Is Gravity Faster than Light?

[Write4U]

Glass half full in a vacuum...
Electromagnetic Harmonics... etc ?

should we be able to see/measure Light being affected by Gravity ?
(theory?) if yes, why can we not determine a speed from that observation ?(postulating question)

The speed of light is a constant. It can slow down travelling through a medium but gravity is not a medium, it is a warp in spacetime. Gravitational pull can bend light but that does not mean light slows down.

I believe someone else made a prior post that passing through a gravity field does not slow anything down. The net momentum is preserved. I'm not sure if I posited this correctly, but that was my understanding. In fact, for massive objects gravity may well be used for a sling shot effect, actually using gravity to speed up an object, except light of course.

Black Holes may be different, but the lensing effect (caused by gravity) does not interfere with the speed of light, IMO..

 

[Write4U]

A non sequitur question. In the standard Feynman path-integral approach, a single particle interferes with itself in taking all possible paths from slits to screen. Any individual screen hit is more or less random but on average, the path(s) generating the least mutual destructive interference are most likely and will correspond to the interference fringes actually observed to build up over a long run of such individual events.

But I thought that as soon as the photon hits the plate its wave function collapses.
So how can you arrive at a conclusion that there are residual interference fringes if the original wave function has collapsed altogether?

Are you proposing that the wave function persists after it has collapsed?

If there is any sign of residual wave function, that would tend to support Bohm's Pilot Wave, which continues independent of the collapse of a particle's own wave function.

 

[RainbowSingularity]

The speed of light is a constant. It can slow down travelling through a medium but gravity is not a medium, it is a warp in spacetime. Gravitational pull can bend light but that does not mean light slows down.

I believe someone else made a prior post that passing through a gravity field does not slow anything down. The net momentum is preserved. I'm not sure if I posited this correctly, but that was my understanding. In fact, for massive objects gravity may well be used for a sling shot effect, actually using gravity to speed up an object, except light of course.

Black Holes may be different, but the lensing effect (caused by gravity) does not interfere with the speed of light, IMO..

Thanks, i am quite interestedin this concept.

soo if the speedof light is not effected by gravitational lensing... is time altered ?
thus is there a different time other than the basic light speed measurement relative to distance and visual anotation of the observer ?
i.e
observer looking through gravitational lensing at a planet...
sees things 1000 years ago
without gravitational lensing sees things 1200 years ago on the same planet ?

 

[Write4U]

Write4U said,
...OTOH photons are much smaller than most particles with rest mass, and easily escape the dense sea of air surrounding the earth. We could draw a relative comparison between a photon surrounded but unaffected by air and smaller massless particles such as fermionsand bosons surrounded but not be affected by the Higgs field....
Write4U, practically every speculative claim in #60 & #62 is wrong or misleading at best. Above is highlighted just a few glaring examples. Far better to ask questions rather than assert falsehoods. Because this is SF it all falls on deaf ears anyway so no great harm is done, but it's always best to consult an authority who really understands the relevant subject matter. Re Higgs:
https://profmattstrassler.com/artic...higgs-field-works-with-math/1-the-basic-idea/

I prefer to do my own research, it makes for better discussions...

Let’s start by discussing “size”. Fermionic particles, for example, quarks, neutrinos, electrons - we aren’t really sure what their sizes are. However, we can know their Compton wavelengths, though. This is the wavelength of a photon with the exact same energy as when the fermionic particle is at rest.

Higher Energy ~ Smaller

One thing about photons we know for certain is that the shorter they are, the more energy they contain. And the longer they are, the less energy contained.

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λ=wavelength" role="presentation" style="display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">λ=wavelengthλ=wavelength

c=" role="presentation" style="display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">c=c=speed of light in a vacuuum

h=" role="presentation" style="display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">h=h=Planck Constant

Based on this, we can make an educated guess on which particles are larger and smaller.

Photons can be any size that is greater than around 6 Planck lengths. So, technically, they take the prize on both ends.

https://www.quora.com/What-is-smaller-a-quark-the-Higgs-boson-or-a-photon

I see no problem with post #61 in that respect. In context of the actual posit ;
More Energy = smaller size, More Mass = larger size. That's wrong?

As to post #62, please explain where it is wrong and why..... Please note that I already qualified it as speculative.
But instead of providing a correct answer, you just repeat my original qualifier. Not very helpful to me or others.

Do you know the answer? Then enlighten me, please......

 

[Q-reeus]

But I thought that as soon as the photon hits the plate its wave function collapses.
So how can you arrive at a conclusion that there are residual interference fringes if the original wave function has collapsed altogether?

Are you proposing that the wave function persists after it has collapsed?

If there is any sign of residual wave function, that would tend to support Bohm's Pilot Wave, which continues independent of the collapse of a particle's own wave function.

Some confusion there. As soon as a photon hits the (absorbing) screen aka 'plate' it is annihilated i.e. ceases to exist. It's position on the screen is partially random, in keeping with the probabilistic nature of QM, but is partially ordered in keeping with the wave interference picture. Hence the observed one-photon-at-a-time speckle pattern that starts out looking completely random, but over time reveals the interference fringes expected of a classical wave picture.
In the Feynman path integral picture, a photon takes all possible paths, through both slits, and thence to the screen. But only paths that have a phase difference close to an integer multiple of the photon wavelength will significantly reinforce and consequently have a high probability of corresponding to a screen hit.

 

[Write4U]

Some confusion there. As soon as a photon hits the (absorbing) screen aka 'plate' it is annihilated i.e. ceases to exist. It's position on the screen is partially random, in keeping with the probabilistic nature of QM, but is partially ordered in keeping with the wave interference picture. Hence the observed one-photon-at-a-time speckle pattern that starts out looking completely random, but over time reveals the interference fringes expected of a classical wave picture.
In the Feynman path integral picture, a photon takes all possible paths, through both slits, and thence to the screen. But only paths that have a phase difference close to an integer multiple of the photon wavelength will significantly reinforce and consequently have a high probability of corresponding to a screen hit.

And what happens after the entire wave function has collapsed when the photon hits the plate.?
You mentioned "residual" wave function. Of what?

 

[Q-reeus]

And what happens after the entire wave function has collapsed when the photon hits the plate.?
You mentioned "residual" wave function. Of what?

Many probably most physicists maintain a photon does not possess a wavefunction since as a massless 'particle' there is no position operator like as applies to massive particles e.g. electron. See last main para p2 this article, whose authors take a contrary view: https://arxiv.org/abs/quant-ph/0604169

At any rate once the photon is annihilated there is obviously nothing left to have any possible wavefunction. The energy-momentum of said photon has been transferred to whatever in the screen absorbed it. Could be an expelled electron, or a chemical change e.g. silver halide molecule decomposing to pure silver etc. - depending on the particular nature of detection screen.

 

[Write4U]

Many probably most physicists maintain a photon does not possess a wavefunction since as a massless 'particle' there is no position operator like as applies to massive particles e.g. electron. See last main para p2 this article, whose authors take a contrary view: https://arxiv.org/abs/quant-ph/0604169

That article assumes a wave function associated with photons. If it did not, it would answer to Bohm's Pilot Wave model.

At any rate once the photon is annihilated there is obviously nothing left to have any possible wavefunction.

Ok, I agree with that (if the wave function is associated with the photon).

The energy-momentum of said photon has been transferred to whatever in the screen absorbed it. Could be an expelled electron, or a chemical change e.g. silver halide molecule decomposing to pure silver etc. - depending on the particular nature of detection screen.

Now we enter secondary or even tertiary conditions as well as changing the chemical composition of the receiving medium. That's changing the experiment and the way you stated it, sounds speculative.
Any evidence of such phenomena?

 

[Q-reeus]

Now we enter secondary or even tertiary conditions as well as changing the chemical composition of the receiving medium. That's changing the experiment.
Any evidence of such phenomena?

How is it 'changing the experiment'? What restrictions are you imposing on the extent of the experiment? Have a careful read through:
https://www.quora.com/In-the-double...ding-is-the-wave-state-maintained-Why?share=1

 

[Write4U]

How is it 'changing the experiment'? What restrictions are you imposing on the extent of the experiment?

Consistency to assure reliable results.

Have a careful read through:
https://www.quora.com/In-the-double...ding-is-the-wave-state-maintained-Why?share=1

I'll need some time to absorb all that. But my intuition tells me that the more complicated a theory (experiment) becomes, the more complicated the proof becomes. The inclusion of mind/brain functions, introduces a completely subjective aspect.

Moreover I already introduced an example of a room filled with photons, but that requires a constant emitter. Turn the emitter (light) off and the room instantly becomes dark. i.e. all wave functions collapse instantly.

I wonder why that poster did not replace the room windows with narrow slits. IMO, that might tell us more about the double slit expariment.

But we are talking about the behavior of single photons. Thus using an example of a (any) lit room does not tell us anything about the double slit experiment.

 

[Q-reeus]

Consistency to assure reliable results.

I'll need some time to absorb all that. But my intuition tells me that the more complicated a theory (experiment) becomes, the more complicated the proof becomes. The inclusion of mind/brain functions, introduces a completely subjective aspect.

Moreover I already introduced an example of a room filled with photons, but that requires a constant emitter. Turn the emitter (light) off and the room instantly becomes dark. i.e. all wave functions collapse instantly.

I wonder why that poster did not replace the room windows with narrow slits. IMO, that might tell us more about the double slit expariment.

But we are talking about the behavior of single photons. Thus using an example of a (any) lit room does not tell us anything about the double slit experiment.

There were several different answers including one where a wavefunction is never part of the picture. None where a lit room is invoked iirc. It was the first one I thought more relevant to your outlook. I suggest thinking about that explanation some more. It may lift at least some of your continued confusion.

 

[exchemist]

Consistency to assure reliable results.



Moreover I already introduced an example of a room filled with photons, but that requires a constant emitter. Turn the emitter (light) off and the room instantly becomes dark. i.e. all wave functions collapse instantly.

This is not what happens. When the emitter is turned off, the last photons travel towards the walls as usual and are duly absorbed, as Q-reeus explained earlier by whatever it is they hit and are absorbed by. There is no instantaneous "collapse". The absorption process is well described by quantum mechanics. Both the travel of the photon and the absorption process take a finite time to occur.

 

[Write4U]

There were several different answers including one where a wavefunction is never part of the picture. None where a lit room is invoked iirc. It was the first one I thought more relevant to your outlook. I suggest thinking about that explanation some more. It may lift at least some of your continued confusion.

1. “If say an electron is being observed, but a device observing is not recording is the wave state maintained? Why?” On the assumption that the off-state device is large and warm enough to cause decoherence (it almost certainly is) then the wave state will not be maintained.​

This makes no sense to me. Is this assuming that when an obervation device is in an off-state it still causes decoherence? Does that mean that the observation device in off-state is somehow still projecting something that will cause decoherence ?​

The right side of the picture depicts a photon detector which is "attached' to the slits by leads and thus draws energy from the photon as it passes the slit, clearly removing its wave function, but replacing it with an electromagnetic "thread", which is a different state altogether.
Seems to me;
Even if you turned photon detector "off", but leave the leadst connected with the slits, the configuration will still produce a field which affects the energy of the photon and possibly decoherence. I should like to know what happens when you remove the leads from the slits altogether. Will the detector still interfere with the photon?

2. When observed - say a photon interacts with the particle to measure it thus causing the collapse. In normal experimental conditions in a lit room aren't photons always interacting? Only on large objects will these interactions cause decoherence, but you have to do the maths to find out exactly how — and this is the phenomenon that those poor engineers trying to maintain coherence in quantum computers are wrestling with.​

https://www.quora.com/In-the-double...ding-is-the-wave-state-maintained-Why?share=1​

Thus, IMO, it is not the observation device itself which is causal to the wave collapse. Its the method, the attached leads, which results in the collapse at the slits.

Can they not use some gas which will show the interference pattern without needing a connection to the slits and thus avoid adding (another field) to the configuration of the slits?

Perhaps this might also reveal what happens in a large area such as a room.

 

[Write4U]

This is not what happens. When the emitter is turned off, the last photons travel towards the walls as usual and are duly absorbed, as Q-reeus explained earlier by whatever it is they hit and are absorbed by. There is no instantaneous "collapse". The absorption process is well described by quantum mechanics. Both the travel of the photon and the absorption process take a finite time to occur.

OK, I can visualize that.
If we were to cover all walls with black photographic plates, would the walls change color to white or show a decoherent scattering of white dots?
How many photons would it take? If we used a very small light source or limit the time of emision, would the walls turn gray?

I am drawing on my experience with B/W photography. Intensity (energy) of light and exposure time both affect the results when developing a negative which determines the black and white shading on the photograph.
I assume this problem of negative to positive has been eliminated by our digital cameras, so that extra step can be ignored.

 

[exchemist]

OK, I can visualize that.
If we were to cover all walls with black photographic plates, would the walls change color to white or show a decoherent scattering of white dots?
How many photons would it take? If we used a very small light source or limit the time of emision, would the walls turn gray?

I am drawing on my experience with B/W photography. Intensity (energy) of light and exposure time both affect the results when developing a negative which determines the black and white shading on the photograph.
I assume this problem of negative to positive has been eliminated by our digital cameras, so that extra step can be ignored.

These are remarkably silly questions for someone claiming to be familiar with photography. I can't be bothered to answer.

 

[Write4U]

These are remarkably silly questions for someone claiming to be familiar with photography. I can't be bothered to answer.

But earlier, I asked a serious question and received no answer either.......

Can they not use some gas which will show the interference pattern without needing a connection to the slits and thus avoid adding (another field) to the configuration of the slits?

Anyone?

 

[Q-reeus]

But earlier, I asked a serious question and received no answer either.......

Write4U said: ↑
Can they not use some gas which will show the interference pattern without needing a connection to the slits and thus avoid adding (another field) to the configuration of the slits?

Anyone?

Can't really follow the logic there but will remark that assuming the gas strongly interacts with a photon initially enroute to the screen (photon absorption then random emission of another photon by a gas molecule), the screen is then irrelevant. No interference pattern could be expected.

 

[Write4U]

Can't really follow the logic there but will remark that assuming the gas strongly interacts with a photon initially enroute to the screen (photon absorption then random emission of another photon by a gas molecule), the screen is then irrelevant. No interference pattern could be expected.

I was thinking of a medium (gas) that does not interfere with the photon and it's interference patterns can actualy be demonstrated and can be measured without interfering with the photon itself.

 

[Q-reeus]

I was thinking of a medium (gas) that does not interfere with the photon and it's interference patterns can actualy be demonstrated and can be measured without interfering with the photon itself.

Can't envisage how that would work. Many double slit experiments have been carried out - in normal air filled lab settings. The air molecules only weakly interact via tiny collective dipole-moment fluctuations. That amount to no more than a very small departure from unity relative permittivity of a perfect vacuum. Since there is no inhomogeneous bias in such permittivity shift, the net effect is zero i.e. experiment might as well be done in perfect vacuum.
And btw this is getting right away from OP topic - agreed?

 

[Write4U]

Can't envisage how that would work. Many double slit experiments have been carried out - in normal air filled lab settings. The air molecules only weakly interact via tiny collective dipole-moment fluctuations. That amount to no more than a very small departure from unity relative permittivity of a perfect vacuum. Since there is no inhomogeneous bias in such permittivity shift, the net effect is zero i.e. experiment might as well be done in perfect vacuum.
And btw this is getting right away from OP topic - agreed?

Thanks for that explanation and your patience...

And I agree, but to me it is valuable info about the nature of light (photons), which is part of the OP question.

Next comes Gravity.......