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Chapter 7: Problem 1

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 \(200 \mathrm{~kg}-\mathrm{wt} / \mathrm{mm}^{2}\).)

Short Answer

Expert verified
Answer: The maximum tension, \(t_{max}\), for which the density remains positive in all frames is given by \(t_{max} = \rho_{0} Ag\), where \(A\) is the cross-sectional area, and \(g\) is the acceleration due to gravity. This expression is the same as that obtained using classical mechanics but accounts for the fact that no reference frames exist where the cable's mass would be reduced below zero.

Step by step solution

01

Identify relevant equations for density and relativistic mass

In order to find the maximum tension that the cable can withstand, we need to explore the dependency between the density and tension in a relativistic frame. The density \(\rho\) in a moving frame can be obtained from the proper density \(\rho_{0}\) as follows: \(\rho = \frac{\rho_{0}}{\gamma}\) Where \(\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}\) is the Lorentz factor and \(v\) is the relative speed between the rest frame and the moving frame. The relativistic mass of the cable can be obtained similarly: \(m = \gamma m_{0}\) Where \(m\) and \(m_{0}\) are the relativistic and proper masses of the cable, respectively.
02

Find the mass per unit length of the cable

In terms of density and cross-sectional area, mass per unit length can be written as: \(\frac{m}{L} = \rho A\) Where \(A\) is the cross-sectional area and \(L\) is the length of the cable. Using the mass per unit length and the tension, we can rewrite the previous expression in terms of acceleration due to gravity \(g\): \(\rho A = \frac{t}{g}\) Since we're looking for the maximum value of \(t\), let's express it in terms of \(\rho\), cross-sectional area \(A\), and gravity \(g\): \(t = \rho Ag\)
03

Relate density and tension in the moving frame

Using the relation \(\rho = \frac{\rho_{0}}{\gamma}\) and \(t = \rho Ag\), we can rewrite the tension in terms of proper density, Lorentz factor, cross-sectional area, and gravity: \(t = \frac{\rho_{0} Ag}{\gamma}\)
04

Find the condition for positive density in all frames

We know that the density must be positive in all frames (\(\rho > 0\)) for the cable not to break. This means that \(\frac{\rho_{0}}{\gamma} > 0\). Since \(\rho_{0}\) and \(\gamma\) are always positive, this condition is always satisfied. Therefore, we need to find the maximum value of \(t\) for which this condition holds true.
05

Find the maximum tension

In order to find the maximum tension that the cable can withstand, we need to analyze when the term inside the parenthesis reaches its minimum value: \(min\left(\frac{1}{\gamma}\right) = min\left(\sqrt{1 - v^2/c^2}\right)\) The minimum value of the term inside the parenthesis occurs when its argument, \(1 - v^2/c^2\), reaches its maximum value, which is 1 (when \(v = 0\)). Therefore: \(min\left(\frac{1}{\gamma}\right) = \sqrt{1 - 0^2/c^2} = 1\) Now plugging this value back into the expression of the tension: \(t_{max} = \rho_{0} Ag\)
06

Compare the maximum tension

The expression for the maximum tension is the same as that obtained using classical mechanics, because it doesn't involve the Lorentz factor. However, in this case, the result accounts for the fact that no reference frames exist where the cable's mass would be reduced below zero. The ultimate breaking tension in actual steel cables is approximately \(200\,\mathrm{kg}-\mathrm{wt} / \mathrm{mm}^{2}\), as given in the problem statement.

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Key Concepts

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

Special Relativity
Understanding the basics of special relativity is essential for tackling problems that involve high velocities, close to the speed of light. This theory, formulated by Einstein in 1905, revolutionized physics by introducing the idea that the laws of physics are consistent for all non-accelerating observers, and that the speed of light in a vacuum is the same no matter the speed at which an observer travels.

One of the dramatic consequences of special relativity is that it alters our perception of space and time - they are no longer absolute but interconnected in what is known as spacetime. Events that are simultaneous in one frame of reference may not be in another if those reference frames are moving relative to each other. Similarly, measurements of time and space, such as lengths and durations, depend on the observer's relative motion. For instance, a steel cable under tension can appear to have different properties, such as density, when viewed from different frames of motion - a phenomenon directly accounted for when calculating relativistic tension.
Lorentz Factor
The Lorentz factor, often denoted by the Greek letter \(\gamma\), plays a pivotal role in understanding relativistic effects and serves as a quantitative measure of time dilation, length contraction, and relativistic mass increase. It is defined mathematically as:\
\
\[\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\]

where \(v\) is the relative velocity between the observer and the object, and \(c\) is the speed of light. The Lorentz factor grows significantly as the velocity approaches the speed of light, exaggerating relativistic effects. In the context of the cable exercise, \(\gamma\) is what dictates how the apparent density of the cable changes with the observer's frame of reference. Knowing that \(t_{max}\) is not influenced by \(\gamma\) for a stationary cable from the observer's perspective helps simplify the problem and use classical mechanics for some calculations.
Relativistic Mass
Relativistic mass refers to the concept that the mass of an object increases with its speed. It is directly tied to the Lorentz factor as the speed of the object approaches the speed of light. The formula for obtaining the relativistic mass \(m\) from its rest mass \(m_{0}\) is:\
\
\[m = \gamma m_{0}\]

This increase in mass implies that it requires more force to accelerate an object as its speed increases. In the exercise, the steel cable's relativistic mass becomes relevant when we calculate the tension in different reference frames. However, the beauty of the maximum tension problem is that under classical conditions, when the cable is at rest relative to the observer, relativistic effects are negligible, and we can deduce the cable's properties without invoking relativistic mass changes. It’s a clear example of how classical and relativistic physics converge under certain circumstances.

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Most popular questions from this chapter

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.

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 \(U^{\mu},{ }_{, v} U^{v}\) \(\left.=A^{\mu}, U_{\mu} A^{\mu}=0 .\right]\)

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 $$
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