Special Relativity Time Dilation/Length Contraction Problem

[Yosty22]

Homework Statement

You are exploring a newly discovered planet when a colleague contracts a deadly alien virus. Left untreated, it will probably kill him in 5 days, but the nearest medical station is 10 light-days away! How can you get him there in time?

Homework Equations

Δt_p=L/v
L=L'/\gamma
\gamma = 1/sqrt(1-v^2/c^2)

The Attempt at a Solution

Intuitively, I read this question and concluded that you could not do it. How could you possibly go 10 light-days in only 5 days? However, I tried to work through it anyway. Given the equation for Δt_p (proper time), I substituted some other values. The L term represents the contracted length, so I replaced L with L'/\gamma into the equation Δt_p=L/v. This gave me Δt_p=[(L'(1-v^2/c^2)^.5)]/v. Doing quite a bit of algebra, I ended up solving for v (the velocity of the ship) and got v=(L'^2/Δt_p^2)^.5.

I converted 5 days into seconds (so that I could work in meters/second) and got it to be 432,000 seconds. I said that this should be my Δt_p term because this is the time measured on the ship, so you need it to be 5 days to save the other astronaut. I then converted 10 light days into meters (again, to keep things simpler for me). I found 10 light days to be 2.592E14m. Plugging this into my previously derived equation, I got the speed necessary for the ship to reach the medical station (including time dilation and length contraction) to be 2c.

Is this supposed to happen, or should there be a true answer that would allow me to get the other passenger to travel 10 light days in a 5 day span (in his frame of reference)?

Any help would be much appreciated.
Thank you.

 

[collinsmark]

Hello Yosty22,

You actually have things set up correctly at some point in your work below, with the possible exception of what gets primed and what does not.

but the final, correct answer is not a velocity 2c. (That's not even possible.) So there obviously was a mistake somewhere.

Before I begin though, let me advise you not to convert things to meter, seconds, or meters per second. Just keep things in terms of light days and days. c = 1 light day/day = 1. That way, gamma is simply \gamma = \frac{1}{\sqrt{1 - v^2}}.

Homework Statement

You are exploring a newly discovered planet when a colleague contracts a deadly alien virus. Left untreated, it will probably kill him in 5 days, but the nearest medical station is 10 light-days away! How can you get him there in time?

Homework Equations

Δt_p=L/v
L=L'/\gamma
\gamma = 1/sqrt(1-v^2/c^2)

The Attempt at a Solution

Intuitively, I read this question and concluded that you could not do it. How could you possibly go 10 light-days in only 5 days? However, I tried to work through it anyway. Given the equation for Δt_p (proper time), I substituted some other values. The L term represents the contracted length, so I replaced L with L'/\gamma into the equation Δt_p=L/v. This gave me Δt_p=[(L'(1-v^2/c^2)^.5)]/v. Doing quite a bit of algebra, I ended up solving for v (the velocity of the ship) and got v=(L'^2/Δt_p^2)^.5.

I converted 5 days into seconds (so that I could work in meters/second) and got it to be 432,000 seconds. I said that this should be my Δt_p term because this is the time measured on the ship, so you need it to be 5 days to save the other astronaut. I then converted 10 light days into meters (again, to keep things simpler for me). I found 10 light days to be 2.592E14m. Plugging this into my previously derived equation, I got the speed necessary for the ship to reach the medical station (including time dilation and length contraction) to be 2c.

Is this supposed to happen, or should there be a true answer that would allow me to get the other passenger to travel 10 light days in a 5 day span (in his frame of reference)?

Any help would be much appreciated.
Thank you.

I'm going to assume that you are traveling along with the passenger. So the general equation we're talking about here is

(delta time) (velocity) = (distance)

where all are measured by you.

The "(delta time)" as measured by your clocks had better by 5 days (or less), otherwise the passenger is going to die. So for the purposes of this exercise, we'll say "(delta time)" is 5 days.

The "(velocity)" is how fast you are going, in units of light days per day. You can think of it as a fraction of the speed of light. For example, if v = 0.8, that's the same thing as v = 0.8c, since c = 1. But you don't know what this velocity is yet. It's what you will be solving for. Once you have your final velocity, feel free to tack a c onto it.

"(distance)" is the distance from your original location to the nearest medical station, as measured by you. Before you start traveling, you know this distance is 10 light days, but that's not the distance you're interested in. Once you start traveling, the measured distance contracts due to Lorentz contractions. As soon as you start moving at your constant velocity, the distance gets scrunched up to (10 light days)/\gamma (if it helps, think of it as you being stationary, and the planet you were previously on, along with the medical facility, moving past you). That's the distance you are interested in, (10 light days)/\gamma.

Now plug those things into your equation and solve for v.

 

[Yosty22]

I went through the calculations again keeping things in terms of light days and light days/day, and I got 2c again. As you said, I let (delta time) be equal to 5 days and I used the lorentz equation L=L'/(gamma). That is, 10 light days/gamma. This gave me (10 Light days * (1-v^2)^.5)/(5 Days). I multiplied by (delta time), divided by 10 light days, and squared it. After some algebra, I got back to the same equation as above: v=(10 light days)^2/(5 days)^2 -- Which again, simplifies to 2c. What am I doing here?

 

[Chestermiller]

I went through the calculations again keeping things in terms of light days and light days/day, and I got 2c again. As you said, I let (delta time) be equal to 5 days and I used the lorentz equation L=L'/(gamma). That is, 10 light days/gamma. This gave me (10 Light days * (1-v^2)^.5)/(5 Days). I multiplied by (delta time), divided by 10 light days, and squared it. After some algebra, I got back to the same equation as above: v=(10 light days)^2/(5 days)^2 -- Which again, simplifies to 2c. What am I doing here?

According to what collinsmark said, your equation should be:

5 v = (10 Light days) * (1-v^2)^.5

 

[kumusta]

In special theory of relativity, one has to keep in mind that there are two sets of quantities: (𝑖) proper quantities - those that are moving with the observer, so no changes in the quantities occur, and (𝑖𝑖) apparent quantities - those that move relative to the observer, so changes in the quantities take place.
In the length contraction formula
$$(1)~~~~~~~L = L_0 \sqrt { 1 - \frac { v^2 } { c^2 } } ~~\to~~~ L = L_0 ~~~ \text{when}~~~~ v = 0 $$ v = 0 $\to$ object whose length is being measured does not move relative to the observer making the measurement $\dots$
v $\neq 0 \to$ object whose length is being measured does move relative to the observer doing the measurement $\dots$
$~~~~~~~~~\to L_0$ on the right hand side of (1) is the proper length $\dots$
$~~~~~~~~~\to L$ on the left hand side of (1) is the apparent length $\dots$
The same thing is true with the time dilation formula
$$(2)~~~~~~~~~~~τ = \frac { τ_0 } { \sqrt {1 - \frac {v^2}{c^2} } } ~~\to~~~ τ = τ_0 ~~~ \text{when}~~~~ v = 0 $$ τ = $τ_0$ when ν = 0 $\to$ $τ_0$ $\text{ is the proper time since the event whose duration is }$ $\text{being measured is at rest relative to the observer }$ $\dots$
τ > $τ_0$ when ν $\neq$ 0 $\to$ τ is the apparent time since the event whose duration is being measured is moving relative to the observer $\dots$
Whether the observer is indeed at rest, or actually moving, does not really matter. What really matters is the relative motion between the observer and the thing being observed. Without any relative motion between them, we get the so-called proper quantities, those with subscripts 0 in (1) amd (2) above.
Now, in this problem, one has to decide whether the 10 light-days distance is a proper distance or an apparent distance. From where (place of origin) is this distance being measured? Does the observer move relative to the place of origin in making the distance measurement? Is the 5 days survival time a proper time or an apparent time? Does the observer (doctor looking after the infected person) move relative to the sick person in critical condition?

 

[PeroK]

There's a large time dilation here between the OP and the latest response: seven and a half years.

 

[Filip Larsen]

There's a large time dilation here between the OP and the latest response: seven and a half years.

Assuming he is still busy trying to save his infected colleague within that 5 day limit he must have found a way to do at least 0.9999983 c. No wonder we haven't heard from him - I'm not even sure Physics Forum will work at such large Doppler shifts!

 

[kumusta]

Using the time dilation formula with $τ_0$ = 5 d $$τ = τ_0(1-{v^2}/{c^2})^{-1/2}~~~\to~~~τ = (5~\rm{d})(1- {\upsilon^2}/{c^2})^{-1/2} $$ Distance of 10 light-days = (10 $\rm{d}$ )c =$~\upsilon$τ = $\upsilon$(5$~\rm{d})(1- {\upsilon^2}/{c^2})^{-1/2}$
$~~~~\to~$(2)c =$~\upsilon(1- ~{\upsilon^2}/{c^2})^{-1/2}~\to~$4(1$- ~{\upsilon^2}/{c^2})$ = ($\upsilon$/c)$^2~\to~$4 = 5($\upsilon$/c)$^2$
$~~~~\to~$($\upsilon$/c) = (4/5)$^{1/2}$ = (0.8)$^{1/2}$ = 0.8944$~\to~$ $\upsilon$ = (0.8944)c$~\dots$
That means traveling at a rocket speed of $\upsilon$ = (0.8944)c, the nearest medical station 10 light-days away could be reached in 5 days to get the wonder drug and thereby save the infected person. In the newly discovered planet, the 5-day-rocket-trip would take a time $$τ = (5~\rm{d})[1- (4/5)]^{-1/2} = (5~\rm{d})[1/5]^{-1/2} = 11.2 ~\rm{d} $$ Nothing was mentioned about solving the problem realistically, accelerating the rocket from rest at the start and then decelerating to finally stop at the nearest medical station, so I assumed that the whole trip would take place at constant speed.
In the rest frame of the moving rocket, with the proper distance $L_0$ = 5 light-days in the stationary frame of the newly discovered planet, the apparent distance traveled is $$L = L_0(1-{v^2}/{c^2})^{1/2} = (10~\rm{ld})[1- (4/5) ]^{1/2} = (10~\rm{ld})[1/5]^{1/2} = 4.5~\rm{ld} $$ or only 4.5 light-days, a contraction in length seen in the moving frame.