MTW Exercise 22.7 -- Calculate the law of local energy conservation for a viscous fluid with heat flow

In summary, Terry was having trouble solving a problem and asked for help. He was told to look at what he had done so far and to see if it was OK. If it was, he was told how to proceed from there.
  • #1
TerryW
Gold Member
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Homework Statement
MTW Ex 22.7(d)
Relevant Equations
(22.16h) - See attachment
I've come to a grinding halt with this and I can't see a way forward.

Can someone please take a look at what I've done so far and let me know if what I have done is OK and then if it is, give me a hint on how to proceed.

First up,

Is ## u \cdot \nabla \cdot T = u_\alpha T^{\alpha\beta}{}_{;\beta} ## ?

If it is, then from (22.16d)

##u_\alpha T^{\alpha\beta}{}_{;\beta} = u_{\alpha}(\rho u^\alpha u^\beta + (p - \zeta \theta)P^{\alpha \beta} -2\eta \sigma ^{\alpha \beta} +q^{\alpha} u^{\beta} +u^{\alpha} q^{\beta})_{;\beta}##

Working through the derivative and the projections, I reach

##u_\alpha T^{\alpha\beta}{}_{;\beta} = \rho \theta + u^{\beta}\rho_{;\beta} + u^i((p - \zeta \theta)_{;i} + \eta u^i(\frac{1}{3} \theta_{;i}) + u_i u^{\beta} q^i{}_{;\beta} = u_i q^i \theta -q^i{}_{;i}##

Where ##\theta = u^{\alpha}{}_{;\alpha}##

Is that OK?

RegardsTerryW
Screenshot 2021-03-10 at 16.24.57.png
 
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  • #2
Hey, sorry you haven't been getting any replies. I must admit I don't have much of a clue either. Here's just some things I was thinking, but I can't guarantee any of its correctness. $$T \nabla \cdot \mathbf{s} = T \partial_{\mu} s^{\mu} = T\partial_{\mu} (nsu^{\mu} + \frac{1}{T} q^{\mu})$$By baron number conservation you have ##\nabla \cdot \mathbf{S} = \partial_{\mu}(n u^{\mu}) = 0##, thus you can re-write this as$$T\nabla \cdot \mathbf{s} = nT u^{\mu} \partial_{\mu} s + \partial_{\mu} q^{\mu} - \frac{1}{T} q^{\mu} \partial_{\mu} T \ \ \ (1)$$The first law of thermodynamics is$$u^{\mu} \partial_{\mu} \rho = \frac{\rho + p}{n} u^{\mu} \partial_{\mu} n + nT u^{\mu} \partial_{\mu} s$$Inserting this into ##(1)## and re-writing ##u^{\mu}\partial_{\mu} n = -n \partial_{\mu} u^{\mu}##gives$$T\nabla \cdot \mathbf{s} = u^{\mu} \partial_{\mu} \rho + (\rho + p) \partial_{\mu} u^{\mu} + \partial_{\mu} q^{\mu} - \frac{1}{T} q^{\mu} \partial_{\mu} T$$Now look at the local energy conservation equation ##\mathbf{u} \cdot \nabla \cdot \boldsymbol{\mathcal{T}} = 0## with$$
\begin{align*}
\boldsymbol{\mathcal{T}} &= - 2 \eta \boldsymbol{\sigma} - \boldsymbol{\zeta} \theta \mathbf{P} + \mathbf{u} \otimes \mathbf{q} + \mathbf{q} \otimes \mathbf{u} + (p+\rho) \mathbf{u} \otimes \mathbf{u} + \rho \mathbf{g} \\ \\

\mathcal{T}^{\mu \nu} &= -2 \eta \sigma^{\mu \nu} - \zeta \theta P^{\mu \nu} + u^{\mu} q^{\nu} + q^{\mu} u^{\nu} + (p + \rho) u^{\mu} u^{\nu} + \rho g^{\mu \nu}

\end{align*}
$$Then you have$$\begin{align*}
\mathbf{u} \cdot \nabla \cdot \boldsymbol{\mathcal{T}} = -2\eta u_{\nu} \partial_{\mu} \sigma^{\mu \nu} - \zeta u_{\nu} \partial_{\mu} (\theta P^{\mu \nu}) + u_{\nu} \partial_{\mu}(u^{\mu} q^{\nu}) + u_{\nu} \partial_{\mu}(q^{\mu} u^{\nu}) \\ + \, u_{\nu} \partial_{\mu}((p + \rho) u^{\mu} u^{\nu}) + u_{\nu} \partial_{\mu}(\rho g^{\mu \nu}) \overset{!}{=} 0
\end{align*}$$where in a local Lorentz frame you have ##\theta = \partial_{\alpha} u^{\alpha}## and also$$\sigma^{\mu \nu} = \frac{1}{2}(P^{\alpha \nu} \partial_{\alpha} u^{\mu} + P^{\alpha \mu} \partial_{\alpha} u^{\nu}) - \frac{1}{3} P^{\mu \nu} \partial_{\alpha} u^{\alpha}$$I'm not sure how to continue. There's some bits and pieces you can tidy up, using e.g. the metric compatibility ##\partial_{\mu} g^{\mu \nu} = 0## and other identities, but I don't know. Can someone else advise?
 
  • #3
Hi Etotheipi,

Thanks for getting back to me on this.

I'll spend a bit of time comparing your working with mine to see if it generates any further ideas

etotheipi said:
Inserting this into (1) and re-writing uμ∂μn=−n∂μuμgives
I hadn't thought of doing this, I just took ##\partial_{\mu} (nu^{\mu}) = 0 ##

etotheipi said:
where in a local Lorentz frame you have θ=∂αuα and also

I know that the Lorentz frame was used earlier to help arrive at a solution but I wonder if it can help here as reducing ##u \cdot \nabla \cdot T## to the Lorentz frame produces a quite simple equation which doesn't look like it will help me get anywhere.

I tried working back starting with ##\sigma_{\alpha \beta} \sigma ^{\alpha \beta}## but that didn't produce anything that looked useful.

I had a chat with my son last night. He read Astrophysics at King's and knew John Stewart. His main comment was the John Stewart was infamous for setting impossible exam questions!

I'll post again if I make any further progress.RegardsTerryW
 
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Related to MTW Exercise 22.7 -- Calculate the law of local energy conservation for a viscous fluid with heat flow

1. What is the law of local energy conservation?

The law of local energy conservation states that energy cannot be created or destroyed, but it can be transformed from one form to another. In the context of a viscous fluid with heat flow, this means that the total energy of the fluid (kinetic, potential, and thermal) must remain constant at any given point in the fluid.

2. How is the law of local energy conservation calculated for a viscous fluid?

The law of local energy conservation can be calculated by considering the various forms of energy in the fluid and their respective rates of change. This includes the kinetic energy due to fluid motion, potential energy due to gravity, and thermal energy due to heat flow. By accounting for these factors, the law of local energy conservation can be expressed as a mathematical equation.

3. What role does viscosity play in the law of local energy conservation for a viscous fluid?

Viscosity, or the resistance of a fluid to flow, plays a crucial role in the law of local energy conservation for a viscous fluid. This is because viscosity affects the rate of energy dissipation in the fluid, which can impact the overall energy balance. In order to accurately calculate the law of local energy conservation, the viscosity of the fluid must be taken into account.

4. How does heat flow impact the law of local energy conservation for a viscous fluid?

Heat flow, or the transfer of thermal energy from one point to another, is an important factor in the law of local energy conservation for a viscous fluid. This is because heat flow can affect the temperature and therefore the thermal energy of the fluid. By considering the rate of heat flow, the law of local energy conservation can be accurately calculated.

5. Why is the law of local energy conservation important in the study of viscous fluids?

The law of local energy conservation is important in the study of viscous fluids because it allows scientists to understand and predict the behavior of these fluids. By accurately calculating the energy balance in a fluid, researchers can gain insights into the flow patterns, heat transfer, and other important properties of viscous fluids. This knowledge can be applied in various fields such as engineering, environmental science, and materials science.

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