Given: Δt’ = t‘2−t‘1 = 1.2s Δx’ = x‘2−x‘1 = 0 The time signal starts and ends at (x’, t‘1) Using the first Lorentz transformation equation we get, Δt = t2 – t1 Δx = x2 – x1 Δt= As the time interval of S’ is fixed, we can consider Δx’ = 0 and the time interval δt becomes: Δt= Substituting Δt’ = 1.2s we get Δt= Δt = 1.6s |
Given Δt = 10-6 sec v = 5 x109 cm/sec ∆t= ∆t' = ∆t / ∆t' = |
A3)
Given parameters are, v = 0.75c c = Speed of light (3.0 x 108 m/s) The formula for length contraction is l0 = or L0 =
L = 10 × √1- (0.75)2 L = 10 × √(o.438) L = 6.618 m. |
The formula for length contraction is l0 = or L0 = L = 5.5 m v = .95c L0 = ? 5.5 = L0×(1- 0.95c2/c2) 5.5 = L0×(0.312) L0 = 5.5/(0.312) = 17.6 m |
Given Δτ=1s v=5830m/s.
Δt =?
We know ∆t= t1 – t2 = Δt=
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Now let us try to find the coordinates of the event (x, y, z, t) in S′ It’s easy to see t′ = t We synchronized the clocks when O′ passed O. Also, evidently, y′ = y and z′ = z, from the figure. We can also see that x = x′ +vt. Thus (x, y, z, t) in S corresponds to (x′, y′, z′, t′) in S′, where x = x′ +vt. y = y’ z = z’ t =t’ That’s how positions transform; these are known as the Galilean Transformations.
The velocity in S′ in the x′ direction is given by u’x
This is obvious anyway it’s just the addition of velocities formula. ux = u’x +v
Since v is constant. That is to say, a’x =ax The acceleration is the same in both frames. This again is obvious as the acceleration is the rate of change of velocity, and the velocities of the same particle measured in the two frames differ by a constant factor-the relative velocity of the two frames. |
The experiment was devised to determine the absolute velocity of earth with respect to ether frame. The simplified plan of the experiment is shown in figure.
Figure: Michelson Morley Experiment
Light from a source S falls on a glass plate G1 inclined at an angle of 45° to the incident beam. The back surface of the plate G1 is half silvered. This plate divides the incident beam into two beams 1 and 2 of equal intensity. The beam 1 (transmitted by the plate G1) goes straight to another glass plate G2 of the same thickness as G1 and inclined at an angle of 45° to the beam. This beam after crossing G2 falls normally on the plate mirror M1 and is reflected back towards glass plate G1. On reaching the back surface of G1 it again suffers 50% reflection. The reflected part enters the telescope T. The beam 2 (reflected by the glass plate G1) moves towards the mirror M2, to fall normally on it. It retraces its path after reflection from M2. When it reaches the back surface of G1, it suffers 50% transmission. The transmitted part goes into the telescope and interferes with the reflected part of the beam 1. Interference fringes are therefore observed in the telescope. Since the two glass plates G1 and G2 are identical then the path of the two beams are equal in glass. Each beam passes three times through a glass plate in its journey from the source to the telescope. If the mirrors M1 and M2 are at the same distance from 0, then the optical path lengths of two beams shall be equal say l.
We know that the earth moves around the sun with an orbital velocity of Since the earth is moving through ether with velocity v it experiences an ether wind of velocity
When the beam 1 transmitted by G1 travels along the positive x direction from 0 to M1, its effective velocity v1 in earth’s frame is v1 = c - v because the ether wind is opposed to the direction of motion of the beam. The time taken by the light to travel from 0 to M1 is t1 = In the return journey from M1 to 0 the ether wind is supporting the beam of light. Therefore its effective velocity v2 in earth's frame is v2 = c + v The time taken by the light for its return journey M1O is t2 = Thus the time of the round trip OM1O is T1 = t1 + t2 T1 = T1 = 2lc ( Since We can expand the above expression by Binomial theorem. This gives T1 =
The beam 2 travels from O to the mirror M2 parallel to y direction. Since it to moving parallel to y direction its effective velocity must be in the y direction. Suppose
The square of magnitude of c2 = u2 + v2 – 2 u v cos 90° c2 = u2 + v2 or u2 = c2 - v2 u = The effective velocity u of the beam is the same during its return journey from M2 to O. Thus time taken to travel from M2 to O is t2' = t1' = Thus the time of round trip for the beam 2 is T2 = 2t1' = T2 = 2l (c2 – v2)-1/2 = Since T2 =
The time difference between the two beams for their round trips is T = T1 – T2 T= T= Thus the optical path difference
Now suppose this whole apparatus is turned through 90° so that the two beams interchange their paths. Thus path OM2, becomes parallel to the incident beam and OM1 becomes perpendicular to the incident beam. Then the two times T1 and T2 will interchange and the path difference between two beams changes from
The change in the optical path difference between the two beams due to 90° rotation of the apparatus
This path difference should lead to the shift of fringes.
Suppose λ is the wavelength of light used. We know that a path difference of one wavelength λ corresponds to shift of one fringe across a cross wire in the field of view of telescope. If
In Michelson Morely experiment the distance l= 11 meters, λ = 5900 Å = 5.9 x 10-7 m v/c = 10-4. Thus
Thus a fringe shift was expected in this experiment if the concept of the ether frame was valid. In actual experiment no fringe shift was observed.
The negative result of Michelson Morley experiment led to the following conclusion. The motion of earth through the ether could not to be detected experimentally. Thus the concept of a preferred ether frame must be discarded. The speed of light in vacuum is same in all frames of reference which are in uniform relative motion. This experiment paved the way for the development of the special theory of relativity by Einstein in the year 1905. |
Lorentz transformation is the relationship between two different coordinate frames that move at a constant velocity and are relative to each other. The name of the transformation comes from a Dutch physicist Hendrik Lorentz. The Galilean transformation nevertheless violates Einstein’s postulates, because the velocity equations state that a pulse of light moving with speed c along the x-axis would travel at speed c−v in the other inertial frame. Specifically, the spherical pulse has radius r=ct at time t in the unprimed frame, and also has radius r′=ct′ at time t' in the primed frame. Expressing these relations in Cartesian coordinates gives x2+y2+z2−c2t2 =0 ………..(1) x′2+y′2+z′2−c2t′2=0 ………..(2) The left-hand sides Equations and can be set equal because both are zero. Because y=y′ and z=z′ so, we obtain x2−c2t2 =x′2−c2t′2 ………..(3) However this cannot be satisfied for nonzero relative velocity v of the two frames if we assume the Galilean transformation results in t=t′ with x=x′+vt′. To find the correct set of transformation equations, assume the two coordinate systems S and S′ The laws of mechanics are invariant under Galilean transformations, whereas electrodynamics and Maxwell’s equations are varying under this transformation. This shows that the velocity of light will have different values for different observers moving with different uniform velocities. But the speed of light is invariant in all inertial frames of reference; hence, Galilean transformations need to be modified. Lorentz deduced the transformation equations, which are in agreement with the results of Michelson–Morley experiment. The Lorentz transformations are deduced as given below. Lorentz transformation equations are developed. Let x, y, z and t be the space and time coordinate in rest system or standard system. The system with x', y', z' moves with velocity 'v' along the x-axis. The origins at t=t'=0 coincide. The most general transformation equations relating the coordinates of an event in two systems can be written as, x’ = Ax +Bt ………..(4) y’ = y ………..(5) z’ = z ………..(6) t’ = Cx +Dt ………..(7)
If they are not linear one system would predict acceleration while in other system velocity was even constant. Because of symmetry, y' and z' axes are left unchanged by the transformation. Constant can be calculated by using the following boundary conditions.
x = vt x’ = 0 Substituting values of x and x' in equation (4) We get 0 = Avt +Bt i.e. B =- Av ………..(8)
x' = vt x = 0 Substituting values of x and x' in equation (4) we get -vt’ =Bt ………..(9) Substituting the value of B from equation (8), we get -vt’ =- Avt ………..(10) In equation (7) put x=0 we get t’ = Dt So equation (10) become -v Dt =- Avt A =D ………..(11)
x= ct x’ = ct’ Substituting values of x and x' in equation (4), we get ct’ = Act +Bt Substituting the value of B from equation (8), we get ct’ = Act +(- Avt) ………..(12) Substituting values of x and D from eq no (11) in eq no.(7) , we get t’ = Cct +At Substituting above value of t' in eq no.(12) , we get c(Cct +At) = Act - Avt or Cc2t +Act =Act - Avt Cc2t =Avt Or C =-A
y= ct, x=0 r’=ct’ where using equation (5) r’2 = x’2+y’2 using equation (5) y’= y= ct and from equation (4) and (7) x’ = Bt = -Avt t’= Dt =-At Therefore r’2 = x’2+y’2 become c2t’2 = (-Avt)2 +(ct)2 Substituting value of t' above equation became c2A2t2 = A2v2t2 +c2t2 (c2 - v2) A2t2 = c2t2 A2 = A = A = Substituting the Values of A, B, C, D in equation (1) and (4) So the Lorentz transformation equations would be, x’= x’= x’= Where And t’ = - t’= Lorentz transformation equations therefore are x’= y’ =y z’ = z t’= Or these can be written as x’= y’ =y z’ = z t’= Where This set of equations, relating the position and time in the two inertial frames, is known as the Lorentz transformation. They are named in honour of H.A. Lorentz (1853–1928), who first proposed them. Interestingly, he justified the transformation on what was eventually discovered to be a fallacious hypothesis. The reverse transformation expresses by the following equations. Simply interchanging the primed and unprimed variables and substituting gives: x= y =y’ z = z’ t= |
Consider a frame S' moving with respect to S with a uniform speed v along x direction figure. Let the rod AB lie along the O'x' axis. The rod is at rest in the frame S' and its length as measured by an observer in S' shall be called the rest length of the rod. To calculate this length the observer in S' will measure the x' coordinates of the two ends A and B. These coordinates shall be independent of the time of observation. If x1' and x2' are the x' coordinates of two ends A and B then the rest length l0 of the rod is l0 = x'2 – x'1 In the coordinate system S the x coordinates of ends A and B will change with time. In order to measure the length of the rod, the observer in S will have to measure simultaneously the coordinates of the ends A and B. If x1 and x2 are the coordinates of ends A and B at the time of observation t then the length of rod as observed in frame S is l= x2 – x1
Figure: Length Contraction From Lorentz equations we have x'2 = x1' = Subtracting these equations, we obtain x2' - x1' = Using eq. (6.57) and (6.58) it becomes l0 = I = Since l < l0
Thus the length of the rod measured from the frame in which it is moving is smaller than its length in a frame in which it is at rest by a factor Note the following points
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Consider a frame S' moving with velocity v with respect to frame S along the x direction Fig (6.6). Suppose the clock is located in the system S' at the point x'. The clock is stationary with respect to S'. Let t1 ' and t2' are two instants of time recorded on this clock by an observer in S'. Then the interval of time as observed in S' is ∆t' = t2' – t1'
Figure: Time dilation Now we wish to find how this time interval appears to an observer in S. This can be done by applying a Lorentz transformation from S' to S with the same value of x'. The transformation equation required is (6.51 d). From this we can write t1 = t2 = Subtracting eq. (6.64) from eq. (6.63) we obtain ∆t= t1 – t2 = Using eq. (6.62) it becomes ∆t= t1 – t2 = Since ∆t > ∆t' This means that the time interval as observed in the moving frame of clock is greater than the One observed in the rest frame of the clock. We can express the same thing by saying that a moving clock appears to run slow. We often say that time has been dilated or lengthened in the moving observer frame. The time as observed by an observer moving with the clock is less than the one observed by a moving observer. We highlight the following points
A stationary clock measures a longer time interval between events occurring in a moving frame of reference than does a clock in the moving frame. |
Let S and S’ be two frames of reference such that S’ is moving with respect to S with a uniform speed v in the positive x direction Figure. Consider a particle P which is moving in the frame S. Suppose its position coordinates at time t are (x,y,z).
Figure
Then its velocity components in the frame S are ux = We wish to find how do these velocity components appear to the observer in the frame S’. If (x’,y’,z’) are the instantaneous position coordinates of the particle P at time t’, then its velocity components in frame S’ are u’x = Consider the Lorentz transformation equations (2) i.e. x’= y’ =y z’ = z t’= Where Take the differentials of both sides of above equations. This gives dx’= dy’ =dy ………….(4) dz’ = dz ………….(5) dt’= Divide equation (3) by (6) u’x = u’x = Using equation (1) it becomes u’x = Now divide equation (4) by (6) u’y = = = Using equation (1) it becomes u’y = Now divide equation (4) by (6) u’z = = = Using equation (1) it becomes u’z = Collecting together equations for u’x ,u’y ,u’z we have u’x = u’y = u’z = These equations allow us to observe the observed velocity in frame S and S’ which is in uniform translation motion. To obtain inverse transformation from S and S’ change dashed quantities to undashed one and vice versa and change v to –v in equation (7), (8) and (9) we get ux = uy = uz = |
To explain the variation of mass with velocity, consider two frames of references S and S'. S' is moving with constant velocity v along X-direction. Let us consider two exactly similar balls A and B each of mass m. Both these balls A and B are moving in opposite direction along X-axis with equal speed u' in frame S'. After collision they merge into one body. Applying the law of conservation of momentum on the collision of the balls in frame S', we have Momentum of balls before collision = Momentum of balls after collision
After collision, the merged mass must be at rest in frame S'. Hence, it moves with velocity v in frame S. Let u1, u2 be the velocities and m1, m2 be the masses of balls A and B, respectively, in frame S. Using the law of addition of velocities, the above velocities can be written as
Applying the law of conservation of momentum on the collision of the balls in frame s, we have m1u1 + m2u2 = (m1 + m2)v ………(3) Substituting u1 and u2 values from Equations (1) and (2), we have
The above equation makes a relationship between the masses of balls in frame S and their velocities in frame S'. Now, to obtain relation between masses of balls and their velocities in frame S, we proceed as follows. Squaring Equation (1)
And using the above equation, the value
Therefore,
Dividing Equation (6) by Equation (5) and taking square root throughout, we have
Now from equation (4) and (7) we get
Suppose, m2 is at rest in frame S, then u2 = 0 and m2 = m0 (say) where m0 is the rest mass of the ball B, then Equation (8) becomes
As both the balls are similar, hence the rest masses of both balls are the same, so we can write the rest mass of m2 is equal to rest mass of m1, that is equal to mo. Then, Equation (9) becomes
Here, m1 is the mass of ball A when it is moving with velocity u1 in frame S. After collision, the merged mass containing mass of ball A moves with velocity v in frame S. In general, if we take the mass of ball A as m, when it is moving with velocity v in frame S, then
Where m0 is the rest mass of the body and m is the effective mass. Equation (11) is the relativistic formula for the variation of mass with velocity. Here, we see some special cases:
When the velocity of the body, v is very small compared to velocity of height, c and then m = m*0
If the velocity of the body v is comparable to the velocity of light c, then The mass of a moving body appears greater than its rest mass.
Suppose the velocity of a body is equal to velocity of light c, then it possess infinite mass. The effective mass of particles has been experimentally verified by using particle accelerators in case of electrons and protons by increasing their velocities very close to velocity of light. |
Derivation of mass energy equation We know that according to classical mechanics, the rate of change of momentum is proportional to force acting upon the body in the direction of motion. F = According to the theory of relativity, mass and velocity are variable so equation (1) after become.
The force F acting on the body makes a displacement dx. We know that work done dw is given by product of force and displacement And also the work is stored as kinetic energy (dε) in the body. So work done can be written as dw = dε = Fdx ...........(3) Substituting force F from Equation (2) in Equation (3). We get
dε = mvdv + v2dm ...........(4) and also velocity v = The relativistic mass m is given by
Squaring on both sides, we get m2 =
Differentiating Equation (5), we get 2mc2dm − 2mv2dm − 2m2vdv = 0 c2dm = v2dm + mvdv …........ (6) Comparing Equations (4) and (6), we get dε = c2dm ............(7) By applying force, the mass of a body increases from m0 to m and hence the kinetic energy. The total energy acquired by the body is obtained by integrating Equation (7) ε = Therefore, increase in kinetic energy is due to the increase in mass of the body. The total energy (E) of the body is the sum of increase in kinetic energy of the body due to motion and the energy at rest (moc2): ∴ E = ε + moc2 Using Equation (8) E = mc2 − moc2 + moc2 = mc2 E = mc2 …………. (9) This shows that increase in energy equivalent to increase in mass. In other words anything that increase energy will increase its relativistic mass. The above equation shows the equivalence between mass and energy also known as Einstein’s mass–energy relation. |
Let us try to find the momentum of massless particle or photon As discussed rest mass of photon is zero. i.e. mo = 0 We know that the relation of energy and momentum (done in previous section) is given by E2 = p2c2 - Substitute the mass of a particle mo = 0 in above equation (1) we have E = pc …………(2) It is just possible that we could allow m0=0 provided the particle always travels at the speed of light c. In this case above equations will not serve to define and so that for massless particle given by E=Pc. Also If ν is the frequency of photon, then its energy is given by E = hν …………(3) Equating equation (2) and (3) pc = hν p = Momentum of massless particle or photon is given by above equation.
19 Question: How relativistic energy and momentum are related to each other? Solution: If m is the relativistic mass of a given particle moving with velocity v then its energy is given by E = mc2 ………. (1) Substitute the value of mass m= E = Squaring both sides we get E2 = The magnitude of momentum of a particle is given by p =mv = Squaring both sides we get p2= = Multiply both sides by c2 equation (4) become p2c2 = subtract equation (5) from equation (3) we get E2 - p2c2 = E2 - p2c2 = E2 - p2c2 = E2 = p2c2 - This is energy momentum relation for a free particle. |
