Bohmian mechanics and special relativity

[Demystifier]

No, because the only justification of that belief would be that taking the number of lattice points to infinity would produce exact QED. But all evidence points to that taking the number of lattice points to infinity would produce QED with zero renormalized charge. The numerical evidence for this is already visible at the lattice sizes that can be simulated today, and the observed closeness to triviality grows with lattice size. Extrapolating to $1.000.000^4$ lattice points or so we would see an almost perfect free theory!

I think you misunderstood triviality. "Almost perfect free theory" means that all effects of interactions are small if the bare coupling constant is $g\sim 1$. But the bare coupling constant does not need to be $g\sim 1$. Instead, it can be $g\gg 1$. By taking $g$ large enough, one can always get a theory which is not almost free. Moreover, one can fit the exact value of $g$ such that the lattice theory predictions agree with the long-distance (low-energy) experiments. (By long distance, I mean distance much larger than the lattice spacing $a$.) It turns out that $g$ must be taken larger when $a$ is smaller, but it is not a problem as long as $a>0$, because then the required bare coupling constant is $g<\infty$.

The problem only appears when $a\rightarrow 0$, because then the required bare coupling constant diverges $g\rightarrow\infty$. In other words, the effects of interaction vanish for any finite value of $g$, which is called triviality.

But from the effective theory point of view, there is no any physical reason to consider $a\rightarrow 0$. One does not expect QED to be a right theory at the Planck distance, but only at distances much larger than that. Hence there is no need to take $a$ smaller than the Planck distance. Hence the triviality is not a problem.

 

[A. Neumaier]

I think you misunderstood triviality. "Almost perfect free theory" means that all effects of interactions are small if the bare coupling constant is $g\sim 1$. But the bare coupling constant does not need to be $g\sim 1$. Instead, it can be $g\gg 1$. By taking $g$ large enough, one can always get a theory which is not almost free.

Please provide a reference for your claim that by taking $g$ large enough, one can always get a theory which is not almost free!

I think you misunderstood triviality. Lattice QED has three free parameters, two coupling constants (the bare electron mass $m_0$ and the bare electron charge $e_0$) and the lattice spacing $a$. To get a physical approximation to QED, one would have to adjust for given lattice spacing $a$ both bare parameters such that the predicted electron mass $m(m_0,e_0,a)$ and the predicted electron charge $e(m_0,e_0,a)$ reproduce the physical electron mass $m_{phys}$ and the physical electron charge $e_{phys}$. These renormalization conditions require that the equations $m(m_0,e_0,a)=m_{phys}$ and $e(m_0,e_0,a)=e_{phys}$ have a solution. Triviality is the statement that
$$e_{max}(a):=\max_{m_0,e_0} e(m_0,e_0,a)\to 0 \mbox{ for }a\to 0,$$
so that for sufficiently small $a>0$ it is impossible to renormalize to nontrivial values of the charge. This is the meaning of the statement

For the renormalized charge and mass we find results which are consistent with the renormalized charge vanishing in the continuum limit.

in the abstract of the Göckeler paper quoted earlier. And the consistency is visible already at the lattice spacings used in their simulation!

Moreover, one can fit the exact value of $g$ such that the lattice theory predictions agree with the long-distance (low-energy) experiments. (By long distance, I mean distance much larger than the lattice spacing $a$.)

Here you assume without proof what has to be shown! Where is the paper or book that shows that at macroscopic distances (much larger than the lattice spacing realized by current computer simulations), lattice QED predictions agree with the long-distance (low-energy) Maxwell equations?

It is unscientific behavior to claim as facts what you don't have proofs for.

It turns out that $g$ must be taken larger when $a$ is smaller

Where is this shown? I think you mean that you hope that this turns out, but the papers I quoted show that this hope is most likely an illusion only.

but it is not a problem as long as $a>0$, because then the required bare coupling constant is $g<\infty$.

Please provide a reference proving this remarkable claim!

But from the effective theory point of view, there is no any physical reason to consider $a\rightarrow 0$. One does not expect QED to be a right theory at the Planck distance, but only at distances much larger than that. Hence there is no need to take $a$ smaller than the Planck distance. Hence the triviality is not a problem.

The problems with your scenario don't begin at a Planck scale $a$ but are already visible at the values of $a$ currently realizable in simulations!

 

[Demystifier]

Please provide a reference for your claim that by taking $g$ large enough, one can always get a theory which is not almost free!

By $g$ I meant the set of all coupling constants, including the mass. (Such a language is common within the general abstract theory of renormalization of field theories.) Is it OK now?

These renormalization conditions require that the equations $m(m_0,e_0,a)=m_{phys}$ and $e(m_0,e_0,a)=e_{phys}$ have a solution.

Do you agree that those equations have a solution for any small but non-zero $a>0$?

 

[A. Neumaier]

By $g$ I meant the set of all coupling constants, including the mass. (Such a language is common within the general abstract theory of renormalization of field theories.)

? The common language has a number of counterterms called Z with various indices.

Is it OK now?

No, not even with $g=(m_0,e_0)$. See my previous post; please read the details!

Do you agree that those equations have a solution for any small but non-zero $a>0$?

No; I stated in my previous post why not. If you have grounds to think that they have, please provide a reference containing a proof.

 

[Demystifier]

If you have grounds to think that they have, please provide a reference containing a proof.

I don't know a reference with a rigorous proof, but I know references which make claims similar to mine. For instance http://www.scholarpedia.org/article/Triviality_of_four_dimensional_phi^4_theory_on_the_lattice
says the following:
"Quantum field theories are only formally defined by their Lagrange density. To extract physical predictions, the theory is regularized (modified) at short distance, an ultraviolet cutoff is imposed. ... A quantum field theory is called trivial, if the above limit invariably leads to a non-interacting renormalized theory ... It is possible that such a theory can still describe Nature if ... the cutoff is not fully removed, but its effects are invisible on a certain limited range of scales to a certain precision. ...Trivial theories can still be useful as effective theories, if there are parameter values with enough interaction to match experiments and at the same time small enough that effects of the unremoved lattice (or other UV cutoff) are so small as to not contradict experiment...In D=4 one can argue that the coupling dies out only logarithmically, gR∝c/|ln(amR)| and the effective scenario is quite viable. Presumably the status of being an effective theory in the above sense is also true for the Standard Model of particle physics, which contains the scalar Higgs sector. ...Historically, all our theories seem to be eventually superseded by a next more comprising theory revealing itself beyond a certain domain of validity. Triviality may now even be seen as a bonus rather than a defect."

The author is an expert in the field:
http://inspirehep.net/search?ln=en&p=find+author+wolff,+u&of=hcs&action_search=Search&sf=earliestdate&so=d

In particular, note the statement above: "Presumably the status of being an effective theory in the above sense is also true for the Standard Model of particle physics". If we ignore the word "presumably", it says that I was essentially right. The word "presumably" indicates that I could have been wrong (mea culpa!), but that there is no proof that I was wrong and that there is a good reason to think that I was right.

Perhaps we need a new thread about whether trivial theories can serve as effective theories, because it seems that there are arguments for both "yes" and "no".

 

[A. Neumaier]

I don't know a reference with a rigorous proof

I'd be satisfied with a proof at the level of rigor of Peskin and Schroeder, say.

but I know references which make claims similar to mine.

Only superficially similar! Wolff first considers a much more general context of QFTs that are somehow regularized by a cutoff. This includes standard continuum QED with the usual textbook cutoff (i.e., not the causal approach). The specialization to the lattice comes later (in scenario B of $\phi^4$ theory). There he says

in D⩾4 we encounter the phenomenon of triviality: we find that $g_R\to 0$ whenever we tune for the continuum limit $am_R→0$. Thus only free non-interacting theories are reached in the strict continuum limit.

which means (unlike your earlier claim) that the tuning of the constants is already taken into account.

For instance http://www.scholarpedia.org/article/Triviality_of_four_dimensional_phi^4_theory_on_the_lattice
says the following:
"Trivial theories can still be useful as effective theories, if there are parameter values with enough interaction to match experiments and at the same time small enough that effects of the unremoved lattice (or other UV cutoff) are so small as to not contradict experiment.

This carefully formulated statement is true since it is only a conditional statement. The problem is that the condition can apparently not be satisfied for lattice QED, while you assume without proof that it can. You mix things up with standard renormalized continuum QED, which is trivial when renormalized with an explicit cutoff, but nevertheless matches experiment quite well. But one may not apply it to lattice QED without first having shown that it also matches experiments quite well!

The paper cited by Wolff for details for $\phi^4$ is

• U. Wolff, Precision check on the triviality of the $\phi^4$ theory by a new simulation method. Physical Review D, 79(2009), 105002.

Triviality sets in already for quite large lattice constants. In the paper just cited one can see from Fig. 1 (which displays $g_{max}(a)$) that one cannot reach $g=20$ at the lattice sizes where the simulation was done. At least for this value of $g$, this clearly disproves your claim that the renormalization equations can be solved for any lattice spacing a>0. From the curve drawn, one actually sees that for any $g>0$ there is a limiting value of the lattice spacing below one cannot solve the equation!

To get a positive result for QED one would need to do similar simulations and check in which range of lattice spacings one can realize the physical values of electron mass and charge, and which accuracy limit this entails.

In particular, note the statement above: "Presumably the status of being an effective theory in the above sense is also true for the Standard Model of particle physics". If we ignore the word "presumably", it says that I was essentially right. The word "presumably" indicates that I could have been wrong (mea culpa!), but that there is no proof that I was wrong and that there is a good reason to think that I was right.

''presumably'' means ''it is conjectured'' or "I believe", not "it is justified by solid arguments".

The word "presumably" indicates that I could have been wrong (mea culpa!), but that there is no proof that I was wrong and that there is a good reason to think that I was right.

You could have looked up the references instead of believing such crucial things upon a word of authority phrased in the subjunctive! And you shouldn't have generalized to statements that are simply not there - making claims for arbitrarily small lattice spacings...

Perhaps we need a new thread about whether trivial theories can serve as effective theories, because it seems that there are arguments for both "yes" and "no".

Since the viability of relativistic Bohmian lattice model depend on the answer of this question, this fits the present thread; so we don't need a new one.

What are the arguments for "yes"? So far you haven't given any, only assertions that it is so.

I only know arguments for "no", except in the asymptotically free case. The latter applies to QCD, where lattice models are indeed heavily used, but neither to QED nor to the standard model, where they are not used at all, except to demonstrate triviality.

 

[PhilDSP]

The effective space-time is defined by its effects on matter. In experiments we never measure the space-time itself, we only measure how matter behaves in it. Hence, from an effective point of view, the effect on matter is all what we really need.

The question remains. How?

Assuming you have valid and accurate classical field equations, couldn't you essentially apply the Lorentz Transformation with the gamma factor separated out?

It seems to be a mystery to many how both the derivation and application of the de Broglie relations can be done without invoking a Lorentz Transformation or space-time. In his monograph "Beyond the Electron" J. J. Thomson sought to understand and resolve that mystery. The first appendix contains a derivation of classical field equations for a moving electron that appear to be valid at all velocities of the electron. One can easily see how the gamma factor of the Lorentz Transformation is "generated" by the relationship of the particle to the fields. It seems a simple step to factor the gamma factor from the field equations and substitute it with a factored Lorentz Transformation so that space-time emerges.

 

[Elias1960]

And the consistency is visible already at the lattice spacings used in their simulation!

This sounds like you have not understood their use of lattice theory.

They have made computations on $12^4$ and $16^4$ lattices (look simply at pic. 1). They can use them for arbitrary small lattice spacings. Because what they have to compute are not particular solutions, so that they don't have to care about the real size of the region where something nontrivial happens in a particular scattering process, but they have to compute how the renormalization works. Renormalization works from the start with different lattices, and they do this too. But to remain on a $16^4$ lattice once they halve the lattice size, they also halve the cube where they make the computations. So, very high momentum will be simply not present if a is yet big. But once the renormalization has reached a state where they become important, it is the low momentum parts will be no longer represented on that fine lattice, on a $16^4$ lattice with very small lattice size they have no place to live. The region of space they would be able to compute with that $16^4$ lattice shrinks together with the lattice size. This is unproblematic for what they do. But that means they have no boundary for the lattice spacings used. They can with this method, if they like, go to $10^{-100}l_{Pl}$ if there would be a point doing such things. In this particular case, they have done what is natural - to go up to the point where the Landau pole looks problematic.

In a real lattice computation for some physical process you have no such freedom. You cannot reduce the lattice computation to a $16^4$ lattice with, say, lattice size $l_{Pl}$ and periodic boundary conditions. At least, this computation would not give you anything interesting for any real scattering.

Here you assume without proof what has to be shown! Where is the paper or book that shows that at macroscopic distances (much larger than the lattice spacing realized by current computer simulations), lattice QED predictions agree with the long-distance (low-energy) Maxwell equations?
It is unscientific behavior to claim as facts what you don't have proofs for.

I would not expect that people would write such papers. Last but not least, something non-trivial should be shown, and such things look like trivialities for physicists.

What you name "unscientific behavior" is simply the difference in standards of mathematical rigorousness between mathematicians and physicists.

In fact, about the cases where some lattice discretization of some equation gives, in the continuum limit, something different than the original equation, physics know too. This is the case of fermion doubling. Nothing similar is known for QED (beyond the fermion doubling itself, of course).

Where is this shown? I think you mean that you hope that this turns out, but the papers I quoted show that this hope is most likely an illusion only.

No, in this case your plausibility argument that this paper would show something is simply based on misunderstanding the method of using the lattice in that paper.