Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 8

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?

Short Answer

Expert verified
Answer: The speed of particle P in the rest frame of observer O will remain constant because the acceleration has no component along the direction of its velocity.

Step by step solution

01

Review four-velocity and four-acceleration

In special relativity, the four-velocity of an object is given by: \(\mathbf{U} = (\gamma c, \gamma\mathbf{v})\) where \(\gamma\) is the Lorentz factor, \(c\) is the speed of light, and \(\mathbf{v}\) is the three-dimensional velocity of the object. Similarly, the four-acceleration of an object is given by: \(\mathbf{A} = (\gamma^3 \mathbf{a} \cdot \mathbf{v}/c, \gamma^2 \mathbf{a} + \gamma^4 (\mathbf{a} \cdot \mathbf{v})\mathbf{v}/c^2)\) where \(\mathbf{a}\) is the three-dimensional acceleration of the object.
02

Inertial observer's four-velocity

In the problem, we are given that the observer O is inertial and has four-velocity U_0. Since the observer is inertial, their velocity relative to the rest frame is zero. Thus, the four-velocity for the observer is simply: \(\mathbf{U}_{0} = (c, \mathbf{0})\)
03

Use the given condition U_0 · A = 0

Now, we are given the condition: \(\mathbf{U}_{0} \cdot \mathbf{A} = 0\) Using the expressions for U_0 and A, we find that: \((c, \mathbf{0}) \cdot (\gamma^3 \mathbf{a} \cdot \mathbf{v}/c, \gamma^2 \mathbf{a} + \gamma^4 (\mathbf{a} \cdot \mathbf{v})\mathbf{v}/c^2) = 0\) This simplifies to: \(c (\gamma^3 \mathbf{a} \cdot \mathbf{v}) = 0\)
04

Find the conclusion about the speed of P

From the previous step, we found \(c (\gamma^3 \mathbf{a} \cdot \mathbf{v}) = 0\). Since c is a nonzero constant (the speed of light), we can infer that \(\gamma^3 \mathbf{a} \cdot \mathbf{v} = 0\). This implies: \(\mathbf{a} \cdot \mathbf{v} = 0\) This dot product represents the projection of the acceleration vector \(\mathbf{a}\) onto the velocity vector \(\mathbf{v}\). In other words, the acceleration of particle P is perpendicular to its velocity in the rest frame of observer O. This means that the speed of P in O's rest frame will remain constant because the acceleration has no component along the direction of its velocity.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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.

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.]

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.
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Electromagnetism

Read Explanation

Scientific Method Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Engineering Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free