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User:Eml4500.f08.team.foskey.ckf/hw6

MATLAB ASSIGNMENT

Electric Pylon Problem Overview
This assignment required the use of a previously created MATLAB file called electric_pylon.m. This code can be seen below. This code creates a two dimensional truss system that resembles an electric pylon. Although the code for the truss system is complex, the code only plots the truss using defined nodal coordinates.

The first part of the assignment involved scaling the height of the truss to a max height of 60m, applying a 1000N load, and calculating the stresses in the elements. The elements under the highest tensile and compressive stresses are identified. The setup of the problem is seen below. In the figure, the truss plot was created using the provided code, and the applied load P was added manually.

The second part of the assignment required the masses of the elements to be accounted for. The half of the mass of each element was lumped to each end. The eigenvalues and eigenvectors were found, and the lowest three were plotted. Also the lowest three vibrational periods were found.

Provided MATLAB Code
The following MATLAB code was provided.

An Explanation Of The Provided Code
Essentially, all the provided code does is plot the truss system. The code is long due to the fact that there are over 90 elements that make up the truss. However, the code is relatively simple.

The code begins by declaring there to be 46 global nodes, and 91 elements. Then is calculates the total number of degrees of freedom. This is accomplished by multiplying the number of nodes by two, because there are two degrees of freedom (horizontal and vertical) for each node.

Next, the code defines the global nodal locations into a position matrix using x-y-z coordinates. For example, the line position(:, 1) = [ 1.5; 0; 0]; gives the location of node 1 as being at the coordinate (1.5, 0, 0). This is repeated for all 46 global nodes.

Then, the code organizes the locations of the nodes into an easier to understand (and use) form. This is done by looping through all of the nodes defining x, y, and z variables for each node. For example, the line y(i) = position(2,i); will define the y coordinate of global node "i" as the second value of the position matrix (which is the y coordinate) in row "i" (which represents node "i"). Again, this is done for all 46 global nodes.

Once the nodal coordinates are defined, the code then assigns a node number to each end of each element. This is done by using a line of code such as node_connect(1, 89) = 39;. In this example line of code, the local node 1 of element 89 is defined as being equal to global node 39. Two lines of code correspond to each element - one line for each local node. This is done for all 91 elements.

Next, the code plots each element. This is done by looping through all of the elements. In each loop, the code obtains the global node numbers for each end of an element, defining these as "node_1" and "node_2". A vector is then defined for each direction going from one node to the other. For example in the line of code xx = [x(node_1),x(node_2)]; , the vector "xx" is defined as a vector going from the x value for node_1 to the x_value of node_2. Once the three vectors ( one each for the x, y and z directions) are defined for an element, they are plotted. This plot will represent one element. The plotting process is repeated for all 91 elements, until the entire truss is plotted.

The provided code ends by titling the plot, and the axis. It should also be noted that the code is designed to be able to plot a three dimensional truss, however all of the z coordinates for the nodes are zero. Therefore the code plots a two dimensional truss.

Additional Functions
The following functions are used in the following codes to solve for certain parts of the truss problem, particularly the loading code.

PlaneTrussElement calculates elemental stiffness matrices NodalSoln calculates the nodal displacements and the reactions of the truss PlaneTrussResults calculates the strain, stress, and axial force in each element

A more complete explanation of these functions can be found here.

PlaneTrussElement.m

NodalSoln.m

PlaneTrussResults.m

The function TransientPlaneTrussElement is used in the lumped mass code. The function creates an elemental stiffness matrix(like the function PlaneTrussElement ), as well as creates a mass matrix for the element. This function was taken from page 568 in the textbook.

TransientPlaneTrussElement.m

MATLAB Code For Loading The Pylon
The following MATLAB code is used to apply a 1000N downward load to the far right tip of the truss. The unloaded and loaded trusses are plotted, and the stresses in the elements are evaluated. The code is commented throughout, and explained further below.

An Explanation Of The Loading Code
The pylon generated by the originally provided code is to be loaded with a 1000N downward force on the right tip. The original, unloaded truss is to be plotted with the deformed truss, and the elements undergoing the highest stresses (tensile and compressive) are to be marked.

The code for loading the pylon can be broken down into three sub-parts: 1) converting the provided code to a usable form for analyzing the truss system, 2) plotting the trusses, and 3) determining the maximum stresses.

In the first part of the loading code (given the file name "solvingpylon1.m") takes the data from the provided code and converts it into variables that will work in a format of solving the truss similar to earlier MATLAB projects (NOTE: see The Two Bar Truss, The Six Bar Truss , and The Three Bar Space Truss) The loading code begins by running the provided code to get all of the data that that code generates. This eliminates the need to retype all of the original code material into a new code. Next, the loading code determines the highest point of the originally created truss and determines a scaling factor. The highest point is determined by setting a dummy variable "keep" equal to zero initially, and then checking the y coordinate value of each node to see if it is greater than the "keep" value. If it is, then that value becomes the new "keep" value, until all nodes have been checked. The scaling factor was found using the form $$ Scaling Factor = \frac{60}{keep} $$. Every node coordinate is then multiplied by the scaling factor, increasing the size of the pylon to a maximum height of 60m.

Next, the loading code is given the parameters of the elements. Each element has a Young's Modulus of 200GPa, and a cross sectional area of 4cm2. The load P is also entered as -1000, because the load is downward (in the negative y direction). The code then creates a 91x2 matrix of the nodal coordinates, with each node represented by a row, the x coordinates in the first column, and the y coordinates in the second row. The total number of displacements is also determined by multiplying the number of nodes by 2, which is the number of degrees of freedom for each node.

The loading program then creates the connectivity array by looping through each element and filling the connectivity array conn with the associated values from the node_connect matrix of the provided code. The location matrix master array lmm is then created. This is accomplished by using the node_connect matrix form the provided code. To fill lmm, one has to realize that each node has two associated degrees of freedom. Using this, the matrix is filled in for elements 1 and 2 like so:

(Note: Only the bold values are actually put into the lmm matrix)

This is accomplished by taking the global node number, decreasing it by one, doubling it, and adding 1 for the horizontal dof, or 2 for the vertical. For example, in element 1, the global nodes associated with it are 1 and 3, and the global dofs are 1, 2, 5 and 6. Looking at the code:

d1= 2*(node_connect(1,i)-1)+1; d2= 2*(node_connect(1,i)-1)+2; d3= 2*(node_connect(2,i)-1)+1; d4= 2*(node_connect(2,i)-1)+2;

lmm(i,1)=d1; lmm(i,2)=d2; lmm(i,3)=d3; lmm(i,4)=d4;

and realizing that node_connect(1,i) will give the global node 1, and that node_connect(2,i) will give 3 (the other global node), it follows that d1, d2, d3 and d4 will equal 1, 2, 5, and 6 respectively. This makes the first row of lmm to be $$ lmm=\begin{bmatrix} 1 & 2 & 5 & 6 \\ \end{bmatrix} $$

From there, the code uses parts of older codes (see note above) to constrain nodes 1 and 2, and to set up the force matrix R. The code then calculates the elemental stiffness matrices and the global stiffness matrix by using the PlaneTrussElement function. The code then calculates the strain, stress and axial force in each element using the NodalSoln and PlaneTrussResults functions.

The second part of the loading code plots the trusses. The code begins by closing the plot created by the provided code. This is why, when running the code, a plot will appear and quickly disappear. The code then creates a matrix of the displaced nodal coordinates defnodes by adding the displacements found in part 1 to the original nodal coordinates. The code then plots the original truss by creating vectors for each element that go from the node at one end, to the node at the other end of an element. This process is repeated for the deformed truss, this time using the coordinates of the dislocated nodes.

In the third part of the loading code, the results matrix from part 1 is examined. Since the second column of the results matrix represents the stress in an element, that value is examined. The stress value of the first element is examined first, and it is checked to be a compressive force or a tensile force (a compressive stress will be negative, and a tensile stress will be positive). The value is then stored the highest stress. The next element's stress is examined, and determined if it is compressive or tensile. The value is then compared to the highest stress. If it is of greater magnitude, it becomes the highest stress, and its element number is remembered. This process continues until all elements have been checked.

The code ends by plotting the elements under the highest stresses in different colors so that they might be easily identified, using the same process explained earlier. The plot is titled, axis are labeled and a legend is created.

MATLAB Code For Lumped Mass
The following code is used to complete the second portion of the assignment. The code is commented throughout, and a more detailed explanation of the code will follow.

An Explanation Of The Lumped Mass Code
The lumped mass code (solvingpylon2.m) begins by running the loading code (solvingpylon1.m) to receive the already calculated values that will be needed. These values include nodal coordinates, and element properties such as cross sectional area and modulus of elasticity. The density of the elements is then given in g/cm3, and then converted to kg/m3. The length of each element is calculated, and the mass of each element is found by multiplying the element's length by the cross sectional area and by the density.

The mass of each element is then divided by two, and each half mass is added to the global node represented by each end of the element. This is done for all of the elements until the entire mass of the truss is concentrated at the nodes.

A mass matrix is then created by using the function TrainsientPlaneTrussElement.

A reduced mass matrix M_bar, and a reduced stiffness matrix K_bar is then created using MATLAB's setdiff function. This function eliminates the values of the matrices that correspond to the constrained points. This is valid due to the principle of virtual work. The eigenvalues and eigenvectors are then found using MATLAB's eig function.

The natural frequencies $$ \omega $$, vibrational frequencies f, and the vibrational periods T are then found using the following formulas: $$ \omega = \sqrt{\lambda} $$

$$f = \frac{\omega}{2\pi}$$

$$T = \frac{1}{f} $$

where $$ \lambda $$ is an eigenvalue.

The code then proceeds to find the three lowest eigenvalues. This is accomplished by setting variables for the lowest eigenvalues (low1, low2, and low3) equal to large numbers, and then looping through all of the eigenvalues, replacing low1, low2, and low3 with any eigenvalues that are encountered that are lower than the previous values. While doing this, the code keeps track of which rows the lowest eigenvalues were in.

The code then organizes the eigenvectors. The calculated eigenvectors are contained in an 88x88 matrix that is very cumbersome. Each column of the eigenvector matrix represents one eigenvector. Since the assignment required the finding of the lowest three eigenvalue-eignevector pairs, the needed eigenvectors are found in the columns that correspond to the rows in which the lowest eigenvalues were in. The code extracts the desired eigenvectors and puts them into column matrices called modes1, modes2, and modes3.

In a similar manner as the determining of the lowest eigenvalues, the lowest three vibrational periods are found.

Since there was a number of things calculated by this code, with a majority of the values being large matrices, only a selected amount of data was deemed important for output. The next section of code outputs the important data. This data consists of the eigenvalues, the three lowest eigenvalues and their corresponding eigenvectors, and the three lowest vibrational periods.

To finish the code, three plots are created. Each plot consists of an undeformed and a deformed truss. The undeformed truss is plotted using the same lines of code from the loading code. The deformed truss is plotted using lines of code similar to the code from the loading code, but with adjustments to the coordinates of the nodes. For this situation, the "deformed" nodal positions are created by adding the values from an eigenvector to the original nodal coordinates. This is done because the eigenvectors represent the natural displacements for the truss system. By adding the natural displacements to the original nodal coordinates, the result is the naturally deformed truss. This is how the truss system will move, even under no loads, due to the natural frequencies associated with the elements. These natural displacements, however, are very small. In order for the natural displacements to be seen on the plot, a scaling factor was used. This scaling factor was 100,000.

Three plots are created from the lumped mass code. Each of these plots represents the deformations that occur as a result of one of the three lowest eigenvectors.

Results
There are no outputted results from the originally provided code.

Running the loading code (solvingpylon.m) generates the following results

Where d is the displacement matrix that shows the displacements for each degree of freedom in meters reactions are the forces at each of the constrained nodes results shows the strain, stress, and axial forces in each element.

trans_elem is the element number of the element under the highest tensile stress, which is max_tens The highest tensile stress encountered was 9.0511 x 106 Pa, or 9.0511 MPa comp_elem is the number of the element under the highest compressive force, which is called max_comp The highest compressive stress encountered was 8.6957 x 106 Pa, or 8.6957 MPa

In the results matrix, the strain, stress, and axial force values are contained in columns 1, 2 and 3 respectively. The rows correspond to the element numbers. For example, elements 1 and 2 are given as follows:

Running the lumped mass code (solvingpylon2.m) results in the following output

Where eigenvalues is a column matrix that is a diagonal matrix of the matrix of eigenvalues low1, low2,& low3 are the three lowest eigenvalues (low1 being the lowest, low3 the highest) modes1, modes2, and modes3 are the three lowest eigenvectors (modes1 being the lowest, modes3 the highest) and lowT1, lowT2, & lowT3 are the three lowest vibrational periods (lowT1 being the lowest, lowT3 being the highest)

Figure From Loading Code
The following figure is created when the M-file solvingpylon1.m is run.

The figure shows the original truss (scaled to a height of 60m) in red dotted lines. The deformed truss is shown in blue solid lines. Arrows point out the elements under the highest stresses. The element under the highest tensile stress is shown in light blue, and the element under the highest compressive stress is shown in green.

Figures From Lumped Mass Code
The following three figures are generated when the lumped mass code is run(solvingpylon2.m). Each figure represents the displacements from one of the three lowest eigenvectors. In each figure, the red dotted lines represent the undeformed truss, and the solid blue lines represent the deformed truss. The x and y axis for all three figures are in terms of meters.

300 M Steel
300 M steel is a type of steel that can be considered a modification of AISI 4340 steel. 300 M steel contains silicon, vanadium and has a greater carbon and molybdenum content than AISI 4340 steel does. 300 M steel is considered to have good strength, toughness, ductility and fracture strength.

The strength of 300 M steel is 280 - 305 ksi (1929 -2100 MPa).

300 M steel is commonly used for applications that require strength near 300 ksi, such as aircraft landing gear.

For more information see: 300M Alloy Steel Material Property Sheet - Metal Suppliers Online

In the electric pylon problem, the highest stress encountered was the tensile stress. It was found to be 9.0511 x 106 Pa, or 9.0511 MPa. If the pylon was constructed of 300 M steel, then the maximum stress is well below the strength of the material.

Since the strength of the material is so much greater than the maximum stress, two observation can be made. First, if the pylon was constructed of 300 M intentionally, then loads much higher than the 1000N force are expected (or combinations of low loads) that would result in much higher stresses encountered in this situation. Second, if 300 M was not specifically required and was chosen without much reason, and if the pylon experiences only low levels of stress, then a different material should have been chosen instead of 300 M steel. A lower grade of steel would have still adequately been able to withstand the applied loads, and would have been less expensive than 300 M steel.

title: User:Eml4500.f08.team.foskey.ckf/hw6


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User:Eml4500.f08.bottle.vitello/HW 6

Principle of Virtual Work (PVW) concerning dynamics of the elastic bar
From the PDE equation derived from the previous homework concerning the elastic bar: $$ \frac{\delta }{\delta x}\begin{bmatrix} (EA)\frac{\delta U}{\delta x} \end{bmatrix}+\textit{f}=m\ddot{u} $$  (1) We can then discrete the above equation with the single degree of freedom equation concerning an elastic spring: $$ \mathbf{M\ddot{d}+Kd=F} $$  (2) which then gives $$ \mathbf{-Kd+F=M\ddot{d}} $$ for multiple degrees of freedom in an elastic bar. Deriving equation (2) from equation (1) gives us: $$ \int_{x=0}^{x=L}{W(x)\begin{Bmatrix} \frac{\delta }{\delta x}\begin{bmatrix} EA\frac{\delta U}{\delta x} \end{bmatrix}+\textit{f}-m\ddot{u} \end{Bmatrix}}dx=0 $$  (3) for all possible weighting functions W(x). Deriving equation (1) to (3) is pretty simple a task to complete, but deriving equation (3) from (1) is a whole new different ball game. Equation (3) can be rewritten as: $$ \int W(x)g(x)dx=0 $$ for all W(x) as before. Since equation (3) is equivalent for all weighting functions, we can then select W(x) = g(x). Then equation (3) becomes: $$ \int g^{2}dx=0\Rightarrow g(x)=0 $$ which then ends our derivation of equation (3).

Integration by Parts (IBP)
Assume we have the following functions: $$r(x)\! $$ and $$ s(x)\! $$ $$ (rs)' = r's + rs' \! $$ where $$ r'=\frac{dr}{dx} \! $$ and $$ s'=\frac{ds}{dx} \! $$ Therefore, $$ \underbrace{\int (rs)'}_{rs}=\int r's + \int rs' \! $$  $$ \int r's = rs-\int rs' \! $$

Applying IBP to PVW
Returning to the continuous PVW, $$ (3)\; \int_{x=0}^{x=L}{W(x)}(\frac{\partial}{\partial x} [EA\frac{\partial u}{\partial x}]+f-m\ddot{u})dx =0 \! $$ for all possible $$ W(x) \! $$ If $$ r(x) = EA\frac{\partial u}{\partial x}\! $$ and $$s(x)=W(x) \!$$ and we integrate by parts we find the following: $$ \int_{x=0}^{x=L}\underbrace_{s(x)} \frac{\partial}{\partial x} \underbrace{[EA\frac{\partial u}{\partial x}]}_{r(x)}dx = [W(EA)\frac{\partial u}{\partial x}]^{x=L}_{x=0}-\int_{x=0}^{x=L}{\frac{dW}{dx}(EA)\frac{\partial u}{\partial x}dx} \! $$  $$ = W(L)\underbrace{(EA(L))\frac{\partial u (L,x)}{\partial x}}_{N(L,t)} - \!$$$$\underbrace{W(0)(EA(0))\frac{\partial u (0,t)}{\partial x}}_{N(0,t)} \!$$$$ - \int_{0}^{L}{\frac{dw}{dx}(EA)\frac{\partial u}{\partial x}}dx \! $$

Boundary Conditions
We now account for boundary conditions. We consider the model problem: At $$ x=0 \! $$ select $$ W(x) \! $$ such that $$ W(0)=0 \! $$ (i.e. kinematically adminisble).

Motivation
Let's reexamine the discrete PVW as applied to the equation on page 10-1. $$w_{6X1}\circ ([]_{6X2}\begin{Bmatrix} d_{3}\\ d_{4}

\end{Bmatrix}_{2X1}-F_{6X1})=0_{1X1} \! $$ $$F^{T}=\begin{bmatrix} F_{1} & F_{2} & F_{3} & F_{4} & F_{5} & F_{6} \end{bmatrix} \! $$  where $$F_{2} \! $$ and $$ F_{3} \! $$ are known and $$ F_{1}, F_{2},  F_{5}, F_{6} \! $$ are unknown.  Since $$ W \! $$ can be selected arbitrary, then we select $$ W \! $$ such that $$ W_{1}=W_{2}=W_{5}=W_{6}=0 \! $$  in order to eliminate equations involving unknown reactions. We are then left with :  $$ K_{2X2}d_{2X1}=F_{2X1} = K\begin{Bmatrix} d_{3}\\ d_{4}

\end{Bmatrix}=\begin{Bmatrix} F_{3}\\ F_{4}

\end{Bmatrix} \! $$

Continuous case PVW cont.
The unknown reaction $$N(0,t)- EA(0)\frac{\partial u (0,t)}{\partial x} \! $$ where $$ N= \! $$ the normal force. $$ W(L)F(t)-\int_{0}^{L}{\frac{dW}{dx}(EA)\frac{\partial u}{\partial x}}dx+\int_{0}^{L}{W(x)[f-m\ddot{u}]dx}=0 \! $$ for all $$ W(x) \! $$ such that $$ W(0)=0 \! $$ $$\Rightarrow \int_{0}^{L}{W(m\ddot{u})dx}+\int_{0}^{L}{\frac{dW}{dx}(EA)\frac{\partial u}{\partial x}dx} \! $$ $$W(L)F(t)+\int_{0}^{L}{wfdx} \! $$ for all $$ w(x) \! $$ such that $$ w(0)=0 \! $$

Continuous PVW vs Discrete Setting
For the stiffness in the elastic bar, we assume displacement u(x) for $$ x_{i}\leq x\leq x_{i+1} \!$$ For the two-bar truss, we derived the undeformed and deformed shape for each section of the truss as shown below. In the image above, there is an implicit assumption of linear interpolation of displacement between the two nodes. Considering the case when there is only axial displacement comes an expression for $$ U(x) \! $$ in terms of $$ d_{i}=U(x_{i}) \!$$ and $$ d_{i+1}=U(x_{i+1}) \!$$ as a linear function on the x-direction. $$ U(x)=N_{i}(x)d_{i}+N_{i+1}(x)d_{i+1} \!$$ where $$ N_{i}(x) \!$$ and $$ N_{i+1}(x) \!$$ are linear functions in the x-direction.

Connecting continuous PVW to the discrete PVW
To connect the two theories we use a Lagrangian interpretation. We use the form of $$ N_{i}(x) \! $$ and $$N_{i+1}(x) \! $$ 1) $$ N_{i}(x) \! $$ and $$N_{i+1}(x) \! $$ are linear, thus any linear combination of $$ N_{i}(x) \! $$ and $$N_{i+1}(x) \! $$ are also linear, specifically $$u(x)= N_{i}(x)d_{i}+ N_{i+1}(x) d_{i+1} \! $$  $$N_{i}(x)=\alpha _{i}+\beta _{i} \! $$  $$N_{i+1}(x)=\alpha _{i+1}+\beta _{i+1} \! $$ where $$ \alpha \! $$ and $$\beta \! $$ are real numbers.  Therefore we get the linear combination:  $$ N_{i}(x)d_{i}+ N_{i+1}(x) d_{i+1} = (\alpha _{i}+\beta _{i})d_{i} + (\alpha _{i+1}+\beta _{i+1})_{i+1} \! $$   $$ = (\alpha _{i}d_{i} +\alpha _{i+1}d_{i+1})+(\beta _{i}d_{i}+\beta _{i+1}d_{i+1})x \! $$  2) Recall the equation  for $$ u(x)\! $$ (the interpolation of $$ u(x) \! $$ consider: $$ \underbrace {N_{i}(x)d_{i}}_{1}+ \underbrace {N_{i+1}(x) d_{i+1}}_{0} \! $$  therefore $$ u(x_{i})=d_{i} \! $$  and $$u(x_{i+1})=d_{i+1} \! $$  We apply the same theory to $$ W(x) \! $$  $$ W(x)=N_{i}(x)W_{i}+N_{i+1}W_{i+1} \! $$

Element Stiffness Matrix for Element i
$$ (\beta ) \! \;$$ $$\int_{x_{i}}^{x_{i+1}}{\underbrace{[N_{i}'W_{i}+N_{i+1}'W_{i+1}]}_{W'(x)}}(EA(x))\underbrace{[N_{i}'d_{i}+N_{i+1}'d_{i+1}]}_{u'(x)}dx \! $$ where $$ N'_{i}=\frac{dN_{i}(x)}{dx} \! $$ and $$N'_{i+1}=\frac{dN_{i+1}(x)}{dx} \! $$ Note: $$ u(x) \! $$ can be written as $$ \underbrace{[N_{i}'N_{i+1}']}_{N(x)_{1X2}} \! $$ Therefore, $$ \frac{du(x)}{dx}=\underbrace{[N_{i}'(x)N_{i+1}'(x)]}_{B(x)_{2X1}}\begin{Bmatrix} d_{1}\\d_{i+1}

\end{Bmatrix}_{2X1} \! $$ Similarly: $$ W(x) = N(x)\begin{Bmatrix} W_{i}\\W_{i+1}

\end{Bmatrix} \! $$ $$ \frac{dW(x)}{dx}=B(x)_{2X1}\begin{Bmatrix} W_{1}\\W_{i+1}

\end{Bmatrix}_{2X1} \! $$ Recall the element dof's: $$\begin{Bmatrix} d_{i}\\d_{i+1}

\end{Bmatrix}= \begin{Bmatrix} d^{i}_{1}\\d^{1}_{2}

\end{Bmatrix}=d^{i} \! $$ and $$\begin{Bmatrix} W_{i}\\W_{i+1}

\end{Bmatrix}= \begin{Bmatrix} W^{i}_{1}\\W^{1}_{2}

\end{Bmatrix}=W^{i} \! $$  $$ (\beta ) = \int_{x_{i}}^{x_{i+1}}{\underbrace{(B(x)W^{(i)})}_{1X1}}\underbrace{(EA)}_{1X1} \underbrace{(\beta d^{(i)})}_{1X1}dx \! $$ $$ = W^{(i)}\circ (k^{(i)}d^{(i)}) \! $$ where our goal is to find $$ k \! $$ We can rewrite $$ (\beta )=\int_{x_{i}}^{x_{i+1}}{(EA) \underbrace{(B(x)W^{(i)})}_{1X1}} \circ \underbrace{(B d^{(i)})}_{1X1}dx  \! $$ as: $$(BW^{(i)})^{T}(Bd^{(i)}) \! $$ Therefore $$ W^{(i)}\circ (\underbrace{\int B^{T}(EA)Bdx)}_{k^{(i)}}d^{(i)} \! $$  $$ k^{(i)}_{2X2}=\int_{x_{i}}^{x_{i+1}}{\underbrace{B(x)^{T}}}_{2X1}\underbrace{(EA)}_{1X1}\underbrace{B(x)}_{1X2}dx  \! $$ We can then transform the coordinantes from $$ x \! $$ to $$ \tilde{x} \! $$ where $$ \tilde{x}= x-x_{i} \! $$ and $$ d\tilde{x} = dx \! $$  $$ k^{(i)}=\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}{B^{T}(\tilde{x})(EA(\tilde{x})B(\tilde{x}) d\tilde{x}} \! $$ Using the k value ubtained in the above equation, we can then compare it to the expression given in the textbook. $$ \frac{E}{L^{(i)}}\frac{(A_{1}+A_{2})}{2}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}=\textbf{k}^{(i)} $$ with $$ \frac{(A_{1}+A_{2})}{2} $$ representing the average area of the truss example. Then a comparison was made with both equations above: $$ \frac{(E_{1}+E_{2})(A_{1}+A_{2})}{4L^{(i)}}\begin{bmatrix} 1 &-1 \\ -1 & 1 \end{bmatrix}=\textbf{k}^{(i)}_{ave} $$ with the part of finding $$ \textbf{k}^{(i)}-\textbf{k}^{(i)}_{ave} $$. As a reminder, the Mean Value Theorem in relation to the centroid is called upon. Mean Value Theorem (MVT)   $$ \int_{x=a}^{x=b}{f(x)dx}=f(\tilde{x})[b-a] $$   for $$ \bar{x} \epsilon [a,b] $$, with epsilon representing as "belongs to" $$ a\leq \bar{x}\leq b $$ $$ \int_{A}{xdA}=\bar{x}\int_{A}{dA}=\bar{x}A $$ $$ \int_{x=a}^{x=b}{f(x)g(x)dx}=f(\bar{x})g(\bar{x})[b-a] $$ with $$ a\leq \bar{x}\leq b $$ But $$ f(\bar{x})\neq \frac{1}{b-a}\int_{a}^{b}{f(x)dx} $$ with $$ \frac{1}{b-a}\int_{a}^{b}{f(x)dx} $$ representing the average value of f.  $$ g(\bar{x})\neq \frac{1}{b-a}\int_{a}^{b}{g(x)dx} $$ with $$ \frac{1}{b-a}\int_{a}^{b}{g(x)dx} $$ representing the average value of g.  end

Homework Solutions
This section covers the solutions to in-class assigned homework problems not addressed in the lecture notes above. Therefore, it is recommended that the reader refer to the above notes if a homework problem in this section has been seemingly omitted.

Principal of Virtual Work and the continuous weighting function integral
Equation (1), a partial differential equation (PDE), was given in lecture 29 and used to derive Equation (2). $$ \frac{\partial }{\partial x}\left[(EA)\frac{\partial u}{\partial x} \right] + f = m\ddot{u} $$             (1) $$ \mathbf{M\ddot{d}}+\mathbf{kd} = \mathbf{F} $$             (2) The partial differential equation, Equation (1), represents a summation of forces for infinitesimally thin sections dx over the bounds 0 to length L.  Each of the terms in Equation (1) can be represented discretely by a matrix of individual points. Equation (1) is just the continuous expression of the same thing. Therefore, every $$m\ddot{u}$$ between 0 and L added up continuously will have very similar results as the matrix expression $$\mathbf{M\ddot{d}}$$. The same is true for the other terms as well. Knowing that Equations (1) and (2) are the continuous and discrete versions of the same system we can derive the continuous Equation (3) and use the discrete case to verify that our process is correct. First, we know that $$\int w(x)g(x)dx = 0$$ for all $$w(x)$$. In the case that $$w(x) = g(x)$$, our above equation becomes $$\int g^{2}(x)dx$$. As long as $$g(x) \geq 0$$ then $$g(x) = 0$$. Therefore, since Equation (1) can also be rearranged to equal zero, Equation (3) is true. $$ 0 = \int_{x = 0}^{x = L}{W(x) \left\{ \frac{\partial }{\partial x}\left[ EA\frac{\partial u}{\partial x}\right] + f - m\ddot{u} \right\} dx} $$             (3) To verify this process, we recall a proof for discrete matrices from lecture 24 that was used to justify weighting matrices. For the case where $$\textbf{W}_{6x1}\cdot (\textbf{Kd}-\textbf{F})_{6x1}=0_{1x1}$$ We use the following proof from Team Bottle HW 5.

Proof: Select W at $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}_{1x6}\ $$ plugging W back into the Principle of Virtual Work equation gives: $$ \begin{bmatrix}1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 1 \cdot \bigg[ \sum_{j=1}^{6}{K_{1j}d_{j}} - F_{1} \bigg]+ 0\cdot \bigg[\sum_{j=1}^{6}{K_{2-6j}d_{j}} - F_{2-6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{1j}d_{j}=F_{1}}\ $$ Following through as before for the rest of the matrices gives these results: $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}_{1x6}\ $$ $$ \begin{bmatrix}0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1j}d_{j}} - F_{1} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{2j}d_{j}} - F_{2} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{3-6j}d_{j}} - F_{3-6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{2j}d_{j}=F_{2}}\ $$ $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix}_{1x6}\ $$ $$ \begin{bmatrix}0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-2j}d_{j}} - F_{1-2} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{3j}d_{j}} - F_{3} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{4-6j}d_{j}} - F_{4-6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{3j}d_{j}=F_{3}}\ $$ $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix}_{1x6}\ $$ $$ \begin{bmatrix}0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-3j}d_{j}} - F_{1-3} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{4j}d_{j}} - F_{4} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{5-6j}d_{j}} - F_{5-6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{4j}d_{j}=F_{4}}\ $$ $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}_{1x6}\ $$ $$ \begin{bmatrix}0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-4j}d_{j}} - F_{1-4} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{5j}d_{j}} - F_{5} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{6j}d_{j}} - F_{6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{5j}d_{j}=F_{5}}\ $$ $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}_{1x6}\ $$ $$ \begin{bmatrix}0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-5j}d_{j}} - F_{1-5} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{6j}d_{j}} - F_{6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{6j}d_{j}=F_{6}}\ $$ Hence, adding up all of these results gives us $$ \textbf{Kd}=\textbf{F} $$.

Finding an expression for k(i)
Given the following equation, find an expression for $$ \mathbf{k^{(i)}} $$. $$ \mathbf{k}^{(i)}_{2x2}=\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}\mathbf{B^{T}}(\tilde{x})(EA)(\tilde{x})\mathbf{B}(\tilde{x})d\tilde{x}$$ It is already known that $$ \mathbf{B}(\tilde{x}) = \begin{bmatrix}N_{i}(\tilde{x}) & N_{i+1}(\tilde{x})\end{bmatrix} $$ $$ \mathbf{B^{T}}(\tilde{x}) = \begin{bmatrix}N_{i}(\tilde{x})\\ N_{i+1}(\tilde{x})\end{bmatrix} $$ Also, from lecture 33, it is known that $$A(\tilde{x})=N^{(i)}_1(\tilde{x})A_1+N^{(i)}_2(\tilde{x})A_2$$ $$E(\tilde{x})=N^{(i)}_1(\tilde{x})E_1+N^{(i)}_2(\tilde{x})E_2$$ Since Ni is assigned to Node 1 and Ni+1 is assigned to Node 2, the following two equations are true. $$N_{i}(x) = N_{1}^{(i)}(\tilde{x}) N_{i+1}(x) = N_{2}^{(i)}(\tilde{x}) $$ Plugging these expressions back into our original $$ \mathbf{k^{(i)}} $$ equation results in $$ \mathbf{k^{(i)}} = \begin{Bmatrix} N_{1}^{(i)}(L)\\ N_{2}^{(i)}(L) \end{Bmatrix} \begin{bmatrix}N_{1}^{(i)}(L) & N_{2}^{(i)}(L)\end{bmatrix} E(L)A(L) - \begin{Bmatrix} N_{1}^{(i)}(0)\\ N_{2}^{(i)}(0) \end{Bmatrix} \begin{bmatrix}N_{1}^{(i)}(0) & N_{2}^{(i)}(0)\end{bmatrix} E(0)A(0) $$ Multiplying the 2x1 and 1x2 matrices together, the equation above becomes $$ \mathbf{k^{(i)}} = \left[   (N_{1}^{(i)}(L))^{2} + (N_{2}^{(i)}(L))^{2}   \right] E(L)A(L) - \left[   (N_{1}^{(i)}(0))^{2} + (N_{2}^{(i)}(0))^{2}   \right] E(0)A(0) $$ According to the definition of N at the point x = 0, N1 = 1, N2 = 0 At the point x = L, N1 = 0, N2 = 1. Therefore, $$ \mathbf{k^{(i)}} = \left[   (0)^{2} + (1)^{2}   \right] \left[(0)E_{1} + (1)E_{2} \right] \left[ (0)A_{1} + (1)A_{2}  \right] - \left[   (1)^{2} + (0)^{2}   \right] \left[(1)E_{1} + (0)E_{2} \right] \left[ (1)A_{1} + (0)A_{2}  \right] $$ $$ \mathbf{k^{(i)}}_{2x2} = \left[ E_{2}A_{2} - E_{1}A_{1} \right] \begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} $$

Formula and MATLAB Code for Linear Interpolation
The conventional formula for linear interpolation is:

$$ N_{i+1}(x)= \frac{x - x_{i}}{x_{i + 1} - x_{i}} $$

This can be programmed in MATLAB as:

Tapered Bar with Constant Modulus of Elasticity
Let $$\displaystyle E_1 = E_2 = E = 70 GPa $$ and let $$\displaystyle A( \tilde{x}) $$ be linear. Also let $$\displaystyle A_1 = 2400 mm^2 $$, $$\displaystyle A_2 = 600 mm^2 $$ and $$\displaystyle L = 300 mm $$.

Given the following equation

$$\displaystyle {E \over L^{(i)}} {(A_1 + A_2) \over 2} \begin{bmatrix} 1 & -1 \\ -1 & 1  \end{bmatrix} = \textbf{k}^{(i)} $$

$$\displaystyle {(70 GPa) \over (300mm)} {(2400mm^2 + 600 mm^2) \over 2} \begin{bmatrix} 1 & -1 \\ -1 & 1  \end{bmatrix} = \textbf{k}^{(i)} $$

$$\displaystyle \textbf{k}^{(i)} = \begin{bmatrix} 714 & -714 \\ -714 & 714  \end{bmatrix} $$

Electric Pylon Analysis
Our team was asked to conduct an analysis of a 91 bar, 60 m high electric pylon. The properties of the pylon are:

Material = 300 M Steel Height = 60 m Young's Modulus = 200 GPa Area = 4 cm2 Density = 7.8 g/cm3

MATLAB was implemented in solving for the displacements, stresses in each bar, the reaction forces, the eigenvectors, and periods of oscillation.

Code
The following MATLAB code was developed and run:

Results
The results array for the axial stress on each bar is: Stresses =

2.1562e+006 -1.3071e+006 1.3468e+006 1.2911e+006 -7.0304e+005 -4.7596e+006 -2.465e+006 1.4236e+006 3.8538e+006 -5.7655e+006 -37882      1755.6  3.8655e+006 2.9395e+006 -3.6988e+006 -5.7661e+006 3.5896e+005 5.242e+006 1.9797e+006 -3.7749e+006 -6.5159e+006 5.5784e+005 6.4388e+005 5.9656e+006 -8.1704e+005 -3.9616e+006 2.32e+006 -8.1307e+005 -5.7556e+006 -28474 -9.9353e+005 -15214 5.9216e+006 55813 -4.8717e+006 1.3915e+006 -3023.3 -5.7588e+006 -4.854e+006 1.3904e+006 -1.8831e+005 2.258e+006 -6.32e+005 -4.0102e+006 1.0133e-007 1.4647e-007 64147 2.7001e+005 2.7149e+005 -2.0778e+006 -4.4271e+006 -5.8777e+006 -7.3549e+006 -8.662e+006 -8.6957e+006 2.5807e-008 -6.1883e-008 -5.397e-008 9.7638e-008 -1.7704e+005 2.2069e+005 -1.9193e+005 1.9887e+005 2.1464e+006 -2.083e+006 2.2895e+006 -2.2643e+006 2.1035e+006 -2.1464e+006 1.4138e+006 -1.3772e+006 1.5835e+006 -3.6928e+006 -97393      -23690        31208  9.0479e+006 9.3927e-008 9.0317e+006 -2.3815e-007 9.0511e+006 -1.3175e+005 -2.9645e+005 1.7966e+006 4.4876e+006 6.6233e+006 7.7755e+006 -1.0741e-008 -2.4017e-008 -1.9214e-008 -1.0741e-008

The highest stresses occur at: MaxComp = -8.6957e+006 CompBar = 55

MaxTens = 9.0511e+006 TensBar = 81

The following is a plot of the deformed (solid black line) vs. undeformed (dashed blue line) pylon structure:

Notice that the bars in which the maximum stresses occur are labeled with arrows.

Effects of Eigenvectors
The lowest eigenvalues and their corresponding oscillation periods are:

lam =  132.16       2468.4       2940.3

T = 0.54654      0.12647      0.11587

Plots of the effects of the three lowest eigenvalues are shown below. The blue dashed line represents the undeformed structure while the solid black line represents the deformed structure.

Statically Determinant?
The final step in this problem was to determine whether the structure can be solved using statics. Since there are only two fixed points and four bars for which reactions must be computed, it can be determined that this problem is statically indeterminate.

Honesty, Innovation, and Ethics, a Brief Commentary
Honesty, Innovation, and Ethics.

These three concepts are fundamental traits of the professional engineer, but easily extend beyond the singular. These are principles which can drive a society and can forge interpersonal relationships without which the world would be subject to a Damoclean verge of chaos.

Without honesty, the integrity of mankind would be shattered, replaced with a nebulous abyss of lies and deceit. By internalizing and reciprocating the concept of honesty, individuals are able to exchange ideas and communications with the knowledge that this information transfer is absolute in self-perceived truth (clearly little "t"). Here, suspicion would only lay not in the wonderment of peril, but in the motivation for the honesty itself. However, this phenomenon is speculated to exist insomuch as the function of integrated human history void of the ideal absolute honesty theorized herein.

Indeed, even Sir Thomas More's Utopia proved that absolute honesty could only be assumed as a satirical tool against those lambasted in metaphor by the book itself. If honesty had this sway, then truth would be self-evident and secrets disavowed as public would be all knowledge if only received upon request. This is within Loc Vu-Quoc's Vision: that is, for the free dissemination of all academic resources via the Wikipedia and Wikiversity interfaces. Here, it is believed that a modern categorical imperative would be enforced (See HW 1) by the willingness of the masses, and hence contributions would be made for the good of mankind to strengthen the collective information pool.

The hitch, however, is that this functions specifically on honesty as a tool to be utilized by all in a Kantian fashion. Reality, though, suggests this is not the universal tendency of human behavior as seen by the necessitation of Wiki admins who tirelessly endeavour to ensure validity of posted material. If the material was always valid, the need for admins would be minimalized significantly; as is, this is unforunately not the case. Henceforth, honesty can only be considered a fluctual tool of mind, something situationally applicable, but never universally so if only due to the fallacies of tabula rasa and the ubiquitous corruptibility of man. More simply, there is no Übermensch.

Innovation, meanwhile, is something which is a limitless pursuit as elusive as Máni to Hati Hróðvitnisson. It is something which inspires and drives, but is ever-presently infinite and inherently unattainable. Innovation is what leads to new ideas and technologies, governments, relationships, and all other forms of human and animal advances. It could be argued that the aspiration of all engineers is to innovate, but it can also be argued that innovation is a function not only on the synthesis of knowledge by the thinker but also of the collective knowledge itself. Ergo, Loc Vu-Quoc's Vision is further realized: through the free and open osmosis of ideas in the international community, especially when governed by the principle of honesty, innovation can occur based upon the synergy intrinsic in the global neural network.

Without innovation, it can be speculated that mankind would have never survived as many generations of procreation as it has managed. Fire, the wheel, and primitive agricultural techniques were innovations to the early mind as modern technology like the iPod, the Nintendo Wii, the iFly VAMP, or Wikipedia are today.

But innovation itself is subject to the concept of ethics, a catchall philosophical term generally finding root in the belief and execution of a "correct" and "proper" behavior of organisms to ensure a "good life". Ethics is exceptionally questionable, as the moralities of good and evil (fundamentally diametric forces defining ethics) can be perceived universally from perspectives arguing favorably to any singular action. For example, if a student is caught plagiarizing from Wikipedia, the common reaction would be to assume the student is exhibiting some form of "evil" as the connotation of the very act is often considered "wrong" by society; conversely, if the action was done in response to the conformities of society as a way of breaking the mold, innovating thought, and pushing the limits of what is convential, how is this different than the "goodly" perceived notions of the American Revolution against King George III? Or further: what if there was a justified reason for the plagiarism, such as the student wasn't able to do the work originally as he had invested time needed to satisfy such criteria in saving a school bus of orphaned children from the cold-hearted wrath of indifferent terrorists threatening their very lives‽

Thus, ethics cannot be objectively considered without understanding the universal. Sometimes this is difficult, and hence open communication that exhibits honesty will aid in a more complete understanding of the situation(s). Yet for the sake of argument based upon the expectations of Loc Vu-Quoc's Vision, ethics should be considered a very heartfelt principle to an engineer, as "proper" moral principles are what would be argued under the Vision as things which will lead to ideal interpersonal communications, and hence refine and improve mankind while leading to innovation through honesty.

Finally, as William Shakespeare once famously noted: "Honesty is the best policy. If I lose mine honor, I lose myself."

Three Bar Truss Problem
This problem is a continuation of a problem introduced in the previous homework assignment where a three bar truss system as seen in the following graphic was to be analyzed for its eigenvectors and eigenvalues using MATLAB. The K matrix for the structure is defined as: where Implying the eigenvectors each correspond to the column vectors composing the V matrix, with the eigenvalues consisting of the diagonal values of the D matrix. Plotting the eigenvectors as deformations to the original structure results in the following series of graphics, each graphic sequentially corresponding the eigenvector column for which the deformation displacements were derived.

MATLAB code for the generation of the plots is as follows:

Effects of Cross-bar to Three Bar System
The addition of a cross-bar to the Three Bar system above results in the following figure:

Where K is defined as:

with eigenvectors and eigenvalues defined similarly to those above as the following:

MATLAB code to model this system can be seen as follows:

Individual Contributions
Eml4500.f08.bottle.vitello 01:16, 10 November 2008 (UTC) Eml4500.f08.bottle.butler 15:55, 17 November 2008 (UTC) Eml4500.f08.bottle.hipps 21:00, 18 November 2008 (UTC) Eml4500.f08.bottle.barnes 22:21, 20 November 2008 (UTC) Eml4500.f08.bottle.loschak 02:47, 21 November 2008 (UTC) Eml4500.f08.bottle.brockmiller 17:34, 21 November 2008 (UTC) Eml4500.f08.bottle.ranto 20:50, 21 November 2008 (UTC)

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Eml4500.f08.bottle.vitello/HW 6


text:
Echinoderm Nervous System Evolution Project/Proposal

This proposal is currently in progress. At present, it is a space to collect thoughts.

1. Choose ONE very specific question about the origin of the echinoderm nervous system, that I would like to answer.

2. How would information about cell type homologies resolve this question?

3. What techniques could be used to gather information about the homology of the cell types?


text: File:Genetically Modified Organisms.pdf

title: File:Genetically Modified Organisms.pdf


text:
Comparative Neuroscience

Introduction
Welcome to the Comparative Neuroscience course at Wikiversity! This course is part of the Department of Neuroscience. We hope you will find these materials helpful to you as you pursue your studies.

Lessons
Lesson 1: Evolution of the Nervous System

Lesson 2: Overview of Mammalian Neuroanatomy Related Resource- MIT's Introduction to Neuroanatomy course, taught by Professor Chris Moore (tactile perception researcher)

Lesson 3:

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Lesson 5: LAB - Virtual Sheep Brain Dissection

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Instructors
If you have questions, feel free to contact any of these individuals.

 * AFriedman

title: Comparative Neuroscience


text:
Template:Social psychology (psychology)

This page serves to point to templates for the Social psychology (psychology) unit of study.

 * SPPP2008S - for substituting onto participants' user pages; it adds:
 * SPPP2008 - for a tranclusion on participants' user pages
 * SPPP2008Withdrawn - replace above templates with this one when a participant withdraws from the unit
 * SPPPW - for substituting to participant talk pages
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 * SPEssay - for essay pages
 * SPLectureDone2008 - for lecture pages (completed)
 * Lecture - for lecture pages (new pages) - could be customised to SPLecture
 * Tutorial - the general WV template for tutorials; could be customised to SPTutorial
 * SPEF - shortcut to Social psychology (psychology)/E-portfolio2008Feedback - has 3 parameters for comments.

title: Template:Social psychology (psychology)


text: Secrets of successful long-term intimate relationships


text:
Template:Social psychology (psychology)/Assessment/Essay

Essay question:


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Statistical thermodynamics

Here we attempt to connect three iconic equations in thermodynamics: (1) the Clausius definition of entropy, (2) the Maxwell-Boltzmann energy distribution, and (3) the various statistical definitions of entropy. Of all the topics in the curriculum of the advanced physics major, thermodynamics is probably the subject presented with the most unanswered questions. To review what most students do learn:

 * 1) Thermometers don't work. A thermometer can only take its own temperature: Zeroth Law of Thermodynamics
 * 2) You can't win.  Energy cannot be created: First Law of Thermodynancs
 * 3) You must lose. Friction is everywhere, friction turns to heat, and you can't use heat: Second Law of Thermodynamics
 * 4) It never ends.  The effort to reach absolute zero never succeeds: Third Law of Thermodynamics
 * 5) Nobody knows what entropy really is... vaguely attributed to John von Neumann.

Three iconic equations in thermodynamics
Consider three iconic equations encountered in the education of a physicist:

Thermodynamic entropy:  In 1862 Clausius proposed that entropy is

 * $$\Delta S = \int \frac {dQ_{rev}}T $$          (1)

where dQ is heat transferred during reversible heat flow (at constant volume). Heat is energy that spontaneously flows when objects of different temperatures are placed in thermal contact.

Maxwell-Boltzmann distribution for the ideal gas:  In 1860 Maxwell produced a symmetry argument that eventually led to a form of the Maxwell-Boltzmann distribution equation for of the speed (not velocity) of atoms in an ideal gas:

 * $$ f_{|v|}(v) = A v^2 \exp \left(- \frac{mv^2}{2k_BT}\right) $$       (2)

where, $$A = 4\pi\sqrt{\left(\frac{m}{2 \pi k_B}\right)^3}$$ is a normalization constant, and kB (or k) is Boltzmann's constant, kB, equals 1.38062 x 10−23 J/K.

Statistical definition of entropy:

(3)    $$S=k_B\ln \Omega$$

Our third equation, originally formulated by Ludwig Boltzmann between 1872 and 1875, is so iconic that it appears on Boltzmann's tombstone. (using $$W$$ instead of the modern $$\Omega$$ for what is often called the 'number of available states') The existence of two definitions (k ln Ω and ʃdQ/T) for the same entity leaves one wondering if they are actually the same thing. Adding to the complexity is the fact that Ω in equation (3) has a number of definitions. For example, in information theory the constant, kB, equals one, and entropy is measured in 'bits' if the base of the logarithm is 2 (instead of base-e as used in physics). Information Ω is the inverse of probability if all probabilities are equal. And if all probabilities are equal, then Ω is the number of possible outcomes. The number of fair coin tosses is its entropy in bits. (Since there are 2N outcomes for N fair coins, and log2N2=N). In physics, entropy can be troublesome to calculate primarily due to issues concerning the appropriate number of equally probable states. Another vexing problem is the fact that the equations of Newtonian and Hamiltonian physics possess time reversal symmetry, while reality apparently does not. Entropy reflects that reality by only increasing as time evolves.

The goal of this essay is to connect equations (1), (2) and (3) by actual calculation.

State variables: Pressure (P), Energy (E), Volume (V), and Temperature (T):

A state variable is a measurable physical property that can be uniquely defined for a substance that is in thermal, mechanical, and chemical equilibrium. We shall refer to this substance as 'the system' and two systems play a key role in this discussion:
 * 1) The monatomic ideal gas.  The ideal gas model tends to fail at lower temperatures or higher pressures, when intermolecular forces and molecular size become important, and more complex ideal gasses, such as the  diatomic molecule require quantum mechanics to understand rotational energy. These complications render such systems unsuitable to be fully modeled.
 * 2) At the opposite extreme, we are investigating any system (however complex) in what is called  thermodynamic equilibrium. One requirement for being in thermodynamic equilibrium is that the pressure and temperature are both uniform throughout the system.  Thermodynamics is often applied as an approximation when these conditions do not hold (for example when calculating the  speed of sound).

Pressure and volume can be measured by observing the force and location of the piston, respectively. Temperature was originally measured with thermometers, but for our purposes the ideal gas thermometer is more convenient. We therefore establish the following operational definition of temperature: The temperature of a substance is such that when placed in a thermal contact with a monatomic heat bath, $$\rm dQ = 0\,$$ where $$\rm dQ\,$$ is heat flow, and temperature is found by solving the equation of state for an ideal monatomic gas:

 * $$PV=nRT=Nk_BT$$    (Equation of state for ideal gas)

Here, N is the number of particles (assumed constant in this introduction to classical thermodynamics), n is the number of moles, and R is called the gas constant. When classical thermodynamics was being constructed in the eighteenth and nineteenth centuries, it was not yet known that one mole consists of 6.022×10 23 atoms. Instead, the mole was defined as equal to the number of carbon atoms in 12 grams of carbon, and the gas constant, R, was an experimentally known parameter equal to 8.314 JK-1mol-1.

We shall assume that energy is a property of all systems of particles and that energy can be uniquely determined by other state variables. This brings our list of state variables to four:
 * Volume (V)
 * Pressure (P)
 * Temperature (T)
 * Energy(E)

Any function of these state variables is also a state variable. Generally speaking, any two state variables is sufficient to determine the system. One exception is energy for an ideal gas, where since energy is directly proportional to temperature, energy and temperature are insufficient for determining the state of an ideal gas.

Heat, work and the heat engine
Heat,$$\rm dQ\,$$, is the flow of energy from a hot to a cold substance. (The fact that heat never flows between objects with the same temperature is part of the zeroth law of thermodynamics.) Work is $$\rm d W =F \rm dx = -P\rm dV\,$$, where the minus sign ensures that dE, dW, and dQ are all positive if they add energy to the system. Since work and heat are the only two ways energy can enter or leave this sealed system, we have,

 * $$\rm dE = \rm dQ +\rm dW $$    First law with differentials; eventually becomes dE=TdS-PdV.

Why heat and work are not state variables
To understand why heat,$$\Delta Q\,$$, and work $$\Delta W\,$$ are not state variables,we introduce the heat engine, which is a cycle that converts heat into work (or it can act as a heat pump if the cycle is reversed.) Since only two state variables are independent, it is possible to fully specify a system as a point on a two dimensional graph, for example on the P-V (pressure-volume) diagram shown in the figure. The other state variable (T) can be depicted using contours on the graph. Contours of constant temperature are called isotherms. For the ideal gas, these contours are hyperbolas, but we are not restricting our discussion to the ideal gas.

''This figure shows pressure (P) as positive, but an equivalent P-V diagram could be constructed for an substance with negative pressure, such as a rubber band. The rubber band is a good example of approximate thermodynamic equilibrium because the system is never properly in chemical equilibrium - the cycles would not exactly close upon themselves as the rubber degrades into a more stable (but less desirable) substance. Nevertheless, one can construct a rubber band heat engine or heat pump: Stretch a thick rubberband and hold it stretched until it cools off to room temperature. Then decompress the rubberband and press it to your lips. It will feel cool because you have created a heat engine in reverse, called a heat pump (or a refrigeration unit).''

Wikipedia refers to the cycle shown in the figure to the right as an Ideal Heat Engine. It is 'ideal' only in the sense that certain aspects are easy for students to calculate. From points 1-2, energy in the form of work leaves the system. No work is done from points 2-3 because the piston remains motionless. Work energy does re-enter the system from points 3-4, but less work is involved because the pressure is lower.

 * The work done by the engine per cycle equals to the area enclosed by the cycle on a P-V diagram.

This equivalence of work performed to area inside the cycle on a P-V plot is a general property of heat engines.

By conservation of energy, the work performed by the engine must equal the energy that enters, and this energy can only be delivered into the system as heat. Since work energy leaves the engine at each cycle, and heat energy enters, work and heat are not state variables. It is meaningless to ask how much 'heat is in a system'. It is best to always write work and heat using notation that reflects this fact, e.g., $$\Delta W$$ and $$\Delta Q$$.

How do we know that entropy even exists?  The Clausius definition of entropy as the integral, dS =ʃdQ/T, defines entropy only as a differential. As we have seen, this does not prove that entropy is a state variable. The next section answers that question.

Carnot's Theorem

The 'ideal heat cycle' described above is easy to understand because work is so easy to calculate. Another heat engine is far more important. The Carnot cycle consists of four legs. Two legs are 'isotherms' with ΔT=0, (temperature remains constant), and two legs are 'adiabats' with ΔQ=0 (no heat flows). It can be shown that the Carnot cycle is the most efficient way to do work by transferring heat from a reservoir at Thot to a reservoir Tcold. But a far more important property of the Carnot cycle is the fact that all substances have exactly the same efficiency when used in a Carnot cycle. This can be proven using the diagram shown to the right. If two substances produced work with different efficiency, one could use the less efficient substance in a heat pump (i.e., air conditioner) and violate the second law of thermodynamics generating useful work from a temperature reservoir without the requirement that heat must be dumped into a colder reservoir. (As discussed below, this would permit the construction of a perpetual motion machine of the second kind.)

The efficiency of a heat engine operating between two temperatures is the ratio net work (per cycle) to the heat flow out of the hot reservoir (per cycle). Ideally, one would want to convert all the heat from the hot reservoir into work and not 'dump' any 'waste heat' into the cold reservoir. The proof that all Carnot cycles have the same efficiency can be found on this Wikipedia article.

Why the Carnot engine is reversible
Key to the proof is the fact that the second engine in the figure to the right is being operated in reverse as a [w: heat pump|] (refrigeration unit). Carnot's theorem does not hold if the engine is not reversible. The Carnot cycle shown in the figure to the left is reversible because heat enters at constant temperature. Strictly speaking, the Carnot cycle is only reversible if the temperature of the substance equals the temperature of the heat reservoir. It may seem odd that two systems (reservoir and substance inside the piston) are exchanging heat when both are at the same temperature, as heat only flows between bodies of unequal temperature. This paradox is resolved by admitting that the two temperatures are not exactly equal, but only approximately equal. And since a heat flow is often directly proportional to the difference in temperature, a true Carnot cycle would almost take an infinite amount of time to accomplish. The Carnot engine is truly an idealized and theoretical construct that can never truly exist. But something arbitrarily close to a Carnot engine can be built, and will operate in reverse in approximately the same fashion.

Why the Carnot engine is important
It is not uncommon for a heat engine to employ a complicated substance (e.g. Freon in an air conditioning unit). Carnot's theorem is a statement about all substances that remain in equilibrium throughout the cycle of a heat engine. If we can fully understand just one simple substance, then we can learn something that is true for all substances in equilibrium. The thing that we learn about will be called entropy. And the substance that we shall (attempt to) fully understand is the monatomic ideal gas.

 * For an understanding of how the Carnot theorem establishes entropy as a state variable, see
 * Entropy and the Carnot cycle

An introduction to motion in phase space
Both the derivation of the Maxwell-Boltzmann equation for energy distribution, as well as the calculation of entropy in would today be called the spirit of information entropy require a deep understanding of how particles move in phase space. Phase space typically has twice the number of dimensions as the 'configuration space' students initially use to visualize motion. For example, instead of viewing motion in the two dimensional (x-y) plane, phase space would include four dimensions (x, y, vx, vy).

 * For a deeper understanding of phase space, see:
 * Phase space

Ergodic hypothesis and second law of thermodynamics

For simplicity we develop ideas with unphysical models. Consider a two dimensional gas of rays for which speed is irrelevant. They never collide, and change direction only after hitting a wall. If the walls are perfectly smooth, most particles would follow paths resembling that shown in (a) of the figure. Such an orbit does not occupy all of its available phase space, and is therefore called non-ergodic. Not only do these paths fail to fill the available phase space, they do not seem uniformly fill the space available to them. But if the smooth surfaces have any flaws, no matter how small, a gas of such particles would eventually fill the area uniformly, as shown in (b). In classical thermodynamics, the assumption is made that particles fill their available phase space uniformly in this fashion. This is called the ergodic hypothesis, and it was originally proposed by L. Boltzmann.

Although we do not yet know why, it is clear that any trick to coax the gas from one of the two connected circles (c) and (d) using an artfully shapted tunnel as shown in the figure would fail. If such a tunnel ever managed to create a significant pressure difference, it could drive a perpetual motion machine of the second kind. An automobile's pistons would be driven by arranging for the atoms to randomly find their way to one side of a piston. The compressed air would expand to power the car. In expanding in this fashion the air would cool. But the exhaust from this engine would be cold air, some of which could be pumped directly into the cab on a hot summer day. The only fuel required is the random motion of atoms that occurs everywhere on earth. The impossibility of ever creating such a device shall be our working definition of the second law of thermodynamics.

The term 'ergodic hypothesis' is somewhat of a misnomer because the word 'hypothesis' often suggests either a controversy or unresolved scientific question. Often it is. For our purposes, it is best to view the 'ergodic hypothesis' as an assumption that may or may not hold for a given system. The reader should note that we have not yet defined ergodic, but instead have only vaguely defined it as to mean something that is somehow 'randomized'. To better understand and define the ergodic hypothesis, we need to better understand the concept of phase space.

For our purposes, the ergodic hypothesis can be stated as follows:

 * All regions of phase space that are accessible to a given particle are equally populated, provided that phase space is expressed in Hamiltonian canonical variables.

Canonical variables, Liouville's theorem, and the ergodic hypothesis
This equality in occupation of phase space is a consequence of Liouville's theorem, and requires that the canonical variables of a Hamiltonian dynamics be used. Liouville's theorem does not prove the Ergodic hypothesis.

Moreover, it likely that no proof of the ergodic hypothesis exists, since it seems to violated for certain Hamiltonians, such as that used in the Fermi-Pasta-Ulam problem. But Liouville's theorem does ensure that once the probability becomes uniform throughout a region of phase space that is accessible to a particle, the ergodic hypothesis will continue to hold.

We shall assume the ergodic hypothesis throughout this introductory look at Statistical thermodynamics. And since this is an introduction to the subject, we shall not prove whether or not the variables are such that Liouville's theorem applies. For example, position and velocity (x,v) are not proper canonical variables, but v is directly proportional to p=mv, phase space in (x,v) obeys Liouville's theorem and we there assume that the ergodic hypothesis applies. On the other hand, as we shall see in the next section, kinetic energy is not a suitable variable for a phase space where the ergodic hypothesis can be applied.

move this statement to a later place in this document
explains why the variables must be canonical. To see the ill consequences of not using canonical variables, read the following discussion of the Maxwell-Boltzmann probability distribution for the ideal gas, and

Probability distribution functions
The probability of having a certain velocity in a one dimensional gas is described by a distribution function. Letting $$Prob$$ denote probability, we have:

 * $$Prob(v_1<v_x<v_2)=\int_{v_1}^{v_2} f_{v}(v)dv \rightarrow f_{v}(v)\Delta v$$      (One dimension)

The 'rightarrow' ($$\rightarrow$$) holds when the range of velocity is so small that $$f_{v}(v)$$ is nearly uniform. The subscript on the probability distribution function, $$f_{v}$$, will permit us to simultaneously talk about probability distributions for different variables.

Two dimensional gas confined to within a circle
The probability density for one particle confined to a circle of radius R can be found by satisfying two criteria: Hence,
 * 1) The integral over all accessible regions of xy space is one.
 * 2) The system is ergodic if proper variables are used; we take x and y to be proper variables without proof.

 * $$f_{xy}(x,y)dxdy = \frac{1}{\pi R^2}\;dxdy$$

It is useful to change variables using $$x = r\cos\theta$$ and $$y = r\sin\theta$$:

 * $$dxdy =

\det\begin{bmatrix} \dfrac{\partial x}{\partial r}   &    \dfrac{\partial x}{\partial \theta}\\ \dfrac{\partial y}{\partial r}   &     \dfrac{\partial y}{\partial \theta} \end{bmatrix}\;drd\theta= \det\begin{bmatrix} \cos\theta   &    -r\sin\theta\\ \sin\theta   &     r\cos\theta \end{bmatrix}\;drd\theta=rdrd\theta $$

Hence,

 * $$f_{r\theta}(r,\theta)drd\theta = \frac{r}{\pi R^2}\;drd\theta$$

Note that rθ space is not uniformly occupied.

Kinetic energy probability distribution
Letting E = ½mv2 equal kinetic energy, we solve for v = (2E/m)1/2the probability distribution in energy can be calculated as follows:

 * $$ f_\epsilon(\epsilon)d\epsilon = 2 f_v(v)dv $$

The peculiar factor of 2 occurs because two values of v contribute to the same kinetic energy. Since this is a one dimensional calculation, it is proper refer to $$\epsilon$$ as kinetic energy for one degree of freedom. (In this way, a free particle in three dimensions has three degrees of freedom.) Using, $$d\epsilon = mvdv$$, and a bit of algebra, it can be shown that :

 * $$ f_\epsilon(\epsilon) = \sqrt{\frac{2}{m\epsilon}} f_v (v)$$

Maxwell's symmetry argument for velocity space
Maxwell's original derivation of the Maxwell-Boltzmann equation relied on the concept that the product of two or more independent events occurring equals the product of the probability of each event. For example the probability of flipping a fair coin twice and obtaining two heads (H-H) is the product of obtain a heads one coin squared:  P(H-H)=P(H)P(H) = ½·½=¼. He reasoned that for N particles, the probabilities distributions for all the particles were independent, and therefore obeyed a similar product rule:

$$ F_{\{ v\}} (v_1, v_2, ... v_N) =\prod_{n=1}^N {f_{v}(v_n)}= f_{v}(v_1)\cdot f_{v}(v_2)\cdot ... \cdot f_{v}(v_N)$$

where,

 * $$ \{ v\} = [v_1, v_2, ... v_N] $$

represents one point in a very large number of dimensions. The use of a capital letter for this 'grand' distribution function reflects the relative complexity of this space. Boltzmann also argued that the 'grand' distribution function depended only on the total energy of the gas:

 * $$ F_{\{ v\}} (v_1, v_2, ... v_N)= F_{\{ v\}}(E)$$,

where $$E = \frac{1}{2}mv_1^2 +\frac{1}{2}mv_2^2 +...+\frac{1}{2}mv_N^2 = \epsilon_1 + \epsilon_2 + ... + \epsilon_N $$

is the total energy. Ordinarily the probability distribution would also depend on density, but an ideal gas is assumed to be sufficiently tenuous that collisions have no impact except that they eventually establish thermal equilibrium. It is the total energy of an ideal gas of N particles that uniquely determines how hot it is.

Carefully redefining functions
The micro distribution functions, $$f(v_n)$$, also depend only on speed, and hence only on the energy of each atom, $$\epsilon_n = mv_n^2/(2m)$$. But extreme care must be used in expressing this fact. It is safest to define a new function, g, as follows:

 * $$ g(\epsilon) = f_v\left (\sqrt{\frac{2\epsilon}{m}}\right )$$.

This function, g, is numerically equal to the micro distribution for a single particle, but as we shall see, is not the distribution function for the kinetic energy of an atom. Also, to be precise, it is necessary to think of $$F$$ as one of two functions:

 * $$ F = F_{\{\epsilon\}} = F(\epsilon_1, \epsilon_2, ... \epsilon_N)= g(\epsilon_1)\cdot g(\epsilon_2)\cdot ... \cdot g(\epsilon_N)$$.

 * $$ F = F_E = F(E) = F(\epsilon_1+ \epsilon_2+ ... +\epsilon_N) $$.

Velocity distribution function for a 1-D ideal gas
Using the first form, we evaluate the variation of F with $$\epsilon_1$$ as:

 * $$\frac{\partial F}{\partial\epsilon_1} = \frac{dg}{d\epsilon_1}\cdot g(\epsilon_2)\cdot....\cdot g(\epsilon_N)

=\frac{1}{g_1}\frac{dg_1}{d\epsilon_1}F$$.

Using the second form,

 * $$ \frac{\partial F}{\partial\epsilon_1} = \frac{dF}{dE}\frac{dE}{d\epsilon_1}=\frac{dF}{dE}$$.

Similarly we can prove that $$ \partial F/\partial\epsilon_2 = dF/dE=\partial F/\partial\epsilon_1$$, which means that two expressions, each (exclusively) involving different variables are equal:

Where $$f_v (v)$$ is evaluated at $$v= 2\epsilon /m)^{1/2}$$.
 * $$ \frac{1}{g_1}\frac{dg_1}{d\epsilon_1}= \frac{1}{g_2}\frac{dg_2}{d\epsilon_2} = -\kappa $$

where we have yet to establish that the constant, $$\kappa$$ is a positive number. But solving this differential equation, we have:

 * $$g(\epsilon)=Ae^{-\kappa\epsilon}$$

Using the ideal gas law to find kappa

The ideal gas law is an observable fact about actual gasses. To establish that $$\kappa=m/(2k_BT)$$ we must perform a theoretical calculation of the pressure using this distribution function. We use the fact that the distribution functions are now known (with $$\kappa$$ as a parameter) and calculate the pressure exerted by this gas by employing the momentum transfer associated with elastic collisions of particles of a known density and velocity distribution off a wall, as shown in the figure. A good proof can be found in the wikibook General Chemistry

Hence we have the probability distribution function for the velocity of a one-dimensional ideal gas expressed in terms of temperature:

 * $$ f_v (v) = A \exp \left(-\epsilon/k_BT\right)$$,

where, $$A=\left( \frac{1}{2 \pi mk_BT} \right)^{1/2}$$, is a normalization constant that can be found by setting $$\int_{-\infty}^\infty f_v dv =1$$.

The importance of using proper phase space variables
If we change variables using the methods described above, we establish that

 * $$ f_\epsilon (\epsilon) \ne A \exp \left(-\epsilon/k_BT\right)$$

In other words, the wrong answer would have resulted if we assumed if this calculation had begun with slightly different (but incorrect) premise:$$f_\epsilon(\epsilon) = $$ $$f_1(\epsilon_1)f_2(\epsilon_2)...f_N(\epsilon_N)$$. This premise is false because the transformation from the permissible phase space variables $$(v_1, ... v_N)$$ to energy variables introduces the factor $$\epsilon^{-1/2}$$.

The velocity distribution if the gas is confined by a potential

If this calculation is repeated for a one dimensional gas confined by a potential well, $$\Phi=\Phi(x)$$, one obtains the following distribution function:

 * $$ f_v (v) = \rm{Constant}\cdot \exp \left(-\frac{\frac{1}{2}mv^2+\Phi(x)}{k_BT}\right)$$.

We leave it to the reader to show that a derivation of this formula follows the methods already introduced for the ideal gas. This is a significant result because the potential can be that of a harmonic oscillator $$(\Phi = \frac{1}{2}k_Sx^2)$$, where $$k_S$$ is the spring constant. Moreover, a Hamiltonian exists that describes two coupled oscillators as two uncoupled oscillators. And, this method can be extended to include a very large number of coupled oscillators. In other words, a linear wave can be viewed as a classical gas of uncoupled harmonic oscillators. It was Plank's attempt to perform an thermodynamic analysis of light waves in thermal equilibrium with a closed box that lead to the breakdown of classical physics.

To be continued
The rest of this essay is presented in outline form only.

Entropy of an ideal gas
Having shown that entropy is a state variable for a monatomic ideal gas, we may use Carnot's theorem to prove that entropy is an ideal gas for all systems (otherwise the difference between Carnot cycles could violate the second law).

Carnot's theorem and the existence of entropy
Carnot's theorm ensures that the integral of dQ/T (for a reversible process) is the same regardless of what substance is inside the piston. This establishes:
 * 1) The Clausius definition of entropy, S=kʃdQ/T, does define entropy as a state variable.
 * 2) The reversible Carnot cycle permits a definition of temperature that is consistent with the definition of temperature introduced with Maxwell-Boltzmann kinetic theory of the ideal gas.

First law of thermodynamics with entropy
Now that the existence of entropy is established we may express the first law of thermodynamics in the following form:
 * $$\rm dE = \rm dQ +\rm dW \rightarrow \rm dE = T \rm dS - P\rm dV$$

Calculating the statistical entropy for a classical ideal gas
This section is the "punch line" to this essay.

The aforementioned Wikipedia link states and resolves the paradox. Following comment was made by a Wikipedia editor on the talk page to that article:
 * See the Wikipedia article on Gibbs Paradox.

 * for what it's worth, a summary of what Gibbs said:

In two different works, published thirty years apart, Gibbs remarked two different ways on how the notion of distinguishability affects entropy. I have been reading Gibbs lately and studying how he saw the issue. For the benefit of future edits on this article, I've included below direct links to Gibbs' remarks, and some interpretation of mine since Gibbs' prose can be painful to read at times.

In neither case did he find there to be an actual problem, since the correct answer fell naturally out of his mathematical formalisms. However it's evident that he saw these things as possibly tricky points for the reader, because in each case he did devote time to carefully talking about the issues.

  On the Equilibrium of Heterogeneous Substances (1876): Internet archive copy of Part I, see section "Considerations relating to the Increase Entropy due to the Mixture of Gases by Diffusion." (starting at page 125 of the scanned file, which is page 227 of the manuscript). The key points:

  Elementary Principles in Statistical Mechanics (1902): Gibbs' 1902 book addresses what is often called "correct Boltzmann counting". This counting is often explained as necessary to obtain extensive thermodynamics, however Gibbs introduces it in a different way (as a tool to avoid overcounting in phase space). The discussion takes place at the beginning and end of Chapter XV (transcription on Wikisource).
 * 1) Gibbs begins by deriving (from thermodynamics) that the entropy increase by mixing two gaseous masses of different kinds, of equal volume and at constant temperature and pressure will be $$(pV/T)\log 2$$, where V is the final volume. When mixing two masses of the same kind, there is no entropy increase.
 * 2) He emphasizes the striking property that the mixing entropy does not depend on which kinds gases are involved, whether they are very different or very similar. They only need to be of distinguishable types.
 * 3) Why entropy should increase in one case and not the other:
 * 4) "When we say that when two different gases mix by diffusion, as we have supposed, the energy of the whole remains constant, and the entropy receives a certain increase, we mean that the gases could be separated and brought to the same volume and temperature which they had at first by means of certain changes in external bodies, for example, by the passage of a certain amount of heat from a warmer to a colder body. But when we say that when two gas-masses of the same kind are mixed under similar circumstances there is no change of energy or entropy, we do not mean that the gases which have been mixed can be separated without change to external bodies. On the contrary, the separation of the gases is entirely impossible. We call the energy and entropy of the gas-masses when mixed the same as when they were unmixed, because we do not recognize any difference in the substance of the two masses. So when gases of different kinds are mixed, if we ask what changes in external bodies are necessary to bring the system to its original state, we do not mean a state in which each particle shall occupy more or less exactly the same position as at some previous epoch, but only a state which shall be undistinguishable from the previous one in its sensible properties. It is to states of systems thus incompletely defined that the problems of thermodynamics relate."
 * 5) Gibbs then goes on to consider the hypothetical situation of having two gases that are distinct (i.e. that can be later separated) but are so similar that for the dynamics of the mixing are exactly the same as the dynamics of mixing two same gases. On the atomic scale these processes would appear identical and yet the entropy would increase in one case and not the other. Gibbs says "In such respects, entropy stands strongly contrasted with energy", meaning that we cannot identify entropy with any kind of microscopic dynamics. Gibbs notes that the apparently irreversible dynamics of diffusion are not special to the distinct gas case; they are also just as well taking place in the mixing of same gases, however there is no entropy increase. To Gibbs, entropy has nothing to do with the motions of atoms but rather has to do with what we (as human observers) are able to distinguish in thermodynamic states.

 * 1) Gibbs says, if an ensemble is meant to represent an actual collection of systems, then the particles in some sense must be indistinguishable because each particle needs to be copied many times. On the other hand, when the ensemble is a probability distribution used to represent the possible states of one system, we might truly speak about unique particles. At this point Gibbs notes there is nothing formally wrong with having distinguishable particles or indistinguishable particles: "The question [whether particles are indistinguishable] is one to be decided in accordance with the requirements of practical convenience in the discussion of the problems with which we are engaged."
 * 2) Classical mechanics can be done equally well with distinguishable or indistinguishable particles. When considering indistinguishable particles, Gibbs notes that for integrating over the state space, the easiest way to do it is to pretend the particles are distinguishable, then divide the phase space integral by the appropriate overcounting factor. Gibbs is not mystified by why he should divide by N!, it is simply a mathematical convenience in order to perform calculations with indistinguishable particles. If one wanted to do it the hard way, the same results could be had by integrating over phase space including each distinguishable state once and only once. "For the analytical description of a specific phase is more simple than that of a generic phase. And it is a more simple matter to make a multiple integral extend over all possible specific phases than to make one extend without repetition over all possible generic phases."
 * 3) Gibbs then goes on to make tons of calculations for the grand canonical ensemble, using the framework of indistinguishable particles. Along the line he mentions why he is focussing on indistinguishable particles: "The interest of the ensemble which has been described lies in the fact that it may be in statistical equilibrium, both in respect to exchange of energy and exchange of particles, with other grand ensembles canonically distributed and having the same values of Θ and of the coefficients μ1, μ2, etc., when the circumstances are such that exchange of energy and of particles are possible, and when equilibrium would not subsist, were it not for equal values of these constants in the two ensembles." It's worth noting that his demonstration of equilibrium would not work out using the framework of distinguishable particles: he demonstrates this in equations (514) and (515).
 * 4) At the end of the chapter he works his way towards the measure of entropy, and finally remarks on the entropy difference (in the canonical ensemble) between considering particles distinguishable or indistinguishable. At the very end of this work (and apparently the very last paragraph he published before death), Gibbs mentions a sort of mixing paradox again, but this time it is regarding extensivity:
 * "For the principle that the entropy of any body has an arbitrary additive constant is subject to limitation, when different quantities of the same substance are concerned. In this case, the constant being determined for one quantity of a substance, is thereby determined for all quantities of the same substance. To fix our ideas, let us suppose that we have two identical fluid masses in contiguous chambers. The entropy of the whole is equal to the sum of the entropies of the parts, and double that of one part. Suppose a valve is now opened, making a communication between the chambers. We do not regard this as making any change in the entropy, although the masses of gas or liquid diffuse into one another, and although the same process of diffusion would increase the entropy, if the masses of fluid were different. It is evident, therefore, that it is equilibrium with respect to generic phases, and not with respect to specific, with which we have to do in the evaluation of entropy, and therefore, that we must use the average of H or of ηgen, and not that of η, as the equivalent of entropy, except in the thermodynamics of bodies in which the number of molecules of the various kinds is constant."

The 1876 work was entirely concerned with thermodynamics. The Gibbs' paradox of 1876 is therefore a pure thermodynamic thing. He does allude to molecular dynamics and probabilities but that is not essential to the argument. Whether entropy increases or not is, in the end, pragmatically determined by whether we need to perform work to return the systems back to their initial thermodynamic states. As to whether the individual particles themselves are distinguishable is not relevant to his argument here: there is anyway no thought to returning each particle back to its original position (returning the systems back to their initial microscopic state) as this would anyway be impossible. Rather it is only required in thermodynamics to return the systems back to their initial thermodynamic state. For what it's worth I think the Jaynes paper does an okay job explaining and extending Gibbs' arguments here.

To sum up Gibbs' remarks in his 1902 work, Gibbs starts out by noting that classical mechanics can be done equally well with distinguishable or indistinguishable particles. Both options can be considered in classical statistical mechanics, and it is really down to the details of the system we are studying, which one we choose. Gibbs then motivates why we must use indistinguishable particles, for the study of statistical thermodynamics, as so: First of all, if they were distinguishable we could not have such an easily defined grand canonical ensemble that is in equilibrium with respect to particle exchange. Secondly, if they were distinguishable then in the canonical ensemble the average logarithm of probability would not give us a quantity behaving like entropy, when the number of particles varies. The Jaynes paper says that Gibbs' 1902 book was "the work of an old man in rapidly failing health" and that Gibbs would have written more on this paradox if he had been in better health. Perhaps true, but I suspect Gibbs didn't want to distract from the rest of the discussion in the chapter.

Are there actually two Gibbs paradoxes? Should they be called "paradoxes"? I would say yes and yes. Gibbs addressed both the mixing-of-very-similar-gasses paradox in thermodynamics, and the mixing-of-identical-gases extensivity paradox in statistical mechanics. These are two different but closely related things, and there very well may be two Gibbs paradoxes, as discussed in the article. Gibbs did not see either as an actual problem, and he never called them paradoxes, yet he did remark on them as possible places where one could make a logical error. In some sense then they really are paradoxes, along the lines of the two-envelope paradox and the Monty Hall paradox: things that can seem on the surface mysterious or counterintuitive yet which have deep down a well-defined, consistent answer.

References and such
Wikipedia...

 * -defines a heat engine here.

 * -describes the Stirling heat engine here.

Topic page
This topic page is for organizing the development of Statistical thermodynamics content on Wikiversity. Please, feel free to improve upon what you see; your contributions will be greatly appreciated.

 * Search for Statistical mechanics at Wikiversity

Wikiversity Resources

 * Topic:Statistical mechanics
 * Statistical mechanics and thermodynamics (Based on probability and the counting of microstates)
 * Statistical thermodynamics (Based on the Clausius definition of entropy)
 * Topic:Advanced_Classical_Mechanics/Phase_Space (a work in progress that is relevant to statistical physics)
 * Basic thermodynamics
 * Thermodynamics and Thermodynamics_and_Equilibrium will soon be merged.
 * Search for Statistical thermodynamics at Wikiversity

Wikipedia/Wikibook Resources

 * Thermodynamics (Wikipedia article)
 * Statistical thermodynamics (Wikipedia article)
 * Category:Thermodynamics (HUGE Category of Wikipedia articles)
 * StatisticalMechanics (wiki-book)
 * Engineering Thermodynamics(wiki-book)

Advanced Resources

 * Thermodynamics OpenCourseWare from the University of Notre Dame. A rigorous and complete upper division course. License is CC BY-NC-SA 3.0. All lectures are PDF files.
 * Kittel and Kroemer's Thermal Physics. A very gentle introduction to an approach to thermodynamics which starts with statistical mechanics and then builds on those rules to derive the thermodynamical equations. This method is often much easier for physics students to understand, because you see a lot of the physical principles involved rather than just memorizing equations.

Miscellaneous
School of Physics

About learning projects and resources
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title: Statistical thermodynamics


text:
User:Tcapps3

Hi! My name is Taylor Capps and I attend UNCC in Charlotte, North Carolina. I hope to earn my elementary education degree and teach either 2nd or 3rd grade. I also work at a print design shop part-time. I am a Christian and I love my church. I also enjoy going to the gym and shopping.

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text: User:EmilyEleven

title: User:EmilyEleven


text:
Wikiversity talk:Main Page/Archive 7

GPRS
I WANT TO STUDY GPRS
 * Hello, thx for your interest. As I see in the search so far we have a few learning resources mentioning this. But you can visit Wikipedia to get more info on the subject: General Packet Radio Service. If you want you can use also an internet search engine to see if there are other facilities who offer such a topic. We would welcome you, if you also create a learning resource here with such a topic. See you, Erkan Yilmaz uses the Chat (try) 10:10, 20 July 2008 (UTC)

Humanities: Why the Chinese character for Middle?
The humanities section has the Chinese character ZHONG, meaning centre, as the symbol associated to it on the left. Does this have any background? Poeloq 03:40, 27 January 2008 (UTC)

I just noticed this too. But then again, the middle ain't too bad, right? -91.42.232.70 22:31, 24 February 2008 (UTC)

Traditionally the Chinese people follow the rules of middle way, which means not be extreme in any kind of activities.

this is connect with chinese culture

It's also the first character in the Chinese word for Chinese. The Jade Knight 15:47, 3 September 2008 (UTC)

This can be traced back long time ago to the ancient China history and the philosophy of life of common people. Being educated and advocating by culture, People have the idea not to do anything going to the extremes, but between them, which are accepted as the supreme conducting rules.--Bobozeke 03:09, 7 September 2008 (UTC)

 * Note: see also: 中庸.Hillgentleman|Talk 05:01, 7 September 2008 (UTC)

In the ancient China, the most developed part is the central region. At that time, people always called China “Zhongyuan”， which means the central part in Chinese. As time past, the central part became the economic, political and cultural center of China, so nowadays, we called the great country "ZhongGuo".

William.

Translation related question - Chinese speaking Wikiversity
I wish to know how to have a Chinese version wikiversity. if there is any help I can do for translation work, I would like to do. and thanks for those who wrote down all the knowledge in this and make it easy for us to learn.
 * 餵, a Chinese language Wikiversity is at the moment incubated at beta wikiversity, see 首页. More info there - see you soon. Erkan Yilmaz uses the Chat (try) 16:31, 8 September 2008 (UTC)

Main Page Success Story
I just want to congratulate the Main Page Workgroup on the excellent results of the Main Page 2.0 project. FInally, we have a front end worth the effort and intent that we all have put into this org. I especially like the Project of the Day section. I believe that is arguably the most important element that this org brings to the internet. Anyway, thanks a bunch. Gustable 02:36, 13 February 2008 (UTC)

Fraternities?
This is a cool idea, is there such a thing as a wikifraternity?
 * Please see also the discussion in the Colloquium, Erkan Yilmaz Wikiversity:Chat  17:18, 27 February 2008 (UTC)

Image Links
How to link image to a wikiversity internal file?????????????? Tushant Jha 07:14, 18 April 2008 (UTC)

 * If the image already exists on the English Wikiversity or on Commons, you can add it to a page by adding  [[Image:Name.jpg]]  (where "Name.jpg" is the full name of the image file). See Help:Media Files for more info. Cormaggio talk 12:11, 18 April 2008 (UTC)

Duplicated welcome
There are two "welcome to wikiversities", one on the header, the other in the introduction. Could these two sections be combined into one, saving some space? And the big purple picture, interesting as it is, may not be entirely useful, given that we already have a logo on the top, and educational pictures below? --Hillgentleman|Talk 22:38, 21 April 2008 (UTC)
 * Thanks for your comments, Hillgentleman. On issue 1 (two welcomes), I know it is redundant and inefficient, but on the other hand, redundancy and inefficiency isn't necessarily bad in educational or introductory contexts. If we changed the word "welcome" as the title of the blue welcome box, what could we change it to (suggestion, please)? On issue 2 (the picture in the top right corner), it is true that it has little purpose other than setting mood, adding interest and various aesthetic layout things. I don't really like the image that much myself - I'm rather neutral about it. On the other hand, if we optimized in the way you suggest, we'd end up with a layout like Portal:Tertiary Education, which looks less like the front page of a site and more like an internal page. A front page needs some extra whitespace, big titles and some real estate inefficiency, I think. But if you have an alternative, by all means let's have a look. --McCormack 08:25, 23 April 2008 (UTC)

Left Justify Text
Large amounts of text (more than one line) should be left justified to make it easier to read.

Example : Click on the image for a full size version which you can freely re-use and modify. Print it and use it for your lessons, integrate it into your pages on Wikiversity, or use it in other learning resources and websites. Use the links below to find more images like this one.

This should be left justified, not centered. Robert Elliott 16:12, 22 April 2008 (UTC)
 * Hi Robert and thanks for your comments about the new design. I have played around with the alignment of everything on the POTD boxes quite a lot, and I also have versions where everything is left-aligned and text wrapped. I eventually decided that if most things were centered (which looks better), then they all had to be centered, as mixing left alignments with centering looked bad. If you like, I can show you a few examples so that you can see what I mean. --McCormack 08:17, 23 April 2008 (UTC)

Article on main page is conveying an error
The WV main page currently features an image portraying the expansion of the universe. It does not state that this expansion is merely theoretic and makes the reader believe it is a very well proven evidence with a certain explanation. It should state that the expansion and the singular point in which the universe allegedly was are all theoretical. EugenSpierer 07:45, 23 April 2008 (UTC)
 * This comment refers to Educational Media Awareness Campaign/Astronomy/POTD 3. I will ask User:Mu301 (possibly has expertise in this field) to review the wording. Could you also state whether your comment is offered from a physicist's or a creationist's point of view (disclosure)? Thank you. --McCormack 08:15, 23 April 2008 (UTC)
 * -Creationist point of view. Though I am merely pointing out a biased text, not stating an opinion. EugenSpierer 08:18, 23 April 2008 (UTC)
 * Thanks for the disclosure! I'm not really an expert in this field, although I was the guilty party who wrote the blurb (rather fast). I'll ask User:Mu301 about this one. --McCormack 08:27, 23 April 2008 (UTC)
 * I've thought about this a little more. The pictures we show include other "disputed" events and "theories", such as the shoes of holocaust victims, climate change and the moon landing - but many more than this. On the whole I do not want to fill our main page with surrenders to a series of "it's just a theory" claims. What we can do instead is add a "discuss this picture" link at the bottom of the box (as we already do with the quote-of-the-day) where people can then voice their opinions at much greater length. --McCormack 08:48, 23 April 2008 (UTC)
 * The WP page discussing the Big Bang theory states that it is only a model by which the universe can be interpreted. However, the Holocaust page does not use such terms and relates to the holocaust as a certain historical genocide. These articles relate to events that are very different in public acceptance - one is highly accepted (the holocaust - even though it has its opponents) and the other has been disputed ever since its conception (the Big Bang theory). Presenting the picture with that blurb may affect any unknowing visitors to the site, and make them certain that the Big Bang is an undisputed historical event. EugenSpierer 09:44, 23 April 2008 (UTC)
 * A discussion link has now been added to all 100 picture pages. I'm sure this discussion can be continued there. --McCormack 10:38, 23 April 2008 (UTC)

Explore Wikiversity category tree default view...
A subtle change that may tweak things in a desirable manner may be to make all the categories in the "Explore Wikiversity" category tree be closed by default. I was just playing around and I did this and it seemed to make the other main category headings stand out substantially more. This could be due to the way I personally perceive things or this may be a common perception among us. Perhaps try it... --Remi 23:36, 11 May 2008 (UTC)

SUL
Hi, I don't know the right place to ask so here goes nothing: I'd like to usurp the account User:Grillo for my unified account. I'm sv:User:Grillo and I have already started the progress. The account in question only has three edits and seems to have no interest in the project. Regards Grillo sv 19:17, 4 June 2008 (UTC)
 * Please see Changing_username.
 * only 3 edits: this "only" can be viewn differently by others, see also here. Erkan Yilmaz uses the Wikiversity:Chat (try) PS: Tag a learning project with completion status !! 19:26, 4 June 2008 (UTC)

84522-37-2 molecular structure
84522-37-2 molecular structure

Hi,I am chelsea,working in a chemical trade platform([link see version history]).I have a problem at present.Could you please supply me with molecular structure of 84522-37-2,please?
 * moved to: Help desk, Erkan Yilmaz uses the Chat (try) 06:37, 18 July 2008 (UTC)

Some sort of create page function on the main page...
Wiktionary has an interesting way of doing this. Perhaps we could tailor something to Wikiversity in the future. Emesee 21:00, 22 August 2008 (UTC)
 * I guess we could add a box like e.g. at Portal:History (see: "Create a new history page") where then users could input the title of the learning resource they want. If it exists, it gets opened to edit. If it doesn't exist a new page is started - filled with some predefined content so the user can add her wished content for the learning resource.
 * A new learning resource can also be created when using the search box on the main page (below the WV icon), when the new user searches and then when she sees "There is no page titled "does this exist". You can create this page." and clicks the link "create this page" she can also create a learning resource. But that link offers no predefined content/help intro for the new users (but could be added). Or to the search box could be added a new option like "Create" ?
 * About predefined content which appears on new learning resources immediately: would we perhaps influence new users too much with adding predefined content ? Perhaps they would bring other ideas into the Wikiversity knowledge pool how to create learning resources. Erkan Yilmaz uses the Chat (try) 06:20, 23 August 2008 (UTC)

 * But there is already an inputbox on the left-hand side. Why would it be useful (instead of confusing) for newcomers to see a page of source wikicode instead of the actual output page? Hillgentleman|Talk 02:10, 9 September 2008 (UTC)

Link to Student Union Page
It has come to my attention that newcomers who maybe students should be redirected to the Student Union page, according to User:JWSchmidt some users have tried to place a such a link from the Main page to the student union but was prevented to do so - I feel that due to it being recently undergone a major update it should now be ready to be mentioned, unfortunately only Custodians can edit the page so could a Custodian please mention the Student union on the main page, I'm rather surprise it wasn't mentioned earlier on the main page. DarkMage 13:09, 16 September 2008 (UTC)

Move this page to Colloquium
This page should be talk about Main Page. The Main page learning project was one-purpose action and is now not suitable for general HP talk. Lets move this to Colloquium and make it about HP as it is elsewhere (e.g. Wikipedia).--Kozuch 19:29, 1 October 2008 (UTC)

title: Wikiversity talk:Main Page/Archive 7