Time evolution of a Jacobian determinant

[Apashanka]

In this paper $J=\frac{\partial f_1(X_1)}{\partial X_1}\frac{\partial f_2(X_2)}{\partial X_2}\frac{\partial f_3(X_3)}{\partial X_3}$ where $f_2(X_2),f_1(X_1),f_3(X_3)$ evolves with time.

Now using this $\dot J=\frac{d}{dt}(\frac{\partial f_1(X_1)}{\partial X_1}\frac{\partial f_2(X_2)}{\partial X_2}\frac{\partial f_3(X_3)}{\partial X_3})$ and using $\frac{d}{dt}=\frac{\partial}{\partial t}+(v•\nabla)$ and $\frac{\partial X_i}{\partial t}=0$ this comes as $3(v•\nabla)J$ but it is given as $J\theta$ where $\theta=\nabla • v$
Will anyone please help me in sort out this...

 

[Apashanka]

Or another one will be to take derivative of $d^3x=Jd^3X$ where $dx,dX$ are the volume element ,but then how to proceed??

 

[Mark44]

Or another one will be to take derivative of $d^3x=Jd^3X$ where $dx,dX$ are the volume element ,but then how to proceed??

As noted in the other thread, taking the derivative of the volume element isn't meaningful.

The snippet you've posted in this problem omits a lot of context, and the notation used is fairly dense, but I believe that you don't have a correct formulation for the Jacobian and its determinant. First, let's unpack some of the notation: $\vec x = \vec f (\vec X)$. As noted in the image, $\vec X$ represents the position of an arbitrary particle in space ($\mathbb R^3$) at time $t_0$, and $\vec x$ represents the position of the same particle at a later time $t$. The function $f$ is the mapping between the two positions.

The Jacobian J should be the determinant of a 3x3 matrix, not a 1x3 matrix as you have written several times. I believe it should look like this:
$J = \begin{vmatrix} \frac{\partial f(X_1)}{\partial X_1} &\frac{\partial f(X_1)}{\partial X_2} & \frac{\partial f(X_1)}{\partial X_3}\\ \frac{\partial f(X_2)}{\partial X_1} &\frac{\partial f(X_2)}{\partial X_2} & \frac{\partial f(X_2)}{\partial X_3} \\ \frac{\partial f(X_3)}{\partial X_1} &\frac{\partial f(X_3)}{\partial X_2} & \frac{\partial f(X_3)}{\partial X_3} \end{vmatrix}$

Also, since the coordinates $X_1, X_2,$ and $X_3$ are independent (I believe), all of the entries off the main diagonal are zero, so the Jacobian determinant reduces to this:
$J = \begin{vmatrix} \frac{\partial f(X_1)}{\partial X_1} & 0 & 0\\ 0 &\frac{\partial f(X_2)}{\partial X_2} & 0 \\ 0 & 0 & \frac{\partial f(X_3)}{\partial X_3} \end{vmatrix}$

So J is just the product of the entries on the main diagonal; i.e., $J = \frac{\partial f(X_1)}{\partial X_1} \cdot \frac{\partial f(X_2)}{\partial X_2} \cdot \frac{\partial f(X_3)}{\partial X_3}$, hence $\dot J = \frac d{dt}\left( \frac{\partial f(X_1)}{\partial X_1} \cdot \frac{\partial f(X_2)}{\partial X_2} \cdot \frac{\partial f(X_3)}{\partial X_3} \right)$, which you show in your previous post.

You also show

and using $\frac{d}{dt}=\frac{\partial}{\partial t}+(v•\nabla)$

I don't know how you came up with what you have for the operator $\frac d{dt}$ -- it seems incorrect to me.

Now that we have an expression for $\dot J$, what remains is to find $\nabla \cdot \vec v$. Wouldn't $\vec v$ be the vector $<\frac {dx_1}{dt}, \frac {dx_2}{dt}, \frac {dx_3}{dt}>$?

 

[Apashanka]

As noted in the other thread, taking the derivative of the volume element isn't meaningful.

The snippet you've posted in this problem omits a lot of context, and the notation used is fairly dense, but I believe that you don't have a correct formulation for the Jacobian and its determinant. First, let's unpack some of the notation: $\vec x = \vec f (\vec X)$. As noted in the image, $\vec X$ represents the position of an arbitrary particle in space ($\mathbb R^3$) at time $t_0$, and $\vec x$ represents the position of the same particle at a later time $t$. The function $f$ is the mapping between the two positions.

The Jacobian J should be the determinant of a 3x3 matrix, not a 1x3 matrix as you have written several times. I believe it should look like this:
$J = \begin{vmatrix} \frac{\partial f(X_1)}{\partial X_1} &\frac{\partial f(X_1)}{\partial X_2} & \frac{\partial f(X_1)}{\partial X_3}\\ \frac{\partial f(X_2)}{\partial X_1} &\frac{\partial f(X_2)}{\partial X_2} & \frac{\partial f(X_2)}{\partial X_3} \\ \frac{\partial f(X_3)}{\partial X_1} &\frac{\partial f(X_3)}{\partial X_2} & \frac{\partial f(X_3)}{\partial X_3} \end{vmatrix}$

Also, since the coordinates $X_1, X_2,$ and $X_3$ are independent (I believe), all of the entries off the main diagonal are zero, so the Jacobian determinant reduces to this:
$J = \begin{vmatrix} \frac{\partial f(X_1)}{\partial X_1} & 0 & 0\\ 0 &\frac{\partial f(X_2)}{\partial X_2} & 0 \\ 0 & 0 & \frac{\partial f(X_3)}{\partial X_3} \end{vmatrix}$

So J is just the product of the entries on the main diagonal; i.e., $J = \frac{\partial f(X_1)}{\partial X_1} \cdot \frac{\partial f(X_2)}{\partial X_2} \cdot \frac{\partial f(X_3)}{\partial X_3}$, hence $\dot J = \frac d{dt}\left( \frac{\partial f(X_1)}{\partial X_1} \cdot \frac{\partial f(X_2)}{\partial X_2} \cdot \frac{\partial f(X_3)}{\partial X_3} \right)$, which you show in your previous post.

You also show I don't know how you came up with what you have for the operator $\frac d{dt}$ -- it seems incorrect to me.

Now that we have an expression for $\dot J$, what remains is to find $\nabla \cdot \vec v$. Wouldn't $\vec v$ be the vector $<\frac {dx_1}{dt}, \frac {dx_2}{dt}, \frac {dx_3}{dt}>$?

$\dot J=\frac{d}{dt}[\frac{\partial f_1}{\partial X_1}\frac{\partial f_2}{\partial X_2}\frac{\partial f_3}{\partial X_3}]$ where the functional form of $f_1(X_1),f_2(X_2),f_3(X_3)$ changes with time but their explicit time dependence is not known...then how to proceed this??

 

[fresh_42]

I do not see how you managed to get a factor $3$ in here. Where does it come from? As I understand it do we have
\begin{equation*}
D_v\Phi = \frac{d_v \Phi}{d t}=(\frac{\partial}{\partial t}+ v \cdot \nabla)(\Phi) = \dot V + (v\cdot \nabla)(\Phi)
\end{equation*}
and the local or comoving behaviour affects only the first term, not the convection. The change in the scalar volume is then as given in the text.

Maybe @Orodruin or someone can explain it better from a physicist's point of view. The framework here is less a differentiation as it is a model for fluid dynamics.

 

[Apashanka]

I do not see how you managed to get a factor 3 in here.

Sorry I didn't get you..

 

[fresh_42]

Sorry I didn't get you..

... and $\frac{\partial X_i}{\partial t}=0$ this comes as $\mathbf{3}(v•\nabla)J$ but it is given as $J\theta$ where $\theta=\nabla • v$
Will anyone please help me in sort out this...

But to be honest, I'm not really an expert here. I just tried to adapt the example in here into a context, where we have time dependent spatial coordinates (and there is no additional factor in sight):
https://www.physicsforums.com/insights/pantheon-derivatives-part-iv/

 

[Apashanka]

But to be honest, I'm not really an expert here. I just tried to adapt the example in here into a context, where we have time dependent spatial coordinates (and there is no additional factor in sight):
https://www.physicsforums.com/insights/pantheon-derivatives-part-iv/

Actually I need some help or hints in proving this...

$\dot J=\frac{d}{dt}[\frac{\partial f_1}{\partial X_1}\frac{\partial f_2}{\partial X_2}\frac{\partial f_3}{\partial X_3}]$ where the functional form of $f_1(X_1),f_2(X_2),f_3(X_3)$ changes with time but their explicit time dependence is not known...then how to proceed this??

 

[fresh_42]

Actually I need some help or hints in proving this...

Have you looked at the example of the river flow on the page I quoted? Now assume that the spatial components are also time dependent. This does not affect the the spatial differentials, only the time differential, i.e. only the comoving part. As we have a change in volume which we examine here, we get the time differential of said volume as the term for local behaviour. This is at least how I see it.

 

[Apashanka]

Have you looked at the example of the river flow on the page I quoted? Now assume that the spatial components are also time dependent. This does not affect the the spatial differentials, only the time differential, i.e. only the comoving part. As we have a change in volume which we examine here, we get the time differential of said volume as the term for local behaviour. This is at least how I see it.

$d^3x=Jd^3X$, since the Jacobian changes with time as the functional form $f_1,f_2,f_3$ changes therefore the volume element $d^3x$ also changes with time,so what's the harm in calculating it's rate of change e.g $\frac{d}{dt}d^3x$ which gives $dv_xdydz+dv_ydxdz+dv_zdxdy=\dot Jd^3X$,now is it possible to prove from this $\dot J=J\nabla•\vec v$

 

[fresh_42]

Do you have a reasonable definition for $\theta$? Also what should $J\nabla v$ be? The gradient times a direction is a scalar, what do you mean by the Jacobian of a scalar?

 

[Apashanka]

Do you have a reasonable definition for $\theta$? Also what should $J\nabla v$ be? The gradient times a direction is a scalar, what do you mean by the Jacobian of a scalar?

$\theta=\vec\nabla•\vec v$ and $\dot J=J(\vec \nabla•\vec v)$

 

[fresh_42]

$\theta=\vec\nabla•\vec v$ and $\dot J=J(\vec \nabla•\vec v)$

That makes no sense. If $\theta$ is the slope, what is the Jacobian of it? A simple time derivative? It is a bit of a disadvantage that your notation neither mentions the vector field nor the point of evaluation or the variable. $J$ as standalone doesn't say very much, and the $J(scalar)$ even less.

 

[Apashanka]

That makes no sense. If $\theta$ is the slope, what is the Jacobian of it? A simple time derivative? It is a bit of a disadvantage that your notation neither mentions the vector field nor the point of evaluation or the variable. $J$ as standalone doesn't say very much, and the $J(scalar)$ even less.

Here $J=\frac{\partial f_1(X_1)}{\partial X_1}\frac{\partial f_2(X_2)}{\partial X_2}\frac{\partial f_3(X_3)}{\partial X_3}$ and $J\theta=\dot J$ ...so what's the problem??

The snap is from the paper by @thomas buchert and @Jurgen ehlers on averaging Newtonian cosmology,here is the link
Web results
Averaging inhomogeneous Newtonian cosmologies
https://arxiv.org/abs/astro-ph/9510056

 

[fresh_42]

...so what's the problem??

The problem is, that we have a $3\times 3$ matrix and no vector. So at least it should be $\operatorname{diag}\left( \frac{\partial f_i}{\partial x_i} \right)$, even better $J(f)$. And is it a function of location, because the differentials are not evaluated, or is it the linear differential of $f$, and if, at which point? Next you defined the result of an inner product, called it $\theta$ and wrote a $J$ in front of it. But what is the Jacobian of a scalar? Zero?

The problem is, that it is mathematically so sloppy, that only people can read it, who are supposed to know what the authors mean. I tried to figure out if you knew what is meant, but obviously my guesswork isn't sufficient.

 

[kimbyd]

You also show I don't know how you came up with what you have for the operator $\frac d{dt}$ -- it seems incorrect to me.

I think it's just a consequence of the chain rule. Written out more explicitly:
$${d \over dt} f(\vec{x}) = {\partial f \over \partial t} + {d\vec{x} \over dt} \cdot \vec{\nabla} f(\vec{x})$$

I do not see how you managed to get a factor $3$ in here. Where does it come from?

It looks like it stemmed from an assumption that all elements of the dot product were identical, thus adding them together results in three times the value of a single element of the dot product. This often occurs for homogeneous fluids, but you have to be careful that you're applying the assumption correctly. I'm not sure I have enough information right now to see whether it was applied correctly here.

 

[pasmith]

The text is somewhat confusing (or at least the parts you have posted are without their surrounding context).

There are two ways of describing fluid flows:

• The more widely used Eulerian specification describes the fluid by the instantaneous velocity $\mathbf{v}$ of the fluid element which at time $t$ is at position $\mathbf{x}$. As time changes, different fluid elements will occupy this position. The independent variables are $\mathbf{x}$ and $t$.
• The Lagrangian specification, used in the post in question, describes the fluid by the position $\mathbf{x}$ at time $t$ of the fluid element originally at position $\mathbf{X}$. The independent variables are $\mathbf{X}$ and $t$.

The two specifications are related by $$\mathbf{v}(\mathbf{x}(\mathbf{X},t),t) = \left.\frac{\partial \mathbf{x}}{\partial t}\right|_{(\mathbf{X},t)}.$$

The material derivative $$\frac{D}{Dt} = \frac{\partial}{\partial t}+ \mathbf{v} \cdot \nabla$$ is relevant to the Eulerian specification, in which partial differentiation with respect to time is at a fixed position occupied by successive different fluid elements. Finding the rate of change with respect to time experienced by a particular fluid element requires the addition of the advective $\mathbf{v} \cdot \nabla$ term. (To be clear, the spatial derivative is with respect to $\mathbf{x}$.) But in the Lagrangian specification, which the text is using here, a particular fluid element is labelled by its initial position $\mathbf{X}$, and partial differentiation with respect to time holding $\mathbf{X}$ constant does "follow the fluid" without the need for an advective term.

The complication here is that when the author writes $\theta = \nabla \cdot \mathbf{v}$ they apparently intend differentiation not with respect to the initial position $\mathbf{X}$ but with respect to the current position $\mathbf{x}$. Allowing for this and using the normal partial time derivative yields the result (at least in the 2D case, which is the only case I could be bothered to check).

Starting with @Mark44's expression for $J$ you can take the partial derivative with respect to time and then simplify some terms using $$\frac{\partial^2 x_i}{\partial X_j\,\partial t} = \frac{\partial v_i}{\partial X_j}$$ since in the Lagrangian specificattion the velocity field is just the partial derivative of the current position with respect to time. You can then use the chain rule $$\frac{\partial v_i}{\partial X_j} = \sum_k \frac{\partial v_i}{\partial x_k} \frac{\partial x_k}{\partial X_j}$$ and you will find some terms will cancel, giving a result equal to $$J \theta = J \sum_i \frac{\partial v_i}{\partial x_i}.$$

 

[Apashanka]

The text is somewhat confusing (or at least the parts you have posted are without their surrounding context).

There are two ways of describing fluid flows:

• The more widely used Eulerian specification describes the fluid by the instantaneous velocity $\mathbf{v}$ of the fluid element which at time $t$ is at position $\mathbf{x}$. As time changes, different fluid elements will occupy this position. The independent variables are $\mathbf{x}$ and $t$.
• The Lagrangian specification, used in the post in question, describes the fluid by the position $\mathbf{x}$ at time $t$ of the fluid element originally at position $\mathbf{X}$. The independent variables are $\mathbf{X}$ and $t$.

The two specifications are related by $$\mathbf{v}(\mathbf{x}(\mathbf{X},t),t) = \left.\frac{\partial \mathbf{x}}{\partial t}\right|_{(\mathbf{X},t)}.$$

The material derivative $$\frac{D}{Dt} = \frac{\partial}{\partial t}+ \mathbf{v} \cdot \nabla$$ is relevant to the Eulerian specification, in which partial differentiation with respect to time is at a fixed position occupied by successive different fluid elements. Finding the rate of change with respect to time experienced by a particular fluid element requires the addition of the advective $\mathbf{v} \cdot \nabla$ term. (To be clear, the spatial derivative is with respect to $\mathbf{x}$.) But in the Lagrangian specification, which the text is using here, a particular fluid element is labelled by its initial position $\mathbf{X}$, and partial differentiation with respect to time holding $\mathbf{X}$ constant does "follow the fluid" without the need for an advective term.

The complication here is that when the author writes $\theta = \nabla \cdot \mathbf{v}$ they apparently intend differentiation not with respect to the initial position $\mathbf{X}$ but with respect to the current position $\mathbf{x}$. Allowing for this and using the normal partial time derivative yields the result (at least in the 2D case, which is the only case I could be bothered to check).

Starting with @Mark44's expression for $J$ you can take the partial derivative with respect to time and then simplify some terms using $$\frac{\partial^2 x_i}{\partial X_j\,\partial t} = \frac{\partial v_i}{\partial X_j}$$ since in the Lagrangian specificattion the velocity field is just the partial derivative of the current position with respect to time. You can then use the chain rule $$\frac{\partial v_i}{\partial X_j} = \sum_k \frac{\partial v_i}{\partial x_k} \frac{\partial x_k}{\partial X_j}$$ and you will find some terms will cancel, giving a result equal to $$J \theta = J \sum_i \frac{\partial v_i}{\partial x_i}.$$

Okk but where is then $\dot J$

 

[pasmith]

Okk but where is then $\dot J$

$\dot J$ is $\frac{\partial J}{\partial t}$ where, as @Mark44 has already said,
$$J = \begin{vmatrix} \frac{\partial x_1}{\partial X_1} &\frac{\partial x_1}{\partial X_2} & \frac{\partial x_1}{\partial X_3}\\ \frac{\partial x_2}{\partial X_1} &\frac{\partial x_2}{\partial X_2} & \frac{\partial x_2}{\partial X_3} \\ \frac{\partial x_3}{\partial X_1} &\frac{\partial x_3}{\partial X_2} & \frac{\partial x_3}{\partial X_3} \end{vmatrix}.$$

The algebra is not difficult, just tedious. You can however see the idea by looking at the 2D case.

 

[Apashanka]

$\dot J$ is $\frac{\partial J}{\partial t}$ where, as @Mark44 has already said,
$$J = \begin{vmatrix} \frac{\partial x_1}{\partial X_1} &\frac{\partial x_1}{\partial X_2} & \frac{\partial x_1}{\partial X_3}\\ \frac{\partial x_2}{\partial X_1} &\frac{\partial x_2}{\partial X_2} & \frac{\partial x_2}{\partial X_3} \\ \frac{\partial x_3}{\partial X_1} &\frac{\partial x_3}{\partial X_2} & \frac{\partial x_3}{\partial X_3} \end{vmatrix}.$$

The algebra is not difficult, just tedious. You can however see the idea by looking at the 2D case.

Here $\dot J=\frac{dJ}{dt}$,$J$ is here the Jacobian determinant which is $\frac{\partial x_1}{\partial X_1}\frac{\partial x_2}{\partial X_2}\frac{\partial x_3}{\partial X_3}$

 

[pasmith]

It is, I think, a sufficiently common abuse of notation for even partial derivatives with respect to time to be indicated by a dot. However the context here absolutely requires a partial derivative: $J$ is a function of time and a vector which indicates an initial position. It cannot sensibly be made a function of time so as to enable a total time derivative to be computed.

In general $\frac{\partial x_i}{\partial X_j}$ will not be diagonal, and its determinant $J$ will not be the product of the diagonal entries. If $\frac{\partial x_i}{\partial X_j}$ is diagonal, or there is a reason why the non-diagonal entries don't contribute to the determinant, then the authors must have made further assumptions which you have not communicated to us.

 

[Apashanka]

It is, I think, a sufficiently common abuse of notation for even partial derivatives with respect to time to be indicated by a dot. However the context here absolutely requires a partial derivative: $J$ is a function of time and a vector which indicates an initial position. It cannot sensibly be made a function of time so as to enable a total time derivative to be computed.

In general $J$ will not be diagonal, and its determinant will not be the product of the diagonal entries. If $J$ is diagonal, or there is a reason why the non-diagonal entries don't contribute to the determinant, then the authors must have made further assumptions which you have not communicated to us.

$x_1(t)=f_1(X_1),x_2(t)=f_2(X_2),x_3=f_3(X_3)$ or,$dx_1dx_2dx_3=\frac{\partial x_1}{\partial X_1}\frac{\partial x_2}{\partial X_2}\frac{\partial x_3}{\partial X_3}dX_1dX_2dX_3$,or $d^3x=Jd^3X$ as defined by author.
The functional form of $f_1(X_1),f(X_2),f(X_3)$ changes with time.
Therefore the Jacobian matrix is of course diagonal where $J$ is the determinant of it.
And if $J$ only depends upon time then $\dot J=\frac{\partial J}{\partial t}$

 

[Apashanka]

The text is somewhat confusing (or at least the parts you have posted are without their surrounding context).

There are two ways of describing fluid flows:

• The more widely used Eulerian specification describes the fluid by the instantaneous velocity $\mathbf{v}$ of the fluid element which at time $t$ is at position $\mathbf{x}$. As time changes, different fluid elements will occupy this position. The independent variables are $\mathbf{x}$ and $t$.
• The Lagrangian specification, used in the post in question, describes the fluid by the position $\mathbf{x}$ at time $t$ of the fluid element originally at position $\mathbf{X}$. The independent variables are $\mathbf{X}$ and $t$.

The two specifications are related by $$\mathbf{v}(\mathbf{x}(\mathbf{X},t),t) = \left.\frac{\partial \mathbf{x}}{\partial t}\right|_{(\mathbf{X},t)}.$$

The material derivative $$\frac{D}{Dt} = \frac{\partial}{\partial t}+ \mathbf{v} \cdot \nabla$$ is relevant to the Eulerian specification, in which partial differentiation with respect to time is at a fixed position occupied by successive different fluid elements. Finding the rate of change with respect to time experienced by a particular fluid element requires the addition of the advective $\mathbf{v} \cdot \nabla$ term. (To be clear, the spatial derivative is with respect to $\mathbf{x}$.) But in the Lagrangian specification, which the text is using here, a particular fluid element is labelled by its initial position $\mathbf{X}$, and partial differentiation with respect to time holding $\mathbf{X}$ constant does "follow the fluid" without the need for an advective term.

The complication here is that when the author writes $\theta = \nabla \cdot \mathbf{v}$ they apparently intend differentiation not with respect to the initial position $\mathbf{X}$ but with respect to the current position $\mathbf{x}$. Allowing for this and using the normal partial time derivative yields the result (at least in the 2D case, which is the only case I could be bothered to check).

Starting with @Mark44's expression for $J$ you can take the partial derivative with respect to time and then simplify some terms using $$\frac{\partial^2 x_i}{\partial X_j\,\partial t} = \frac{\partial v_i}{\partial X_j}$$ since in the Lagrangian specificattion the velocity field is just the partial derivative of the current position with respect to time. You can then use the chain rule $$\frac{\partial v_i}{\partial X_j} = \sum_k \frac{\partial v_i}{\partial x_k} \frac{\partial x_k}{\partial X_j}$$ and you will find some terms will cancel, giving a result equal to $$J \theta = J \sum_i \frac{\partial v_i}{\partial x_i}.$$

Thanks @pasmith ,the result is coming true