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Chapter 15: Problem 59

If one defines a variable mass \(m_{\text {var }}=\gamma m\), then the relativistic momentum \(\mathbf{p}=\gamma m \mathbf{v}\) becomes \(m_{\text {var }} \mathbf{v}\) which looks more like the classical definition. Show, however, that the relativistic kinetic energy is not equal to \(\frac{1}{2} m_{\mathrm{var}} v^{2}\)

Short Answer

Expert verified
Relativistic and classical kinetic energies differ due to relativistic effects and the Lorentz factor \(\gamma\).

Step by step solution

01

Define Variables and Formulas

Let's begin by defining the variables and important concepts. We have the relativistic mass given by the variable \(m_{\text{var}} = \gamma m\) where \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\) is the Lorentz factor. The classical momentum is \(m_{\text{var}} \mathbf{v}\). We need to analyze the kinetic energy expression \(\frac{1}{2} m_{\text{var}} v^2\) and see how it compares to the relativistic kinetic energy.
02

Relativistic Kinetic Energy Formula

The relativistic kinetic energy is given by the formula \(T = (\gamma - 1)mc^2\). This formula results from the difference between the total energy \(E = \gamma mc^2\) and the rest energy \(E_0 = mc^2\).
03

Compare Classical and Relativistic Kinetic Energies

The classical kinetic energy formula is \(T_{\text{classical}} = \frac{1}{2}m_{\text{var}} v^2\). Substituting \(m_{\text{var}} = \gamma m\), we have:\[T_{\text{classical}} = \frac{1}{2}\gamma m v^2\]Compare this expression with the relativistic kinetic energy \((\gamma - 1)mc^2\). Note that they are fundamentally different due to the presence of the Lorentz factor \(\gamma\) and the \(c^2\) term in the relativistic formula.
04

Analyze Differences in Formulas

The classical kinetic energy \(\frac{1}{2} \gamma mv^2\) does not account for the relativistic effects encapsulated in \(\gamma\). As the speed \(v\) approaches the speed of light \(c\), \(\gamma\) becomes significantly larger, leading to discrepancies between both kinetic energy expressions. Classically, \(\gamma\) is not considered, leading to fundamentally different results as \(v\) increases.

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Key Concepts

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

Lorentz Factor
The Lorentz factor, often denoted as \( \gamma \), is a crucial concept in the realm of relativity. It is defined as \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \). This factor becomes significant especially when dealing with objects moving at speeds comparable to the speed of light, \( c \). The key idea behind the Lorentz factor is that it accounts for how time and space measurements change for objects moving at high velocities.

  • When an object moves much slower than the speed of light, \( \gamma \approx 1 \) and relativistic effects are negligible.
  • As the object's speed \( v \) increases towards \( c \), \( \gamma \) increases, leading to more noticeable relativistic effects.
This factor appears in several relativistic equations, like in the expression of relativistic momentum \( \mathbf{p} = \gamma m \mathbf{v} \) and relativistic energy \( E = \gamma mc^2 \). It shows how these physical quantities differ from their classical counterparts as speed increases.
Relativistic Mass
Relativistic mass is a concept that takes into account the increase in mass that a body experiences as its velocity approaches the speed of light. It is expressed as \( m_\text{var} = \gamma m \), where \( m \) is the rest mass and \( \gamma \) is the Lorentz factor.

This term \( m_\text{var} \) emphasizes that the effective inertial mass of an object is velocity-dependent. As a result:
  • The faster an object moves, the more massive it seems, due to \( \gamma \).
  • At speeds approaching \( c \), \( \gamma \) increases and so does the relativistic mass.
This concept helps in explaining why accelerating objects to the speed of light is energetically prohibitive since their mass would effectively become infinite, requiring infinite energy to continue acceleration.
Classical Mechanics
Classical mechanics, established by Isaac Newton, describes the motion of objects using laws that assume time and space are absolute. This field is effective at low velocities compared to the speed of light.

In classical mechanics, concepts like momentum and kinetic energy have simpler forms:
  • Momentum \( \mathbf{p} = m \mathbf{v} \)
  • Kinetic energy \( T_\text{classical} = \frac{1}{2} m v^2 \)
These formulas are adequate when speeds are much lower than \( c \). However, as speeds increase, these expressions diverge from reality, requiring relativistic mechanics to account for effects such as time dilation and length contraction. Consequently, formulas incorporating the Lorentz factor are essential for accurate predictions at high velocities.
Speed of Light
The speed of light, denoted by \( c \), is a constant with a value of approximately 299,792,458 meters per second. It is fundamental in Einstein's theory of relativity, serving as the ultimate speed limit in the universe.

Several key points concerning the speed of light include:
  • Nothing with mass can travel at or faster than the speed of light, as this would require infinite energy.
  • Light travels at \( c \) in a vacuum, and this speed is invariant鈥攎eaning all observers, regardless of their relative motion, measure the speed of light as \( c \).
The speed of light plays a pivotal role in relativistic equations, notably those involving the Lorentz factor and relativistic energy, illustrating how \( c \) impacts time and space perception at high velocities. Understanding \( c \) enables a deeper grasp of how the universe operates at a fundamental level.

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Most popular questions from this chapter

A particle \(a\) traveling along the positive \(x\) axis of frame \(\mathcal{S}\) with speed 0.5c decays into two identical particles, \(a \rightarrow b+b,\) both of which continue to travel on the \(x\) axis. (a) Given that \(m_{a}=2.5 m_{b},\) find the speed of either \(b\) particle in the rest frame of particle \(a .\) (b) By making the necessary transformation on the result of part (a), find the velocities of the two \(b\) particles in the original frame S.

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 particle of mass \(m\) and charge \(q\) moves in a uniform, constant magnetic field \(\mathbf{B}\). Show that if \(\mathbf{v}\) is perpendicular to \(\mathbf{B},\) the particle moves in a circle of radius $$r=|\mathbf{p} / q B|$$ [This result agrees with the nonrelativistic result \((2.81),\) except that \(\mathbf{p}\) is now the relativistic momentum \(\mathbf{p}=\gamma m \mathbf{v} .]\)

The relativistic kinetic energy of a particle is \(T=(\gamma-1) m c^{2} .\) Use the binomial series to express \(T\) as a series in powers of \(\beta=v / c .\) (a) Verify that the first term is just the non relativistic kinetic energy, and show that to lowest order in \(\beta\) the difference between the relativistic and non relativistic kinetic energies is \(3 \beta^{4} m c^{2} / 8 .\) (b) Use this result to find the maximum speed at which the non relativistic value is within \(1 \%\) of the correct relativistic value.

We have seen that the scalar product \(x \cdot x\) of any four-vector \(x\) with itself is invariant under Lorentz transformations. Use the invariance of \(x \cdot x\) to prove that the scalar product \(x \cdot y\) of any two four-vectors \(x\) and \(y\) is likewise invariant.
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