Lancaster and Blundell Chapter 1

I will be documenting my answers to exercises in this textbook.

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(1.1) Snell’s law from the principle of least time

Speeds and indices of refraction , with angles of incidence and transmission to the normal of the interface . The light travels on the left of the interface in the direction of the interface normal from , and on the right. The total time is $$T=\frac{1}{v_1}\sqrt{d_1^2+(y_1-y)^2}+\frac{1}{v_2}\sqrt{d_2^2+(y-y_2)^2}$$

To get the minimum, we write .

$$\frac{dT}{dy}=\frac{1}{v_1}\frac{y-y_1}{\sqrt{d_1^2+(y_1-y)^2}}+\frac{1}{v_2}\frac{y-y_2}{\sqrt{d_2^2+(y_2-y)^2}}=0$$

Assuming WLOG that , this is exactly Snell’s law.

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(1.2) Finding functional derivatives

$$H[f]=\int G(x,y)f(y)dy$$

We are asked to evaluate:

$$\frac{\delta H[f]}{\delta f(z)}=\lim_{\epsilon\rightarrow 0}\frac1\epsilon\left(\int G(x,y)(f(y)+\epsilon \delta(z-y)-f(y))\right)dy=G(x,z)$$

$$I[f]=\int_{-1}^1 f(x)dx$$

We are asked to evaluate:

$$\frac{\delta^2 I[f^3]}{\delta f(x_0)\delta f(x_1)}$$

Differentiating with respect to first,

$$\frac{\delta I[f^3]}{\delta f(x_1)}=\lim_{\epsilon\rightarrow 0}\frac1\epsilon\left(\int_{-1}^1 (f(x)+\epsilon\delta(x-x_1))^3-f(x)^3\right)dx$$

From the binomial expansion, the only term which is nonzero after taking the limit is :

$$=\int_{-1}^13f(x)^2\delta(x-x_1)dx=3f(x_1)^2$$

with the condition that .

Differentiating with respect to ,

$$\frac{\delta^2 I[f^3]}{\delta f(x_0)\delta f(x_1)}=\lim_{\epsilon\rightarrow 0}\frac1\epsilon\left(3(f(x_1)+\epsilon\delta(x_1-x_0))^2-f(x_1)^2\right)$$

$$=6f(x_1)\delta(x_1-x_0)$$

Since is if , we could also have written . Functional derivatives are commutative, as expected. Again, we require that .

$$J[f]=\int\left(\frac{\partial f}{\partial y}\right)^2dy$$

We are asked to evaluate:

$$\frac{\delta J[f]}{\delta f(x)}=\lim_{\epsilon\rightarrow 0}\frac1\epsilon\int\left(\left[\frac{\partial}{\partial y}\left(f(y)+\epsilon \delta(x-y)\right)\right]^2-\left(\frac{\partial f}{\partial y}\right)^2\right)dy$$

$$=\lim_{\epsilon\rightarrow 0}\frac1\epsilon\int\left(2\frac{\partial f}{\partial y}\epsilon\frac{\partial}{\partial y}\delta(x-y)\right)$$

Integrating by parts,

$$=2\frac{\partial f}{\partial y}\delta(x-y)\bigg\vert_{\text{boundaries}}-2\int \frac{\partial^2 f}{\partial y^2}\delta(x-y)dy=-2\frac{\partial^2 f}{\partial y^2}(x)$$

with the usual caveats that the boundary term is , and is within the integration region.

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(1.3) Consider the functional . Show that $$\frac{\delta G[f]}{\delta f(x)}=\frac{\partial g(x,f)}{\partial f}$$

$$\frac{\delta G[f]}{\delta f(x)}=\frac{\partial g(x,f)}{\partial f}=\lim_{\epsilon\rightarrow 0}\frac1\epsilon\int\left(g(y,f+\epsilon\delta(x-y))-g(y,f)\right)dy$$

We perform a Taylor expansion in the second variable to first order in , since the higher order terms have limit .

$$=\lim_{\epsilon\rightarrow 0}\frac1\epsilon\int\left(g(y,f)+\epsilon\delta(x-y)\frac{\partial g(y,f)}{\partial f}-g(y,f)\right)dy=\frac{\partial g(x,f)}{\partial f}$$

Now consider the functional and show that $$\frac{\delta H[f]}{\delta f(x)}=\frac{\delta g}{\delta f}-\frac{d}{dx}\frac{\partial g}{\partial f’}$$ Similarly for .

These facts are fairly obvious from the last part of the previous exercise, since requires an integration by parts, and we just perform the Taylor expansion in the new arguments.

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(1.4) Show that $$\frac{\delta \phi(x)}{\delta \phi(y)}=\delta(x-y)$$ and $$\frac{\delta \dot\phi(t)}{\delta\phi(t_0}=\frac{d}{dt}\delta(t-t_0)$$

$$\frac{\delta \phi(x)}{\delta \phi(y)}=\lim_{\epsilon\rightarrow 0}\frac1\epsilon(\phi(x)+\epsilon\delta(x-y)-\phi(x))=\delta(x-y)$$

$$\frac{\delta \dot\phi(t)}{\delta\phi(t_0)}=\lim_{\epsilon\rightarrow 0}\frac1\epsilon\frac{d}{dt}(\phi(t)+\epsilon\delta(t-t_0)-\phi(t))=\frac{d}{dt}\delta(t-t_0)$$

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(1.5) For a three-dimensional elastic medium, the potential energy is $$V=\frac{\mathcal T}{2}\int d^3x(\nabla \psi)^2$$ and the kinetic energy is $$T=\frac{\rho}{2}\int d^3x\left(\frac{\partial \psi}{\partial t}\right)^2$$ Derive the wave equation.

The principle of stationary action gives us

$$\frac{\delta S[\psi]}{\delta\psi(x,t)}=\frac{\delta}{\delta\psi(x,t)}(T-V)=0$$

The derivative for gives

$$-\rho\frac{\partial^2 \psi}{\partial t^2}$$

The derivative for gives

$$-\mathcal T\nabla^2\psi$$

Imposing stationary action,

$$\nabla^2\psi=\frac{\rho}{\mathcal T}\frac{\partial^2 \psi}{\partial t^2}$$

and the speed of propagation is .

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(1.6) Show that if is given by $$Z_0[J]=\exp\left(-\frac12\int d^4xd^4yJ(x)\Delta(x-y)J(y)\right)$$ with then $$\frac{\delta Z_0[J]}{\delta J(z_1)}=-\left[\int d^4y\Delta(z_1-y)J(y)\right]Z_0[J]$$

As usual, we write

$$\frac{\delta Z_0[J]}{\delta J(z_1)}=\lim_{\epsilon\rightarrow 0}\frac1\epsilon\exp\left(-\frac12\int d^4xd^4y\left((J(x)+\epsilon\delta(z_1-x))\Delta(x-y)(J(y)+\epsilon\delta(z_1-y))-J(x)\Delta(x-y)J(y)\right)\right)$$

Ignoring the order term which dies when we take the limit, we indeed get the quoted answer above.