Vocabulary

Gaussian Intergral

\[\int_{-\infty}^{\infty} e^{-ax^2} \mathrm dx = \sqrt{\frac{\pi}{a}}\]

A comprehensive description of how this is done is here.

Basicly, we just transform the problem to the \(\int e^{-x} \mathrm dx\) problem.

Behavior of Functions

Note

To see the true nature of graphs, make quantities dimensionless. This is also true for theoretical derivations. Dimensionless equations can reveal more.

  1. Boltzmann factor

    The most important and weirdest function in statmech

    ../_images/boltzfactor1.png

    Fig. 1 The nth derivative of this function is always 0 at x=0, for all finite n. Then how does it rise? The only thing I can say is that we are actually dealing with infinite n.

    Professor Kenkre: sleeping lion!

  2. Tanh(x)

    ../_images/tanh.jpg
  3. \(1-exp(-x)\)

    ../_images/exp1.jpg
  4. \(cosh(1/x)-1/x\)

    ../_images/cosh1.jpg
  5. \(1/(1+1/x)\)

    ../_images/fraction1.jpg

    Examples in Physics

    An example of this \(1/(1+1/x)\) is the modified gas model.

    \[P (V - b) = N k T\]

    We can find out \(1/V\), which is

    \[\frac{1}{V} = \frac{1}{b+\frac{N k T}{P}}\]

    Now we can plot out \(\frac{1}{V} ~ P\) and it shows a behavior just like \(1/(1+1/x)\).

Fourier Transform

Fourier transform for continuous equation is

\[\begin{split}\frac{\partial}{\partial x} e^{ikx}=ike^{ikx} &\implies \frac{\partial}{\partial x} \to ik \\ \frac{\partial^2}{\partial x^2} e^{ikx} = -k^2 e^{ikx} & \implies \frac{\partial^2}{\partial x^2} \to -k^2\end{split}\]

Laplace Transform

Laplace transform is a transform of a function \(f(t)\) to a function of \(s\),

\[L[f(t)] = \int_0^\infty f(t) e^{ - s t} dt .\]

Some useful properties:

  1. \(L[\frac{d}{dt}f(t)] = s L[f(t)] - f(0)\);
  2. :math:`L[frac{d^2}{dt^2}f(t) = s^2 L[f(t)] - s f(0) - frac{d f(0)}{dt} `;
  3. \(L[\int_0^t g(\tau) d\tau ] = \frac{L[f(t)]}{s}\);
  4. \(L[\alpha t] = \frac{1}{\alpha} L[s/\alpha]\);
  5. \(L[e^{at}f(t)] = L[f(s-a)]\);
  6. \(L[tf(t)] = - \frac{d}{ds} L[f(t)]\).

Some useful results:

  1. \(L[1] = \frac{1}{s}\);
  2. \(L[\delta] = 1\);
  3. \(L[\delta^k] = s^k\);
  4. \(L[t] = \frac{1}{s^2}\);
  5. \(L[e^{at}]= \frac{1}[s-a]\).

A very nice property of Laplace transform is

\[\begin{split}L_s [e^{at}f(t)] &= \int_0^\infty e^{-st} e^{-at} f(t) dt \\ & = \int_0^\infty e^{-(s+a)t}f(t) dt \\ & = L_{s+a}[f(t)]\end{split}\]

which is very useful when dealing with master equations.

Two useful results are

\[L[I_0(2Ft)] = \frac{1}{\sqrt{ \epsilon^2 - (2F)^2 }}\]

and

\[L[J_0[2Ft]] = \frac{1}{\sqrt{\epsilon^2 + (2F)^2}},\]

where \(I_0(2Ft)\) is the modified Bessel functions of the first kind. \(J_0(2Ft)\) is its companion.

Using the property above, we can find out

\[L[I_0(2Ft)e^{-2Ft}] = \frac{1}{\sqrt{(\epsilon + 2F)^2 - (2F)^2}} .\]

Functions that will saturate

\[1-e^{-\alpha x} \tanh(x) \cosh(\frac{1}{x}) - \frac{1}{x}\]

Legendre Transform

The geometrical of physical meaning of Legendre transformation in thermodynamics can be illustrated by the following graph.

Legendre Transform made clear

Fig. 2 Legendre transform

For example, we know that entropy \(S\) is actually a function of temperature \(T\). For simplicity, we assume that they are monotonically related like in the graph above. When we are talking about the quantity \(T \mathrm d S\) we actually mean the area shaded with blue grid lines. Meanwhile the area shaded with orange line means \(S \mathrm d T\).

Let’s think about the change in internal energy which only the thermal part are considered here, that is,

\[\mathrm d U = T \mathrm d S .\]

So internal energy change is equal to the the area shaded with blue lines. Now think about a three dimensional graph with a third axis of internal energy which I can’t show here. Notice that the line of internal energy is on the plane which is vertical to \({T, S}\) plane and contains the line black line in the graph above. The change of internal energy with an increase of \(\mathrm dS\) is the value that the line of internal energy goes up.

Now we do such a transform that we actually remove the internal energy from \(\mathrm d ( T S )\), which finally gives us Helmholtz free energy,

\[\mathrm d A = S \mathrm d T .\]

It’s obvious that after this Legendre transform, the new area is the part shaded with orange lines.

Now the key point is that \(S(T)\) is a function of \(T\). So if we know the blue area then we can find out the orange area, which means that the two function \(A(T)\) and \(U(S)\) are identical. Choosing one of them for a specific calculation is a kind of freedom and won’t change the final results.

Refs & Note

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