* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Future Circular Collider wikipedia, lookup

Derivations of the Lorentz transformations wikipedia, lookup

Renormalization group wikipedia, lookup

ALICE experiment wikipedia, lookup

Noether's theorem wikipedia, lookup

Quantum vacuum thruster wikipedia, lookup

ATLAS experiment wikipedia, lookup

Dirac equation wikipedia, lookup

Monte Carlo methods for electron transport wikipedia, lookup

Old quantum theory wikipedia, lookup

Symmetry in quantum mechanics wikipedia, lookup

Angular momentum operator wikipedia, lookup

Eigenstate thermalization hypothesis wikipedia, lookup

Compact Muon Solenoid wikipedia, lookup

Photon polarization wikipedia, lookup

Electron scattering wikipedia, lookup

Relativistic quantum mechanics wikipedia, lookup

Theoretical and experimental justification for the Schrödinger equation wikipedia, lookup

Transcript

Relativistic Momentum In classical mechanics, the momentum of a particle is defined as a product of its mass and its velocity, m v . In an isolated system of particles, with no net force acting on the system, the total momentum of the system remains the same. However , we can see from a simple though experiment that the quantity Σmivi is not conserved in isolated system. As was shown before, the Galilean transformations leads to same acceleration in different inertial frames: ax’ = ax and forces are also the same in both frames. However, according to Lorentz transformations, acceleration is not the same in two such reference frames. If a particle has acceleration a, and velocity ux in frame S, it’s acceleration in S’, obtained by computing dux’/dt’ is: a ' x ax vux 1 2 c 3 3 We consider two observers: observer A in reference frame S and observer B in frame S’, which is moving to the right in the x direction with speed v with respect to the frame S. Each has a ball of the mass m. One observer throws his ball up with a speed u0 relative to him and the other throws his ball down with a speed u0 relative to him, so that each ball travel a distance L, makes an elastic collision with the other ball, and returns. Classically, each ball has vertical momentum of magnitude mu0. Since the vertical components of the momenta are equal and opposite, the total vertical momentum are zero before the collision. The collision merely reverses the momentum of each ball, so the total vertical momentum is zero after the collision. Relativistically, however, the vertical components of the velocities of the two balls as seen by the observers are not equal and opposite. Thus, when they are reversed by the collision, classical momentum is not conserved. As seen by A in frame S, the velocity of his ball is uAy=+u0. Since the velocity of ball B in frame S’ is u’Bx=0, uBy’=-u0, the y component of the velocity of ball B in frame S is uBy=-u0/γ. So, the Σ miui is conserved only in the approximation that u<<c. We will define the relativistic momentum p of a particle to have the following properties: 1. In collisions , p is conserved. 2.As u/c approaches 0, p approaches mu. We will show that quantity mu p mu u2 1 2 c is conserved in the elastic collision and we take this equation for the definition of the relativistic momentum of a particle. Conservation of the Relativistic Momentum p mu u2 1 2 c One interpretation of this equation is that the mass of an object increases with the speed. Then the quantity mrel m u2 1 2 c is called the relativistic mass. The mass of a particle when it is at rest in some reference frame is then called its rest mass . mrel= γmrest Conservation of the Relativistic Momentum The speed of ball A in S is u0, so the y component of its relativistic momentum is: p Ay mu0 u02 1 2 c The speed of B in S is more complicated. Its x component is v and its y component is –u0/γ. 2 v 2 2 2 2 u B u Bx u By v u 0 1 2 c 2 2 u0 v 2 2 2 u B v u0 2 c 2 Using this result to compute √1-(u2B/c2), we obtain 2 B 2 2 2 0 2 2 u v u 2 v 1 1 2 u0 4 c c c c v 2 u02 1 2 1 2 c c and 2 B 2 2 2 0 2 u u u v 1 1 1 2 1 1 c c c c 2 0 2 Conservation of the Relativistic Momentum The y component of the relativistic momentum of ball B as seen in S is therefore mu0 muBy pBy u B2 u02 1 1 2 1 2 c c pBy mu0 u02 1 2 c Conservation of the Relativistic Momentum Since pBy=-pAy’ the y component of the total momentum of the two balls is zero. If the speed of each ball is reversed by the collision, the total momentum will remain zero and momentum will be conserved. p mu with 1 2 u 1 2 c Relativistic momentum as given by equation p mu u2 1 2 c versus u/c, where u is speed of the object relative to an observer. The magnitude of momentum p is plotted in units of mc. The dashed line shows the classical momentum mu. Relativistic Energy In classic mechanic: The work done by the net force acting on a particle equals the change in the kinetic energy of the particle. In relativistic mechanic: The net force acting on a particle to accelerate it from rest to some final velocity is equal to the rate of change of the relativistic momentum. The work done by the net force can then be calculated and set equal to the change of kinetic energy. Relativistic Energy As in classical mechanic, we will define the kinetic energy as the work done by the net force in accelerating a particle from rest to some final velocity uf. uf uf uf uf mu dp K Fnet ds ds udp ud 2 dt u 0 0 0 0 1 2 c where we have used u=ds/dt. d mu 1 u2 c 2 m d 1 2u m u 2 2 c 2 1 u m 2 c u2 1 c2 m 1 2 1 u c2 3 1 1 u2 c2 u 1 2 u 1 c2 3 3 2 u 1 c2 2 du m 1 u 2 c 1 du u2 1 c2 1 du 2 u 1 2 c 1 du 3 2 u 1 2 c 3 2 du Relativistic Energy Substituting the last result in the equation for kinetic energy we obtain: uf uf 2 mu u du m1 2 K ud 2 c u 0 0 1 2 c 1 2 2 2 mc 1 mc mc u2 1 2 c 3 2 udu Relativistic Energy K mc2 u2 1 2 c mc 2 The second part of this expression, mc2, is independent of the speed and called the rest energy Eo of the particle. E0=mc2 The total relativistic energy: E K mc 2 mc2 1 u c 2 2 Relativistic Energy The work done by unbalanced force increases the energy from the rest energy to the final energy mc2 u2 1 2 c where mrel is the relativistic mass. mrel c 2 m 2 u 1 2 c Relativistic Energy We can obtain a useful expression for the velocity of a particle by multiplying equation for relativistic momentum by c2: p pc 2 mu 2 u 1 2 c 2 mc u 2 u 1 2 c Eu u pc c E Energies in atomic and nuclear physics are usually expressed in units of electron volts (eV) or megaelectron volts (MeV): 1eV = 1.602 x 10-19J A convenient unit for the masses of atomic particles is eV/c2 or MeV/c2, which is the rest energy of the particle divided by c2. The expression for kinetic energy K mc 2 2 mc 2 u 1 2 c does not look much like the classical expression ½(mu2). However when u<<c, we can approximate 1/√1-(u2 /c2) using the binomial expansion: (1+x)n =1 + nx + n(n-1)x2/2 + …….≈ 1 + nx Then u2 1 2 2 u c 1 2 c 1 1 2 1 u2 1 2 c2 and when v<<c 2 1 1 1u 2 2 2 K mc 1 mc 1 1 mu 2 2 2 2 c 1 u 2 c The equation for relativistic momentum: p mu u2 1 2 c and equation for the total energy: E K mc2 mc2 1 u2 c 2 can be combined to eliminate the speed u: E2 = p2c2 +(mc2)2 E2 = p2c2 +(mc2)2 We received the relation for total energy, momentum, and rest energy. If the relativistic energy of a particle is much grater than its rest energy mc2, the second term on the right side of the last equation can be neglected, giving the useful approximation: E ≈ pc (for E>>mc2)