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diff --git a/content/know/concept/cauchy-strain-tensor/index.pdc b/content/know/concept/cauchy-strain-tensor/index.pdc new file mode 100644 index 0000000..2994674 --- /dev/null +++ b/content/know/concept/cauchy-strain-tensor/index.pdc @@ -0,0 +1,332 @@ +--- +title: "Cauchy strain tensor" +firstLetter: "C" +publishDate: 2021-03-31 +categories: +- Physics +- Continuum physics + +date: 2021-03-31T09:43:40+02:00 +draft: false +markup: pandoc +--- + +# Cauchy strain tensor + +**Strain** quantifies the deformation of a solid object. +If the body has been deformed, e.g. by pulling or bending, +its constituent particles have moved a bit. +Let $\va{X}$ be the original location of a particle, +and $\va{x}$ its new location after the deformation. +We can thus define the **displacement field** $\va{u}$: + +$$\begin{aligned} + \va{u} + \equiv \va{x} - \va{X} +\end{aligned}$$ + +We restrict ourselves to **infinitesimal strain**, +where $\va{u}$ is so tiny that the material's properties are unchanged, +and a **slowly-varying strain**, +where the particle's neighbourhood has been distorted, +but not completely changed. + +A key challenge when quantifying deformation +is that we need to somehow exclude movements of the *entire* body: +for example, you can bend a twig in your hands while walking or dancing, +but we are only interested in the twig's shape change, +not in your movements. +The above definition of $\vu{u}$ includes both, +so we should be careful how we extract the strain from it. + + +## Definition + +We use the **Eulerian description** of deformation, +where the new position $\va{x}$ is the reference, +and the old position $\va{X}$ is expressed as a function of $\va{x}$: + +$$\begin{aligned} + \va{u}(\va{x}) + \equiv \va{x} - \va{X}(\va{x}) +\end{aligned}$$ + +Let us choose two nearby points in the deformed solid, +and call them $\va{x}$ and $\va{x} + \va{a}$, +where $\va{a}$ is a tiny vector pointing from one to the other. +Before the displacement, these points respectively had these positions, +where we define $\va{A}$ as the "old" version of $\va{a}$: + +$$\begin{aligned} + \va{X} = \va{X}(\va{x}) + \qquad + \va{X} + \va{A} = \va{X}(\va{x} + \va{a}) +\end{aligned}$$ + +Because the new positions $\va{x}$ are our reference, +we would like to write $\va{A}$ without $\va{X}$. +To do so, we use the definition of $\va{u}(\va{x})$, yielding: + +$$\begin{aligned} + \va{A} + &= \va{X}(\va{x} + \va{a}) - \va{X}(\va{x}) + \\ + &= \big( \va{x} + \va{a} - \va{u}(\va{x} + \va{a}) \big) - \big( \va{x} - \va{u}(\va{x}) \big) + \\ + &= \va{a} - \va{u}(\va{x} + \va{a}) - \va{u}(\va{x}) +\end{aligned}$$ + +Using the fact that $\va{a}$ is tiny by definition, +we expand the middle term to first order in $\va{a}$: + +$$\begin{aligned} + \va{u}(\va{x} + \va{a}) + \approx \va{u}(\va{x}) + a_x \pdv{\va{u}}{x} + a_y \pdv{\va{u}}{y} + a_z \pdv{\va{u}}{z} + = \va{u}(\va{x}) + \va{a} \cdot \nabla \va{u}(\va{x}) +\end{aligned}$$ + +With this, we can now define the "shift" $\delta\va{a}$ +as the difference between $\va{a}$ and $\va{A}$ like so: + +$$\begin{aligned} + \delta{\va{a}} + \equiv \va{a} - \va{A} + = \va{a} \cdot \nabla \va{u}(\va{x}) +\end{aligned}$$ + +In index notation, we write this expression as follows, +with $\nabla_j = \pdv*{x_j}$ simply being the partial derivative +with respect to the $j$th coordinate: + +$$\begin{aligned} + \delta a_i + = \sum_{j} a_j \nabla_j u_i +\end{aligned}$$ + +Where $\nabla_j u_i$ are called the **displacement gradients**, +and are just one step away from the desired definition of strain. +Note that these gradients are dimensionless, +so we can more formally define a *slowly-varying* displacement $\va{u}(\va{x})$ +as one where $|\nabla_j u_i| \ll 1$. + +Now, to solve the problem of macroscopic movements, +we take another tiny vector $\va{b}$ starting in the same point $\va{x}$ as $\va{a}$. +Here is the trick: if the whole body is uniformly translated or rotated, +the scalar product $\va{a} \cdot \va{b}$ is unchanged, +but if there is a non-uniform distortion, it changes. +We thus define the scalar product's difference like so: + +$$\begin{aligned} + \delta(\va{a} \cdot \va{b}) + \equiv \va{a} \cdot \va{b} - \va{A} \cdot \va{B} +\end{aligned}$$ + +Where $\va{B}$ is the old version of $\va{b}$. +Since these vectors are all tiny, we apply the product rule: + +$$\begin{aligned} + \delta(\va{a} \cdot \va{b}) + &= \delta\va{a} \cdot \va{b} + \va{a} \cdot \delta\va{b} +\end{aligned}$$ + +It is more informative to switch to index notation here. +Inserting $\delta\va{a}$ and $\delta\va{b}$ yields: + +$$\begin{aligned} + \delta(\va{a} \cdot \va{b}) + &= \sum_{i} \delta{a}_i \: b_i + \sum_{i} \delta{b}_i \: a_i + \\ + &= \sum_{ij} \nabla_j u_i \: a_j b_i + \sum_{ij} \nabla_j u_i \: a_i b_j + \\ + &= \sum_{ij} \big( \nabla_i u_j + \nabla_j u_i \big) \: a_i b_j +\end{aligned}$$ + +At last, we define the **Cauchy infinitesimal strain tensor** $\hat{u}$ +such that it has $u_{ij}$ as components: + +$$\begin{aligned} + \boxed{ + u_{ij} + \equiv \frac{1}{2} \big( \nabla_j u_i + \nabla_i u_j \big) + } +\end{aligned}$$ + +Which allows us to rewrite the shift of the scalar product in the following compact way: + +$$\begin{aligned} + \delta(\va{a} \cdot \va{b}) + &= 2 \sum_{ij} u_{ij} a_i b_j + = 2 \va{a} \cdot \hat{u} \cdot \va{b} +\end{aligned}$$ + +The Cauchy strain tensor $\hat{u}$ is a second-order tensor, +and can alternatively be expressed like so: + +$$\begin{aligned} + \boxed{ + \hat{u} + \equiv \frac{1}{2} \big( \nabla \va{u} + (\nabla \va{u})^\top \big) + } +\end{aligned}$$ + +Where $\top$ is the transpose. Being defined from the scalar product, +all macroscopic movements of the body are removed from the tensor, +which turns out to make it symmetric, i.e. $u_{ij} = u_{ji}$. + + +## Geometry + +So far we have used Cartesian coordinates, +but we can choose any three vectors $\va{a}$, $\va{b}$ and $\va{c}$, +and **project** $\hat{u}$ onto this basis. +For example, the component $u_{ab}$ then becomes: + +$$\begin{aligned} + \boxed{ + u_{ab} + = \frac{\va{a} \cdot \hat{u} \cdot \va{b}}{\big|\va{a}\big| \big|\va{b}\big|} + } +\end{aligned}$$ + +And so forth, for the other eight components. +The basis in which $\hat{u}$ is diagonal is the one formed by its eigenvectors, +and their directions are the **principal axes of strain** +at that point in the solid. +Because $\hat{u}$ is symmetric, such a basis always exists. + +Given a vector $\va{a}$, its relative length change +due to the deformation is simply given by: + +$$\begin{aligned} + \boxed{ + \frac{\delta|\va{a}|}{|\va{a}|} + = u_{aa} + } +\end{aligned}$$ + +To find the angle change $\delta\theta$ +between two vectors $\va{a}$ and $\va{b}$, +we start with the product rule: + +$$\begin{aligned} + \delta(\va{a} \cdot \va{b}) + = \delta(\big|\va{a}\big| \big|\va{b}\big| \cos\theta) + = \delta\big|\va{a}\big| \big|\va{b}\big| \cos\theta + + \big|\va{a}\big| \delta\big|\va{b}\big| \cos\theta + - \big|\va{a}\big| \big|\va{b}\big| \sin\theta \: \delta\theta +\end{aligned}$$ + +We isolate this for $\delta\theta$, using the fact that +$\delta(\va{a} \cdot \va{b}) = 2 \big|\va{a}\big| \big|\va{b}\big| u_{ab}$ +thanks to the projection $u_{ab}$: + +$$\begin{aligned} + \delta\theta + = \frac{\delta\big|\va{a}\big| \big|\va{b}\big| \cos\theta + + \big|\va{a}\big| \delta\big|\va{b}\big| \cos\theta + - 2 \big|\va{a}\big| \big|\va{b}\big| u_{ab}} + {\big|\va{a}\big| \big|\va{b}\big| \sin\theta} +\end{aligned}$$ + +By recognizing the length change $\delta|\va{a}|/|\va{a}| = u_{aa}$, +we arrive at the following expression: + +$$\begin{aligned} + \boxed{ + \delta\theta + = \frac{(u_{aa} + u_{bb}) \cos\theta - u_{ab}}{\sin\theta} + } +\end{aligned}$$ + +Now, everything so far has been about tiny vectors, +so the change of the line element $\dd{\va{l}}$ +is easy to express using the displacement field $\va{u}$: + +$$\begin{aligned} + \boxed{ + \delta(\dd{\va{l}}) + = \dd{\va{l}} \cdot \nabla \va{u} + = (\nabla \vec{u})^\top \cdot \dd{\va{l}} + } +\end{aligned}$$ + +Next, we calculate the change of the differential volume element $\dd{V}$ +by treating it as the volume of a tiny parallelepiped +spanned by $\va{a}$, $\va{b}$ and $\va{c}$: + +$$\begin{aligned} + \delta(\dd{V}) + = \delta(\va{a} \cross \va{b} \cdot \va{c}) + &= \delta\va{a} \cross \va{b} \cdot \va{c} + \va{a} \cross \delta\va{b} \cdot \va{c} + \va{a} \cross \va{b} \cdot \delta\va{c} + \\ + &= (\va{a} \cdot \nabla\va{u}) \cross \va{b} \cdot \va{c} + + \va{a} \cross (\va{b} \cdot \nabla\va{u}) \cdot \va{c} + + \va{a} \cross \va{b} \cdot (\va{c} \cdot \nabla\va{u}) +\end{aligned}$$ + +We can reorder the factors like so +(write it out in index notation if you are not convinced): + +$$\begin{aligned} + \delta(\dd{V}) + &= (\va{a} \cdot \nabla) \va{u} \cross \va{b} \cdot \va{c} + + (\va{b} \cdot \nabla) \va{a} \cross \va{u} \cdot \va{c} + + (\va{c} \cdot \nabla) \va{a} \cross \va{b} \cdot \va{u} +\end{aligned}$$ + +By applying a couple of vector identities, +we can rewrite this more compactly as follows: + +$$\begin{aligned} + \delta(\dd{V}) + &= \Big( \va{b} \cross \va{c} (\va{a} \cdot \nabla) \cross \va{b} + + \va{c} \cross \va{a} (\va{b} \cdot \nabla) + + \va{a} \cross \va{b} (\va{c} \cdot \nabla) \Big) \cdot \va{u} + \\ + &= (\va{a} \cross \va{b} \cdot \va{c}) (\nabla \cdot \va{u}) +\end{aligned}$$ + +Here, we recognize the definition of $\dd{V}$, +leading to the following infinitesimal volume change: + +$$\begin{aligned} + \boxed{ + \delta(\dd{V}) + = \nabla \cdot \va{u} \dd{V} + } +\end{aligned}$$ + +Finally, for the surface element $\dd{\va{S}} = \va{a} \cross \va{b}$, +we use that the volume element $\dd{V} = \va{c} \cdot \dd{\va{S}}$: + +$$\begin{aligned} + \delta(\dd{V}) + = \delta(\va{c} \cdot \dd{\va{S}}) + = \delta\va{c} \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}}) + = (\va{c} \cdot \nabla\va{u}) \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}}) +\end{aligned}$$ + +By comparing this to the previous result for $\delta(\dd{V})$, +we arrive at the following equation: + +$$\begin{aligned} + \nabla \cdot \va{u} (\va{c} \cdot \dd{\va{S}}) + = (\va{c} \cdot \nabla\va{u}) \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}}) +\end{aligned}$$ + +Since $\va{c}$ is dot-multiplied at the front of each term, +we remove it, and isolate the rest for $\delta(\dd{\va{S}})$: + +$$\begin{aligned} + \boxed{ + \delta(\dd{\va{S}}) + = \big( (\nabla \cdot \va{u}) \va{1} - \nabla \va{u} \big) \cdot \dd{\va{S}} + } +\end{aligned}$$ + + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. diff --git a/content/know/concept/cauchy-stress-tensor/index.pdc b/content/know/concept/cauchy-stress-tensor/index.pdc index 6c7c97a..a26e2a8 100644 --- a/content/know/concept/cauchy-stress-tensor/index.pdc +++ b/content/know/concept/cauchy-stress-tensor/index.pdc @@ -171,7 +171,7 @@ $$\begin{aligned} \end{aligned}$$ For some people, this equation may be more enlightening in index notation, -where $\vec{\sigma}_i$ is the $i$th row of the tensor (implicitly transposed): +where $\nabla_j = \pdv*{x_j}$ is the partial derivative with respect to the $j$th coordinate: $$\begin{aligned} F_{s, i} diff --git a/content/know/concept/euler-equations/index.pdc b/content/know/concept/euler-equations/index.pdc new file mode 100644 index 0000000..37d2fea --- /dev/null +++ b/content/know/concept/euler-equations/index.pdc @@ -0,0 +1,187 @@ +--- +title: "Euler equations" +firstLetter: "E" +publishDate: 2021-03-31 +categories: +- Physics +- Fluid mechanics +- Fluid dynamics + +date: 2021-03-31T19:04:17+02:00 +draft: false +markup: pandoc +--- + +# Euler equations + +The **Euler equations** are a system of partial differential equations +that govern the movement of **ideal fluids**, +i.e. fluids without viscosity. +There exist several forms, depending on +the surrounding assumptions about the fluid. + + +## Incompressible fluid, uniform density + +In a fluid moving according to the velocity vield $\va{v}(\va{r}, t)$, +the acceleration felt by a particle is given by +the **material acceleration field** $\va{w}(\va{r}, t)$, +which is the [material derivative](/know/concept/material-derivative/) of $\va{v}$: + +$$\begin{aligned} + \va{w} + \equiv \frac{\mathrm{D} \va{v}}{\mathrm{D} t} + = \pdv{\va{v}}{t} + (\va{v} \cdot \nabla) \va{v} +\end{aligned}$$ + +This infinitesimal particle obeys Newton's second law, +which can be written as follows: + +$$\begin{aligned} + \va{w} \dd{m} + = \va{w} \rho \dd{V} + = \va{f^*} \dd{V} +\end{aligned}$$ + +Where $\dd{m}$ and $\dd{V}$ are the particle's mass volume, +and $\rho$ is the fluid density, which we assume, in this case, to be constant in space and time. +Then the **effective force density** $\va{f^*}$ represents the net force-per-particle. +By dividing the law by $\dd{V}$, we find: + +$$\begin{aligned} + \rho \va{w} + = \va{f^*} +\end{aligned}$$ + +Next, we want to find another expression for $\va{f^*}$. +We know that the overall force $\va{F}$ on an arbitrary volume $V$ of the fluid +is the sum of the gravity body force $\va{F}_g$, +and the pressure contact force $\va{F}_p$ on the enclosing surface $S$. +Using Gauss' theorem, we then find: + +$$\begin{aligned} + \va{F} + = \va{F}_g + \va{F}_p + = \int_V \rho \va{g} \dd{V} - \oint_S p \dd{\va{S}} + = \int_V (\rho \va{g} - \nabla p) \dd{V} + = \int_V \va{f^*} \dd{V} +\end{aligned}$$ + +Where $p(\va{r}, t)$ is the pressure field, +and $\va{g}(\va{r}, t)$ is the gravitational acceleration field. +Combining this with Newton's law, we find the following equation for the force density: + +$$\begin{aligned} + \va{f^*} + = \rho \va{w} + = \rho \va{g} - \nabla p +\end{aligned}$$ + +Dividing this by $\rho$, +we get the first of the system of Euler equations: + +$$\begin{aligned} + \boxed{ + \va{w} + = \frac{\mathrm{D} \va{v}}{\mathrm{D} t} + = \va{g} - \frac{\nabla p}{\rho} + } +\end{aligned}$$ + +The last ingredient is **incompressibility**: +the same volume must simultaneously +be flowing in and out of an arbitrary enclosure $S$. +Then, by Gauss' theorem: + +$$\begin{aligned} + 0 + = \oint_S \va{v} \cdot \dd{\va{S}} + = \int_V \nabla \cdot \va{v} \dd{V} +\end{aligned}$$ + +Since $S$ and $V$ are arbitrary, +the integrand must vanish by itself everywhere: + +$$\begin{aligned} + \boxed{ + \nabla \cdot \va{v} = 0 + } +\end{aligned}$$ + +Combining this with the equation for $\va{w}$, +we get a system of two coupled differential equations: +these are the Euler equations for an incompressible fluid +with spatially uniform density $\rho$: + +$$\begin{aligned} + \boxed{ + \frac{\mathrm{D} \va{v}}{\mathrm{D} t} + = \va{g} - \frac{\nabla p}{\rho} + \qquad \quad + \nabla \cdot \va{v} + = 0 + } +\end{aligned}$$ + + +## Incompressible fluid, variable density + +The above form is straightforward to generalize to incompressible fluids +with non-uniform spatial densities $\rho(\va{r}, t)$. +In other words, these fluids are "lumpy" (variable density), +but the size of their lumps does not change (incompressibility). + +To update the equations, we demand conservation of mass: +the mass evolution of a volume $V$ +is equal to the mass flow through its boundary $S$. +Applying Gauss' theorem again: + +$$\begin{aligned} + 0 + = \dv{t} \int_V \rho \dd{V} + \oint_S \rho \va{v} \cdot \dd{\va{S}} + = \int_V \dv{\rho}{t} + \nabla \cdot (\rho \va{v}) \dd{V} +\end{aligned}$$ + +Since $V$ is arbitrary, the integrand must be zero. +This leads to the following **continuity equation**, +to which we apply a vector identity: + +$$\begin{aligned} + 0 + = \dv{\rho}{t} + \nabla \cdot (\rho \va{v}) + = \dv{\rho}{t} + \va{v} \cdot \nabla \rho + \rho (\nabla \cdot \va{v}) +\end{aligned}$$ + +Thanks to incompressibility, the last term disappears, +leaving us with a material derivative: + +$$\begin{aligned} + \boxed{ + 0 + = \frac{\mathrm{D} \rho}{\mathrm{D} t} + = \dv{\rho}{t} + \va{v} \cdot \nabla \rho + } +\end{aligned}$$ + +Putting everything together, Euler's system of equations +now takes the following form: + +$$\begin{aligned} + \boxed{ + \frac{\mathrm{D} \va{v}}{\mathrm{D} t} + = \va{g} - \frac{\nabla p}{\rho} + \qquad + \nabla \cdot \va{v} + = 0 + \qquad + \frac{\mathrm{D} \rho}{\mathrm{D} t} + = 0 + } +\end{aligned}$$ + + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. diff --git a/content/know/concept/material-derivative/index.pdc b/content/know/concept/material-derivative/index.pdc index 36113cc..af65ca0 100644 --- a/content/know/concept/material-derivative/index.pdc +++ b/content/know/concept/material-derivative/index.pdc @@ -93,7 +93,7 @@ $$\begin{aligned} } \end{aligned}$$ -Where the advective term is to be evaluated in the following way: +Where the advective term is to be evaluated in the following way in Cartesian coordinates: $$\begin{aligned} \va{v} \cdot \nabla \va{U} |