Length contraction Length measured differs from frame to

Length contraction Length measured differs from frame to frame – another consequence of relativistic effect Gedankan experiment again! 1

Two observers: O on Earth, O’ traveling to and from Earth and alpha centauri with speed u Total distance between Earth - alpha centauri – Earth, according to O (Earth observer), = L 0 O sees O’ return to Earth after Dt 0 Observer O’ in a spaceship is heading AC with speed u and returns to Earth after Dt’ according to his clock 2

Use some simple logics… In O: 2 L 0 = u. Dt 0 In O’: 2 L 0’ = u. Dt 0’ Due to time dilation effect, Dt 0’ is shorter than Dt 0 , i. e. Dt 0 > Dt 0’ Dt 0 is related to Dt 0’ via a time dilation effect, Dt 0’ = Dt 0 /g , hence L 0’ / L 0 = Dt 0’ /Dt 0 = 1 / g , or 3

L 0’ = L 0 / g L 0 is defined as the proper length = length of object measured in the frame in which the object (in this case, the distance btw Earth and AC) is at rest L 0’ is the length measured in the O’ frame, which is moving wrp to the object The length of a moving objected is measured to be shorter than the proper length – length contraction 4

If an observer at rest wrp to an object measures its length to be L 0 , an observer moving with a relative speed u wrp to the object will find the object to be shorter than its rest length by a foctor 1 / g. 5

A stick moves to the right with a speed u. (a) The stick as viewed by a frame attached to it (b) The stick as seen by an observer in a frame at rest relative to the stick. The length measured in the rest frame is shorter than the proper length by a factor 1/ g 6

Length contraction only happens along the direction of motion In 3 -D, the length contraction effect is a shortening of length plus a rotation 7

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An observer on Earth sees a spaceship at an altitude of 435 moving downward toward the Earth with a speed of 0. 97 c. What is the altitude of the spaceship as measured by an observer in the spaceship? Solution One can consider the altitude seen by the stationary (Earth) observer as the proper length (say, L'). The observer in the spaceship should sees a contracted length, L, as compared to the proper length. Hence the moving observer in the ship finds the altitude to be L = L' / g = 435 m x [1 - (0. 97)2]-1/2 = 106 9 m

Relativistic kinematics 10

Lorentz Transformation All inertial frames are equivalent All physical processes analysed in one frame can also be analysed in other inertial frame and yield consistent results A transformation law is required to related the space and time coordinates from one frame to another 11

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O' frame uses {x', y', z‘; t‘} to denote the coordinates of an event, whereas O frame uses {x, y, z; t} How to related {x', y', z', t‘} to {x, y, z; t} In Newtonian mechanics, we use Galilean transformation But GT must not be valid when u c because it is not consistent with the constancy of the light speed postulate The relativistic version of the transformation law is given by Lorentz transformation 13

Derivation of Lorentz transformation 14

Consider a rocket moving with a speed u (O' frame) along the xx' direction wrp to the stationary O frame A light pulse is emitted at the instant t' = t =0 when the two origins of the two reference frames coincide the light signal travels as a spherical wave at a constant speed c in both frames After some times t, the origin of the wave centered at O has a radius r = ct, where r 2 = x 2 + y 2 + z 2 15

From the view point of O', after some times t' the origin of the wave, centered at O' has a radius r' = ct‘ , (r’ )'2 = (x)‘ 2 + (y)2 + (z)2 y'=y, z' = z (because the motion of O' is along the xx‘) axis – no change for y, z coordinates The transformation from x to x’ (and vice versa) must be linear, i. e. x’ x From the view point of O: x = ct corresponds to x ’ = 0, so we assume the form x’ = k(x - ct ); k some proportional constant; Likewise, from the view point of O’, x’ = -ct’ corresponds to x = 0, so we set x = k(x’ + ct’ ) 16

With r = ct , r’ = ct ’, x = k(x’ + ct’ ), x’ = k(x - ct), we solve for {x', t'} in terms of {x, t } to obtain 17

the constant k is identified as the Lorentz factor, g Note that, now, the length and time interval measured become dependent of the state of motion (in terms of g) – in contrast to Newton’s viewpoint Lorentz transformation reduces to Galilean transformation when u << c (show this yourself) 18

To recap the LT given in the previous analysis Which relates x’ , t’ to x, t’ , for which O’ is moving at velocity +u wrp to O 19

I see O’ moving with a velocity +u I measures the coordinates of M as {x, t} O +u Object M O’ 20

From the view point of O’ Equivalently, we could also perform the analysis from the view point of O’ that O is moving in the –u direction. 21

From the view point of O’ I measures the coordinates of M as {x’, t’} -u O I see O moving with a velocity -u Object M O’ We wish to express x, t in terms of x’, t’ 22

Mathematically, this simply means making the permutation: The two transformations above are equivalent; use which is appropriate in a given question 23

Length contraction can be recovered from the LT Consider the rest length of a ruler as measured in frame O’ is L’ = Dx’ = x’ 2 - x’ 1 (the proper length) measured at the same instant in that frame, hence t’ 2 = t’ 1 What is the length of the rule as measured by O? The length in O, according the LT is L’ = Dx’ = x’ 2 - x’ 1 = g [(x 2 - x 1) – u(t 2 -t 1)] The length of the ruler in O is simply the distance btw x 2 and x 1 measured at the same instant in that frame, hence t 2 = t 1, hence L’ = g L 24

How would you recover time dilation from the LT? DIY as an exercise during Raya 25

Lorentz velocity transformation M is moving with a velocity +ux from my point of view I see M moving with a velocity +ux’ O I see O’ moving with a velocity +u O Object M +u O’ How to relate the velocity in the O’ (u’x) frame to that of the O frame (ux)? 26

By definition, ux = dx/dt, u’x = dx’/dt’ The velocity in the O’ frame can be obtained by taking the differentials of the Lorentz transformation, 27

Combining where we have made used of the definition ux = dx/dt 28

Compare the Lorentz transformation of velocity with that of Galilean transformation of velocity: GT reduces to LT in the limit u << c 29

Please try to understand the definition of ux , u’x , u so that you wont get confused 30

LT is consistent with the constancy of speed of light in either O or O’ frame, the speed of light seen must be the same, c Say object M is moving with speed of light as seen by O, i. e. ux = c According to LT, the speed of M as seen by O’ is 31

That is, in either frame, both observers agree that the speed of light they measure is the same, c = 3 x 108 m/s In contrast, according to GT, the speed of light seen by O’ would be Which is inconsistent with constancy of speed of light postulate 32

To recap the LT given in the previous analysis relates u’x to ux in which O’ is moving with +u wrp to O, 33

From the view point of O’ Equivalently, we could also perform the analysis from the view point of O’ that O is moving in the –u direction. We would be able to express ux in terms of u’x in the same spirit as we derive u’x in terms of ux 34

From the view point of O’ I see M moving with a velocity +ux’ -u I see O moving with a velocity -u O Object M O’ I see M moving with a velocity ux 35

Mathematically, this simply means making the permutation: The two transformations above are equivalent; use which is appropriate in a given question 36

Example Rocket 1 is approaching rocket 2 on a head-on collision course. Each is moving at velocity 4 c/5 relative to an independent observer midway between the two. With what velocity does rocket 2 approaches rocket 1? C. f. In GT, their relative speed would just be 4 c/5 + 4 c/5 = 1. 6 c – which violates constancy of speed of light postulate. See how LT handle this situation: 37

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The observer in the middle is the stationary frame, O Choose rocket 1 as the moving frame O‘ Call the velocity of rocket 2 as seen from rocket 1 u’x. This is the quantity we are interested in Frame O' is moving in the +ve direction as seen in O, so u = +4 c/5 The velocity of rocket 2 as seen from O is in the -ve direction, so ux = - 4 c/5 Now, what is the velocity of rocket 2 as seen from frame O', u ’x = ? (intuitively, u ’x must be in the negative direction) 39

Use the LT i. e. the velocity of rocket 2 as seen from rocket 1 (the moving frame, O’) is – 40 c/41, which means that O’ sees rocket 2 moving in the –ve direction (to the left in the picture), as expected. 40

Read other examples from the texts and also the lecture notes 41

Selamat Hari Raya Aidilfitri 42