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Specialy theory of relativity

  • Thread starter rojasharma
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what effect would increasing the speed of the spacecraft have on the astronaut's pulse as measured by the astronaut and by the earth observer? why. I think in both measurements, an increased pulse rate will be found?..
 

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536
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The pulse is only an example of a clock in general. Any type of clock. You can look for a description in the textbook that probably phrases it as "the rate that time passes" or the "the speed of a clock", in one frame of reference, compared to another frame of reference. Consider which frame of reference undergoes acceleration.
 
66
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the question is refering to heart beat..per min.
 
Dick
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As mikelepore said, it's the same thing as the rate of a clock.
 
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so...the astronaut will observe ...her pulse being slowed down?....:Sstill confused
 
536
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Maybe you have a formula for an interval of time as it is measured by different observers, some formula that begins: t = ......
 
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t=to/(sqrt1-v^2/c^2)
 
536
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You have two people, an astronaut and an earth observer. You also have two times, t and t0. Your textbook must explain how to assign which to which. In fact, most books use a traveler on a spaceship as the author's favorite example. Then you have the astronaut moving at speed v, which is less than c. That sqrt is in the denominator and not the numerator, and that must have some effect the answer.
 
Dick
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so...the astronaut will observe ...her pulse being slowed down?....:Sstill confused
v is the relative velocity between the observer and the observed. The astronaut isn't moving relative to herself. In that case v=0.
 
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I know. but in general ..what happens? when the speed of the spacecraft is increased. will the earth based observer's measurement be more beats per min or less than b4? and what about astronaut, will she experience the frequency of her pulse to be more or less than b4?
 
Dick
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If you say "I know.", what is it you know? Will the speed increase change the speed of the astronaut relative to herself? For the other question what does your formula t=to/(sqrt1-v^2/c^2) tell you? t0 is the rest frame interval (and it's fixed). t is the observed interval. Does t get bigger or smaller as v approaches c?
 
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t gets bigger...
 
Dick
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Good. Now what does that tell you about the original question? t0 is the interval between beats measured by the astronaut. t is the interval between beats measured from earth.
 
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to is same while t increases....?
 
Dick
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t0 is the interval between beats that the astronaut herself measures. That doesn't depend on v, right? Right? You said "I know". I remember.
 
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righttt........i think i said i know to somthing else....
 
Dick
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righttt........i think i said i know to somthing else....
I'd be curious what that something is. A moving observer never observes her own heart beating slow or her own clocks running slow. She can consider herself 'at rest'. At v=0. That's why it's called the "Specialy theory of relativity". Every observer can consider their own time reference as unchanging.
 
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i was replying to someone else..when i said "i know" lol....i finally get it ....:)
 
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thankyou
 

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