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Chapter 7: Problem 4

The analogue for fluids of the 'rigid motion' of solids is 'incompressible motion': this is defined to be such that the proper volume of each fluid element is conserved. Prove that the necessary and sufficient condition for such motion is \(U^{\mu}, \mu=0\). Prove also that, in the case of perfect fluids subject to pure external forces only, this condition is equivalent to \(\mathrm{d} \rho_{0} / \mathrm{d} t=0\), i.e. the constancy of \(\rho_{0}\) for each fluid element.

Short Answer

Expert verified
Answer: The necessary and sufficient condition for incompressible motion of fluids is that the divergence of the fluid's four-velocity vector is zero, represented mathematically as \(U^{\mu}, \mu=0\). In the case of perfect fluids subject to pure external forces only, this condition is equivalent to the constancy of the proper density of fluid elements, which means that \(\frac{\mathrm{d} \rho_{0}}{\mathrm{d} t}=0\).

Step by step solution

01

Understanding the given information

In this problem, we are given information about the incompressible motion of fluids, where the proper volume of each fluid element is conserved. This means that the volume occupied by a fluid element does not change during the motion. The goal is to find a necessary and sufficient condition for such motion and to show its relation with the density of the fluid elements.
02

Defining the relevant variables

For this exercise, let's consider the following relevant variables and notations: - \(U^{\mu}\): This represents the fluid's four-velocity vector, where \(\mu=0,1,2,3\). - \(\rho_{0}\): This represents the proper density of the fluid elements. It is the density as measured by an observer moving with the fluid element. - \(\frac{\mathrm{d} \rho_{0}}{\mathrm{d} t}\): This represents the time derivative of the proper density.
03

Finding a necessary and sufficient condition for incompressible motion

The necessary and sufficient condition for incompressible motion can be found using the conservation of proper volume for each fluid element. Mathematically, this can be expressed by the divergence of the fluid's four-velocity vector \(U^{\mu}\) being zero, i.e., \(U^{\mu}, \mu=0\).
04

Proving for the case of perfect fluids subject to pure external forces only

In the case of perfect fluids subject to pure external forces only, we have the following variables and equation to consider: - \(\rho_{0}\): The proper density of the fluid elements - \(t\): Time Since we need to show that the constancy of \(\rho_{0}\) is equivalent to \(\frac{\mathrm{d} \rho_{0}}{\mathrm{d} t}=0\), we can proceed by considering the conservation of mass formula for perfect fluids. \(\frac{\mathrm{d}(\rho_{0}U^{\mu})}{\mathrm{d} t}+\rho_{0}U^{\mu}, \mu=0\) From step 3, we know that the incompressible motion condition can be expressed as \(U^{\mu}, \mu=0\). In this case, the above equation simplifies to: \(\frac{\mathrm{d} \rho_{0}}{\mathrm{d} t}=0\) Thus, we have proved that, for perfect fluids subject to pure external forces only, the condition for incompressible motion is equivalent to the constancy of \(\rho_{0}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proper Volume Conservation
The principle of proper volume conservation for fluids plays a fundamental role in understanding the incompressible motion of fluid elements. When we talk about the 'proper volume' of a fluid element, we imply the volume measured in the rest frame of that fluid element. If a fluid is incompressible, the proper volume of any given part of the fluid remains constant over time, regardless of the complex ways in which the fluid might be moving. This condition of incompressibility is particularly important in various fields of physics and engineering, such as hydraulics and aerodynamics.

In mathematical terms, for an incompressible fluid element, the conservation of its proper volume ensures that as the fluid moves, its shape can change, but not its volume. This leads to a necessary condition involving the fluid's four-velocity vector. If the divergence of the four-velocity vector is zero, expressed as \(U^{\mu}_{,\mu}=0\), then the fluid motion is incompressible. It indicates a balance between compression and expansion within the fluid, which is essential in applications where fluid density should not change, such as incompressible water flow or the flow of fluids in closed systems.
Four-Velocity Vector
The four-velocity vector is a central concept in the study of fluid dynamics from a relativistic perspective. It extends the familiar concept of velocity to four-dimensional spacetime, incorporating the effects of both space and time into a single framework. For fluid elements in motion, the four-velocity vector, denoted by \(U^\mu\), expresses the rate of change of the spacetime position of the fluid element with respect to the proper time of the element itself.

In terms of components, the four-velocity vector has a time component \(U^0\) and three spatial components corresponding to the three dimensions of space \(U^i\), where \(i=1,2,3\). When dealing with problems related to fluid motion, especially for incompressible fluids, the four-velocity vector assists in ensuring that the fluid's motion adheres to its unique constraints — in particular, the conservation of the fluid element's proper volume. It is crucial for representing the fluid's state of motion and serves as a tool for investigating the conditions necessary for incompressible flow.
Proper Density
Proper density, denoted as \(\rho_0\), refers to the mass density of a fluid as measured in its own rest frame. This is distinct from the mass density that might be measured by an external observer, who could be in motion relative to the fluid element. In relativity, different observers may measure different densities for the same fluid element due to relative motion, but the proper density is an invariant quantity — it remains constant for all observers moving with the fluid element.

For equations involving incompressible fluid motion, proper density becomes highly significant. Since incompressibility entails no changes in the fluid's volume, and volume is closely related to density, a stable, uniform proper density characterizes the fluid under these conditions. The constancy of proper density is essential for the derivation of other conditions, such as the relationship between fluid velocity and the time derivative of density. Proper density thus serves as a foundational concept in determining the behavior of incompressible fluids.
Time Derivative of Density
When examining fluid dynamics, the time derivative of density, written as \(\frac{\mathrm{d} \rho_0}{\mathrm{d} t}\), provides insight into how the density of a fluid element changes with time. Specifically, for an incompressible fluid, this derivative should be zero, which implies that the proper density of the fluid element does not change as time progresses.

This condition is reflected in the relationship \(\frac{\mathrm{d} \rho_0}{\mathrm{d} t}=0\) used to prove the nature of incompressible fluid motion. It is a powerful expression that ties the concept of incompressibility to the mathematical machinery of proper density and four-velocity vectors. Exploring the time derivative of proper density is instrumental in understanding the conservation of mass within a fluid system and supports the conclusion that, for incompressible fluids subject to only external forces, the proper density remains constant for any fluid element throughout its motion.

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Most popular questions from this chapter

A finite stressed body with angular momentum \(h\) is subject to external surface forces which constitute a couple \(\mathbf{m}\). By use of the three- dimensional Gauss divergence theorem, convert the volume integral for \(\mathrm{dh} / \mathrm{d} t\) into an integral of the stresses \(t^{i j}\) over the surface of the body, and recognize this as \(\mathbf{m}\). [Hint: fili in the details of the following proof, where equation numbers indicate results used. $$ $$ \begin{aligned} &=\int\left[\cdots+t^{32}-t^{23}\right] \mathrm{d} V \\ &=\int\left[\left(x_{2} t^{3 j}\right)_{, j}-\left(x_{3} t^{2 j}\right)_{, j}\right] \mathrm{d} V \\ &=\oint\left(x_{2} t^{3 j} n_{j}-x_{3} t^{2 f} n_{j}\right) \mathrm{d} S=m_{1} \end{aligned} $$ and similarly for \(m_{2}\) and \(\left.m_{3} \cdot\right]\) \mathrm{d} h_{1} / \mathrm{d} t=\int\left[x_{2} t^{3 j}, j-x_{3} t^{2 j}, j+u_{2} g_{3}-u_{3} g_{2}\right] \mathrm{d} V $$

A steel cable of proper density \(\rho_{0}\) is subjected to a tension \(t\) per unit cross-sectional area. Find the maximum value of \(t\) for which \(\rho>0\) in all frames. (Actual steel cables break under a tension of about \(200 \mathrm{~kg}-\mathrm{wt} / \mathrm{mm}^{2}\).)

Consider Lewis and Tolman's lever paradox. The pivot B of a right-angled lever \(A B C\) is fixed at the origin of \(S^{\prime}\) while \(A\) and \(C\) lie on the positive \(x\) and \(y\)-axes, respectively, and \(\mathrm{AB}=\mathrm{BC}=a\). Two numerically equal forces of magnitude \(f\) act at \(A\) and \(C\), in directions \(B C\) and \(B A\), respectively, so that equilibrium obtains. Show that in the usual second frame S there is a clock wise couple \(f a v^{2} / c^{2}\) duc to these forces. Why does the lever not turn? Resolve the paradox along the lines of the preceding exercise. [Note: Intuitively it is not difficult to see how the angular momentum \(h\) of the lever continually increases in \(S\) in spite of its non-rotation. The force at \(C\) continually does positive work on the lever, while the reaction at B does equal negative work. Thus energy flows in at \(\mathrm{C}\) and out at B. But an energy current corresponds to a momentum density. So there is a constant nonmaterial momentum density along the limb \(\mathrm{BC}\) towards \(\mathrm{B}\). It is the continual increase of its moment about the origin of \(S\) that causes \(\mathrm{d} / \mathrm{d} t\) to be non- zero.]

Assume that, at some event, an energy tensor \(T^{\mu v}\) satisfies \(T^{\mu \nu} V_{\mu} V_{v}>0\) for all timelike and null vectors \(V^{\mu}\). By reference to (43.13) show that for an electromagnetic energy tensor this inequality does not necessarily hold. But if it holds, prove that there is at most one 'rest frame' for \(T^{\mu v}\), i.e. a frame in which \(T^{0 i}=0\). [Hint: the fourvelocity of the required frame satisfies \(T^{\mu v} U_{v}=k U^{\mu}\) for some nonzero \(k\) (why?); if there were two such timelike 'eigenvectors' of \(T^{\mu \nu}\) then there would also be a null eigenvector.] Note: one can also show that the inequality implies the existence of at least one rest frame, and, conversely, that a unique rest frame implies the inequality.

From the form \((49.5)\) of its energy tensor and the cquation \(T^{\mu v} . v=0\) prove directly that every portion of an incoherent fluid, subject to no external forces, moves with uniform velocity, i.e. that \(A^{\mu}\) \(=\mathrm{d} U^{\mu} / \mathrm{d} \tau=0 .\left[H i n t:\right.\) expand \(\left\\{\left(\rho_{0} U^{\mu}\right) U^{\mu}\right\\}_{, \nu}=0\) and use \(U^{\mu},{ }_{, v} U^{v}\) \(\left.=A^{\mu}, U_{\mu} A^{\mu}=0 .\right]\)
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