Question

Suppose an event is measured to be at a = (0,-2, 3, 5) in one reference frame. Find the components of this event in another reference frame that is moving with a speed of 0.85 relative to the original frame, in the positive x-direction

Answer 1

Answer:

(2.3529, -3.7966, 3,5)

Explanation:

We can do this simply by taking the Lorentz transformations:

Taking the coordinates

$t \\x \\y\\\z$

in one reference frame, and taking the second reference frame moving at speed v in the x-direction of the original frame, the Lorentz transformations are:

$ct' = \frac{1}{\sqrt{ 1 - \frac{v^2}{c^2}}}  ( ct - \frac{v}{c} x )$

$x' = \frac{1}{\sqrt{ 1 - \frac{v^2}{c^2}}}  ( x' - \frac{v}{c} ct )$

$y' = y$

$z' = z$

For our problem

$t= 0\\x=-2\\y=3\\z=5\\v=0.85 c$

So, the transformation will give

$ct' = \frac{1}{ \sqrt{ 1 - 0.85^2 }}  ( 0 - 0.85 * (-2) )$

$ct' =  \frac{1.7}{0.5267}$

$ct' =  2.3529$

$x' = \frac{1}{\sqrt{ 1 - 0.85^2}} ( (-2) - 0.85 * 0 )$

$x' =  \frac{- 2}{0.5267}$

$x' =  -3.7966$

$y' = 3$

$z' = 5$

Answer 2

Answer:

The coordinates of the same event in this moving frame are a' = (3.226, -3.7966, 3, 5).

Explanation:

Given:

• The event in the first reference frame is a = (0,-2, 3, 5).

• The speed of the second reference frame with the respect to the first reference frame = 0.85.

The coordinates of the event are given as (t, x, y, z)

Therefore, for the first frame, the event a has coordinates:

t = 0

x = -2

y = 3

z = 5.

The coordinates of the same event in the moving frame as given by the Lorentz transformation as

$\rm t'=\gamma(t-\dfrac{vx}{c^2})\\ x'=\gamma (x-vt).\\y'=y.\\z'=z.$

where,

$\rm \gamma = \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}$

In natural system of units, c = 1.

Therefore,

$\rm \gamma = \dfrac 1{\sqrt{1-v^2}}=\dfrac{1}{\sqrt{1-0.85^2}}=1.898.$

$\rm t'=1.898(0-0.85\times(-2))=3.226\\ x'=1.898(-2-0\times 0.85)=-3.7966.\\y'=3.\\z'=5.$

The coordinates of the same event in this moving frame are a' = (3.226, -3.79, 3, 5).