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# I What is the best theory why our vacuum may be in the edge of metastability?

1. Apr 4, 2018

### jtlz

There seems to be many meanings of Naturalness so it's good to sort the subtleness of each meaning. In the paper depicting the vacuum metastability illustration in the initial message.. the conclusion in the paper started with the sentence "One of the most important questions addressed by the LHC is naturalness.". I found three good reference about naturalness after you emphasized opposite of naturalness is having a theory.. (I thought naturalness means exclusively having equations that gives the values or relationship.. and I mentioned these references so in the future I can refer to these if I forgot them):

http://backreaction.blogspot.com/2018/02/what-does-it-mean-for-string-theory.html

"That the mass be natural means, roughly speaking, that getting masses from a calculation should not require the input of finely tuned numbers".

and https://arxiv.org/pdf/1501.01035.pdf

"In implementing ’t Hooft’s notion of naturalness, we have so far considered symmetries of a sort familiar from quantum mechanics, generated by a charge operator which is a scalar under rotations. But there is another type of symmetry, allowed by general principles of quantum mechanics and relativity, where the symmetry generators are spinors. This symmetry is known as supersymmetry. We will consider it, ﬁrst, as a global symmetry, but the symmetry can be elevated to a local, gauge symmetry."
[...]
"it is still possible that nature is “natural”, in the sense of ’t Hooft. Future runs of the LHC might provide evidence for supersymmetry, warped extra dimensions, or some variant of technicolor. But the current experimental situation raises the unsettling possibility that naturalness may not be a good guiding principle. Indeed, naturalness is in tension with another principle: simplicity. Simplicity has a technical meaning: the simplest theory is the one with the smallest number of degrees of freedom consistent with known facts. Contrast, for example, the minimal Standard Model, with its single Higgs doublet, with supersymmetric theories, with their many additional ﬁelds and couplings. So far, the experimental evidence suggests that simplicity is winning. The observed Higgs mass is in tension with expectations from supersymmetric theories, but also technicolor and other proposals."

----

If the masses of the superpartners are very high.. what mechanism in superstring theory besides Kane's (his 0.5 TeV bino was already excluded) that can solve the Hiearchy Problem without 't Hooft notion of naturalness or natural supersymmetry (In the sense of the above paragraph at low masses)?
I'm interested for now in theory that can solve for them (instead of Multiverse or real fine tuning between quadratic radiative corrections and the bare mass being put there on purpose by design (if these were the mechanism chosen by nature. Then we should have new fields that can maintain the constants values without possibly any formulas).

And is the solution to the Hiearchy Problem independent of the Vacuum metastability issue (to what extend can solution of each be solution of the other)? Thank you.

2. Apr 4, 2018

### Urs Schreiber

No, naturalness is a principle imposed in the absence of any known equation for the actual numbers. In the absence of an explanation, cancellations of large numbers against each other seems a weird coincidence, and naturalness means that weird coincidences should not happen. It's a qualitative principle akin to the principles of the times of "natural philosophy" before physics became a maths-based science.

What you call "Kane's theory" is called the $G_2$-MSSM, which became a serious contender for realistic model building with the result of Acharya-Witten 01.
This model does not predict explicit numbers without feeding in some concrete assumptions on the precise nature of the compactification (if it did we would be living in a Douglas Adams novel), but -- and that's really the point that Kane et al. have been driving home over the years -- it does make generic predictions that hold irrespective of the detailed numbers that describe the choice of compactfication.

One of these generic predictions is that there is an exponential hierarchy between the Planck scale and the gravitino mass scale, hence the susy breaking scale, hence the electroweak scale. This argument is due to Acharya-Kane-Kumar 12, section V.A.2 (pages 10-11) , and it is not hard to follow:

First, the higher gauge symmetry of the supergravity C-field implies "shift symmetry" of its KK-modes $\Phi_j$. But the perturbative superpotential must be a holomophic function of the $\Phi_j$ and under shift symmetry this is only possible if perturbatively it vanishes identically. As a result, one deduces that in these models the superpotential consist entirely of non-perturbative contributions, such as membrane instantons, which are known to break the shift symmetry. But these non-perturbative contributions are negative exponentials in the instanton contributions, and hence imply that the gravitino mass (hence the susy breaking scale) is exponentially smaller than the Planck mass. This is the required exponential hierarchy.

This is not a mathematical proof, but it is a decent scientific argument based on an actual theory, and it makes clear that exponential hierarchies between the susy breaking scale and the Planck scale have a good scientific explanation from first principles, just as the exponential scale between the power of a supernova and that of an ordinary star does, both of which may look suprprisingly "unnatural" to the mathematically un-aided observer.

Last edited: Apr 4, 2018
3. Apr 4, 2018

### jtlz

What I was asking in your reply was the following especially the unknown contribution (glad to find this today as this vocalizes my concern):

https://profmattstrassler.com/artic...ics-basics/the-hierarchy-problem/naturalness/
"Fig. 5: Summing up the energy from the quantum fluctuations of known fields (schematically shown, upper row) up to the maximum energy scale vmax(down to the minimum distance scale) where the Standard Model still applies, and adding to this contributions from unknown effects from still higher energies and shorter distances (schematically shown, middle row), we must somehow find what experiment tells us is true: that the Higgs field’s average value is 246 GeV and the Higgs particle’s mass is 125 GeV/c2. If vmax is much larger than 500 GeV, this requires a very precise cancellation between the known and unknown sources of energy, one that is highly atypical of quantum theories."

Do you have a more mathematical version of it in your site in terms of renormalization group flow and Higgs self-coupling, etc that described the same stuff? If Vmax is just 500 GeV instead of planck scale.. does it mean the Higgs self-coupling cover up to 500 GeV only instead of up to planck scale? What stuff must the LHC produce so Vmax is up to 500 GeV only? Matt wrote in the commentary after the article:

"The obvious solution is indeed that vmax is near 500 GeV. And if that is true, the LHC will discover as-yet unknown particles, and other predictions of the Standard Model will fail as well. The strongest evidence against it — inconclusive at this time — is that the LHC has not yet discovered any such non-Standard-Model particles, and there are no known deviations from the Standard Model at the current time. Arguably we should have seen subtle deviations already by now. But I will get to this issue soon."

4. Apr 4, 2018

### Urs Schreiber

The problem with this folklore story is that it is trying to nail jelly to the wall by arguing about the size of quantum corrections in the presence of renormalization freedom.

The fact of the matter is that the space of renormalization choices is an affine space (i.e. like a vector space, but with no origin singled out) which means first of all that there is no absolute concept of "size" of a quantum correction. This only appears once one fixes a renormalization scheme, which is like a choice of coordinate chart. It has no physical meaning. Even if we fix a renormalization scheme (which is implicitly assumed in discussions such as you quote) then it still remains a fact that there is an arbitrary freedom in choosing renormalization constants, large or not.

In conclusion, to make progress on these kinds of matters, one needs more theoretical input than just low energy effective perturbative quantum field theory with its arbitrary renormalization freedom, or otherwise one is going in circles forever. As Kane 17 points out, "we should look harder for a theory that does provide a UV-completion".

Notice how the solution of the hierarchy problem observed by Acharya-Kane-Kumar 12, section V.A.2 (p. 10-11) deals with this issue: They invoke a UV-completion that goes beyond perturbation theory. In that theory one knows 1) that the superpotential is protected against perturbative renormalization freedom and 2) the form of the non-perturbative corrections is known. Namely these are exponentials in the inverse coupling. This yields the exponential hierarchy that is to be explained.

You see, this works not by long story-telling and analogies and showing colorful pictures, but by a logical deduction from a theoretical framework.
(Not fully mathematically rigorous, but fairly solid by the standards of phenomenology.)

5. Apr 5, 2018

### protonsarecool

To come back to the original question, here is another very recent calculation regarding meta-stability of our universe:

Scale-invariant instantons and the complete lifetime of the standard model
Anders Andreassen, William Frost, and Matthew D. Schwartz
Phys. Rev. D 97, 056006 – Published 12 March 2018
https://journals.aps.org/prd/abstract/10.1103/PhysRevD.97.056006

6. Apr 5, 2018

### jtlz

I read and reread Matt arguments trying to understand his viewpoints. I think he got discouraged with the LHC non-null results in 2013.. and he wrote in separate page in August 2013 https://profmattstrassler.com/2013/08/27/a-first-stab-at-explaining-naturalness/

"This in turn is why so many particle physicists have long expected the LHC to discover more than just a single Higgs particle and nothing else… more than just the Standard Model’s one and only missing piece… and why it will be a profound discovery with far-reaching implications if, during the next five years or so, the LHC experts sweep the floor clean and find nothing more in the LHC’s data than the Higgs particle that was found in 2012."

It's happening now.. it's 5 years from his pronouncement. Only Higgs and nothing else is "Profound discovery with far-reaching implications". There is possibility there may not even be a UV complete theory. To be on topic, where is Matt page about Vacuum Metastability? He should emphasize it too.

Anyway. If AdS/CFT is UV complete. But it doesn't describe our spacetime. Is this duality just toy model or does it completely describe a hidden sector that can act as holographic surface. Is there any reference about this search for actual surface or it will remain as toy model for centuries to come? And can AdS/CFT explain the why of the vacuum metastability?

7. Apr 5, 2018

### jtlz

Here's Matt page about vacuum metastability:

https://profmattstrassler.com/articles-and-posts/particle-physics-basics/theories-and-vacua/

Metastability relies on the Higgs mass of ~125 GeV.
Equally important to know is why is the Higgs vev 246 GeV. Can this be computed from first principle.
And also why the Higgs is turned on.

I guess pondering on them and solving for them simultaneously may be more encompassing. What I want to know is if 125 GeV was metastable.. what range of the Higgs field vev (it's 246 Gev but what if it's say 3 TeV) can still affect the metastability? (or no direct relationship or independent?)

8. Apr 5, 2018

### Urs Schreiber

We have given you pointers to detailed discussion of this. If you want to go beyond scratching the surface, you'll need to eventually pass from reading blogs to reading real scientific documents. That will also allow you to incrementally improve your questions.

From AFS18 the best present data on the stability issue the seems to be the following:

9. Apr 5, 2018

### jtlz

After watching this video it seems more likely the Correct Fix (above) would shed light on the vacuum metastability too. Yes. I'll read all scientific papers from now on having seen the Bird's eye view. Thanks a lot for helping.

10. Apr 6, 2018

### Urs Schreiber

Sure. Notice that the "Correct Fix" on the left is "having a theory".

Concerning the remainder of the video: Maybe our times serve to raise the subconcious archetype of crisis, but really in natural science it is not a "crisis" to discover a phenomenon that looks non-random (non-generic). On the contrary, this is what drives science, and the left column of that table is a triumphal testimony of that.

One part of the hep community says that the LHC results reveal a "nightmare scenario" where nothing interesting is discovered; but when something interesting is discovered after all, the other part calls this a "crisis". Worse, these two parts of the community seem to have non-empty intersection. This smells of hysteria more than of sober scientific discourse.

The truth is that the LHC results both confirm established theory and at the same time show clearly and unambiguously a new physical effect that is not explained by established theory. That's really the dream scenario of natural science. Instead of rambling on about chance and likeliness, this should make the scientific community turn to its core task, which is to produce theories and iteratively improve on them.

Here we need a theory that explains why the vaccuum would sit on the verge of instabiity, but not beyond. I am aware of two good contestants:

One is maybe I) asymptotic safety, the other is II) any theory in which a) field moduli such as the Higgs potential are themselves dynamical and b) they are prevented from crossing into the realm of instability, by some principle or symmetry

About I) I wish the analysis of theory and data had been made more intelligibly, see #8, #20 above.

About II): This is the theory that people like Gordon Kane are pointing out (Kane 18, "Clue 4"). It may turn out wrong, but at least it is a theory. The sensible thing to do in science is to investigate this theory further and check if it holds water.

Or else come up with a better theory. But we do need to talk about theories and not get lost in informal handwaving about probabilities, likeliness, genericity. The universe is neither likely nor generic, instead it is exceptional in its existence and its properties. In the past natural science, in the modern guise of maths-based theoretical physics, has managed to understand to a large extent why this is so. There is no reason to give up on this success story now.

11. Apr 6, 2018

### jtlz

It is not far from logic that the following needs to be solved at the same time (or at least to ponder on them simultaneously):

1. Why is the Higgs mass and vev low for SM coupled to gravity.. Matt mentioned how it should be tuned to 1/10^-30 (Naturalness problem)
2. Hierarchy Problem (why are all particles not planck size mass)
3. Vacuum Metastability (why is the Higgs mass 125 GeV bordering on the edge of metastability
4. Cosmological Constant Problem (disagreement between the observed values of vacuum energy density and theoretical large value of ZPE suggested by QFT)

You mentioned the first 3 solutions to this thread. May I know what is your take on the 4th problem (cosmological constant problem) and possible solution and if there is (any) connection to the first 3 problems especially the vacuum metastability?

Do you have a single model that can explain all the 4? How many separate models must they need to at least to explain them? 2? 4?

Wiki last paragraphs on the cosmological constant problem is:
"In the case of the Higgs mechanism, the usual Higgs vacuum expectation value in the instant-form vacuum is replaced by a constant scalar background field - a "zero mode" with kμ=0. The phenomenological predictions are unchanged using the LF formalism. Since the Higgs zero mode has no energy or momentum density, it does not contribute to the cosmological constant.

The small non-zero value of the cosmological constant must then be attributed to other mechanisms; for example a slight curvature of the shape of the universe (which is not excluded within 0.4% (as of 2017)[14][15][16]) could modify the Higgs field zero-mode, thereby possibly producing a non-zero contribution to the cosmological constant."

Could there be one model or solution to all of them at same time? It's weird they come in company.. maybe they have a common source or origin.

12. Apr 7, 2018

### Urs Schreiber

On the one hand:

A) in perturbative quantum field theory also the cosmological constant is subject to renormalization freedom (a basic point often neglected, detailed discussion is referenced here) so that all we can do is measure it.

On the other hand:

B) the approach of appealing to non-perturbative effects thought to be known in string theory, which has become a small industry since KKLT 03, has been argued to be performed too carelessly, see the well written recent article
• Ulf Danielsson, Thomas Van Riet,
"What if string theory has no de Sitter vacua?"
(arXiv:1804.01120)
These authors argue that the "several hundred papers on the topic" of which "unfortunately there is no review" (p. 7) are all inconclusive, because none of them take care to really check the assumptions being made; and in fact likely all wrong, since closer inspection shows that these assumptions are quite dubious.

(Sometimes physicists question the point of rigorous mathematical argument. This is a good example for why it's useful: It simply means not to leave big gaps in your argument that later make the whole conclusion collapse. Large swaths of contemporary physics arguments are bound to eventually collapse this way, until the general attitude of the community changes.)

The conclusion is (p. 26) "that string theory has not made much progress on the problem of the cosmological constant during the last 15 years [since] string theorists have not been up to the challenge."

So we don't know how B) pans out. Until we do, we are stuck with A).

Last edited: Apr 7, 2018
13. Apr 7, 2018

### jtlz

Thanks. So the CC is not related to vacuum metastability. So let's get back to the latter exclusively.

Do you have models where instead of just the ground state and the false vacuum.. there are many false vacuums instead? This makes better sense. You see, the universe is so exceptional in its existence and properties, so incredibly designed for life and all that it doesn't make a lot of sense the ground state is its abyss. To illustrate it. It's like our universe is Disneyland. It doesn't make sense to have tactical nukes in the middle of Disneyland able to destroy it at any moment (or possibility for false vacuum to tunnel to a destructive true vacuum). Instead, there must be many false vacuums.. with each having properties as incredible as our universe or compatible with life in Ascended forms. Or imagine all the Disneyland on earth are located in one location but only one of them can manifest at one time and tunneling to false vacuum means tunneling to different Disneyland scenery and design.

14. Apr 7, 2018

### jtlz

This one sentence caused me an hour of reading of your references. About your statement "This means that apart from the freedom of choosing a classical comsological constant in the Einstein-Hilbert action as above, its perturbative quantization (perturbative quantum gravity) introduces renormalization freedom to the value of the cosmological constant."

and on your Quantum Gravity page reference
"Thus, this idea seems to be very unlikely. At the same time, it is clear that present experiments cannot directly probe the effects of that supposed UV-completion of Einstein-gravity.

Approaches to a full quantization of gravity therefore roughly fall into two different strategies
1.One assumes that the Einstein-Hilbert action is indeed the effective QFT that approximates a “UV-completion”, a more fundamental theory valid at all energies. This is the approach taken for instance in string theory.
2.One assumes that by some other fact that has been overlooked, one can make sense of a non-perturbative quantization of the EH action at arbitrary energies after all. This is for instance the case in speculations that EH-gravity has a UV fixed point?."

Which of the two above has more weight (if they are not equal) to the renormalization freedom of the CC. I mean what if gravity and QFT were just emergence of a third theory (this was mentioned by Smolin and other). The renormalization freedom of the CC stays valid?

15. Apr 8, 2018

### mitchell porter

My understanding is that in the G2-MSSM, the scale of the Higgs vev is generated and protected by specific mechanisms, such as you describe. But protection of the Higgs mass from large corrections still relies on low-scale supersymmetry (just like other susy models). And so in that regard, the G2-MSSM conforms to Strassler's reasoning. Strassler says that the scale "vmax" at which the SM ceases to be valid can't be too high, or else there will be unacceptable finetuning; and in these G2 compactifications of M theory, the MSSM is indeed supposed to replace the SM, not too far above the weak scale.

As for the cosmological constant, Bobkov argued that a version of the Bousso-Polchinski mechanism can work within the G2-MSSM.

16. Apr 9, 2018

### jtlz

To tame Vmax without fine tuning.. it seems supersymmetry (like MSSM) is really required not far from weak scale.. but then I wonder what stuff can do it too without supersymmetry.. anyone got any idea? Perhaps some quantum gravity stuff that can affect outside the planck scale too? Remember above 10^-18 are scales that are beyond particles.. maybe space is composed of something that can tame it? solving it and CC at same time and even the Hierarchy Problem?

Also what's weird beside having the mass so low is why its on the edge of metastability.. maybe the 4 are related somehow if you can use some mechanism of quantum gravity that can affect outside planck scale.

17. Apr 13, 2018

### mitchell porter

These comments have been bothering me for over a week now. To really address them authoritatively, I would have to review a lot of renormalization theory, but for now I'll just state my thoughts and see what comes of it.

There is a kind of field theory that is renormalizable up to arbitrarily high scales. The Lagrangian only contains renormalizable terms. This is the original concept of a renormalizable field theory.

Then there are field theories which also contain non-renormalizable terms with coefficients that have a dependence on a physical cutoff scale, above which the theory is not defined. This is called an effective field theory.

I feel like Urs's comments apply most directly to a field theory that is renormalizable in the original, unrestricted sense. This actually includes the standard model, because, as explained here, all the corrections can be absorbed into the renormalization.

However, Strassler is, by hypothesis, treating the standard model as an effective theory, valid only up to a physical cutoff scale. And he's saying that in that case, the higher the cutoff scale, the greater the finetuning required in order to keep the Higgs mass low.

Surely Strassler's reasoning can be made as rigorous as anything else in quantum field theory. We might have to use the "standard model effective field theory", in which the non-renormalizable terms are added to the SM lagrangian. And we might quantify the finetuning by looking at how restricted are the coefficients of the non-renormalizable terms (which codify the effects of unknown BSM physics), if we want the Higgs to stay light.

My complaint is that Urs treats strictly renormalizable field theory, and a UV-complete theory like string theory, as the only useful forms of reasoning, when in fact effective field theory is enormously useful. And a technical question would be, just how great are the "freedoms", the ambiguities, in effective field theory? I think they would be far more reasonable in size.

As evidence, I would point to the discussion of renormalon-induced uncertainties in the top quark pole mass, in QCD. No-one there is saying that the top quark pole mass can be anything at all; the uncertainty is pretty small. And I think that would be far more characteristic of the ambiguities in EFT quantities.

18. Apr 14, 2018

### Urs Schreiber

I don't see which distinction you are meaning to invoke. The perspective of effective field theory is one way of several to look at the issue of renormalization freedom. All these are different perspectives on the same thing. This is explained in the Subsection titled Wilson-Polchinski effective QFT flow within the PF-Insights on Mathematical QFT -- Renormalization.

Independently of that, it may be worthwhile to recall the following subtlety in the terminology "renormalizable":

The technical meaning of "renomalizable" just means that of the a priori infinitely many renormalization constants, it so happens that only a finite number appear. If a theory happens to be non-renormalizable in this sense, it just means that there are infinitely many choices to be made in computing any quantity at arbitrary energy. That may be physically disappointing, but it is not mathematically inconsistent.

This point was established way back by Epstein-Glaser 73, but somehow the community goes through cycles of forgetting and rediscovering it. Once place where this is rediscovered is

J. Gomis and Steven Weinberg,
"Are nonrenormalizable gauge theories renormalizable?",
Nucl. Phys. B 469 (1996) 473
(arXiv:hep-th/9510087)

Namely we can just keep making choices of renomralization constants on and on. In Epstein-Glaser 73 this infinite list of choices is organized by loop order, while in effective field theory it is organized by energy scale, but the principle is the same in both cases.

What is good about the Epstein-Glaser perspective of renormalization is that it gives one crystal clear picture of what the space of choices is.

19. Apr 16, 2018

### mitchell porter

The difference between a renormalizable field theory with a cutoff that can go to infinity, and a nonrenormalizable field theory that is still predictive but only up to a finite cutoff. As a reference, let me cite Mark Srednicki's QFT text, which introduces the distinction at the start of chapter 18, and develops effective field theory in chapter 29. The fine-tuning problem is mentioned halfway through chapter 29. The physical interpretation of a nonrenormalizable theory as an effective action valid only up to a finite cutoff begins just before eqn 29.37 ("The Wilson scheme also allows us...").

I'm just saying that this is the context in which Strassler's argument is made, and it is a valid argument given its premises: if there is an EFT with a light scalar, the higher the cutoff, the more tuning is required. (Maybe it could be formalized with Kevin Costello's framework.)


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# Swinging Vine

1. Jan 14, 2004

### MikeH

Swinging Vine(HELP ASAP)

All help is apprieciated.

Backwoods Allister hunts deer with the following weapon...a rock attached to a 2m vine. He holds the end above his head at a point 2m above the ground, when the above ground angle reaches 60 degrees with the vertical the rock breaks off and flies towards the deer. At what minimum distance from Allister can the deer stand with no danger of a direct hit?

Last edited: Jan 19, 2004
2. Jan 14, 2004

### Jupiter

I think it depends on how tall the deer is. But all in all this is a simple trig problem. Why don't you give it a shot first?

3. Jan 14, 2004

### MikeH

Well here's what I have so far. I used trig to find that the rock will be 1m above the ground and that the radius of the circle is 1.7m. From here I'd need to find the speed? Then use projectile? I'm kinda lost here.

4. Jan 14, 2004

### NateTG

Try applying conservation of energy.

5. Jan 14, 2004

### MikeH

How so?

6. Jan 14, 2004

### NateTG

The change in kinetic energy=the change in potential energy. That will give you the velocity of the rock. From there it's straight kinematics.

7. Jan 15, 2004

### MikeH

I know KE = 1/2mv^2
I know that the change in GPE = mg(delta)h
How would one write the change in KE?
1/2mv^2 - 1/2mvf^2 ??

8. Jan 17, 2004

### MikeH

Anyone?

9. Jan 18, 2004

### HallsofIvy

Staff Emeritus
Your original problem does not give enough information to answer this. You would need to know how fast he is swing the rock.

10. Jan 18, 2004

### MikeH

There must be some way, my teacher wouldn't have given a question that wasn't possible to answer.

I need to work to this formula: dh = vht

To find t(time).
dv= 1/2gt^2
t = (square root of)(1/2gdv)
t = 2.21 seconds

Now if I try to use Conservation of Energy.

KE(i) + GPE(i) = KE(f) (There is no GPE since it will be on the reference point)
.5mv(i)^2 + mgh = .5mv(f)^2
Masses cancel
.5v(i)^2 + gh = .5v(f)^2

This give you any ideas?

Last edited: Jan 18, 2004
11. Jan 19, 2004

### NateTG

You'd make my life a whole lot easier if you used LaTeX, but that's not really something you can deal with right now.

Ok, so:
$$KE_0+PE_0=KE+PE$$

Now, the change in height is:
$$2mg\sin{60}=\sqrt{3}mg$$

So, we have:
$$\frac{1}{2}mv_{launch}^2=\frac{1}{2}mv_0^2+\sqrt{3}mg$$
Some cancelations give:
$$v_{launch}=\sqrt{v_0^2+2\sqrt{3}}g$$

Whatever the initial velocity that you're dropping the weight with is is $$v_0$$. You might see some hint for that somewhere.

From there you can apply the normal kinematic formulae.

12. Jan 19, 2004

### MikeH

I don't understand what you mean. V(launch) is the speed that it leaves the rope and V(o) is the speed that it hits the ground with?

13. Jan 19, 2004

### NateTG

$$v_0$$ is the speed at the top of the swing.

14. Jan 19, 2004

### MikeH

I've been looking at this as swinging around him horizontally, I guess I've been looking at it the wrong way all along.

15. Jan 19, 2004

### MikeH

How did you find the change in height?
$$2mg\sin{60}=\sqrt{3}mg$$

I'm not really following this question too well. The more I think about it the more I get confused. Hopefully I get a snowday tomorrow so I don't have to pass it in. I need the help ASAP!
Thanks

Last edited: Jan 19, 2004
16. Jan 20, 2004

### NateTG

Maybe that should be cosine. You should be able to use trig to find the change in height.

format: plain


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# Motivating the Bessel translation operator

In a paper I am reading on the Hankel transform (this paper to be exact), I've come across a somewhat peculiar definition for a generalized translation operator. The operator is designed with a convolution theorem for the Hankel transform in mind. The Hankel transform, as given in the paper, is defined to be (for suitable $f$)

$$\mathcal{F}_{\alpha}f(y) = \int_0^{\infty} j_{\alpha}(2\pi xy)f(x)\,d\mu_{\alpha}(x),$$

where $d\mu_{\alpha}(x) = \frac{2\pi^{\alpha+1}}{\Gamma(\alpha+1)}x^{2\alpha+1}$ and

$$j_{\alpha}(x) = \Gamma(\alpha+1)\sum_{l=0}^{\infty} \frac{(-1)^l}{l!\Gamma(\alpha+l+1)}\left(\frac{x}{2}\right)^{2l}.$$

I understand the philosophy behind defining such a generalized translation however the intuition for how it it should be defined in this case is lost on me. Here's the definition that is given in the paper:

Definition Let $f\in C^2(\mathbb{R}^+)$, define the generalized Bessel translation operator $$T^{\alpha}_y f(x) = u(x,y), \quad x,y\in\mathbb{R}^+$$ where $u(x,y)$ is a solution of the following Cauchy problem $$\left(\frac{\partial^2}{\partial x^2} + \frac{2\alpha+1}{x}\frac{\partial}{\partial x}\right)u(x,y) = \left(\frac{\partial^2}{\partial y^2} + \frac{2\alpha+1}{y}\frac{\partial}{\partial y}\right)u(x,y)$$ with initial conditions $u(x,0) = f(x)$ and $\frac{\partial}{\partial x}u(x,0) = 0$.

With this definition, one can define a generalized convolution:

$$(f\ast_{\alpha} g)(y) = \int_0^{\infty} T^{\alpha}_y f(x)g(x)\,dx$$

for suitable $f$ and $g$. Then we have that $\mathcal{F}_{\alpha}(f\ast_{\alpha}g) = \mathcal{F}_{\alpha}f\cdot\mathcal{F}_{\alpha}g$ as desired.

The function $j_{\alpha}$ is an eigenfunction of the operator $D_{\alpha} = \frac{\partial^2}{\partial x^2}+\frac{2\alpha+1}{x}\frac{\partial}{\partial x}$ (with eigenvalue $-1$) so its inclusion in the definition for the generalized translation is reasonable to me. Especially considering that translation in the Fourier case is nothing more than the exponential of the derivative operator (for which the Fourier kernel is an eigenfunction). However beyond this triviality, it isn't clear to me why we should define generalized translation in such a way.

After some thought, I decided to see what happens in the Fourier case. Since $j_{\alpha}$ is an eigenfunction of $D_{\alpha}$, it stood to reason that we should look to the following PDE in the Fourier case:

$$\frac{\partial}{\partial x}u(x,y) = \frac{\partial}{\partial y}u(x,y) \tag{1}$$

since the exponential is an eigenfunction of the derivative operator and subject to the "initial" condition (if we view this as an evolution equation) $u(x,0) = f(x)$. The method of characteristics gives us that $u(x,y) = g(x+y)$ for some $g$. Based on our initial condition, it must be the case that $u(x,y) = f(x+y)$ and so the usual displacement operator naturally falls out of these considerations. We could also Fourier transform $(1)$ and again the same result pops out. This approach seems more promising for the case of Bessel translation operator since the method of characteristics isn't applicable.

This is a pretty interesting way to view displacement operators I feel, but very surprising. I would have naturally expected the Bessel translation operator should be defined by $j_{\alpha}(D_{\alpha})$ (or something quite similar) as a parallel to the Fourier displacement operator which can be viewed as $\exp(D)$. However after some thought, I can understand why the PDE approach is more desirable than considering $j_{\alpha}(D_{\alpha})$. The analysis of this operator is hardly trivial and very technical, whereas the PDE definition is very natural and easily analyzed (by comparison).

1. The generalized translation operator helps to move singularity for PDE from zero to any value.
2. It helps to define generalized positively defined functions and so on, analogous to use of standard shift in classical harmonic analysis. Now it is important for Dunkl operators.
3. Via this PDE form it was originally defined by J. Delsarte and in books of B.M.Levitan. Levitan construct general theory of generalized translation operators on groups.
4. Google gives 9 480 000 references for "generalized translation operator " -enough to satisfy any curiosity and questions, not so?
• I'm well aware with why we come up with generalized translation. It allows for a generalization of a convolution theorem, positive-definite functions, reproducing kernel Hilbert space related ideas, etc. That was not my question. My question was why we chose that particular definition but in light of my answer above, it is quite natural and reasonable. Initially one might have been tempted to do something like $j_{\alpha}(D_{\alpha})$ which I think is what Delsarte does but that operator is quite difficult to analyze; whereas the translation defined by Levitan is easy to analyze. – Cameron Williams Apr 2 '15 at 23:10
• It is also easy to generalize in this particular PDE form, just change a pair of differential operators $(B_\alpha,D^2)$ to another pair. – Sergei Apr 3 '15 at 4:13


text:
# Vector space with continuum basis as an inverse limit of finite dimensional vector spaces

Let $$K$$ be a field and $$Vec_K$$ - category of all vector spaces over $$K$$. I want to prove that every vector space with continuum basis is an inverse limit of countable inverse system of finite-dimensional vector spaces. $$\dots \leftarrow V_i \leftarrow V_j \leftarrow \dots$$

I need a hint about how I can construct such inverse system since all I came up with gave me a limit with countable basis. Thanks in advance!

• Just to make it clear, a "continuum basis" for $V$ means that a minimal set of algebraic generators for $V$ has cardinality $2^{\aleph_0}$? – Fosco Dec 25 '19 at 19:54
• @Fosco yes, I meant exactly it – Gleb Chili Dec 25 '19 at 19:56
• You can obtain such a $V$ from a filtered co limit of cardinality $2^{\aleph_0}$, of finite dimensional spaces. Do you want a limit instead? Was it a typo? – Fosco Dec 25 '19 at 19:59
• @Fosco I want to show that every continuum basis space is a projective limit of countable inverse system of finite-dimensional spaces in the following sense: en.wikipedia.org/wiki/Inverse_limit – Gleb Chili Dec 25 '19 at 20:24
• Yes, my question was meant to avoid the possible misunderstanding of confusing projective limit (i.e. limit) VS inductive limit (i.e. colimit). Did you find the statement quoted somewhere? I see an evident reason why every space is a COlimit of finite spaces, I have never seen a similar result for limits – Fosco Dec 25 '19 at 20:27

This is not true without additional hypotheses on $$K$$. To compute the inverse limit of a system of finite-dimensional vector spaces, you can just observe that it is dual to the direct limit of the dual system (this is immediate from the universal properties; see for instance this answer). In particular, the inverse limit of any system of finite-dimensional vector spaces is a dual vector space, so is isomorphic to $$K^S$$ for some set $$S$$. But the dimension of $$K^S$$ is always $$|K^S|$$ if $$S$$ is infinite (see this answer on MO). In particular, if $$|K|>2^{\aleph_0}$$, then any inverse limit of finite-dimensional vector spaces which is infinite-dimensional must have dimension greater than $$2^{\aleph_0}$$.

On the other hand, if $$|K|\leq 2^{\aleph_0}$$, then it is true (and it is true not just for some sequential inverse limit of finite-dimensional vector spaces but every sequential inverse limit of finite-dimensional vector spaces which is infinite-dimensional). Indeed, for sequential systems, the colimit of the dual system must be countable-dimensional, and so if it is infinite-dimensional the inverse limit will be isomorphic to $$K^{\mathbb{N}}$$ which has dimension $$|K|^{\aleph_0}=2^{\aleph_0}$$ since $$|K|\leq 2^{\aleph_0}$$. For a very concrete example, let $$V_n=K^n$$ where the maps $$V_{n+1}\to V_n$$ are the projections that drop the last coordinate, so the inverse limit is $$K^{\mathbb{N}}$$ in a very natural way.


text:
# Prove the convergence of $\prod\limits^{n}_{k=1}{\left(1+\frac{k}{n^p}\right)}$ and Find Its Limit

Suppose $$p> 1$$ and the sequence $$\{x_n\}_{n=1}^{\infty}$$ has a general term of $$x_n=\prod\limits^{n}_{k=1}{\left(1+\frac{k}{n^p}\right)} \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space n=1,2,3, \cdots$$ Show that the sequence $$\{x_n\}_{n=1}^{\infty}$$ converges, and hence find $$\lim_{n\rightarrow\infty}{x_n}$$ which is related to $$p$$ itself.

I have been attempted to find the convergence of the sequence using ratio test but failed. The general term has a form of alike the $$p$$-series. And also the question seems difficult to find its limit because the denominator is of $$p^{th}$$ power. How do I deal it?

• There's no way it converges when $p=1$ – mathworker21 Nov 22 '18 at 14:19
• What if the case for $p>1$? I have edited the question. – weilam06 Nov 22 '18 at 14:21
• $\log(x_n) \le \sum_{k=1}^n \log(1+\frac{k}{n^p}) \le \sum_{k=1}^n \frac{k}{n^p} = \frac{1}{2}n^{2-p}$, which goes to $0$ if $p > 2$. Idk about $1 < p \le 2$. – mathworker21 Nov 22 '18 at 14:23

We have that

$$\prod\limits^{n}_{k=1}{\left(1+\frac{k}{n^p}\right)}=e^{\sum^{n}_{k=1}{\log \left(1+\frac{k}{n^p}\right)}}$$

and

$$\sum^{n}_{k=1}{\log \left(1+\frac{k}{n^p}\right)}=\sum^{n}_{k=1} \left(\frac{k}{n^p}+O\left(\frac{k^2}{n^{2p}}\right)\right)=$$$$=\frac{n(n-1)}{2n^{p}}+O\left(\frac{n^3}{n^{2p}}\right)=\frac{1}{2n^{p-2}}+O\left(\frac{1}{n^{2p-3}}\right)$$

therefore the sequence converges for $$p\ge 2$$

• for $$p=2 \implies x_n \to \sqrt e$$
• for $$p>2 \implies x_n \to 1$$

and diverges for $$1.

Refer also to the related

• How do you attempt $\sum\limits^{n}_{k=1}{\log {(1+\frac{k}{n^p})}^{n^p}}$~$\frac{n^2}{2}$? – weilam06 Nov 22 '18 at 15:06
• For n large $(1+k/n^p)^{n^p}\to e^k$ and $\sum k=n(n+1)/2$. – gimusi Nov 22 '18 at 15:23
• @weilam06 In my answer I've given the key fact to find a complete solution, I can add some more detail to make thinkg more clear and rigorous. – gimusi Nov 22 '18 at 15:56
• That's okay. I understand that $\frac{1}{2n^{p-2}}$ is a $p-$series. – weilam06 Nov 22 '18 at 16:08
• @weilam06 That'snice. I've just modified a little bit the key step in a more precise way! Bye – gimusi Nov 22 '18 at 16:13


text:
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# The figure above represents a triangular field.

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The figure above represents a triangular field. [#permalink]  14 Oct 2017, 15:04
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The figure above represents a triangular field. What is the minimum distance, in meters, that a person would have to walk to go from point A to a point on side BC?

[Reveal] Spoiler: OA
60

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Re: The figure above represents a triangular field. [#permalink]  14 Oct 2017, 20:47
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The figure above represents a triangular field. What is the minimum distance, in meters, that a person would have to walk to go from point A to a point on side BC?

[Reveal] Spoiler: OA
60

Here the two sides AB and AC are equal, so we can conclude that

the$${\triangle} ABC$$ is isosceles.

Now the minimum length from A to BC will be the perpendicular line.

Since the $${\triangle} ABC$$ is isosceles and perpendicular line from A to BC will meet at the midpoint D (as shown in the figure)

Now consider the $${\triangle} ABD$$

AB = 100 , BD =80 and we need to find AD

Using Pythagoras theorem

AD = $$\sqrt{(10000 - 6400)}$$ = 60

or we can see that the sides of $${\triangle} ABD$$ are in the ratio 3:4:5

= 60:80:100

Ans = 60
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Re: The figure above represents a triangular field.   [#permalink] 14 Oct 2017, 20:47
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## Ash90 one year ago i need help solving this replace r and g with the following and find the time-out. r=91-7* sqrt(t) g=14*t

1. Ash90

@AkashdeepDeb

2. Ash90

idk what to do with this?

3. AkashdeepDeb

Do we just have to find the time in this question? And is any other equation given? :)

4. Ash90

yes we just need to find the time in this equation but I have to write it out by showing my work. let me check and see if anything else is given...

5. Ash90

that's all nothing else is need right now.

6. Ash90

what this is is a programm that when you do the calculations the colors are set up to do what you need them to.

7. AkashdeepDeb

That's cool! Well, if you have to find the value of t in terms of r or in terms of g, you can do that. But if you want to find the value of t, in terms of real numbers, [Although, I don't think it is possible] I suggest you tag someone else. The equations you have posted are called Parametric equations, you can also express r in terms of g here but I am not sure how to go about calculating the sole value of 't'. :)

8. Ash90

could i show you what i have and maybe you can help from there?

9. Ash90

|dw:1397796459525:dw|

10. Ash90

|dw:1397796552699:dw|

11. AkashdeepDeb

Yes, I can help you from there! But is it given that r=g ? Because you have equated it over there! :D

12. Ash90

well i think what i had to do was set them equal to each other if i'm not mistaking i think thats what the teacher wanted.

13. AkashdeepDeb

Okay. So if r=g $$91-7 \sqrt{t} = 14t$$ $$91-14t = 7\sqrt{t}$$ $$Squaring ~~both ~~sides ~~we~~ get:$$ $$(91-14t)^2 = 49t$$ $$196t^2 + 8281 - 2548t = 49t$$ $$196t^2 - 2597t + 8281$$ $$~~~~~~~~~~~~Where~~[a = 196, ~b = -2597,~c = 8281]$$ Use the quadratic equation below to solve for $$t$$: $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ Got it? :)

14. Ash90

$t=\frac{ 53+\sqrt{105} }{ 8 }, t=\frac{ 53-\sqrt{105} }{ 8 }$

15. AkashdeepDeb

Yes, if you got that by calculating the above equation, then yes. :)

16. Ash90

yay!

17. Ash90

can we do one more?

18. AkashdeepDeb

Sure!

19. Ash90

okay thanks.

20. Ash90

$r=7*t^2-51*t+92$

21. Ash90

for this one i thought using quadratic formula right of hand would be a good idea

22. AkashdeepDeb

You cannot always do that. The quadratic formula can only be used when f(x) = ax^2 + bx + c = 0 [And zero only] As it gives us the zeroes of the polynomial/expression. What you could do is write it like this: 7t^2 - 51t + 92 - r = 0 Now apply the quadratic formula, considering, [c = 92-r] Getting this? :) You may also try this by the completing-the-square method, although I am not a 100% sure if it'll work here.

23. Ash90

|dw:1397798167883:dw|

24. Ash90

|dw:1397798377673:dw|

25. Ash90

thanks for the help

26. AkashdeepDeb

:)


text:
# Long-time support

LaTeX files with a bibliography (or an index) require several compiler passes to resolve all references for building:

latex paper.tex
bibtex paper.aux
latex paper.tex
latex paper.tex

Tedious, isn't it? For that reason, wrappers have been invented, automating this procedure. The veterans on this field are latexmk and rubber, but there are several others, including a new kid on the block called latexrun that sounds truly promising.

I've used rubber for many years, despite the fact that its development had stalled nine years ago. Rubber has worked for me since scientific publishers are ... ahem ... extremely conservative: they will probably stick to the pdflatex/bibtex workflow for the next few centuries (or forever, whatever comes earlier).

However, for documents not intended for submission to a publisher, I've long ago switched to XeLateX/Biber. For example, my curriculum vitae is to be compiled with these engines rather than with their prehistoric ancestors. Since it is technically identical to its template, you can download Adrien's CV to test the wrapper you want to use.

The following commands should compile this CV in a single run, including the bibliography:

latexmk --xelatex --synctex=1 --silent cv.tex

rubber --module xelatex --module biber --synctex cv.tex

latexrun --latex-cmd xelatex --bibtex-cmd biber cv.tex

In July 2015, only latexmk managed to do that, while rubber 1.2 failed despite my manual addition of XeLaTeX and Biber support. I was only mildly disappointed: after all, rubber had become lame and grey, and I realized that it was time to leave it behind. I was surprised, though, that latexrun, the newly developed contender, failed as well.

Now, to my absolute astonishment, the rubber project suddenly became alive again this year. After nine years of inactivity, rubber 1.3 was published in October, and 1.4 just a few days ago. Rubber 1.4 passed the test above with flying colors, and twice as fast as latexmk. What's more, SyncTeX support, which I previously smuggled in manually, has now been added to the rubber feature list as well.

Rubber thus manages to once again take the crown of LaTeX wrappers, with latexmk just seconds behind (literally) because of its inferior performance. Latexrun in its present state leaves much to be desired, but I'll keep an eye on its further development (if any ;) ).


text:
## Journal of Symbolic Logic

### Possible PCF Algebras

#### Abstract

There exists a family $\{B_\alpha\}_{\alpha < \omega_1}$ of sets of countable ordinals such that: (1) $\max B_\alpha = \alpha$, (2) if $\alpha \in B_\beta$ then $B_\alpha \subseteq B_\beta$, (3) if $\lambda \leq \alpha$ and $\lambda$ is a limit ordinal then $B_\alpha \cap \lambda$ is not in the ideal generated by the $B_\beta, \beta < \alpha$, and by the bounded subsets of $\lambda$, (4) there is a partition $\{A_n\}^\infty_{n = 0}$ of $\omega_1$ such that for every $\alpha$ and every $n, B_\alpha \cap A_n$ is finite.

#### Article information

Source
J. Symbolic Logic, Volume 61, Issue 1 (1996), 313-317.

Dates
First available in Project Euclid: 6 July 2007

https://projecteuclid.org/euclid.jsl/1183744942

Mathematical Reviews number (MathSciNet)
MR1380692

Zentralblatt MATH identifier
0878.03036

JSTOR


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