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Annales Geophysicae An interactive open-access journal of the European Geosciences Union
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Ann. Geophys., 36, 37–46, 2018
https://doi.org/10.5194/angeo-36-37-2018
Ann. Geophys., 36, 37–46, 2018
https://doi.org/10.5194/angeo-36-37-2018

Regular paper 12 Jan 2018

Regular paper | 12 Jan 2018

Uncertainties in the heliosheath ion temperatures

Uncertainties in the heliosheath ion temperatures
Klaus Scherer1,2, Hans Jörg Fahr3, Horst Fichtner1,2, Adama Sylla1, John D. Richardson4, and Marian Lazar1 Klaus Scherer et al.
• 1Institut für Theoretische Physik IV: Weltraum- und Astrophysik, Ruhr-Universität Bochum, Germany
• 2Research Department, Plasmas with Complex Interactions, Ruhr-Universität Bochum, Germany
• 3Argelander Institut für Astronomie, Universität Bonn, Germany
• 4Kavli Institute for Astrophysics and Space Sciences, MIT, Cambridge, MA 02139, USA

Correspondence: Klaus Scherer (kls@tp4.rub.de)

Abstract

The Voyager plasma observations show that the physics of the heliosheath is rather complex and that the temperature derived from observation particularly differs from expectations. To explain this fact, the temperature in the heliosheath should be based on κ distributions instead of Maxwellians because the former allows for much higher temperature. Here we show an easy way to calculate the κ temperatures when those estimated from the data are given as Maxwellian temperatures. We use the moments of the Maxwellian and κ distributions to estimate the κ temperature. Moreover, we show that the pressure (temperature) given by a truncated κ distribution is similar to that given by a Maxwellian and only starts to increase for higher truncation velocities. We deduce a simple formula to convert the Maxwellian to κ pressure or temperature. We apply this result to the Voyager 2 observations in the heliosheath.

Keywords. Space plasma physics (kinetic and MHD theory)

1 Introduction

Knowledge about the temperature of an astrophysical plasma is of significance for the correct hydrodynamical treatment of various plasma processes like heat conduction, wave propagation, compression, or charge exchange. The temperature of a space plasma is, of course, not directly measurable but must be derived indirectly. If the velocity distribution function of a plasma constituent is known, the temperature can be computed as a second velocity moment. If not, assumptions have to be made about this distribution function. Consequently, an uncertainty regarding the latter will translate into an uncertainty regarding the temperature. For simplicity, unmeasured velocity distributions are often assumed to be Maxwellians. One must be aware that the derived “Maxwellian temperature” is an assumption and might be rather different from the actual thermodynamically relevant temperature of the considered plasma .

One such example is the plasma in the inner heliosheath, i.e., the region of the heliosphere between the solar wind termination shock and the heliopause. The temperatures for the inner heliosheath can, in contrast to those determined for the upstream solar wind , only be derived from Voyager 2 measurements under the Maxwellian assumption (e.g., Richardson and Wang2012). From the modeling of fluxes of energetic neutral atoms (ENAs) it is known, however, that the distribution function of protons in this region is probably not a Maxwellian but a so-called κ distribution with the parameter κ<2 (e.g., Heerikhuisen et al.2008). Consequently, the temperature of the proton population should be computed from these non-Maxwellian distributions. In the present paper we quantify these alternative κ temperatures and discuss their significant difference from the Maxwellian ones. Before both temperatures are discussed in detail, we briefly review the κ distributions and the physical reason for their importance downstream of the termination shock.

2 κ distributions

The κ distributions are an often-used tool for the quantitative treatment of non-Maxwellian plasmas as is described in the reviews by and . In the following, we assume that the actual distribution function is isotropic in the bulk velocity frame because the protons undergo efficient pitch angle scattering at Alfvén waves (MHD wave turbulences), leading to this form of isotropy. Furthermore, without the capabilities to measure 2-D or 3-D distributions and their anisotropy implicitly, the observations are always integrated over all pitch angles and are thus isotropic. Thus, the definition of the isotropic κ distribution reads

$\begin{array}{}\text{(1)}& {f}_{\mathit{\kappa }}\left(\mathbit{v}\right)=n\left(\mathbit{r}\right)\frac{\mathrm{\Gamma }\left(\mathit{\kappa }+\mathrm{1}\right)}{{\left(\sqrt{\mathit{\pi }}\sqrt{\mathit{\kappa }}\mathrm{\Theta }\right)}^{\mathrm{3}}\mathrm{\Gamma }\left(\mathit{\kappa }-\frac{\mathrm{1}}{\mathrm{2}}\right)}{\left(\mathrm{1}+\frac{{v}^{\mathrm{2}}}{\mathit{\kappa }{\mathrm{\Theta }}^{\mathrm{2}}}\right)}^{-\left(\mathit{\kappa }+\mathrm{1}\right)},\end{array}$

where Θ is a fitting parameter, which normalizes the speed; often it is identified with a thermal speed. In the limit κ→∞, these distributions converge to the Maxwellian distribution

$\begin{array}{ll}{f}_{\mathrm{m}}\left(\mathbit{v}\right)& ={n}_{\mathrm{0}}\left(\mathbit{r}\right){\left(\frac{m}{\mathrm{2}\mathit{\pi }kT}\right)}^{\mathrm{3}/\mathrm{2}}\mathrm{exp}\left(-\frac{m{\mathbit{v}}^{\mathrm{2}}}{\mathrm{2}kT}\right)\phantom{\rule{2em}{0ex}}\text{or}\\ \text{(2)}& & ={n}_{\mathrm{0}}\left(\mathbit{r}\right){\left(\mathit{\pi }{v}_{\mathrm{T}}^{\mathrm{2}}\right)}^{-\mathrm{3}/\mathrm{2}}\mathrm{exp}\left(-\frac{{v}^{\mathrm{2}}}{{v}_{\mathrm{T}}^{\mathrm{2}}}\right),\end{array}$

where the fitting parameter is usually the “thermal speed” ${v}_{\mathrm{T}}\equiv \sqrt{\mathrm{2}kT/m}$, with the temperature T, the proton mass m, and the Boltzmann constant k.

Both distribution functions are normalized to the number density n(r) so that

$\begin{array}{}\text{(3)}& \int {f}_{m,\mathit{\kappa }}{\mathrm{d}}^{\mathrm{3}}v=\mathrm{4}\mathit{\pi }\underset{\mathrm{0}}{\overset{\mathrm{\infty }}{\int }}{v}^{\mathrm{2}}{f}_{m,\mathit{\kappa }}\mathrm{d}v=n\left(\mathbit{r}\right).\end{array}$

Furthermore, for the following, we normalize all speeds (v) to the solar wind speed usw=436 km s−1, which corresponds to a proton energy of 1 keV.

In order to fit the thermal core of the Maxwellian, it is required that the speed Θ is equal to the thermal speed Θ=vp, as shown in Appendix A. The latter relation corresponds to the preferable of two alternatives to interpret κ temperatures, as is discussed in detail in .

2.1 κ distributions induced by pick-up protons

Due to the different velocity–space processes acting upon solar wind protons when moving out from 1 AU to great radial solar distances, the resulting distribution function is far from being an equilibrium distribution in the form of a shifted Maxwellian. This is theoretically evident mainly from the fact that the solar wind proton plasma is permanently loaded with newly injected pick-up protons. This, in connection with wave-driven diffusion processes, keeps the resulting ion distribution function off a Maxwellian equilibrium shape. The transport of such pick-up ions was the subject of many investigations in the past; see Fahr (1973), , and for an ever-increasing sophistication in treating the exact form of this pick-up ion incorporation process . Also from in situ observations it has been clearly recognized that pick-up protons show a typical core distribution below the velocity injection border and an extended power-law tail above . This then justified the theoretical endeavor of to treat the evolving joint solar wind ion distribution as a κ distribution with a κ parameter evolving with radial distance. These authors could show with the help of an adequate pressure-moment transport equation that the resulting ion κ distribution evolves into a highly nonthermal distribution with κ≤2.0 by the time the solar wind plasma arrives at the termination shock.

From this finding it became evident that the passage of such a nonequilibrium proton distribution over the solar wind termination shock would generate a distribution downstream of the shock with strong nonequilibrium signatures. There are good reasons given by the Liouville theorem (see ) that this downstream distribution function will also be a kappa function obeying the relation

$\begin{array}{}\text{(4)}& {f}_{\mathrm{2}}\left(v\right)=s\cdot {f}_{\mathrm{1}}\left(\frac{v}{\sqrt{s}}\right),\end{array}$

where the indices “1” and “2” indicate upstream and downstream quantities, respectively, and where s denotes the shock compression ratio. This then shows that downstream of the shock the following kappa function has to be expected:

$\begin{array}{}\text{(5)}& {f}_{\mathrm{2}}\left(v\right)=A\phantom{\rule{0.25em}{0ex}}s{\left(\mathrm{1}+\frac{{v}^{\mathrm{2}}}{{\mathit{\kappa }}_{\mathrm{1}}s{\mathrm{\Theta }}_{\mathrm{1}}^{\mathrm{2}}}\right)}^{-\left({\mathit{\kappa }}_{\mathrm{1}}+\mathrm{1}\right)}.\end{array}$

This expresses the fact that downstream one must expect a kappa distribution with κ identical to its upstream value κ1, but with an increased core width Θ2=sΘ1.

The above finding raises the question of how in situ low-energy ion measurements, e.g., by Voyager 2, should be interpreted in terms of velocity moments of the distribution function. Solar wind ion data downstream of the shock are, in general, not displayed as spectral flux ion data, but as plasma parameters (n, flow speed, and T) derived from a fit of the measured fluxes to a convected isotropic Maxwellian distribution; see, e.g., . The question arises on the basis of whether the underlying distribution function is a Maxwellian and, if not, how the actual plasma parameters differ from those derived using a Maxwellian fit. A recently published paper by demonstrated how different the values of the velocity moments can be if based on Maxwellian or on κ distributions. In the following part of the paper we shall thus focus on the specific question of how Voyager 2 heliosheath plasma parameters displayed in the literature change values when interpreted on the basis of nonthermal ion kappa distributions.

3 Maxwellian temperatures along the Voyager 2 trajectory

have shown that during the years 2008 to 2011 (denoted by t8 and t11 in the following text) the heliosheath proton temperature measured along the Voyager 2 trajectory has fallen from T(t8)=140 000 K to T(t11)=40 000 K (see Fig. 1). This temperature decrease occurred while the measured proton density remained fairly constant at an average value of $〈{n}_{p}〉\simeq \mathrm{0.001}\phantom{\rule{0.125em}{0ex}}{\mathrm{cm}}^{-\mathrm{3}}$. The temperature decrease during the Voyager 2 time period of t11t8=3 yr thus indicates a nonadiabatic origin. We begin by investigating this temperature decrease in a first-order view studying the temperature in a plasma volume comoving with the plasma bulk velocity usw being subject to ongoing charge exchange reactions with cold LISM H atoms.

Figure 1Daily (points) and 11-day running averages (lines) of the radial speed, density, and temperature observed at V2. Taken from .

In this volume a proper time τ is counted (i.e., the time of the comoving clock), and we want to describe the temperature change within the box as a function of τ. For an adequate estimate we use a simple thermodynamic transport equation describing the proton temperature as the result of charge-exchange-related energy exchanges between the plasma and the neutral gas in the following form:

$\begin{array}{}\text{(6)}& \frac{d}{\mathrm{d}\mathit{\tau }}\left({n}_{p}kT\right)=-{n}_{p}{n}_{\mathrm{H}}{\mathit{\sigma }}_{\text{ex}}{v}_{\text{rel}}\cdot \left(kT-k{T}_{\mathrm{0}}\right),\end{array}$

with np≃const (see the assumption of incompressible heliosheath plasma flow as made in Fahr et al.2016), nH≃const, kTkT0, and T(τ)∼O(105K) (see Richardson and Wang, 2012) denoting the actual, local ion temperature, T0 denoting the thermalized heliosheath pick-up ion temperature downstream of the shock, and ${T}_{\mathrm{0}}\simeq \left(\mathrm{3}/\mathrm{2}K\right)m{U}_{\mathrm{0}}^{\mathrm{2}}=O\left({\mathrm{10}}^{\mathrm{3}}K\right)$ denoting the temperature of the newly injected ions originating from the cold LISM H atoms (where U0 is the bulk speed of the proton fluid in the heliosheath). This equation can evidently be simplified to

$\begin{array}{}\text{(7)}& \frac{\mathrm{1}}{T}\frac{d}{\mathrm{d}\mathit{\tau }}\left(T\right)=-{n}_{\mathrm{H}}{\mathit{\sigma }}_{\text{ex}}{v}_{\text{rel}},\end{array}$

and with ${v}_{\text{rel}}\simeq \sqrt{\mathrm{8}\mathit{\pi }kT/m}$ leads to

$\begin{array}{}\text{(8)}& \frac{\mathrm{1}}{{T}^{\mathrm{3}/\mathrm{2}}}\frac{d}{\mathrm{d}\mathit{\tau }}\left(T\right)=-{n}_{\mathrm{H}}{\mathit{\sigma }}_{\text{ex}}\sqrt{\mathrm{8}\mathit{\pi }k/m},\end{array}$

yielding with σex=const

$\begin{array}{}\text{(9)}& \frac{\mathrm{1}}{\sqrt{T}}-\frac{\mathrm{1}}{\sqrt{{T}_{\mathrm{8}}}}={n}_{\mathrm{H}}{\mathit{\sigma }}_{\text{ex}}\sqrt{\frac{\mathrm{2}\mathit{\pi }k}{m}}\left(\mathit{\tau }-{\mathit{\tau }}_{\mathrm{8}}\right),\end{array}$

where ${\mathit{\tau }}_{\mathrm{8}}={t}_{\mathrm{8}}=$ January 2008, the date when Voyager 2 started moving within the heliosheath. Adopting a time period $\left({\mathit{\tau }}_{\mathrm{11}}-{\mathit{\tau }}_{\mathrm{8}}\right)\simeq \left({t}_{\mathrm{11}}-{t}_{\mathrm{8}}\right)=\mathrm{9.6}×{\mathrm{10}}^{\mathrm{7}}$ s, we find

$\begin{array}{}\text{(10)}& \sqrt{{T}_{\mathrm{11}}/{T}_{\mathrm{8}}}=\frac{\mathrm{1}}{\mathrm{1}+{n}_{\mathrm{H}}{\mathit{\sigma }}_{\text{ex}}\sqrt{\frac{\mathrm{2}\mathit{\pi }k{T}_{\mathrm{8}}}{m}}\left({t}_{\mathrm{11}}-{t}_{\mathrm{8}}\right)}=\frac{\mathrm{1}}{\mathrm{1.19}},\end{array}$

assuming $\sqrt{\frac{\mathrm{2}\mathit{\pi }k{T}_{\mathrm{8}}}{m}}=\mathrm{100}$ km s−1 and ${n}_{\mathrm{H}}\mathit{\sigma }\approx {\mathrm{10}}^{-\mathrm{16}}\phantom{\rule{0.125em}{0ex}}{\mathrm{cm}}^{-\mathrm{1}}$. This yields ${T}_{\mathrm{11}}={T}_{\mathrm{8}}/{\mathrm{1.119}}^{\mathrm{2}}\approx \mathrm{98}\phantom{\rule{0.125em}{0ex}}\mathrm{863}$ K. This estimate has to be compared with temperatures of about 40 000 K measured by Voyager 2 in 2011 and indicates that an essential part of the ion cooling may be ascribed to ongoing charge exchange reactions removing high-energy ions and replacing them by low-energetic ones.

Because the heliosheath plasma is convected along streamlines that originate at different points on the termination shock, it must be kept in mind that Voyager 2 crosses different streamlines (see Fig. 2) when moving through the inner heliosheath. On these different streamlines the solar wind evolves differently, as has been quantified in . Furthermore, we assumed that the locally prevailing temperature is a κ temperature Tκ resulting from an incompressible plasma flow:

$\begin{array}{}\text{(11)}& {T}_{\mathit{\kappa }}\left(s\right)=\frac{m}{\mathrm{2}k}{\mathrm{\Theta }}^{\mathrm{2}}\left(s\right)\frac{\mathit{\kappa }\left(s\right)}{\mathit{\kappa }\left(s\right)-\mathrm{3}/\mathrm{2}}.\end{array}$

In the following, after discussing the variation in κ and κ temperature along the Voyager 2 trajectory, we derive how the Maxwellian temperatures can be translated in κ temperatures.

4 Heliosheath plasma along the Voyager 2 trajectory

Figure 2Some streamlines originating at different points at the termination shock are shown with different colors. The Voyager 2 trajectory is the black straight line crossing the various streamlines on which the solar wind has evolved differently .

Figure 3κ evolution for three different momentum diffusion coefficients: ${D}_{\mathrm{0}}={\mathrm{10}}^{-\mathrm{10}},{\mathrm{10}}^{-\mathrm{9}},{\mathrm{10}}^{-\mathrm{8}}$ (red, green, and blue curve, respectively) along the Voyager 2 trajectory. See for details.

Figure 4The κ temperature derived from Eq. (9) of using the results shown in Fig. 3

Figure 5Sketch of a realistic situation for fitting the data (histogram, i.e., currents in the instrument cups). The black line shows a shifted Maxwellian and three shifted κ distributions with κ=1.6 (red), κ=3 (blue), and κ=10 (green). The numbers on the x axis are the channel numbers, which can be translated to energy or velocity. The y axis is in arbitrary units. The distribution functions are multiplied with the velocity, which is approximately a representation of the currents; see .

Figure 6The truncated moments ${M}_{\mathit{\kappa },x}^{\mathrm{0}}$ (a) and ${M}_{\mathit{\kappa },x}^{\mathrm{2}}$ (b) for different κ values. The logarithmic x axis displays the speed normalized to 100 km s−1 and the y axis is in arbitrary units. See text for details. The dotted vertical line indicates a speed of roughly 200 km s−1, which is about the solar wind speed in the heliosheath. A speed about 30 units corresponds to that of protons as measured in the lowest-energy channel of the LECP instrument onboard Voyager.

In we developed a numerical procedure to calculate the ion pressure evolution along arbitrarily selected flow lines of the heliosheath plasma flow on the basis of a pressure transport equation in which it was assumed that at each location in the plasma flow the underlying distribution function can be approximated by a κ distribution $f\left(s,v\right)={f}_{\mathit{\kappa }}\left(s,v\right)$. This method allowed us to estimate the evolution of the κ parameter along each streamline.

In Fig. 3 we show how the κ parameter varies in turn along the Voyager 2 trajectory (displayed as heliocentric distance in AU). Depending on the magnitude of the velocity diffusion coefficient D0 , one can see along the Voyager 2 trajectory different κ values (green, red, or blue curves). The more efficient the velocity diffusion process, the lower the resulting κ values at Voyager 2. Directly connected with these κ values are the associated κ temperatures Tκ along the Voyager 2 trajectory (see Eq. 11) as shown in Fig. 4.

5 κ temperatures along the Voyager 2 trajectory

We use the standard coordinate system for distribution functions, namely the space coordinates in the solar frame and the velocity coordinates in the rest frame (bulk frame) of the fluid. We assume isotropic distributions as defined in Sect. 1.1 above in each case.

Figure 7The estimated κ temperature (blue lines) using Eq. (17) and the κ parameter along the Voyager 2 trajectory as calculated by . The lower red curves are the Voyager data form NSSDC (http://omniweb.gsfc.nasa.gov/coho/form/voyager2.html). From top to bottom (a–c), the diffusion coefficients as given in change from ${D}_{\mathrm{0}}={\mathrm{10}}^{-\mathrm{8}}$, 10−9, 10−10cm2 s−3, respectively.

5.1 The data fitting procedure

A problem now unavoidably arises when using observations like those from the plasma instrument onboard Voyager 2 with only a few data points in a limited energy range to fit a distribution function.

The physical parameters like density, temperature, and velocity are obtained from the currents of the different cups in the PLS instrument onboard Voyager 2. These currents correspond to a kinetic energy (momentum) from which the parameters are derived assuming a best-fit Maxwellian . Such a best fit is shown in Fig. 5 using data given by , in which the black (step-like) lines show the measurements and the result of the best fit with a Maxwellian. We have also plotted results using κ distributions with different κ values. It can be seen that the representations by the alternative κ distributions give similar results in the measurement range but deviate from the Maxwellian fit at higher energies (channels). So, in the range not covered by data the tails of the distribution functions can contribute remarkably to the higher moments, especially to the pressure (temperature). Thus the latter can strongly differ from that obtained by using Maxwellians.

5.2 Truncated distribution functions

Based on the above discussion we only have available truncated distribution functions that have to deliver the fit. In the case of a Maxwellian this is not a problem. For the κ distribution, however, the core for different κ values can be fitted quite well, while the very different tails that are not taken into account into the fit imply that particles far off the core contribute remarkably to the temperature moments. It can be seen in Fig. 5 that the fits to the κ distribution can easily produce substantial differences in the expected values for different κ. This is shown in Fig. 6 in which we have plotted the truncated zeroth- and second-order moments normalized to the number density. That is,

$\begin{array}{}\text{(12)}& {M}_{\mathit{\kappa },x}^{\mathrm{0}}=\mathrm{4}\mathit{\pi }\underset{\mathrm{0}}{\overset{x}{\int }}{v}^{\mathrm{2}}{f}_{\mathit{\kappa }}\mathrm{d}v\phantom{\rule{2em}{0ex}}{M}_{\mathit{\kappa },x}^{\mathrm{2}}=\mathrm{4}\mathit{\pi }\underset{\mathrm{0}}{\overset{x}{\int }}{v}^{\mathrm{4}}{f}_{\mathit{\kappa }}\mathrm{d}v.\end{array}$

It can be seen in the left panel of Fig. 6 that for different sets of (κ) for small normalized speed values, the different κ distribution zeroth-order moments are approximately the same, but strongly differ in the second-order moments for small κ 2 above a truncation speed (see right panel), which is usually higher than the range covered by observations. These incomplete distribution functions are the true problem when interpreting observational data, while the problem faced by considering complete (idealized) distribution functions appears more academic. Measured distributions are constrained by the instrument limitations, for instance given by the energy range detected by the instrument for which the highest-energy channel may define a truncation limit of the distribution function and its moments.

We have extended the range of the x axis to match the lowest channel of the LECP instrument onboard Voyager 2, which measures energetic protons in the energy range of ≈40 keV corresponding to roughly 3000 km s−1. It becomes obvious from Fig. 6 that these particles do not essentially contribute to the number density for different κ values, but they heavily influence the second-order moment (the pressure). Because the contribution of the high-energy particles to the number density is so small, it is not an easy task to determine the κ value. It will require more advanced models, which go far beyond the scope of this paper to determine the κ distribution from data of different instruments.

5.3 Analytic estimate of the κ distribution temperature

From the second moments we can find a relation between the pressures of a κ distribution Pκ and that of a Maxwellian Pm (see Eq. A10).

$\begin{array}{}\text{(13)}& {P}_{\mathit{\kappa }}=\frac{\mathrm{2}\mathit{\kappa }}{\mathrm{2}\mathit{\kappa }-\mathrm{3}}\phantom{\rule{0.125em}{0ex}}\frac{{\mathrm{\Theta }}^{\mathrm{2}}}{{v}_{\mathrm{T}}^{\mathrm{2}}}{P}_{\mathrm{m}}\end{array}$

Now we can either calculate the Maxwellian pressure from the observed Maxwellian temperature Tm by using

$\begin{array}{}\text{(14)}& {P}_{\mathrm{m}}=nk{T}_{\mathrm{m}}\end{array}$

with the Boltzmann constant k and comparing the pressure terms or, we define the κ temperature Tκ by using

$\begin{array}{}\text{(15)}& {T}_{\mathit{\kappa }}=\frac{{P}_{\mathit{\kappa }}}{nk}\end{array}$

and compare the “temperatures”. The result is the same, except for absolute numbers. With this definition we find

$\begin{array}{}\text{(16)}& {T}_{\mathit{\kappa }}=\frac{\mathrm{2}\mathit{\kappa }}{\mathrm{2}\mathit{\kappa }-\mathrm{3}}\phantom{\rule{0.125em}{0ex}}\frac{{\mathrm{\Theta }}^{\mathrm{2}}}{{v}_{\mathrm{T}}^{\mathrm{2}}}{T}_{\mathrm{m}},\end{array}$

which is identical to Eq. (11).

Furthermore, because for κ→∞ the κ temperature must be equal to that of the Maxwellian, we find that vT and finally, since Θ is independent of κ,

$\begin{array}{}\text{(17)}& {T}_{\mathit{\kappa }}=\frac{\mathrm{2}\mathit{\kappa }}{\mathrm{2}\mathit{\kappa }-\mathrm{3}}{T}_{\mathrm{m}}\end{array}$

depends only on κ and can be easily calculated from the Maxwellian temperature knowing the corresponding κ. Thus one can use the temperatures derived from a Maxwellian distribution and calculate from that the κ temperatures; κ and Θ can be estimated elsewhere, for instance by solving a kinetic transport equation as done in .

That also means that the presented temperatures derived with a Maxwellian as available from Omniweb (http://omniweb.gsfc.nasa.gov/coho/form/voyager2.html) must be taken with care because they should translate into κ distribution temperatures along the Voyager 2 trajectory through the inner heliosheath. With Eq. (17) this is an easy task knowing κ.

In Fig. 7 we have plotted the Voyager 2 temperature data between 84 and 112 AU, together with those obtained from the κ parameter of using Eq. (17). In the left panel of Fig. 7 the κ parameter reaches the limiting value of 1.5, and to avoid numerical problems it was set to κ=1.5001 beyond r≈98 AU. Thus beyond 98 AU the model by is not applicable for such high values of the diffusion coefficient (${D}_{\mathrm{0}}\approx {\mathrm{10}}^{-\mathrm{8}}$cm2 s−1), while it can nicely explain the nonadiabatic behavior of the temperature between ≈88 and ≈98 AU.

This shows that the temperature highly depends on the underlying distribution function, and thus temperature data have to be taken with care because they can easily lead to erroneous interpretations of the data. This holds true in general.

6 Conclusions

The above study shows how Maxwellian temperatures derived from Voyager 2 measurements can easily be translated into corresponding κ temperatures. We have also shown (like Nicolaou and Livadiotis2016) that the temperature of a plasma highly depends on the underlying distribution function and differs from that obtained by a Maxwellian. This implies the need for knowledge of the underlying distribution function because otherwise the temperature is not well defined.

The recently presented study by is not helpful in the “data-relevant” aspects discussed here because the authors compare moments calculated on the basis of κ  or Maxwell distributions; however, they are taken from an infinite velocity range, while in reality data only support moments in a very limited velocity range. This is the important aspect that has to be taken seriously in these matters.

Nevertheless, after deriving a κ distribution from theoretical considerations it is an easy task to determine the corresponding κ temperature when a Maxwellian temperature is given. From IBEX observations it might be possible to obtain the κ value from which the temperature can be estimated. Note that previous attempts to do so use a κ-independent temperature, which is a concept under debate .

The procedure discussed in the present paper may not only be applied to spacecraft observations but also to all observations in which temperature is derived with the assumption that the underlying distribution function is a Maxwellian.

Data availability
Data availability.

The Voyager data used for Fig. 7 can be accessed at http://omniweb.gsfc.nasa.gov/coho/form/voyager2.html (OMNIWeb, 2018).

Appendix A: Moments of the distribution function

From the moments of a distribution function the macroscopically observable quantities can be derived (for example, Goedbloed and Poedts2004, and many others). To calculate the moments one has to integrate the distribution function times some power α of the speed. For the Maxwell distribution the integrals are well known and can be found elsewhere (e.g., Gradshteyn and Ryzhik2007). For the κ distribution they can also be found in a more general form in , Nr 3.251.2:

$\begin{array}{ll}I& =\underset{\mathrm{0}}{\overset{\mathrm{\infty }}{\int }}{x}^{\mathit{\mu }-\mathrm{1}}\left(\mathrm{1}+\mathit{\beta }{x}^{p}{\right)}^{-\mathit{\nu }}\mathrm{d}x\\ \text{(A1)}& & =\frac{\mathrm{1}}{p}{\mathit{\beta }}^{-\mathit{\mu }/p}\frac{\mathrm{\Gamma }\left(\mathit{\mu }/p\right)\mathrm{\Gamma }\left(\mathit{\nu }-\mathit{\mu }/p\right)}{\mathrm{\Gamma }\left(\mathit{\nu }\right)},\end{array}$

where Γ is the gamma function. With

$\begin{array}{}\text{(A2)}& \mathit{\mu }=\mathrm{3}+\mathit{\alpha },\phantom{\rule{1em}{0ex}}\mathit{\beta }=\frac{\mathrm{1}}{\mathit{\kappa }{\mathrm{\Theta }}^{\mathrm{2}}},\phantom{\rule{1em}{0ex}}\mathit{\nu }=\mathit{\kappa }+\mathrm{1},\phantom{\rule{1em}{0ex}}p=\mathrm{2}\end{array}$

we have the integral

$\begin{array}{ll}{I}_{\mathit{\alpha }}& \equiv \underset{\mathrm{0}}{\overset{\mathrm{\infty }}{\int }}{v}^{\mathit{\alpha }+\mathrm{2}}{\left(\mathrm{1}+\frac{{v}^{\mathrm{2}}}{\mathit{\kappa }{\mathrm{\Theta }}^{\mathrm{2}}}\right)}^{-\mathit{\kappa }-\mathrm{1}}\mathrm{d}v\\ \text{(A3)}& & =\frac{\mathrm{1}}{\mathrm{2}}{\left(\sqrt{\mathit{\kappa }}{\mathrm{\Theta }}^{\mathrm{2}}\right)}^{\frac{\mathrm{3}+\mathit{\alpha }}{\mathrm{2}}}\frac{\mathrm{\Gamma }\left(\frac{\mathrm{3}+\mathit{\alpha }}{\mathrm{2}}\right)\mathrm{\Gamma }\left(\mathit{\kappa }-\frac{\mathrm{1}+\mathit{\alpha }}{\mathrm{2}}\right)}{\mathrm{\Gamma }\left(\mathit{\kappa }+\mathrm{1}\right)},\end{array}$

and thus ${I}_{\mathrm{0}}\phantom{\rule{0.33em}{0ex}}g\left(\mathit{\kappa },\mathrm{\Theta }\right)=\mathrm{1}$, where

$g\left(\mathit{\kappa },\mathrm{\Theta }\right)=n\left(\mathbit{r}\right)\frac{\mathrm{\Gamma }\left(\mathit{\kappa }+\mathrm{1}\right)}{{\left(\sqrt{\mathit{\pi }}\sqrt{\mathit{\kappa }}\mathrm{\Theta }\right)}^{\mathrm{3}}\mathrm{\Gamma }\left(\mathit{\kappa }-\frac{\mathrm{1}}{\mathrm{2}}\right)}$

is the normalization factor of the κ distribution defined in Eq. (1).

With the above integral we can thus easily calculate the following moments of the κ distribution of fκ and compare them to those for a Maxwell distribution fm.

In the comoving reference frame the first-order moment M1 vanishes. Furthermore, we assume that the stress tensor will be described by an isotropic pressure and thus the dyadic vv can be contracted to a scalar M2 as $\mathbit{v}\otimes \mathbit{v}\to {v}^{\mathrm{2}}$. In addition we calculate the most probable speed by using

$\begin{array}{}\text{(A4)}& {M}^{\mathrm{1}}=\frac{\mathrm{1}}{n\left(\mathbit{r},t\right)}\int vf\left(\mathbit{r},\mathbit{v},t\right){\mathrm{d}}^{\mathrm{3}}v={v}_{\mathrm{T}}.\end{array}$

We find the following for the different moments.

$\begin{array}{}\text{(A5)}& & {M}_{\mathrm{m}}^{\mathrm{0}}={M}_{\mathit{\kappa }}^{\mathrm{0}}=n\left(\mathbit{r},t\right)\text{(A6)}& & {M}_{\mathrm{m}}^{\mathrm{1}}=\frac{\mathrm{2}}{\sqrt{\mathit{\pi }}}\sqrt{\frac{\mathrm{2}kT}{m}}\equiv \frac{\mathrm{2}}{\sqrt{\mathit{\pi }}}{v}_{\mathrm{T}}\text{(A7)}& & {M}_{\mathrm{m}}^{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}{v}_{\mathrm{T}}^{\mathrm{2}}n\left(\mathbit{r},t\right)\text{(A8)}& & {M}_{\mathit{\kappa }}^{\mathrm{1}}=\frac{\mathrm{2}}{\sqrt{\mathit{\pi }}}\mathrm{\Theta }\frac{\mathrm{1}}{\left(\mathit{\kappa }-\mathrm{1}\right)\sqrt{\mathit{\kappa }}}\frac{\mathrm{\Gamma }\left(\mathit{\kappa }+\mathrm{1}\right)}{\mathrm{\Gamma }\left(\mathit{\kappa }-\frac{\mathrm{1}}{\mathrm{2}}\right)}\text{(A9)}& & {M}_{\mathit{\kappa }}^{\mathrm{2}}=\frac{\mathrm{3}\mathit{\kappa }{\mathrm{\Theta }}^{\mathrm{2}}}{\mathrm{2}\mathit{\kappa }-\mathrm{3}}n\left(\mathbit{r},t\right)\end{array}$

The pressure ratio is

$\begin{array}{}\text{(A10)}& {P}_{\mathit{\kappa }}=\frac{\mathrm{2}\mathit{\kappa }}{\mathrm{2}\mathit{\kappa }-\mathrm{3}}\phantom{\rule{0.125em}{0ex}}\frac{{\mathrm{\Theta }}^{\mathrm{2}}}{{v}_{\mathrm{T}}^{\mathrm{2}}}{P}_{\mathrm{m}}.\end{array}$

Finally, replacing the velocity in the fluid rest frame by that in the observer frame, which is $\mathbit{v}=\mathbit{u}-\mathbit{w}$ where u is the fluid bulk velocity and w the thermal velocity, one easily finds that the zeroth moment remains, the first moment gives the bulk speed, and in the second moment the ram pressure appears as an additional term.

Competing interests
Competing interests.

The authors declare that they have no conflict of interest.

Acknowledgements
Acknowledgements.

The work of Hans Jörg Fahr and Klaus Scherer was partly carried out within the framework of the bilateral BMBF–NRF project “Astrohel” (01DG15009) funded by the Bundesministerium für Bildung und Forschung. Responsibility for the contents of this work lies with the authors. Adama Sylla acknowledges support via DFG project FI706/21-1. John D. Richardson was supported under NASA contract 959203 from the Jet Propulsion Laboratory to the Massachusetts Institute of Technology.
The topical editor, Margit Haberreiter, thanks two anonymous referees for help in evaluating this paper.

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text:
Change of basis of the vector

So I understand that the change of basis of the matrix is $X' = U^{\dagger}XU$. But how about if I just want to change the basis of a vector $v$.

Is $v' = U^{\dagger}v$ the right one? Why does the $U$ disappear here?

• You are using a lot of symbols without letting us know what any of them mean. – Gerry Myerson Sep 23 '16 at 7:18
• – Ka-Wa Yip Sep 23 '16 at 7:33
• Presumably $U$ is unitary and $\dagger$ is the adjoint. – Omnomnomnom Sep 23 '16 at 11:49
• The key is to understand that $X'v'$ should produce the same result as $Xv$, but that output $Xv$ of $X$ should be expressed with respect to the new basis. – Omnomnomnom Sep 23 '16 at 11:53
• @Omnomnomnom thank you. does that mean $v' = X'^{\dagger}Xv$? – Ka-Wa Yip Oct 6 '16 at 22:06

format: plain


text:
Hello, my teacher assigned a homework problem about a series-parralel circut. I was wonder if possible if you could help me find the Resistance Total, Current Total, Current Drop(of each resistor), and Voltage Drop (of each resisitor). I have no clue on how to do these things. It would be a nice help if you showed me step by step of how to get each answer.

Hello, my teacher assigned a homework problem about a series-parralel circut. I was wonder if possible if you could help me find theResistance Total, Current Total, Current Drop(of each resistor), and Voltage Drop (of each resisitor). I have no clue on how to do these things. It would be a nice help if you showed me step by step of how to get each answer.


text:
# Proving limit exists

• November 2nd 2009, 04:12 PM
eXist
Proving limit exists
Well the problem reads: Suppose $f: (a, b) \rightarrow R$ and $\lim_{x\to\infty}f(x_n)$ exists for every sequence $x_n$ (where n goes from $1 \to \infty$) in $(a, b)$ such that $\lim_{x\to\infty}x_n = a$. Prove that $\lim_{x\to a^+}f(x_n)$ exists.

I'm going to assume that $\lim_{x\to a^+}f(x_n)$ does not exist and prove it by contradiction. So then we still have $\lim_{x\to\infty}x_n = a$. But since the sequence $x_n$ lies in the interval $(a, b)$ the $\lim_{x\to\infty}x_n = a$ can't exist because we assumed that right hand limit: $\lim_{x\to a^+}f(x_n)$ does not exist, so the limit as anything approaches a in the interval $(a, b)$ can't exist. But that is a contradiction since we are given $\lim_{x\to\infty}x_n = a$.

I'm not sure if this is right because it seems a little short. Can someone let me know if this is the right approach? Thanks, Chad.

Edit: Maybe I have to use Bolzano-Weierstrass and say that $x_n$ has a convergent sub sequence that tends to a as x approaches infinity, not really sure that makes a difference though since we already have a sequence that converges to a.
• November 2nd 2009, 04:52 PM
Jose27
Quote:

Originally Posted by eXist
Well the problem reads: Suppose $f: (a, b) \rightarrow R$ and $\lim_{x\to\infty}f(x_n)$ exists for every sequence $x_n$ (where n goes from $1 \to \infty$) in $(a, b)$ such that $\lim_{x\to\infty}x_n = a$. Prove that $\lim_{x\to a^+}f(x_n)$ exists.

I'm going to assume that $\lim_{x\to a^+}f(x_n)$ does not exist and prove it by contradiction. So then we still have $\lim_{x\to\infty}x_n = a$. But since the sequence $x_n$ lies in the interval $(a, b)$ the $\lim_{x\to\infty}x_n = a$ can't exist because we assumed that right hand limit: $\lim_{x\to a^+}f(x_n)$ does not exist, so the limit as anything approaches a in the interval $(a, b)$ can't exist. But that is a contradiction since we are given $\lim_{x\to\infty}x_n = a$.

I'm not sure if this is right because it seems a little short. Can someone let me know if this is the right approach? Thanks, Chad.

Edit: Maybe I have to use Bolzano-Weierstrass and say that $x_n$ has a convergent sub sequence that tends to a as x approaches infinity, not really sure that makes a difference though since we already have a sequence that converges to a.

You have to assume that the limit $c=f(x_n)$ is the same for any sequence $x_n \rightarrow a$ (otherwise the conclusion would not be valid for obvious reasons). Assume that $\lim_{x\rightarrow a^+ } f(x)$ doesn't exists, in particular there would exist an $\epsilon >0$ such that for every $0<\delta (=1/n)$ we have $x_{\delta }-a < \delta$ and $\vert f(x_{\delta })-c \vert \geq \epsilon$

Edit: Your argument fails because you assume that because $\lim_{x\rightarrow a^+ } f(x)$ doesn't exist this implies that the limit $f(x_n)$ doesn't exist for any sequence converging to a.
• November 2nd 2009, 05:46 PM
eXist
So if I assume that the limit: $\lim_{x\to a^+}f(x)$ does not exist, that changes the fact that every sequence converges to a? I might be miss understanding you.

I understand that a must be a cluster point since every sequence in the interval $(a, b)$ converges to it, and therefor the limit at a from the right must exist since the terms of the sequence are all larger than a since they are in the interval $(a, b)$.

I see why this is true, I'm just struggling to write it down I guess.


text:
# How do you find the product of (8r-1)(r-6)?

Aug 13, 2017

$\left(8 r - 1\right) \left(r - 6\right) = 8 {r}^{2} - 49 r + 6$

#### Explanation:

If you find it helpful, you can use the FOIL mnemonic to remember all the combinations of terms to multiply:

$\left(8 r - 1\right) \left(r - 6\right) = {\overbrace{\left(8 r\right) \left(r\right)}}^{\text{First"+overbrace((8r)(-6))^"Outside"+overbrace((-1)(r))^"Inside"+overbrace((-1)(-6))^"Last}}$

$\textcolor{w h i t e}{\left(8 r - 1\right) \left(r - 6\right)} = 8 {r}^{2} - 48 r - r + 6$

$\textcolor{w h i t e}{\left(8 r - 1\right) \left(r - 6\right)} = 8 {r}^{2} - 49 r + 6$


text:
The number line shown contains three points R, S, and T, whose coordinates have absolute values r, s, and t, respectively.  Which of the following equals the average (arithmetic mean) of the coordinates of the points R, S, and T ?

s

s+t-r

~$\frac{r-s-t}{3}$~

~$\frac{r+s+t}{3}$~

~$\frac{s+t-r}{3}$~


text:
## Files in this item

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## Description

 Title: ROTATIONAL SPECTROSCOPY OF VIBRATIONALLY EXCITED N2H+ and N2D+ UP TO 2 THZ Author(s): Yu, Shanshan Contributor(s): Amano, Takayoshi; Elliott, Ben; Daly, Adam M; Crawford, Timothy J; Drouin, Brian; Pearson, John Subject(s): Astronomy Abstract: Terahertz absorption spectroscopy was employed to extend the measurements on the pure rotational transitions of N$_2$H$^+$, N$_2$D$^+$ and their $^{15}$N-containing isotopologues in the ground state and first excited vibrational states for the three fundamental vibrational modes. In total 88 new pure rotational transitions were observed in the range of 0.7--2.0~THz. The observed transition frequencies were fit to experimental accuracy, and the improved molecular parameters were obtained. The new measurements and predictions will support the analysis of high-resolution astronomical observations made with facilities such as SOFIA and ALMA where laboratory rest frequencies with uncertainties of 1 MHz or smaller are required for proper analysis of velocity resolved astrophysical components. Issue Date: 25-Jun-15 Publisher: International Symposium on Molecular Spectroscopy Citation Info: ACS Genre: CONFERENCE PAPER/PRESENTATION Type: Text Language: English URI: http://hdl.handle.net/2142/79155 Date Available in IDEALS: 2016-01-05


text:
Introduction to Neural Networks

Programming Lab 5: Spiking neural networks

Spiking neural networks (SNNs)

Today, we will scale up our simulations from single neurons to networks. This will allow us to simulate the dynamics of the networks of neurons in a small patch of cortex, using networks of LIF neurons. We will study the spiking network simulation below:

from brian2 import *

# network parameters
N = 1000; Ne = int( 0.8*N ); Ni = int( 0.2*N );

# membrane parameters
C = 200 * pF
taum = 20 * msecond
gL = C / taum
El = -70 * mV
Vth = -50 * mV

# synapse parameters
Ee = 0 * mvolt
Ei = -80 * mvolt
taue = 5 * msecond
taui = 5 * msecond

# excitatory and inhibitory weights
g = 4 							# balance between excitation and inhibition
we = 0.5 * nS 						# excitatory synaptic weight
wi = ( 7 * g * (we/nS) ) * nS 				# inhibitory synaptic weight

# input parameters
input_rate = 2*Hz					# Poisson input rate

# membrane equation
eqs = Equations('''
dv/dt = ( gL*(El-v) + ge*(Ee-v) + gi*(Ei-v) ) * (1./C) : volt
dge/dt = -ge*(1./taue) : siemens
dgi/dt = -gi*(1./taui) : siemens''')

# setup network
P = NeuronGroup( N, model=eqs, threshold='v>Vth', reset='v=El', refractory=5*ms )
Pe = P[0:Ne]; Pi = P[Ne:]

# setup excitatory and inhibitory connections
Ce = Synapses( Pe, P, on_pre='ge+=we' )
Ce.connect( True, p=0.01 )
Ci = Synapses( Pi, P, on_pre='gi+=wi' )
Ci.connect( True, p=0.01 )

# setup Poisson input
Ie = PoissonInput( P, 'ge', Ne, input_rate, weight=we )

# initialize network
P.v = randn( len(P) ) * 5 * mV - 70 * mV

# record spikes + rates (excitatory population)
M = SpikeMonitor( Pe )
R = PopulationRateMonitor( Pe )

# run simulation
run( 1 * second, report='text' )

# Make plot
figure( figsize=(15,5) )
xlabel( 'time (ms)' ); ylabel( 'cell id' )
plot( M.t/ms, M.i, '.k' )
show()

This parameter regime creates a low-rate, irregular spiking regime called the asynchronous-irregular state. These highly disordered states are an important model in computational neuroscience, as they recreate the firing patterns of neurons in cortex. Specifically, during spontaneous waking activity, neurons fire with low rate (< 10 Hz) and high irregularity ($$C_V \approx 1$$). We will study how this state emerges as a function of the balance between excitation and inhibition.

The most important parameters determining the state of the network are g (the balance of excitation and inhibition) and input_rate (the firing rate of the external inputs). We can study the mean firing rate in the network using the PopulationRateMonitor object in Brian (R in the code above). Try plotting the firing rate as a function of time for one run of the simulation and study how that matches up with the spike train raster. Also try running the network for different numbers of neurons, to understand how this affects the spiking state.

Finally, we can study how states of regular, highly synchronous spiking emerge in the network at different values of g and input_rate. To quantify the irregularity in the network, we can calculate $$C_V$$ using the code below:

def coefficient_of_variation( spike_monitor, N ):
# calculate Cv of inter-spike intervals
cv = np.zeros( N )
for ii in range(N):
tau = diff( spike_monitor.spike_trains()[ii] )
cv[ii] = std( tau ) / mean( tau )

If you need to run multiple simulations of the network, you can put the network setup and running code into a function

Experiments: (1) Simulate the network with increasing values of input_rate and observe how the network state changes. (2) Simulate the network for increasing values of g and again observe the change in network state. (3) Quantify the dependence on both control parameters, either in a one- or two-dimensional scan across values.


text:
# Scrooge and the Ethical Case for Consumption Taxation

Most of the popular discussion about tax fairness focuses on how much money people make. But it has always seemed to me that it is when people spend money on themselves that they incur whatever debt to society the principles of taxation ought to imply.

• Suppose first that I made  $10 billion, and gave all but$100,000 of it away to take care of the poor. Should I really be taxed on the $10 billion that I no longer have, but already gave away, or only on the$100,000 that I then actually spend on myself? And with that much of the task of taking care of the poor taken off of the shoulders of the government, won’t it have enough money from taxing everyone’s spending to take care of the other necessary tasks of government?
• Next, suppose I save the money. Then it is still indeterminate whether I will eventually spend the money on myself or give it away. Shouldn’t the government wait until it is clear whether I will spend the money or give it away to take care of the poor before the government taxes it?  Here, notice also that in saving the money for later use, I am making resources available for building factories or other places of business, which employ people.
• Now, suppose I give the money to my children. Then shouldn’t the government wait to see whether the children spend the money on themselves or give it away to take care of the poor before the government taxes it?

The essence of this argument is that it is only at the moment of consumption spending that I appropriate money for myself. Until then, I am only acting as a steward for those resources whose ultimate use has not yet been determined. And if I give money to my children, it is only at the moment of consumption spending that my children actually appropriate money for themselves.

For the rest of the argument, click through to Steven Landsburg’s essay  “What I Like About Scrooge: In Praise of Misers.” Here is a short excerpt:

If you build a house and refuse to buy a house, the rest of the world is one house richer. If you earn a dollar and refuse to spend a dollar, the rest of the world is one dollar richer—because you produced a dollar’s worth of goods and didn’t consume them.

Postscript. Most of the tricky issues that remain after saying this much fall under the heading of defining “consumption spending.” But I maintain that income is much harder to define than “consumption spending” is.

There is one more point to be made about aggregate demand. If I am going to spend money on something sooner or later, I am being more helpful to society if I spend the money during a recession than if I spend it during a boom. But it would be even more helpful if during the recession I gave it away to take care of the poor (which would also add to aggregate demand, and alleviate suffering). When the economy is at the natural level of output or above, there is no social value in adding to aggregate demand, since that creates inflationary pressures that need to be counteracted. The benefit to those I employ is balanced out by the harm to those left unemployed because of the measures that will be taken to avoid inflation.


text:
# Dice Roll Probabilities

I'm trying to figure out the probabilities for the following casino game:

You and the dealer each roll a pair of dice and the person with the highest individual die roll wins. If its a tie, you win.

First, what is the probability you win?

Second, given that you've won, whats the probability that the game resulted in a tie?

--

Here are my thoughts:

For the first - the expected value of your die roll is 4.25; which means you will win unless the dealer rolls a 5 or 6, giving you a 2/3 chance of winning.

For the second I'm not sure how to think about it.

• Expected values really don't let you compute probabilities. – Thomas Andrews Jul 4 '15 at 0:32
• In what sense is the expected value of your die roll $4.25$? The expected value of your highest individual roll is $161/36\approx4.47$ if I calculated correctly. Also, ignoring for a minute the problem that Thomas pointed out, why is there a $2/3$ chance that the dealer rolls a $5$ or $6$? The probability that the higher of two die rolls is $5$ or $6$ is $5/9$. – joriki Jul 4 '15 at 0:45
• Thank you for sharing how you are doing the work, that's the expectation on this site: share what you are trying/have tried, and be specific about your questions. Good work! – FundThmCalculus Jul 4 '15 at 0:51

Suppose $2$ dice are rolled, say by you. We will find the probability that the "larger" result is $i$, for $i=1$ to $6$.

The probability that the larger is $1$ is $\frac{1}{36}$. The probability that they are both $\le 2$ is $\frac{4}{36}$, so the probability the larger is $2$ is $\frac{4-1}{36}$. The probability they are both $\le 3$ is $\frac{9}{36}$, so the probability the larger is $3$ is $\frac{9-4}{36}$. And so on.

So the probability the larger is $i$, for $i=1$ to $6$, is $$\frac{1}{36},\quad\frac{3}{36},\quad \frac{5}{36}, \quad\frac{7}{36},\quad\frac{9}{36}, \quad\frac{11}{36}.\tag{1}$$

Now suppose that you and the dealer each toss two dice. From (1), the probability the higher of your tosses is equal to the higher of the dealer's tosses (that is, the probability $t$ of a tie) is given by $$t=\frac{1}{36^2}\left(1^2+3^2+5^2+7^2+9^2+11^2\right).$$

Let $p$ be the probability you win. Then $p-t$ is the probability your number is bigger than the dealer's. By symmetry this is the same as the probability the dealer's number is bigger than yours. Thus $p+p-t=1$, and therefore $p=\frac{1+t}{2}$. We know $t$, so we know $p$.

As to the conditional probability, by the usual conditional probability calculation, or by inspection, the probability you won through a tie, given that you won, is $\frac{t}{p}$.

Remark: Presumably the real game is that if there is a tie the casino wins. A casino that lets customers win with probability $\gt \frac{1}{2}$ will not be long in business.

• Perhaps I stated the problem unclearly, but each person rolls 2 die and takes the larger of his role. For ex: I could roll 1 and 4 and the dealer could roll 2 and 4. My highest was a 4 and his was a 4 so our rolls tie, and I win. – user3168139 Jul 4 '15 at 1:10
• The problem was stated clearly, and that is exactly the problem that I solved. – André Nicolas Jul 4 '15 at 1:12

Denote $a,b$ are the results of my dices with $6 \ge a>b \ge 1$ We consider $6$ cases:

1. If the largest result of my dice is $1$, or $a=1$ then $b=1$. Hence, the dealer's results must less than or equal to $1$. We obtain $1$ way to determine the dealer's results such that the condition is satisfied. Thus, there are $1 \times 1$ ways that I won when $1$ is my largest result.
2. If $a=2$ then $b=1$ or $b=2$. Now we need to determine the dealer's results so that I could win. This means, the largest result of the dealer is $2$ or $1$. There are $2$ ways that the dealer gets the same result in two dices ($11$ and $22$). We consider the case that the result in two dices is different. We get $\frac{(2-1) \times 2}{2}=1$ ways. This means the dealer has $1+2=3$ ways to get the result. Therefore, there are $2 \times 3$ ways so that I won when my largest result is $2$
3. If $a=3$ then $b=1,2$ or $3$. Similarly, there are $3$ ways so that the dealer can get the same result in 2 dices ($11,22$ and $33$) and $\frac{(3-1) \times 3}{2}=3$ ways in order to get the different results in 2 dices. This means the dealer has $3+3=6$ ways to get the result. Hence, there are $6 \times 3$ ways so that I won when my largest result is $3$.
4. We obtain the formula $n \times \left(n+ \frac{(n-1)n}{2} \right)=\frac{n^2(n+1)}{2}$, which is the number of ways that I won when my largest result is $n$.
5. In conclusion, the probability to win is $$\dfrac{\sum_{n=1}^6 \frac{n^2(n+1)}{2}}{\left[ 6+ \frac{(6-1)6}{2} \right]^2}= \dfrac{\sum_{n=1}^6 \frac{n^2(n+1)}{2}}{21^2}$$


text:
• 论文 •

### 四川短尾鼩的生境选择及对人为干扰的响应

1. 四川师范大学生命科学学院
• 出版日期:2017-08-14 发布日期:2017-07-05

### Habitat selection and the response to human disturbances by Anourosorex squamipes

ZONG Hao，XIANG Dabing, WU Sisi, LI Jiang，REN Xi，LI Wanyu，FU Changkun，WANG Qiong，CHEN Shunde

1. College of Life Science，Sichuan Normal university
• Online:2017-08-14 Published:2017-07-05

Abstract: The population sizes of Anourosorex squamipes in twelve different types of habitats was surveyed during April to September in 2014. According to the order of trapping success and the degree of human disturbance of the different habitats, we analyzed trends of habitat selection of Anourosorex squamipes. Simulated population disturbance was executed by the way of artificial capture. The investigation consisted of population size, age group and sex ratio for different habitats specified as one of four typical sorts (grassland, abandoned pear garden, bush-land and landscape planted woods) after 1, 2, 3 and 4 months of human disturbance, It is shown that the trapping success for Anourosorex squamipes in the 12 habitats presents a sequence from large to small for five grades as follows: forest area ＞ human habitat environment or abandoned pear garden ＞landscape planted woods or bush-land or abandoned vineyard or grassland ＞  Eucalyptus woods or vegetable field ＞ bamboo woods or vineyard or rape field. It was found that Anourosorex squamipes select their habitat more likely in a sequence as follows: habitats with intermediate disturbance and environmental heterogeneity, habitats with high disturbance and low environmental heterogeneity, habitats with low disturbance and high environmental heterogeneity, and habitats with excessive disturbance or environmental heterogeneity or both. Moreover, the population of Anourosorex squamipes in all the habitats was reduced significantly at the beginning of human disturbance, and then could be restored to the normal level in 2 months. More specifically, the average sex ratio was increased by 3.52% , the pregnancy ratio changed slightly, the rate of adult groups was decreased by 1.71%, the rate of old groups decreased by 7.53%, the rate of young group increased by 2.33% and the sub-adult group increased by 6.91% after the human disturbance. Nevertheless, the increasing of rate of young group and sub-adult group suggests that Anourosorex squamipes populations would maintain a continuous growth trend in a long period.


text:
Automatic generation of crystal structure descriptions

## Project description

Robocrystallographer is a tool to generate text descriptions of crystal structures. Similar to how a real-life crystallographer would analyse a structure, robocrystallographer looks at the symmetry, local environment, and extended connectivity when generating a description. The package includes utilities for identifying molecule names, component orientations, heterostructure information, and more…

## Usage

Robocrystallographer can be used from the command-line or from a python API. The package integrates with the Materials Project to for allow generation of structure descriptions directly from materials project ids. For example, to generate the description of SnO 2 (mp-856), one can simply run:

robocrystallographer mp-856

Alternatively, a structure file can be specified in place of an mp-id. Robocrystallographer supports the same file formats as pymatgen, including the Crystallographic Information Format (CIF), and common electronic structure package formats such as POSCAR.

### Python Interface

The two core classes in robocrystallographer are:

• StructureCondenser: to condense the structure into an intermediate JSON representation.

• StructureDescriber: to turn the condensed structure into a text description.

The format of the intermediate JSON representation is detailed on the Condensed Structure Format page page. A minimal working example for generating text descriptions is simply:

from robocrys import StructureCondenser, StructureDescriber

condenser = StructureCondenser()
describer = StructureDescriber()
condensed_structure = condenser.condense_structure(structure)
description = describer.describe(condensed_structure)

Both classes have many options for customising the output of the structure descriptions. More information is provided in the Module Documentation page.

### Example Output

An example of the output generated by robocrystallographer for SnO 2 (mp-856) is given below:

SnO2 is Rutile structured and crystallizes in the tetragonal P4_2/mnm space group. The structure is three-dimensional. Sn(1) is bonded to six equivalent O(1) atoms to form a mixture of edge and corner-sharing SnO6 octahedra. The corner-sharing octahedral tilt angles are 51°. All Sn(1)–O(1) bond lengths are 2.09 Å. O(1) is bonded in a trigonal planar geometry to three equivalent Sn(1) atoms.

## Installation

Robocrystallographer can be installed using pip:

pip install robocrystallographer

Robocrystallographer requires Python 3.5+. The OpenBabel package is required to determine molecule names. This is an optional requirement but its use is recommended for best results. If you are using the Conda package management system, OpenBabel can be installed using:

conda install -c openbabel openbabel

Robocrystallographer is released under a modified BSD license; the full text can be found here.

## Project details

### Source Distribution

robocrys-0.1.0.tar.gz (3.8 MB view hashes)

Uploaded source

### Built Distribution

robocrys-0.1.0-py3-none-any.whl (3.8 MB view hashes)

Uploaded py3


text:
# Changeset 265

Ignore:
Timestamp:
Apr 11, 2005, 5:46:00 PM (17 years ago)
Message:

doc added and some changes of variable names

Location:
trunk/src
Files:
2 edited

### Legend:

Unmodified
 r256 /// /// Function calculating the score for the best aligning of two /// sequences. In the dot_matrix \a A, elament \f$A_{ij} \f$ is /// score how well element i in the first sequence match to element /// j in the other sequence. In the aligning the elements from the /// two sequences are matched or a gap is added in one of the /// sequences. The latter is penalized with \a gap, which means if /// \a gap is large there will be no gaps in the best aligning (if /// the sequences have the same length). \a path contains what /// elements in \A that is used in the best path. In case of /// degeneracy only one path is returned. /// Function performing alignment following the Needleman-Wunch /// algorithm. The algorithm starts from the dot-matrix, \a s, where /// \f$s_{ij} \f$ describes how well element \f$i \f$ in the first /// sequence match to element \f$j \f$ in the second sequence. A /// path through this matrix corresponds to an alignment of the two /// sequences, in which a diagonal step over a matrix element /// corresponds to a match between the corresponding sequnce /// elements and a vertical or a horizontal step corresponds to /// inserting a gap in one of the sequnces. The function maximizes /// \f$\sum s_{ij}-n \f$ \a gap, where the first sum goes over all /// matches and the second term is a penalty term for inserting \f$/// n \f$ gaps. Default is to have the \a gap cost set to zero. The /// vector \a path contains information about which elements that /// were matched. If the first element in the first sequence was /// matched to the first element in the second sequence, the last /// element in \a path is pair(0,0). /// /// @return score from the best aligning /// @return the global maximum alignment score. /// double NeedlemanWunsch(const gslapi::matrix& A, double NeedlemanWunsch(const gslapi::matrix& s, std::vector >& path, const double gap);


text:
# Conditional Expectation (orthogonal projection): Why unique minimizer?

Let $(\Omega, \mathcal{A}, P)$ be a probability space and consider a square-integrable random variable $X$ and a sub-sigma-Algebra $\mathcal{F}$.

Denote $Z:=E[X \mid \mathcal{F}]$.

Conditional expectation is the unique minimizer (orthogonal projection), meaning that

$E[(X-Z')^2] \geq E[(X-Z)^2]$ for all $Z' \in L^2(\Omega, \mathcal{F},P)$.

Now we have seen the proof of this, deriving:

$E[(X-Z')^2] \geq E[(X-Z)^2] + E[(Z-Z')^2]$

Now it is written that we have equality iff $Z=Z'$. But why? Couldn't it be that $E[(Z-Z')^2]=0$ but $Z\neq Z'$?

Since $(Z-Z')^2 \geq 0$, its expectation is zero if and only if $Z=Z'$ with probability $1$. We sometimes just write this as $Z=Z'$; in particular, $Z$ and $Z'$ are indeed equal if you identify them with their $L^2$ (or any $L^p$) equivalence classes.


text:
# new to linux...makefiles & kdevelop

This topic is 3329 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

##### Share on other sites
I tried this before and I ended up using the automake facilities of KDevelop. It seemed the easiest way. The project contains of source files spread across different directories, using sources which are synchronized over svn with an MSVC project (I didn't set up the svn control as part of the KDevelop project).

You will need to create a new automake based project.
Then use the automake tool.

Read from the link above onwards to first add new files and become familiar with the concept.
The menu entry "Create automake and friends" should take care of building the makefiles in the background.

Try to link against libraries by looking at the Library FAQ. Add some code to use the linked library/ies. Compile this simple new project.
Try to add a subfolder and the files contained in it.

When this works try a new automake project and add your existing sources using the automake manager.

Please report if you can make any progress.

##### Share on other sites
Quote:
 all of the makefile tutorials I have found are saying you have to list every single .cpp file in your project. This is a bit of work since I have about 40-50 .cpp files, all in 3 or 4 different directories.

Yes, you have to list them, or just use a special rule that will apply to all files with the same extension:
.cpp.o:	$(CPP) -c$(ANY_OPTIONS_YOU_LIKE) $*.cpp.c.o:$(CC) -c $(ANY_OPTIONS_AGAIN)$*.c

This will just apply the commands above to all .cpp and .c files producing .o files.
[biased comment]But, frankly, I'd just create a project with all files with CodeBlocks or QTCreator and let it roll..[/biased comment]

##### Share on other sites
It shouldn't take you to long to add files from just 3,4 directories. Multiple file add is possible afaik.

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# How do you find the slope that is perpendicular to the line  y = x + 7?

Mar 19, 2018

See a solution process below:

#### Explanation:

The equation in the problem is in slope intercept form. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$

Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.

$y = \textcolor{red}{1} x + \textcolor{b l u e}{7}$

Therefore the slope is: $\textcolor{red}{m = 1}$

Let's call the slope of a perpendicular line: $\textcolor{b l u e}{{m}_{p}}$

The slope of a line perpendicular to a line with slope $\textcolor{red}{m}$ is the negative inverse, or:

$\textcolor{b l u e}{{m}_{p}} = - \frac{1}{\textcolor{red}{m}}$

Substituting the slope for the line in the problem gives:

$\textcolor{b l u e}{{m}_{p}} = \frac{- 1}{\textcolor{red}{1}} = - 1$


text:
Lemma 15.70.1. Let $R$ be a ring. Let $M$, $N$ be $R$-modules.

1. Given an $R$-module map $\varphi : M \to N$ the following are equivalent: (a) $\varphi$ factors through a projective $R$-module, and (b) $\varphi$ factors through a free $R$-module.

2. The set of $\varphi : M \to N$ satisfying the equivalent conditions of (1) is an $R$-submodule of $\mathop{\mathrm{Hom}}\nolimits _ R(M, N)$.

3. Given maps $\psi : M' \to M$ and $\xi : N \to N'$, if $\varphi : M \to N$ satisfies the equivalent conditions of (1), then $\xi \circ \varphi \circ \psi : M' \to N'$ does too.

Proof. The equivalence of (1)(a) and (1)(b) follows from Algebra, Lemma 10.77.2. If $\varphi : M \to N$ and $\varphi ' : M \to N$ factor through the modules $P$ and $P'$ then $\varphi + \varphi '$ factors through $P \oplus P'$ and $\lambda \varphi$ factors through $P$ for all $\lambda \in R$. This proves (2). If $\varphi : M \to N$ factors through the module $P$ and $\psi$ and $\xi$ are as in (3), then $\xi \circ \varphi \circ \psi$ factors through $P$. This proves (3). $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
# A steel wire of unifrom cross-section 1 mm^(2) is heated to 70^(@)C and strechted by tying it two ends rigidly. Calculate the change in tension on the wire when temperature falls form 70^(@)C to 35^(@)C

Updated On: 17-04-2022

Get Answer to any question, just click a photo and upload the photo
and get the answer completely free,

Text Solution

70N72N74N77N

F = YA* Delta t


text:
Standard Equation of a Circle Calculator

Created by Krishna Nelaturu
Reviewed by Aleksandra Zając, MD
Last updated: Oct 18, 2022

This standard equation of a circle calculator is helpful to calculate the standard form of a circle equation using its center coordinates and radius, or any other form of the circle equation. With this simple tool in your hand, you can easily find the circle equation in any form you want!

• How to write a circle equation in standard form?
• How to convert the standard form into parametric or general form and vice versa.
• FAQ

How to write a circle equation in standard form?

The standard equation of a circle is given by:

$\left(x-a\right)^2 + (y-b)^2 = r^2$

where:

• $(x,y)$ - The coordinates of any point on the circle;
• $(a,b)$ - The coordinates of the center of the circle; and
• $r$ - The radius of the circle.

We can use this equation to find the standard form from its center and radius or vice versa.

How to convert the standard equation into parametric form

The parametric form of a circle equation is given by:

\begin{align*} x &= a + r\cos(\alpha)\\ y &= b + r\sin(\alpha)\\ \end{align*}

where:

• $(x,y)$ - The coordinates of any point on the circle;
• $(a,b)$ - The coordinates of the center of the circle;
• $r$ - The radius of the circle; and
• $\alpha$ - The angle subtended by the point $(x,y)$ at the circle's center $(a,b)$.

It is a straightforward conversion between these two forms; the additional parameter $\alpha$ is not essential for this conversion, so long as the other variables are known.

How to convert the standard equation into general form

The general form of the circle equation is an expansion of its standard form. It can be expressed as:

$x^2 + y^2 + Dx+ Ey+ F = 0$

where:

• $(x,y)$ - The coordinates of any point on the circle;
• $D$ - The sum of the coefficients of the x-terms;
• $E$ - The sum of the coefficients of the y-terms; and
• $F$ - The sum of the constant terms.

Note that the right-hand side (RHS) of this equation has to be zero. Bring every term to the left-hand side (LHS) and simplify.

As this is the expansion of the standard form, we can complete the squares of this expanded form to arrive at the standard equation and establish the following relationships between the various parameters in these two forms:

\begin{align*} D &= -2a\\ E &= -2b\\ F &= a^2+b^2-r^2\\ \end{align*}

We can use these equations to convert between the standard form and the general form of a circle equation.

How to use this standard equation of a circle calculator

You can use this calculator for more than one thing:

• Enter the standard equation of a circle to obtain the center, radius, and circle equation in other forms.
• Give the center and radius of a circle to simultaneously determine its equation in all three forms.
• Enter the equation of a circle in parametric or general form to calculate its center, radius, and equation in the standard form.

In addition, this calculator will also determine other properties of the circle, like its area and circumference.

FAQ

What is the equation of a circle with a center (0,0) and radius of 7?

x2+y2 = 49. To find this equation, follow these steps:

1. Insert the center coordinates in the place of (a,b) in the standard form of a circle equation (x-a)2 + (y-b)2 = r2. This gives (x-0)2 + (y-0)2 = r2.
2. Substitute the value of radius in the place of r in this equation. This gives x2+y2 = 72.
3. Evaluate this equation to get the equation of the circle, x2+y2 = 49.

How do you determine if a point lies on a circle?

To determine whether a point P(px,py) lies on a circle (x-a)2 + (y-b)2 = r2, follow these steps:

1. Substitute the coordinates of the point P(px,py) in place of x and y in LHS of the circle equation to get (px-a)2 + (py-b)2.
• If (px-a)2 + (py-b)2 = r2, then the point P(px,py) lies on the circle.
• If (px-a)2 + (py-b)2 ≠ r2, then the point P(px,py) does not lie on the circle.
Krishna Nelaturu
x-coordinate
y-coordinate
Standard form: (x - A)² + (y - B)² = C
A
B
C
Parametric form: x = A + r cos(α), y = B + r sin(α)
A
B
r
General form: x² + y² + Dx + Ey + F = 0
D
E
F
Other circle properties
R
Diameter
Area
Circumference
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text:
# Using Conversion Factors

In Europe, gasoline efficiency is measured in $\mathrm{km/L}$. If your car's gas mileage is $35.0~\mathrm{mi/gal}$, how many liters of gasoline would you need to buy to complete a 142-km trip in Europe? Use the following conversions: $1~\mathrm{km} = 0.6214~\mathrm{mi}$ and $1~\mathrm{gal} = 3.78~\mathrm{L}$. Express your answer numerically in liters.

I am bit confused as to how to compute my last step.

My conversion pathway:

$$\frac{35.0\ \textrm{mi}}{1\ \textrm{gal}}×\frac{1\ \textrm{km}}{0.6214\ \textrm{ mi}}×\frac{1\ \textrm{gal}}{3.78\ \textrm{L}}= 15\ \textrm{km/L}$$ So: $$15\ \textrm{km/L}×142/\textrm{km}=2115\ \textrm{km}$$

Am I right?

• Note that Europe generally uses litres/100 km and not a mileage per litre. – Jan Oct 30 '15 at 17:18

You did the conversion to km/L right, but I wouldn't round to 15 right away. In your last step you multiplied, should have divided, 142 km / 15 km/L = X L. Just keep track of units and that will help you determine how to do the math.

• So I should round the number and determine the significant figures when I have my final calculation? – Cetshwayo Sep 12 '14 at 18:34
• Why am I dividing 142 by 15? Please explain? – Cetshwayo Sep 12 '14 at 18:40
• In general, don't round until the end, that will help preserve accuracy. And you divide 142 km by 15km/L because you want L as your answer. Dividing km by km/L is the same as multiplying km by L/km. This puts a km on each side of the fraction and they cancel out, leaving L. – user137 Sep 12 '14 at 18:55


text:
# Undefined control sequence error \end{gather}

\documentclass{article}
\usepackage{mathtools,amssymb}

\begin{document}

\begin{gather}
\begin{split}

(1-\underset{(0.034)}{0.060}{\textit{B}}^{12}-\underset{(0.032)}{0.123}{\textit{B}}^{2(12)}){y}_t &=(1-\underset{(0.016)}{0.878}\textit{B}^{12})(1-\underset{(0.012)}{0.969}\textit{B}){e}_t\\

\end{split}

\notag\\

\end{gather}

\end{document}

• Welcome to TeX.SX! Please make your code compilable (if possible), or at least complete it with \documentclass{...}, the required \usepackage's, \begin{document}, and \end{document}. That may seem tedious to you, but think of the extra work it represents for TeX.SX users willing to give you a hand. Help them help you: remove that one hurdle between you and a solution to your problem. Jun 15 '17 at 0:24

Don't use blank lines in math mode, post a complete MWE, and please don't tell me that Box-Ljung is still being taught......

\documentclass{article}
\usepackage{mathtools,amssymb}

\begin{document}

\begin{gather}
\begin{split}
(1-\underset{(0.034)}{0.060}{\textit{B}}^{12}-\underset{(0.032)}{0.123}{\textit{B}}^{2(12)}){y}_t &=(1-\underset{(0.016)}{0.878}\textit{B}^{12})(1-\underset{(0.012)}{0.969}\textit{B}){e}_t\\
\end{split}
\notag\\