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As an observer moves through space with position \(\mathbf{x}(t),\) the four- vector \((\mathbf{x}(t), c t)\) traces a path through space-time called the observer's world line. Consider two events that occur at points \(P\) and \(Q\) in space-time. Show that if, as measured by the observer, the two events occur at the same time \(t,\) then the line joining \(P\) and \(Q\) is orthogonal to the observer's world line at the time \(t\); that is, \(\left(x_{P}-x_{Q}\right) \cdot d x=0,\) where \(d x\) joins two neighboring points on the world line at times \(t\) and \(t+d t\).

Short Answer

Expert verified
The line joining \(P\) and \(Q\) is orthogonal to the observer's world line at time \(t\) because their inner product equals zero.

Step by step solution

01

Set Up the Problem

We are given two events at points \(P\) and \(Q\) in space-time that occur at the same time \(t\) as measured by an observer. The observer's world line is represented by the four-vector \((\mathbf{x}(t), ct)\). We need to show that the line joining \(P\) and \(Q\) is orthogonal to the observer's world line at time \(t\).
02

Calculate the Difference Vector

Since the events occur at the same time \(t\), the time components are equal. The spatial part of the four-vector difference is \(\mathbf{x}_P - \mathbf{x}_Q\). Thus, we form the space-time difference \((\mathbf{x}_P - \mathbf{x}_Q, 0)\).
03

Derive the Differential Element of the World Line

The differential element of the observer's world line at time \(t\) is \(d\mathbf{x} = \left(\frac{d\mathbf{x}}{dt}dt, cdt\right)\). It indicates a small change along the trajectory of the observer through space-time.
04

Use the Orthogonality Condition

To be orthogonal, the inner product of the difference vector \((\mathbf{x}_P - \mathbf{x}_Q, 0)\) and the differential element \(d\mathbf{x}\) must be zero. This gives us the condition:\[(\mathbf{x}_P - \mathbf{x}_Q) \cdot \frac{d\mathbf{x}}{dt}dt + 0 \cdot cdt = 0.\]This simplifies to:\[(\mathbf{x}_P - \mathbf{x}_Q) \cdot d\mathbf{x} = 0.\]
05

Conclude with Orthogonality

The condition \((\mathbf{x}_P - \mathbf{x}_Q) \cdot d\mathbf{x} = 0\) confirms that the line segment joining points \(P\) and \(Q\) is orthogonal to the observer's world line at time \(t\).

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Key Concepts

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

Space-time
Space-time is a four-dimensional framework that combines the three dimensions of space with the one dimension of time into a single continuum. This concept is groundbreaking as it merges space and time, which traditionally were considered separately. A point in space-time is called an "event," described by four coordinates: three spatial coordinates \(x, y, z\) and one temporal coordinate, often expressed as \(ct\), where \(c\) is the speed of light.

Space-time considers how events are sequenced and measured relative to an observer's position and time, hence the incorporation of time into the concept. Imagine each point in this 4D matrix (3D space + time) represents an event. It allows us to map out not just where something happens, but when it happens too, giving us a deep understanding of motion and causality in the universe.

The concept was popularized by Albert Einstein’s theory of relativity, which shows that space and time are not constant but are interwoven and affected by the relative motion of observers and the presence of mass.
Observer's World Line
An observer's world line is the path traced in space-time as an observer or object moves. Picture it as a trail left by the moving observer over time through the fabric of space-time. It is represented mathematically by a four-vector \(\vec{x}(t), ct\), comprising the position vector \(\vec{x}(t)\) and the time component \(ct\).

Every event that the observer experiences is a point along this line. The world line describes how the observer's position in space changes over time, playing an essential role in understanding motion from a relativistic perspective. It's like a timeline of all positions taken by the observer in both space and time.

A key feature of the world line is its use in analyzing events in relativity. For example, if two events occur simultaneously on an observer's clock, they are said to have the same temporal component on the observer’s world line. This assists in visualizing events that may be simultaneous from one frame of reference but not another.
Orthogonality
In geometry and vector analysis, orthogonality signifies perpendicularity. Two vectors are orthogonal if their inner product is zero. Intuitively, this means the vectors are at right angles to each other. This concept extends to space-time, involving the four-vector representation.

To consider orthogonality in space-time, we'll look at the line joining two points, or events, \(P\) and \(Q\). The exercise shows that this line is orthogonal to the observer's world line when both events occur simultaneously for that observer. If the spatial difference between two events \(\mathbf{x}_P - \mathbf{x}_Q\) is perpendicular to a small segment on the observer’s world line, the vectors are orthogonal at that moment.

This orthogonality condition is expressed through the inner product, stating that the dot product (inner product in four-vector terms) between the spatial difference and the world line's directional derivative is zero. This condition is crucial in relativity, where it verifies whether certain events are connected perpendicular to the observer's trajectory.
Inner Product
The inner product, often called a dot product when dealing with vectors in regular three-dimensional space, measures the resemblance or angle between them. For two vectors \(\mathbf{a}\) and \(\mathbf{b}\), given by \(\mathbf{a} \cdot \mathbf{b}\) in three-dimensional space, it results in a scalar. If the product is zero, the vectors are orthogonal. This principle extends into four-dimensional space-time.
In space-time, events are best described by four-vectors, and their inner product accommodates both spatial and temporal elements. Understanding the inner product in this context is crucial for analyzing relations between different events for an observer. When two events are orthogonal at a specific time on the observer’s world line, their spatial difference's inner product with the world line's differential element is zero.

This form of inner product helps determine whether events lie perpendicular to the world line in the four-dimensional space-time construct. Such calculations underpin crucial concepts in relativity, allowing for clearer insights into the nature of simultaneity and separation of events as perceived by observers.

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Most popular questions from this chapter

The muons created by cosmic rays in the upper atmosphere rain down more-or- less uniformly on the earth's surface, although some of them decay on the way down, with a half-life of about \(1.5 \mu\) s (measured in their rest frame). A muon detector is carried in a balloon to an altitude of \(2000 \mathrm{m}\), and in the course of an hour detects 650 muons traveling at \(0.99 c\) toward the earth. If an identical detector remains at sea level, how many muons should it register in one hour? Calculate the answer taking account of the relativistic time dilation and also classically. (Remember that after \(n\) half-lives, \(2^{-n}\) of the original particles survive.) Needless to say, the relativistic answer agrees with experiment.

Verify directly that \(x^{\prime} \cdot y^{\prime}=x \cdot y\) for any two four- vectors \(x\) and \(y,\) where \(x^{\prime}\) and \(y^{\prime}\) are related to \(x\) and \(y\) by the standard Lorentz boost along the \(x_{1}\) axis.

A mad physicist claims to have observed the decay of a particle of mass \(M\) into two identical particles of mass \(m,\) with \(M<2 m .\) In response to the objections that this violates conservation of energy, he replies that if \(M\) was traveling fast enough it could easily have energy greater than \(2 m c^{2}\) and hence could decay into the two particles of mass \(m\). Show that he is wrong. [He has forgotten that both energy and momentum are conserved. You can analyse this problem in terms of these two conservation laws, but it is much simpler to go to the rest frame of \(M .]\)

Let \(\Lambda_{\mathrm{B}}(\theta)\) denote the 4 \(\times 4\) matrix that gives a pure boost in the direction that makes an angle \(\theta\) with the \(x_{1}\) axis in the \(x_{1} x_{2}\) plane. Explain why this can be found as \(\Lambda_{\mathrm{B}}(\theta)=\Lambda_{\mathrm{R}}(-\theta) \Lambda_{\mathrm{B}}(0) \Lambda_{\mathrm{R}}(\theta)\) where \(\Lambda_{\mathrm{R}}(\theta)\) denotes the matrix that rotates the \(x_{1} x_{2}\) plane through angle \(\theta\) and \(\Lambda_{\mathrm{B}}(0)\) is the standard boost along the \(x_{1}\) axis. Use this result to find \(\Lambda_{\mathrm{B}}(\theta)\) and check your result by finding the motion of the spatial origin of the frame \(\mathcal{S}\) as observed in \(\mathcal{S}^{\prime}\).

In non relativistic mechanics, the energy contains an arbitrary additive constant \(-\) no physics is changed by the replacement \(E \rightarrow E+\) constant. Show that this is not the case in relativistic mechanics. [Hint: Remember that the four-momentum \(p\) is supposed to transform like a four- vector.]

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