Relativity Notes Sem 2 FH23 PDF

# Relativity Dr. Suma Jebin FCRIT, Vashi ## Introduction - Classical mechanics regarded space and time to be absolute and separate entities. - It assumed the moments of time and time intervals are supposed to be identical in all frames of reference. - Similarly, lengths are assumed to be identica...

# Relativity Dr. Suma Jebin FCRIT, Vashi ## Introduction - Classical mechanics regarded space and time to be absolute and separate entities. - It assumed the moments of time and time intervals are supposed to be identical in all frames of reference. - Similarly, lengths are assumed to be identical in all frames. - An analysis of these concepts at high velocities revealed that the length of moving bodies contract in the direction of motion and the clocks in motion slow down. - The length, position, size, time and motion are relative. - The theory which deals with the relativity of motion and rest is called theory of relativity. - It is divided into two parts - special theory and general theory. - The special theory of relativity deals with objects and systems which are either moving at a constant speed with respect to one another or are at rest. - The general theory deals with objects or systems which are speeding up or slowing down with respect to one another. ## Frame of Reference - A well defined coordinate system with respect to which the motion of a body is described is called as frame of reference. - For describing the motion of bodies on the earth, we choose a frame of reference rigidly connected to the earth. - In investigation of the Earth's motion, we attach the coordinate system to the sun. - The frame of reference is selected in such a way that the laws of nature may become fundamentally simple in that frame of reference. - There are two types of frames of reference namely - Inertial or unaccelerated frame of reference - Non-inertial or accelerated frame of reference. ### Inertial Frame of Reference: - A frame of reference is said to be inertial when bodies in this frame obey Newton's law of inertia and other laws of Newtonian mechanics. - In this frame, a body is not acted upon by external force. It is at rest or moves with a constant velocity. - The surface of the earth is almost an inertial frame. ### Non-inertial Frame of reference: - A frame of reference is said to be non-inertial frame when a body is not acted upon by any external force. It is accelerated. - In this frame the Newton's laws are not valid. - A ball placed on the floor of a train will move to the rear if the train accelerates forward even though no forces act on it. - A coin placed on a rotating turntable will slide to the periphery though no visible force pushes it away from the centre. ## Galilean Transformations - The transformations which are used to transform the coordinates of position and time from one inertial frame to the other are called as Galilean transformation. - Let us consider two frames of reference s and s'. - Let the velocity of s' along x' direction relative to s be v. - Consider an event happening at P at any particular time t'. - Let the co-ordinates of 'P' with respect to s be (x, y, z, t) and with respect to s' be (x', y', z', t'). - For the sake of simplicity, let us choose our axes sothat x & x' are parallel to v. Let y' & z' be parallel to y & z respectively. - Let us also count the time from the instant at which the origins O and O' coincide. - From figure: - x' = x - vt - y' = y - z' = z - t' = t - Equations are known as Galilean transformation for co-ordinates. ### Velocity Transformation: - Let (x, y, z, t) and (x', y', z', t') are the co-ordinates as observed by the observers in system s and s' respectively. - Then Galilean transformation for co-ordinates is given by - x' = x - vt - y' = y - z' = z - t' = t - Let us assume that the body moves parallel to x-axis with velocity ux. - Differentiating egn © we get - dx' = dx - vdt - Velocity ux = dx/dt = (dx - vdt)/dt = ux - v. - Therefore Galilean velocity transformation eqns is given by - ux' = ux - v - uy' = uy - uz' = uz - A person in a moving railcar throws a ball with a speed u in the direction of motion of the railcar, where u is measured by the person in the railcar. If v is the velocity of the moving railcar, the speed of the ball relative to a stationary observer on the ground will be u + v. ### Acceleration Transformation: - Let us consider (x, y, z, t) and (x', y', z', t') are the co-ordinates as observed by the observer in system s and s' respectively. - Then the velocity transformation equation is given by - ux' = ux - v - uy' = uy - uz' = uz - Let us consider the body moves along x-axis of frame with acceleration. - Differentiating egn ① - ax' = d(ux)/dt = d(ux - v)/dt = ax. - Keeping vx is constant - ax' = ax - ay' = ay - az' = ax - Thus the particle possesses the same acceleration in inertial frames moving relative to each other with constant velocity v. - Multiplying eqns ③ by m. - max' = max - F' = F - egn ③ shows that forces do not change from one inertial frame to another which shows that Newton's laws are invariant with respect to Galilean transformation. ## s' moving relative to s along any direction - Galilean transformation for co-ordinates - x' = x - vxt - y' = y - vyt - z' = z - vzt - t' = t - Galilean transformation for velocity - ux' = ux - vx - uy' = uy - vy - uz' = uz - vz - Galilean transformation for acceleration - ax' = ax - ay' = ay - az' = ax ## Einstein Theory of Relativity (special theory of Relativity) - Einstein suggested that all the inertial frames are completely equivalent. - The special theory is based on the following postulates: - **All physical laws are the same in all inertial frames of reference which are moving with constant velocity relative to each other.** - **The speed of light in vacuum is the same in every inertial frame.** ## Lorentz Transformations - Galilean transformation equations are not valid as the value of c increases. - The transformation equations that apply for all speeds up to velocity of light 'c' and incorporate the invariance of speed of light were developed by H.A. Lorentz. They are known as Lorentz transformations. ### Coordinate transformations - Consider two inertial frames of reference s and s'. - Let s’ is moving with a velocity v with respect to s in the x' direction. - Let the axes of s and s' coincide at t=t’=0. - Let a pulse of light be generated at t=0 (from the origin). - Now consider the situation when the pulse reaches a point P. - Let the position and time co-ordinates of P measured by observers in s and s’ are (x, y, z, t) and (x’, y’, z’, t’) respectively. - Consider the light pulse travels in the form of spherical wavefront at the speed, ‘c’ in both s and s’. - Therefore the point p lies in the spherical surface of radius r = ct in s and r = ct’ in s’. - r = ct - r’ = ct’ - For the frame s: - r^2 = x^2 + y^2 + z^2 = c^2t^2 - For the frame s’: - r'^2 = x'^2 + y'^2 + z'^2 = c^2t'^2 - Since the system moves only in x direction, the coordinates y and z in y’ and z’ do not change: - y’ = y - z’ = z - Then: - x = ct - x’ = ct’ - The transformation equation of x & x’ can be written as: - x’ = λ (x - vt) - inverse relations - x = λ (x’ + vt’) - Using equations above: - ct’ = λ (ct - vt) - ct = λ (ct’ + vt’) - Multiplying equations above: - c^2t’t = [λ (ct-vt)][λ (ct’+vt’)] - c^2t’t = [λ^2 (c-v)][λ^2 (c+v)] - c^2t’t = λ^4 (c-v)(c+v) - λ^4 = (c-v)(c+v) / c^2 = c^2 - v^2 / c^2 = 1 – v^2/c^2 - λ = (1 – v^2/c^2)^(1/2) - Using equations above: - x’= (x - vt)/(1 – v^2/c^2)^(1/2) - Using equations above: - t' = (t - (vx/c^2))/(1 – v^2/c^2)^(1/2) - Thus Lorentz co-ordinates transformation is given by: - x’ = (x - vt)/(1 – v^2/c^2)^(1/2) - y’ = y - z’ = z - t’ = (t - (vx/c^2))/(1 – v^2/c^2)^(1/2) ### Lorentz inverse transformations are given by: - x = (x’ + vt’)/(1 - v^2/c^2)^(1/2) - y = y’ - z = z’ - t = (t’ + (vx’/c^2))/(1 - v^2/c^2)^(1/2) ### Velocity Transformation: - Let us consider two frames of reference s and s’ - s’ is moving relative to s with velocity v. - Then the Lorentz transformation for co-ordinates is given by: - x’ = (x - vt)/(1 – v^2/c^2)^(1/2) - y’ = y - z’ = z - t’ = (t - (vx/c^2))/(1 – v^2/c^2)^(1/2) - Let us assume that the body moves along x-direction with velocity ux. - Differentiating egn ① - dx’ = (dx - vdt)/(1 – v^2/c^2)^(1/2) - Velocity ux’ = dx’/dt’ = (dx - vdt)/(dt - (v/c^2) dx) x (1 – v^2/c^2)^(1/2) - ux’ = (ux - v)/(1 - (uxv/c^2)) - Therefore Lorentz velocity transformation eqns is given by: - ux’ = (ux - v)/(1 - (uxv/c^2)) - uy’ = uy - uz’ = uz ## Length Contraction: - Consider two frames of reference s and s’. - Let s’ is moving with velocity v relative to s along positive x-direction. - Let a rod be placed in s’ along x-axis. - If x1’ and x2’ are the co-ordinates of the ends of the rod, the length of the rod in system s’ is: - l’ = x2’ - x1’ - If x1 and x2 are the co-ordinates of the ends of the rod, the length of the rod in s is: - l = x2 - x1 - According to Lorentz transformation equation, we have: - x2 = (x2’ - vt)/(1 – v^2/c^2)^(1/2) - x1 = (x1’ - vt)/(1 – v^2/c^2)^(1/2) - Substituting equations above: - l’ = ((x2’ - vt)/(1 – v^2/c^2)^(1/2) – (x1’ - vt)/(1 – v^2/c^2)^(1/2) ) - l’ = (x2’ - x1’)/(1 – v^2/c^2)^(1/2) - l’ = l/(1 – v^2/c^2)^(1/2) - l = l’(1 – v^2/c^2)^(1/2) - Thus, the length of the rod moving with velocity v relative to the observer is contracted by a factor (1 – v^2/c^2)^(1/2) in the direction of motion. This is known as Lorentz-Fitzgerald Contraction. ## Time Dilation: - Consider two frames of references s and s’. - Let s’ is moving with a velocity v with respect to s in the positive x-direction. - Suppose a clock is situated in the system s at position x and given signals at intervals Δt. - Δt = t2 - t1 - If this interval is observed by an observer in system s’, then the interval Δt’ recorded by him is given by: - Δt’ = t2’ - t1’ - From Lorentz transformations, we have: - t’ = (t - (vx/c^2))/(1 – v^2/c^2)^(1/2) - t1’ = (t1 - (vx/c^2))/(1 – v^2/c^2)^(1/2) - t2’ = (t2 - (vx/c^2))/(1 – v^2/c^2)^(1/2) - Substituting above: - Δt’ = ((t2 - (vx/c^2))/(1 – v^2/c^2)^(1/2) - (t1 - (vx/c^2))/(1 – v^2/c^2)^(1/2) ) - Δt’ = (t2 - t1)/(1 – v^2/c^2)^(1/2) - Δt’ = Δt/(1 – v^2/c^2)^(1/2) - Δt = Δt’(1 – v^2/c^2)^(1/2) - In egn ⑤, Δt’>Δt as (1 – v^2/c^2)^(1/2) less than unity. - The time interval between two events recorded by a moving clock appears to be greater than the time interval between the same events recorded by the clock when it is at rest. This is called time dilation. ## Mass-Energy Relation: - According to classical mechanics, the mass of a body does not change with velocity. It is independent of velocity. - But according to Einstein, the mass of the body in motion is different from the mass of the body at rest. - The relative formula for the variation of mass with velocity is given by: - m = m0/(1 – v^2/c^2)^(1/2) - Where: - m → relative mass or effective mass - m0 → rest mass - v → velocity - c → velocity of light - The force F = d(mv)/dt as rate of change of momentum is given by: - F = ma + (dm/dt)v - When the particle is displaced through a distance dx by the application of force, the increase in kinetic energy is given by: - dEk = Fdx = mvdx + v^2dm dx ­- dEk = mvdv + v^2dm - Squaring and differentiating egn above: - m^2c^2 - m0^2c^2 = m^2c^2/(1 – v^2/c^2) – mo^2c^2 - c^2(2mdm) - v^2(2mdm) - m^2(2vdv) = 0 ­- c^2dm - v^2dm - mvdv = 0 - Equating egn ⑤ & ⑥ - dEk = c^2dm - Consider the body is at rest initially and by the application of force it acquires a velocity v. Let the mass of the body increases from m0 to m. - Then the total kinetic energy is - Ek = ∫dEk = ∫c^2dm = c^2(m - m0) - The total energy of a moving particle is the sum of its kinetic energy and the energy at rest. - E = Ek + moc^2 = (cm - mo) + moc^2 = mc^2 - moc^2 + moc^2 = mc^2 - egn ⑨ gives the universal equivalence between mass and energy. ## Formulae - Galilean transformation - x’ = x - vt - y’ = y - z’ = z - t’ = t - Inverse Galilean transformation - x = x’ + vt’ - y’ = y - z’ = z - t’ = t - Lorentz transformation - x’ = (x - vt)/(1 – v^2/c^2)^(1/2) - y’ = y - z’ = z - t’ = (t - (vx/c^2))/(1 – v^2/c^2)^(1/2) - Inverse Lorentz transformation - x = (x’ + vt’)/(1 - v^2/c^2)^(1/2) - y = y’ - z = z’ - t = (t’ + (vx’/c^2))/(1 - v^2/c^2)^(1/2) - Length Contraction - l’ = l(1 – v^2/c^2)^(1/2) - Time Dilation - T = T0/(1 – v^2/c^2)^(1/2) - Einstein’s mass-energy relation - E = mc^2