Spinoff from the "Matter is excitations in a field" thread.

[koti]

First off, kudos to Stringjunky for this great find:

 

At 26:37, Sean Carroll mentions a spinning top which cannot revolve slower than a single 360 degree turn once in 100 mln times the age of the universe. Could someone explain the principal behind this? Where does this matter vs spacetime limitation come from?

Edited May 10, 2017 by koti

[Strange]

This was to explain that the spin of particles is quantised; i.e. the amount of angular momentum they have must be in discrete increments. To show how small those increments are, he used the example of the rate at which a top would spin if it had a spin of "1".

 

(I don't know if the spin of a macroscopic object like that would actually be quantised or not, though...)

[Mordred]

First off, kudos to Stringjunky for this great find:

 

At 26:37, Sean Carroll mentions a spinning top which cannot revolve slower than a single 360 degree turn once in 100 mln times the age of the universe. Could someone explain the principal behind this? Where does this matter vs spacetime limitation come from?

 

 

 

excellent question and one that can be answered under QM.

This was to explain that the spin of particles is quantised; i.e. the amount of angular momentum they have must be in discrete increments. To show how small those increments are, he used the example of the rate at which a top would spin if it had a spin of "1".

 

(I don't know if the spin of a macroscopic object like that would actually be quantised or not, though...)

 

yes to the first and yes to the spin of a macroscopic objects. Mathematically particle spin is mathematically the equivalent of rotational angular momentum.

 

So lets put this all together.

 

For a Quantum system the angular momentum is an Observable. Now what does that mean under QM? well an observable requires action. Observable also means that with the hypothetical ideal instruments we can measure this action. (Action equates to kinematic motion).

 

So from the above under QM we need to associate this observable a Hermitean operator.

 

[latex]\hat{L}=\hat{X}*\hat{P}[/latex]

 

[latex] L=r*p=\begin{pmatrix}i&j&k\\x&y&z\\p_x&p_y&p_z\end{pmatrix}[/latex]

 

the three Cartesian components are [latex] L_x=yPz-zPy, L_y=zPx-xPz, L_z=zPy-yPx[/latex]

 

https://en.wikipedia.org/wiki/Angular_momentum

 

where [latex] \hat{P}=i\hbar\triangledown[/latex]

 

the definitions of the position and momentum operators above along with being Observable. (All operators in QM and QFT are observable).

 

we can see from the last equation we are indeed quantizing the spin of a macroscopic top under QM formalism We cannot measure any value that is not quantized. Any instrument requires an interaction and all interactions require action and action requires discrete units of quanta as defined by their QFT /QM operators. (other operator examples creation, annihilation operators)

​from the above details we can derive

 

[latex] \hat{L}_x=i\hbar(y\frac{\partial}{\partial_z}-\frac{\partial}{\partial_y})[/latex]

[latex]\hat{L}_y=i\hbar(z\frac{\partial}{\partial_x}-\frac{\partial}{\partial_z})[/latex]

[latex]\hat{L}_z=i\hbar(x\frac{\partial}{\partial_y}-x\frac{\partial}{\partial_z})[/latex]

Edited May 11, 2017 by Mordred

[koti]

This was to explain that the spin of particles is quantised; i.e. the amount of angular momentum they have must be in discrete increments. To show how small those increments are, he used the example of the rate at which a top would spin if it had a spin of "1".

 

(I don't know if the spin of a macroscopic object like that would actually be quantised or not, though...)

I understand that it is an example meant to illustrate that spin is quantized under QM and I am familiar with these very basics of QM however it's shocking to me that a macroscopic object could actually abide by these principles.

 

 

 

excellent question and one that can be answered under QM.

 

yes to the first and yes to the spin of a macroscopic objects. Mathematically particle spin is mathematically the equivalent of rotational angular momentum.

 

So lets put this all together.

 

For a Quantum system the angular momentum is an Observable. Now what does that mean under QM? well an observable requires action. Observable also means that with the hypothetical ideal instruments we can measure this action. (Action equates to kinematic motion).

 

So from the above under QM we need to associate this observable a Hermitean operator.

 

[latex]\hat{L}=\hat{X}*\hat{P}[/latex]

 

[latex] L=r*p=\begin{pmatrix}i&j&k\\x&y&z\\p_x&p_y&p_z\end{pmatrix}[/latex]

 

the three Cartesian components are [latex] L_x=yPz-zPy, L_y=zPx-xPz, L_z=zPy-yPx[/latex]

 

https://en.wikipedia.org/wiki/Angular_momentum

 

where [latex] \hat{P}=i\hbar\triangledown[/latex]

 

the definitions of the position and momentum operators above along with being Observable. (All operators in QM and QFT are observable).

 

we can see from the last equation we are indeed quantizing the spin of a macroscopic top under QM formalism We cannot measure any value that is not quantized. Any instrument requires an interaction and all interactions require action and action requires discrete units of quanta as defined by their QFT /QM operators. (other operator examples creation, annihilation operators)

​from the above details we can derive

 

[latex] \hat{L}_x=i\hbar(y\frac{\partial}{\partial_z}-\frac{\partial}{\partial_y})[/latex]

[latex]\hat{L}_y=i\hbar(z\frac{\partial}{\partial_x}-\frac{\partial}{\partial_z})[/latex]

[latex]\hat{L}_z=i\hbar(x\frac{\partial}{\partial_y}-x\frac{\partial}{\partial_z})[/latex]

Ah Mordred... you and your beautiful latex math. I kinda hate you for it - I think I told you that in that other thread few months ago

Okay, although I had to google "Hermitian Operator", what you wrote is more or less clear to me. What is unclear to me is what Sean Carroll and yourself are doing here - using QM formalism to describe behavior of a macroscopic spinning top. Unless I'm being dense here and this is just an illustration and the spinning top could actually take an infinite time to revolve 360 degrees ?

Edited May 11, 2017 by koti

[Mordred]

Well no it wouldn't be infinite. The reason is we have already established the top must be rotating so eventually the top can undergo a single rotation.

 

What I have done is apply an educated guess based on QFT on the logic behind Sean Carroll's spinning top analogy under QFT action treatment. Remember were also using field waves so the Compton wavelength is involved as well as eugenstates.

 

The difference between one eugenstate the next eugenstate is ie if your at some arbitrary groundpoint if you add a creation operator or subtract an annihilation operator the plus or minus eugenstates will be + or -

 

[latex]\Delta E=\hbar w [/latex]

 

Under QM treatments energy is in discrete packets. Unlike GR or classical where you can have any arbitrary energy value.

Edited May 11, 2017 by Mordred

[koti]

Well no it wouldn't be infinite. The reason is we have already established the top must be rotating so eventually the top can undergo a single rotation.

 

What I have done is apply an educated guess based on QFT on the logic behind Sean Carroll's spinning top analogy under QFT action treatment. Remember were also using field waves so the Compton wavelength is involved as well as eugenstates.

 

The difference between one eugenstate the next eugenstate is ie if your at some arbitrary groundpoint if you add a creation operator or subtract an annihilation operator the plus or minus eugenstates will be + or -

 

[latex]\Delta E=\hbar w [/latex]

 

Under QM treatments energy is in discrete packets. Unlike GR or classical where you can have any arbitrary energy value.

If we were to try to revolve a stationary top by a smaller distance than the Planck length we would in effect move it by the Planck length anyway. On the other hand, we also cannot move the stationary top through spacetime by an increment of the Planck length in a longer period of time than the Planck time unit thus the time limit in which the top (or any other piece of matter independent of its size) cannot exceed in its movement. Is it as simple as that and I just missed it initially ?

Edited May 12, 2017 by koti

[KipIngram]

First please let me echo the kudos for Mordred that I've seen many express. Dude, your explanations are fantastically lucid - thanks for being here. If I can even approach your level before I'm gone I'll be ecstatic. And yes - I'm also jealous of your smooth and easy use of LaTex.

 

I didn't fret too much over the macroscopic top; I figured it was likely that the quantization had to apply to it too, literally, but I mostly took that as just an analogy to make it easier for the audience to picture what aspects of particles he was talking about.

 

Mordred, I do have a question; something that I was hoping the video would address but it didn't. I get it that particles are excitation "packets" in the field; that makes total sense. But what keeps them that way? What aspect of the physics keeps those wave packets from spreading out? Is it something about the quantization that just doesn't allow it, or what exactly?

If we were to try to revolve a stationary top by a smaller distance than the Planck length we would in effect move it by the Planck length anyway. On the other hand, we also cannot move the stationary top through spacetime by an increment of the Planck length in a longer period of time than the Planck time unit thus the time limit in which the top (or any other piece of matter independent of its size) cannot exceed in its movement. Is it as simple as that and I just missed it initially ?

 

That makes it sound like quantization of space and time are required, but I thought that was still debated?

Oh - so after I sent that message above I went off to Google, and ran into this statement on another site:

 

"Have a google on electron knot, and there's people out there saying the electron is stable because it is in essence a "knot" of field."

 

So that means little or nothing to me beyond the vague concept at my current level of knowledge, but it sounds reasonable.

 

Comments?

[Mordred]

Your close but although the Planck length and time are indeed involved its easier to simply realize the following. What you posted isn't fundamentally incorrect its just easier and more accurate to understand with the following

 

In physics, a quantum (plural: quanta) is the minimum amount of any physical entity involved in an interaction.

 

the Planck constant itself is a quantum of action. This is why the minimal observable action or otherwise is in units of quanta.

 

The minimal action of a spinning body from the above will equal the relations of the Planck constant.

 

[latex]6.62607004*10^{-34} m^2/kg / s[/latex]

 

1 action or 1 quantum of action is the lowest possible rate of spin in this case in accordance to the rotation operators I defined earlier.

 

Another key note is under QFT or QM all units are discrete regardless if the system is quantum or macroscopic. (one example being spinfoam under LQC loop quantum cosmology) in this case length,width, volume, area, momentum, energy, etc are all discrete units.

 

(key side note from video. Recall Sean mentions a wave is the probability distribution function) never forget this key note. It is the probability of finding the position and momentum of a particle.(for example. All quantum numbers have a wavefunction. This includes angular momentum..

 

sidenote spacetime itself is discrete under QM/QFT

First please let me echo the kudos for Mordred that I've seen many express. Dude, your explanations are fantastically lucid - thanks for being here.

your welcome just noticed your cross post.

I didn't fret too much over the macroscopic top; I figured it was likely that the quantization had to apply to it too, literally, but I mostly took that as just an analogy to make it easier for the audience to picture what aspects of particles he was talking about.

 

Mordred, I do have a question; something that I was hoping the video would address but it didn't. I get it that particles are excitation "packets" in the field; that makes total sense. But what keeps them that way? What aspect of the physics keeps those wave packets from spreading out? Is it something about the quantization that just doesn't allow it, or what exactly?

 

See my last post in particular what a wavefunction is defined as. (probability distribution) One key difference to understand between QM and classical physics.

 

QM is the statistical probability of a measurable quantity whereas classical attempts exact determination of a measurable quantity. Waves wavefunctions included.

 

So putting the above together visualize a highly irregular siusoidal wave just like you would see under an oscilloscope.

 

When you have a sharp spike of 1 quanta or greater (but always using discrete confined by the Compton wave length that is your most probalistic position and momemtum (in essence your most likely particle location) though its a bit more complex than this heuristic descriptive aka Heisenburg uncertainty as one example)

 

Which is one of the key reasons for QM being probabilistic. Requiring statistical mathematics (including all terminology ie superposition).

 

The last part on electron knotis an accurate descriptive but its not the entire story there are no smaller Leptons for the electron to decay into and the conservation of lepton number applies.

 

However its a bit OT to get into too much detail. I recommend a new thread to discuss Electron knots in greater detail

 

Art Hobson has an excellent write up including the grizzly mathematics. note it is on arxiv working from phone atm.

 

" There are no Particles there is only fields"

 

 

https://www.google.ca/url?sa=t&source=web&rct=j&url=https://arxiv.org/pdf/1204.4616&ved=0ahUKEwitn8Lhq-nTAhVIZFAKHdNnC4wQFggcMAA&usg=AFQjCNEqAKaDGcbyMG2ax22sA9BakBSaTQ&sig2=DavQ6RLz8gqGAchmIr7C4g

 

Example quote..

 

quantized fields have certain particle-like appearances: quanta are unified bundles of field that carry energy and momentum and thus "hit like particles;" quanta are discrete and thus

countable. But quanta are not particles; they are excitations of spatially unbounded fields".

 

There is considerable detail on how this applies to the double slit experiment.

Edited May 12, 2017 by Mordred

[Sensei]

The minimal action of a spinning body from the above will equal the relations of the Planck constant.

 

[latex]6.62607004*10^{-34} m^2/kg / s[/latex]

 

1 action or 1 quantum of action is the lowest possible rate of spin in this case in accordance to the rotation operators I defined earlier.

Planck constant has unit

[math]J * s = \frac{kg * m^2}{s^2} * s = \frac{kg * m^2}{s}[/math]

[Mordred]

good catch screwed up the SI sequence. Also should have used the SI notation.

 

[latex]kg* m^2*s^{-1}[/latex]

Edited May 12, 2017 by Mordred

[koti]

Your close but although the Planck length and time are indeed involved its easier to simply realize the following. What you posted isn't fundamentally incorrect its just easier and more accurate to understand with the following

 

In physics, a quantum (plural: quanta) is the minimum amount of any physical entity involved in an interaction.

 

the Planck constant itself is a quantum of action. This is why the minimal observable action or otherwise is in units of quanta.

 

The minimal action of a spinning body from the above will equal the relations of the Planck constant.

 

[latex]6.62607004*10^{-34} m^2/kg / s[/latex]

 

1 action or 1 quantum of action is the lowest possible rate of spin in this case in accordance to the rotation operators I defined earlier.

 

Another key note is under QFT or QM all units are discrete regardless if the system is quantum or macroscopic. (one example being spinfoam under LQC loop quantum cosmology) in this case length,width, volume, area, momentum, energy, etc are all discrete units.

 

(key side note from video. Recall Sean mentions a wave is the probability distribution function) never forget this key note. It is the probability of finding the position and momentum of a particle.(for example. All quantum numbers have a wavefunction. This includes angular momentum..

 

 

Frankly Mordred, my explanation in post #6 as crude as it is, is easy for me to understand. It's not that Planck length and time are involved implicating other factors - they are the fundamental and only reason for the spinning top not capable of reaching a longer, single rotation spinning time than the stated 100mln x the age of the universe. I had an issue with this analogy which I no longer have so definitely +1 for your explanations.

[Mordred]

Your welcome glad to be of help. As the other Planck units derive from the same principles using Planck length and time is in essence the same thing as applying the Planck constant.

Edited May 12, 2017 by Mordred

[KipIngram]

That Art Hobson paper linked above is fantastic.