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

[harrylin]

[..]
thwle got to feeling a little frustrated and rashly promised to demonstrate that a unique frame or reference in which light proagates at the same speed in all directions not only exists but is discernable from other frames of reference.

Now, thwle has had the audacity to perform on his promise: See the new thread "Absolute Rest" under special and general relativity.

And now harrylin wonders what happened to thwle!

For not only did he suddenly stop posting (could be the flu or a little trip), but also there is no such thread and, to top it off, it is impossible to contact him!
and I can't even find mention of that possibility in the FAQ.

 

[Dale]

And now harrylin wonders what happened to thwle!

For not only did he suddenly stop posting (could be the flu or a little trip), but also there is no such thread and, to top it off, it is impossible to contact him!
and I can't even find mention of that possibility in the FAQ.

If you look back at one of his old posts you can see that his name is crossed out. That means that the account has been canceled, either by him or by an admin.

 

[Dale]

But since previous attempts to go over the math, did not bridge for me the gap between words and math, I am trying to remember things also by using words.

You have said this many times. Frankly, it is sounding more and more like an excuse, but that may be partly my own bias where people objecting to doing the math always seems like one of my kids trying to avoid doing their schoolwork.

Say you need to formalize in math, what is considered a physically impossible situation, regarding the crystal arrangement, that the detectors in the moving apparatus, don't turn on together, is that considered:

Just do it and see what you get. Simply follow the steps I laid out above. You will see that it all works out correctly.

 

[harrylin]

DaleSpam thank you,
I might go over the math as you and harrylin suggest.

But since previous attempts to go over the math, did not bridge for me the gap between words and math, I am trying to remember things also by using words.

Did you re-read our conversation as I suggested?

And my question in words now, is this:

Say you need to formalize in math, what is considered a physically impossible situation, regarding the crystal arrangement, that the detectors in the moving apparatus, don't turn on together, is that considered:

1. Impossible to formalize as a mathematical transformation.

2. Plainly going back to the Galilean transformation.

3. It is possible, and it is not a Galilean transformation, but this kind of transforation is called... (If this kind of option exists, and
it does not appear in the crystal arrangment, does in appear, in any physical phenomenon at all? not as a mathematical
approximation, or as a mathematical approximation as the Galilean transformation is considered to be?)

4. Neither, I don't understand (even in words) what a transformation is, or some other answer?

Thanks,
Roi.

"Say you need to formalize in math, what is considered a physically impossible situation,"

I consider that a faulty proposition. Only physically possible situations should be considered, for example the last drawing you made, it describes what happens in practice. The puzzle is then of the kind that we described, and we gave you the mathematical solution of relativity. As a matter of fact, we gave you:

1. the correct formulation of the problem and
2. the answer and
3. how to do the calculation

From your post it's not clear to me if you know the Galilean transformation or if you only know the sound bite "Galilean transformation". If it is the last, don't despair!

- Can you solve Pythogoras and so on? Then you know enough mathematics.
- Do you know how to calculate the time you need to go from A to B on a boat that sails at a certain speed on a river which flows at another speed? If so, then you know enough physics. Then you can also do the basic special relativity calculations! Harald

 

[harrylin]

If you look back at one of his old posts you can see that his name is crossed out. That means that the account has been canceled, either by him or by an admin.

Thank you! I thought that the line through his name was a fancy way he wrote it, I had not noticed before that it was not crossed out.

 

[John232]

And now harrylin wonders what happened to thwle!

For not only did he suddenly stop posting (could be the flu or a little trip), but also there is no such thread and, to top it off, it is impossible to contact him!
and I can't even find mention of that possibility in the FAQ.

If he was talking like that i can assure you he just got banned. In effect, that was what the M&M experiment did, and we know that the Earth is not at rest. I don't even think they picked up any of the acceleration that would cause curvature in the beam of light in the M&M experiment or both beams in each direction where curved by the same amount, I am not sure...

 

[harrylin]

[..]
One could say that spacetime is the medium that allows light to travel, but then spacetime is warped so that light always travels at the same speed.
[..]
Spacetime has this strange connection with photons that forces it to be bent and warped just so that the photon can say it is traveling the speed of light relative to everything else at the same time.

You can look it that way and use Minkowski's spacetime concept. Or you can stick with Lorentz's stationary ether concept with light waves (although photonic ones), which just as much complies with special relativity and experimental results.

Then the strangeness is gone and there is no warping (in special relativity), instead the relativistic behaviour is straightforward and easy to understand. Too boring perhaps?

 

[John232]

You can look it that way and use Minkowski's spacetime concept. Or you can stick with Lorentz's stationary ether concept with light waves (although photonic ones), which just as much complies with special relativity and experimental results.

Then the strangeness is gone and there is no warping (in special relativity), instead the relativistic behaviour is straightforward and easy to understand. Too boring perhaps?

I am not sure I am following you. How does that lead to no warping? I was just saying that they thought light had to have had a meduim to travel in because of its wavelike properties. It would have to have something to wave through as what they thought and is the reason why they came up with the aether theory. If aether was to equal spacetime then there would have to be warping of spacetime to comply with results. Any aether theory that attempted to comply with results would have to warp in much the same way as spactime would in relativity. Spacetime also affects waves traveling across the universe, they get stretched out along with the space they are traveling in. So to me spacetime and aether is the same thing, but the aether has to be warped by relativistic standards.

 

[harrylin]

I am not sure I am following you. How does that lead to no warping? I was just saying that they thought light had to have had a meduim to travel in because of its wavelike properties. It would have to have something to wave through as what they thought and is the reason why they came up with the aether theory. If aether was to equal spacetime then there would have to be warping of spacetime to comply with results. Any aether theory that attempted to comply with results would have to warp in much the same way as spactime would in relativity. Spacetime also affects waves traveling across the universe, they get stretched out along with the space they are traveling in. So to me spacetime and aether is the same thing, but the aether has to be warped by relativistic standards.

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!

 

[Dale]

So to me spacetime and aether is the same thing

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.

 

[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!