Reynolds Transport Theorem Derivation

Reynolds transport theorem can be derived using some mathematical relations. The derivation yields an equation governing the balance of a given quantity in a given control volume.

Frames of Reference and Derivative Definitions

Illustration of the changes undergone by a single fluid particle $ \xi $. The rate of change of an intensive property $ \mathcal{F} $ of the fluid particle is equal to the substantial (material) derivative $ \frac{d \mathcal{F}}{dt} $.

To begin, the initial position of a fluid particle (not a particle in the molecular sense, but a particle in a macroscopic, continuum sense) may be written as a random variable $ \boldsymbol{\xi} $

The location of this particular fluid particle at a later time is determined, first, by its initial location $ \boldsymbol{\xi} $, and second, by the time that has passed, $ t $

For a control volume with a fixed position $ \boldsymbol{x} $, multiple fluid particles, each with a different initial position $ \boldsymbol{\xi}_i $, pass through $ \boldsymbol{x} $ at different times $ t_i $. The rate of change, keeping $ \boldsymbol{x} $ constant, of an intensive fluid property $ \mathcal{F} $ is the partial derivative $ \frac{\partial \mathcal{F}}{\partial t} $.

This later position can be written as $ \boldsymbol{x}(\boldsymbol{\xi},t) $

We will assume that, given a location $ \boldsymbol{x} $, we can backtrack and find the fluid particle's initial location $ \boldsymbol{\xi} $

This means, the function $ \boldsymbol{x}(\boldsymbol{\xi},t) $ is assumed to be invertible.

In other words, we assume we can find a function $ \boldsymbol{\xi} = \boldsymbol{\xi}(\boldsymbol{x},t) $ that is continuous and single-valued (i.e. invertible).

The necessary and sufficient condition for invertability is for a non-vanishing Jacobian to exist:

$ J = \frac{ \partial \boldsymbol{x} }{ \partial \boldsymbol{\xi} } $

Now, let's consider some property field $ F $. This property field evolves with the state of the fluid, and can be written one of two ways, each with a unique interpretation:

$ \mathcal{F}(\boldsymbol{x},t) = \mathcal{F}( \boldsymbol{\xi}(\boldsymbol{x},t), t ) $

This way of writing $ F $ can be interpreted as follows: The value of the property $ F $ at the spatial/temporal location $ (\boldsymbol{x},t) $ is the value appropriate to the particle located at $ (\boldsymbol{x},t) $

Alternatively,

$ \mathcal{F}(\boldsymbol{\xi},t) = \mathcal{F}( \boldsymbol{x}(\boldsymbol{\xi},t), t ) $

This way of writing $ \mathcal{F} $ can be interpreted as: The value seen by the particle $ \boldsymbol{\xi} $ at time $ t $ is the value of the property $ \mathcal{F} $ at the position the particle occupies at that time $ t $

In keeping with these interpretations, two different temporal derivatives can be written.

$ \displaystyle{\frac{\partial}{\partial t}} $ : The derivative with respect to time, keeping $ x $ constant (i.e., the derivative at a fixed point)

$ \displaystyle{\frac{d}{dt}} $ : The derivative with respect to time, keeping $ \boldsymbol{\xi} $ constant (i.e. moving with the particle)

$ \displaystyle{\frac{D}{Dt}} $ : material derivative; same as $ \frac{d}{dt} $

Keep in mind that the material derivatives are not partial derivatives, because $ \boldsymbol{\xi} $ (the initial particle position) is a constant for a given fluid particle

Position and Velocity

The material derivative of the position of the fluid particle is the velocity:

$ \mathcal{F} = x_i $

$ v_i = \frac{dx_i}{dt} = \frac{\partial}{\partial t} x_i (\xi_1, \xi_2, \xi_3, t) \vert_{ \boldsymbol{\xi} } $

In other words, holding $ \boldsymbol{\xi} $ constant (that is, considering a given fluid particle), the rate of change of the particle's current position $ x_i $ is the velocity.

$ v = \frac{ d \boldsymbol{x} }{ dt } $

This material derivative, $ \frac{dx}{dt} $, is for a given $ \boldsymbol{\xi} $

Derivative Relationships

The two derivatives can be related. First, to repeat what these derivatives physically mean:

$ \left( \frac{ \partial \mathcal{F} }{ \partial t } \right)_{\boldsymbol{x}} $ : the rate of change of $ \mathcal{F} $ at a FIXED POINT $ \boldsymbol{x} $

$ \left( \frac{ d \mathcal{F} }{ d t } \right)_{\boldsymbol{\xi}} $ : the rate of change of $ \mathcal{F} $ for a given particle $ \boldsymbol{\xi} $. When $ \boldsymbol{\xi} $ is kept constant, then $ x $ (a function of $ \xi $) is $ x(\xi) $ for a given particle $ \xi $!

Next, the derivatives of F with respect to time can be equated at a particular spatial and temporal location, and the chain rule used, to get the relationship between these two derivatives:

$ \begin{array}{rcl} \displaystyle{ \frac{ d \mathcal{F} }{dt} } &=& \displaystyle{ \frac{\partial}{\partial t} \left( \mathcal{F}(\boldsymbol{\xi},t) \right) } \\ &=& \displaystyle { \frac{\partial}{\partial t} \left( \mathcal{F}( \boldsymbol{x}( \boldsymbol{\xi},t ), t ) \right) } \\ &=& \displaystyle{ \frac{ \partial \mathcal{F} }{\partial x_i} \left( \frac{d x_i}{dt} \right)_{\boldsymbol{\xi}} + \left( \frac{ \partial \mathcal{F} }{\partial t} \right)_{x} } \\ &=& \displaystyle{ \frac{ \partial \mathcal{F} }{ \partial t } + v_i \frac{ \partial \mathcal{F} }{ \partial x_i } } \end{array} $

This can be written conveniently as

$ \frac{ d \mathcal{F} }{ dt } = \frac{ \partial \mathcal{F} }{ \partial t } + ( v \cdot \nabla ) \mathcal{F} $

Practically, in order to obtain the paths of fluid particles from the velocity field, an ordinary differential equation must be solved for each fluid particle:

$ \frac{ d x_i }{ dt } = v_i ( x_1, x_2, x_3, t) $

subject to the initial condition $ x_i = \xi_i $ at $ t=0 $.

To solve for acceleration of the particle, another ordinary differential equation must be solved for each fluid particle:

$ \boldsymbol{a} = \frac{ d \boldsymbol{v} }{ dt } = \frac{ \partial \boldsymbol{v} }{ \partial t } + (\boldsymbol{v} \cdot \nabla) \boldsymbol{v} $

in steady flow, the accumulation term goes away, making the acceleration:

$ \boldsymbol{a} = ( \boldsymbol{v} \cdot \nabla ) \boldsymbol{v} $

Dilation and the Euler Expansion Formula

Changing coordinates from $ \boldsymbol{\xi} $ to $ \boldsymbol{x} $ will inevitably cause a volume change:

$ dV = \frac{ \partial (x_1, x_2, x_3) }{ \partial (\xi_1, \xi_2, \xi_3) } d \xi_1 d \xi_2 d \xi_3 = J dV_0 $

where $ J $ is the Jacobian.

$ J = \frac{ dV }{ dV_0 } $

And the Jacobian is a matrix that looks like this:

$ J = \displaystyle{ \left| \begin{array}{ccc} \frac{ \partial x_1 }{ \partial \xi_1 } & \frac{ \partial x_1 }{ \partial \xi_2 } & \frac{ \partial x_1 }{ \partial \xi_3 } \\ \frac{ \partial x_2 }{ \partial \xi_1 } & \frac{ \partial x_2 }{ \partial \xi_2 } & \frac{ \partial x_2 }{ \partial \xi_3 } \\ \frac{ \partial x_3 }{ \partial \xi_1 } & \frac{ \partial x_3 }{ \partial \xi_2 } & \frac{ \partial x_3 }{ \partial \xi_3 } \end{array} \right| } $

The initial volume of the fluid is $ d \xi_1 d \xi_2 d \xi_3 = dV_0 $ at $ t=0 $

Motion is continuous, so the volume cannot break up... another way to state that is, $ 0 < J < \infty $ (required so that neither $ J $ nor $ J^{-1} $ vanish, and the mapping from $ \xi $ to $ x $ and vice-versa are continuous and smooth).

The natural question to ask is how the volume dilation changes with time - that is, $ \frac{ dJ }{ dt } $

First, the time derivative of the terms in the Jacobian matrix can be simplified:

$ \frac{d}{dt} \left( \frac{\partial x_i}{\partial \xi_j} \right) = \frac{\partial}{\partial \xi_j} \frac{d x_i}{d t} = \frac{ \partial v_i }{ \partial \xi_j } $

where the second step is possible because $ \frac{d}{dt} $ holds $ \boldsymbol{\xi} $ constant.

It was specified above that the velocity is a function of location, $ \boldsymbol{v} = \boldsymbol{v}( x_1, x_2, x_3 ) $. This can be plugged into the relation $ \frac{ \partial v_i }{ \partial \xi_j } $, and the chain rule used, to yield:

$ \frac{ \partial v_i }{ \partial \xi_j } = \frac{ \partial v_i }{ \partial x_1 } \frac{ \partial x_1 }{ \partial \xi_j } + \frac{ \partial v_i }{ \partial x_2 } \frac{ \partial x_2 }{ \partial \xi_j } + \frac{ \partial v_i }{ \partial x_3 } \frac{ \partial x_3 }{ \partial \xi_j } $

which can be generalized to the result:

$ \frac{ d v_i }{ dt } = \frac{ d v_i }{ d x_k } \frac{ d x_k }{ dt } $

This expression gives a way to write the time derivative of the terms in the Jacobian matrix.

The derivative of the determinant of the Jacobian is the sum of three terms; each term is the Jacobian matrix, with only one row differentiated. Thus,

$ \frac{dJ}{dt} = \displaystyle{ \left| \begin{array}{ccc} \frac{d}{dt} \frac{ \partial x_1 }{ \partial \xi_1 } & \frac{d}{dt} \frac{ \partial x_1 }{ \partial \xi_2 } & \frac{d}{dt} \frac{ \partial x_1 }{ \partial \xi_3 } \\ \frac{ \partial x_2 }{ \partial \xi_1 } & \frac{ \partial x_2 }{ \partial \xi_2 } & \frac{ \partial x_2 }{ \partial \xi_3 } \\ \frac{ \partial x_3 }{ \partial \xi_1 } & \frac{ \partial x_3 }{ \partial \xi_2 } & \frac{ \partial x_3 }{ \partial \xi_3 } \end{array} \right| } + \displaystyle{ \left| \begin{array}{ccc} \frac{ \partial x_1 }{ \partial \xi_1 } & \frac{ \partial x_1 }{ \partial \xi_2 } & \frac{ \partial x_1 }{ \partial \xi_3 } \\ \frac{d}{dt} \frac{ \partial x_2 }{ \partial \xi_1 } & \frac{d}{dt} \frac{ \partial x_2 }{ \partial \xi_2 } & \frac{d}{dt} \frac{ \partial x_2 }{ \partial \xi_3 } \\ \frac{ \partial x_3 }{ \partial \xi_1 } & \frac{ \partial x_3 }{ \partial \xi_2 } & \frac{ \partial x_3 }{ \partial \xi_3 } \end{array} \right| } + \displaystyle{ \left| \begin{array}{ccc} \frac{ \partial x_1 }{ \partial \xi_1 } & \frac{ \partial x_1 }{ \partial \xi_2 } & \frac{ \partial x_1 }{ \partial \xi_3 } \\ \frac{ \partial x_2 }{ \partial \xi_1 } & \frac{ \partial x_2 }{ \partial \xi_2 } & \frac{ \partial x_2 }{ \partial \xi_3 } \\ \frac{d}{dt} \frac{ \partial x_3 }{ \partial \xi_1 } & \frac{d}{dt} \frac{ \partial x_3 }{ \partial \xi_2 } & \frac{d}{dt} \frac{ \partial x_3 }{ \partial \xi_3 } \end{array} \right| } $

Using the identity derived above, it can be shown that in the derivative of the first row, only the terms with k=1 survive, since the coefficients of the other terms have a coefficient that is a determinant of a matrix with two rows the same.

For this reason, the determinant of the first matrix (in the expression for $ \frac{dJ}{dt} $ above) has a value of

$ \frac{ \partial v_1 }{ \partial x_1 } J $

and the others have similar values. This makes the final value of the time-derivative of the Jacobian:

$ \frac{dJ}{dt} = \left( \frac{d v_1}{d x_1} + \frac{ d v_2 }{d x_2} + \frac{ d v_3 }{ d x_3 } \right) J $

which can also be written,

$ \displaystyle{ \frac{1}{J} \frac{dJ}{dt} = div \mathbf{v} } $

or,

$ \displaystyle{ \frac{ d ln J }{ dt } = div \mathbf{v} } $

This is also called the Euler Expansion Formula.

For incompressible fluids, the fluid particle's volume will not change due to compression or dilution, so the Jacobian is zero (that is, the fluid particle volume is always equal to the initial fluid particle volume)

This means that, for a fluid particle,

$ div \mathbf{v} = 0 $

Reynolds' Transport Theorem

All of the above can be put together to derive Reynolds' transport theorem

Starting with a function $ \mathcal{F} $, and a closed volume $ V(t) $ moving with the fluid, then the amount of $ \mathcal{F} $ may be written using a balance equation:

$ F(t) = \iiint_{V(t)} \mathcal{F}(\boldsymbol{x},t) dV $

The interest is in the material derivative of this quantity, $ \frac{dF}{dt} $

The integral is over a varying volume $ V(t) $

So the integral $ \iiint_{V(t)} $ and the derivative $ \frac{d}{dt} $ don't commute

But if the integration were with respect to a volume in $ \boldsymbol{\xi} $-space, they would

The volume $ \boldsymbol{x} = \boldsymbol{x}(\boldsymbol{\xi},t) $ and the Jacobian $ dV = J dV_0 $ allow us to do this

So the volume integral can be converted as follows:

$ \begin{array}{rcl} \frac{d}{dt} \iiint_{V(t)} \mathcal{F}(\boldsymbol{x},t) dV &=& \frac{d}{dt} \iiint_{V_0} \mathcal{F}(\boldsymbol{x}(\boldsymbol{\xi},t)) J dV_0 \\ &=& \iiint_{V_0} \left[ \frac{d \mathcal{F}}{dt} J + \mathcal{F} \frac{dJ}{dt} \right] dV_0 \\ &=& \iiint_{V_0} \left[ \frac{d \mathcal{F}}{dt} + \mathcal{F} \frac{1}{J} \frac{dJ}{dt} \right] J dV_0 \end{array} $

Next, the Euler Expansion formula can be used,

$ = \iiint_{V_0} \left[ \frac{d \mathcal{F}}{dt} + \mathcal{F} \left( \nabla \cdot \mathbf{v} \right) \right] J dV_0 $

and transforming from $ V_0 $ back to $ V $ (and noting that $ \frac{dF}{dt} = \iiint_{V(t)} \mathcal{F}(\mathbf{x},t) dV $), it becomes Reynolds Transport Theorem:

$ \frac{dF}{dt} = \iiint_{V(t)} \frac{d \mathcal{F}}{dt} + \mathcal{F} \left( \nabla \cdot \mathbf{v} \right) dV $

Now the definition of the substantial derivative can be used to put this in different forms

$ \frac{d}{dt} = \frac{ \partial }{ \partial t } + \mathbf{v} \cdot \nabla $

to get

$ \frac{d}{dt} \iiint_{V(t)} \mathcal{F}(\mathbf{x},t) dV = \iiint_{V(t)} \left[ \frac{ \partial \mathcal{F} }{ \partial t } dV + \nabla \cdot \left( \mathcal{F} \mathbf{v} \right) \right] dV $

Green's Theorem can be applied to this last divergence term, so that

$ \frac{dF}{dt} = \iint_{S(t)} \mathcal{F} \mathbf{v} \cdot \mathbf{n} dS $

which provides another form of Reynolds Transport Theorem,

$ \frac{dF}{dt} = \iiint_{V(t)} \frac{ \partial \mathcal{F} }{\partial t} dV + \iint_{S(t)} \mathcal{F} \mathbf{v} \cdot \mathbf{n} dS $