August 6, 2010

Thermodynamics and Wick Rotation

Posted by John Baez My grad student Mike Stay is a busy guy with many interests. He has a full-time job at Google working on Caja, which is a ‘source-to-source translator for securing Javascript-based web content’. He’s working with me on monoidal bicategories and their applications to computer science. And on the side, he’s been pondering some questions about thermodynamics and the idea of ‘Wick rotation’.

I asked him to post those latter questions here. Can you help him out?

guest post by Mike Stay

In A Spring in Imaginary Time, John Baez showed how statics in $n$ dimensions turns into dynamics in $n-1$ dimensions. Consider a spring whose endpoints are fixed. The height of the spring is given by a function

$q(s):\mathbb{R} \to \mathbb{R}.$

The energy $E$ of the spring is given by

$E = \int \left[\frac{k}{2}\left(\frac{d q(s)}{d s}\right)^2 + V(s)\right]ds.$

where we interpret the data as follows:

$\array{s & unitless parametrization \\ q(s) & height of the spring at s, in meters \\ k & spring constant, in k g/s \\ V(s) & gravitational potential energy density at s \\ T(s) = \frac{k}{2}\left(\frac{d q(s)}{d s}\right)^2 & stretching potential energy density at s \\ S = \int (T + V)(s) d s & total energy in the spring }$

If we allow $s$ to be complex—that is, let $s = x + it$—and take $q$ and $V$ to be complex-analytic functions, then we can set $x = 0$ and reinterpret the equation as follows:

$\array{t & time, in seconds \\ q(it) & the height of the particle at time t, in meters \\ m & mass, in k g \\ V(i t) & potential energy at time t \\ T(i t) = \frac{m}{2}\left(\frac{d q(i t)}{i d t}\right)^2 & minus kinetic energy at time t \\ S = \int (T + V)(i t) i d t& \\ = -i \int \left[\frac{m}{2}\left(\frac{\partial q(i t)}{\partial t}\right)^2 -V(i t)\right] dt & -i times the total action of the path }$

If we let the potential energy be proportional to the height, then we find that the critical point $\delta S = 0$ is where $q(s) \propto s^2$; along the real and imaginary axes, it’s a parabola, opening upward for a hanging spring and downward for a thrown particle.

There’s another substitution we could do, however; Wick rotation relates quantum mechanics to statistical mechanics by replacing $i t / \hbar$ with $\beta/k_B$. If we do the same thing in this equation, the interpretation is less clear to me:

$\array{\beta & inverse temperature, in K^{-1} \\ q(\beta) & height of something at invtemp \beta, in meters \\ r & bits / m^2 K \\ T(\beta) = \frac{r}{2}\left(\frac{d q(s)}{d \beta}\right)^2 &"kinetic" energy \\ V(\beta) & potential energy \\ S = \int (T + V)(\beta) d \beta & total bits of entropy in the path }$

where one bit = $k_B ln(2) \; [J/K]$.

What kind of system is governed by this equation? Does the fact that $r$ has units of information per square meter Kelvin have anything to do with the holographic principle? What does it mean for there to be entropy in a single path? When $\delta S = 0,$ does this maximize the entropy?

Say we have a system with microstates indexed by $j$. The entropy of a distribution $p$ is

$S = - \sum_j p(j) ln p(j).$

We’re told that the average energy is $E$; the distribution that maximizes entropy subject to that constraint is the Gibbs distribution

$p(j) = \frac{exp(-\beta E)}{Z},$

where

$Z = \sum_j exp(-\beta E_j)$

and

$\beta = \frac{\partial S}{\partial E}.$

Wick rotation suggests there ought to be an analogous situation in which

$i t = \frac{\partial S}{\partial E};$

where $S$ is action. If so, then action should be a function of some distribution-like thing. Is it the wave function?

We can extend the analogy between quantum and statistical mechanics:

$\array{wave function modulo global phase & & distribution modulo multiplicative constant \\ \langle Q \rangle = \langle\phi| exp(-H i t/\hbar) Q|\phi\rangle & & \langle Q \rangle = \langle\phi| exp(-H \beta/k_B) Q|\phi\rangle}$

In the second line, $\phi$ is the uniform distribution; in the quantum case the equation only holds when $Q$ is an observable that commutes with $H$, while in the statistical case, it’s any observable.

What’s the complete picture? Do we get pairs of conjugate observables whose product has units of entropy? Can we talk about perturbation theory and Feynman diagrams for a statistical ensemble? How far does the analogy go? What are the important differences?

Posted at August 6, 2010 3:43 AM UTC

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Re: Thermodynamics and Wick Rotation

A very partial answer to your question “Can we talk about perturbation theory and Feynman diagrams for a statistical ensemble?” is “yes, as for any integral”. More precisely, recall that in Dyson’s original work on Feynman diagrams, he rejected Feynman’s idea that the diagrams were pictures of physical events in favor of a purely formal mathematical interpretation: the diagrams for Dyson do nothing more nor less than keep track of the terms of an asymptotic expansion of an integral. Let $f$ be any real-valued smooth function on $\mathbb{R}^n$, $C \subseteq \mathbb{R}^n$ compact and the closure of its interior, and suppose that $f$ has a unique critical point in $C$. Then as $\epsilon \to 0$, the asymptotics of $\int_C \exp( i\epsilon^{-1} f)$ depend only on the values of $f$ in a formal neighborhood of its critical point, i.e. only on the Taylor coefficients. When the critical point is a minimum, the same statement holds upon replacing $i$ by $-1$, and the dependence on the Taylor coefficients is the same. In particular, when the critical point is nondegenerate, and now letting $\epsilon$ be pure imaginary or negative as the case may be, we have:

$\epsilon \log \int \exp( \epsilon^{-1} f) = \sigma + f^{(0)} - \frac{\epsilon}{2} \log \det f^{(2)} + \sum_{\text{connected diagrams }\Gamma} \frac{\operatorname{ev}(\Gamma) \, \epsilon^{\beta(\Gamma)}}{|\operatorname{Aut}\Gamma|} + O(\epsilon^\infty)$

Here $f^{(k)}$ is the $n$th Taylor coefficient of $f$ at the critical point (a symmetric tensor in $(\mathbb{R}^n)^{\otimes k}$; $\sigma$ depends only on $\epsilon, n$ and the signature of $f^{(2)}$; a diagram is a graph with all vertices trivalent or higher; the evaluation $\operatorname{ev}(\Gamma)$ of a diagram $\Gamma$ places a weight on each vertex, a balloon on each edge, and reads it from top to bottom as a “string” diagram in Vect where the vertical edges as $\mathbb{R}^n$, the caps are the $(-f^{(2)})^{-1}$, and an $k$-valent vertex is $f^{(k)}$; and the first Betti number $\beta(\Gamma)$ is the dimension of the first homology of $\Gamma$ (one plus the number of edges minus the number of vertices). The sum is of course a Baez-and-Dolan “groupoid integral” over the groupoid of all connected diagrams. The LHS is not analytic in $\epsilon$; the RHS makes it clear that the failure of analyticity is in $\sigma$ alone. Exponentiating, we have $\exp\sigma = (2\pi\epsilon)^{n/2} i^{-\nu(f^{(2)})}$, where $\nu(f^{(2)})$ is the Morse index of $f$ at the critical point, i.e. the number of negative eigenvalues of $f^{(2)}$, i.e. the maximal dimension of any subspace of $\mathbb{R}^n$ on which $f^{(2)}$ is negative-definite, if I’m remembering the signs correctly, which I probably am not. It’s worth emphasizing that upon exponentiating the sum of connected diagrams, you get the sum of all diagrams, still weighted in the groupoid-integral sense.

(Incidentally, it’s tempting to interpret $f^{(0)}$ as the value of the connected Feynman diagram with a unique 0-valent vertex. In fact, remembering that $f^{(1)} = 0$, you can try to consider the sum of all connected diagrams, without the condition that all vertices be trivalent. When you do this, you should evaluate a $k$-valent vertex as $f^{(k)}$, and evaluate an edge not as $-(f^{(2)})^{-1}$ but as “$0^{-1}$” — more precisely, add and subtract from $f^{(2)}$ some metric on $\mathbb{R}^n$, and use the inverse metric for a cap and the rest for a bivalent vertex. Then there are infinitely many diagrams per Betti number, and the sums do not converge, but formally they equal what is above. Actually, if all functions were analytic, and if all reasonable series converged, then you could pick any point in $C$, and do the same thing, now allowing 1-valent vertices to represent $f^{(1)}$.)

Thus, since one formalism for statistical mechanics is in a “path integral” sense, it is perfectly reasonable to employ Feynman diagrams for computing asymptotics. But I don’t believe that the diagrams depict pictures of physical events in statistical mechanics. Similarly, from the path-integral formalism for quantum mechanics, you can compute asymptotics via Feynman diagrams — this is implicit in the work by e.g. DeWitt-Morette and also Kleinert, and I have some papers on the arXiv that make the diagrammatic approach very explicit. But again the diagrams do not depict physical events. It is only for formal QFT on a Riemannian manifold that perturbs a free theory that it is reasonable to interpret the pictures as events.

Posted by: Theo on August 6, 2010 7:12 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

the evaluation ev($\Gamma$) of a diagram $\Gamma$ places a weight on each vertex, a balloon on each edge

Is a balloon a negative weight, like a helium balloon? Or something else?

Posted by: Mike Stay on August 6, 2010 7:17 PM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

I have some papers on the arXiv that make the diagrammatic approach very explicit.

I’d love to read them! Could you post links? (Or if you’d prefer to remain anonymous to the rest of the cafe, mail the links to metaweta@gmail.com?)

Posted by: Mike Stay on August 7, 2010 11:59 PM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Posted by: Mike Stay on August 17, 2010 1:16 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Disclaimer: I am a lawyer.

OK, I’m not getting this. It looks like there’s an analogy being claimed between a spring and a particle in what appears to be a gravitational field, i.e.

s is supposed to be a “unitless parameterization” but it really looks like it does have units, i.e. time, otherwise q(s) which is “height of the spring at s, in meters” would be without meaning, and “allowing s to be complex” letting “s = x + it” would change the unitless parameter into a parameter with the units of (complex) time, if x=0. c = a + bi is a complex number, so maybe it would be fair to say that s = (ax + bi)t, then you could set a to 0 and get rid of the real part, and set b to unity, so you’d end up with s = it, which would have units of “complex time”, whose physical meaning is a mystery to me.

q(s) is a distance, in meters.

The total energy of a spring is comprised of potential and kinetic energy. If the endpoints are fixed, and the spring is under tension, i.e. it’s stretched, then there’s no kinetic energy term because nothing’s moving. All the energy is potential energy, part of it due to the action of the gravitational field on the mass of the spring and part of it due to the tendency of the spring to resist deformation (the spring constant).

if the endpoints are fixed q(s) doesn’t change. The potential energy of a spring due to the deformation of the spring is T(x) = (1/2)(k)x**2. In the equation for the “stretching potential energy density at s”,
T(s) = (1/2)(k)(dq(s)/ds)**2, which seems to me to be kind of strange, since it suggests that dq(s)/ds, the rate of change of a distance with respect to s, which either has units of time or “complex time”, is somehow equivalent to a distance. Perhaps the expression for “stretching potential energy density at s” should be T(s) = (1/2)(k)(q(s))**2 instead.

Of course, I’m just a lawyer remembering physics from long ago, and I might have missed something…

Posted by: streamfortyseven on August 6, 2010 8:52 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Yes, the units in the analogy above are troublesome. The proper units for $s$ in the first example are “meters” and for $d q/d s$ are “meters per meter”, that is, “meters up per meter across”. If we use $y$ for “meters up” and $x$ for meters across, then [$d q/d s$] = [$y/x$] instead of .

If we do that, then the spring constant $k$ has to have units [$k g x/s^2$]; $T$ and $V$ become energy densities with units [$J/x$] so that $\int(T+V)(s) d s$ is an energy.

Now if we set $s = x + i t$, then we’re effectively saying that imaginary meters are seconds; see also Toby Bartels’ explanation (scroll down to “Time is imaginary”).

Posted by: Mike Stay on August 6, 2010 5:00 PM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

scroll down to “Time is imaginary”

Posted by: Toby Bartels on August 6, 2010 5:29 PM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

I see in the “Time is imaginary” paper that the Minkowski arclength element is defined as ds = sqrt((dx)**2 + (dy)**2 + (dz)**2 - (dt)**2) = sqrt((dx)**2 + (dy)**2 + (dz)**2 + (idt)**2) … then Bartels states “instead of working with t, work with i*t, instead of working with momentum p, work with i*p… Velocity is imaginary, the magnetic field is imaginary and so on…”

what it looks like to me is that the elements dx, dy, and dz appear to be fundamentally dissimilar to dt, the former measured in meters, the latter in seconds. I study aikido, and my sensei stated one day that there is no such thing as time without movement; time and movement are inseparable… so maybe in spacetime, in the 4 dimensional euclidean metric, the arclength ds is measured in “meter*seconds”, where dx, dy, and dz, which we think of as being solely extensive in space and moving along mutually orthogonal spatial axes really have a time quality as well, and where dt has a space quality, instead of being solely extensive in time and moving along a time axis. So maybe the imaginary nature of the time element dt is a way to make dt mutually orthogonal with dx, dy, and dz, reflecting a “conjugate space” which is intimately connected with “real space”.

So when you set s = x + it, s really is in units of “meter*seconds” … you’re not quite equating meters with seconds, you’re saying that meters and seconds are inseparable in some fundamental physical way, like particles and waves… and so there’s no such thing as a “meter” or a “second”, there’s just “meter*seconds” or some such.
================================
Disclaimer: I Am A Lawyer

Posted by: streamfortyseven on August 6, 2010 9:59 PM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

sorry, it’s not “meters” which are being equated with “seconds”, it’s “imaginary meters”, which I take are distances along complex axes ix, iy, and iz which are mutually orthogonal to each other and also to a real time axis t…

so now we have Minkowski space, with:
x, y, z: real mutually orthogonal spatial axes and
it : complex axis mutually orthogonal to the spatial axes x, y, and z

and

conjugate Minkowski space, with:
ix, iy, iz: imaginary mutually orthogonal spatial axes and
t : real axis mutually orthogonal to the spatial axes ix, iy, and iz.

It’s a weird space because x, y, and z are just components of some indivisible spatial quantity, so maybe we should say something like
ds = sqrt((dr)**2 + (idt)**2) where dr = sqrt((dx)**2 + (dy)**2 + (dz)**2)

At this point, I’m just trying to visualize and figure out how this all works out, so excuse my fumbling…

Posted by: streamfortyseven on August 6, 2010 10:18 PM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Only someone who doesn’t care about units could write something like $x + i t$ without getting a cringe in their stomach. When working with meters and seconds, you must write $x + i c t$, but after wrangling with mathematicians long enough we might agree to use space and time units such that

$c = 1 \frac{distance unit}{time unit}.$

In this case, the term $i t$ in $x+i t$ secretly has a unit-full constant $c$ that converts the time units to distance units without changing the numerical value.

But when we do this, you have to remember we are no longer talking about meters and seconds, which I happen to think is more of a travesty than carrying around $c$ everywhere in your calculations.

The Minkowski metric should be

$\eta = d x\otimes d x + d y\otimes d y + d z\otimes d z - c^2 d t\otimes d t.$

Note the $c$’s!

Dropping $c$’s from calculations MIGHT have made sense 100 years ago when typesetting in journals was a pain, but today? There is no point and only confuses things.

Posted by: Eric Forgy on August 7, 2010 1:05 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Well, I think part of the point of Toby’s note was to say that

(1)$1 s = 3\times 10^8 i m,$

that is, we can choose units of time so that

(2)$1 time = i length$

and then length and time aren’t different units.

But the thermal stuff doesn’t even use $i$, so it should be far less controversial; what kind of system is governed by the equation when we use $\beta$ instead of $s$?

Posted by: Mike Stay on August 7, 2010 1:22 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Or put another way, he is saying we can define a new time variable

$\tau = i c t$

(note the $c$!) such that

$\eta = d x\otimes d x + d y\otimes d y + d z\otimes d z + d\tau\otimes d\tau.$

That is fine. Whether you write $c$ or fill in the number $3\times 10^8$, doesn’t matter. But even I am too lazy to write $3\times 10^8$ :)

Posted by: Eric Forgy on August 7, 2010 1:39 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

It’s kind of fun to occasionally resurrect some old conversation from a decade or more ago since everything still exists on SPR :)

If I knew then what I know now and I read Toby’s article back then, I would have tried to convince him that the best thing to overcome his concerns would be to forget about “Cartesian” coordinates $x$, $y$, $z$, and $t$ altogether. The reason the metric looks funny with a seemingly arbitrary “choice” of metric coefficient $\pm \delta_{i,j}$ is due to unnatural choice of Cartesian coordinates in the first place.

Nature doesn’t like Cartesian coordinates.

Nature likes light-like coordinates (say that 10 times fast).

Sticking to 2-spacetime dimensions, we have 2 Cartesian coordinates $x$ and $t$ and light rays can travel in one of two directions

$d v^+ = d x + c d t$

and

$d v^- = d x - c d t.$

We can invert these relations giving

$d x = \frac{1}{2} \left(d v^+ + d v^-\right)$

and

$d t = \frac{1}{2c} \left(d v^+ - d v^-\right).$

With these coordinates, the metric looks like

$\eta = \frac{1}{2} \left( d v^+\otimes d v^- + d v^-\otimes d v^+\right).$

Mystery solved.

The problem isn’t whether time should be real or imaginary, but whether we should be using Cartesian coordinates that split spacetime artificially into space and time at all.

This realization was the “EUREKA!” moment for me (thanks to Urs) in my work on discrete/finitary models of spacetime.

In early attempts, I was trying to discretize space while time was an afterthought, but it turns out the only natural way to get things to work out is to treat them together. This amounts to abandoning Cartesian coordinates in favor of light-like coordinates. Edges in discrete spacetime need to be light-like. Not Cartesian.

Posted by: Eric Forgy on August 7, 2010 2:09 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Eric wrote:

Only someone who doesn’t care about units could write something like $x+i t$ without getting a cringe in their stomach.

Or $x + i y$. We all know that horizontal distances are measured in furlongs, while vertical distances are measured in fathoms! We need to write $x + i c y$ where $c = 0.00909090909$ furlongs/fathom.

(I’ve heard of a crackpot named Schmeinstein who says vertical distances and horizontal distances are just two aspects of the same concept. He claims you can ‘rotate’ one to the other. But that’s absurd, because then one person’s ‘up’ could be another person’s ‘east’, which makes no sense.)

Dropping $c$’s from calculations MIGHT have made sense 100 years ago when typesetting in journals was a pain, but today? There is no point and only confuses things.

Okay. I’ll call a press conference and tell reporters from all the major newspapers and television networks that every theoretical physicist working on quantum field theory and general relativity is doing things wrong. They need to switch back to writing in the $c$’s! It’s too confusing to leave them out!

Seriously: I bet this issue is not very relevant to Mike’s original questions. He could have written in a bunch of $c$’s to make you happy, and still asked those questions.

It is indeed fun to see discussions like those from the old sci.physics.research days. Digressions are part of the fun. But I wanna see more people take a crack at Mike’s questions!

Posted by: John Baez on August 7, 2010 2:35 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Note, my reply was to a reasonable set of elementary questions from streamfortyseven. The $c$’s don’t bother me (obviously I’ve been doing things long enough), but they only confuse the uninitiated (as observed here).

There really is no point to leave them out except for conformity with some social norm, i.e. that is the way its always been around here.

Maybe, for the sake of this conversation, we don’t care about the uninitiated and that would be fine.

Posted by: Eric Forgy on August 7, 2010 2:51 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Eric wrote:

The $c$’s don’t bother me…

Good — I’m glad you were just feigning botherment when you said

Dropping $c$’s from calculations MIGHT have made sense 100 years ago when typesetting in journals was a pain, but today? There is no point and only confuses things.

An excellent acting job!

Seriously, I’m glad you’re telling streamfortyseven some stuff that might help him. I’m all for helping out the uninitiated. But part of getting initiated is learning that theoretical physicists who take relativity seriously — special or general — tend not to include $c$’s in their calculations — not until the end, when they hand the answers over to the experimentalists. The disgust you feigned at ‘leaving out the $c$’s’ is not an attitude I’d encourage — if only because beginners with that attitude tend to be mocked by experts, as I am illustrating here.

But back to the main show!

Posted by: John Baez on August 7, 2010 3:18 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Surely one of the fundamental insights of relativity theory is that space and time are really two different aspects of the same thing, and therefore they really must be measured in the same units? So leaving out the $c$ is not just a convention or a convenience or a matter of “getting initiated;” it has real conceptual content.

Posted by: Mike Shulman on August 9, 2010 8:09 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Mike wrote:

So leaving out the $c$ is not just a convention or a convenience or a matter of “getting initiated;” it has real conceptual content.

I agree. If I were feeling argumentative — and on the internet, who isn’t? — I might argue that the realization that space and time are the same thing is part of getting initiated when it comes to relativity.

But anyway, yeah: space and time are the same thing, and you have to really bend over backwards to measure them in different units, as my example involving furlongs and fathoms was supposed to indicate.

Posted by: John Baez on August 9, 2010 8:52 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Everything should be made as simple as possible, but not simpler.

I believe someone famous once said that :)

Space and time are part of the same thing, but they are not the same in the way that furlongs and fathoms are the same. The parable of furlongs and fathoms for making a point about time units is an example of making things simpler than possible.

I like to play, “What is more fundamental?”

Sure, coordinate independence is an important lesson from relativity, but is that the most fundamental thing? To me, the most fundamental thing is causality. The light-cone is the fundamental thing that makes space and time “different”. You can play around with different space and time units all you want, but you can never mess with the light cone. So inserting a $c$ to pay homage to this distinction is less of a sin than confusing furlongs and fathoms.

Posted by: Eric on August 9, 2010 9:52 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Hey, wait a minute! You played with your super powers! :)

My response was to a curt two sentences of yours, but when I posted my comment, I see a lot more content that makes my response seem out of place. What happened? :)

Anyway, I understand if you are concerned about the discussion deviating from the original topic of the post. But if that is the case, a better approach to bring it back would be to write something ON TOPIC rather than complain about the deviation.

I’m sure you have a lot you could say on the subject and I’d love to hear it (we all would) :)

Posted by: Eric Forgy on August 7, 2010 3:18 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Eric wrote:

Hey, wait a minute! You played with your super powers! :)

Sorry, I often rewrite my $n$-Café comments three or four times to bring them to a devastating level of wit and elegance. (I don’t actually talk like this, in case anyone is wondering.)

Anyway, I understand if you are concerned about the discussion deviating from the original topic of the post.

I mainly just enjoy teasing engineers who get upset at how physicists work in units where time and space have the same dimensions — and it’s been a long time since I’ve had the chance.

Posted by: John Baez on August 7, 2010 3:59 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

and from a dimensional analysis standpoint including the c’s explicitly makes things work out, since c is a velocity in meters/second, so

ds = sqrt((dx(m))(dx(m)) +
(dy(m))(dy(m)) +
(dz(m))(dz(m)) +
c**2(m**2/s**2)*(idt(s))(idt(s)))

so ds ends up with units in meters, and so we don’t have the weird stuff happening with space and time after all, maybe we do, but it’s fixed by putting c in the expression. the speed of light is vital, it’s not a dimensionless scalar, for the dimensional analysis to work out it *has* to be in there, else everything’s weird. So instead of looking at i*t, we’re looking at i*c*t, as in

s(m) = x (m) + i*c(m/s)*t(s) , so now everything’s cool.

Posted by: streamfortyseven on August 7, 2010 3:33 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

hey John, having read your comment above w/r/t physicists, engineers, being initiated, and c, please use your superpowers to obliterate my comment above… thanks. … now you see why I use a pseudonym…

Posted by: streamfortyseven on August 7, 2010 4:25 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Hey streamfortyseven,

I don’t think there was anything needing obliteration in your comments. Questions like yours and the subsequently knowledge gained usually helps someone reading this.

Plus, it is never a good idea to start playing with the spacetime continuum :)

Posted by: Eric Forgy on August 7, 2010 5:25 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

I’ll just note that the geometric algebra people would say that the standard complex number representation is a non-obvious way of picking a weird “scalar and bivector” basis rather than a vector basis for $\mathbb{C}$ (and then making certain identifications between elements that don’t co-incide in other dimensions).

http://en.wikipedia.org/wiki/Geometric_algebra#Complex_numbers

In such a view, the difference in units doesn’t look at all strange. (Just an observation, not an attempt to say this is actually a better way of looking at things.)

Posted by: bane on August 8, 2010 10:11 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

s is supposed to be a “unitless parameterization” but it really looks like it does have units, i.e. time, otherwise q(s) which is “height of the spring at s, in meters” would be without meaning

It sounds like you’re making the mistake that I made when I first read the post—I thought that we were looking at the familiar “spring with a mass on the end of it that bobs up and down.”

What we’re actually looking at is a long, floppy spring hanging between two points. Because the middle of the spring isn’t supported by anything, gravity makes it sag, pulling the spring into a curved shape. We’re trying to predict the shape of the spring.

The parameter s specifies a point on the spring. For example, s could range from 0 to 1, with 0 being the left end of the spring, 1 being the right end of the spring, 0.5 being the center, 0.2 being a fifth of the way from the left end, etc…

I hope that makes things clearer!

p.s. Please read the word “string” as “spring” anywhere it may appear in the text above. If you study physics long enough, there comes a time when the word “string” is closer to the tip of your tongue than the word “spring.” That’s when you know you’ve lost touch with reality.

Posted by: Aaron F. on August 10, 2010 5:08 PM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Oops! I somehow missed the whole string of replies above. Sorry about the redundant post!

Posted by: Aaron F. on August 10, 2010 5:12 PM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Hi Mike,

I was going to try to make amends for getting a bit distracted by saying something about your original post, but ran into a slight speed bump. So here is a soft ball for you.

You write down

$E_{spring} = \frac{k}{2} \int \left(\frac{d q(s)}{d s}\right)^2 ds.$

Can you derive this?

I’ve gone through two pages of doodles so far and it is not yet obvious.

Posted by: Eric Forgy on August 7, 2010 6:44 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

“Spring in Imaginary Time” was a homework assignment from the fall 2006 QG seminar; you can see five derivations there!

Posted by: Mike Stay on August 7, 2010 6:54 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Mike is referring to Week 2 here.

However, in none of the 5 solutions do any of you actually derive the expression. You all take it for granted.

I’d like to know how it was derived. I’ll post my derivation if I can find the time.

Hooke’s law says:

$F_{spring} = - k (x - x_{eq})$

where $x_eq$ is the length of the spring at equilibrium. Since this is a conservative force, we can integrate it to find the potential energy

$V_{spring} = \frac{k}{2} (x - x_{eq})^2.$

It should be a fairly simple derivation to get

$V_{spring} = \frac{k}{2} \int \left(\frac{d q(s)}{d s}\right)^2 d s,$

but, like I said, it is not obvious after 2 pages. However, I’ll be the first to admit the part of my brain that used to do these kinds of calculations in my sleep is now full of cobwebs. I’ll report back if/when I get it if no one else beats me to it :)

For now, I suppose I pose it as a homework assignment for anyone reading this.

Posted by: Eric Forgy on August 7, 2010 10:04 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

A couple pages later…

Still no answer on my side, but here is what I’m thinking.

If we make the physically questionable assumption that the equilibrium length of the spring is “0”, i.e. it cannot be compressed, then the math works out a little easier and we can write

$V_{spring} = \frac{k}{2} \mathcal{l}^2$

where $\mathcal{l}$ is the length of the spring.

We know that

$\mathcal{l} = \int_0^1 \left|\frac{d q(s)}{d s}\right | d s$

so we can use a handy integration trick

$\mathcal{l}^2 = \int_0^1 \int_0^1 \left|\frac{d q(s')}{d s'}\right | \left|\frac{d q(s'')}{d s''}\right | d s' d s''$

and we have

$V_{spring} = \frac{k}{2} \int_0^1 \int_0^1 \left|\frac{d q(s')}{d s'}\right | \left|\frac{d q(s'')}{d s''}\right | d s' d s''$

but then getting from here to

$V_{spring} = \frac{k}{2} \int_0^1 \left| \frac{d q(s)}{d s}\right|^2 d s$

is still not obvious to me unless there is a Dirac delta function I left out somewhere :)

I would NOT be surprised if I am making some elementary error somewhere and will only be slightly embarrassed when it is pointed out to me, but understanding the physics of what is going on is far more important to me than any potential embarrassment suffered :)

PS: This is kind of fun to do physics again after so many years. I miss it :)

Posted by: Eric Forgy on August 7, 2010 11:12 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

I’ve had a chance to sleep on this and I think I’m done with calculations for the time being. The derivation above seems simple enough and I don’t see any room for error (which doesn’t mean there is no error), so although the homework assignment had the word “Spring” in it and mentioned Hooke’s law, I don’t think that potential energy formula actually corresponds to the physics of spring governed by Hooke’s law.

The reason I questioned it in the first place is that it is “too cute” and looks too much like kinetic energy for me to never have noticed it before.

Now I have some questions about the physics of the gravitational potential energy formula given.

Physically, the spring should have a mass $m$, but somewhat unphysically, the problem seems to assume the equilibrium length of the spring is “0”, meaning that, at rest, with nothing tugging on it, the mass density of the spring at equilibrium will be infinite.

If we ignore this physically unsatisfying property for a moment and assume we’re always tugging on it at least a little bit so its length is not zero, then its mass is still $m$, but the mass density will be a function of how it is stretched. In portions that are stretched more, the mass density will be lower and vice versa.

The gravitational potential energy at a point $q(s)\in\mathbb{R}^3$ depends on the mass at that point. The expression

$V_{gravity}(q(s)) = m g z(q(s))$

assumes the mass is the same at all points along the curve, but this can’t be for a spring for reasons described above.

There are two issues with the posing of this homework problem that may affect the desired interpretations of the original post:

• The “spring” is not really a spring governed by Hooke’s law, but it seems interesting and not totally unrelated. What is it?

• The gravitational potential energy would not correspond physically to the gravitional potential energy of a spring, but it might to an inelastic “rope”.

Then again, maybe we can just focus on the dynamic version for which there are no physical issues aside from the usual ones (e.g. Newtonian vs relativistic, etc) and its Wick rotation and forget about the original static spring-like situation that contains some questionable physical assumptions.

Posted by: Eric Forgy on August 8, 2010 2:12 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Could you move these questions to the nForum? They’re worth answering, but not here; this thread is for answering my questions, which take the answers to yours for granted.

Posted by: Mike Stay on August 8, 2010 2:23 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

I’ll just bow out of this thread, no problem. No need to take it to the n-Forum. I was trying to help.

My point was that the physical interpretation of the original homework problem that was taken for granted seems to be incorrect. If I were the TA for the class, I’d mark all 5 solutions down a notch for not noticing :)

Before taking things for granted and moving on, my suggestion is to get a handle on what was being assumed first. Interpreting incorrect physics is not a very useful exercise.

As I said, you could simply ignore the spring problem because your questions can be posed as relating the dynamic problem to the thermodynamic problem with no real need for the original questionable static problem.

Posted by: Eric Forgy on August 8, 2010 3:04 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

I’m sorry; that was too rude of me. John was doing me a favor by posting these questions, and it’s up to the moderators to guide the flow of the conversation.

Posted by: Mike Stay on August 8, 2010 4:51 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

This is the model I’ve got for the “spring”. We have an array of parallel rods, separated by distance $\Delta x$ with masses on bearings that slide freely up and down between them:

                      +   |   +-m-+
|   |   |   |
... +-m-+   |   | ...
|   |   |   +
|   +-m-+   |


                      +   |   +-m-+
s   |   s   s
... +-m-+   s   s ...
|   s   s   +
|   +-m-+   |


The total stretching energy in the spring is

(1)$T = k/2 \sum_{j=0}^{N+1} (\Delta q_j)^2.$

Then we let $N\to \infty$.

Posted by: Mike Stay on August 9, 2010 7:25 PM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

I think Eric’s point is worth noting, namely that the ‘spring energy’ in this homework problem is not the usual energy of any ordinary sort of spring you can buy or even manufacture. The ‘stretching’ part of this energy is also different from the usual stretching energy of the strings considered in string theory, which is simply proportional to the arclength

$\int (\dot{q}(s) \cdot \dot{q}(s))^{1/2} d s$

rather than what we’ve got here

$\int \dot{q}(s) \cdot \dot{q}(s) d s$

But I think it’s likely these issues aren’t terribly relevant to Mike’s questions, which could be asked of any number of models.

Whoops — I’ve gotta go shopping for a dishrack now, but later sometime I’ll try to say something about Mike’s questions….

Posted by: John Baez on August 8, 2010 5:08 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

I still don’t have time to say something really substantial — just a small comment on ‘Hamiltonians for springs’.

Eric wrote:

Still no answer on my side, but here is what I’m thinking.

If we make the physically questionable assumption that the equilibrium length of the spring is “0”, i.e. it cannot be compressed, then the math works out a little easier and we can write

$V_{spring} = \frac{k}{2} \mathcal{l}^2$

And I’m not talking about the fact that it’s hard to buy a spring whose equilibrium length is zero.

The problem I’m thinking about is different. It’s that this potential $V_{spring}$ isn’t local. It’s not the integral over the spring of some locally computable quantity. As you note,

$V_{spring} = \frac{k}{2} \left(\int_0^1 \left|\frac{d q(s)}{d s}\right | d s\right)^2$

and there’s no way to rewrite this as a single integral over the spring. You can of course rewrite it as a double integral over the spring, but that just makes it clear how nonlocal this potential is. Each point along the spring is interacting with each other point to produce an energy density proportional to the product of the ‘speeds’ at the two points.

(Here by ‘speed’ I mean the amount you move along the spring per change in the parameter; this has nothing to do with distance per time until we Wick rotate.)

The potential I proposed:

$V = \frac{k}{2} \int \frac{d q(s)}{d s} \cdot \frac{d q(s)}{d s} d s$

while not realistic for springs we actually buy in the store, is local — and as Mike points out, we can ‘derive’ it from a kind of microscopic model of a spring.

So I don’t think it’s insane to study this sort of potential. And don’t forget, there’s an entire industry devoted to studying relativistic quantized loops of spring that have an equilibrium length of zero: Spring Theory.

Posted by: John Baez on August 10, 2010 3:41 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Local or not, like it or not,

$V_{spring} = \frac{k}{2} \mathcal{l}^2$

is the potential for a spring. To eliminate the $\mathcal{l}_{eq} = 0$ issue, we can write more explicitly

$V_{spring} = \frac{k}{2} \left(\mathcal{l}-\mathcal{l}_{eq}\right)^2.$

If you don’t like the non-localness of this, it is not an issue with my formula, but an issue with springs that are subject to Hooke’s law. If we’re not talking about this, we’re not talking about springs, and that is one point I’ve been trying to make.

I agreed with Mike’s model via email after a brief glance, but now I’m not too sure it has the correct continuum limit. I’ll need to double check. If someone beats me to it, I won’t be heartbroken.

Even if Mike’s model does have the correct continuum limit, you’d have to admit this is FAR from the model of a spring. It is more like a mattress :)

PS: Mike and I agreed via email that it is worth trying to interpret the physics of the static problem correctly first as a step to understand the thermodynamics. I thought it was a shame to discuss this via email since it is an interesting problem, so asked if we could move it to the n-Forum and Mike suggested moving it back here to the n-Cafe instead since no one yet has taken up the original questions.

PPS: As an alternative to trying to interpret this “mattress” problem, we could switch to a different system that does have well-defined and intuitive static and dynamics versions and then look at the thermodynamic version.

If we did this correctly, we’d probably end up re-discovering well known symmetries in string theory… err… I mean spring theory :)

PPPS: Starting with the double integral version of the spring potential, you could end up with your potential if there was a Dirac delta function, i.e. the difference between your potential and a spring potential is a Dirac delta. Still trying to interpret that…

Posted by: Eric on August 10, 2010 5:53 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Eric wrote:

Local or not, like it or not,

$V_{spring} = \frac{k}{2} \mathcal{l}^2$

is the potential for a spring.

No it’s not: not for a spring that’s tracing out an arbitrary path $q(s)$. It can’t possibly be, because it’s nonlocal, and nature is local.

If you don’t like the non-localness of this, it is not an issue with my formula, but an issue with springs that are subject to Hooke’s law.

Hooke never claimed that his law applied to a spring stretched out along an arbitrary twisting and turning path. His law only says that if you hold two ends of a spring and stretch it so it has length $\mathcal{l}$, its potential energy is proportional to $\mathcal{l}^2$. And note: this follows from my formula

$V \propto \int \frac{d q}{d s} \cdot \frac{d q}{d s} d s$

Say you had a spring with this potential energy. If you hold the two ends of the spring at a distance $\mathcal{l}$ from each other, and see what it does, it will minimize energy $V$ by tracing out a straight line — and you’ll see that $V$ is proportional to $\mathcal{l}^2$.

So my law is a generalization of Hooke’s law from straight strings to strings that trace out arbitrary paths. Obviously it’s an idealization in various ways, most notably that it describes springs that are happiest (have the least energy) when their length is zero. But it’s nice in two other ways: it’s local, and it’s quadratic in $q$, which gives it the mathematically tractable charm of Hooke’s original law.

$V \propto \ell^2 = \int \int |\frac{d q(s_1)}{d s_1}| |\frac{d q(s_2)}{d s_2}| d s_1 d s_2$

is more complicated to work with, and also physically impossible, since it requires ‘spooky nonlocal interactions’ between different portions of the spring.

Posted by: John Baez on August 10, 2010 6:52 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Oops! I wrote a response, but lost it due to an errant window close. Probably a good thing because it contained some errors :)

Anyway, non-locality and spooky action at a distance doesn’t bother me much at this stage of the conversation since we are talking about the non-relativistic situation. After all $V = m g h$ is spooky as is

$V_{gravity} = -G\frac{m}{r}.$

Anyway, in the absence of an external potential, the solution to the variational problem is

$\frac{d^2 q}{d s^2} = 0.$

In particular, this means

$\left|\frac{d q}{d s}\right| = constant$

so the length of our string is given by

$\mathcal{l} = \int_{s_1}^{s_2} \left|\frac{d q}{d s}\right| d s = \left|\frac{d q}{d s}\right| \Delta s.$

Coming back to your potential, you have

$V = \frac{k}{2} \int_{s_0}^{s_1} \left|\frac{d q}{d s}\right|^2 d s = \frac{1}{2} k\frac{\mathcal{l}^2}{\Delta s}.$

That extra factor of $\Delta s$ is what concerned me about the continuum limit of Mike’s model. The units don’t quite work out and it seems to be parameterization dependent.

While googling, I found an interesting homework assignment:

Prove that the potential energy of a light elastic string of natural length $\mathcal{l}_{eq}$ and modulus $k$ when stretch to a length $\mathcal{l}$ is $\frac{1}{2} k \frac{(\mathcal{l}-\mathcal{l}_{eq})^2}{\mathcal{l}_{eq}}$.

Note this is nonlocal and spooky.

For the record, I like your potential. I think it is interesting. I’m just trying to understand it :)

Posted by: Eric on August 10, 2010 11:36 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

I know this is a year too late, but no one seems to have answered Eric’s question about deriving the stated potential for the spring, or commented on his “extra” factor of $\Delta s$. So, I thought I’d give it a shot.

First of all, this spring potential *is* for a spring that has a relaxed length of zero. Its not hard to generalize this to the more physical case, but the result isn’t as pretty, so I’m not going to type it up.

As a warm up, take a straight, evenly stretched spring with constant $k$ and stretch $L$. This has energy $U = 1/2 k L^2$. Split it up evenly into n pieces. Each piece will have a stretch of $L/n$. This has $U = n (k'/2) (L/n)^2 = (1/2) (k'/n) L^2$. Since the two energies must be equal, we find that $k' = n k$. We’ll see that this $n$ cancels the extra $\Delta s$ mentioned before.

Now, let’s tackle the full problem. Let $s$ run uniformly from 0 to 1 over an evenly stretched spring. Now discritize into n chunks so that $s_i = i\Delta s$ where $\Delta s = 1/n$. Take $q_i = q(s_i)$. The length of the ith chunk of spring is then $\Delta q_i = q_i - q_{i-1}$. So, the total energy of the spring is:

$U = (1/2) (nk) \sum_{i=1}^{n} |\frac{\Delta q_i}{\Delta s}|^2 \Delta s^2$

Now, after using $n \Delta s = 1$, we can take n to infinity to get

$U = (1/2) k \int_{0}^{1} |\frac{dq}{ds}|^2 ds$.

Posted by: Brian on September 21, 2011 8:03 PM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Posted by: Eric on September 22, 2011 4:15 PM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Eric wrote:

Anyway, non-locality and spooky action at a distance don’t bother me much at this stage of the conversation since we are talking about the non-relativistic situation.

Okay, if you think each piece of a spring is able to interact with each other piece, no matter how far apart, to produce an energy density equal to

$\left|\frac{d q(s_1)}{d s_1}\right| \left|\frac{d q(s_2)}{d s_2}\right|$

then I want to know the mysterious mechanism whereby this occurs.

Until I hear that story, I’ll stick with my belief that the energy density of a stretched spring is pretty accurately described as an integral over the spring of some locally computed quantity. Indeed, people who study strain energy density functions seem to agree with me here.

But if a hundred theorists aren’t enough to convince you, do an experiment! Take a spring, stretch it, and hold the two ends fixed. Then: grab a point in the middle and move it towards one end of the spring. Since the total length of the spring remains unchanged, you claim that the energy will remain unchanged as well, so this process will take no work. I claim that it will take work.

This is an easy experiment to do. In fact, I think you only need to imagine doing it to realize that I’m right.

And in fact my formula predicts that it will take just as much work as if we treat the whole spring as made of two ‘sub-springs’, and compute the energy as a sum of two parts, each individually computed using Hooke’s law.

Remember, my formula for the potential energy

$V \propto \int \frac{d q}{d s} \cdot \frac{d q}{d s} d s$

reduces to Hooke’s potential in the case of a straight spring with constant ‘velocity’ $d q / d s$. So, in the experiment I’m talking about, I can compute the potential energy as a sum of two parts, each computed a la Hooke. The locality built into my integral is another way of saying that a big spring can be thought of as a bunch of little springs glued end to end. You, on the other hand, treat the whole spring as a single thing and say is proportional to its overall length squared. That’s nonlocal.

Posted by: John Baez on August 11, 2010 6:12 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

I keep reading over your statement of the problem w/r/t substituting it/hbar with beta/k sub B. You talk about q(beta) being the height of something at inverse temperature beta, in meters, and then introduce r as “bits/m**K” and then ask about the Gibbs/Boltzmann entropy. Shouldn’t you be talking about the Shannon entropy instead of the Gibbs/Boltzmann entropy?

Posted by: streamfortyseven on August 7, 2010 8:13 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

I’ve got to figure out itex. I don’t mean r is “bits/m**K”, I mean it’s “bits/((m**2)(K))”. Arrgh. The rest of my comment still stands…

Posted by: streamfortyseven on August 7, 2010 8:17 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Select a filter that uses itex. Wrap your math in dollar signs. Exponentiation is a caret: ^. So you want to write

$\$ bits / (m^2 K) $\$

and the output will be

$bits/(m^2K)$.

Regarding Shannon entropy and Gibbs/Boltzmann entropy: they’re the same thing, physically. See Information Distance by Bennett, Gacs, Li, Vitanyi, & Zurek.

Posted by: Mike Stay on August 7, 2010 8:41 PM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

I don’t really have enough background in this area to contribute much, but perhaps I can ask a question and maybe stimulate further thoughts with a failed attempt.

First, I was a bit confused by the use of V(s) and V(it) when I was thinking it ought to be V(q(s)) and V(q(it)). Unless q is fixed, it seems to me that they’re different concepts. Was dropping the q intentional, and if so, how does it help?

Looking at the thermodynamic analogy, the problem seems to require that the temperature varies through space, that there is an exchange between internal thermal energy and an external position-based potential energy, and that this is mediated by some analogy to Hooke’s law that relates a temperature gradient to energy density.

To exchange thermal for gravitational potential energy immediately suggests buoyant fluid flow, in which fluid that is made warmer than its neighbourhood, expands and rises. The expansion causes the temperature to reduce, the rising causes the potential energy to increase. If you were to place hot and cold heat reservoirs above and below such a fluid, the temperature would vary with height along some curve from one endpoint to the other decided by a variational principle.

The neat analogy is spoilt by the fact that the expansion also changes the pressure, so there is internal “spring” energy present, too. And any relationship here between temperature gradient and Hooke’s law is opaque.

So I’m thinking, while the squared derivative form of kinetic energy of a free particle is pretty fundamental, Hooke’s law isn’t. It’s an accidental property of certain materials in certain configurations. (Non-linear springs are easy to construct.) So is it just a convenient coincidence that when the dynamic variational principle is transformed to statics, there just so happens to be a well-known example that matches it?

Posted by: Nullius in Verba on August 7, 2010 1:34 PM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

First, I was a bit confused by the use of $V(s)$ and $V(i t)$ when I was thinking it ought to be $V(q(s))$ and $V(q(it))$. Unless $q$ is fixed, it seems to me that they’re different concepts. Was dropping the $q$ intentional, and if so, how does it help?

If $U$ is the potential in terms of the position $U(q(s))$, then we can define $V(s) = U(q(s))$. $V$ is more general than $U$.

Looking at the thermodynamic analogy, the problem seems to require that the temperature varies through space

I don’t think so; it can vary through time, but time is irrelevant once the temperature changes slowly enough to be considered adiabatic; whether it takes an hour or a month doesn’t matter.

So is it just a convenient coincidence that when the dynamic variational principle is transformed to statics, there just so happens to be a well-known example that matches it?

Well, Hooke’s law is just a linear approximation of some nonlinear behavior; it happens to hold over a useful range in steel springs. But we can take a linear approximation of any nonlinear system in a small enough range, so in that sense it was inevitable.

Posted by: Mike Stay on August 7, 2010 8:58 PM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

What I meant was that q is a variable in this context. V(q,s) would be another way of putting it. The problem is that if s is simply a parametrization, you can change variables to any other parametrization without changing the physics, but when expressing it as a function of s alone this has to be implicitly understood.

There’s nothing invalid about doing so, but for the purposes of seeking out analogies, V(q) leads you to look for analogous potentials that are a function of whatever q is analogous to, while V(s) leads one to look for a potential that depends on s. Is the potential we are looking for a function of inverse temperature, or a function of the height of something?

I figured it out pretty quickly and mentally re-inserted the q, but wondered if I was right to do so. I thought there could be a reason for doing it, where the analogy could be to a more general potential for some reason. But it’s a minor point, not worth pursuing.

I did also consider that whatever it is might vary through time - I was thinking of things like Carnot cycles - but q(beta) is explicitly described as the height of something in metres. With time as a function of temperature, the potential would have to be made purely time dependent.

I take your point about linear approximations (so long at the non-linear function is smooth). For that matter, the kinetic energy formula is an approximation for low speeds. But that suggests that the analogy might be to such a linearisation, rather than anything realistic. It changes the mental space being searched.

Posted by: Nullius in Verba on August 8, 2010 11:04 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

I am not a physicist. What this reminds me of is Erik Verlindes’s paper on gravity as an entropic force.

http://staff.science.uva.nl/~erikv/page20/page20.html

I guess that the statistical equivalent to a spring held fixed at both ends is an ideal polymer (or ideal chain) held fixed at both ends.

http://en.wikipedia.org/wiki/Ideal_chain

Posted by: Graham on August 8, 2010 12:27 PM | Permalink | Reply to this

Leineker, Queiroz, Santana, Siqueira; Re: Thermodynamics and Wick Rotation

Thermofied Dynamics for Twisted Poincare-Invariant Field Theories: Wick Theorem and S-matrix
Authors: Marcelo Leineker, Amilcar R. Queiroz, Ademir E. Santana, Chrystian de Assis Siqueira
Subjects: High Energy Physics - Theory (hep-th); Mathematical Physics (math-ph)

Poincare invariant quantum field theories can be formulated on non-commutative planes if the statistics of fields is twisted. This is equivalent to state that the coproduct on the Poincare group is suitably twisted. In the present work we present a twisted Poincare invariant quantum field theory at finite temperature. For that we use the formalism of Thermofield Dynamics (TFD). This TFD formalism is extend to incorporate interacting fields. This is a non trivial step, since the separation in positive and negative frequency terms is no longer valid in TFD. In particular, we prove the validity of Wick’s theorem for twisted scalar quantum field at finite temperature.

Posted by: Jonathan Vos Post on August 9, 2010 5:38 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation I am a big fan of imaginary time in classical mechanics, ever since I learnt of the cool fact that the imaginary period of Jacobi’s elliptic functions can be physically understood of the period of the pendulum when gravity goes up instead of down (see John’s notes here for a nice account). What I’ve never understood, and something I’d like to get a grip on to tackle Mike’s questions, is how to interpret the arguments living in the rest of the complex plane physically. In other words, we have a physical understanding for $q(t)$ and $q(it)$, but how does one physically interpret $q(a+ib)$?

Posted by: Bruce Bartlett on August 10, 2010 8:36 AM | Permalink | Reply to this

Re: Thermodynamics and Wick Rotation

Posted by: David Corfield on August 11, 2010 3:54 PM | Permalink | Reply to this

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