Lecture 15: The Reynolds Transport Theorem
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Overview
Steven M. Wise develops the Reynolds transport theorem by pulling spatial integrals back to material coordinates, using the deformation-gradient Jacobian and its time evolution, and then applying the divergence theorem. He also explains how boundary integrals transform, establishes equivalent criteria for isochoric motion, and derives the Leibniz integral rule for moving intervals as a one-dimensional case of Reynolds transport.
Key takeaways
- Changing a spatial volume integral to material coordinates multiplies its integrand by J = det F, and admissible motions ensure J is positive.
- The Jacobian evolves according to J̇ = J div v, so velocity divergence measures the instantaneous local rate of volume change.
- Reynolds transport separates the rate of change of an integral into local field variation and flux across the moving boundary.
- The cofactor G = JF⁻ᵀ transforms material surface normals into the corresponding spatial surface-area vectors.
- Isochoric motion is characterized by J = 1 and, for motion from the reference configuration, equivalently by div v = 0.
- The Leibniz rule for an interval with moving endpoints follows from Reynolds transport, with endpoint contributions f(b,t)b′(t) and −f(a,t)a′(t).
Chapters
- For a motion map φ, a spatial domain Ωₜ is the image of a material domain Ω.
- Transforming a spatial scalar-field integral to Ω introduces the Jacobian J = det F, where F = ∇ₓφ.
- Admissible motions have J > 0, so the change-of-variables formula needs no absolute-value sign.
- If J equals 1 at a given time, changing variables in the volume integral gives |Ωₜ| = ∫Ω J dX = |Ω|.
- The set may change shape under the motion map, but its volume remains unchanged when its Jacobian is one.
- Differentiating J = det F uses the determinant derivative identity and the tensor inner product expressed as a trace.
- The motion map gives Ḟ = (∇v)ₘF; multiplying by F⁻¹ reduces the determinant-rate expression to tr((∇v)ₘ).
- The resulting identity is J̇ = J(div v)ₘ, linking local volume change to the spatial velocity divergence.
- For a moving spatial region Ωₜ, the theorem states d/dt ∫Ωₜ Q dv = ∫Ωₜ(Q̇ + Q div v) dv.
- Using Q̇ = ∂Q/∂t + ∇Q·v gives the equivalent form ∫Ωₜ ∂Q/∂t dv + ∫∂Ωₜ Q(v·n) da.
- The theorem accounts for both the changing field and the changing integration region.
- Write Ωₜ = φₜ(Ω) and transform the integral to the fixed material region Ω using QₘJ.
- Because Ω is fixed, the time derivative passes through the integral and the product rule differentiates QₘJ.
- Converting the resulting terms back to spatial coordinates uses Q̇ and the Jacobian identity J̇ = J(div v)ₘ.
- Expand the material form of Q̇ as ∂Q/∂t + ∇Q·v.
- Combine ∇Q·v and Q div v as div(Qv), using the divergence product rule.
- The divergence theorem turns ∫Ωₜ div(Qv) dv into the boundary flux ∫∂Ωₜ Q(v·n) da.
- The cofactor field is G = JF⁻ᵀ, and it transforms the material normal into the spatial surface-area vector.
- A spatial vector-field flux W·n pulls back to an integral involving Wₘ·(GN) over ∂Ω.
- A tensor traction integral involving Tn pulls back with TₘGN, preserving the vector-valued integrand.
- A motion is isochoric when every material region and its image have equal volume at every time.
- The pointwise Jacobian criterion is J = det F = 1 throughout the material configuration.
- With the initial configuration fixed, the equivalent velocity criterion is div v = 0 throughout the moving spatial domain.
- Map the fixed interval [0,1] to the moving interval [a(t), b(t)] with φ(X,t) = a(t) + (b(t) − a(t))X.
- The endpoint velocities are a′(t) and b′(t); the outward normals are −1 at a(t) and +1 at b(t).
- Applying Reynolds transport yields d/dt ∫ₐ(t)^ᵇ(t) f(x,t) dx = ∫ₐᵇ ∂f/∂t dx + f(b,t)b′(t) − f(a,t)a′(t).
- When both endpoints are fixed, the boundary terms vanish and only the integral of the partial time derivative remains.
Transcript excerpt
0:04 Welcome back to continue mechanics. This is class number 15. We're still in chapter 4. This is kinematics and um this is part four of our study in chapter 4. Today we're going to talk about a bunch of things. First of all, change of variables. How do we change variables whenever we're uh trying to integrate uh on a change an integral from a spatial description to one to a material description. Then we're going to talk about Reynolds transport theorem and we're going to learn or we're going to use what we learned in the change of variable section to prove a very famous result Reynolds transport theorem. Then we're going to talk about how boundary integrals transform when we uh change
0:53 them from material descriptions to spatial and and vice versa. Uh finally we're going to talk about isoturic or volume preserving flows. Uh that's going to be the final uh theoretical result that we want to cover in today's lecture. And then finally, I want to show you that the linenets rule of integration that you learned in calculus is actually um easily derived from the Reynolds transport theorem. It's not ever proved that way in your elementary uh calculus class or your advanced calculus class, but uh it actually is a result of the Reynolds transport theorem, which we'll discuss today. So, first of all, let's talk about change of variables. The idea is we are calculating an integral on a spatial or in a spatial
1:45 description and we want to transform that to one of a material description. So how do we do it? Well, we need to know something about the motion map. So suppose that fi is our motion map. uh let b of t be the mapping of b some set under the motion map. So this
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Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Steven M. Wise.