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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.

Short Answer

Expert verified
The dot product is invariant under Lorentz transformation: \( x' \cdot y' = x \cdot y \).

Step by step solution

01

Understand the Four-Vectors

A four-vector in special relativity is represented as \( x = (ct, x_1, x_2, x_3) \) where \( c \) is the speed of light and \( t \) is time. Similarly, \( y = (ct', y_1, y_2, y_3) \). The dot product in four-vector notation is defined as \( x \cdot y = c^2 t t' - x_1 y_1 - x_2 y_2 - x_3 y_3 \).
02

Apply Lorentz Transformation

For a standard Lorentz boost along the \( x_1 \) axis, the transformations are: \( x'_0 = \gamma(ct - \beta x_1) \), \( x'_1 = \gamma(x_1 - \beta ct) \), and \( x'_2 = x_2 \), \( x'_3 = x_3 \). Similarly for \( y \), \( y'^0 = \gamma(ct' - \beta y_1) \), \( y' = \gamma(y_1 - \beta ct') \), and \( y'_2 = y_2 \), \( y'_3 = y_3 \). Here, \( \beta = \frac{v}{c} \) and \( \gamma = \frac{1}{\sqrt{1-\beta^2}} \).
03

Compute the Transformed Dot Products

Compute the dot products for \( x' \cdot y' \) using the transformed coordinates: \( x' \cdot y' = (\gamma^2 (c^2tt' - x_1y_1) + \beta^2 c^2 t t' - \beta \gamma (c t y_1 + c t' x_1) + x_2 y_2 + x_3 y_3 \). Simplify using \( \gamma^2 (1-\beta^2) = 1 \).
04

Simplify and Verify Equivalence

Upon simplification, show \( x' \cdot y' = x \cdot y \). Using the relation \( \gamma^2 (1-\beta^2) = 1 \), the expanded terms cancel out the extra \( \beta \) and \( \gamma \) terms, finally revealing \( x' \cdot y' = c^2tt' - x_1y_1 - x_2y_2 - x_3y_3 = x \cdot y \).

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Key Concepts

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

Four-Vectors
In the realm of physics, especially special relativity, a four-vector is a profound concept that extends the classical three-dimensional vector by adding an extra dimension for time. Think of it as a package containing two parts: time and spatial components. For instance, a four-vector for an event can be expressed as \( x = (ct, x_1, x_2, x_3) \), where \(ct\) represents the time component (with \(c\) as the speed of light) and \(x_1, x_2, x_3\) represent the spatial components in three-dimensional space.
  • The time component is crucial because it links the time dimension with the spatial dimensions through the speed of light \(c\).
  • This framework allows one to handle transformations between different reference frames in a mathematically consistent way.
Understanding how these components interact through transformations, such as the Lorentz transformation, is essential for studying the laws of physics in different inertial frames.
Special Relativity
Special relativity is a cornerstone theory in physics, proposed by Albert Einstein, that revolutionized our understanding of space and time. A critical postulate of special relativity is that the laws of physics are the same in all inertial reference frames, and that the speed of light is constant in a vacuum for all observers, regardless of their motion relative to the light source.
  • This means that measurements of time and space are relative to the observer's state of motion.
  • The theory introduces concepts such as time dilation, length contraction, and the constancy of the speed of light.
Special relativity is not just theoretical; it has practical implications. For example, GPS satellites account for relativistic effects to provide accurate positioning data. By utilizing the four-vectors and transformations from this theory, one can translate events from one frame to another seamlessly.
Dot Product
The dot product in the context of four-vectors plays a similar role to its three-dimensional counterpart but with a twist. Instead of merely taking the sum of the products of corresponding components, the time component is handled differently due to the Minkowski metric used in special relativity.

For two four-vectors \( x \) and \( y \), their dot product is defined as \( x \cdot y = c^2tt' - x_1y_1 - x_2y_2 - x_3y_3 \). Here, the temporal component contributes positively, while the spatial elements are subtracted.
  • This formulation preserves the spacetime interval (a constant quantity) between different events, no matter the observer's motion.
  • The invariance of the dot product under Lorentz transformations is a testament to the robustness of the formulation in different reference frames.
In essence, this invariance reflects how spacetime geometry remains consistent across different observers, making it a pivotal concept in relativity.
Lorentz Boost
The Lorentz boost is a specific type of Lorentz transformation applied when changing reference frames moving at constant velocity relative to each other, particularly along one axis. For example, a boost along the \(x_1\) axis adjusts the coordinates of events in time and space according to:
  • \( x'_0 = \gamma(ct - \beta x_1) \)
  • \( x'_1 = \gamma(x_1 - \beta ct) \)
  • \( x'_2 = x_2 \) and \( x'_3 = x_3 \)
Here, \(\beta = \frac{v}{c}\) is the velocity as a fraction of the speed of light, and \(\gamma = \frac{1}{\sqrt{1-\beta^2}}\) is the Lorentz factor accounting for time dilation and length contraction.
Such a transformation shows how observers moving relative to each other perceive coordinates and time differently but remain consistent with the principles of relativity. Lorentz boosts allow us to quantify these changes precisely, ensuring physical phenomena have the same descriptions in different inertial frames.

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Most popular questions from this chapter

Show that the four-velocity of any object has invariant length squared \(u \cdot u=-c^{2}\)

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 \(M .]\)

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