16 hours ago, studiot said:

And there is a fundamental issue between GR and particle physics that no one has thus far brought up.

Well, one of these issues would be that the energy-momentum tensor, being the source term in the Einstein equations, cannot be promoted (to the best of my knowledge) to a Hermitian operator in the QM sense.

On 1/15/2025 at 5:21 AM, Markus Hanke said:

Well, one of these issues would be that the energy-momentum tensor, being the source term in the Einstein equations, cannot be promoted (to the best of my knowledge) to a Hermitian operator in the QM sense.

My comments only take into account the effects of rotations of not the direct effects on quantities (differential or otherwise) as projections or components. Rotations by themselves do not lead to the cross products involved.

Can anyone verify this information? Does this then corroborate my claim about GR violating HUP? Specifically, but not isolated to, the metric tensor within GR and its application of calculus to conjugate pairs, which are linked by HUP.

We have conjugate pairs in the metric tensor.

Edited June 7Jun 7 by AbstractDreamer

6 minutes ago, AbstractDreamer said:

Can anyone verify this information? Does this then corroborate my claim about GR violating HUP?

We have conjugate pairs in the metric tensor.

Classical GR (or any other non-quantised theory) obviously violates HUP, as coordinates and their canonical momenta commute.

You need coordinates, a Lagrangian, and canonically-conjugate momenta that do not commute with the former, for the HUP to be satisfied. There is no Planck's constant, nor non-commuting operators in GR.

1 minute ago, joigus said:

Classical GR (or any other non-quantised theory) obviously violates HUP, as coordinates and their canonical momenta commute.

You need coordinates, a Lagrangian, and canonically-conjugate momenta that do not commute with the former, for the HUP to be satisfied. There is no Planck's constant, nor non-commuting operators in GR.

I wouldn't say it was "obvious". When I first mentioned this, most of the response were not in agreement. Instead, most of the responses were trying to nit pick irrelevances and mistakes that I made, instead of getting straight to the point.

Going back to the OP. Here is a "flaw".

1 hour ago, AbstractDreamer said:

I wouldn't say it was "obvious". When I first mentioned this, most of the response were not in agreement. Instead, most of the responses were trying to nit pick irrelevances and mistakes that I made, instead of getting straight to the point.

Going back to the OP. Here is a "flaw".

When you first mentioned this you said things that were flat-out wrong, and it’s hard to answer a question based on such misconceptions unless those misconceptions are first cleared up. They were fundamental, rather than being irrelevancies.

47 minutes ago, swansont said:

When you first mentioned this you said things that were flat-out wrong, and it’s hard to answer a question based on such misconceptions unless those misconceptions are first cleared up. They were fundamental, rather than being irrelevancies.

You're going to have to quote me where is said things were "flat-out wrong".

I still maintain the overall gist of what I was trying to say is valid - that GR violates HUP - despite having to talk about irrelevancies like what x and y represent, what isn't superposition, whether infinitesimals are irrelevant.

I left the conversation 4 months ago with a sense that I was entirely wrong. A sense that you and others contributed towards. A sense that was entirely false.

I had to go and find proof that conjugate pairs existed in the metric tensor because no-one here was willing to overtly support my suspicion of HUP violation. Your silence on that point is louder than all the minor mistakes I may have made.

15 minutes ago, AbstractDreamer said:

You're going to have to quote me where is said things were "flat-out wrong".

I still maintain the overall gist of what I was trying to say is valid - that GR violates HUP - despite having to talk about irrelevancies like what x and y represent, what isn't superposition, whether infinitesimals are irrelevant.

I left the conversation 4 months ago with a sense that I was entirely wrong. A sense that you and others contributed towards. A sense that was entirely false.

I had to go and find proof that conjugate pairs existed in the metric tensor because no-one here was willing to overtly support my suspicion of HUP violation. Your silence on that point is louder than all the minor mistakes I may have made.

Minor mistakes? You made some extraordinarily silly statements, including that calculus violates the Uncertainty Principle (when QM uses calculus all the time), that pressure and volume are non-commuting operators and so on. You thus gave other readers every reason to think you had no idea what you were talking about.

Where I said calculus contradicts HUP, was in the context of GR and observables, NOT as a mathematical tool in itself.

Whether I gave reason for other readers to think I have no idea what I'm talking about is again irrelevant. Lets say I factually had no idea what I'm talking about, not just readers thinking that I don't. How does that change anything?

Does that justify your responses in COMPETELY failing to acknowledge the most important point and focus SOLEY on my "extraordinarily silly statements"?

I have just shown GR metric tensor contains conjugate variables in it's differential equations, in direct violation of HUP. Why didn't you just tell me this - IN ADDITION to correcting all my silly statements? Why focus on the silly statements?

Just now, AbstractDreamer said:

Can anyone verify this information? Does this then corroborate my claim about GR violating HUP? Specifically, but not isolated to, the metric tensor within GR and its application of calculus to conjugate pairs, which are linked by HUP.

We have conjugate pairs in the metric tensor.

Once again this shows your failure to understand Physics and Mathematics.

In particular the meaning of the word 'conjugate'

I have alredy explained in some detail the meaning in relation to HUP, which is not the same as the meaning in your extracts.

Please tell us that you have not spent the last 5 months hunting for 'conjugate pairs' without understanding what that means ?

Quantities, Variables or other entities P and Q are said to be conjugate if when combined according to specified rules produce a specified result.

These quantites may or may not be physical quantities for instance x and 1/x are conjugate when the specified result is the number 1 or the identity in certain sets.

Such relationships are so common that other particular names are given to the conjugacy for instance, adjunct, inverse, complex conjugate, canonical and so on.

Please go back and consider the specification for Heisenberg I gave you, and then show me where those particular quantities appear in General Relativity.

As an example from Thermodynamics I could declare

"Thermodynamics is flawed because it is incompatible with Time"

Of course this is not true since the Three Laws of Thermo do not mention Time at all.

50 minutes ago, AbstractDreamer said:

Where I said calculus contradicts HUP, was in the context of GR and observables, NOT as a mathematical tool in itself.

Whether I gave reason for other readers to think I have no idea what I'm talking about is again irrelevant. Lets say I factually had no idea what I'm talking about, not just readers thinking that I don't. How does that change anything?

Does that justify your responses in COMPETELY failing to acknowledge the most important point and focus SOLEY on my "extraordinarily silly statements"?

I have just shown GR metric tensor contains conjugate variables in it's differential equations, in direct violation of HUP. Why didn't you just tell me this - IN ADDITION to correcting all my silly statements? Why focus on the silly statements?

Because what you said was wrong. It is trivially obvious that GR does not take into account the Uncertainty Principle. That is not because calculus is incompatible with it, which is the reason you gave, but because GR models phenomena for which the Uncertainty Principle is irrelevant. You then followed up your wrong assertion with further silly statements about non-commuting operators, for which I called you an idiot. I see no reason to revise that assessment.

44 minutes ago, AbstractDreamer said:

Where I said calculus contradicts HUP, was in the context of GR and observables, NOT as a mathematical tool in itself.

That’s not how your very first post in the thread reads. We can all read it. You talked about GR being flawed because of infinitesimals, followed by a very incorrect summary of the uncertainty principle. You followed that up with “Calculus is in direct contradiction with Uncertainty Principle”

Direct contradiction. Not via GR.

Pretty much every new post had an error that needed to be corrected.

Like I said, we can all read it. There will not be any rewriting of history.

Anyway, GR does not violate what YOU presented as being the HUP, and I think that was answered.

“Calculus assumes and necessarily requires the logical leap-of-faith that if you break a curve into infinitesimally small parts, then each part is a straight line.  An infinitesimal change in y with respect to an infinitesimal change in x.   Calculus claims it knows the value of both x and y, at infinitesimality.”

Quantum uncertainty principle claims that at infinitesimality, observables are in superposition.  You cannot know fully and simultaneously both the values of y and x.”

(note that you are the one who presented x and y as being the variables associated with a straight line, rather than y being momentum, as you later argued)

4 hours ago, AbstractDreamer said:

I wouldn't say it was "obvious".

Let me put it this way: It's trivial. GR trivially violates QM, therefore, HUP.

Is that any better?

Trivial it is. You can bet your life's savings on that.

Some further notes.

Commutativity is more complicated than just commutable and non-commutable.

For matrix multiplication

Some matrices will commute with their transpose.
some will commute with their complex conjugate matrix.
But some will not.

but reversal of the order may make MN = - NM ( this is called anticommuting.)

Some pairs of matrices will multiply as MN just fine but cannot be multiplied at all as NM simply because they have incompatible rows and columns.

In classical wave theory there is an uncertainty principle (not the HUP) which reads

[math]\Delta t\Delta \omega \ge \frac{1}{2}[/math]

Where delta t is the RMS duration of a wave packet and delta omega the RMS bandwidth.

To prove this you start with Parsevals theorem introduce the Schwarz inequality and several pages of calculus.

Maxwells Reciprocal theorem has an interesting form of commutation

In a linear-elastic system a unit load applied at A produces a deflection at B which is the same deflection as the unit load applied at B produces at A.

10 hours ago, studiot said:

Some further notes.

Commutativity is more complicated than just commutable and non-commutable.

For matrix multiplication

Some matrices will commute with their transpose.
some will commute with their complex conjugate matrix.
But some will not.

but reversal of the order may make MN = - NM ( this is called anticommuting.)

Some pairs of matrices will multiply as MN just fine but cannot be multiplied at all as NM simply because they have incompatible rows and columns.

In classical wave theory there is an uncertainty principle (not the HUP) which reads

ΔtΔω≥12

Where delta t is the RMS duration of a wave packet and delta omega the RMS bandwidth.

To prove this you start with Parsevals theorem introduce the Schwarz inequality and several pages of calculus.

Maxwells Reciprocal theorem has an interesting form of commutation

In a linear-elastic system a unit load applied at A produces a deflection at B which is the same deflection as the unit load applied at B produces at A.

That inequality is interesting. Does that imply that the persistence of a (classical) wave packet is shorter, the wider the range of Fourier components that make it up, i.e. its rate of dispersion is greater?

I suppose in that case the extreme example would be a “packet” with only one component, which would then persist indefinitely. But then it would extend throughout space so would not be a localised packet any more. Which raises the question of whether there is another inequality relating the degree of the packet’s extension in space to bandwidth. Do you know?

(I recall my old tutor, who had been a keen amateur radio buff in his youth, told us QM was easy to grasp for anyone who was a radio engineer.)

Edited June 8Jun 8 by exchemist

15 hours ago, AbstractDreamer said:

that GR violates HUP

To make a long story really short - GR is a purely classical theory (by design), whereas the HUP concerns observables of quantum systems. Thus the HUP does not apply to GR, or to any other classical theory, for that matter.

2 hours ago, exchemist said:

That inequality is interesting. Does that imply that the persistence of a (classical) wave packet is shorter, the wider the range of Fourier components that make it up, i.e. its rate of dispersion is greater?

There is an inverse relationship between the standard deviation width of a function and the standard deviation width of its Fourier transform. The value of the product of the two standard deviations depends on the function (hence the inequality) with the minimum value (where the inequality becomes the equality) achieved by the Gaussian function whose Fourier transform is also the Gaussian function.

2 hours ago, exchemist said:

I suppose in that case the extreme example would be a “packet” with only one component, which would then persist indefinitely. But then it would extend throughout space so would not be a localised packet any more. Which raises the question of whether there is another inequality relating the degree of the packet’s extension in space to bandwidth. Do you know?

It can be proven that the Fourier transform of a function with compact support does not have compact support. To prove this, note that any function with compact support is the product of some function with the rectangular function. Therefore, the Fourier transform of any function with compact support is the convolution of some function (with or without compact support) with the sinc function, and therefore does not have compact support.

The converse is not necessarily true. For example, both the Gaussian function and its Fourier transform (also the Gaussian function) are functions without compact support.

2 minutes ago, KJW said:

There is an inverse relationship between the standard deviation width of a function and the standard deviation width of its Fourier transform. The value of the product of the two standard deviations depends on the function (hence the inequality) with the minimum value (where the inequality becomes the equality) achieved by the Gaussian function whose Fourier transform is also the Gaussian function.

It can be proven that the Fourier transform of a function with compact support does not have compact support. To prove this, note that any function with compact support is the product of some function with the rectangular function. Therefore, the Fourier transform of any function with compact support is the convolution of some function (with or without compact support) with the sinc function, and therefore does not have compact support.

The converse is not necessarily true. For example, both the Gaussian function and its Fourier transform (also the Gaussian function) are functions without compact support.

Yes but now you have to explain what compact support is. I’m only a chemist. 🙂 what does this mean physically for a wave packet?

1 minute ago, exchemist said:

Yes but now you have to explain what compact support is. I’m only a chemist. 🙂 what does this mean physically for a wave packet?

"Compact support" means that the function is zero everywhere outside of some finite domain.

1 minute ago, KJW said:

"Compact support" means that the function is zero everywhere outside of some finite domain.

OK, thanks. But I assume that, while a wave packet has amplitude centred around a specific location, the amplitude declines asymptotically on either side, i.e. does not strictly go to zero elsewhere. So this compact support issue is not really what I am asking about.

Just now, exchemist said:

That inequality is interesting. Does that imply that the persistence of a (classical) wave packet is shorter, the wider the range of Fourier components that make it up, i.e. its rate of dispersion is greater?

I suppose in that case the extreme example would be a “packet” with only one component, which would then persist indefinitely. But then it would extend throughout space so would not be a localised packet any more. Which raises the question of whether there is another inequality relating the degree of the packet’s extension in space to bandwidth. Do you know?

(I recall my old tutor, who had been a keen amateur radio buff in his youth, told us QM was easy to grasp for anyone who was a radio engineer.)

In more physical terms than KJW's answer

If F(w) is the fourier transform of f(t) then (1/a)F(w/a) is the fourier transform of f(at) where a is a positive constant scale factor which changes the duration of the waveform in time.

If a> 1 the waveform f(at) is compressed in comparison with the waveform f(t).

The spectrum is then more spread out.

Which is the maths underlying what I think is your question and correct observation. +1

This is known a a similarity relation and makes the process scale independant.

Note the 1/2 is dimensionless so the product delta-t delta-omega is dimensionless and is the condition for a similarity transformation.

The meaning depends upon the function chosen for f(t) so KJW's gaussian function has minimal spectral width, in contrast to the width of a rectangular pulse ( a la Dirac) which has infinite spectral width

Just now, Markus Hanke said:

To make a long story really short - GR is a purely classical theory (by design), whereas the HUP concerns observables of quantum systems. Thus the HUP does not apply to GR, or to any other classical theory, for that matter.

The maths we are discussing is pure maths but it not only leads to the classical uncertainty in radio waves that exchemist mentioned it also leads to the HUP.

3 hours ago, exchemist said:

(I recall my old tutor, who had been a keen amateur radio buff in his youth, told us QM was easy to grasp for anyone who was a radio engineer.)

As far as the HUP is involved, sure. Remember AM radio? There were times that radio stations started broadcasting shortly before the program started. At that moment, you could tune very precisely, namely at the narrow point where the radio became nearly silent. As soon as the program started, there was no precise point anymore, you could already hear the program if you did not tune in exactly: the bandwidth had increased, because the frequency was not precise anymore.

The extreme case were very short pulses: lightning. One could hear that over more or less all AM bands, so the frequencies were extremely spread out.

I am wondering about the unequality Studiot mentioned: compare with 'Küpfmüller's uncertainty principle (1924)':

delta f . delta t > 1/2

Replace f with E/h and you get:

delta E . delta t > h/2

Pretty close to the HUP, isn't it?

Basic knowledge for radio amateurs, I would say.

Exactly.
R Feynman and my 3rd year Physics Prof taught me in 1980 that the HUP is an inherent quality of all wave motion, be it radio waves or the deBroglie waves of quantum particles.

As for the OP, every theoretical model has 'flaws', because they are not the actual thing/effect being modelled.
But we usually call them limitations of the model, certainly not 'flaws', because every model is useful within its limits

Edited June 8Jun 8 by MigL

1 hour ago, exchemist said:

OK, thanks. But I assume that, while a wave packet has amplitude centred around a specific location, the amplitude declines asymptotically on either side, i.e. does not strictly go to zero elsewhere. So this compact support issue is not really what I am asking about.

The Heisenberg uncertainty principle relates the standard deviations of conjugate pairs of variables. But a variable having finite standard deviation does not necessarily have compact support (eg Gaussian function). So the Heisenberg uncertainty principle doesn't really say anything about the support of conjugate pairs of variables. However, I have indicated that at least one of a conjugate pair of variables must be without compact support. How this impacts on the physics is unclear since it is reasonable to assume that physical variables are bounded in value.

To complete the connection between Fourier transforms and differential operators with regards to conjugate variables that I started earlier in this topic:

[math]\text{Let }F(\xi) = \displaystyle \int_{-\infty}^{\infty} f(x)\ \exp(-2\pi i\ \xi x)\ dx \ \ \ \ ;\ \ \ \ f(\pm \infty ) = 0[/math]

[math]\text{Then }\displaystyle \int_{-\infty}^{\infty} \dfrac{d}{dx}f(x)\ \exp(-2\pi i\ \xi x)\ dx[/math]

[math]= -\displaystyle \int_{-\infty}^{\infty} f(x)\ \dfrac{d}{dx}\exp(-2\pi i\ \xi x)\ dx[/math]

[math]= 2\pi i\ \xi\ \displaystyle \int_{-\infty}^{\infty} f(x)\ \exp(-2\pi i\ \xi x)\ dx[/math]

[math]= 2\pi i\ \xi\ F(\xi)[/math]

[math]\text{Therefore }\dfrac{d}{dx} \equiv 2\pi i\ \xi[/math]

[Please refresh browser window if the above LaTex doesn't render]

Edited June 8Jun 8 by KJW