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Relativistic Momentum In classical mechanics, the momentum of a particle is defined as a product of its mass and its velocity,. 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 Σm i v i is not conserved in isolated system.

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As was shown before, the Galilean transformations leads to same acceleration in different inertial frames: a x ’ = a x 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 u x in frame S, it’s acceleration in S’, obtained by computing du x ’/dt’ is: However, according to Lorentz transformations, acceleration is not the same in two such reference frames. If a particle has acceleration a, and velocity u x in frame S, it’s acceleration in S’, obtained by computing du x ’/dt’ is:

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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.

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One observer throws his ball up with a speed u 0 relative to him and the other throws his ball down with a speed u 0 relative to him, so that each ball travel a distance L, makes an elastic collision with the other ball, and returns.

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Classically, each ball has vertical momentum of magnitude mu 0. 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.

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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 u Ay =+u 0. Since the velocity of ball B in frame S’ is u ’ Bx =0, u By ’=-u 0, the y component of the velocity of ball B in frame S is u By =-u 0 /γ. So, the Σ m i u i is conserved only in the approximation that u<

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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 is conserved in the elastic collision and we take this equation for the definition of the relativistic momentum of a particle.

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Conservation of the Relativistic Momentum One interpretation of this equation is that the mass of an object increases with the speed. Then the quantity 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. m rel = γm rest 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. m rel = γm rest

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Conservation of the Relativistic Momentum

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The speed of ball A in S is u 0, so the y component of its relativistic momentum is: The speed of B in S is more complicated. Its x component is v and its y component is –u 0 /γ.

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Using this result to compute √1-(u 2 B /c 2 ), we obtain and

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Conservation of the Relativistic Momentum The y component of the relativistic momentum of ball B as seen in S is therefore

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Conservation of the Relativistic Momentum Since p By =-p Ay’ 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. with

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Relativistic momentum as given by equation 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.

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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.

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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 u f. where we have used u=ds/dt.

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Relativistic Energy Substituting the last result in the equation for kinetic energy we obtain:

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Relativistic Energy The second part of this expression, mc 2, is independent of the speed and called the rest energy E o of the particle. E 0 =mc 2 E 0 =mc 2 The total relativistic energy:

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Relativistic Energy The work done by unbalanced force increases the energy from the rest energy to the final energy where is the relativistic mass.

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Relativistic Energy We can obtain a useful expression for the velocity of a particle by multiplying equation for relativistic momentum by c 2 :

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Energies in atomic and nuclear physics are usually expressed in units of electron volts (eV) or mega- electron volts (MeV): 1eV = x J A convenient unit for the masses of atomic particles is eV/c 2 or MeV/c 2, which is the rest energy of the particle divided by c 2. A convenient unit for the masses of atomic particles is eV/c 2 or MeV/c 2, which is the rest energy of the particle divided by c 2.

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The expression for kinetic energy does not look much like the classical expression ½(mu 2 ). However when u<

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Then Then and when v<

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The equation for relativistic momentum: and equation for the total energy: can be combined to eliminate the speed u: E 2 = p 2 c 2 +(mc 2 ) 2

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E 2 = p 2 c 2 +(mc 2 ) 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 mc 2, the second term on the right side of the last equation can be neglected, giving the useful approximation: If the relativistic energy of a particle is much grater than its rest energy mc 2, the second term on the right side of the last equation can be neglected, giving the useful approximation: E ≈ pc(forE>>mc 2 ) E ≈ pc (for E>>mc 2 )

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