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Suppose you have a collection of particles, all moving in the \(x\) direction, with energics \(E_{1}, E_{2}, E_{3}, \ldots\) and momenta \(p_{1}, p_{2}, p_{3}, \ldots\) Find the velocity of the center of momentum frame, in which the total momentum is zero.

Short Answer

Expert verified
The velocity is \(v = \frac{p_{total}c^2}{E_{total}}\)."

Step by step solution

01

Understanding the Center of Momentum Frame

The center of momentum frame is a reference frame in which the total momentum of the system is zero. To find this frame, we must determine the velocity at which an observer sees the total momentum as zero.
02

Defining Total Energy and Momentum

The total energy of the particles is the sum of individual energies, given by \(E_{total} = E_1 + E_2 + E_3 + \ldots \). The total momentum is likewise the sum of momenta, \(p_{total} = p_1 + p_2 + p_3 + \ldots \).
03

Using Lorentz Transformation for Momentum

In order to find the velocity of the frame where the total momentum is zero, we use the Lorentz transformation for momentum. The momentum of a particle in a moving frame is given by \(p' = \gamma(p - vE/c^2)\), where \(c\) is the speed of light, \(v\) is the velocity of the moving frame compared to the original frame, and \(\gamma\) is the Lorentz factor \(\gamma = \frac{1}{\sqrt{1-v^2/c^2}}\).
04

Setting the Total Momentum to Zero

Set \(p'_{total} = \gamma(p_{total} - vE_{total}/c^2) = 0\). Since \(\gamma > 0\), we then have \(p_{total} = vE_{total}/c^2\).
05

Solving for the Velocity

Rearrange and solve for \(v\): \[v = \frac{p_{total}c^2}{E_{total}}\] This equation gives the velocity of the frame where the total momentum is zero.

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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 the theory of relativity, crucial for understanding how measurements of space and time change between different inertial frames. It establishes how to convert the coordinates of an event as perceived in one inertial frame to another that moves at a constant velocity relative to the first. This is particularly important when dealing with velocities close to the speed of light.
When considering momentum in a relativistic context, the Lorentz transformation helps adjust the observed momentum of particles in different frames. The equation for transforming momentum under Lorentz conditions is given by:
  • \[ p' = \gamma (p - vE/c^2) \]
Here, \( p' \) represents the momentum in the new frame, \( v \) is the relative velocity of the frame, \( c \) is the speed of light, and \( \gamma \) is the Lorentz factor \( \gamma = \frac{1}{\sqrt{1-v^2/c^2}} \).
By applying this transformation, you can determine how the momentum changes as you move from one inertial frame to another, which is essential in finding the center of momentum frame.
total momentum
Total momentum in a system of particles is the sum of all individual momenta. For a series of particles moving linearly, the total momentum \( p_{total} \) can be expressed as:
  • \[ p_{total} = p_1 + p_2 + p_3 + \ldots \]
This total momentum is a vector quantity, meaning it has both magnitude and direction. Recognizing that momentum is conserved in an isolated system allows us to find reference frames where particular conditions, such as zero total momentum, are met.In the problem of finding the center of momentum frame, understanding total momentum helps define the velocity at which this frame is achieved. By setting the total transformed momentum to zero, we can solve for the velocity \( v \) that makes \( p'_{total} = 0 \), leading to the equation:
  • \[ p_{total} = vE_{total}/c^2 \]
where \( E_{total} \) is the sum of all the energies of the particles. This provides a direct connection between the system's total momentum and the velocity of the center of momentum frame.
relativistic velocity
Relativistic velocity comes into play when considering speeds that approach the speed of light, requiring the theory of relativity for accurate calculations. It's not merely the classic speed definition but involves nuanced changes highlighted by relativistic effects.Unlike classical velocity addition, relativistic velocity considers the limits and effects as objects approach the speed of light. The speed is defined concerning an observer within a reference frame, and the effects of relativity change how we calculate it.In the context of the center of momentum frame, the relativistic velocity \( v \) is derived where total momentum is zero. This special velocity reflects the perspective in which the system's net linear momentum vanishes, calculated by the equation:
  • \[ v = \frac{p_{total}c^2}{E_{total}} \]
This equation highlights how both energy and momentum define the frame's velocity, incorporating the relativistic speed nuances. Understanding this concept allows dealing with relativistic particles, clarifying why speed appears differently from different reference frames and emphasizing the relativity principles in motion.

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Most popular questions from this chapter

(a) Draw a space-time diagram representing a game of catch (or a conversation) between two people at rest, \(10 \mathrm{ft}\) apart. How is it possible for them to communicate, given that their separation is spacelike? (b) There's an old limerick that runs as follows: There once was a girl named Ms. Bright. Who could travel much faster than light. She departed one day, The Einsteinian way, And rcturned on the previous night. What do you think? Even if she could travel faster than the speed of light, could she return before she set out? Could she arrive at some intermediate destination before she set out? Draw a space-time diagram representing this trip.

You may have noticed that the four-dimensional gradient operator \(\partial / \partial x^{\mu}\) functions like a covariant 4 -vector-in fact, it is often written \(\partial_{\mu}\), for short. For instance. the continuity equation, \(\partial_{\mu} J^{\mu}=0\), has the form of an invariant product of two vectors. The corresponding contravariant gradient would be \(\partial^{\mu} \equiv \partial x_{\mu} .\) Prove that \(\partial^{\mu} \phi\) is a (contravariant) 4 -vector, if \(\phi\) is a scalar function, by working out its transformation law, using the chain rule.

(a) Construct a tensor \(D^{\mu v}\) (analogous to \(F^{\mu v}\) ), out of \(\mathbf{D}\) and \(\mathbf{H}\). Use it to express Maxwell's equations inside matter in terms of the free current density \(J_{f}^{\mu} .\) [Answer: \(D^{01} \equiv c D_{x}\), \(D^{12} \equiv H_{2}\), etc.; \(\left.\partial D^{\mu v} / \partial x^{\nu}=J_{f}^{\mu} .\right]\) (b) Construct the dual tensor \(H^{\mu v}\) (analogous to \(G^{\mu v}\) ). [Answer: \(H^{01} \equiv H_{x}, H^{12} \equiv-c D_{z}\), etc.] (c) Minkowski proposed the relativistic constitutive relations for linear media: $$ D^{\mu v} \eta_{v}=c^{2} \epsilon F^{\mu v} \eta_{v} \text { and } H^{\mu v} \eta_{v}=\frac{1}{\mu} G^{\mu v} \eta_{v} $$ where \(\epsilon\) is the proper \(^{20}\) permittivity. \(\mu\) is the proper permeability, and \(\eta^{\mu}\) is the 4 -velocity of the material. Show that Minkowski's formulas reproduce Eqs. \(4.32\) and \(6.31\), when the material is at rest. (d) Work out the formulas relating \(\mathbf{D}\) and \(\mathrm{H}\) to \(\mathbf{E}\) and \(\mathbf{B}\) for a medium moving with (ordinary) velocity \(\mathbf{u}\).

An electromagnetic plane wave of (angular) frequency \(\omega\) is traveling in the \(x\) direction through the vacuum. It is polarized in the \(y\) direction, and the amplitude of the electric field is \(E_{0}\). (a) Write down the electric and magnetic fields, \(\mathbf{E}(x, y, z, t)\) and \(\mathbf{B}(x, y, z, t) .\) [Be sure to define any auxiliary quantities you introduce, in terms of \(\omega, E_{0}\), and the constants of nature.] (b) This same wave is observed from an inertial system \(\overline{\mathcal{S}}\) moving in the \(x\) direction with speed u relative to the original systcm \(\mathcal{S}\). Find the electric and magnetic fields in \(\overline{\mathcal{S}}\), and express them in terms of the \(\overline{\mathcal{S}}\) coordinates: \(\overline{\mathbf{E}}(\bar{x}, \bar{y}, \bar{z}, \bar{t})\) and \(\overline{\mathbf{B}}(\bar{x}, \bar{y}, \bar{z}, \bar{t})\). [Again, be sure to define any auxiliary quantities you introduce.] (c) What is the frequency \(\bar{\omega}\) of the wave in \(\bar{S}\) ? Interpret this result. What is the wavclength \(\bar{\lambda}\) of the wave in \(\overline{\mathcal{S}}\) ? From \(\bar{\omega}\) and \(\bar{\lambda}\), determine the speed of the waves in \(\overline{\mathcal{S}}\). Is it what you expected? (d) What is the ratio of the intensity in \(\overline{\mathcal{S}}\) to the intensity in \(\mathcal{S}\) ? As a youth, Einstein wondered what an electromagnetic wave would look like if you could run along beside it at the spced of light. What can you tell him about the amplitude, frequency, and intensity of the wave, as \(v\) approaches \(c\) ?

Two charges \(\pm q\) approach the origin at constant velocity from opposite directions along the \(x\) axis. They collide and stick together, forming a neutral particle at rest. Sketch the electric field before and shortly after the collision (remernber that electromagnetic "news" travels at the speed of light). How would you interpret the field after the collision, physically? \(^{19}\)

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