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Prove the following useful result, called the zero-component theorem: Let \(q\) be a four-vector, and suppose that one component of \(q\) is found to be zero in all inertial frames. (For example, \(q_{4}=0\) in all frames.) Then all four components of \(q\) are zero in all frames.

Short Answer

Expert verified
All four components of the four-vector are zero in all inertial frames.

Step by step solution

01

Understanding the Four-Vector

A four-vector in relativity is a vector with four components: \( (q_1, q_2, q_3, q_4) \). These components can change under Lorentz transformations between different inertial frames, just as regular vectors might change under rotations.
02

Assumption and Given Condition

We are given that the fourth component of the four-vector \( q \, (q_4) \) is zero in all inertial frames: \( q_4 = 0 \). We need to show that if one component is always zero in any frame, then all components are zero.
03

Lorentz Transformation Properties

Recall that Lorentz transformations relate the components of four-vectors in different inertial frames. Mathematically, this is: \( q'_i = \sum_j \Lambda_{ij} q_j \), where \( \Lambda_{ij} \) is the Lorentz transformation matrix. This transformation preserves the nature of the four-vector, specifically it is intended to preserve the interval (or scalar product).
04

Scalar Product of the Four-Vector

The scalar product (or norm) of a four-vector is invariant under Lorentz transformations. For \( q \), this is expressed as \( q \cdot q = q_1^2 + q_2^2 + q_3^2 - q_4^2 = 0 \), given that \( q_4 = 0 \). Thus, the scalar product simplifies to \( q_1^2 + q_2^2 + q_3^2 = 0 \).
05

Conclusion from the Invariant Scalar Product

Since the sum \( q_1^2 + q_2^2 + q_3^2 = 0 \) and each term is a non-negative square, the only solution is that each must individually be zero: \( q_1 = 0 \), \( q_2 = 0 \), \( q_3 = 0 \). Hence in any inertial frame, all components are zero.

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Key Concepts

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

Four-Vector
In the realm of relativity, a four-vector is a special type of vector that combines both space and time into a single entity with four components. These components are usually written as \((q_1, q_2, q_3, q_4)\). Each part of the four-vector corresponds to a specific aspect of an event or location in spacetime: the first three (\(q_1, q_2, q_3\)) relate to space, while the fourth \(q_4\) correlates to time.

The importance of four-vectors arises because they remain consistent under Lorentz transformations—mathematical operations that allow us to shift perspectives between different inertial frames. While the individual components of a four-vector can change when observed from various inertial frames, the entire set of four-components transforms coherently. These stable relationships permit the preservation of physical laws, keeping them applicable across different frames.
Lorentz Transformation
Inertial frames in relativity are reference frames in which objects move with constant velocity without any outside force acting upon them. When two inertial frames are in relative motion, transitioning from one to another requires a Lorentz transformation.

These transformations are a mathematical set of rules that modify the space and time components of a four-vector as you change the viewpoint from one inertial frame to another. Mathematically, they are described as: \[ q'_i = \sum_j \Lambda_{ij} q_j \]where \( \Lambda_{ij} \) represents the Lorentz transformation matrix.
  • They ensure the invariance of the speed of light across all frames.
  • They provide consistency in the way laws of physics manifest, regardless of the observer's motion.
  • They preserve the scalar product, keeping the nature of physical phenomena constant.
This preservation means four-vectors, when transformed, retain their essence, making Lorentz transformations central to understanding relativistic physics.
Scalar Product
The scalar product, also known as the dot product for four-vectors, holds a unique place because it stays invariant under Lorentz transformation. This means that no matter how the four-vector components change between inertial frames, the scalar product's value remains constant.

For a four-vector \( q = (q_1, q_2, q_3, q_4) \), the scalar product is given by:\[ q \cdot q = q_1^2 + q_2^2 + q_3^2 - q_4^2 \] When one component, such as \( q_4 \), is zero across all frames, the scalar product simplifies further to:\[ q_1^2 + q_2^2 + q_3^2 = 0 \]
This equation implies that individually each square must also be zero, thus:
  • \( q_1 = 0 \)
  • \( q_2 = 0 \)
  • \( q_3 = 0 \)
Therefore, if one component of a four-vector is zero everywhere, the rest must follow suit.
Inertial Frames
Inertial frames of reference are essential for understanding motion and the laws of physics in relativity. They are perspectives in which objects either remain at rest or continue moving at a constant velocity unless acted upon by a force. All observers in inertial frames will agree on the laws of physics, thanks to the concept of relativity.

When different inertial frames observe the same physical phenomenon, they apply Lorentz transformations to translate the observations between the frames. This ensures:
  • All physical laws hold true regardless of the observer's motion.
  • The constancy of the speed of light is upheld.
  • The scalar product of four-vectors remains unchanged.
Understanding inertial frames helps illustrate the effects of relativity and why certain values, even as fundamental as time or distance, can vary based on the observer's motion and perspective.

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Most popular questions from this chapter

Consider two events that occur at positions \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\) and times \(t_{1}\) and \(t_{2} .\) Let \(\Delta \mathbf{r}=\mathbf{r}_{2}-\mathbf{r}_{1}\) and \(\Delta t=t_{2}-t_{1} .\) Write down the Lorentz transformation for \(\mathbf{r}_{1}\) and \(t_{1}\), and likewise for \(\mathbf{r}_{2}\) and \(t_{2},\) and deduce the transformation for \(\Delta \mathbf{r}\) and \(\Delta t\). Notice that differences \(\Delta \mathbf{r}\) and \(\Delta t\) transform in exactly the same way as \(\mathbf{r}\) and \(t\). This important property follows from the linearity of the Lorentz transformation.

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 \(\mathcal{S}^{\prime}\).

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 \(t+d t\).

(a) Find the 3 \(\times 3\) matrix \(\mathbf{R}(\theta)\) that rotates three- dimensional space about the \(x_{3}\) axis, so that \(\mathbf{e}_{1}\) rotates through angle \(\theta\) toward \(\mathbf{e}_{2}\). (b) Show that \([\mathbf{R}(\theta)]^{2}=\mathbf{R}(2 \theta),\) and interpret this result.

Prove that if \(x\) is time-like and \(x \cdot y=0,\) then \(y\) is space-like.

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