Jump to content

The Nature of Time


scuddyx

Recommended Posts

2 hours ago, scuddyx said:

As mass-less particles travel at the speed of light I make the assertion that they would experience zero time.  Is there an error with this logic?

Are photons the only massless particles?  Also, consensus is that a photon is massless but isn't there still the possibility that it has a miniscule undiscovered mass in which case it travels less than c?

Link to comment
Share on other sites

15 minutes ago, MigL said:

and no.

Better tell these guys.

https://www.researchgate.net/publication/321589753_Are_photons_really_massless

"There are good theoretical reasons to believe that the photon mass should be exactly zero, but there is no experimental proof of this belief. Physicists have not stopped on assumption of massless."

https://www.sciencedirect.com/science/article/pii/S0370269316301083

"When setting an upper limit on the photon mass..." 

Why have such a fancy way of saying zero?  :)

I only mentioned a photon having mass as it would certainly get rid of the singularity, right?  I don't know anybody that likes dividing by zero.

Link to comment
Share on other sites

Of course; we can't measure zero.
We can only set an upper limit on the mass,

A lot of Physics would no longer be valid if photons had mass and didn't move at exactly c .
You do know that 'c' is an integral part of Maxwell's equations which govern electromagnetism, and the propagation speed of light comes out of those equations ?
If photons had mass, that means they could be accelerated, and STOPPED. Call me when you  you re-write Maxwell's equations to incorporate a variable speed of propagation.

Edited by MigL
Link to comment
Share on other sites

3 hours ago, scuddyx said:

...but he has been influenced by a wide range of spiritual works including Buddhism (Wiki).  Tolle's philosophy is based on the nature of time. 

Hard to comment without knowing his work, but it sounds like he was talking about the conscious experience of time, rather than the actual physics of it. If you want to study the physics of it, then do so. And if you want to use your experience of time in your spiritual practice, then do so. But chasing them both, thinking they are the same thing - it looks like you're chasing your tail - which is fine as long as you know you're not going to get anywhere. 

Link to comment
Share on other sites

3 hours ago, Huckleberry of Yore said:

Better tell these guys.

https://www.researchgate.net/publication/321589753_Are_photons_really_massless

"There are good theoretical reasons to believe that the photon mass should be exactly zero, but there is no experimental proof of this belief. Physicists have not stopped on assumption of massless."

https://www.sciencedirect.com/science/article/pii/S0370269316301083

"When setting an upper limit on the photon mass..." 

Why have such a fancy way of saying zero?  :)

I only mentioned a photon having mass as it would certainly get rid of the singularity, right?  I don't know anybody that likes dividing by zero.

Another way of stating the first part of MigL’s answer is that error bars are a necessary part of any experimental result.

Link to comment
Share on other sites

On 1/14/2020 at 11:07 PM, studiot said:

I really don't understand this.

I omitted the diagram from my post so I went back in plenty of time and uploaded it as an edit.
I checked that the diagram was in place, but now it is gone.

time1.jpg.da585df08553d49444cc463523864215.jpg

 

This should help make sense of my post.

Thanks for the diagram.  Does it assume Classical Time pre-exists and then looks at the variation in V?

Perhaps Time is an emergent phenomenon that can be explained from the decoherence of the wave function created at the Big Bang.

Time is real and exists but not as a hard arena that Classical Physics assumes. 

Link to comment
Share on other sites

1 hour ago, scuddyx said:

 Perhaps Time is an emergent phenomenon that can be explained from the decoherence of the wave function created at the Big Bang.

!

Moderator Note

You have claimed this several times without explanation. It seems you have some alternative take on this, so I have moved it to speculations. Now: either expand on what you mean and answer questions that have been put to you, or this gets locked.

 
Link to comment
Share on other sites

2 hours ago, scuddyx said:

Thanks for the diagram.  Does it assume Classical Time pre-exists and then looks at the variation in V?

Perhaps Time is an emergent phenomenon that can be explained from the decoherence of the wave function created at the Big Bang.

Time is real and exists but not as a hard arena that Classical Physics assumes. 

 

@swansont scuddyx promised to answer me and has kept his word so please let the thread run at least a little longer.

 

First let me say thank you for continuing the conversation and explain tha V represents the quantity etc under scrutiny for change.
I know I labelled a time axis but I would like to consider the implications of not having a formal axis at all and then show that we can't do without it.

This is quite different from an emergent phenomenon (as I understand the term) which arises as a result of specific and particular circumstances with a particular collection of objects.
It cannot be linked to other phenoma taking place elsewhere (ie is independent of them).

A good example is the action of an arch bridge.
The massive strength of an arch bridge only occurs with a particular configuration of the arch stones and only when they are all there.
The arch will not stand up, let alone support load, untill all parts are in place. If one is removed or missing the arch will collapse.
A variation on this which shows that the remarkable emergent phenomenon is the fact that the shape causes the apllied force to be diverted through exactly one right angle is the ring.
Load a part ring radially and it will bend. Apply that same load to a complete ring and the radial force will be turned into a circumferential one.

To return to my diagrams.

Say we repeatedly sample the variable under scrutiny, V.

When change occurs we expect continuity in V.
That is there is always some value for V (including zero) and there is never a sample for which there is no value.
There are no gaps in the continuum, and every value is distinct and different so V1 is different from V2 is different from V3 (although it may be the same as V1)

Now my plots show how we can distinguish V1, V2 and V3 using the second property I mentioned, that of order.

If V1 = V3 but not V2 we have two changes using the order V1 V2 V3 but only one change using the order V1 V3 V2.

So our running variable distinguished this difference.

 

Consider a 50kg lump of Uranium slowly degrading through fission.

V1 = 50kg

V2 = slightly less than 50 kg because of radioactive decay.

V3 = slighly less again.

and so on

But if all this decay occured together as one big change, not as a lot of small ones we would have a nuclear explosion instead of a slowly degrading lump.

Link to comment
Share on other sites

3 hours ago, swansont said:
!

Moderator Note

You have claimed this several times without explanation. It seems you have some alternative take on this, so I have moved it to speculations. Now: either expand on what you mean and answer questions that have been put to you, or this gets locked.

 

Hi, I am keen to explain my ideas further.  I asked this forum if there is anything wrong with my initial premise - Mass-less particles, travelling at the speed of light, experience zero time.  The discussions appear to drift into related topics.  I can't move to the next line of explanation until this premise is accepted or rejected.  Thanks. 

Link to comment
Share on other sites

9 minutes ago, scuddyx said:

Hi, I am keen to explain my ideas further.  I asked this forum if there is anything wrong with my initial premise - Mass-less particles, travelling at the speed of light, experience zero time.  The discussions appear to drift into related topics.  I can't move to the next line of explanation until this premise is accepted or rejected.  Thanks. 

It was pointed out that that is not a valid frame of reference and results in division by zero if you try to use it as such.

Link to comment
Share on other sites

3 hours ago, studiot said:

 

@swansont scuddyx promised to answer me and has kept his word so please let the thread run at least a little longer.

 

First let me say thank you for continuing the conversation and explain tha V represents the quantity etc under scrutiny for change.
I know I labelled a time axis but I would like to consider the implications of not having a formal axis at all and then show that we can't do without it.

This is quite different from an emergent phenomenon (as I understand the term) which arises as a result of specific and particular circumstances with a particular collection of objects.
It cannot be linked to other phenoma taking place elsewhere (ie is independent of them).

A good example is the action of an arch bridge.
The massive strength of an arch bridge only occurs with a particular configuration of the arch stones and only when they are all there.
The arch will not stand up, let alone support load, untill all parts are in place. If one is removed or missing the arch will collapse.
A variation on this which shows that the remarkable emergent phenomenon is the fact that the shape causes the apllied force to be diverted through exactly one right angle is the ring.
Load a part ring radially and it will bend. Apply that same load to a complete ring and the radial force will be turned into a circumferential one.

To return to my diagrams.

Say we repeatedly sample the variable under scrutiny, V.

When change occurs we expect continuity in V.
That is there is always some value for V (including zero) and there is never a sample for which there is no value.
There are no gaps in the continuum, and every value is distinct and different so V1 is different from V2 is different from V3 (although it may be the same as V1)

Now my plots show how we can distinguish V1, V2 and V3 using the second property I mentioned, that of order.

If V1 = V3 but not V2 we have two changes using the order V1 V2 V3 but only one change using the order V1 V3 V2.

So our running variable distinguished this difference.

 

Consider a 50kg lump of Uranium slowly degrading through fission.

V1 = 50kg

V2 = slightly less than 50 kg because of radioactive decay.

V3 = slighly less again.

and so on

But if all this decay occured together as one big change, not as a lot of small ones we would have a nuclear explosion instead of a slowly degrading lump.

Thanks for the thorough explanation of your diagram.  I will read it carefully and get back to you soon as I can.

Link to comment
Share on other sites

25 minutes ago, scuddyx said:

Hi, I am keen to explain my ideas further.  I asked this forum if there is anything wrong with my initial premise - Mass-less particles, travelling at the speed of light, experience zero time.  The discussions appear to drift into related topics.  I can't move to the next line of explanation until this premise is accepted or rejected.  Thanks. 

The problem, to me, is that you keep saying it's decoherence, and then not explaining what you mean by that. And that seems decoupled from the issue of massless particles not having their own frame of reference.

 

Link to comment
Share on other sites

17 minutes ago, Strange said:

It was pointed out that that is not a valid frame of reference and results in division by zero if you try to use it as such.

I explained this as the limit as you get arbitrary close to the speed of light. 

As with Differentiation we are not dividing by zero but considering the limit.

Link to comment
Share on other sites

v=c is not a valid reference frame as you have seperation distance [math]ds^2=0 [/math] however a less than c reference frame can measure the interval between emitter and receiver and determine that massless particles propogate at c.

 So to state that time doesn't  occur for massless particles is nonsense or that the massless particle can exist at every location in the universe based on the seperation distance is obviously nonsense.

 Hence v=c is not a valid reference frame.

Link to comment
Share on other sites

3 hours ago, swansont said:

The problem, to me, is that you keep saying it's decoherence, and then not explaining what you mean by that. And that seems decoupled from the issue of massless particles not having their own frame of reference.

 

Decoherence is the collapse of the wave function in quantum mechanics. No time or space is experienced by particles as their wave function propagates – until the wave function is said to collapse.  Consequently, two entangled particles described by a wave function, appear to communicate instantaneously even if they are light years apart.  Decoherence is the key to understanding time.  Other particles couple or interfere with the Higgs Field.  The more they couple, the more decoherence happens, and more time is created for them.  This slows them down from the speed of light.  The particles that experience the most decoherence slow down the most.  This appears has if they have more inertia.  The ones with most inertia appear to us as having the greatest mass.  This is how mass arises and varies between particles

2 hours ago, Mordred said:

v=c is not a valid reference frame as you have seperation distance ds2=0 however a less than c reference frame can measure the interval between emitter and receiver and determine that massless particles propogate at c.

 So to state that time doesn't  occur for massless particles is nonsense or that the massless particle can exist at every location in the universe based on the seperation distance is obviously nonsense.

 Hence v=c is not a valid reference frame.

Clocks travelling at 10% light speed would slow down by 0.5%.  At 99.9999% light speed clocks will slow down by 99.86%.  Taking this towards the limit - the photons of light 'experience' zero time (instantaneous travel time).  There is no absolute time frame of reference.  Where is the error in this logic? 

Link to comment
Share on other sites

The error is as I described v=c is not a valid reference frame and nothing one can state will change that.

Obviously time must pass for massless particles as those particles can and do change location. They can change in wavelength Ie redshift and they can decay to other particles.

So obviously time must be passing for massless particles. To state otherwise is wrong

Link to comment
Share on other sites

3 hours ago, scuddyx said:

Thanks for the thorough explanation of your diagram.  I will read it carefully and get back to you soon as I can.

Thanks for taking time to explain the diagram. I agree with what you are saying.

In quantum mechanics time and space does not exist between emission and measurement of particles (for them).  This easily explains the double slit experiment, spooky action at a distance and even the relativistic twins paradox.

7 minutes ago, scuddyx said:

Decoherence is the collapse of the wave function in quantum mechanics. No time or space is experienced by particles as their wave function propagates – until the wave function is said to collapse.  Consequently, two entangled particles described by a wave function, appear to communicate instantaneously even if they are light years apart.  Decoherence is the key to understanding time.  Other particles couple or interfere with the Higgs Field.  The more they couple, the more decoherence happens, and more time is created for them.  This slows them down from the speed of light.  The particles that experience the most decoherence slow down the most.  This appears has if they have more inertia.  The ones with most inertia appear to us as having the greatest mass.  This is how mass arises and varies between particles

Clocks travelling at 10% light speed would slow down by 0.5%.  At 99.9999% light speed clocks will slow down by 99.86%.  Taking this towards the limit - the photons of light 'experience' zero time (instantaneous travel time).  There is no absolute time frame of reference.  Where is the error in this logic? 

 

Link to comment
Share on other sites

3 hours ago, scuddyx said:

I explained this as the limit as you get arbitrary close to the speed of light. 

Yes, you can develop an intuitive sense that photons "experience no time" (which is a pretty meaningless statement) but it is not mathematically useful or consistent. You cannot cannot base a theory on this because it will result in division by zero.

1 minute ago, scuddyx said:

In quantum mechanics time and space does not exist between emission and measurement of particles (for them).  This easily explains the double slit experiment, spooky action at a distance and even the relativistic twins paradox.

Please provide the mathematics that support this claim. (Otherwise I will request this thread is closed for lacking any scientific content.)

Also, we measure the effects of the double-slit experiment, entanglement, etc. in our frame of reference, not that of the photon. So invoking the (non-existent) photon frame of reference explains nothing.

Link to comment
Share on other sites

10 minutes ago, Mordred said:

The error is as I described v=c is not a valid reference frame and nothing one can state will change that.

Obviously time must pass for massless particles as those particles can and do change location. They can change in wavelength Ie redshift and they can decay to other particles.

So obviously time must be passing for massless particles. To state otherwise is wrong

Time passes for mass less particles - for instance it takes 8 minutes for them to travel from the sun.

However, they don't experience time themselves. 

There is no absolute time frame of reference.

Photons can undergo red shifts and decay into other particles - but not during flight - they need to decohere first.  The first suspicion that neutrinos had mass was when they were observed changing state between source and detector.

Link to comment
Share on other sites

No as stated v=c is not a reference frame. End of discussion this is well established under GR. It is a singularity condition where the math no longer accurately describes the physics. A photon is a field excitation with a Compton wavelength you cannot have a wavelength without time.

Edited by Mordred
Link to comment
Share on other sites

20 minutes ago, scuddyx said:

In quantum mechanics time and space does not exist between emission and measurement of particles (for them).  This easily explains the double slit experiment, spooky action at a distance and even the relativistic twins paradox.

Actually it does under the Uncertainty principle.

The emission time of a photon can be related to the line width of the spectral line produced (or the other way round)

This is very important in spectroscopy as it controls the resolution available.

 

I would suggest that much of your difficulty arises from your attempts to introduce QM into Relativity.

I do not know of any successful attempts in that direction.

Introducing relativity into QM has been managed, for instance the relativistic Schrodinger equation was derived by Dirac in the 1930s.

But marrying the two is still a long way off.

Link to comment
Share on other sites

17 minutes ago, Strange said:

Yes, you can develop an intuitive sense that photons "experience no time" (which is a pretty meaningless statement) but it is not mathematically useful or consistent. You cannot cannot base a theory on this because it will result in division by zero.

Please provide the mathematics that support this claim. (Otherwise I will request this thread is closed for lacking any scientific content.)

Also, we measure the effects of the double-slit experiment, entanglement, etc. in our frame of reference, not that of the photon. So invoking the (non-existent) photon frame of reference explains nothing.

Indeed the double-slit experiment, entanglement, etc. look spooky in our frame of reference.  However they are easy explained if space and time are assumed not to exist for particles before the wavefunction collapses.  In QM time and space can't be treated classically.  There is no absolute frame of reference in QM or classical physics.

Currently this is a thought experiment like Einstein's lift analogy - his maths for GR came later.

7 minutes ago, studiot said:

Actually it does under the Uncertainty principle.

The emission time of a photon can be related to the line width of the spectral line produced (or the other way round)

This is very important in spectroscopy as it controls the resolution available.

 

I would suggest that much of your difficulty arises from your attempts to introduce QM into Relativity.

I do not know of any successful attempts in that direction.

Introducing relativity into QM has been managed, for instance the relativistic Schrodinger equation was derived by Dirac in the 1930s.

But marrying the two is still a long way off.

A lot of time has been spent trying to marry QM and GR by trying to quantize GR.  I prefer to work the other way round assuming both theories are correct but QM is the more fundamental. Sean Carroll (CalTech Physics Professor) is developing a theory of space as an emergent QM property.  I am working on time as an emergent QM property.

https://www.preposterousuniverse.com/blog/2016/07/18/space-emerging-from-quantum-mechanics/

Link to comment
Share on other sites

20 minutes ago, scuddyx said:

However they are easy explained if space and time are assumed not to exist for particles before the wavefunction collapses.

Please show this to be true, in appropriate mathematical detail.

20 minutes ago, scuddyx said:

In QM time and space can't be treated classically. 

They can be and, in fact, they are. It is almost like you don't know what you are talking about.

22 minutes ago, scuddyx said:

I am working on time as an emergent QM property.

Then show us the mathematical model.

Link to comment
Share on other sites

Guest
This topic is now closed to further replies.
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.