In yesterday's post, I totally screwed up when identifying the middle part of the spectrum as low frequency. It is not. Please ignore what I said or better take it as a warning what happens when you don't double check.

Apologies to everybody that I stirred up!

## Friday, June 16, 2017

## Thursday, June 15, 2017

### Some DIY LIGO data analysis

**UPDATE: After some more thinking about this, I have very serious doubt about my previous conclusions. From looking at the power spectrum, I (wrongly) assumed that the middle part of the spectrum is the low frequency part (my original idea was, that the frequencies should be symmetric around zero but the periodicity of the Bloch cell bit me). So quite to the opposite, when taking into account the wrapping, this is the high frequency part (at almost the sample rate). So this is neither physics nor noise but the sample rate. For documentation, I do not delete the original post but leave it with this comment.**

Recently, in the Arnold Sommerfeld Colloquium, we had Andrew Jackson of NBI talk about his take on the LIGO gravitational wave data, see this announcement with link to a video recording. He encouraged the audience to download the freely available raw data and play with it a little bit. This sounded like fun, so I had my go at it. Now, that his paper is out, I would like to share what I did with you and ask for your comments.

I used mathematica for my experiments, so I guess the way to proceed is to guide you to an html export of my (admittedly cleaned up) notebook (Source for your own experiments here).

The executive summary is that apparently, you can eliminate most of the "noise" at the interesting low frequency part by adding to the signal its time reversal casting some doubt about the stochasticity of this "noise".

I would love to hear what this is supposed to mean or what I am doing wrong, in particular from my friends in the gravitational wave community.

## Thursday, June 08, 2017

### Relativistic transformation of temperature

Apparently, there is a long history of controversy going back to Einstein an Planck about the proper way to deal with temperature relativistically. And I admit, I don't know what exactly the modern ("correct") point of view is. So I would like to ask your opinion about a puzzle we came up during yesterday's after colloquium dinner with Erik Verlinde:

Imagine a long rail of a railroad track. It is uniformly heated to a temperature T and is in thermodynamic equilibrium (if you like a mathematical language: it is in a KMS state). On this railroad track travels Einstein's relativistic train at velocity v. From the perspective of the conductor, the track in front of the train is approaching the train with velocity v, so one might expect that the temperature T appears blue shifted while behind the train, the track is moving away with v and so the temperature appears red-shifted.

Following this line of thought, one would conclude that the conductor thinks the rail has different temperatures in different places and thus is out of equilibrium.

On the other hand, the question of equilibrium should be independent of the observer. So, is the assumption of the Doppler shift wrong?

A few remarks: If you are worried that Doppler shifts should apply to radiation then you are free to assume that both in front and in the back, there are black bodies in thermal contact with the rail and thus exhibiting a photon gas at the same temperature as the rail.

You could probably also make the case for the temperature transforming like the time component of a four vector (since it is essentially an energy). Then the transformed temperature would be independent of the sign of v. This you could for example argue for by assuming the temperature is so high that in your black body photon gas you also create electron-positron pairs which would be heavier due to their relativistic speed relative to the train and thus requiring more energy (and thus temperature) for their creation.

A final remark is about an operational definition of temperature at relativistic speeds: It might be difficult to bring a relativistic thermometer in equilibrium with a system if there is a large relative velocity (when we define temperature as the criterium for two systems in contact to be in equilibrium). Or to operate a heat engine between he front part of the rail and the back while moving along at relativistic speed and then arguing about the efficiency (and defining the temperature that way).

Thanks for all your comments. We also had some further discussions here and I would like to share my conclusions:

1) It probably boils down to what exactly you mean when you say ("temperature"). Of course, you want that his at least in familiar situations agrees with what thermometers of this type or another measure. (In the original text I had hinted at two possible definitions that I learned about from a very interesting paper by Buchholz and Solveen discussing the Unruh effect and what would actually be observed there: Either you define temperature that the property that characterises equilibrium states of systems such there is no heat exchange when you bring in contact two systems of the same temperature. This is for example close to what a mercury thermometer measures. Alternatively, you operate a perfect heat engine between two reservoirs and define your temperatures via

$$\eta = \frac{T_h - T_c}{T_h}.$$

This is for example hinted at in the Feynamn lectures on physics.

One of the commentators suggested using the ratio of eigenvalues of the energy momentum tensor as definition of temperature. Even though this might give the usual thing for a perfect fluid I am not really convinced that this generalises in the right way.

2) I would rather define the temperature as the parameter in the Gibbs (or rather KMS) state (it should only exist in equilibrium, anyway). So if your state is described by density matrix $\rho$, and it can be written as

$$\rho = \frac{e^{-\beta H}}{tr(e^{-\beta H})}$$

then $1/\beta$ is the temperature. Obviously, this requires the a priori knowledge of what the Hamiltonian is.

For such states, under mild assumptions, you can prove nice things: Energy-entropy inequalities ("minimisation of free energy"), stability, return to equilibrium and most important here: passivity, i.e. the fact you cannot extract mechanical work out of this state in a cyclic process.

2) I do not agree that it is out of the question to have a thermometer with a relative velocity in thermal equilibrium with a heat bath at rest. You could for example imagine a mirror fixed next to the track and in thermal equilibrium with the track. A second mirror is glued to the train (and again in thermal equilibrium, this time with a thermometer). Between the mirrors is is a photon gas (black body) that you could imagine equilibrating with the mirrors on both ends. The question is if that is the case.

3) Maybe rails and trains a a bit too non-spherical cows, so lets better look at an infinitely extended free quantum gas (bosons or fermions, your pick). You put it in a thermal state at rest, i.e. up to normalisation, its density matrix is given by

$$\rho = e^{-\beta P^0}.$$

Here $P^0$ is the PoincarĂ© generator of time translations.

Now, the question above can be rephrased as: Is there a $\beta'$ such that also

$$\rho = e^{-\beta' (\cosh\alpha P^0 + \sinh \alpha P^1)}?$$

And to the question formulated this way, the answer is pretty clearly "No". A thermal state singles out a rest frame and that's it. It is not thermal in the moving frame and thus there is no temperature.

It's also pretty easy to see this state is not passive (in the above sense): You could operate a windmill in the slipstream of particles coming more likely from the front than the back. So in particular, this state is not KMS (this argument I learned from Sven Bachmann).

4) Another question would be about gravitational redshift: Let's take some curve space-time and for simplicity assume it has no horizons (for example, let the far field be Schwarzschild but in the center, far outside the Schwarzschild radius, you smooth it out. Like the space-time created by the sun). Make it static, so it contains a timeline Killing vector (otherwise no hope for a thermal state). Now prepare a scalar field in the thermal state with temperature T. Couple to it a harmonic oscillator via

$$ H_{int}(r) = a^\dagger a + \phi(t, r) (a^\dagger + a).$$

You could now compute a "local temperature" by computing the probability that the harmonic oscillator is in the first excited state. Then, how does this depend on $r$?

**Update one day later:**Thanks for all your comments. We also had some further discussions here and I would like to share my conclusions:

1) It probably boils down to what exactly you mean when you say ("temperature"). Of course, you want that his at least in familiar situations agrees with what thermometers of this type or another measure. (In the original text I had hinted at two possible definitions that I learned about from a very interesting paper by Buchholz and Solveen discussing the Unruh effect and what would actually be observed there: Either you define temperature that the property that characterises equilibrium states of systems such there is no heat exchange when you bring in contact two systems of the same temperature. This is for example close to what a mercury thermometer measures. Alternatively, you operate a perfect heat engine between two reservoirs and define your temperatures via

$$\eta = \frac{T_h - T_c}{T_h}.$$

This is for example hinted at in the Feynamn lectures on physics.

One of the commentators suggested using the ratio of eigenvalues of the energy momentum tensor as definition of temperature. Even though this might give the usual thing for a perfect fluid I am not really convinced that this generalises in the right way.

2) I would rather define the temperature as the parameter in the Gibbs (or rather KMS) state (it should only exist in equilibrium, anyway). So if your state is described by density matrix $\rho$, and it can be written as

$$\rho = \frac{e^{-\beta H}}{tr(e^{-\beta H})}$$

then $1/\beta$ is the temperature. Obviously, this requires the a priori knowledge of what the Hamiltonian is.

For such states, under mild assumptions, you can prove nice things: Energy-entropy inequalities ("minimisation of free energy"), stability, return to equilibrium and most important here: passivity, i.e. the fact you cannot extract mechanical work out of this state in a cyclic process.

2) I do not agree that it is out of the question to have a thermometer with a relative velocity in thermal equilibrium with a heat bath at rest. You could for example imagine a mirror fixed next to the track and in thermal equilibrium with the track. A second mirror is glued to the train (and again in thermal equilibrium, this time with a thermometer). Between the mirrors is is a photon gas (black body) that you could imagine equilibrating with the mirrors on both ends. The question is if that is the case.

3) Maybe rails and trains a a bit too non-spherical cows, so lets better look at an infinitely extended free quantum gas (bosons or fermions, your pick). You put it in a thermal state at rest, i.e. up to normalisation, its density matrix is given by

$$\rho = e^{-\beta P^0}.$$

Here $P^0$ is the PoincarĂ© generator of time translations.

Now, the question above can be rephrased as: Is there a $\beta'$ such that also

$$\rho = e^{-\beta' (\cosh\alpha P^0 + \sinh \alpha P^1)}?$$

And to the question formulated this way, the answer is pretty clearly "No". A thermal state singles out a rest frame and that's it. It is not thermal in the moving frame and thus there is no temperature.

It's also pretty easy to see this state is not passive (in the above sense): You could operate a windmill in the slipstream of particles coming more likely from the front than the back. So in particular, this state is not KMS (this argument I learned from Sven Bachmann).

4) Another question would be about gravitational redshift: Let's take some curve space-time and for simplicity assume it has no horizons (for example, let the far field be Schwarzschild but in the center, far outside the Schwarzschild radius, you smooth it out. Like the space-time created by the sun). Make it static, so it contains a timeline Killing vector (otherwise no hope for a thermal state). Now prepare a scalar field in the thermal state with temperature T. Couple to it a harmonic oscillator via

$$ H_{int}(r) = a^\dagger a + \phi(t, r) (a^\dagger + a).$$

You could now compute a "local temperature" by computing the probability that the harmonic oscillator is in the first excited state. Then, how does this depend on $r$?

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