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Since the four-velocity \(u=\gamma(\mathbf{v}, c)\) is a four-vector its transformation properties are simple. Write down the standard Lorentz boost for all four components of \(u\). Use these to deduce the relativistic velocity- addition formula for v.

Short Answer

Expert verified
The relativistic velocity-addition formula is \( v'_x = \frac{v_x - v}{1 - \frac{vv_x}{c^2}} \).

Step by step solution

01

Understanding the Four-Velocity

The four-velocity is given by \( u = \gamma(\mathbf{v}, c) \), where \( \gamma \) is the Lorentz factor, \( \mathbf{v} \) is the three-velocity, and \( c \) is the speed of light. Recall \( \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \). The components of four-velocity in the rest frame (\( S \)) are \( (\gamma v_x, \gamma v_y, \gamma v_z, \gamma c) \).
02

Lorentz Boost Formula

The standard Lorentz boost in the x-direction relates the coordinates between two inertial frames (\( S \) and \( S' \)): \( x' = \gamma (x - vt) \), \( t' = \gamma (t - \frac{vx}{c^2}) \), with \( \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \). Adjust these formulas for velocities: \( v'_x = \frac{v_x - v}{1-\frac{vv_x}{c^2}} \), \( v'_y = \frac{v_y}{\gamma(1-\frac{vv_x}{c^2})} \), \( v'_z = \frac{v_z}{\gamma(1-\frac{vv_x}{c^2})} \).
03

Applying Lorentz Transformation to Four-Velocity

Apply the Lorentz transformation to each component of the four-velocity. The four-velocity \( u \) transforms as: \( u'_x = \gamma (u_x - \frac{v}{c} u_t) = \frac{\gamma}{\gamma(1-\frac{v}{c}v_x/c)}(v_x - v) \), \( u'_y = \frac{u_y}{\gamma(1-\frac{v}{c}v_x/c)} \), \( u'_z = \frac{u_z}{\gamma(1-\frac{v}{c}v_x/c)} \), \( u'_t = \gamma (u_t - \frac{v}{c} u_x) \).
04

Deriving the Relativistic Velocity Addition Formula

To find the transformed velocities in the \( S' \) frame, using the components of \( u' \), we solve for \( v'_x = \frac{u'_x}{u'_t} \), leading to \( v'_x = \frac{v_x - v}{1 - \frac{vv_x}{c^2}} \). Similarly, for the y and z components, \( v'_y = \frac{v_y}{\gamma (1-\frac{vv_x}{c^2})} \), \( v'_z = \frac{v_z}{\gamma (1-\frac{vv_x}{c^2})} \). These results present the relativistic velocity addition formulas for each component.

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Key Concepts

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

Lorentz transformation
The Lorentz transformation is an essential concept in understanding how observations of time and space vary for observers in different inertial frames of motion. This transformation provides the mathematical equations that relate the coordinates of events as measured in one inertial frame to those in another, particularly those moving at a constant velocity relative to each other.
For example, consider two inertial frames, typically labeled as \( S \) and \( S' \), moving with a relative velocity \( v \). The transformation for the x-component is given by:
  • \( x' = \gamma (x - vt) \)
  • \( t' = \gamma (t - \frac{vx}{c^2}) \)
Here, \( \gamma \) is the Lorentz factor \( \left( \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \right) \). These equations ensure that the speed of light \( c \) is constant in all frames, a cornerstone of Einstein's theory of relativity.
four-velocity
To better understand relativistic motion, we introduce the four-velocity, which is a four-vector and extends the concept of velocity into the realm of relativity. The four-velocity of an object is given as \( u = \gamma(\mathbf{v}, c) \), where \( \gamma \) is the Lorentz factor. This factor scales the three-dimensional velocity \( \mathbf{v} \) along with an additional component, the speed of light \( c \), forming a four-dimensional spacetime vector.

The components in the rest frame \( S \) are:
  • \( (\gamma v_x, \gamma v_y, \gamma v_z, \gamma c) \)
These components reflect both spatial and temporal aspects of an object's motion, ensuring that the speed of light remains constant across different frames.
Utilizing four-velocity in calculations allows for a more straightforward application of Lorentz transformations, reducing complex, multi-step problems into manageable equations.
inertial frames
Inertial frames are reference frames where Newton's first law — an object's uniform motion will continue unless acted upon by a force — holds true. These frames move at constant velocity relative to each other, meaning that they are not accelerating.
  • In these frames, the laws of physics appear the same, and measurements of time and space can vary based on relative motion.
When dealing with problems involving relativity, we often analyze scenarios from the perspectives of different inertial frames. For example, the transformation formulas such as those given by the Lorentz equations allow us to translate velocities and positions observed in one frame (\( S \)) to another (\( S' \)).

It's the distinctive property of relativity that the speed of light remains constant in all inertial frames, requiring us to adjust our conventional understanding of time and space, leading to phenomena such as time dilation and length contraction.
Lorentz boost
The Lorentz boost is a specific type of Lorentz transformation where the transformation matrix aligns with the direction of relative velocity between two inertial frames. This mathematical tool is crucial for transitioning between frames moving at constant velocity with respect to each other.
  • For instance, consider a boost in the x-direction: using the Lorentz factor \( \gamma \), the transformed coordinates \( (x', t') \) are obtained as \( x' = \gamma (x - vt) \) and \( t' = \gamma (t - \frac{vx}{c^2}) \).
Applying a Lorentz boost provides a formula for determining how velocities shift between frames, such as with the relativistic velocity addition formula. This formula addresses how velocities in x, y, and z directions transform between frames. By using a Lorentz boost, we find:
  • \( v'_x = \frac{v_x - v}{1-\frac{vv_x}{c^2}} \)
  • For transverse components, \( v'_y = \frac{v_y}{\gamma(1-\frac{vv_x}{c^2})} \) and \( v'_z = \frac{v_z}{\gamma(1-\frac{vv_x}{c^2})} \).
These calculations allow for accurate descriptions of motion within relativistic physics frameworks, offering deeper insights into how objects behave at high velocities.

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Most popular questions from this chapter

A useful form of the quotient rule for three-dimensional vectors is this: Suppose that a and b are known to be three-vectors and suppose that for every orthogonal set of axes there is a \(3 \times 3\) matrix T with the property that \(\mathbf{b}=\mathbf{T}\) a for every choice of \(\mathbf{a},\) then \(\mathbf{T}\) is a tensor. (a) Prove this. (b) State and prove the corresponding rule for four-vectors and four-tensors.

What is the factor \(\gamma\) for a speed of \(0.99 c\) ? As observed from the ground, by how much would a clock traveling at this speed differ from a ground-based clock after one hour (one hour as measured by the latter, that is)?

One way to create exotic heavy particles is to arrange a collision between two lighter particles $$a+b \rightarrow d+e+\cdots+g$$ where \(d\) is the heavy particle of interest and \(e, \cdots, g\) are other possible particles produced in the reaction. (A good example of such a process is the production of the \(\psi\) particle in the process \(\left.e^{+}+e^{-} \rightarrow \psi, \text { in which there are no other particles } e, \cdots, g .\right)\) (a) Assuming that \(m_{d}\) is much heavier that any of the other particles, show that the minimum (or threshold) energy to produce this reaction in the CM frame is \(E_{\mathrm{cm}} \approx m_{d} c^{2} .\) (b) Show that the threshold energy to produce the same reaction in the lab frame, where the particle \(b\) is initially at rest, is \(E_{\mathrm{lab}} \approx m_{d}^{2} c^{2} / 2 m_{b} .\) (c) Calculate these two energies for the process \(e^{+}+e^{-} \rightarrow \psi,\) with \(m_{e} \approx 0.5 \mathrm{MeV} / c^{2}\) and \(m_{\psi} \approx 3100 \mathrm{MeV} / c^{2} .\) Your answers should explain why particle physicists go to the trouble and expense of building colliding-beam experiments.

(a) Show that if a body has speed \(v < c\) in one inertial frame, then \(v < c\) in all frames. [Hint: Consider the displacement four-vector \(d x=(d \mathbf{x}, c d t),\) where \(d \mathbf{x}\) is the three-dimensional displacement in a short time \(d t .]\) (b) Show similarly that if a signal (such as a pulse of light) has speed \(c\) in one frame, its speed is \(c\) in all frames.

For any two objects \(a\) and \(b\), show that the scalar product of their four- velocities is \(u_{a} \cdot u_{b}=\) \(-c^{2} \gamma\left(v_{\mathrm{rel}}\right),\) where \(\gamma(v)\) denotes the usual \(\gamma\) factor, \(\gamma(v)=1 / \sqrt{1-v^{2} / c^{2}},\) and \(v_{\mathrm{rel}}\) denotes the speed of \(a\) in the rest frame of \(b\) or vice versa.

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