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Special Relativity Time Dilation/Length Contraction Problem

  1. Jan 14, 2014 #1
    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations

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

    3. 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 Δtp (proper time), I substituted some other values. The L term represents the contracted length, so I replaced L with L'/\gamma into the equation Δtp=L/v. This gave me Δtp=[(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/Δtp^2)^.5.

    I converted 5 days in to seconds (so that I could work in meters/second) and got it to be 432,000 seconds. I said that this should be my Δtp 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 in to meters (again, to keep things simpler for me). I found 10 light days to be 2.592E14m. Plugging this in to 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.
     
  2. jcsd
  3. Jan 14, 2014 #2

    collinsmark

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    Homework Helper
    Gold Member

    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 [itex] \gamma = \frac{1}{\sqrt{1 - v^2}}. [/itex]

    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. :wink: 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)/[itex] \gamma [/itex] (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)/[itex] \gamma [/itex].

    Now plug those things into your equation and solve for v. :smile:
     
    Last edited: Jan 14, 2014
  4. Jan 14, 2014 #3
    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?
     
  5. Jan 14, 2014 #4
    According to what collinsmark said, your equation should be:

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