Does Relativistic Time Dilation Affect an Astronaut's Pulse Rate?

[lamerali]

Hi I'm not sure if i am doing this correctly, can someone please check my answer! THANKS A LOT :D

Homework Statement

An Astronaut traveling at 0.90 c, with respect to Earth, measures his pulse adn 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) calaculare the astronaut's pulse, as measured by an Earth-based observer.
d) What effect, if any, would increasing the speed of the sapacecraft have eon the astronaut's pulse as measured by the astronaut and by the Earth observer? Why?

Homework Equations

$$\Delta$$ t = $$\frac{\Delta t_o}{\sqrt{1 - \frac{ v^2 }{ c^2 }}}$$

The Attempt at a Solution

a)
(1 min / 70 pulses) x (60 sec) = 0.85 s
Therefore the astronaut feels a pulse every 0.85 s.

b)
$$\Delta$$ t = $$\frac{\Delta t_o}{\sqrt{1 - \frac{ v^2 }{ c^2 }}}$$
$$\Delta$$ t_o = $$\Delta$$ t x $$\sqrt{1 - \frac{ v^2}{c^2}}$$
$$\Delta$$ t_o = (0.85) x $$\sqrt{1 - \frac{ 0.90c^2}{c^2}}$$
= 6 seconds

Therefore the time required for one pulse to occur as measured by an Earth-based observer is 6 seconds.

c) (60 seconds / 6 seconds) = 10 beats per second
Therefore the astronauts pulse as measured by an Earth-based observer is 10 beats per minute.

d) As the speed of the spacecraft increases the astronaut's pulse will increase from the frame reference of the astronaut. From the frame reference of the Earth observer the pulse of the astronaut will decrease.

 

[tiny-tim]

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.
a)
(1 min / 70 pulses) x (60 sec) = 0.85 s
Therefore the astronaut feels a pulse every 0.85 s.

b)
$$\Delta$$ t = $$\frac{\Delta t_o}{\sqrt{1 - \frac{ v^2 }{ c^2 }}}$$
$$\Delta$$ t_o = $$\Delta$$ t x $$\sqrt{1 - \frac{ v^2}{c^2}}$$
$$\Delta$$ t_o = (0.85) x $$\sqrt{1 - \frac{ 0.90c^2}{c^2}}$$
= 6 seconds

Therefore the time required for one pulse to occur as measured by an Earth-based observer is 6 seconds.

Hi lamerali!

Just looking at a) and b) …

a) is fine, of course!

b) I don't see how you got from .85 to 6 … and you forgot to square the 0.9.

 

[lamerali]

Hey tiny-tim :D

I did forget to square the 0.9 so my new answer is 0.37 s...does that sound more realistic?

 

[tiny-tim]

Hey tiny-tim :D

I did forget to square the 0.9 so my new answer is 0.37 s...does that sound more realistic?

Hey lamerali

Shouldn't it be more?

The Earth observer regards the astornaut's clock as slow, so his pulse is slow, which means the pulse takes longer?

 

[lamerali]

isn't 0.37 seconds the time for ONE pulse to occur, his actual beat would be 162 beats per minute. if this is incorrect do you know where i am going wrong?

 

[tiny-tim]

isn't 0.37 seconds the time for ONE pulse to occur, his actual beat would be 162 beats per minute. if this is incorrect do you know where i am going wrong?

When the astronaut returns, he will be younger than he should be, so he will have had less pulses, so his pulses must have taken longer.

I think your √ factor should be 1/√.

 

[eok20]

tiny-tim is correct. You have the correct formula but are using it wrong. In general, $\Delta t_0$ represents the proper time interval, i.e. the time difference between two events as measured in a reference frame where the events occurs at the same place. In this example, this is the spaceship's frame.

 

[lamerali]

OHH ok so sould it be

= (0.85) / √ (1-(0.9c^2)/c^2
= 2 s

therefore each pulse takes 2 seconds and his pulse is (60s / 2s) = 30 pulses per minute