# lecture5 0709

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Published on November 13, 2007

Author: sanay

Source: authorstream.com

Slide1:  The Twin Paradox Homer and Ulysses are identical twins. Ulysses travels at a constant high speed to a star and returns to Earth while Homer remains at home. Homer’s clock: it takes Ulysses 5 y to get to the star and 5 y to come back. Thus Homer is 10 years old when Ulysses comes back. If v= 0.8c, then the time interval recorded by Ulysses is 6 years (10/). Ulysses on the other hand should expect Homer to have aged only 3.6 years (6/). Slide2:  Once each year, each twin will send a light signal to the other. Slide4:  The Pole and Barn Paradox A runner carries a 10 m pole (rest length) toward a 5 m long barn. The runner carrying the pole at speed v enters the barn and at some instant the farmer sees the pole completely contained in the barn: a 10-m pole into a 5-m barn!! The minimum speed v: For the runner, the pole is 2.5 m long. How can a 2.5 m barn enclose a 10 m pole ????? Slide5:  From the farmers’s point of view Slide6:  From the runner’s point of view Relativistic Momentum:  By considering a collision between two bodies, we can infer that the mass of a body appears to vary with its velocity. Frank (fixed or stationary system) is at rest in system K holding a ball of mass m. Mary (moving system) holds an identical ball in system K that is moving in the x direction with velocity v with respect to system K. Each of them projects his/her sphere with a speed u (as judge by himself/herself) in a direction perpendicular to x (again as judge by himself/herself) so that a collision takes place (u << v) Relativistic Momentum Slide10:  A B Although observers in both frames see the same event, they disagree about the length of time the particle thrown from the other frame requires to make the collision and return. Relativistic Momentum:  Rather than abandon the conservation of linear momentum, let us look for a modification of the definition of linear momentum that preserves both it and Newton’s second law. To do so requires reexamining mass to conclude that: Relativistic Momentum Relativistic momentum (2.48) Relativistic Momentum:  Some physicists like to refer to the mass in Equation (2.48) as the rest mass m0 and call the term m = m0 the relativistic mass. In this manner the classical form of momentum, mv, is retained. The mass is then imagined to increase at high speeds. Most physicists prefer to keep the concept of mass as an invariant, intrinsic property of an object. We adopt this latter approach and will use the term mass exclusively to mean rest mass. Relativistic Momentum Relativistic Second Law:  Relativistic Second Law Due to the new idea of relativistic mass, we must now redefine the concepts of work and energy. Therefore, we modify Newton’s second law to include our new definition of linear momentum, and force becomes: Slide14:  EXAMPLE: Find the acceleration of a particle of mass m and velocity v when it is acted upon by the constant force F, where F is parallel to v. Mass and Energy: where E0=mc2 comes from:  The work W12 done by a force to move a particle from position 1 to position 2 along a path is defined to be where K1 is defined to be the kinetic energy of the particle at position 1. (2.55) Mass and Energy: where E0=mc2 comes from For simplicity, let the particle start from rest under the influence of the force and calculate the kinetic energy K after the work is done. Relativistic Kinetic Energy:  The limits of integration are from an initial value of 0 to a final value of . The integral in Equation (2.57) is straightforward if done by the method of integration by parts. The result, called the relativistic kinetic energy, is (2.57) (2.58) Relativistic Kinetic Energy Relativistic Kinetic Energy:  Equation (2.58) does not seem to resemble the classical result for kinetic energy, K = (1/2) mu2. However, if it is correct, we expect it to reduce to the classical result for low speeds. Let’s see if it does. For speeds u << c, we expand in a binomial series as follows: where we have neglected all terms of power (u/c)4 and greater, because u << c. This gives the following equation for the relativistic kinetic energy at low speeds: which is the expected classical result. We show both the relativistic and classical kinetic energies in Figure 2.31. They diverge considerably above a velocity of 0.6c. (2.59) Relativistic Kinetic Energy Relativistic and Classical Kinetic Energies:  Relativistic and Classical Kinetic Energies Total Energy and Rest Energy:  Total Energy and Rest Energy We rewrite Equation (2.58) in the form The term mc2 is called the rest energy and is denoted by E0. This leaves the sum of the kinetic energy and rest energy to be interpreted as the total energy of the particle. The total energy is denoted by E and is given by (2.63) (2.64) (2.65) Momentum and Energy:  We square this result, multiply by c2, and rearrange the result. We use Equation (2.62) for 2 and find Momentum and Energy Momentum and Energy (continued):  Momentum and Energy (continued) The first term on the right-hand side is just E2, and the second term is E02. The last equation becomes We rearrange this last equation to find the result we are seeking, a relation between energy and momentum. or Equation (2.70) is a useful result to relate the total energy of a particle with its momentum. The quantities (E2 – p2c2) and m are invariant quantities. Note that when a particle’s velocity is zero and it has no momentum, Equation (2.70) correctly gives E0 as the particle’s total energy. (2.71) (2.70) Computations in Modern Physics:  Computations in Modern Physics We were taught in introductory physics that the international system of units is preferable when doing calculations in science and engineering. In modern physics a somewhat different, more convenient set of units is often used. The smallness of quantities often used in modern physics suggests some practical changes. Units of Work and Energy:  Units of Work and Energy Recall that the work done in accelerating a charge through a potential difference is given by W = qV. For a proton, with the charge e = 1.602 10-19 C being accelerated across a potential difference of 1 V, the work done is W = (1.602  10-19)(1 V) = 1.602  10-19 J The work done to accelerate the proton across a potential difference of 1 V could also be written as W = (1 e)(1 V) = 1 eV Thus eV, pronounced “electron volt,” is also a unit of energy. It is related to the SI (Système International) unit joule by the 2 previous equations. 1 eV = 1.602  10-19 J Other Units:  Other Units Rest energy of a particle: Example: E0 (proton) Atomic mass unit (amu): Example: carbon-12 Mass (12C atom) Mass (12C atom) A particle has a kinetic energy of 62 MeV and a momentum of 335 MeV/c. Find its mass (in MeV/c2) and speed (as a fraction of c). :  A particle has a kinetic energy of 62 MeV and a momentum of 335 MeV/c. Find its mass (in MeV/c2) and speed (as a fraction of c). Find the mass (in GeV/c2) of a particle whose total energy is 4.00 GeV and whose momentum is 1.45 GeV/c. Find the total energy of this particle in a reference frame in which its momentum is 2.00 GeV/c. :  Find the mass (in GeV/c2) of a particle whose total energy is 4.00 GeV and whose momentum is 1.45 GeV/c. Find the total energy of this particle in a reference frame in which its momentum is 2.00 GeV/c. The mass of a system is generally DIFFERENT than the sum of the masses of its parts.:  The mass of a system is generally DIFFERENT than the sum of the masses of its parts.

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October 21, 2020

October 21, 2020

October 21, 2020

October 21, 2020

October 21, 2020

October 21, 2020

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