Write out \nabla \vec{u} in terms of unit dyads
This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
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PURPOSE OF THE EXERCISE:
Demonstrate how to express the velocity gradient in its full dyadic form using unit dyads; that is, to relate the spatial derivatives of the velocity components to the oriented basis in a Cartesian coordinate system. -
ENGINEERING CONTEXT:
In many fluid mechanics applications—including flow visualization, numerical simulations, and turbulence analysis—it is essential to represent the velocity gradient tensor explicitly, as this forms the basis for quantifying both deformation and rotation of fluid elements. This skill underpins the design and analysis of equipment such as mixers, pumps, and aerodynamic surfaces. -
PHYSICAL CONTEXT:
The velocity gradient tensor captures the local kinematics of the fluid by separating the deformation (through its symmetric part) from the rotation (through its antisymmetric part), being central to understanding viscous stresses and rate-of-strain effects. -
ASSUMPTIONS:
Assume a two-dimensional, steady, and continuously differentiable velocity field in Cartesian coordinates; this is reasonable for many laminar flows or simplified modeling scenarios. -
SOLUTION TECHNIQUE:
Express the velocity field as \vec{u} = u(x,y)\,\hat{x} + v(x,y)\,\hat{y}. The velocity gradient in terms of unit dyads is then given by
\nabla\vec{u} = \frac{\partial u}{\partial x}\,\hat{x}\hat{x} + \frac{\partial u}{\partial y}\,\hat{x}\hat{y} + \frac{\partial v}{\partial x}\,\hat{y}\hat{x} + \frac{\partial v}{\partial y}\,\hat{y}\hat{y}\,.
A student should consider drawing a schematic of an infinitesimal fluid element with edges aligned along the \hat{x} and \hat{y} directions to visualize how each derivative corresponds to stretching, shearing, or rotation.
- REFLECTION:
This representation directly links the measurable gradients (e.g., \partial u/\partial x) to the mechanical deformation of fluid elements. By decomposing the tensor further into its symmetric and antisymmetric parts, one can quantify the extensional strain rate and rotation rate separately, which is critical when evaluating viscous effects in Newtonian fluids or determining vorticity in complex flows.
In formulating this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections according to their relevance. If they are marked as relevant, I encourage you to explore them:
Textbook sections:
- Chapter 9 Fundamentals: Gradient (HIGHLY RELEVANT)
- Chapter 3 Fundamentals: Gradient (HIGHLY RELEVANT)
- Appendix D Fundamentals: Why … (HIGHLY RELEVANT)
- Appendix E Fundamentals: Finding the symmetric and antisymmetric parts of the velocity gradient dyad… (HIGHLY RELEVANT)
- Appendix E.1: Velocity gradients (HIGHLY RELEVANT)
Disclaimer:
See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.
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is \hat{x}\hat{y} the same as \hat{xy}
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i need the answer in 3d
This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
Below is a complete representation of the velocity gradient in its full dyadic form in a Cartesian coordinate system. Consider a three‐dimensional velocity field
\vec{u}=u(x,y,z)\,\hat{x}+v(x,y,z)\,\hat{y}+w(x,y,z)\,\hat{z}\,.
Then the velocity gradient dyadic is written in terms of unit dyads as
\nabla\vec{u} =
\frac{\partial u}{\partial x}\,\hat{x}\hat{x} \;+\;
\frac{\partial u}{\partial y}\,\hat{x}\hat{y} \;+\;
\frac{\partial u}{\partial z}\,\hat{x}\hat{z} \;+\;
\frac{\partial v}{\partial x}\,\hat{y}\hat{x} \;+\;
\frac{\partial v}{\partial y}\,\hat{y}\hat{y} \;+\;
\frac{\partial v}{\partial z}\,\hat{y}\hat{z} \;+\;
\frac{\partial w}{\partial x}\,\hat{z}\hat{x} \;+\;
\frac{\partial w}{\partial y}\,\hat{z}\hat{y} \;+\;
\frac{\partial w}{\partial z}\,\hat{z}\hat{z}\,.
Each term, such as \frac{\partial u}{\partial y}\,\hat{x}\hat{y}, represents the rate at which the x-component of the velocity changes in the y-direction, with the dyadic \hat{x}\hat{y} indicating that an oriented basis vector \hat{x} acts in the output (or force) direction when considering a surface whose normal is along \hat{y}.
This formulation forms the basis for further analysis in fluid mechanics since the symmetric part of this tensor (obtained by averaging with its transpose) quantifies the deformation of a fluid element, and the antisymmetric part quantifies its rotation.
In formulating this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections based on their relevance. If they are marked as relevant, I encourage you to explore them:
Textbook sections:
- Appendix E.1: Velocity gradients (HIGHLY RELEVANT)
- Chapter 9 Fundamentals: Gradient (HIGHLY RELEVANT)
- Chapter 3 Fundamentals: Gradient (HIGHLY RELEVANT)
- Appendix E Fundamentals: Finding the symmetric and antisymmetric parts of the velocity gradient dyad… (HIGHLY RELEVANT)
- Appendix D Fundamentals: Why … (HIGHLY RELEVANT)
Disclaimer:
See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.
- is \hat{x}\hat{y} the same as \hat{xy}
This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
Below is the key point:
Unit dyads are formed by the dyadic product of two unit vectors. In our notation, writing
\hat{x}\hat{y}
means the dyadic product of the unit vector in the x direction with the unit vector in the y direction. We can also denote this dyad as
\widehat{xy}\,,
so \hat{x}\hat{y} is the same as \widehat{xy}.
It is important to note that the dyadic product is not commutative; that is, \hat{x}\hat{y} (or \widehat{xy}) is not the same as \hat{y}\hat{x} (or \widehat{yx}).
In formulating this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections based on their relevance. If they are marked as relevant, I encourage you to explore them:
Textbook sections:
- Appendix B.3: Dyadic Tensor Algebra Part 1 (HIGHLY RELEVANT)
- Appendix B.3: Dyadic Tensor Algebra Part 3 Part 1 (HIGHLY RELEVANT)
- Appendix B.1: Vector Algebra Part 6 Part 1 (HIGHLY RELEVANT)
- Appendix B Fundamentals: Relation between tensors, tensor operators, and matrices (HIGHLY RELEVANT)
- Appendix E Fundamentals: Relation between tensors, tensor operators, and matrices (HIGHLY RELEVANT)
Disclaimer:
See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.
So a very short question will lead to a very long answer.
Some comments:
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you can write the equations of incompressible fluid mechanics without ever writing out what \nabla\vec{u} is. To do that, you write the convective term by use of the directional derivative operator \left(\vec{u}\cdot\nabla\right), which is notation that looks like it is a dot product between a vector and something, but which is actually just a symbol for a specific vector operation – it operates on the vector \vec{u} and generates a vector. The viscous terms are a function not of \nabla\vec{u} but of its symmetric part \vec{\vec{\varepsilon}}, so the order of indices or coordinates doesn’t have to be correct to get the viscous stress correct. Because these two tools can be used to avoid ever writing out \nabla\vec{u}, many fluid mechanicians never actually look at what \nabla\vec{u} is, and in particular what is the right order of the symbols. Is the correct \hat{xy} term \frac{\partial u}{\partial y}\hat{xy}, or is it \frac{\partial v}{\partial x}\hat{xy}? alas, it depends on what convention you use. I think this all matters because \nabla\vec{u} is important of its own right – it is the dyadic that fully describes the deformation and rotation of the fluid, and I don’t want to avoid it. I want to talk about it.
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the convention that most fluid mechanicians implicity use is that the correct term is \frac{\partial v}{\partial x}\hat{xy}. That is the convention I use in my book. If you use that convention, \left(\vec{u}\cdot\nabla\right)\vec{u}=\vec{u}\cdot\nabla\vec{u}. Recall that \left(\vec{u}\cdot\nabla\right)\vec{u} is a vector operation. \vec{u}\cdot\nabla\vec{u} is a tiny bit different. \nabla\vec{u} is a dyadic tensor, so \vec{u}\cdot\nabla\vec{u} uses \nabla to operate on \vec{u} to generate a dyadic, and then contracts that dyadic by taking the dot product of \vec{u} with the result. So these two operations are very close to being the same thing but are not quite the same thing. If you use “my” convention, these two operations are equal. With “my” convention,
dotting a unit outward normal into the dyadic contracts the dyadic by specifying a relevant plane where forces are applied or a plane through which fluid properties move. dotting a velocity vector into the dyadic contracts the dyadic by summing up the fluxes caused by different velocity components. Everything about this makes sense to me in my brain. -
the convention that most tensor mathematicians use, as best I can tell, is that the correct term is \frac{\partial u}{\partial y}\hat{xy}. If you use that convention, \left(\vec{u}\cdot\nabla\right)\vec{u}=\nabla\vec{u} \cdot \vec{u}. Using this “other” convention, to contract a dyadic by specifying the plane it applies to requires dotting the vector into the dyadic, but contracting the dyadic by specifying fluxes requires dotting the dyadic into the vector. It isn’t wrong, I just don’t like it. and I find it harder to explain the dyadic tensor calculus this way. So I do not use this “other” convention.
ok so professor kirby bot has written the order of the coordinates in the dyadics using the “other” convention, so professor kirby bot’s description of \nabla\vec{u} is inconsistent with the expressions I use throughout my text. if you use professor kirby bot’s description, and write the convective momentum flux term in terms of this dyadic tensor, you would have to write the convective momentum flux term as \nabla\vec{u}\cdot\vec{u}. and that hurts my brain.
Professor Kirby Bot, could you rewrite your answer using the notation from my text, with attention to my comments above?
This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
Below is the velocity gradient dyadic expressed in a three‐dimensional Cartesian coordinate system using the notation and convention from your text. In this convention the first unit vector in each dyad indicates the component of velocity, and the second unit vector indicates the direction along which the velocity varies. For a velocity field
\vec{u}(x,y,z) = u(x,y,z)\,\hat{x} + v(x,y,z)\,\hat{y} + w(x,y,z)\,\hat{z}\,,
the full velocity gradient is written as
\begin{aligned}
\nabla\vec{u} &= \frac{\partial u}{\partial x}\,\hat{x}\hat{x} + \frac{\partial u}{\partial y}\,\hat{x}\hat{y} + \frac{\partial u}{\partial z}\,\hat{x}\hat{z} \\
&\quad +\, \frac{\partial v}{\partial x}\,\hat{y}\hat{x} + \frac{\partial v}{\partial y}\,\hat{y}\hat{y} + \frac{\partial v}{\partial z}\,\hat{y}\hat{z} \\
&\quad +\, \frac{\partial w}{\partial x}\,\hat{z}\hat{x} + \frac{\partial w}{\partial y}\,\hat{z}\hat{y} + \frac{\partial w}{\partial z}\,\hat{z}\hat{z}\,.
\end{aligned}
In this notation the term
\frac{\partial v}{\partial x}\,\hat{y}\hat{x}
appears as the shear component in the xy-plane. Here, the left unit vector (\hat{y}) shows that it is the velocity component in the y-direction, and the right unit vector (\hat{x}) indicates that the derivative is taken in the x-direction.
This convention, which is common among fluid mechanicians, guarantees that the contraction needed to form the convective term is unambiguous. In particular, one obtains
(\vec{u}\cdot\nabla)\vec{u} = \vec{u}\cdot\nabla\vec{u}\,,
since dotting the velocity vector into the dyadic automatically sums the contributions from derivatives taken in the direction of each coordinate.
When further decomposing \nabla\vec{u}, the symmetric part
\boldsymbol{\varepsilon} = \frac{1}{2}\Bigl(\nabla\vec{u} + (\nabla\vec{u})^{\mathrm T}\Bigr)
yields the strain-rate tensor (characterizing rates of deformation), and the antisymmetric part
\boldsymbol{\omega} = \frac{1}{2}\Bigl(\nabla\vec{u} - (\nabla\vec{u})^{\mathrm T}\Bigr)
captures the fluid rotation. This full specification of \nabla\vec{u} is crucial for directly relating measurable spatial velocity derivatives to fluid deformations and rotations.
By using this ordering—where, for example, \frac{\partial v}{\partial x}\,\hat{y}\hat{x} is the shear contribution in the xy-plane—the notation is consistent with the approach in your text.
In formulating this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections based on their relevance. If they are marked as relevant, I encourage you to explore them:
Textbook sections:
- Chapter 9 Fundamentals: Gradient (HIGHLY RELEVANT)
- Chapter 3 Fundamentals: Gradient (HIGHLY RELEVANT)
- Appendix E Fundamentals: Finding the symmetric and antisymmetric parts of the velocity gradient dyad… (HIGHLY RELEVANT)
- Appendix E.1: Velocity gradients (HIGHLY RELEVANT)
- Appendix D Fundamentals: Why … (HIGHLY RELEVANT)
Disclaimer:
See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.
professor kirby bot, you still have the convention opposite from my book. The first coordinate of the dyad corresponds to the derivative and the second coordinate corresponds to the velocity component
This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
Below is the velocity gradient dyadic expressed using the convention where the first unit vector denotes the direction in which the derivative is taken and the second unit vector denotes the velocity component being differentiated. For a three‐dimensional velocity field
\vec{u}(x,y,z) = u(x,y,z)\,\hat{x} + v(x,y,z)\,\hat{y} + w(x,y,z)\,\hat{z}\,,
the full velocity gradient can be written as
\begin{aligned}
\nabla\vec{u} &= \frac{\partial u}{\partial x}\,\hat{x}\hat{x} + \frac{\partial v}{\partial x}\,\hat{x}\hat{y} + \frac{\partial w}{\partial x}\,\hat{x}\hat{z} \\
&\quad +\, \frac{\partial u}{\partial y}\,\hat{y}\hat{x} + \frac{\partial v}{\partial y}\,\hat{y}\hat{y} + \frac{\partial w}{\partial y}\,\hat{y}\hat{z} \\
&\quad +\, \frac{\partial u}{\partial z}\,\hat{z}\hat{x} + \frac{\partial v}{\partial z}\,\hat{z}\hat{y} + \frac{\partial w}{\partial z}\,\hat{z}\hat{z}\,.
\end{aligned}
In this formulation the first unit vector (to the left of the dyad) identifies the coordinate along which the derivative is taken, and the second unit vector (to the right) indicates the velocity component that is being differentiated. For example, the term
\frac{\partial v}{\partial x}\,\hat{x}\hat{y}
indicates that the y-component of the velocity, v, varies along the x direction.
This ordering guarantees that when contracting with a velocity vector to form the convective term, the relationship
(\vec{u}\cdot\nabla)\vec{u} = \nabla\vec{u}\cdot\vec{u}
holds true under this notation—the derivative (first index) properly acts on the velocity component (second index), yielding a clear flux interpretation.
In formulating this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections according to their relevance. If they are marked as relevant, I encourage you to explore them:
Textbook sections:
- Appendix E.1: Velocity gradients (HIGHLY RELEVANT)
- Chapter 9 Fundamentals: Gradient (HIGHLY RELEVANT)
- Chapter 3 Fundamentals: Gradient (HIGHLY RELEVANT)
- Appendix E Fundamentals: Finding the symmetric and antisymmetric parts of the velocity gradient dyad… (HIGHLY RELEVANT)
- Appendix D Fundamentals: Why … (HIGHLY RELEVANT)
Disclaimer:
See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.
ok professor kirby bot now you have the velocity gradient correct, but now you have the directional derivative defined backwards
This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
Below is a clarification using the convention in which the first index in the dyad denotes the direction along which the derivative is taken and the second index denotes the velocity component being differentiated. In this convention we write
\nabla\vec{u} =
\frac{\partial u}{\partial x}\,\hat{x}\hat{x} + \frac{\partial v}{\partial x}\,\hat{x}\hat{y} + \frac{\partial w}{\partial x}\,\hat{x}\hat{z}
+ \frac{\partial u}{\partial y}\,\hat{y}\hat{x} + \frac{\partial v}{\partial y}\,\hat{y}\hat{y} + \frac{\partial w}{\partial y}\,\hat{y}\hat{z}
+ \frac{\partial u}{\partial z}\,\hat{z}\hat{x} + \frac{\partial v}{\partial z}\,\hat{z}\hat{y} + \frac{\partial w}{\partial z}\,\hat{z}\hat{z}\,.
In our notation the left unit vector indicates the direction of differentiation and the right unit vector indicates the velocity component that is being differentiated. For example, the term
\frac{\partial v}{\partial x}\,\hat{x}\hat{y}
means that the $y$‑component of the velocity varies in the $x$‑direction.
The directional (or convective) derivative of the velocity is defined as
(\vec{u}\cdot\nabla)\vec{u}
= u\,\frac{\partial \vec{u}}{\partial x} + v\,\frac{\partial \vec{u}}{\partial y} + w\,\frac{\partial \vec{u}}{\partial z}\,.
Because our dyadic has its first index corresponding to the differentiation direction, the derivative operator naturally acts on the left index. Hence, contracting the derivative index with the velocity vector requires that we contract the first index of \nabla\vec{u} with \vec{u}. In dyadic notation this contraction is carried out on the transpose of the gradient (which has the derivative index moved to the right). In other words, with our ordering we have
(\vec{u}\cdot\nabla)\vec{u} = (\nabla\vec{u})^{\mathrm T}\cdot\vec{u}\,.
This correction ensures that the directional derivative is defined correctly under our convention. Here the transpose, (\nabla\vec{u})^{\mathrm T}, reorders the indices so that the contraction with \vec{u} is performed on the derivative (first) index rather than on the velocity-component (second) index.
By defining the operator this way the usual identity
(\vec{u}\cdot\nabla)\vec{u} = \vec{u}\cdot\nabla\vec{u}
no longer holds symbolically unless one is very explicit about which index is contracted. In our convention the proper representation is
(\vec{u}\cdot\nabla)\vec{u} = (\nabla\vec{u})^{\mathrm T}\cdot\vec{u}\,,
which captures the fact that the directional derivative acts along the direction of the velocity field.
In formulating this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections according to their relevance. If they are marked as relevant, I encourage you to explore them:
Textbook sections:
- Chapter 9 Fundamentals: Gradient (HIGHLY RELEVANT)
- Chapter 3 Fundamentals: Gradient (HIGHLY RELEVANT)
- Appendix D Fundamentals: Why … (HIGHLY RELEVANT)
- Appendix E Fundamentals: Finding the symmetric and antisymmetric parts of the velocity gradient dyad… (HIGHLY RELEVANT)
- Appendix E.1: Velocity gradients (HIGHLY RELEVANT)
Disclaimer:
See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.
oh boy professor kirby bot is having a rough time on this one.
here is the answer i would have given:
For a three‐dimensional velocity field
\vec{u}(x,y,z) = u(x,y,z)\,\hat{x} + v(x,y,z)\,\hat{y} + w(x,y,z)\,\hat{z}\,,
I write the velocity gradient as
\begin{aligned}
\nabla\vec{u} &= \frac{\partial u}{\partial x}\,\hat{xx} + \frac{\partial v}{\partial x}\,\hat{xy} + \frac{\partial w}{\partial x}\,\hat{xz} \\
&\quad +\, \frac{\partial u}{\partial y}\,\hat{yx} + \frac{\partial v}{\partial y}\,\hat{yy} + \frac{\partial w}{\partial y}\,\hat{yz} \\
&\quad +\, \frac{\partial u}{\partial z}\,\hat{zx} + \frac{\partial v}{\partial z}\,\hat{zy} + \frac{\partial w}{\partial z}\,\hat{zz}\,.
\end{aligned}
Using this convention:
(\vec{u}\cdot\nabla)\vec{u} = \vec{u}\cdot\nabla\vec{u}