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.