Invisibility of the Lorentz Contraction

Discussions on classical and modern physics, quantum mechanics, particle physics, thermodynamics, general and special relativity, etc.

Rotating Ring of telescopes: bangstrom scenario

Postby DJ_Juggernaut on July 16th, 2018, 11:03 pm 

bangstrom: I considered your rotating telescope setup. You are correct that the aberration angle will be the same for all telescopes, if a ring of telescopes rotate or circle around a point source. None of the telescopes point to the center. But that's not my scenario. You could start a new thread and I will respond there.
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Re: Three locations

Postby bangstrom on July 17th, 2018, 1:25 am 

DJ_Juggernaut » July 16th, 2018, 9:55 pm wrote:bangstrom,

JMPs graphic is correct. His interpretation is incorrect. There are three positions to consider here:

Laser location 0
Laser location 1
Laser location 2

He conflates, laser location 1 with laser location 0. That's the source of his (and possibly your) confusion. This graphic highlighting these three locations should help fix the confusion. Let me know if it worked for you.


Try this explanation. There is only one position to consider here:
Laser location 0.

The position of the laser at the instant of emission is l,0. The position of the telescope at the instant of absorption is L,2. The true path of the photon is l,0 to L,2. The "false" position of the laser from the telescope is l,0. Positions l,2 and l,3 are irrelevant to the observations because it makes no difference where the laser went after its one emission.
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False location - what does it mean?

Postby DJ_Juggernaut on July 17th, 2018, 3:57 am 

bangstrom wrote:The true path of the photon is l,0 to L,2. The "false" position of the laser from the telescope is l,0.

No, the true path is shown in bold lines. l1 to L2. The false path is called false because it did not come from there. Hence it is shown in dotted lines. And is called "apparent" for that reason. Whereas, the photon did come from l1 to L2. You cannot describe the path of a photon arriving from l1 to L2 as 'apparent or false path' because that would be a "false" statement. Would it not?

Three locations Graphic
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Re: False location - what does it mean?

Postby bangstrom on July 17th, 2018, 12:32 pm 

DJ_Juggernaut » July 17th, 2018, 2:57 am wrote:
bangstrom wrote:The true path of the photon is l,0 to L,2. The "false" position of the laser from the telescope is l,0.

No, the true path is shown in bold lines. l1 to L2. The false path is called false because it did not come from there. Hence it is shown in dotted lines. And is called "apparent" for that reason. Whereas, the photon did come from l1 to L2. You cannot describe the path of a photon arriving from l1 to L2 as 'apparent or false path' because that would be a "false" statement. Would it not?

Three locations Graphic


The telescope doesn’t see the light coming straight down from l1 because the telescope is moving to the right as the light is moving down. The telescope doesn’t see the laser light coming from where the laser is at the instant the light arrives. The telescope sees the light coming from where the laser was at some time in the past when the light was emitted so the telescope can see light coming from l0 but not l1.

The same is true of celestial objects. We don’t see the stars where they are now but where they were years ago. We may still see stars that went nova centuries ago because light from their nova is still on its way to Earth.
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Re: False location - what does it mean?

Postby DJ_Juggernaut on July 17th, 2018, 2:01 pm 

bangstrom wrote:The telescope doesn’t see the light coming straight down from l1 because the telescope is moving to the right as the light is moving down.

It does enter a telescope. Bradley discovered this a while ago. I showed you an animation earlier. You might wanna give it another look. All that matters is, do you agree with the angle of the incoming photon in the laser frame. If you tell me its angle, we can tell you the tilt in the telescope frame.

So, what is the angle of the photon in the laser frame?
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Aberration

Postby DJ_Juggernaut on July 17th, 2018, 2:31 pm 

bangstrom wrote:The telescope sees the light coming from where the laser was at some time in the past when the light was emitted so the telescope can see light coming from l0 but not l1.

l0 and l1 are coordinates. If light was emitted at l1, then all frames agree that a photon was emitted at l1. You can't change this just because you move the coordinates.

If l0 and l1 were stationary, and a photon was emitted at l1, then all frames agree with this. Where they were where they would be, is meaningless. This is why we label diagrams. Once a photon is emitted at l1. We say it was emitted at l1. If it was absorbed at L2. All frames agree it was absorbed at L2. At the time of absorption if a telescope points at l0, then all frames agree that it points at l0.

Moving the coordinates that were stationary in a scenario makes no difference. The events that occurred do not change when you change frames. Do you dispute the angle of the photon in the source's frame? Maybe that is the source of your confusion.

Without question, it is 90 degrees in the source's frame. This is why the equation works in both frames. This is why the telescope is tilted in both the frames. If it were not 90 degrees, the tilting angle would be different.

Questions for you:

What is θ? (in laser frame)
What is v? (relative velocity)

If θ = 90 degrees, the tilting angle of the telescope ϕ is given by tan(ϕ) = v/c. True in all frames.

What do you disagree with here?

Aberration - true and false positions
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Relativistic beaming does not occur

Postby DJ_Juggernaut on July 17th, 2018, 4:57 pm 

In the source frame, a sphere of light expands spherically, whether the source moves or not. The value of θ in the aberration formula considers the angle of photons as viewed in the source's frame. Which means, the sphere of light looks the same or expands the same, whether the source moves relative to an observer or not.

θ = angle of a photon in the source's frame.
v = relative velocity between source and a telescope.
ϕ = tilting angle of the telescope.

Therefore, in my laser setup with a ring of telescopes, the value of θ for each laser is fixed whether the lasers move or not. This means, relativistic beaming does not occur.

Lasers and a ring of telescopes
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Re: Aberration

Postby bangstrom on July 17th, 2018, 7:35 pm 

DJ_Juggernaut » July 17th, 2018, 1:31 pm wrote:
Questions for you:

What is θ? (in laser frame)
What is v? (relative velocity)

If θ = 90 degrees, the tilting angle of the telescope ϕ is given by tan(ϕ) = v/c. True in all frames.

What do you disagree with here?

Aberration - true and false positions


I disagree with the position of the laser at the instant of the flash. It should be at l,0 instead of l,1.

If a duck hunter sees two ducks flying from his left to right and he fires 90 degrees straight ahead at the first duck, He won’t hit the first duck because it will have moved on by the time the shot reaches its position but he might hit the second duck if the second duck flies into the 90 degree path of the shot. This is the point where the first duck was when the shot was fired.

If the light flash strikes the telescope at L2 then the light must have come from a forward position which is l,0. If the laser source is stationary, the laser will be at l,1 and 90 degrees from the telescope at the moment the light strikes at L2 but the light will have come from l,0.
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Re: Aberration

Postby DJ_Juggernaut on July 18th, 2018, 2:06 am 

bangstorm: The telescope is a duck. Your gun fires from l1 at 90 degrees. Since the duck is moving to the right from L1, you aim where the duck would be, at 90 degrees. It would be at L2. Your 90 degree bullet would hit the duck. Or telescope

--

With regards to telescope and lasers: Photon was emitted at l1, your telescope was at L1 at the time of emission. After emission, it reaches L2. The telescope will have moved to l2. It's all there in the graphic. If you want to change the diagram. Make a new one. I will comment on that. As far as my diagram goes. It's accurate.
Last edited by DJ_Juggernaut on July 18th, 2018, 2:34 am, edited 2 times in total.
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Re: Invisibility of the Lorentz Contraction

Postby DJ_Juggernaut on July 18th, 2018, 2:27 am 

What you might be missing is this. The sphere of light scenario is usually viewed from the perspective of a stationary source. It's always spherical in its frame. It's the telescopes that move in through this expanding sphere. This is why θ is the same whether you move or not. Or which way you move, right or left. Pick any ray in the sphere of light. Get its θ, you can work out the tilting angle ϕ of your telescope if you're moving. If you're moving to the right, you tilt your telescope to your right. If you're moving to the left, the telescope is tilted to the left.

Relativistic beaming does not occur for any observer. The value of θ is fixed for all observers moving relative to an expanding sphere of light.
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Re: Invisibility of the Lorentz Contraction

Postby bangstrom on July 18th, 2018, 3:19 am 

DJ_Juggernaut » July 18th, 2018, 1:27 am wrote:What you might be missing is this. The sphere of light scenario is usually viewed from the perspective of a stationary source. It's always spherical in its frame. It's the telescopes that move in through this expanding sphere. This is why θ is the same whether you move or not. Or which way you move, right or left. Pick any ray in the sphere of light. Get its θ, you can work out the tilting angle ϕ of your telescope if you're moving. If you're moving to the right, you tilt your telescope to your right. If you're moving to the left, the telescope is tilted to the left.

Relativistic beaming does not occur for any observer. The value of θ is fixed for all observers moving relative to an expanding sphere of light.


I get where you are coming from now. I read the 90 degree straight line from l,1 to L2 as indicating that you were firing the laser for the telescope when it was at L2 but it was actually at L1 prior to reaching L2 so you were giving it lead.
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Relativistic beaming does not occur

Postby DJ_Juggernaut on July 18th, 2018, 8:29 am 

bangstrom wrote:I get where you are coming from now.

I take it that you are now able to make sense of this JMP graphic then?
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Re: Invisibility of the Lorentz Contraction

Postby bangstrom on July 18th, 2018, 1:33 pm 

Yes, the graphic and the spherical frame.
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Relativisitic Law of Reflection

Postby DJ_Juggernaut on July 18th, 2018, 10:22 pm 

Now that we agree that θ is the same for all frames, think that a 90 degree (θ) photon lands on a 45 degree mirror that is moving to the right. What is the angle of reflection? Isn't it the same as the angle of incidence?

Angle of incidence = 45 degrees
Angle of reflection = ?

Isn't it the same? If so, Relativistic Law of Reflection, is not correct. Right?
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Re: Invisibility of the Lorentz Contraction

Postby bangstrom on July 19th, 2018, 2:09 am 

A mirror moving at relativistic speed would appear rotated beyond the 90 degree angle that it would have if stationary. That's a complicated problem but why do you think the angle of incidence would not be the same as the angle of reflection?
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Invisibility of relativistic law of reflection

Postby DJ_Juggernaut on July 19th, 2018, 6:17 am 

bangstrom wrote:A mirror moving at relativistic speed would appear rotated beyond the 90 degree angle that it would have if stationary. That's a complicated problem...

Not sure about that. The rotation you speak of, is probably a misinterpretation of aberration of light. Relativistic beaming does not occur. Θ is fixed for all frames, whether the source frame moves or not. Or if the observe frame moves or not. Θ is Θ. However, ϕ changes with relative velocity. As I've said before, ϕ is the aberration angle, aka, the tilting angle of a moving telescope.

If the rotation you speak of is, due to relativistic beaming, it's false.

bangstrom wrote:but why do you think the angle of incidence would not be the same as the angle of reflection?

It would be equal in the 45 degree mirror frame. But the source frame would disagree and say the 45 degree mirror angle is not 45 degrees, due to length contraction.

Therefore, they both disagree on the angle of reflection, (while agreeing the angle of incidence is the same). Mirror frame says, the angle of reflection is 45 degrees and the source frame says, the angle of reflection is greater than 45. Who is right?

I suspect an observation made in the 45 degree mirror frame must hold for all frames. Hence, relativistic law of reflection must be wrong.

Also, 'relativistic law of reflection' is a manifestation of 'length contraction'. One can't be right or wrong without the other. Add to that you can't "see" length contraction. So the odds stack up. I suspect they are both wrong.
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Re: Invisibility of the Lorentz Contraction

Postby bangstrom on July 19th, 2018, 1:38 pm 

A 90 degree angle in any frame should be the one that reflects light back from its apparent source or a beam of light from the apparent source should reflect light back to the real source.
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Re: Invisibility of the Lorentz Contraction

Postby DJ_Juggernaut on July 19th, 2018, 9:37 pm 

I agree with your definition of what a 90 degree photon is. But this is a simple problem, even though this really involves "seeing". What is Θ in the laser frame? This is the angle of incidence for the 45 degree mirror frame and the source frame. Even Einstein's aberration formula agrees with this. So,

In the mirror frame: (no length contraction)
Angle of reflection = 45

In the source frame: (there is length contraction)
Angle of reflection = greater than 45

Who is right? I prefer the mirror frame observation over the source frame's indirect invisible inference.
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Re: Invisibility of the Lorentz Contraction

Postby bangstrom on July 19th, 2018, 10:49 pm 

There is no length contraction at the local level of either frame. The source frame appears contracted and in motion from the (inertially at rest) perspective of the moving telescope. It works both ways. There is no preferred frame of reference.
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Re: Invisibility of the Lorentz Contraction

Postby DJ_Juggernaut on July 20th, 2018, 4:38 am 

I understand what you're saying. I guess an experiment here would be decisive. I looked up an experiment performed by Michelson in 1913. He reported that the angle of reflection on a moving mirror is equal to the angle of incidence.

It appears therefore that within the limit of error of experiment (say 2 per cent) the velocity of a moving mirror is without influence on the velocity of light reflected from its surface.

That should settle it then.
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Re: Invisibility of the Lorentz Contraction

Postby bangstrom on July 21st, 2018, 2:15 am 

Unfortunately we don’t have a mirror or any other mechanism that can tell us if an object is truly in motion or not. All motion remains relative just as Galileo said it was.

On a personal and unrelated note, I usually don’t support discussions involving photons because I think photons are purely imaginary unless they are understood as the amount of energy in a light related event and nothing more. I don’t care for photon theory where light travels through space as either a particle or a wave but photon theory usually works as an explanation for light and that is its value. But it can be misleading.
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Invisibility of a photon

Postby DJ_Juggernaut on July 21st, 2018, 7:14 am 

bangstrom » July 21st, 2018, 2:15 am wrote:Unfortunately we don’t have a mirror or any other mechanism that can tell us if an object is truly in motion or not. All motion remains relative just as Galileo said it was.

Galileo's relativity does not apply to photons. Theoretically you can determine who is truly moving with a photon and a telescope. If your telescope is truly moving relative to a source, it needs tilting. If the source is truly moving, relative to the telescope, the telescope does not need tilting.

bangstrom wrote:On a personal and unrelated note, I usually don’t support discussions involving photons because I think photons are purely imaginary unless they are understood as the amount of energy in a light related event and nothing more. I don’t care for photon theory where light travels through space as either a particle or a wave but photon theory usually works as an explanation for light and that is its value. But it can be misleading.

Photons are kind of imaginary, already. All elementary particles in the standard model, including, a photon have a dimension of: zero. How we still "see" a photon is just mind-boggling.
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Re: Invisibility of a photon

Postby bangstrom on July 22nd, 2018, 5:30 am 

DJ_Juggernaut » July 21st, 2018, 6:14 am wrote:
Galileo's relativity does not apply to photons. Theoretically you can determine who is truly moving with a photon and a telescope. If your telescope is truly moving relative to a source, it needs tilting. If the source is truly moving, relative to the telescope, the telescope does not need tilting.

Photons are kind of imaginary, already. All elementary particles in the standard model, including, a photon have a dimension of: zero. How we still "see" a photon is just mind-boggling.


Galileo’s relativity doesn’t apply to imaginary things like photons but it does apply to material objects like a signal source and a receiver. Either one can serve as an “at rest” source which makes the other an “in motion” receiver.

We don’t “see” photons. We see electrons losing and gaining energy but we never see anything existing or happening between a signal and its reception. Light appears to be electrons interacting by direct resonance with the loss of energy from one atom resulting in the instant gain of energy in another atom no matter how remote.

Space and time, as we perceive it, does not exist for light but all distances in our observer-space are also observer-times in the ratio of the dimensional constant c so what is instant and direct at the light level is observed by us as a separation in distance d and an observer-time of d/c. Quantum mechanics does not require a direct physical contact for one particle to affect another so the photon intermediary is unnecessary.
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Re: Invisibility of a photon

Postby DJ_Juggernaut on July 22nd, 2018, 9:49 pm 

bangstrom wrote:Galileo’s relativity doesn’t apply to imaginary things like photons but it does apply to material objects like a signal source and a receiver. Either one can serve as an “at rest” source which makes the other an “in motion” receiver.

That's a bingo. A material object's motion relative to an incoming or an outgoing photon of some angle (90 degrees), gives you your object's true velocity.

bangstrom wrote:Light appears to be electrons interacting by direct resonance with the loss of energy from one atom resulting in the instant gain of energy in another atom no matter how remote.

Interesting theory. You have a link on this or something?
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Re: Invisibility of a photon

Postby bangstrom on July 24th, 2018, 6:14 am 

DJ_Juggernaut » July 22nd, 2018, 8:49 pm wrote:
bangstrom wrote:Galileo’s relativity doesn’t apply to imaginary things like photons but it does apply to material objects like a signal source and a receiver. Either one can serve as an “at rest” source which makes the other an “in motion” receiver.

That's a bingo. A material object's motion relative to an incoming or an outgoing photon of some angle (90 degrees), gives you your object's true velocity.


The information obtained by the “moving” object would be a simple vector in space so how would they convert that to a velocity?

DJ_Juggernaut » July 22nd, 2018, 8:49 pm wrote:
bangstrom wrote:Light appears to be electrons interacting by direct resonance with the loss of energy from one atom resulting in the instant gain of energy in another atom no matter how remote.

Interesting theory. You have a link on this or something?


The most complete form of the theory is John Cramer's "Transactional Interpretation of Quantum Mechanics." TIQM.

The following article by Carver Mead is a short summary with history.

http://cns.caltech.edu/people/faculty/m ... hotons.pdf

10. A “photon” can be best viewed as a transaction between two atoms on the same light cone. Such a transaction requires an exquisite degree of phase matching between the quantum states of the two coupled atoms, and is therefore a rare event. Photon statistics result from the chance correlations of atomic wave function phases.

11. The “photon” transaction can be viewed as a brief entanglement of the quantum states of the two participating atoms.
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Postby DJ_Juggernaut on July 24th, 2018, 11:39 am 

bangstrom wrote:The information obtained by the “moving” object would be a simple vector in space so how would they convert that to a velocity?

A telescope and two parallel mirrors. Θ = 90 degrees. The tilt angle gives you the velocity v by tan(Θ - ϕ) = v/c

bangstrom wrote:11. The “photon” transaction can be viewed as a brief entanglement of the quantum states of the two participating atoms.

IMO, entanglement stems from a misinterpretation of the Stern-Gerlach experiment.
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Re:

Postby bangstrom on July 24th, 2018, 1:48 pm 

DJ_Juggernaut » July 24th, 2018, 10:39 am wrote:
A telescope and two parallel mirrors. Θ = 90 degrees. The tilt angle gives you the velocity v by tan(Θ - ϕ) = v/c

Where are the two mirrors and how does a moving object know its true angle from the light source?
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Re: Re:

Postby DJ_Juggernaut on July 25th, 2018, 4:02 am 

bangstrom » July 24th, 2018, 1:48 pm wrote:Where are the two mirrors and how does a moving object know its true angle from the light source?

The telescope is on one of the mirrors. If the mirrors are moving, the telescope needs to be tilted. All observers agree that Θ = 90 degrees and that the tilt angle gives the velocity v of the mirrors relative to the path of a photon, via tan(Θ - ϕ) = v/c.

If the mirrors are not moving relative to the path of a 90 degree photon, there is no need for the tilting of the telescope. All observers will agree with this too.
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What is up or down?

Postby DJ_Juggernaut on July 25th, 2018, 8:14 am 

DJ_Juggernaut wrote:IMO, entanglement stems from a misinterpretation of the Stern-Gerlach experiment.

An unrelated note: I see the Stern-Gerlach experiment as a manifestation of the Lorentz force. A charged particle moving in a magnetic field. It will either go up or down. And that's that. Nothing new there, imo.
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Re: What is up or down?

Postby bangstrom on July 26th, 2018, 3:01 am 

DJ_Juggernaut » July 25th, 2018, 7:14 am wrote: IMO, entanglement stems from a misinterpretation of the Stern-Gerlach experiment.

An unrelated note: I see the Stern-Gerlach experiment as a manifestation of the Lorentz force. A charged particle moving in a magnetic field. It will either go up or down. And that's that. Nothing new there, imo.


The Stern-Gerlach experiment demonstrates that two electrons generated by the same event will have opposite spin states and that is nothing new.

The question is, ‘Where was the spin state decided?’ Was it decided when the two electrons first went their separate ways or was it decided when the spin state of the first electron was measured? Bell’s inequality is a statistical test to determine where the spin state was decided and the results show that the spin state was determined when the first electron was observed rather than being present from the start.

This is counter intuitive and contrary to the EPR thought experiment because it demonstrates that observing a particle in one location can instantly determine the quantum state its entangled partner even though they may be far apart. This is evidence of quantum entanglement when observing the state of one particle simultaneously determines the state of another particle.

Bell’s inequality test has been used for a number of different reproducible experiments involving entanglement besides the S-G experiments and the results are statistically valid so I don’t see a reason to doubt them. Entanglement explains how light appears to be prescient of its destination and it best explains the results of quantum experiments such as tests of Wheeler’s delayed choice or quantum erasure.
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