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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.]

Short Answer

Expert verified
Adding a constant to energy in relativistic mechanics changes the invariant mass, affecting physical laws.

Step by step solution

01

Understand Non-Relativistic Mechanics

In non-relativistic mechanics, the total energy includes kinetic and potential energy. The zero point of potential energy can be adjusted by adding a constant, which does not affect the physical predictions of the system, since only energy differences matter.
02

Consider Relativistic Energy-Momentum Relation

In relativistic mechanics, energy and momentum are components of the four-momentum vector: \[ p^\mu = (\frac{E}{c}, \mathbf{p}) \]where \(E\) is the energy, \(\mathbf{p}\) is the momentum, and \(c\) is the speed of light. This vector must transform like a four-vector under Lorentz transformations.
03

Transform Four-Momentum under Lorentz Transformations

Under a Lorentz transformation, a four-vector \(p^\mu\) transforms as:\[ p'^\mu = \Lambda^\mu_{\ u} p^u \]where \(\Lambda^\mu_{\ u}\) is the Lorentz transformation matrix. This preserves the four-vector's scalar product \(p^\mu p_\mu\).
04

Analyze the Impact of Adding a Constant to Energy

Adding a constant to energy changes the four-momentum from \( (\frac{E}{c}, \mathbf{p}) \) to \( (\frac{E+\text{constant}}{c}, \mathbf{p}) \). This alters the scalar product:\[ (\frac{E+\text{constant}}{c})^2 - \mathbf{p}^2 eq (\frac{E}{c})^2 - \mathbf{p}^2 \]Thus, adding a constant affects the invariant mass, violating the requirement for it to be a Lorentz scalar.
05

Conclusion based on Scalar Invariance

The scalar invariance of the four-momentum's product implies that energy differences are meaningful, and absolute energies are fixed. Arbitrary constants cannot be added to energy in relativistic mechanics without altering physical predictions.

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Key Concepts

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

Relativistic Energy-Momentum Relation
In the realm of relativistic mechanics, energy and momentum are intricately linked through the concept of four-momentum, denoted as \( p^\mu = \left( \frac{E}{c}, \mathbf{p} \right) \). Here, \( E \) represents energy, \( \mathbf{p} \) denotes momentum, and \( c \) is the speed of light. Unlike in non-relativistic mechanics where energy and momentum are treated separately, they are unified into a single four-dimensional vector. This unification is important for preserving key physical properties under transformations.
Why is this link crucial? Because it retains the consistency of physical laws for observers in different inertial frames. The relationship is determined by the relativistic equation:
\[ E^2 = (pc)^2 + (m_0 c^2)^2 \] Here, \( m_0 \) is the invariant mass, a term we'll discuss further. This equation is a cornerstone of relativity, illustrating that changes to energy affect momentum and vice versa.
  • The energy term encodes not just kinetic energy but also rest mass energy.
  • Momentum remains coupled with energy, contrary to classical physics.
This holistic view helps us understand why, unlike in classical mechanics, adding an arbitrary constant to energy in relativistic scenarios changes the entire physical context.
Lorentz Transformation
The Lorentz transformation is a set of linear equations that describe how measurements of space and time change for observers in different inertial frames moving relative to each other at a constant velocity. This transformation is fundamental in Einstein's theory of Special Relativity.
A four-vector, such as four-momentum \( p^\mu = ( \frac{E}{c}, \mathbf{p} ) \), transforms according to the Lorentz transformation equations:\[ p'^\mu = \Lambda^\mu_{\ u} p^u \] Where \( \Lambda^\mu_{\ u} \) represents the transformation matrix, tailored to the velocity between the frames.
  • The transformation ensures that the laws of physics remain the same in all inertial frames.
  • It binds spatial and temporal coordinates, illustrating how they can't be independently changed without affecting the other.
  • In turn, the concepts of energy and momentum transform as parts of a single entity — the four-momentum.
Understanding Lorentz transformation is critical because it explains how the physical quantities are observed differently in different frames, without altering their fundamental characteristics. When we attempt to modify energy alone by adding a constant, it disrupts the delicate balance maintained by the Lorentz transformation.
Invariant Mass
Invariant mass, also known as rest mass, is a fundamental quantity retained across different frames of reference in relativistic physics. It is called 'invariant' because it remains constant regardless of an observer's relative motion. In simpler terms, invariant mass does not change even if the energy and momentum of a body do.
Mathematically, the invariant mass is part of the four-momentum scalar product:
\[ (p^\mu p_\mu) = \left( \frac{E}{c} \right)^2 - \mathbf{p}^2 = (m_0 c^2)^2 \] From this relation, it is apparent that altering energy by some constant changes the scalar product and thereby the invariant mass as well.
  • Invariant mass represents the 'true' mass as it is experienced in any frame of reference.
  • This mass is a key measure independent of the body's speed or energy.
  • In a broader sense, invariant mass provides a single-valued measure of a body's energy content when it is at rest.
Thus, any changes made to energy must respect the invariance of mass, emphasizing how energy differences alone are significant in relativistic physics. This key concept shows why you can't just add an arbitrary constant to the energy without altering the physical predictions.

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Most popular questions from this chapter

A positive pion decays at rest into a muon and neutrino, \(\pi^{+} \rightarrow \mu^{+}+\nu .\) The masses involved are \(m_{\pi}=140 \mathrm{MeV} / c^{2}, m_{\mu}=106 \mathrm{MeV} / c^{2},\) and \(m_{\nu}=0 .\) (There is now convincing evidence that \(m_{\nu}\) is not exactly zero, but it is small enough that you can take it to be zero for this problem.) Show that the speed of the outgoing muon has \(\beta=\left(m_{\pi}^{2}-m_{\mu}^{2}\right) /\left(m_{\pi}^{2}+m_{\mu}^{2}\right) .\) Evaluate this numerically. Do the same for the much rarer decay mode \(\pi^{+} \rightarrow \mathrm{e}^{+}+\nu,\left(m_{\mathrm{e}}=0.5 \mathrm{MeV} / c^{2}\right)\)

As a meter stick rushes past me (with velocity v parallel to the stick), I measure its length to be \(80 \mathrm{cm} .\) What is \(v ?\)

(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.

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.

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 \(t+d t\).

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