Special Relativity: Super-Novae light on earth

[Destroxia]

Homework Statement

A nova is the sudden, brief brightening of a star. Suppose Earth astronomers see two novas occur simultaneously, one in the constellation Lyra. Both nova are the same distance from Earth, $2.5 \times 10^3 [cy]$, and are in exactly opposite direction from Earth. Observers on board an aircraft flying at $1000 [\frac {km} {h}]$ on a line from Orion toward Lyra see the same novas, but note they are not simulatneous.

(a) For the observers on the aircraft, how much time separates the novas?

(b) Which one occurs first? (Assume Earth is an inertial reference frame)

I know someone has asked this question on here before, but my issue is more of a conceptual one, than it is a calculations difficulty (Although, the conceptual argument is impeding my calculations).

Homework Equations

Givens:
$D_E = 2.5 \times 10^3 [cy]$
$v_A= 1000 [\frac {km} {h}]$

Equations:
$\gamma = \frac {1} {\sqrt{1- \frac {v^2} {c^2}}}$
$x' = \gamma (x-vt)$
$t' = y(t- \frac {vx} {c^2})$

The Attempt at a Solution

[/B]
I'm REALLY having a hard time tackling special relativity. From my understanding, from the observer on airships perspective, the airship is at rest, while the super-novae are shifting at 1000 [km/h]... since this velocity is so small $\gamma$ must go to 1? Then we are just left with the above equations, without the gamma.

So the first question, (a), I have the equation $t' = (t- \frac {vx} {c^2})$, I have to find some kind of way to replace the t value, as I can't use the equation without it. Am I understanding correctly that if we take time to be 0 at the event I can cancel that out? In which case I would be left with $t' = - \frac {vx} {c^2}$ . In order to evaluate this $t'$, I need the original x, not the x' ? Which is just the distance from one novae to the Earth times 2...

I'm also very confused as to whether, or not "how much time separates the novas?" means how much distance of time is between the two novae, or the time DIFFERENCE between the two events, it is incredibly unclear with that wording, but continuing with my thought process...

$t' = - \frac {vx} {c^2}$

I'm not sure if this is the correct expression for the time "separation", or not. Some critique on my thought process, and my equations would be much appreciated.

 

[jbriggs444]

You are asked "how much time separates the two novas". That is a question about the difference in the time coordinate for the one nova and the time coordinate for the other nova (in the airplane frame). You can calculate the time coordinate for both in the Earth frame.

 

[Chestermiller]

Homework Statement

A nova is the sudden, brief brightening of a star. Suppose Earth astronomers see two novas occur simultaneously, one in the constellation Lyra. Both nova are the same distance from Earth, $2.5 \times 10^3 [cy]$, and are in exactly opposite direction from Earth. Observers on board an aircraft flying at $1000 [\frac {km} {h}]$ on a line from Orion toward Lyra see the same novas, but note they are not simulatneous.

(a) For the observers on the aircraft, how much time separates the novas?

(b) Which one occurs first? (Assume Earth is an inertial reference frame)

I know someone has asked this question on here before, but my issue is more of a conceptual one, than it is a calculations difficulty (Although, the conceptual argument is impeding my calculations).

Homework Equations

Givens:
$D_E = 2.5 \times 10^3 [cy]$
$v_A= 1000 [\frac {km} {h}]$

Equations:
$\gamma = \frac {1} {\sqrt{1- \frac {v^2} {c^2}}}$
$x' = \gamma (x-vt)$
$t' = y(t- \frac {vx} {c^2})$

The Attempt at a Solution

[/B]
I'm REALLY having a hard time tackling special relativity. From my understanding, from the observer on airships perspective, the airship is at rest, while the super-novae are shifting at 1000 [km/h]... since this velocity is so small $\gamma$ must go to 1? Then we are just left with the above equations, without the gamma.

So the first question, (a), I have the equation $t' = (t- \frac {vx} {c^2})$, I have to find some kind of way to replace the t value, as I can't use the equation without it. Am I understanding correctly that if we take time to be 0 at the event I can cancel that out? In which case I would be left with $t' = - \frac {vx} {c^2}$ . In order to evaluate this $t'$, I need the original x, not the x' ? Which is just the distance from one novae to the Earth times 2...

I'm also very confused as to whether, or not "how much time separates the novas?" means how much distance of time is between the two novae, or the time DIFFERENCE between the two events, it is incredibly unclear with that wording, but continuing with my thought process...

$t' = - \frac {vx} {c^2}$

I'm not sure if this is the correct expression for the time "separation", or not. Some critique on my thought process, and my equations would be much appreciated.

Most of your assessment is correct. All you need to do now is to write $\Delta t' = - \frac {v\Delta x} {c^2}$, since in the Earth frame of reference (neglecting the movement of the Earth itself, which certainly must be significant), $\Delta t=0$.