Solving Astronaut Problem: Time Dilation, Pulse Calculation

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An astronaut traveling at 0.90c measures a pulse of 70 beats per minute, equating to 0.857 seconds per pulse in his frame. For an Earth observer, time dilation calculations show the pulse duration as approximately 0.37 seconds, but this is incorrectly derived; moving clocks run slow, meaning the Earth observer should measure a longer duration. Consequently, the astronaut's pulse rate as observed from Earth is calculated to be about 160.6 beats per minute. Increasing the spacecraft's speed would further increase the pulse rate observed from Earth, while the astronaut's own pulse remains unchanged. The discussion highlights the importance of correctly applying the principles of time dilation and the Lorentz transformation in relativistic scenarios.
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An astronaut traveling at 0.90c, with respect to Earth, measures his pulse and finds it to be 70 beats per minute.
a) Calculate the time required for one pulse to occur, as measured by the astronaut.
b) Calculate the time required for one pulse to occur, as measured by an Earth-based observer.
c) Calculate the astronaut’s pulse, as measured by an Earth-based observer.
d) What effect, if any, would increasing the speed of the spacecraft have on the astronaut’s pulse as measured by the astronaut and by an Earth-based observer? Why?

this is my solution

a. the time for one pulse to occur in the astronauts frame is 1 minute/70 beats per minute or 1/70 minute or 60/70 seconds which in decimal form is 0.857 seconds.

b. the formula for time dilation is part of the Lorentz transformation is:

t = t0/(1 - (v^2/c^2))^1/2
= 60/70/[1 - (.81/1)]^1/2
= 60/70[.19]^1/2
= 60/70(0.435889894354067)
= 0.37 seconds
So they are about .37 seconds apart form the frame of reference of the Earthling

c. Pulse is 1/.37 seconds or 2.677 beats per seconds which is:
2.67651689515656 x 60 = 160.6 beats per minute

d. As the speed of the astronaut increases the astronauts pulse will also increase from the frame of reference of the Earthling. As v approaches c the denominator or the Lorentz transformation approaches 0 so the whole thing goes to infinity.

is that right?
 
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salsabel said:
= 60/70(0.435889894354067)
= 0.37 seconds
You may want to recheck your calculations between these two steps. Your error throws your answer to (c) off and means that the conclusion you draw in (d) is incorrect. Does your answer to (d) make sense intuitively?
 
salsabel said:
this is my solution

a. the time for one pulse to occur in the astronauts frame is 1 minute/70 beats per minute or 1/70 minute or 60/70 seconds which in decimal form is 0.857 seconds.
Good.

b. the formula for time dilation is part of the Lorentz transformation is:

t = t0/(1 - (v^2/c^2))^1/2
= 60/70/[1 - (.81/1)]^1/2
= 60/70[.19]^1/2
= 60/70(0.435889894354067)
= 0.37 seconds
So they are about .37 seconds apart form the frame of reference of the Earthling
The answer is backwards. (Remember the rule: Moving clocks are measured to run slow.) You made a mistake in line 3: You should be dividing, not multiplying.

c. Pulse is 1/.37 seconds or 2.677 beats per seconds which is:
2.67651689515656 x 60 = 160.6 beats per minute
Again, backwards.

d. As the speed of the astronaut increases the astronauts pulse will also increase from the frame of reference of the Earthling. As v approaches c the denominator or the Lorentz transformation approaches 0 so the whole thing goes to infinity.
Again, backwards. (And what about from the astronaut's perspective?)
 
thanks
it makes sense to me
 
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