We provide a further algebraic proof that the lines of entrapment conditions for inclusions calculated with the formula of Guiraud and Powell (2006) are not thermodynamic isomekes and therefore do not represent exactly lines of possible entrapment conditions.Isomekes, inclusions, EosFit-Pinc, thermodynamicsZhong et al. (2020) argue that the solution for entrapment conditions of inclusions presented by Guiraud and Powell (2006) is as equally valid as that proposed by Angel et al. (2017). In addition they propose a third calculation route based on the logarithmic definition of strain and a change in reference conditions. For the example that Zhong et al. (2020) show in their Figure 1, all three methods yield calculated entrapment pressures at metamorphic temperatures that differ by a small amount that is larger than numerical rounding error in the calculations. However, if the three solutions yield three different sets of entrapment pressures for a given temperature and the same input parameters, then the three different results cannot all be simultaneously thermodynamically correct. The essential question is therefore: which one is thermodynamically correct?The thermodynamic concept behind the determination of conditions of entrapment of inclusions within host minerals is the isomeke (Rosenfeld and Chase 1961; Adams et al. 1975). The isomeke is a thermodynamically defined line along which the fractional volume changes of two phases (host and inclusion) remain equal but non-zero. Therefore, if two P-T points, a and b, lie on the same isomeke of the host h and inclusion i their volumes at these two points are related exactly by:The consequence, for host-inclusion calculations, is that an inclusion that perfectly fits the hole it occupies in a host crystal at the time of entrapment will continue to do so at all P-T conditions along the isomeke line through entrapment conditions, without developing additional stresses, regardless of the slope of the isomeke. The inclusion pressure therefore remains equal to the external pressure along this entrapment isomeke (e.g., Angel et al. 2015). And inclusions trapped at different points along the same host-inclusion isomeke will exhibit the same residual pressure as one another when examined at room conditions (or any other P-T point away from the entrapment isomeke). The consequence is that unique entrapment conditions cannot be inferred from the measurement of the residual pressure in an inclusion alone under the assumption that the phases are isotropic; only the entrapment isomeke (a line in P-T space) can be inferred.Therefore, one can determine which model or models for inclusion entrapment pressures are valid by testing whether or not they predict entrapment conditions that all lie on a single isomeke from a single value of Pinc. We first apply this condition to the equation of Guiraud and Powell (2006) to prove that it does not represent an isomeke. Then we confirm the algebraic analysis by simply calculating from the individual EoS of the phases the ratios in Equation 1 along the predicted line of entrapment.The equation given by Guiraud and Powell (2006) that relates the Ptrap and Ttrap to Pinc at room temperature when the host pressure is zero (and vice versa), is:We now multiply the first term on the right by the ratio of volumes of the inclusion at one point on the isomeke at room temperatureand the second by the same volume ratio (also of unity) for the host. Then the Guiraud and Powell (2006)Equation 2 becomes:The second line in Equation 3 is just labels that we have assigned to the corresponding terms in the full equation to make the following explanation clearer. The condition that Pfoot,Tend and Ptrap,Ttrap both lie on the same isomeke is, from Equation 1, the condition [B] = [D].If we have measured Pinc at room temperature and we know Gh, we can solve Equation 3 for Ptrap= Pfoot and Ttrap = Tend to define one point on the entrapment isomeke, at Pfoot,Tend. At these conditions, Equation 3 is algebraically identical to the solution given in Angel et al. (2017), and the terms [A], [C], and [E] are now fixed. With these terms fixed as defining one point on the entrapment isomeke, there are no other solutions to Equation 3 unless [B] ≠ [D], which implies a violation of the isomeke condition of Equation 1 which in turn shows that Guiraud and Powell (2006) equation cannot represent an isomeke. The same conclusion can be reached another way: if [B] = [D] then we can write Equation 3 as [A] = [B]([C] – [E]). Since[A], [C], and [E] are fixed, this tells us there is only one value of [B] that simultaneously satisfies both the isomeke condition and the Guiraud and Powell (2006) equation; again a demonstration that the Guiraud and Powell (2006) equation cannot represent an isomeke.Figure 1c of Zhong et al. (2020) shows as an example the calculated entrapment conditions for a quartz inclusion in almandine garnet with a residual inclusion pressure of 0.6 GPa when the host garnet is at room conditions. Table 1a shows entrapment pressures calculated with the Guiraud and Powell (2006) and Angel et al. (2017) models, at room T and 750 °C. The table shows that the fractional volume changes of the host and inclusion change along the entrapment line calculated with the Guiraud and Powell (2006) model, confirming that it is not an isomeke. The differences are larger when the contrast between the bulk moduli of the host and inclusion is smaller, for example for zircon in pyrope (Table 1b). While the differences in these two examples may not be geologically significant, the errors for other host-inclusion pairs cannot be predicted without explicit calculation.The proof can also be performed with the same conclusion but with more complex working for P-T points other than Pfoot,Tend. This shows that we previously erred in stating that the relaxation has to be calculated isothermally; this should have been obvious from the fact that the solution of Pinc was presented in Angel et al. (2017) in terms of force balance at the final conditions. Therefore, we agree with Zhong et al. (2020) that this is not the cause of the difference between the Guiraud and Powell (2006) and Angel et al. (2017) solutions that they show in their Figure 1. However, we also want to note that, contrary to the statements in Zhong et al. (2020), there is no restriction in this analysis or that of Angel et al. (2017) to the final conditions being at room conditions.We have proved that the lines of entrapment conditions calculated by the method of Guiraud and Powell (2006) are not thermodynamically correct because they are not exactly isomekes. The source of this difference was shown in Equations A6 and A9 of Angel et al. (2017) and is the factor that we here callthe fractional volume change of the two phases along the entrapment isomeke. When this factor is close to unity, the solution of Guiraud and Powell (2006) is a close approximation to the thermodynamically correct solution. The same is true for the logarithmic basis proposed as an alternative by Zhong et al. (2020). Factors that further reduce the accuracy of the Guiraud and Powell (2006) approximation include hosts with small shear moduli (Eq. 2) and systems that have a high contrast in values of the pressure derivatives of their bulk moduli (i.e., K′) leading to strongly curved isomekes, such as often occur in mixed-phase inclusions (e.g., Musiyachenko et al. 2020). Furthermore, the differences in the calculated entrapment conditions between the approach of Guiraud and Powell (2006) and the thermodynamic isomekes are further magnified when unique entrapment conditions are being inferred from the anisotropy of strains of the trapped inclusions (e.g., Alvaro et al. 2020). On the other hand, the approach of Angel et al. (2017) is thermodynamically correct for all cases because it calculates points on the entrapment isomeke from Pfoot,Tend by explicitly enforcing the isomeke condition. We therefore recommend that all host-inclusion calculations are based on the thermodynamically correct basis of the isomeke, regardless of the value of the final Ptrap.Our current research on host-inclusion systems is supported by the European Research Council under the European Union’s Horizon 2020 research and innovation program grant agreement 714936 TRUE DEPTHS to Matteo Alvaro at the University of Pavia. The work reported in the paper under discussion was supported by ERC starting grant 307322 to Fabrizio Nestola, and by the MIUR-SIR grant “MILE DEEp” (RBSI140351) to Matteo Alvaro.

中文翻译：

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“ EosFit-Pinc：用于包含主体的弹性热压法的简单GUI” —回复Zhong等。

我们提供了进一步的代数证明，证明使用Guiraud和Powell（2006）的公式计算的夹杂物的夹杂条件线不是热力学等温线，因此不能精确地表示可能的夹杂条件线。等。（2020年）认为，Guiraud和Powell（2006年）提出的包裹体包裹条件的解决方案与Angel等人提出的解决方案同样有效。（2017）。此外，他们根据应变的对数定义和参考条件的变化提出了第三种计算途径。对于钟等人的例子。（2020）在他们的图1中显示，这三种方法在变质温度下产生的计算出的包封压力相差很小，比计算中的数值舍入误差大。但是，如果对于给定的温度和相同的输入参数，这三种溶液产生三组不同的包封压力，则这三种不同的结果不可能同时在热力学上都正确。因此，基本问题是：哪个在热力学上是正确的？确定包裹体中夹杂物的夹带条件的热力学概念是同位异构体（Rosenfeld and Chase 1961; Adams et al。1975）。等能量线是热力学定义的一条线，沿着该线，两相（主体和夹杂物）的体积分数变化保持相等，但不为零。因此，如果两个PT点a和b，位于主体h的同一个等值线上，并且夹杂物i在这两个点上的体积确切地由以下方式关联：对于主体夹杂物的计算结果是，一个夹杂物恰好适合当时它在主体晶体中占据的孔沿等面线的所有PT条件，通过等量线的夹带将继续如此，而不会产生额外的应力，而与等面线的斜率无关。因此，夹杂物压力保持等于沿该夹带等当点的外部压力（例如，Angel等人，2015）。当在室温条件下（或远离夹杂同位素的任何其他PT点）进行检查时，沿着同一主体-夹杂物同位素在不同点处捕获的夹杂物将表现出彼此相同的残余压力。结果是，在相为各向同性的假设下，无法仅通过对夹杂物中的残余压力进行测量就可以得出独特的截留条件。只能推断出夹杂物等深线（在PT空间中的一条线），因此，可以通过测试它们是否可以从单个值预测所有同一个同形异物上的夹杂条件来确定哪个模型或哪些夹杂物夹杂物压力模型有效的。我们首先将此条件应用于Guiraud和Powell（2006）的方程式，以证明它不代表等值线。然后，我们可以通过简单地从各个相的EoS计算沿预测的夹带线的等式1中的比率，从而确定代数分析。Guiraud和Powell（2006）给出的等式在室温下，当主压力为零时将Ptrap和Ttrap与Pinc关联（反之亦然）为：我们现在将右边的第一项乘以在室温下，同构异构体上某一点的夹杂物，对于宿主，第二个夹杂物具有相同的体积比（也等于1）。然后Guiraud和Powell（2006）的等式2变为：等式3中的第二行只是我们分配给完整等式中相应项的标签，以使以下解释更加清楚。根据公式1，Pfoot，Tend和Ptrap，Ttrap都位于同一个等当上的条件是条件[B] = [D]。如果我们在室温下测量Pinc并知道Gh，则可以求解公式3对于Ptrap = Pfoot和Ttrap = Tend定义陷害同位点上的一个点，趋向于Pfoot。在这些条件下，等式3在代数上与Angel等人给出的解相同。（2017），并且[A]，[C]和[E]这个词现在已固定。固定这些术语以定义诱捕同位点上的一个点后，方程[3]没有其他解决方案，除非[B]≠[D]，这意味着违反了方程1的同位条件，这反过来表明Guiraud和Powell（ 2006年）方程不能代表等值线。可以通过另一种方式得出相同的结论：如果[B] = [D]，则可以将等式3写为[A] = [B]（[C] – [E]）。因为[A]，[C]和[E]是固定的，所以这告诉我们只有[B]的一个值同时满足等当面条件和Guiraud and Powell（2006）方程；再次证明了Guiraud和Powell（2006）方程不能代表等值线。Zhong等人的图1c。（2020）举例说明了当主石榴石在室温下时，铝金刚石石榴石中石英夹杂物的计算截留条件，残留夹杂物压力为0.6 GPa。表1a显示了由Guiraud和Powell（2006）和Angel等人计算得出的截留压力。（2017）模型，在室温T和750°C下。该表显示，使用Guiraud和Powell（2006）模型计算的宿主的体积分数变化和夹杂物沿截留线的变化，证实这不是同位异构。当主体和夹杂物的体积模量之间的对比度较小时，例如锆石中的锆石（表1b），则差异较大。尽管这两个示例的差异在地质上可能并不重要，如果没有显式计算，则无法预测其他主体-包含对的错误。也可以以相同的结论进行证明，但对Pfoot，Tend以外的PT点进行更复杂的工作。这表明我们先前错误地指出松弛必须等温地计算。从Pinc的解决方案已在Angel等人的论文中发现这一点应该是显而易见的。（2017）在最终条件下的兵力平衡方面。因此，我们同意钟等人的观点。（2020年），这不是Guiraud和Powell（2006年）与Angel等人之间差异的原因。（2017）在图1中显示的解决方案。但是，我们还想指出的是，与Zhong等人的说法相反。（2020年），这种分析或Angel等人的分析都没有限制。（2017）的最终条件是在室温下。我们已经证明，根据Guiraud和Powell（2006）的方法计算得出的夹带条件线在热力学上不是正确的，因为它们不是完全相同的。Angel等人的方程A6和A9中显示了这种差异的来源。（2017），这就是我们在此称为夹带同位异构体的两相体积分数变化的因素。当这个因素接近于统一时，Guiraud和Powell（2006）的解非常接近于热力学正确的解。Zhong等人提出作为对数基础的对数基础也是如此。（2020）。进一步降低Guiraud和Powell（2006）逼近精度的因素包括剪切模量较小的主体（Eq。2）和系统中，其体积模量（K'）的压力导数的值具有很高的对比度，导致强烈弯曲的等值线，例如经常出现在混合相夹杂物中（例如Musiyachenko等人2020）。此外，当根据被困夹杂物的应变各向异性来推导独特的夹带条件时，Guiraud和Powell（2006）的方法与热力学等当之间的计算夹带条件之间的差异将进一步放大（例如，Alvaro等人，2020年）。 ）。另一方面，Angel等人的方法。（2017）在所有情况下在热力学上都是正确的，因为它通过明确实施等温条件从Pfoot，Tend计算了夹带等温点。因此，我们建议所有宿主夹杂物的计算均基于同位异构体的热力学正确基础，无论最终诱集装置的价值如何。我们目前对宿主夹杂物的研究得到了欧盟Horizo​​n 2020的欧洲研究委员会的支持研究和创新计划向Pavia大学的Matteo Alvaro授予了714936 TRUE DEPTHS协议。正在讨论中的论文中报告的工作得到了Fabrizio Nestola的ER322起始拨款307322的支持以及MIUR-SIR的Matteo Alvaro的“ MILE DEEp”（RBSI140351）的资助。欧洲研究委员会根据欧盟Horizo​​n 2020研究与创新计划授予协议714936 TRUE DEPTHS授予Pavia大学的Matteo Alvaro，以支持我们目前对宿主包含系统的研究。正在讨论中的论文中报告的工作得到了Fabrizio Nestola的ER322起始拨款307322的支持以及MIUR-SIR的Matteo Alvaro的“ MILE DEEp”（RBSI140351）的资助。欧洲研究委员会根据欧盟Horizo​​n 2020研究与创新计划授予协议714936 TRUE DEPTHS授予Pavia大学的Matteo Alvaro，以支持我们目前对宿主包含系统的研究。正在讨论中的论文中报告的工作得到了Fabrizio Nestola的ER322起始拨款307322的支持以及MIUR-SIR的Matteo Alvaro的“ MILE DEEp”（RBSI140351）的资助。