Relativity - Time Dilation - Solving for t

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Homework Help Overview

The problem involves time dilation in the context of special relativity, specifically examining how a passenger's heartbeat is perceived by an observer on Earth as the spaceship travels at a significant fraction of the speed of light (0.6c).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between period and frequency, noting that the heartbeat rate measured in the spaceship differs from that measured on Earth due to time dilation. There are attempts to convert heartbeat frequency into period and vice versa, with some questioning the relevance of time units used.

Discussion Status

Participants are actively engaging with the problem, exploring the implications of time dilation and the relationship between frequency and period. Some have suggested calculations based on the Lorentz factor (γ), while others are clarifying definitions of time intervals. There is no explicit consensus yet, but productive lines of reasoning are emerging.

Contextual Notes

There is some confusion regarding the definition of "initial" time and whether it should be referred to as "proper" time. Participants are also navigating the conversion between different time units (minutes to seconds) and its impact on the final answer.

PeachBanana
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Homework Statement



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

Homework Equations



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

The Attempt at a Solution



Δt initial = 60
γ = 1.25

(60)(1.25) = 75 bpm

not an option
 
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PeachBanana said:

Homework Statement



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

Homework Equations



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

The Attempt at a Solution



Δt initial = 60
γ = 1.25

(60)(1.25) = 75 bpm

not an option

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:
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.
 
PeachBanana said:
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.

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:
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?
 
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)?
 
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.
 
PeachBanana said:
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.

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.)
 
Oh, wow. Should I be thinking about γ = 1.25? Because...60 / 1.25 = 48.
 
  • #10
PeachBanana said:
Oh, wow. Should I be thinking about γ = 1.25? Because...60 / 1.25 = 48.

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).
 
  • #11
For clarification: Was I incorrect in stating Δt initial was 60?
 
  • #12
PeachBanana said:
For clarification: Was I incorrect in stating Δt initial was 60?

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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