Physics, asked by sanjaybansal1141, 11 months ago

in which condition the expression of position under Lorenz transformation is equivalents to expression under Galileans transformation​

Answers

Answered by 00001919
0

In physics, the Lorentz transformations are a one-parameter family of linear transformations from a coordinate frame in spacetime to another frame that moves at a constant velocity (the parameter) relative to the former. The respective inverse transformation is then parametrized by the negative of this velocity. The transformations are named after the Dutch physicist Hendrik Lorentz.

The most common form of the transformation, parametrized by the real constant {\displaystyle v,}v, representing a velocity confined to the x-direction, is expressed as[1]

{\displaystyle {\begin{aligned}t'&=\gamma \left(t-{\frac {vx}{c^{2}}}\right)\\x'&=\gamma \left(x-vt\right)\\y'&=y\\z'&=z\end{aligned}}}{\displaystyle {\begin{aligned}t'&=\gamma \left(t-{\frac {vx}{c^{2}}}\right)\\x'&=\gamma \left(x-vt\right)\\y'&=y\\z'&=z\end{aligned}}}

where (t, x, y, z) and (t′, x′, y′, z′) are the coordinates of an event in two frames, where the primed frame is seen from the unprimed frame as moving with speed v along the x-axis, c is the speed of light, and {\displaystyle \gamma =\textstyle \left({\sqrt {1-{\frac {v^{2}}{c^{2}}}}}\right)^{-1}}{\displaystyle \gamma =\textstyle \left({\sqrt {1-{\frac {v^{2}}{c^{2}}}}}\right)^{-1}} is the Lorentz factor. When speed v is much smaller than c, the Lorentz factor is negligibly different from 1, but as v approaches c, {\displaystyle \gamma }\gamma  grows without bound. The value of v must be smaller than c for the transformation to make sense.

Expressing the speed as {\displaystyle \beta ={\frac {v}{c}},}{\displaystyle \beta ={\frac {v}{c}},} an equivalent form of the transformation is[2]

{\displaystyle {\begin{aligned}ct'&=\gamma \left(ct-\beta x\right)\\x'&=\gamma \left(x-\beta ct\right)\\y'&=y\\z'&=z.\end{aligned}}}{\displaystyle {\begin{aligned}ct'&=\gamma \left(ct-\beta x\right)\\x'&=\gamma \left(x-\beta ct\right)\\y'&=y\\z'&=z.\end{aligned}}}

may you understood

Answered by aakulaarjun2005
1

SORRY! I don't know this answer. Since i am school student, I don't know this mechanism

Similar questions