Question

prove that x2+y2+z2=c2t2 is invariant under lorentz transformation

Answer 1

Answer:

The Lorentz transformation equations are, x'= x-vt/√1-v2/c2 y'= y ; z'= z. Substituting these values of x' y' z' and t' in R.H.S of eq. ... Hence x2+y2+z2-c2t2 is invariant under Lorentz transformation.

Explanation:

Answer 2

Concept Introduction:

The relationship between two distinct coordinate frames that are moving relative to one another at a constant speed is known as a Lorentz transformation. Dutch scientist Hendrik Lorentz is credited with coining the term of the transformation. There are two frames of reference: Inertial Frames, which relate to motion that has a constant speed.

Given:

The Lorentz transformation equations are, x'= x-vt/√$1$-v$2$/c$2$ y'= y ; z'= z.

t'= t-vx/c$2$/√1-v$2$/c$2$

To Find:

We have to prove that  x$2$+y$2$+z$2$-c$2$t$2$  is invariant under lorentz transformation

Solution:

According to the problem,

Substituting these values of x' y' z' and t' in R.H.S of eq.[1],

We have,

x'$2$+y'$2$+z'$2$-c$2$t'$2$ = (x-vt) $2$/($1$-v$2$/c$2$) + y$2$+z$2$-c$2$ [t-(vx/c$2$)]$1$/-v$2$/c$2$,

= $1$/($1$-v$2$/c$2$)[x$2$+v$2$t$2$-$2$xvt-c$2$ (t$2$+x$2$v$2$/c$4$-$2$t xv/c$2$)]+y$2$+z$2$,

= $1$/($1$-v$2$/c$2$)[x$2$ ($1$-v$2$/c$2$)-c$2$t$2$ ($1$-v$2$/c$2$)+y$2$+z$2$,

= (x$2$-c$2$t$2$)+y$2$+z$2$ = L.H.S of eq. [$1$].

Hence x$2$+y$2$+z$2$-c$2$t$2$ is invariant under Lorentz transformation.

Final Answer:

x$2$+y$2$+z$2$-c$2$t$2$  is invariant under Lorentz transformation.

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