#zettel #physics #radiation/detection-and-imaging/czt-detectors sourced from: [[Shockley-Ramo Theorem]] and [[@heReviewShockleyRamo2001]] # Proof of the [[Shockley-Ramo Theorem]] ## Conservation of Energy (and Proof of the Theorem) Another key component of the Shockley-Ramo theorem is the idea that the energy in the system is conserved, as previously noted. As a note, Jim considers THIS concept in conjunction with the idea of superposition as the two major concepts related to Shockley-Ramo. The conservation of energy in this system is what proves the Shockley-Ramo Theorem. The energy density $w$ of an electric field through a medium, where the electric displacement is $\vec{D} = \varepsilon \cdot \vec{E}$, is $w = \frac{1}{2} \vec{E} \cdot \vec{D} = \frac{1}{2} \varepsilon E^2 $ The total energy in this electric field is the integral of energy density over the entire volume $\tau$, and from conservation of energy, this total energy of the field can only change by: 1. Energy exchanges between the electric field and moving charge $q$ in the form of kinetic energy 2. Energy exchanges between the field and power supplies. If we have a device configuration with a space charge $\rho$, as well as a singular charge $q$ in a detector with grounded electrodes, we have the third case which can play a role as well to an extent. Let's consider this space and determine the amount of work done by the electric field when a charge $q$ moves from $\vec{x_i}$ to $\vec{x_f}$. $ = \int_{\vec{x_i}}^{\vec{x_f}} q \vec{E'_1} \cdot d\vec{x} \tag{1} $ We note that $\vec{E'}$ is simply the the electric field, $\vec{E}$ excluding $qs own field. As the charge moves from $\vec{x_i}$ to $\vec{x_f}$, the induced charge on the electrodes in the space will redistribute accordingly, while remaining with a 0 potential as all electrodes are grounded. This therefore means that no work is done on the moving induced charges by the power supplies. From conservation of energy __the work done on the charge $q$ must come from the energy stored in the field__. Because of this, we can say that: $ \int_{\vec{x_i}}^{\vec{x_f}} q \vec{E'_1} \cdot d\vec{x} = \frac{1}{2} \int_{\tau} \varepsilon \left(E^2_{1i} - E^2_{1f} \right) d\tau $ where $E_{1i}$ and $E_{1f}$ are the electric fields in the grounded electrode system with space charge $\rho$ and charge $q$, when $q$ is at $\vec{x_i}$ and $\vec{x_f}$, respectively. From the superposition principle, we know that the total electric field is found by summing these electric field terms together such that $\vec{E} = \vec{E_0} + \vec{E_1}$ (this is derived from the case in which electrodes are all at individual voltages, $V_i$). We can use this sum to expand $\eqref{1}$ to say: $ \int_{\vec{x_i}}^{\vec{x_f}} q \vec{E'_1} \cdot d\vec{x} \implies \int_{\vec{x_i}}^{\vec{x_f}} q \left( \vec{E_0} + \vec{E'_1} \right) \cdot d\vec{x} \tag{2} $ <br/> When the induced charge $\Delta Q_i$ on each electrode moves between the electrode and the ground as a result of $q$ moving, the total work is a sum of the induced charges times the voltage on each respective electrode of the $K$ total electrodes: $ Work = \sum_{i=1}^{K} V_i\Delta Q_i \tag{3} $ This is all fine and good, but we need to consider how the conservation of energy plays a part in this whole calculation. As a result, we need to consider the TOTAL work done on the system from $\eqref{eq3}$ and what is removed from it, the calculation of the work done by the energy stored in the field from $\eqref{2}$. (Work done by power supplies - energy absorbed by the moving charge $q$) <br/> This difference should be equal to the increase of energy stored in the field. $ \sum_{i=1}^{K} V_i\Delta Q_i - \int_{\vec{x_i}}^{\vec{x_f}} q \left( \vec{E_0} + \vec{E'_1} \right) \cdot d\vec{x} = \frac{1}{2} \int_{\tau} \varepsilon \left[ \left(\vec{E_0} + \vec{E_{1f}}\right)^2 - \left(\vec{E_0} + \vec{E_{1i}}\right)^2 \right] d\tau \tag{4} $ From the right hand side of $\eqref{4}$, we can apply Green's First Identity, which is as shown: $ \int_V \left( \phi_1 \nabla^2 \phi_0 + \nabla\phi_0 \cdot \nabla\phi_1 \right) dV = \oint_S \phi_1 \nabla \phi_0 \cdot d\vec{S} $ <br/> From the case in which all electrodes are grounded but a charge $q$ and a space charge $\rho$ exist, we can say that $\phi_1 |_S = 0$ (this sets a boundary condition that all electrode surfaces are grounded) within $\tau$ (the detector volume). This therefore lets us use Green's First Identity to say $\int_{\tau} \vec{E_0} \cdot \vec{E_1} d\tau = \int_{\tau} \nabla\phi_0 \cdot \nabla\phi_1 $ $ = \oint_S \phi_1 \nabla \phi_0 \cdot d\vec{S} - \int_{\tau} \phi_1 \nabla^2 \phi_0 d\tau = 0 $ <br/> From this conclusion: $ \int_{\tau} \varepsilon \left[ \left(\vec{E_0} + \vec{E_{1f}}\right)^2 - \left(\vec{E_0} + \vec{E_{1i}}\right)^2 \right] d\tau = \int_{\tau} \varepsilon \left(E^2_{1i} - E^2_{1f} \right) d\tau \tag{5} $ (note that the $\frac{1}{2}$ term from both sides of the equation which were present in both sides in previous statements has been cancelled out). Equation $\eqref{5}$ simply says that when $q$ moves from $\vec{x_i}$ to $\vec{x_f}$, the change of energy of the electric field is the same between when there are potentials on electrodes and when the potentials are grounded. This means that the energy stored in the field is the sum of $\vec{E_0}$ and $\vec{E_1}$, and that $\vec{E_0}$ is held constant. With appropriate substitutions of Equation $\eqref{5}$ and the equation between $\eqref{1}$ and $\eqref{2}$ which I neglected to label out of poor pre-planning (oops :]), we arrive at the end result: the Shockley-Ramo Theorem: $ \sum_{i=1}^{K} V_i\Delta Q_i = \int_{\vec{x_i}}^{\vec{x_f}} q \vec{E_0} \cdot d\vec{x} = -q \left[\phi_0(\vec{x_f}) - \phi_0(\vec{x_i}) \right] \tag{6} $ where $\vec{E_0}$ and $\phi_0$ correspond to the electric field and potential when electrode $L$ is biased at a unit potential (which is dimensionless), all other potentials are grounds, and all charges are removed. This implies that al the potentials that $\phi_0$ may be will be between 0 and 1, hence the equipotential lines may instead be viewed as weights on a total voltage. we simply choose the unit potential ($V = 1$) for simplicity in calculation and simulation. The maximum induced charge, which is $-q$ on the weighting electrode $L$ comes when the charge $q$ is infinitely close to the electrode. This is a result of the [[Method of Images (Mirror Charge Technique)|Method of Images technique]] previously outlined. By setting $\phi_0(\vec{x}) = 0$ in $\eqref{eq6}$, we see that the induced charge on electrode $L$ is: $Q_L = -q \phi_0(\vec{x}) $ which completes the proof. $\square$