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Chapter 4: Problem 2

Prove that, for any straight-line segment in spacetime, \(|\Delta s|\) \(=\int|\mathrm{d} s|\), where \(\Delta s^{2}\) is defined by \((21.1)\) and \(\mathrm{d} s^{2}\) is similarly defined in terms of the differentials. This result, of course, is necessary for the consistency of the notation. [Hint: let the segment have equation \(\left.x^{\mu}=A^{\mu} \theta+B^{\mu}, 0 \leqslant \theta \leqslant 1 .\right]\)

Short Answer

Expert verified
A: In the solution, we arrived at the integral \(|\Delta s| = \int_{0}^{1} \sqrt{C} \; |\mathrm{d} \Theta|\). After evaluating the integral, we found that \(|\Delta s| = \sqrt{C}\). This result demonstrates the consistency of the notation used for the proper length in spacetime.

Step by step solution

01

Write down the relevant definitions and notation.

We are given that the proper length \(\Delta s^{2}\) is defined by \((21.1)\), which states that for two events with coordinates \(x^{\mu}\) and \(y^{\mu}\) in spacetime, \(\Delta s^{2} = - (x^0 - y^0)^2 + (x^1 - y^1)^2 + (x^2 - y^2)^2 + (x^3 - y^3)^2 = - |\Delta x^0|^2 + |\Delta x^1|^2 + |\Delta x^2|^2 + |\Delta x^3|^2\) Similarly, we have the differential, \(\mathrm{d}s^2\) defined as: \(ds^2 = - (dx^0)^2 + (dx^1)^2 + (dx^2)^2 + (dx^3)^2\) And we are given the hint to consider the straight-line segment defined by: \(x^{\mu} = A^{\mu}\Theta + B^{\mu}, \quad 0 \le \Theta \le 1\)
02

Write the straight-line segment equation in terms of differentials.

Taking differentials on both sides of the equation given in the hint, we have: \(d x^{\mu} = A^{\mu} d\Theta\) Substituting this into the expression for \(\mathrm{d}s^2\), we get: \(ds^2 = - (A^0 d\Theta)^2 + (A^1 d\Theta)^2 + (A^2 d\Theta)^2 + (A^3 d\Theta)^2\) Factoring out \(d\Theta^2\) from the expression, we get: \(ds^2 = (- (A^0)^2 + (A^1)^2 + (A^2)^2 + (A^3)^2)d\Theta^2 = C d\Theta^2\) Where \(C=- (A^0)^2 + (A^1)^2 + (A^2)^2 + (A^3)^2\) is a constant term.
03

Set up the integral to find \(|\Delta s|\)

We know that \(|\Delta s| = \int|\mathrm{d} s|\). So, using the expression for \(ds^2\) derived in the previous step, we get: \(|\Delta s| = \int_{0}^{1} \sqrt{C} \; |\mathrm{d} \Theta|\)
04

Calculate the integral

Now that we have a simple integral, we can integrate with respect to \(\Theta\) easily. As the \(\sqrt{C}\) is positive, we can write: \(|\Delta s| = \sqrt{C} \int_{0}^{1} \; \mathrm{d} \Theta = \sqrt{C}(\Theta |_0^1) = \sqrt{C}\) Here, we have obtained the desired result which shows that \(|\Delta s| = \int|\mathrm{d} s|\). This demonstrates the consistency of the notation used.

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Most popular questions from this chapter

An inertial observer \(\mathrm{O}\) has four-velocity \(\mathrm{U}_{0}\) and a particle \(\mathrm{P}\) has (variabie) four-acceleration \(\mathbf{A}\). If \(\mathbf{U}_{0} \cdot \mathbf{A}=0\), what can you conclude about the speed of P in O's rest frame?

A particle moves rectilinearly with constant proper acoeleration x. If \(\mathbf{U}\) and \(\mathbf{A}\) are its four-velocity and four-acceleration, \(\tau\) its proper time, and units are chosen to make \(\mathrm{c}=1\), prove that \((\mathrm{d} / \mathrm{d} \tau) \mathbf{A}=\alpha^{2} \mathbf{U}\). [Hint: Exercise II(14).] Prove, conversely, that this equation, without the information that \(\alpha\) is the proper acceleration, or constant, implies both these facts. [Hint: differentiate the equation \(\mathbf{A} \cdot \mathbf{U}=0\) and show that \(\alpha^{2}=-\mathbf{A} \cdot \mathbf{A}\). And finally show, by integration, that the equation implies rectilinear motion in a suitable inertial frame, and thus, in fact, hyperbolic motion. Consequently \((\mathrm{d} / \mathrm{d} \tau) \mathbf{A}=\alpha^{2} \mathbf{U}\) is the tensor equation characteristic of hyperbolic motion.

(i) Prove the zero-component lemma for four-vectors: if a fourvector \(V^{\mu}\) has a particular one of its four components zero in \(a l l\) inertial frames then the entire vector must be zero. [Hint: suppose first \(V^{i} \equiv 0\); if there is a frame in which \(V^{2} \neq 0\) or \(V^{3} \neq 0\), we can rotate the spatial axes to make \(V^{t} \neq 0\); if there is a frame in which \(V^{0} \neq 0\), we can apply a Lorentz transformation to make \(V^{1} \neq 0\); and so on.] (ii) If \(A_{\mu v}\) is a symmetric tensor and \(A_{00} \equiv 0\), prove \(A_{\mu \nu}=0\). Is this result true for the vanishing of any one component of \(A_{\mu v}\) ?

1\. Theorem: A transformation which transforms a metric \(g_{\mu r} \mathrm{~d} x^{n} \mathrm{~d} x^{\nu}\) with constant coefficients into a metric \(g_{\mu^{\prime} v^{\prime}} \mathrm{d} x^{\mu^{\prime}} \mathrm{d} x^{\gamma^{\prime}}\) with constant coefficients must be linear. Fill in the details of the following proof: \(g_{\mu v} \mathrm{~d} x^{\mu} \mathrm{d} x^{\nu}=g_{\mu^{\prime} v^{\prime}} \mathrm{d} x^{\mu^{\prime}} \mathrm{d} x^{\nu^{\nu}}=g_{\mu^{\prime} v^{\prime}} p_{\mu}^{\mu^{\prime}} p_{v}^{v^{\prime}} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}\), so \(g_{\mu \nu}=g_{\mu^{\prime} v^{\prime}} p_{\mu \nu}^{\|^{\prime}} p_{v}^{v^{\prime}}\) Differentiating with respect to \(x^{\phi}\) we get \(g_{\mu^{\prime} v} p_{\mu d} p_{v}^{v^{*}}+g_{\mu^{*} v} p_{\mu}^{\sigma_{v}} p_{v_{\sigma}}^{v^{*}}=0\) (i). Interchange \(\mu\) and \(\sigma\) to form equation (ii). Again, in (i) interchange and \(\sigma\) to form (iii). Subtract (iii) from the sum of (i) and (ii). Then \(2 g_{\mu^{*} v^{\prime}} p_{\mu \sigma}^{k^{*}} p_{v}^{v^{*}}=0\). Transvect first by \(p_{\sigma^{*}}^{v}\) and then by \(g^{v^{*} \sigma^{*}}\) and obtain pus \(=0\). This proves linearity.

The aggregate of events considered by an inertial observer to be simultaneous at his time \(t=t_{0}\) is said to be the observer's instantancous three-space \(t=t_{0}\). Show that the join of any two events in such a space is orthogonal to the observer's worldline, and that, conversely, any two events whose join is orthogonal to the observer's worldline are considered simultaneous by him.
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