In this lecture we cover the very basics of supersonic shock fronts and the physics behind them and look at applying these relations to an example of a supernova explosion.
Shocks occur when there is an object traveling faster than the fluid velocity. The fluid velocity as we recall described from before is the change in the pressure (pressure wave) in response to the changes in density. This we define as: \[v_s^2 = \frac{\partial P}{\partial \rho}\] Since there is not enough time for the fluid to "react" to the shock as its traveling at supersonic speeds this introduces a jump condition and a boundary introduces two areas called a pre-shock (\(\rho_1, P_1,v_1,T_1 \cdots\)) and post-shock (\(\rho_2, P_2,v_2,T_2 \cdots\)). The goal: is to be able to relate pre and post shock quantities with one another. Often times in real life problems one might only have access to one or the other measurements and thus can reveal additional insight about the physics that occur. To do this we introduce a simplification call the Rankine-Hugoniot Relations
There are a number of assumptions. Firstly we assume the system is in steady state as most shocks tends to be meaning there is no accumulation of mass, momentum or energy in any particular area. We also assume it is adiabatic and thus has the Adibatic equation of state. This gives the first relation of mass \[\] Mass Relation \[\frac{\partial \rho}{\partial t} + \nabla \rho v = 0\] Steady state so no time derivative \[ \nabla \rho v = 0\] Assume this is 1-D problem... \[ \frac{\partial }{\partial x}\rho v = 0\] Notice there is no spatial change and thus across the shock front we have: \[\boxed{\rho_1 v_1 = \rho_2 v_2}\] Momentum Relation \[\frac{\partial v}{\partial t} + v \cdot \nabla v = \frac{-1}{\rho} \nabla P - \nabla \Phi\] Assume it steady state and that there is no gravtiy and it is 1-D \[v_x \cdot \frac{\partial }{\partial x}v_x = \frac{-1}{\rho} \frac{\partial }{\partial x} P \] \[\rho v_x\frac{\partial }{\partial x}v_x =-\frac{\partial }{\partial x} P \] Recall that \(\frac{\partial }{\partial x}\rho v = 0\) so we can add "0" into the equation which is \(v_x\frac{\partial }{\partial x}\rho v_x \) \[v_x\frac{\partial }{\partial x}\rho v_x + \rho v_x\frac{\partial }{\partial x}v_x =-\frac{\partial }{\partial x} P \] Reverse the product rule to get: \[\frac{\partial }{\partial x}\rho v_x^2 =-\frac{\partial }{\partial x} P \] \[\frac{\partial }{\partial x}[\rho v_x^2 + P] = 0 \] This means that \[\boxed{\rho_1 v_1^2 + P_1 = \rho_2 v_2^2 + P_2}\] Energy relation Recall that this is adiabatic and this \(P = k\rho^\gamma\). We also recall that internal energy is the following from first law \[dQ = d\epsilon + P dV\] Assume it is adiabatic \[0 = d\epsilon + P dV\] \[d\epsilon = -P dV\] Integrate \[\epsilon = - \int P dV\] \[\epsilon = - \int k\rho^\gamma dV\] We recall that this was proportional to \(V^{-\gamma}\) call that constant \(A\) \[\epsilon = - \int kA(V)^{-\gamma} dV\] \[\epsilon = -\frac{1}{-\gamma +1} k A(V)^{-\gamma+1} \] \[\epsilon = \frac{1}{\gamma - 1} k A(V)^{-\gamma}V \] Rewrite things interms of pressure and density \[\epsilon = \frac{1}{\gamma - 1} \frac{P}{\rho}\] Now recall the energy conservation \[\frac{d}{dx}\left(\rho(\frac{1}{2}v^2 + \epsilon) +P \right)v = - \rho \dot Q_{cool}\] Recall its adiabatic \[\frac{d}{dx}\left(\rho\frac{1}{2}v^2 + \rho\epsilon +\frac{P}{\rho} \right)v = 0\] \[\frac{d}{dx}\left(\rho\frac{1}{2}v^2 + \frac{\rho}{\gamma -1} \frac{P}{\rho}+\frac{P}{\rho} \right)v = 0\] Collect the terms \[\frac{d}{dx}\left(\rho\frac{1}{2}v^2 + \frac{\gamma}{\gamma -1} \frac{P}{\rho}\right)v = 0\] This means that \[\rho_1 v_1\left(\frac{1}{2}v_1^2 + \frac{\gamma}{\gamma -1} \frac{P_1}{\rho_1}\right) = \rho_2 v_2\left(\frac{1}{2}v_2^2 + \frac{\gamma}{\gamma -1} \frac{P_2}{\rho_2}\right)\] Based on mass equations we can simplify both sides \[\left(\frac{1}{2}v_1^2 + \frac{\gamma}{\gamma -1} \frac{P_1}{\rho_1}\right) = \left(\frac{1}{2}v_2^2 + \frac{\gamma}{\gamma -1} \frac{P_2}{\rho_2}\right)\] We then recall that the definition of speed of sound is \[v_s^2 = \frac{\partial P}{\partial \rho} = k\frac{\partial \rho^\gamma}{\partial \rho} = \frac{\gamma P}{\rho}\] So we can replace again \[\boxed{\left(\frac{1}{2}v_1^2 + \frac{1}{\gamma -1}v_{1s}^2\right) = \left(\frac{1}{2}v_2^2 + \frac{1}{\gamma -1} v_{1s}^2\right)}\] Now our goal is to isolate and combine these equations such that we can relate each of these functions. Step 1: define conserved quantity \[\boxed{j \equiv \rho_1 v_1 = \rho_2 v_2}\] Step 2: momentum equation \[\rho_1 v_1^2 + P_1 = \rho_2 v_2^2 + P_2\] \[\rho_1 v_1^2 - \rho_2 v_2^2 = + P_2- P_1\] Recall \(\rho_1 v_1 = \rho_2 v_2 \Rightarrow v_1 = \frac{\rho_2 v_2}{\rho_1}\) \[\rho_1 (\frac{\rho_2 v_2}{\rho_1})^2 - \rho_2 v_2^2 = P_2- P_1\] \[\frac{\rho_2^2 v_2^2}{\rho_1}- \rho_2 v_2^2 = P_2- P_1\] \[\rho_2^2 v_2^2 (\frac{1 }{\rho_1} - \frac{1}{\rho_2}) = P_2- P_1\] \[\rho_2^2 v_2^2 = \frac{P_2- P_1}{\frac{1 }{\rho_1} - \frac{1}{\rho_2} }\] Recall that this is just the square \[\boxed{j^2 = \frac{P_2- P_1}{\frac{1 }{\rho_1} - \frac{1}{\rho_2} }}\] Step 3: Energy equations \[\boxed{\frac{\rho_2}{\rho_1} = \frac{(\gamma -1)P_1 + (\gamma +1)P_2}{(\gamma +1)P_1 + (\gamma -1)P_2} = \frac{v_1}{v_1}}\] This allows us to obtain the following equations that tell us about the relation within and outside the shock front (with a lot of algebra..) \[\boxed{\frac{\rho_2}{\rho_1} =\frac{v_1}{v_2} = \frac{\gamma +1}{\gamma -1 + \frac{2}{M^2_1}}} \] \[\boxed{\frac{P_2}{P_1} = \frac{2\gamma M_1^2 - (\gamma -1)}{\gamma +1 }} \] \[\boxed{M_2^2 = \frac{(\gamma -1)M_1^2+2}{2\gamma M_1^2- (\gamma -1) }} \]
Strong Shock Limit \(M_1 >> 1\) For the first equation \(\frac{2}{M^2_1} \to 0\) \[\boxed{\frac{\rho_2}{\rho_1} =\frac{v_1}{v_2} = \frac{\gamma +1}{\gamma -1}} \] We see the first term dominates \[\boxed{\frac{P_2}{P_1} = \frac{2\gamma M_1^2 }{\gamma +1 }} \] We see the domiate terms are the power \[M_2^2 = \frac{(\gamma -1)M_1^2}{2\gamma M_1^2 } \] \[\boxed{M_2^2 = \frac{\gamma -1}{2\gamma }} \] We also reconize that the temperature doesnt change throughout the front and thus can get the temperature as well \[\boxed{T_2 = \frac{2(\gamma -1)}{(\gamma +1)^2}\frac{\mu m_p}{k_B}v_1^2}\] This is from idea gas law.
These expressions are quite simple. Consider a blast wave traveling out in a sphere outwards radially. We want to know how does the shock front of the blast travel as related with energy etc. A blast wave is when all the material is concentrated in a thing shell which is the following: \[\frac{4}{3} \pi R^3 \rho_1 = \frac{4}{3} \pi \rho_2((R+D)^3 -R^3 ) \] If it is small thin shell then \[\frac{4}{3} \pi R^3 \rho_1 = \frac{4}{3} \pi \rho_2 R^3(\frac{(R+D)^3}{R^3} - 1) \] \[\frac{4}{3} \pi R^3 \rho_1 = \frac{4}{3} \pi \rho_2 R^3((1+\frac{D}{R})^3 - 1) \] \[\frac{4}{3} \pi R^3 \rho_1 \approx \frac{4}{3} \pi \rho_2 R^3((1+\frac{3D}{R}) - 1) \] \[\frac{4}{3} \pi R^3 \rho_1 \approx 4 \pi \rho_2 R^2 D \] Rearrange for the thickness \[D \approx \frac{R}{3} \frac{\rho_1}{\rho_2} = \frac{\gamma -1}{3(\gamma +1)}R\]
We see that the shell is thin. We can calculate some of its properties: \[\] Kinetic Energy \[\frac{1}{2} m v_1^2 =\frac{1}{2}\frac{4}{3} \pi R^3 \rho_1 \cdot \dot R^2 \propto \rho_1R^5 t^{-2}\] Internal Energy \[PV =\frac{1}{\gamma-1 }\frac{2}{\gamma +1}\rho_1v_1^2 \cdot 4\pi R^2 D\propto \rho_1R^5t^{-2}\] Total Energy \[E \propto \rho_1 R^5 t^{-2}\]
Solving we have the following: \[\boxed{R(t) = A (\frac{E}{\rho})^{1/5}t^{2/5}}\]
This only works when the following is satisfied:
We can then apply our problem to the death of stars in a supernova explosion. Let us try to solve for the radius of the explosion. \[\] Energy: can be derived from observation \[E = 10^{51}ergs \] Mass: can also be derived from observations such as binaries. In this case lets say its a massive star \[M = 10M_\odot\] Solve for velocity \[E = \frac{1}{2}Mv^2 \Rightarrow v = \sqrt{2E/M} = 5000km/s \approx 0.02c\] Let us say that the density of the shock is the same as the interstellar medium which we make out to be \[\rho = 10^{-24}g/cm^{3}\] This gives us the radius of expansion to be \[R(t) =0.3 pc (\frac{E}{10^{51}ergs})^{1/5} (\frac{\rho}{10^{-24}g/cm^{3}})^{-1/5} (\frac{t}{3\times 10^{7}s})^{2/5}\]
The interesting element about the Sedov-Taylor Expressions is the existence of self similar solutions. These solutions are places where there is no characteristic length scale meaning the solutions are scale invariant. For our equations you can choose to define a unitless quantity \(\xi\) where we have \[\xi = \frac{r}{R(t)}\] We can get the following unitless velocity , density,and pressure with unitless dependence: \[V(\xi) = \frac{5tv}{2r}, \quad G(\xi) = \frac{\rho}{\rho_0}, \quad Z(\xi) = \frac{25 t^2 c^2}{4r^2}\] By substituting these self-similar variables into the governing equations will lead to three ordinary differential equations which are really hard to solve so I will just not do them....