have the light clock detectors synchronized as they were at the beginning, no matter what the velocity is?

 

[Dale]

For the type of device on a jet, the size could be no more than, say 10 meters long?
It takes light to cross 10 meters ~0.00000003 seconds? at 0.00001C, time dilation factor is ~1.00000000004. Multiplication of these numbers gives ~10^-9.

Well, a jet flies at more like 10^-6 c where the time dilation factor is 1 + 5 10^-13 or 2 parts in 10^12. This is orders of magnitude less than the current precision of atomic clocks (~1 parts in 10^17).

If such an experiment was not done, then of course it all comes back, to my effort to understand ,why is it so clear, that there is no need to build such an experiment?

Why is it so clear, that this device will always have the light clock detectors synchronized as they were at the beginning, no matter what the velocity is?

It is very clear because this device operates purely on EM principles, and the time dilation of EM phenomena is well studied and well established.

When you mentioned that you had done calculations I was thinking something more along the lines of a thourough analysis of the device itself. Specifically, in the rest frame you have the following:
A) button push, electrical transmission to light, optical transmission to crystal, transmission through crystal, optical transmission to detector, electrical transmission to comparator
B) button push, electrical transmission to light, optical transmission to mirror, reflection, optical transmission to detector, electrical transmission to comparator

Based on assumptions about the geometry of the device and the speed of the signals in the electrical wires and the crystal you should be able to determine if the comparator registers a synchronous activation or not. Then by using standard relativity (time dilation, length contraction, velocity addition, etc.) you should be able to boost the experiment to a different reference frame and determine if the comparator registers a synchronous activation or not. From first principles you are guaranteed that the answers will be the same, so if you get a different answer then it is probably due to numerical precision errors.

 

[harrylin]

Please see attached diagram,

I still don't understand, why there is no way to explain, in very general terms, or words, and without having light going both ways, but only one way, why this arrangement will not work as I think it will, while in the moving frame.

I know you are trying to explain to me many times, why both beams will always arrive together, no matter what the velocity is...But I still don't understand why.

Is there no middle way between understanding in words, and understanding in math?

Understanding in only words or in only math is half-baked, and often mistaken!
A good understanding needs all the help possible, from words, drawings and math.

But I already answered your question with math and words in #131! It is impossible in your drawing that both signals go in only one way (have you ever heard of Escher?). PS I hope that you did notice that the signal from the button to the bottom light is going in the other direction...

Please make sure that you now understand the time dilation of a signal that is going in two directions, as I explained in #131.

Or at least, that you now understand why it takes longer to return in the moving system.

 

[Buckleymanor]

How could a propagation speed "contract"? Instead, fields are maintained at the speed of light.

This proposition necessarily followed from special relativity, which was the outcome of the above considerations: in order to explain such null effects as of MMX, all material objects and fields have to contract in the direction of motion if we assume that the speed of light is constant in all directions.

Note that you have mentioned that fields have to contract.So why can you rule out light speed, once it has propagated.
I can't see a clear distinction between the two types of fields or how experimentialy one is drawn.

 

[harrylin]

Note that you have mentioned that fields have to contract.So why can you rule out light speed, once it has propagated.
I can't see a clear distinction between the two types of fields or how experimentialy one is drawn.

Light speed is not a field, just as sound speed is not grass!

 

[Buckleymanor]

Light speed is not a field, just as sound speed is not grass!

So light propagation is not a field even though its part of the electromagnetic spectrum and can't be said to contract.
Allthough electromagnetic fields maintained at lightspeed which hold matter together can?
Sorry I am just too dim to understand why one state can and one can't.

 

[harrylin]

So light propagation is not a field even though its part of the electromagnetic spectrum and can't be said to contract.
Allthough electromagnetic fields maintained at lightspeed which hold matter together can?
Sorry I am just too dim to understand why one state can and one can't.

A state of motion is not a state of being... For example, do you think that a contracted loudspeaker can affect the speed of sound? Inversely, if a loudspeaker were made up of sound waves (don't think about how it could be done), would a moving loudspeaker have the same shape as one in rest?

 

[roineust]

harrylin, Dalespam, Buckleymanor.

Thanks for trying to put things into words.
Please Look at the new diagram here.

I hope that I am not bringing the question into places, that can not be explained by words, no matter what...

Roi.

 

[harrylin]

harrylin, Dalespam, Buckleymanor.

Thanks for trying to put things into words.
Please Look at the new diagram here.

I hope that I am not bringing the question into places, that can not be explained by words, no matter what...

Roi.

In the top picture, do you mean that the clock adds a delay time?

This situation is very different from before, as the two detectors are not at the same place along the direction of motion.

In relativity, the simultaneity of distant clocks is "relative": when you bring a system in motion the clocks that appeared synchronous now do not anymore appear in sync - just as you found.

What people do for a standard independent measurement system is to adjust the clocks in the moving system so that the light appears to take the same time both ways. Then your signals will appear to arrive at the same time again for observers in the moving system.

And if you try to calculate it, don't forget that also the moving system is contracted. It's a complicated calculation, but in words what happens is that you make the one-way-speed equal to the average two-way speed. And because the average two-way speed works out, then also the one-way speed works out.

 

[Dale]

Thanks for trying to put things into words.
Please Look at the new diagram here.

I hope that I am not bringing the question into places, that can not be explained by words, no matter what...

If I understand correctly then the device is designed such that in the frame where the device is stationary the light arrives simultaneously and the clocks are synchronized so the light is detected as simultaneous.

If that is correct then in the frame where it is moving the light arrives at different times (not simultaneous) but the clocks are also not synchronized. The de-synchronization of the clocks exactly offsets the time difference between the arrival of the light so the light is detected as simultaneous even though it is not.