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Chapter 4: Problem 11

We shall say that three particles move codirectionally if their three- velocities are parallel in some inertial frame. Prove that the necessary and sufficient condition for this to be the case is that the four-velocities U, V, W of these particles be linearly dependent. [IInt: the component specializations of Section \(22(v)\) may help.]

Short Answer

Expert verified
Answer: The condition for particles to move codirectionally in terms of their four-velocities is that these four-velocities (U, V, and W) must be linearly dependent. In other words, there must exist constants a, b, and c (not all zero) such that aU + bV + cW = 0. The particles' velocities must also be proportional, as in ku = v and lw = v, where k and l are constants.

Step by step solution

01

Define Four-velocities of the particles

We are given three particles, and their four-velocities are U, V, and W. Recall that the four-velocity of a particle in an inertial frame has the form (c, v), where c is the speed of light and v is a three-velocity vector. We can write the four-velocities U, V, W as: U = (c, u) V = (c, v) W = (c, w) where u, v, and w are the three-velocities of the particles.
02

Check if four-velocities U, V, W are linearly dependent

Linear dependence means that there exist constants a, b, and c (not all zero) such that aU + bV + cW = 0. Let's plug the four-velocities: a(c, u) + b(c, v) + c(c, w) = (0, 0) Breaking this into components, we get: ac + bc + cc = 0 Equation (1) a(u) + b(v) + c(w) = 0 Equation (2)
03

Solve Equation (1) to find the relationship between a, b, and c.

We have to solve Equation (1) to get the constraint for a, b, and c. From Equation (1), we can rewrite as: a + b + c = 0 Since the particles are moving codirectionally, their velocities must be proportional, that is ku = v and lw = v, thus a ≠ b ≠ c.
04

Express three-velocities in terms of a linear relationship

Using the relationship we found in Step 3, which is a + b + c = 0, we can rearrange it and get the expressions of u and w in terms of v: ku = v, lw = v. Equation (2) now becomes: a(ku) + bv + clw = a(kv) + bv + clv = 0
05

Verify that the expression satisfies the condition for linear dependence

Check if the resulting equation from Step 4 is satisfied, that is, a(kv) + bv + clv = 0. Factor out v, we get : v(ak + b + cl) = 0 Now, either v = 0, or ak + b + cl = 0. Thus, if v = 0, the three-velocities u, v, and w are parallel. If ak+b+cl=0, we can see that a+b+c=0 from step 3. Both these expressions are the necessary and sufficient conditions for the four-velocities U, V, and W to be linearly dependent. Therefore, the exercise is proven, and if the four-velocities U, V, and W are linearly dependent, the particles move codirectionally.

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Most popular questions from this chapter

Consider Euclidean three-space and in it the group of transformations consisting of rotations and translations of the right-handed orthogonal axes of \(x^{i}\). These we shall call RT transformations, and they are clearly linear. Three-tensors are objects that behave as tensors under this transformation group [cf. Exercises \(A(2), A(10) ;\) translations have no effect on tensor components]. Evidently RT transformations (augmentud by \(t^{\prime}=t\) ) form a subgroup of the Poincaré group. Prove: (i) if an RT transformation is applied to a four-vector \(A^{\mu}=\left(A^{0}\right.\), a),\(A^{0}\) transforms as a scalar and a as a threevector; (ii) if an RT transformation is applied to a four-tensor \(T^{\mu v}\), then \(T^{00}\) transforms as a scalar, \(T^{0 i}\) and \(T^{i 0}\) transform as threevectors, and \(T^{i j}\) transforms as a three-tensor. Note that corresponding to any three- vector \(b^{i}\) we can define a 'dual' tensor \(b_{i j}=\varepsilon_{i j k} b^{k}\), so that \(\left(b^{1}, b^{2}, b^{3}\right)=\left(b_{23}, b_{31}, b_{12}\right)\). Accordingly, the \(b\) s in the preceding exercise transform as a three-vector under RT transformations, as the notation suggests.

All vectors in this problem are presumed to be real and non-zero. Let \(T, S, N, V\), respectively denote timelike, spacelike, null, and general vectors. Prove: (i) any \(V\) orthogonal to a \(T\) or \(N\) (other than the \(N\) itself ) is an \(S\); (ii) the sum of two \(T_{5}\), or of a \(T\) and an \(N\), which are isochronous (i.e. both pointing into the future or both into the past) is a \(T\) isochronous with them; (iii) the sum [difference] of two isochronous \(N s\) is a \(T[S]\), or, in the case of two parallel \(N s\), an \(N\); (iv) every \(T[S]\) is expressible as the sum [difference] of two isochronous \(N \$$; (v) the scalar product of two \)T_{\$}\(, or of a \)T\( and an \)N\(, which are isochronous, is positive; that of two isochronous \)N s\( is positive, unless they are parallel, in which case it is zero. [Hint: the component specializations of Section \)22(\mathrm{v})$ may help; so may a spacetime diagram, to organize ideas.]

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.

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.

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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