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Homework Help: Relativity - Time Dilation - Solving for t

  1. Dec 10, 2012 #1
    1. The problem statement, all variables and given/known data

    A spaceship approaches Earth with a speed of 0.6c. A
    passenger in the spaceship measures his heartbeat as 60 beats
    per minute. What is his heartbeat rate according to an
    observer who is rest relative to Earth?

    1. 48 beats per minute
    2. 56 beats per minute
    3. 65 beats per minute
    4. 69 beats per minute

    2. Relevant equations

    Δt = γ * Δt initial
    γ = 1 / [1-(v^2/c^2)]^1/2

    3. The attempt at a solution

    Δt initial = 60
    γ = 1.25

    (60)(1.25) = 75 bpm

    not an option
  2. jcsd
  3. Dec 10, 2012 #2


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    Hint: If an event is timed in the frame of the ship to be 1s in duration (this is the proper time), what is the duration as measured in the earth frame?

    What is the relationship between period (has units of time) and frequency (like a heartbeat rate)?

    Can you now work out the answer?

    EDIT: And I have to say you're rather lucky that 75bpm was not an option, because it's wrong. If I was setting the question, I would've tricked people with that option.
    Last edited: Dec 10, 2012
  4. Dec 10, 2012 #3
    Period and frequency are inversely related. So since heart beat is frequency, 60 bpm, period would be 1/60 min. My next problem is the SI unit of time is not minutes but if I convert I get 1/3600 seconds.
  5. Dec 10, 2012 #4


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    Don't overcomplicate it. You know the factor by which time measurements in the two frames are related (you've already worked it out). Now can you figure out the factor by which frequencies in the two frames are related?
    Last edited: Dec 10, 2012
  6. Dec 10, 2012 #5
    I still don't think I quite understand...I think I might be stuck on the whole minutes to seconds thing.

    Edit: Does it even matter that it's in minutes since the final answer is in minutes too?
  7. Dec 10, 2012 #6
    60 beats per minute is one beat per second. If it takes 1 second between beats in the rest frame, how many seconds are measured to occur between the same two beats by observers in the other frame? If it takes t seconds between beats in the other frame, how many beats occur in 60 seconds (in terms of t)?
  8. Dec 10, 2012 #7
    I think after re-reading and thinking and google searching it has dawned on me, haha. So we know 60 beats per minute means one beat / one second which is the same as...

    1 / 60 s.
  9. Dec 10, 2012 #8


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    How is one beat per second equal to 1/60s? The latter implies 1 beat per 60 seconds!

    The guy on the ship measures 1 beat per second.

    The guy on earth sees the same beat but times it as ??? seconds.

    What's "???"

    Now, what heart rate does the earth guy measure?

    (You shouldn't even have to care about minutes and seconds, all you need is the factor that relates time measurements in one frame to the other.)
  10. Dec 10, 2012 #9
    Oh, wow. Should I be thinking about γ = 1.25? Because...60 / 1.25 = 48.
  11. Dec 10, 2012 #10


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    Of course. You already calculated γ = 1.25. So a proper time duration of 1s (ship frame) is measured as 1.25s from earth. Therefore the heartbeat is correspondingly slower (more time between beats), and it's as you calculated (in beats per minute).
  12. Dec 10, 2012 #11
    For clarification: Was I incorrect in stating Δt initial was 60?
  13. Dec 10, 2012 #12


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    You need to define your Δt "initial". In this case, it's not obvious. If you meant that Δt "initial" is the proper time taken for 60 beats of the ship guy's heart, then yes, that's correct.

    The reason I put the "initial" in quotes is because I don't think it's a good descriptor. Better to call it "proper".
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