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\prime =ct\prime $ at time
$t\prime $ in the primed frame. Expressing these relations in Cartesian coordinates gives
This cannot be satisfied for nonzero relative velocity
v of the two frames if we assume the Galilean transformation results in
$t=t\prime $ with
$x=x\prime +vt\prime .$
To find the correct set of transformation equations, assume the two coordinate systems
S and
$S\prime $ in
[link] . First suppose that an event occurs at
$(x\prime ,0,0,t\prime )$ in
$S\prime $ and at
$(x,0,0,t)$ in
S , as depicted in the figure.
Suppose that at the instant that the origins of the coordinate systems in
S and
$S\prime $ 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\prime $ to be at
$x=vt.$ With the help of a friend in
S , the
$S\prime $ observer also measures the distance from the event to the origin of
$S\prime $ and finds it to be
$x\prime \sqrt{1-{v}^{2}\text{/}{c}^{2}}.$ This follows because we have already shown the postulates of relativity to imply length contraction. Thus the position of the event in
S is
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
$S\prime .$ Simply interchanging the primed and unprimed variables and substituting gives:
Spacecraft
$S\prime $ is on its way to Alpha Centauri when Spacecraft
S passes it at relative speed
c /2. The captain of
$S\prime $ 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
Identify the known:
$\text{\Delta}t\prime ={t}_{2}\prime -{t}_{1}\prime =1.2\phantom{\rule{0.2em}{0ex}}\text{s};\phantom{\rule{0.2em}{0ex}}\text{\Delta}x\prime =x{\prime}_{2}-x{\prime}_{1}=0.$
Identify the unknown:
$\text{\Delta}t={t}_{2}-{t}_{1}.$
Express the answer as an equation. The time signal starts as
$\left(x\prime ,{t}_{1}\prime \right)$ and stops at
$\left(x\prime ,{t}_{2}\prime \right).$ Note that the
$x\prime $ coordinate of both events is the same because the clock is at rest in
$S\prime .$ Write the first Lorentz transformation equation in terms of
$\text{\Delta}t={t}_{2}-{t}_{1},$$\text{\Delta}x={x}_{2}-{x}_{1},$ and similarly for the primed coordinates, as:
unpolarized light refers to a wave collection which has an equal distribution of electric field orientations for all directions
pro
In a grating, the angle of diffraction for second order maximum is 30°.When light of wavelength 5*10^-10cm is used. Calculate the number of lines per cm of the grating.
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