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The lorentz transformation equations

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 = c t at time t in the unprimed frame, and also has radius r = c t at time t in the primed frame. Expressing these relations in Cartesian coordinates gives

x 2 + y 2 + z 2 c 2 t 2 = 0 x 2 + y 2 + z 2 c 2 t 2 = 0 .

The left-hand sides of the two expressions can be set equal because both are zero. Because y = y and z = z , we obtain

x 2 c 2 t 2 = x 2 c 2 t 2 .

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 + v t .

To find the correct set of transformation equations, assume the two coordinate systems S and S in [link] . First suppose that an event occurs at ( x , 0 , 0 , t ) in S and at ( x , 0 , 0 , t ) in S , as depicted in the figure.

The axes of frames S and S prime are shown. S has axes x, y, and z. S prime is moving to the right with velocity v and has axes x prime, y prime and z prime. S and S prime are aligned along the horizontal x and x prime axes and are separated by a distance v t. An event on the horizontal x and x prime axes is indicated by a point which is a distance x from the y z plane of the S frame and a distance x prime from the y prime, z prime plane of the S prime frame.
An event occurs at ( x , 0, 0, t ) in S and at ( x , 0 , 0 , t ) in S . The Lorentz transformation equations relate events in the two systems.

Suppose that at the instant that the origins of the coordinate systems in S and S coincide, a flash bulb emits a spherically spreading pulse of light starting from the origin. At time t , an observer in S finds the origin of S to be at x = v t . With the help of a friend in S , the S observer also measures the distance from the event to the origin of S and finds it to be x 1 v 2 / c 2 . This follows because we have already shown the postulates of relativity to imply length contraction. Thus the position of the event in S is

x = v t + x 1 v 2 / c 2

and

x = x v t 1 v 2 / c 2 .

The postulates of relativity imply that the equation relating distance and time of the spherical wave front:

x 2 + y 2 + z 2 c 2 t 2 = 0

must apply both in terms of primed and unprimed coordinates, which was shown above to lead to [link] :

x 2 c 2 t 2 = x 2 c 2 t 2 .

We combine this with the equation relating x and x to obtain the relation between t and t :

t = t v x / c 2 1 v 2 / c 2 .

The equations relating the time and position of the events as seen in S are then

t = t + v x / c 2 1 v 2 / c 2 x = x + v t 1 v 2 / c 2 y = y z = z .

This set of equations, relating the position and time in the two inertial frames, is known as the Lorentz transformation    . They are named in honor 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 correct theoretical basis is Einstein’s special theory of relativity.

The reverse transformation expresses the variables in S in terms of those in S . Simply interchanging the primed and unprimed variables and substituting gives:

t = t v x / c 2 1 v 2 / c 2 x = x v t 1 v 2 / c 2 y = y z = z .

Using the lorentz transformation for time

Spacecraft S is on its way to Alpha Centauri when Spacecraft S passes it at relative speed c /2. The captain of S sends a radio signal that lasts 1.2 s according to that ship’s clock. Use the Lorentz transformation to find the time interval of the signal measured by the communications officer of spaceship S .

Solution

  1. Identify the known: Δ t = t 2 t 1 = 1.2 s ; Δ x = x 2 x 1 = 0 .
  2. Identify the unknown: Δ t = t 2 t 1 .
  3. Express the answer as an equation. The time signal starts as ( x , t 1 ) and stops at ( x , t 2 ) . Note that the x coordinate of both events is the same because the clock is at rest in S . Write the first Lorentz transformation equation in terms of Δ t = t 2 t 1 , Δ x = x 2 x 1 , and similarly for the primed coordinates, as:
    Δ t = Δ t + v Δ x / c 2 1 v 2 c 2 .

    Because the position of the clock in S is fixed, Δ x = 0 , and the time interval Δ t becomes:
    Δ t = Δ t 1 v 2 c 2 .
  4. Do the calculation.
    With Δ t = 1.2 s this gives:
    Δ t = 1.2 s 1 ( 1 2 ) 2 = 1.6 s.

    Note that the Lorentz transformation reproduces the time dilation equation.
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Questions & Answers

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Gatkuoth Reply
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SALONI
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Gatkuoth
torque=rFSintheta
SALONI
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RRGN
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ashwani
Torque is moment of Force if a force is applied at some finite distance from COM of a body it produce rotation. For pure rotation we need to apply Couple not Torque...
Researchers
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the speed of light in a
amanuel Reply
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All matter is composed of two sets of three dimensions. The first set (1,2,3) decay with a positive charge. The second set (4,5,6) decay with a negative charge. As they decay, they create space (7 8,9) dimensions.
John
Two sets of (1,2,3,4,5,6) dimensions create a proton, a neutron, and an electron. This is the primordial atom.
John
A 10kg mass lift to a height of 24m and release. what is the total energy of the system
ADEPOJU Reply
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ADEPOJU
E=Mgh=10*10*24=2400J
Adamu
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Natanim Reply
Atoms are single neutral particles. Molecules are neutral particles made of two or more atoms bonded together.
Manfred
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A body quadruples its momentum when its speed doubles.What was the initial speed in units of c.i.e..what was u/c ?
Lekshmi Reply
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a thermodynamic quantity equivalent to the total heat content of a system
RAMLA
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bloch
What is the meaning of Nuclear Fission?
Benita Reply
nuclear fission. are atom split apart which releases energy
Komolafe
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Peacekamei Reply
عند قذف جسم إلى أعلى بسرعة إبتدائية فإنه سيصل إلى ارتفاع معين (أقصى ارتفاع) ثم يعود هابطاً نحو سطح الأرض .   إذا قُذِفَ جسم إلى أعلى ووجد أن سرعته 18 م / ث عندما قطع 1/4 المسافة التي تمثل أقصى ارتفاع سيصله فالمطلوب إيجاد السرعة التي قُذِف بها بالمتر / ث . إن هذه السرعة هي واحدة من الإجابات التالية
Aml Reply
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Ayebanifesunday Reply
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kenneth
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John
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John
This creation of new space is "Dark Energy".
John
The first two sets of three dimensions, 1 through 6, are "Dark Matter".
John
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John
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John
There are three sets of three neutrons, 9.
John
A free neutron decays into a proton, an electron, and a neutrino.
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There are three sets of five neutrinoes, 15.
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Neutrinoes are two dimensional.
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A positron is composed of the first three dimensions.
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An electron is composed of the second three dimensions.
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Vinod
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SALONI
What is photoelectric
Hsssan Reply
light energy (photons) through semiconduction of N-P junction into electrical via excitation of silicon purified and cristalized into wafers with partially contaminated silicon to allow this N-P function to operate.
Michael
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Photoelectric emission is the emission of electrons on a metal surface due to incident rays reflected on it
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If you lie on a beach looking at the water with your head tipped slightly sideways, your polarized sunglasses do not work very well.Why not?
Rakhi Reply
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Jallal
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Practice Key Terms 4

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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