Physics 3 for Electrical Engineering

Ben Gurion University of the Negev www.bgu.ac.il/atomchip,www.bgu.ac.il/nanocenter Physics 3 for Electrical Engineering Lecturers: Daniel Rohrlich, Ron Folman Teaching Assistants: Daniel Ariad, Barukh Dolgin Week 2. Special relativity – kinematic paradoxes • addition of velocities • relativistic mass and momentum • relativistic energy • energy-momentum 4-vector • decay and scattering in relativity • nuclear energy Sources: Feynman Lectures I, Chap. 15, Sects. 5-6 and Chap. 16; Tipler and Llewellyn, Chap. 1, Sect. 6 and Chap. 2, Sects. 1-4

Kinematic paradoxes The “Twin Paradox” You go out into space for one year at 0.9 of the speed of light. When you come back, how old will your twin brother be? Is there a paradox here? How would you solve it?

The “Triplet Paradox” Grumpy lives in Haifa; Dumpy and Jump live in Tel Aviv. They are identical triplets with synchronized watches. One day, Jump takes a high-speed (v = c/2) train to visit Grumpy. Grumpy and Dumpy expect Jump’s watch to run slow during the train ride. But Jump expects their watches to run slow! He reasons, “When the train left the station, I saw that my watch was still synchronized with Dumpy’s. So it was still synchronized with Grumpy’s watch. Now I am sitting and they are moving fast relative to me! So their watches must run slower than mine!” Who is right???

Event 1: Jump looks at his watch and at Dumpy’s watch. He sees t1′ = 0 and t1 = 0. Let their location (Tel Aviv) be x1′ = x1 = 0. Summary of Event 1: t1′ = t1 = 0, x1′ = x1 = 0. Event 2: Grumpy looks at his watch and sees t2 = 0. Let his location (Haifa) be x2 = 85 km. Summary of Event 2 : t2 = 0, x2 = 85 km.

Event 1: Jump looks at his watch and at Dumpy’s watch. He sees t1′ = 0 and t1 = 0. Let their location (Tel Aviv) be x1′ = x1 = 0. Summary of Event 1: t1′ = t1 = 0, x1′ = x1 = 0. Event 2: Grumpy looks at his watch and sees t2 = 0. Let his location (Haifa) be x2 = 85 km. Summary of Event 2 : t2 = 0, x2 = 85 km. Hence ∆t = t2 – t1 = 0, ∆x = x2 – x1 = 85 km.

Event 1: Jump looks at his watch and at Dumpy’s watch. He sees t1′ = 0 and t1 = 0. Let their location (Tel Aviv) be x1′ = x1 = 0. Summary of Event 1: t1′ = t1 = 0, x1′ = x1 = 0. Event 2: Grumpy looks at his watch and sees t2 = 0. Let his location (Haifa) be x2 = 85 km. Summary of Event 2 : t2 = 0, x2 = 85 km. Hence ∆t = t2 – t1 = 0, ∆x = x2 – x1 = 85 km. What is ∆t′ ?

“The barn paradox” The barn is 5 m long to the farmer. The pole is 10 m long to the runner and 5 m long to the farmer, i.e. we have g = 2 and so b = v/c = 0.866. According to the farmer, the pole fits into the barn at a single instant. But the runner cannot agree! The pole is 10 m long and the barn is only 2.5 long!

What are the events? Event 1: the rear end of the pole reaches the entrance to the barn. Event 2: the front tip of the pole reaches the far end of the barn. In the farmer’s reference frame, Event 1 occurs at time t1 and at place x1; Event 2 occurs at time t2 and at place x2 ; Dt = t2 – t1 = 0 and Dx = x2 – x1= 5 m. In the runner’s reference frame: Dt′ = g(Dt – vDx/c2) = (2)(-0.866)(5 m) / c = -28.9 ns, Dx′ = g(Dx – vDt) = 10 m.

What are the events? Event 1: the rear end of the pole reaches the entrance to the barn. Event 2: the front tip of the pole reaches the far end of the barn. For the runner, the two events are not simultaneous! What does it mean that Dt′ is negative?

What are the events? Event 1: the rear end of the pole reaches the entrance to the barn. Event 2: the front tip of the pole reaches the far end of the barn.

Further applications of Lorentz transformations • Dx′ = g(Dx – vDt) • Dy′ = Dy • Dz′ = Dz • Dt′ = g(Dt – vDx/c2) How do we derive time dilation from the Lorentz transformations?

cDt/2 L cDt/2 ☺ vDt If Alice (primed system) moves along the x-axis with velocity v relative to Bob (unprimed system), then • Dx′ = g(Dx – vDt) • Dy′ = Dy • Dz′ = Dz • Dt′ = g(Dt – vDx/c2) How do we derive time dilation from the Lorentz transformations?

cDt/2 L cDt/2 ☺ vDt If Alice (primed system) moves along the x-axis with velocity v relative to Bob (unprimed system), then • Dx′ = g(Dx – vDt) • Dy′ = Dy • Dz′ = Dz • Dt′ = g(Dt – vDx/c2) For Alice, the clock stays in one place. Hence Dx′ = 0.

cDt/2 L cDt/2 ☺ vDt If Alice (primed system) moves along the x-axis with velocity v relative to Bob (unprimed system), then • Dx′ = g(Dx – vDt) Hence Dx = vDt • Dt′ = g(Dt – vDx/c2)= g(Dt – v2Dt/c2) = g(Dt/g2 ) = Dt/g 0 = For Alice, the clock stays in one place. Hence Dx′ = 0.

Further applications of Lorentz transformations • Dx′ = g(Dx – vDt) • Dy′ = Dy • Dz′ = Dz • Dt′ = g(Dt – vDx/c2) How do we derive length contraction from the Lorentz transformations?

If the runner (primed system) moves along the x-axis with velocity v relative to the farmer (unprimed system), then • Dx′ = g(Dx – vDt) • Dy′ = Dy • Dz′ = Dz • Dt′ = g(Dt – vDx/c2) How do we derive length contraction from the Lorentz transformations?

If the runner (primed system) moves along the x-axis with velocity v relative to the farmer (unprimed system), then • Dx′ = g(Dx – vDt) • Dy′ = Dy • Dz′ = Dz • Dt′ = g(Dt – vDx/c2) For the farmer, Dx = L and Dt = 0. For the runner, Dx′ = L′.

If the runner (primed system) moves along the x-axis with velocity v relative to the farmer (unprimed system), then • Dx′ = g(Dx – vDt) = gL • Dy′ = Dy • Dz′ = Dz • Dt′ = g(Dt – vDx/c2) L′ = For the farmer, Dx = L and Dt = 0. For the runner, Dx′ = L′.

Addition of velocities: ux= dx/dt ; uy= dy/dt ; uz= dz/dt ux′= dx′/dt′ ; uy′= dy′/dt′ ; uz′= dz′/dt′ For inertial reference frames moving with relative velocity v along the x–axis, ux′= (ux–v)/(1–vux/c2) uy′= uy /g(1–vux/c2) uz′= uz /g(1–vux/c2) Thus you can never surpass the speed of light e.g. by shooting a 0.9c bullet from a 0.9c jet: (0.9c+0.9c)/[1+(0.9c) · (0.9c)/c2] = (1.80/1.81)c = 0.9945c

Derivation: apply the Lorentz transformations to the differentials, and substitute them into ux′, uy′ and uz′ to obtain

Relativistic mass and momentum We now move from kinematics (pure description of motion) to dynamics (description of forces). We will discover that if momentum (mv) is conserved in one inertial reference frame, it is not conserved in another! But conservation of momentum is a law of physics. And Einstein postulated that the laws of physics are the same in all reference frames….

Two identical particles collide in their center-of-mass frame: y uy The blue particle has mass m and speed v and so does the red particle. In the center-of-mass frame, total momentum is 0 before and after the collision. x uy ux

Two identical particles collide in their center-of-mass frame: y uy The blue particle has mass m and speed v and so does the red particle. In the center-of-mass frame, total momentum is 0 before and after the collision. x uy ux But now we go to a reference frame moving to the right with speed ux:

Two identical particles collide in their center-of-mass frame: y uy The blue particle has mass m and speed v and so does the red particle. In the center-of-mass frame, total momentum is 0 before and after the collision. x uy y ux uy′ But now we go to a reference frame moving to the right with speed ux: x uy uy′ uy′ uy uy′ ux′ = 0

Since uy′ +uy′ ≠ 0, the y-component of momentum is not conserved… y uy′ x uy uy′ uy′ uy uy′ ux′ = 0

Since uy′ +uy′ ≠ 0, the y-component of momentum is not conserved…unless we redefine momentum! y uy′ x uy uy′ uy′ uy uy′ ux′ = 0

Since uy′ +uy′ ≠ 0, the y-component of momentum is not conserved…unless we redefine momentum! Let’s try p = m f(v) v with f(0)=1. y uy′ x uy uy′ uy′ uy uy′ ux′ = 0

y Boost left ux Boost right ux x ux y uy" uy′ y x ux"= 0 x uy′ Boost right v = ux" uy" ux′ = 0

Addition of velocities with v = ux" and : uy" uy" uy" uy′ vux"/ c2 y uy" = – uy′ uy′ y x ux"= 0 x uy′ Boost right v = ux" uy"= – uy′ ux′ = 0

So now if we assume p = m f(v) v, then and in the limit uy"→ 0 we replace ux" with v to obtain and therefore p = mγv .

Relativistic momentum: p = gmv (or “relativistic mass”?) v/c . gmv/ mv v/c

Relativistic energy We accelerate a body using a force F = dp/dt :

If we accelerate the body from rest to a final speed v, we find that the change in kinetic energy of the body is For small v/c the kinetic energy is W = (γ–1) mc2 ≈ mv2/2. But from W we do not learn what the total energy is. Could there really be a huge “rest energy” mc2 ?

Here’s the answer: consider a head-on collision between two bodies of mass m and relativistic momentum ±mγv forming a body of mass M at rest: Momentum is conserved. Now let’s look at the collision in a frame that moves perpendicular to v with a small, nonrelativistic velocity u: From conservation of momentum in the u-direction, we have where the factor γ' differs slightly from γ = because there is a component of velocity u added to v. m m m m

Now divide both sides by u; in the limit u → 0 we get where γ depends only on v. We see that the body at rest having mass Mcontains the kinetic energy of the colliding particles! A body at rest contains energy! Since Mc2 = 2γmc2 the energy E of a body of mass m and speed v must equal E = γmc2. m m

Energy-momentum 4-vector E = mγc2, so E and pc form a 4-vector u = (E, pc) and (Note E and pc have the same units.) The energy-momentum 4-vector transforms like any other 4-vectorunder Lorentz transformations. Hence both the energy E and the momentum p are conserved in each inertial reference frame.

Decay and scattering in relativityA fateful example A heavy particle of mass M decays into two lighter particles of mass m. What is the velocity of each of the lighter particles? How much kinetic energy is released? Let us treat this problem in the rest frame of the heavy particle. Energy-momentum conservation: (Mc2, 0) = (mγ1c2, mγ1v1c) + (mγ2c2, mγ2v2c) = (E1, p1c) + (E2, p2c) Hence p1= – p2 and γ1= γ2 = γ from momentum conservation, and Mc2 = 2mγc2 from energy conservation. Therefore γ = M/2m and since we get .

So in the rest frame of the heavy particle, the lighter particles come out back to back with opposite velocities and equal speeds There is no kinetic energy initially. The final kinetic energy is More generally, if the heavy particle decays into several lighter particles of masses mi, the kinetic energy released is

Application: fission of U235 into Ba141 and Kr92 and also two neutrons yields how much kinetic energy? mass (a.m.u.) U235 Ba141 Kr92 n Kinetic energy yield = (235.043923 – 140.914411 – 91.926156 – 2 × 1.008665) a.m.u. × (1.660539× 10–27 kg/a.m.u.) × (2.997925 × 108 m/s)2 = 2.77629 × 10–11 J

One gram of U235 contains 2.56 × 1021 atoms of U235, so fission of one gram of pure U235 yields (2.56 × 1021 ) × (2.78 × 10–11 J) = 7.12 × 1010 J of kinetic energy. By way of comparison, consider two cars, each with 1500 kg mass, at speed 90 km/hour = 25 m/s, in a head-on collision. The total kinetic energy lost in the collision is (3000 kg) × (25 m/s)2 = 1.875 × 106 J. So fission of one gram of U235 has the destructive power of about 40,000 such collisions.

The bomb dropped on Hiroshima caused about 800 grams of U235 to undergo fission…

...and about 80,000 deaths on that day, two to three times as many within five years.

Relativistic scattering: another fateful example Two deuterium (1H2) nuclei collide and produce a helium-3 (2He3) nucleus and a neutron. What are the velocities and energies of the helium nucleus and the neutron?

The easiest way to solve this problem is to go to the “center-of-momentum” frame, i.e. the inertial frame in which the total momentum is zero:

In this case the problem reduces to the previous problem, with the initial mass M of the decaying particle replaced by the initial energy of the two deuterium nuclei divided by c2.

What is the minimum amount of kinetic energy produced by this scattering process (in the center-of-momentum frame)?

What is the minimum amount of kinetic energy produced by this scattering process (in the center-of-momentum frame)? It is [2 × m(deuterium) – m(helium-3) – m(neutron)] × c2

Note: In fission of uranium-235, the fraction of mass converted into energy is

Physics 3 for Electrical Engineering