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# Lesson 6

Side-Angle-Side Triangle Congruence

## 6.1: Information Overload? (5 minutes)

### Warm-up

Students are familiar with the idea that only some information is needed to uniquely determine a triangle, and this warm-up emphasizes that not every piece of information you can measure about two triangles is needed to prove the triangles are congruent. This activity gives students a chance to read a complete proof and makes a great point of reference as they write their own proofs.

### Launch

Invite students to use highlighters or colored pencils to match the given information to the proof statement.

### Student Facing

Highlight each piece of given information that is used in the proof, and each line in the proof where that piece of information is used.

Given:

• $$\overline {AB} \cong \overline {DE}$$
• $$\overline {AC} \cong \overline {DF}$$
• $$\overline {BC} \cong \overline {EF}$$
• $$\angle A \cong \angle D$$
• $$\angle B \cong \angle E$$
• $$\angle C \cong \angle F$$

Proof:

1. Segments $$AB$$ and $$DE$$ are the same length so they are congruent. Therefore, there is a rigid motion that takes $$AB$$ to $$DE$$.

2. Apply that rigid motion to triangle $$ABC$$. The image of $$A$$ will coincide with $$D$$, and the image of $$B$$ will coincide with $$E$$.

3. We cannot be sure that the image of $$C$$ coincides with $$F$$ yet. If necessary, reflect the image of triangle $$ABC$$ across $$DE$$ to be sure the image of $$C$$, which we will call $$C’$$, is on the same side of $$DE$$ as $$F$$. (This reflection does not change the image of $$A$$ or $$B$$.)

4. We know the image of angle $$A$$ is congruent to angle $$D$$ because rigid motions don’t change the size of angles.

5. $$C’$$ must be on ray $$DF$$ since both $$C’$$ and $$F$$ are on the same side of $$DE$$, and make the same angle with it at $$D$$.

6. Segment $$DC’$$ is the image of $$AC$$ and rigid motions preserve distance, so they must have the same length.

7. We also know $$AC$$ has the same length as $$DF$$. So $$DC’$$ and $$DF$$ must be the same length.

8. Since $$C’$$ and $$F$$ are the same distance along the same ray from $$D$$, they have to be in the same place.

9. We have shown that a rigid motion takes $$A$$ to $$D$$, $$B$$ to $$E$$, and $$C$$ to $$F$$; therefore, triangle $$ABC$$ is congruent to triangle $$DEF$$.

### Student Response

For access, consult one of our IM Certified Partners.

### Activity Synthesis

The goal of this discussion is for students to be clear that not every piece of given information was used in the proof, and so some information wasn’t needed to be sure that the triangles are congruent. This discussion gives another opportunity to connect these ideas:

• Not every piece of information you can measure about two triangles is needed to prove the triangles are congruent.
• Not every piece of information you can measure about one triangle is needed to make an exact copy of that triangle.

## 6.2: Proving the Side-Angle-Side Triangle Congruence Theorem (15 minutes)

### Activity

In this activity, students prove the Side-Angle-Side Triangle Congruence Theorem. They are encouraged to use an example of a proof that they can repurpose for this situation. This puts the focus on looking for and making use of structure (MP7), rather than coming up with the right words to justify that their sequence of transformations works.

Because students are asked to draw their own triangles, some students will use different transformations than other students. Monitor for students whose proofs do and do not require reflection steps.

Making dynamic geometry software available gives students an opportunity to choose appropriate tools strategically (MP5).

### Launch

Representing, Conversing: MLR7 Compare and Connect. Use this routine to help students develop the mathematical language of geometric proofs. After students write their proof that triangle $$LMN$$ is congruent to triangle $$PQR$$, invite them to create a visual display of their proof. Then ask students to quietly circulate and read at least two other visual displays in the room. Give students quiet think time to consider what is the same and what is different about their proofs. Next, ask students to find a partner to discuss what they noticed. Listen for and amplify the language students use to convey the idea that a transformation exists to take a segment to a congruent segment. Also, listen for and amplify the language students use to justify why each of the points in the triangles will coincide after the sequence of rigid motions.
Design Principle(s): Cultivate conversation
Action and Expression: Develop Expression and Communication. To help get students started, display sentence frames such as “I know _____ is congruent to _____ because….”
Supports accessibility for: Language; Organization

### Student Facing

1. Two triangles have 2 pairs of corresponding sides congruent, and the corresponding angles between those sides are congruent. Sketch 2 triangles that fit this description and label them $$LMN$$ and $$PQR$$, so that:
• Segment $$LM$$ is congruent to segment $$PQ$$
• Segment $$LN$$ is congruent to segment $$PR$$
• Angle $$L$$ is congruent to angle $$P$$
2. Use a sequence of rigid motions to take $$LMN$$ onto $$PQR$$. For each step, explain how you know that one or more vertices will line up.
3. Look back at the congruent triangle proofs you’ve read and written. Do you have enough information here to use a proof that is like one you saw earlier? Use one of those proofs to guide you in writing a proof for this situation.

### Student Response

For access, consult one of our IM Certified Partners.

### Student Facing

#### Are you ready for more?

It follows from the Side-Angle-Side Triangle Congruence Theorem that if the lengths of 2 sides of a triangle are known, and the measure of the angle between those 2 sides is known, there can only be one possible length for the third side.

Suppose a triangle has sides of lengths of 5 cm and 12 cm.

1. What is the longest the third side could be? What is the shortest it could be?
2. How long would the third side be if the angle between the two sides measured 90 degrees?

### Student Response

For access, consult one of our IM Certified Partners.

### Anticipated Misconceptions

If students are searching too far back, point students toward the proof in the warm-up activity Information Overload. The goal is for students to understand and adapt that proof to this situation, so help students find the proof relatively quickly so they can have time to engage in productive struggle as they try to understand and adapt it.

### Activity Synthesis

The goal of this discussion is to continue to emphasize that proofs using transformations are generalized statements that work for all triangles that match the given criteria, rather than just one specific drawing.

Select students whose triangles require translation and rotation but not reflection to share their drawings and the steps in their transformations. Record a proof these triangles are congruent.

Then, select a student whose triangles also require reflection. Rather than writing a separate proof for that case, ask students what we could add or change in our existing proof to account for that case. Add an ‘if needed’ reflection step and explain when you might need to reflect and why that reflection will work.

Explain that we now have a general theorem that any triangles in which you know two pairs of corresponding sides are congruent and the corresponding angles between them are congruent must be congruent, and that we don’t have to show the transformations anymore if we can just show that we’ve met the Side-Angle-Side Triangle Congruence Theorem.

Point out the concluding statement. Add this to the display of sentence frames for proofs. This display should be posted in the classroom for the remaining lessons within this unit. You will add to the display throughout the unit. An example template is provided in the blackline masters for this lesson.

Conclusion Statement:

• We have shown that a rigid motion takes $$\underline{\hspace{1in}}$$ to $$\underline{\hspace{1in}}$$, $$\underline{\hspace{1in}}$$ to $$\underline{\hspace{1in}}$$, and $$\underline{\hspace{1in}}$$ to $$\underline{\hspace{1in}}$$; therefore, triangle $$\underline{\hspace{1in}}$$ is congruent to triangle $$\underline{\hspace{1in}}$$.

Ask students to add the Side-Angle-Side Triangle Congruence Theorem to their reference charts as you ceremoniously add it to the class reference chart:

Side-Angle-Side Triangle Congruence Theorem: In two triangles, if two pairs of congruent corresponding sides and the pair of corresponding angles between the sides are congruent, then the two triangles are congruent. (Theorem)

## 6.3: What Do We Know For Sure About Isosceles Triangles? (15 minutes)

### Activity

A major component of looking for and making use of structure (MP7) in geometry is drawing and defining auxiliary lines. In this lesson, a dialogue introduces students to the thinking that mathematicians do when they add lines to figures. In this proof, the line of symmetry is defined as the bisector of angle $$P$$. Because the line of symmetry of an isosceles triangle has so many properties, students may want to define it as the perpendicular bisector of $$AB$$, or as the line from $$P$$ to the midpoint of $$AB$$, which are all the same auxiliary line. The case of the perpendicular bisector will appear in the cool-down, and students haven‘t proven a necessary property of perpendicular bisectors yet. Students don‘t have the Side-Side-Side Triangle Congruence Theorem, so the midpoint option won‘t result in a complete proof, either.

As students work, listen for students who have chosen those different definitions, and how they think about using the Side-Angle-Side Triangle Congruence Theorem in each case.

### Launch

Display an isosceles triangle with base angles $$A$$ and $$B$$, and other vertex $$P$$.

Invite two students to act out this dialogue:

Kiran: I’m stumped on this proof.

Mai: What are you trying to prove?

Kiran: I’m trying to prove that in an isosceles triangle, the two base angles are congruent. So in this case, that angle $$A$$ is congruent to angle $$B$$.

Mai: Let’s think of what geometry ideas we already know are true.

Kiran: We know if two pairs of corresponding sides and the corresponding angles between the sides are congruent, then the triangles must be congruent.

Mai: Yes, and we also know that we can use reflections, rotations, and translations to prove congruence and symmetry . . . . The isosceles triangle you’ve drawn makes me think of symmetry. If you draw a line down the middle of it, I wonder if that could help us prove that the angles are the same? [Mai draws the line of symmetry of the triangle and labels the intersection of AB and the line of symmetry Q].

Kiran: Wait, when you draw the line, it breaks the triangle into two smaller triangles. I wonder if I could prove those triangles are congruent using Side-Angle-Side Triangle Congruence?

Mai: It’s an isosceles triangle, so we know that one pair of corresponding sides is congruent. [Mai marks the congruent sides].

Kiran: And this segment in the middle here is part of both triangles, so it has to be the same length for both. Look. [Kiran draws the two halves of the isosceles triangle and marks the shared sides as congruent].

Mai: So we have two pairs of corresponding sides that are congruent. How do we know the angles between them are congruent?

Kiran: I’m not sure. Maybe it has to do with how we drew that line of symmetry?

Tell students: Mathematicians call additional lines auxiliary lines, because auxiliary means “providing additional help or support.” Ask students what properties Mai might have used to draw the line of symmetry. (It’s perpendicular to the base. It bisects the base. It bisects angle $$P$$.) If no student mentions it, draw their attention to the bisected angle. If students want to use a different property, invite them to attempt to write their own proof rather than fill in the blanks (there are several ways to get close to a proof but then get stuck since they have limited theorems available to prove something).

Representation: Provide Access for Perception. Display or provide students with a physical copy of the acted out dialogue. Check for understanding by inviting students to rephrase dialogue in their own words. Consider keeping the display of dialogue visible throughout the activity.
Supports accessibility for: Language; Memory

### Student Facing

Mai and Kiran want to prove that in an isosceles triangle, the 2 base angles are congruent. Finish the proof that they started. Draw the auxiliary line and define it so that you can use the Side-Angle-Side Triangle Congruence Theorem to complete each statement in the proof.

Draw $$\underline{\hspace{1in}}$$.

Segment $$PA$$ is congruent to segment $$PB$$ because of the definition of isosceles triangle.

Angle $$\underline{\hspace{1in}}$$ is congruent to angle $$\underline{\hspace{1in}}$$ because $$\underline{\hspace{1in}}$$.

Segment $$PQ$$ is congruent to itself.

Therefore, triangle $$APQ$$ is congruent to triangle $$BPQ$$ by the Side-Angle-Side Triangle Congruence Theorem.

Therefore, $$\underline{\hspace{1in}}$$.

### Student Response

For access, consult one of our IM Certified Partners.

### Activity Synthesis

The goal of this discussion is to have students explain why, in order to use the Side-Angle-Side Triangle Congruence, it was helpful to define the auxiliary line as an angle bisector.

• Why did Kiran and Mai use the angle bisector as their auxiliary line? (So they would have congruent angles at point $$P$$ between two congruent sides.)
• Could they have used a different definition for the auxiliary line? (The perpendicular bisector of $$AB$$ almost works, but we would have to prove that point $$P$$ is on that line before we could use the Side-Angle-Side Triangle Congruence Theorem.)

Ask students to add the Isosceles Triangle Theorem to their reference charts as you add it to the class reference chart:

Isosceles Triangle Theorem: In an isosceles triangle, the base angles are congruent. (Theorem)

Speaking: MLR8 Discussion Supports. To help students respond to the questions for discussion, provide sentence frames such as: “Kiran and Mai used the angle bisector as their auxiliary line because _____” and “Kiran and Mai could (or could not have) used a different definition for the auxiliary line because _____.” As students share their responses, press for details by asking students why defining the auxiliary line as an angle bisector is essential in order to apply the Side-Angle-Side Triangle Congruence Theorem.
Design Principle(s): Support sense-making; Optimize output (for explanation)

## Lesson Synthesis

### Lesson Synthesis

Display one problem at a time. Give students quiet think time for each problem and ask them to give a signal when they have an answer and a strategy. Keep all problems displayed throughout the talk. Follow with a whole-class discussion. The purpose of this Math Talk is to elicit strategies and understandings students have related to adding auxiliary lines, and to do some auxiliary line error analysis.

“What auxiliary line would you add to the diagram to help you use the Side-Angle-Side Triangle Congruence Theorem to prove that, in a square, the diagonals form $$45^{\circ}$$ angles with the sides?” (Draw diagonal $$HF$$. Then you can use the Side-Angle-Side Triangle Congruence Theorem because all the sides are equal and all the right angles are equal. Then because the two triangles are congruent, the angles formed on each side of the diagonal must be equal, so they must be half of $$90^{\circ}$$, which is $$45^{\circ}$$.)

“What auxiliary line would you add to the diagram to help you use the Side-Angle-Side Triangle Congruence Theorem to prove that, in a quadrilateral with both pairs of opposite sides congruent and one pair of opposite angles congruent, opposite sides are parallel?” (Draw segment $$BC$$, creating two triangles $$ABC$$ and $$DCB$$ that are congruent by the Side-Angle-Side Triangle Congruence Theorem. $$BC$$ is also a transversal for lines $$BA$$ and $$DC$$, and the congruent triangles make congruent alternate interior angles, so the lines are parallel.)

“This is an example of how not to draw an auxiliary line.
Conjecture: Given any two points on a circle and any point outside the circle, the two points on the circle are the same distance to the exterior point. Tyler says, “I know that radii of a circle are congruent, and $$LI$$ is congruent to itself. So then I drew the angle bisector from $$I$$ through $$L$$ and now I can use the Side-Angle-Side Triangle Congruence Theorem to prove that triangles $$ILK$$ and $$ILJ$$ are congruent.” Can that really be true? If not, what went wrong?” (Tyler says to draw the angle bisector from $$I$$ through $$L$$, but how does he know that angle bisector of angle $$JIK$$ will go through $$L$$?)

## 6.4: Cool-down - Revisiting Perpendicular Bisectors (5 minutes)

### Cool-Down

For access, consult one of our IM Certified Partners.

## Student Lesson Summary

### Student Facing

If all pairs of corresponding sides and angles in 2 triangles are congruent, then it is possible to find a rigid transformation that takes corresponding vertices onto one another. This proves that if 2 triangles have all pairs of corresponding sides and angles congruent, then the triangles must be congruent. But, justifying that the vertices must line up does not require knowing all the pairs of corresponding sides and angles are congruent. We can justify that the triangles must be congruent if all we know is that 2 pairs of corresponding sides and the pair of corresponding angles between the sides are congruent. This is called the Side-Angle-Side Triangle Congruence Theorem.

To find out if 2 triangles, or 2 parts of triangles, are congruent, see if the given information or the diagram indicates that 2 pairs of corresponding sides and the pair of corresponding angles between the sides are congruent. If that is the case, we don’t need to show and justify all the transformations that take one triangle onto the other triangle. Instead, we can explain how we know the pairs of corresponding sides and angles are congruent and say that the 2 triangles must be congruent because of the Side-Angle-Side Triangle Congruence Theorem.

Sometimes, to find congruent triangles, we may need to add more lines to the diagram. We can decide what properties those lines have based on how we construct the lines (An angle bisector? A perpendicular bisector? A line connecting 2 given points?). Mathematicians call these additional lines auxiliary lines because auxiliary means “providing additional help or support.” These are lines that give us extra help in seeing hidden triangle structures.


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## anonymous one year ago Calculate the number of moles of helium that contains 3.25 x 10^21 atoms.

1. anonymous

Without numbers, can you figure this out. Always start with given value $\frac{ atoms}{ 1}\times \frac{ mol }{ atoms }$

2. anonymous

3.25 x 10^ 21 atom x (1mol/6.022 x 10 23 atom)

3. anonymous

Right?

4. anonymous

$3.25 x 10^{21} atom x \frac{ 1 mol }{ 6.022 x 10^{23} atom }$

5. anonymous

Yes, that is right. Sorry I was helping someone.

6. anonymous

And I get .539687811

7. anonymous

Units would be in moles of helium.

8. anonymous

So .539687811 mol He?

9. anonymous

It has to be 3 sf since the given value only has 3. Remember that your answer has a 0 in the front. It does not count as a sf.

10. anonymous

.540 mol He?

11. anonymous

Yes Nice rounding!

12. anonymous

Thanks

13. anonymous

No problem!

14. anonymous

It's saying the answer is wrong. I did it right...Right?

15. anonymous

I did not do the calculations since you got the right conversion. You did the calculation. Do the calculation again.

16. anonymous

I think you put 3.25*10^23 instead of 3.25*10^21

17. anonymous

I did it again and got the same answer.

18. anonymous

Write down what you calculate then.

19. anonymous

$3.25\times10^{21} atomtimes \frac{ 1mol }{ 6.022 \times10^{23 } }atom$

20. anonymous

3.25 x 10^21 atom x

21. anonymous

Dont forget parentheses. (3.25*10^21)/(6.022*10^23)

22. anonymous

Right

23. anonymous

Got .00540 mol He

24. anonymous

So what did you do wrong earlier?

25. anonymous

Forgot parentheses or wrong exponent?

26. anonymous

I put it in the calculator without parentheses

27. anonymous

Another way is multiply 3.25*10^21 and lets called that x then multiply 6.023*10^23 and call that y. Then just do x/y if you arent comfortable with parentheses.

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# For what values of $\alpha$ does this integral converge?

For what real values of $\alpha$ does this integral converge?

$$\int_0^\infty {x^\alpha e^{-x} dx}$$

And what's the usual approach to this kind of problem?

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en.wikipedia.org/wiki/Gamma_function –  Jack Nov 25 '12 at 21:18
@Jack: But my textbook says it's only convergent when $\alpha>-1$, which differs with what Wikipedia says ($\alpha\not\in\mathbb{Z}_{(-\infty,-1]}$). –  Voldemort Nov 25 '12 at 21:27

## 3 Answers

Note: $$\frac{1}{e} \int_0^1 {x^\alpha dx}<\int_0^1 {x^\alpha e^{-x} dx} <\int_0^1{x^\alpha dx}$$ Thus, $\int_0^1 {x^\alpha e^{-x} dx}$ converges iff $\int_0^1{x^\alpha dx}$ does which happens for $\alpha>-1$. $\int_1^\infty {x^\alpha e^{-x}dx} < \int_1^\infty {x^\alpha ( x^N/N!)^{-1} dx}$ (where $N>\alpha +2$) does always converge.

-
+1. Neat. ${}{}$ –  user17762 Nov 25 '12 at 21:34
But why in the Gamma Function article wikipedia says that it can also be defined at all points except $\mathbb{Z}_{(-\infty,-1]}$? –  Voldemort Nov 25 '12 at 21:37
Wikipedia says that you can continue the function defined by this to an analytical function defined everywhere except on the negative integers but not that the integral itself converges everywhere. –  Dominik Nov 25 '12 at 21:41

Note that for $\alpha > -1$, there exists $X(\alpha)$ such that $x^{\alpha} e^{-x} \leq e^{-x/2}$ for all $x > X(\alpha)$.

Hence, $$\int_0^{\infty} x^{\alpha} e^{-x} dx \leq \int_0^{X(\alpha)} x^{\alpha} e^{-x} dx + \int_{X(\alpha)}^{\infty} e^{-x/2} dx = \underbrace{\int_0^{X(\alpha)} x^{\alpha} e^{-x} dx}_{< \infty} + 2e^{-X(\alpha)/2}$$

For $\alpha \leq -1$, the integral diverges, which can be shown as follows. Let $\alpha = -1-\beta$ $$\int_0^{\infty} x^{\alpha} e^{-x} dx = \underbrace{\int_0^{\epsilon} \dfrac{e^{-x}}{x^{1+\beta}} dx}_{I} + \underbrace{\int_{\epsilon}^{\infty} \dfrac{e^{-x}}{x^{1+\beta}} dx}_{II}$$ Note that $$II \leq \displaystyle \int_{\epsilon}^{\infty} \dfrac1{x^{1+\beta}} dx < \infty$$ Note that $$I \geq \int_0^{\epsilon} \dfrac{e^{-\epsilon}}{x^{1+\beta}} dx = \infty$$ Hence, the integral converges for $\alpha > 1$ and diverges for $\alpha \leq -1$.

The integral you are interested in, wherever it converges, is called the Gamma function. The Gamma function is defined for $\alpha \leq -1$ as well through analytic continuation of the above integral. However, the Gamma function still has poles (blows to $\infty$) at negative integers.

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I would split the interval into $[0,1] \cup (1,\infty)$.

In $(1,\infty)$, $e^x$ dominates $x^{\alpha}$ as $x \to \infty$, so $\int_1^\infty {x^\alpha e^{-x} dx}$ is convergent for all $\alpha$. (To make this precise: given $\alpha \in \mathbb R$, we can find $x_0$ such that $x^{\alpha} < e^{x/2}$ for all $x \ge x_0$. So $0 \le x^{\alpha}e^{-x} \le e^{-x/2}$ for all $x \ge x_0$, and the integral converges because the integral of $e^{-x/2}$ converges.)

And in $[0,1]$, $0.3 < e^{-x} \le 1$, so the convergence of the integral is determined only by the convergence of $\int_0^1 {x^\alpha dx}$, which you can work out for yourself.

-


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# Open statistical ensemble

Open statistical ensemble

Open statistical ensemble (OSE) corresponds to a physical system, which exchanges energy with the environment, being with it in thermal equilibrium, exchanges particles with the environment for a given chemical potential and correctly takes into account the surface terms.

The expression for the partition function of the ensemble is similar to the partition function of the grand canonical ensemble (GCE), with the replacement of the Boltzmann factor in terms of the series on the correlation functions of special form.

The coefficient of surface tension, included in the partition function corresponds to the interface of fluid and hard solid, due to the strict respect of probability and potential limitations.

Unlike the GCE, for OSE average number of particles in a given volume is strictly coincides with the volume term. Also in contrast to the GCE, the correlation functions of the OSE strictly satisfy the requirement of translational invariance.

The expression for the general term of the distribution has the form

$p^v_m = \frac{z^m}{m! \Upsilon_v} \sum_{t=0}^\infty \frac{z^t}{t!} \int \left [ \prod_{i = 1}^m \prod_{j = m+1}^{m+t}\psi^v_i\chi^v_j \right ] {\mathcal B}^{(m,t)}_{1...m+t} d\boldsymbol{r}_1...d\boldsymbol{r}_{m+t},$

where $p^v_m$ is the probability to find $~m$ particles in volume $~v$, $~m \geq 1$, $~z$ is activity, $~\Upsilon_v$ - OSE partition function. $~\psi^v_i$ and $~\chi^v_j = 1 - \psi^v_j$ - characteristic functions, equal to unity inside and outside the system, respectively, and $~{\mathcal B}^{(m,t)}_{1...m+t}$ - partial localization factors, generalize the notions of the Boltzmann and Ursell factors and contain them as extreme cases.

The first term of the series corresponds to the distribution of the GCE with the accuracy up to normalizing factors - partition functions.

Summing the series we obtain the expression

$p^v_m = \frac{1}{m! \Upsilon_v} \int \left [ \prod_{i = 1}^m \psi^v_i \right ] \varrho^{(m)}_{G,1...m}(\chi^v) d\boldsymbol{r}_1...d\boldsymbol{r}_m,$

where $~\varrho^{(m)}_{G,1...m}$ - correlation function, depending on $~z$, $~\chi^v$ and expressed through a series of $~{\mathcal B}^{(m,t)}_{1...m+t}$. The last expression is equivalent to the GCE distribution with the replacement

$\varrho^{(m)}_{G,1...m} \approx z^m \exp(-\beta U^{m}_{1...m})$

and corresponding renormalization, where $~U^{m}_{1...m}$ - $~m$-particle interaction potential, $~1/\beta = k_B T$, $~k_B$ - Boltzmann constant, $~T$ - temperature. This expression shows that the GCE is a low-density approximation of the OSE.

For the partition function of the OSE, we have the expression

$\Upsilon_v = \exp { \sum_{t=1}^\infty \frac{z^t}{t!} \int \left [1 - \prod_{i = 1}^{t} \chi^v_i \right ] {\mathcal U}^{(t)}_{1...t} d\boldsymbol{r}_1...d\boldsymbol{r}_t },$

unlike the partition function of GCE

$\Xi_v = \exp { \sum_{t=1}^\infty \frac{z^t}{t!} \int \left [ \prod_{i = 1}^{t} \psi^v_i \right ] {\mathcal U}^{(t)}_{1...t} d\boldsymbol{r}_1...d\boldsymbol{r}_t },$

where $~{\mathcal U}^{(t)}_{1...t}$ - Ursell factors. Collapsed series of activities for $~\Upsilon_v$ we obtain the alternative representation

$~\Upsilon_v = \exp{\beta [ vP(z,T) + a\sigma (z,T) ]},$

where $~P(z,T)$ is the pressure, $~\sigma (z,T)$ - coefficient of surface tension on the interface of the fluid and hard solid, $~a$ - the surface bounding the system.

It should be stressed that an open system does not singled out, and the surface tension is created due to the fluctuation component of the partition function.

The last expression is exactly consistent with the probability of the formation of the hole volume $~v$ in the fluid

$~p^v_0 = \exp-{\beta [ vP(z,T) + a\sigma (z,T) ]},$

is determined from thermodynamic considerations

$~p^v_0 \propto \exp{( -\beta R_{min})},$

where $~R_{min}$ - minimum work of formation of such fluctuations.

## Some properties of the OSE

• Scale invariance. In contrast to grand canonical ensemble, an open statistical ensemble satisfies the scale invariance requirement: general term of the included subsystem distribution corresponds to that of the original system.
• Application to small systems. This distribution may be applied to however small volumes including those less than a molecule size. In this case it degenerates into a Bernoulli distribution with $~p=\varrho v$.
• Separation of fluctuations. When volume is much greater than the size of the molecule squared deviation of the number of particles is divided into bulk and surface terms.

## References

• Zaskulnikov V. M., Open statistical ensemble and surface phenomena: arXiv:0911.3106
• Zaskulnikov V. M., Open statistical ensemble: new properties (scale invariance, application to small systems, meaning of surface particles, etc.): arXiv:1004.0896v1

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• Entropy — This article is about entropy in thermodynamics. For entropy in information theory, see Entropy (information theory). For a comparison of entropy in information theory with entropy in thermodynamics, see Entropy in thermodynamics and information… …   Wikipedia

• Density matrix — Mixed state redirects here. For the psychiatric condition, see Mixed state (psychiatry). In quantum mechanics, a density matrix is a self adjoint (or Hermitian) positive semidefinite matrix (possibly infinite dimensional) of trace one, that… …   Wikipedia