# Understanding Spatial Description: A Mathematical Perspective

• roshangomez
In the Eulerian description, ##\vec{x}## always refers to a point in space and the velocity field ##\vec{v}(t,\vec{x})## tells you the velocity of the fluid element which is at ##\vec{x}## at time ##t##.f

#### roshangomez

I understand the mathematical difference, that the independent variables are the "reference position vector+time" for material description, and "current position vector+time" for the spatial description. But I can't seem to wrap my head around the concept of the spatial description.

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Think of a unit cell. For lagrangian, the cell moves and deforms and no material leaves/enters the cell. For eulerian, material flows in and out of the non-moving, non-deforming cell.

Think of a point. For lagrangian, it moves along with the material. For eulerian, it is fixed in space.

• Demystifier and roshangomez
Think of a unit cell. For lagrangian, the cell moves and deforms and no material leaves/enters the cell. For eulerian, material flows in and out of the non-moving, non-deforming cell.

Think of a point. For lagrangian, it moves along with the material. For eulerian, it is fixed in space.
Thank you for that. Makes sense now.

In the Lagrangian description you describe the motion of material fluid cells. In fluid mechanics it's convenient to simply use the positions ##\vec{x}_0## of your fluid cells at initial time ##t=0## to label each fluid element. Then the motion of the fluid element which was at ##\vec{x}_0## at time ##t=0## at time ##t## is at ##\vec{x}(t,\vec{x}_0)##. The velocity of this fluid element is of course simply ##\vec{v}_0(t,\vec{x}_0) = \partial_t \vec{x}(t,\vec{x}_0)##, where the time derivative is taken at fixed ##\vec{x}_0##, i.e., for the fixed fluid element which was at ##\vec{x}_0## at time ##t_0##. The same is true for the acceleration of this material fluid element, ##\vec{a}_0(t,\vec{x}_0)=\partial_t \vec{v}_0(t,\vec{x}_0)##.

The Eulerian description is a field description, i.e., ##\vec{x}## labels a fixed point in space, and you describe the fluid motion by the velocity field ##\vec{v}(t,\vec{x})##, which tells you the velocity of the fluid element which is at ##\vec{x}## at time ##t##.

Now obviously ##\vec{v}_0(t,\vec{x}_0)=\vec{v}(t,\vec{x}(t,\vec{x}_0))## and thus
$$\vec{a}(t,\vec{x}_0) =\partial_t \vec{v}_0 = \partial_t \vec{v} + [\partial_t \vec{x}(t,\vec{x}_0) \cdot \vec{\nabla}] \vec{v} = \partial_t \vec{v} + (\vec{v} \cdot \vec{\nabla}) \vec{v}.$$
This defines the "material time derivative", i.e., for any quantity ##f(t,\vec{x})## described in the Eulerian description the time derivative corresponding to a fixed material fluid element is
$$\mathrm{D}_t f(t,\vec{x}) = \partial_t f(t,\vec{x}) + (\vec{v} \cdot \vec{\nabla}) f(t,\vec{x}).$$

• • Demystifier and roshangomez
In the Lagrangian description you describe the motion of material fluid cells. In fluid mechanics it's convenient to simply use the positions ##\vec{x}_0## of your fluid cells at initial time ##t=0## to label each fluid element. Then the motion of the fluid element which was at ##\vec{x}_0## at time ##t=0## at time ##t## is at ##\vec{x}(t,\vec{x}_0)##. The velocity of this fluid element is of course simply ##\vec{v}_0(t,\vec{x}_0) = \partial_t \vec{x}(t,\vec{x}_0)##, where the time derivative is taken at fixed ##\vec{x}_0##, i.e., for the fixed fluid element which was at ##\vec{x}_0## at time ##t_0##. The same is true for the acceleration of this material fluid element, ##\vec{a}_0(t,\vec{x}_0)=\partial_t \vec{v}_0(t,\vec{x}_0)##.

The Eulerian description is a field description, i.e., ##\vec{x}## labels a fixed point in space, and you describe the fluid motion by the velocity field ##\vec{v}(t,\vec{x})##, which tells you the velocity of the fluid element which is at ##\vec{x}## at time ##t##.

Now obviously ##\vec{v}_0(t,\vec{x}_0)=\vec{v}(t,\vec{x}(t,\vec{x}_0))## and thus
$$\vec{a}(t,\vec{x}_0) =\partial_t \vec{v}_0 = \partial_t \vec{v} + [\partial_t \vec{x}(t,\vec{x}_0) \cdot \vec{\nabla}] \vec{v} = \partial_t \vec{v} + (\vec{v} \cdot \vec{\nabla}) \vec{v}.$$
This defines the "material time derivative", i.e., for any quantity ##f(t,\vec{x})## described in the Eulerian description the time derivative corresponding to a fixed material fluid element is
$$\mathrm{D}_t f(t,\vec{x}) = \partial_t f(t,\vec{x}) + (\vec{v} \cdot \vec{\nabla}) f(t,\vec{x}).$$
This is exactly what I needed...Thanks a bunch • vanhees71
In the Lagrangian frame of reference, the observer is moving along at the fluid velocity. In the Eulerian frame of reference, the observer is stationary and the fluid is flowing past.

No, in the Lagrangian description ##\vec{x}(t,\vec{x}_0)## are the components of the position vector at time ##t## of the fluid element which has been at ##\vec{x}_0## at the initial time ##t_0##. These components usually refer to an inertial frame of reference (or, if you deal with aerodynamics for meteorology on Earth, in the rotating frame fixed at the location of an observer at rest on Earth).

No, in the Lagrangian description ##\vec{x}(t,\vec{x}_0)## are the components of the position vector at time ##t## of the fluid element which has been at ##\vec{x}_0## at the initial time ##t_0##. These components usually refer to an inertial frame of reference (or, if you deal with aerodynamics for meteorology on Earth, in the rotating frame fixed at the location of an observer at rest on Earth).
Sorry, that is the opposite of the way I learned it. https://en.wikipedia.org/wiki/Lagrangian_and_Eulerian_specification_of_the_flow_field

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Where is the contradiction? The observer, who follows a fluid cell in the Lagrangian description uses the coordinates of a fixed (usually inertial) reference frame. He is not co-moving with the fluid cell as you seem to suggest.

Where is the contradiction? The observer, who follows a fluid cell in the Lagrangian description uses the coordinates of a fixed (usually inertial) reference frame. He is not co-moving with the fluid cell as you seem to suggest.
What do the words "observer follows an individual fluid parcel as it moves through space and time." from the Wiki article mean to you?

For me it means what the math says: In the Lagrange description, the observer is sitting at rest in some inertial frame of reference using Cartesian coordiantes ##\vec{x}(t,\vec{x}_0)## to follow the position of a fluid parcel (which I'm used to call a fluid cell, but I guess that's synonymous) which has been at position ##\vec{x}_0## at the initial time ##t_0## (the components of the position vector referring to the same inertial frame of reference). The observer is not co-moving with the fluid parcel in the standard Lagrangian formulation.

For me it means what the math says: In the Lagrange description, the observer is sitting at rest in some inertial frame of reference using Cartesian coordiantes ##\vec{x}(t,\vec{x}_0)## to follow the position of a fluid parcel (which I'm used to call a fluid cell, but I guess that's synonymous) which has been at position ##\vec{x}_0## at the initial time ##t_0## (the components of the position vector referring to the same inertial frame of reference). The observer is not co-moving with the fluid parcel in the standard Lagrangian formulation.
I don't know what to say except that this is not the way that I learned it.

Hm, that's how I learned it. For a very concise description, see, e.g.,

K. Hutter, Y. Wang, Fluid and Thermodynamics, vol. 1, Sect. 3.1

I agree with @vanhees71, in the Lagrangian description the origin of your coordinate system can still be the same as for Eulerian. But at each time step you compute the new coordinates of the parcels that you are concerned with, and also its new mass, temperature, drag and what have you.

For Eulerian you don't compute new coordinates, just the new values for the variables that you are solving for at each location.

Some computations combine the two. They use Eulerian for the bulk flow (e.g. water around a ship propeller) and release some particles (e.g. cavitation bubbles) at some locations using the Lagrangian description for which they then solve the Rayleigh-Plesset equation. This can be a one-way coupling (bulk flow to parcels) or a two-way coupling (parcels also having an influence on the bulk flow). You can then also choose to have the bulk flow in steady state or not, etc.

• vanhees71
What do the words "observer follows an individual fluid parcel as it moves through space and time." from the Wiki article mean to you?
"Observer follows" and the "observer moves along with" mean two different things. The observer can follow ('look at' or 'track with radar' or whatever) without moving itself.

• vanhees71
The (fictitious) observer is indeed the same in both descriptions. You can think of the Lagrangian description in the way that at ##t_0## you put some color on a specific fluid parcel located at ##\vec{x}_0## and then you follow this fluid parcel by looking, where the colored parcel is at time ##t##, and this position by definition is ##\vec{x}(t,\vec{x}_0)##.

The Eulerian description is rather looking at local quantities at a fixed point in time. E.g., in the Eulerian description ##\vec{v}(t,\vec{x})## is the velocity of the fluid parcel, which is located at ##\vec{x}## at time ##t##.

The connection between the two equivalent descriptions of fluid motions is given by the map ##\vec{x}_0 \mapsto \vec{x}(t,\vec{x}_0)##. E.g., the velocity field is uniquely described either by
$$\tilde{\vec{v}}(t,\vec{x}_0)=\partial_t \vec{x}(t,\vec{x}_0).$$
That's the velocity of the fluid parcel that was at ##t_0## at ##\vec{x}_0## at time ##t##.

Assuming that the mapping is a diffeomorphism you can also define the field ##\vec{x}_0(t,\vec{x})##. That's the position ##\vec{x}_0## of the fluid parcel that is at time ##t## at position ##\vec{x}## at time ##t_0##.

The Eulerian fluid-velocity field is then related to the Lagrangian one by
$$\vec{v}(t,\vec{x}) = \tilde{\vec{v}}[t,\vec{x}_0(t,\vec{x})]$$
or in the other direction
$$\tilde{v}(t,\vec{x}_0) = \vec{v}[t,\vec{x}(t,\vec{x}_0)].$$
From this you get the acceleration of the individual fluid parcel, which you need in Newton's equation of motion,
$$\tilde{\vec{a}}(t,\vec{x}_0)=\partial_t \tilde{\vec{v}}(t,\vec{x}_0) = \partial \vec{v}[t,\vec{v}(t,\vec{x}_0)] + [\underbrace{\partial_t \vec{x}(t,\vec{x}_0)}_{=\tilde{\vec{v}}(t,\vec{x}_0)} \cdot \vec{\nabla} \vec{v}(t,\vec{x})|_{\vec{x}=\vec{x}(t,\vec{x}_0)}].$$
In the Eulerian description from this you get immediately
$$\vec{a}(t,\vec{x})=\mathrm{D}_t \vec{v}(t,\vec{x}) = \partial_t \vec{v}(t,\vec{x}) + [\vec{v}(t,\vec{x}) \cdot \vec{\nabla}] \vec{v}(t,\vec{x}).$$

• Arjan82
In the Eulerian description from this you get immediately
$$\vec{a}(t,\vec{x})=\mathrm{D}_t \vec{v}(t,\vec{x}) = \partial_t \vec{v}(t,\vec{x}) + [\vec{v}(t,\vec{x}) \cdot \vec{\nabla}] \vec{v}(t,\vec{x}).$$

The ##\mathrm{D}_t ## is often called the material derivative.

• vanhees71
The ##\mathrm{D}_t ## is often called the material derivative.
I regard that as the Lagrangian description of the time derivative, as reckoned by an observer traveling along with a parcel of fluid. The Eulerian time derivative is just ##\frac{\partial}{\partial t}##, as reckoned by a stationary observer.

"Observer follows" and the "observer moves along with" mean two different things. The observer can follow ('look at' or 'track with radar' or whatever) without moving itself.
In my judgment, it's the same thing.

I regard that as the Lagrangian description of the time derivative, as reckoned by an observer traveling along with a parcel of fluid. The Eulerian time derivative is just ##\frac{\partial}{\partial t}##, as reckoned by a stationary observer.
Again no! Both the Lagrangian and the Eulerian description work in a fixed inertial frame (or sometimes, when you apply hydro to the physics of the atmosphere/meteorology in the rotating frame fixed on a point on Earth). Neither of these use the very complicated non-inertial frame of an observer fixed to a fluid element.

An exception is cosmology, where the most simple observer is an observer co-moving with the "cosmological substrate", i.e., an observer at rest wrt. the standard reference frame of the FLRW spacetime.

• Arjan82 and Demystifier
Again no! Both the Lagrangian and the Eulerian description work in a fixed inertial frame (or sometimes, when you apply hydro to the physics of the atmosphere/meteorology in the rotating frame fixed on a point on Earth). Neither of these use the very complicated non-inertial frame of an observer fixed to a fluid element.

An exception is cosmology, where the most simple observer is an observer co-moving with the "cosmological substrate", i.e., an observer at rest wrt. the standard reference frame of the FLRW spacetime.
I am guessing that you have spent time with relativistic fluids. For those of us who are slow, this is a formulation, not a frame, question. It is a question of whether we use current ##\vec x## or initial ##\vec X## position configurations. The term “frame” is used to mean this, not the observer’s frame of reference.

(In the eulerian formulation we have ##\vec x## which are fixed in space. Material flows through these points. In the lagrangian formulation, we have ##\vec X## which are allowed to move. Material flows with these points.)

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Again no! Both the Lagrangian and the Eulerian description work in a fixed inertial frame (or sometimes, when you apply hydro to the physics of the atmosphere/meteorology in the rotating frame fixed on a point on Earth). Neither of these use the very complicated non-inertial frame of an observer fixed to a fluid element.

An exception is cosmology, where the most simple observer is an observer co-moving with the "cosmological substrate", i.e., an observer at rest wrt. the standard reference frame of the FLRW spacetime.
My background is in rheology, which includes non-Newtonian fluid mechanics. In these disciplines, we routinely make use of moving observers and even observers who rotate with the average velocity of the fluid parcel (Jaumann derivative) and deform with the fluid parcel (upper- and lower covective derivatives).

Call the material derivative D/Dt what you want, but do you disagree that $$\frac{DT}{Dt}=\frac{\partial T}{\partial t}+v\centerdot \nabla T$$ can be interpreted the derivative of temperature with respect to time as measured by an observer traveling with the fluid parcel?

Call the material derivative D/Dt what you want, but do you disagree that $$\frac{DT}{Dt}=\frac{\partial T}{\partial t}+v\centerdot \nabla T$$ can be interpreted the derivative of temperature with respect to time as measured by an observer traveling with the fluid parcel?
No, I agree with this.

I (now) think this is more a word-game than any real disagreement. The material derivative computes changes in some variable as perceived by the moving fluid parcel. That is indeed 'as moving with the parcel'.

But, in all cases I'm familiar with, you still observe (or reference to) the parcel from some fixed reference frame, not moving along with it, apparently you do some more fancy stuff .

• vanhees71 and Chestermiller
My background is in rheology, which includes non-Newtonian fluid mechanics. In these disciplines, we routinely make use of moving observers and even observers who rotate with the average velocity of the fluid parcel (Jaumann derivative) and deform with the fluid parcel (upper- and lower covective derivatives).

Call the material derivative D/Dt what you want, but do you disagree that $$\frac{DT}{Dt}=\frac{\partial T}{\partial t}+v\centerdot \nabla T$$ can be interpreted the derivative of temperature with respect to time as measured by an observer traveling with the fluid parcel?
In Newtonian physics that's the material time derivative with respect to (absolute Newtonian) time. In relativistic fluid dynamics you's rather take the time derivative of the temperature field wrt. the proper time of the fluid cell, i.e., ##\mathrm{D}_{\tau} T(x)=u^{\mu} \nabla_{\mu} T(x)##, where ##u^{\mu}## is the four-flow-velocity of the fluid cell (##u_{\mu} u^{\mu}=1##), and ##T(x)## is the temperature as measured in a local inertial rest frame of this fluid cell.

I also think now that we quarrel just about words and not about the physics content. The importance to choose a local inertial rest frame is also only important in relativistic fluid dynamics, not so much in the Newtonian case.

• Arjan82