Question

The star Delta goes supernova. One year later and 2.0 ly away, as measured by astronomers in the galaxy, star Epsilon explodes. Let the explosion of Delta be at $x_D$= 0 and $t_D$= 0. The explosions are observed by three spaceships cruising through the galaxy in the direction from Delta to Epsilon at velocities

$v_1$= 0.30c, $v_2$= 0.50c, and $v_3$= 0.70c. All three spaceships, each at the origin of its reference frame, happen to pass Delta as it explodes.

a. What are the times of the two explosions as measured by scientists on each of the three spaceships?

b. Does one spaceship find that the explosions are simultaneous? If so, which one?

c. Does one spaceship find that Epsilon explodes before Delta? If so, which one?

d. Do your answers to parts b and c violate the idea of causality? Explain

Short answer

(a) $t_E\text{'}=0.42y,t_E\text{'}\text{'}=0,t_E\text{'}\text{'}\text{'}=0.56y.$

(b) Second spaceship finds explosions are simultaneous.

(c) Third spaceship finds the explosion before delta.

(d) No, explosion are far enough.

Step-by-step solution

Given information

We have given that, the star Delta goes supernova. One year later and 2.0 ly away, as measured by astronomers in the galaxy, star Epsilon explodes. Let the explosion of Delta be at $x_D$= 0 and $t_D$= 0. The explosions are observed by three spaceships cruising through the galaxy in the direction from Delta to Epsilon at velocities $v_1$= 0.30c, $v_2$= 0.50c, and$v_3$ = 0.70c. All three spaceships, each at the origin of its reference frame, happen to pass Delta as it explodes.

Part(a) Step 1.

Let S be the earth’s frame of reference frame, and S′, S″, and S″′ be the frame of reference of the three spaceships cruising through the galaxy in the direction from Delta to Epsilon at velocities $v_1$= 0.3c , $v_{2=}$0.5c , and

$v_3$ = 0.7 relative to the earth’s frame.

Step 2.

In frame S, $x_D$= 0 ly, $t_D$= 0 y, $x_E$= 2 ly, and $t_E$= = 1 ly. In moving reference frames, role="math" localid="1653208902765" $t{\text{'}}_D=t\text{'}{\text{'}}_D=t\text{'}\text{'}{\text{'}}_D$= 0 y.

We will use the Lorentz transformation to calculate the time at which Epsilon explodes.

In frame S′,

role="math" localid="1653215755007" $t{\text{'}}_E=\frac{t_E-{\displaystyle \raisebox{1ex}{$x_E{v}_1$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}{\sqrt{1-{\displaystyle \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}$

$t{\text{'}}_E=\frac{1y-(2ly)(0.3c)/c^2}{\sqrt{1-{(0.3)}^2}}$

$t{\text{'}}_E=0.42y$

For $t\text{'}\text{'}$:

$t_E\text{'}\text{'}=\frac{t_E-x_E{v}_2/c^2}{\sqrt{1-{\displaystyle \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}$

$t_E\text{'}\text{'}=\frac{1y-(2ly)(0.5)/c^2}{\sqrt{1-{(0.5)}^2}}$

$t_E\text{'}\text{'}=0y$

For $t_E\text{'}\text{'}\text{'}$:

$t_E\text{'}\text{'}\text{'}=\frac{t_E-x_E{v}_1/c^2}{\sqrt{1-{\displaystyle \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}$

$t_E\text{'}\text{'}\text{'}=\frac{1y-(2ly)(0.7c)/c^2}{\sqrt{1-{(0.7)}^2}}$

role="math" localid="1653225708620" $t_E\text{'}\text{'}\text{'}=-0.56y$

Part(b) Step 1 : 

The time at which Epsilon explosion observed by the scientist on the second spaceships $t_E\text{'}\text{'}=0$y.

Step 2.

So, the spaceship 2 finds that the explosions are simultaneous.

Part(c) Step 1.

The time at which Epsilon explosion observed by the scientist on the third spaceship is$t_E\text{'}\text{'}\text{'}=-0.56$.

Step 2.

So, the spaceship 3 finds that Epsilon explodes before Delta.

Part(d) Step 1.

Our answers to parts b and c do not violate the idea of causality.

Step 2.

The explosions are far enough that Delta can have no causal influence on Epsilon. Thus there’s no difficulty if Epsilon explodes before Delta in some reference frames.