Matter/energy Interface

fahad

12th October 2008 - 11:19 PM

Hi,I'm kinda stuck with this question.Pls help me out..

An astronaut travelling 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 the earth's observer?Why?

**I know am supposed to use the formula
<>t{relativistic time} = <>t{proper time} * sq.root[1-v^2/c^2] bt I'm kinda stuck with the values.plz help me out..

bm1957

13th October 2008 - 08:21 AM

This doesn't look too complicated. I'm assuming you can answer a)?

For b), use your equation with v=0.9c, and the answer should pop out quite simply.

For c), well that's just a) backwards once you've done b).

d) is asking you to interpret what the numbers you've just calculated really mean. You'll have to extrapolate your results.

Hope that helps!!!

fahad

13th October 2008 - 03:31 PM

Well,i gave it a try:

{a} 70 beats ----60s
01 beats ---- ?

1/70 * 60 = 0.857s

= 0.86s

{b} <>t{relativistic time} = <>t{proper time} * sq.root[1-v^2/c^2]
0.857 / sq root [1 - {0.9c}^2/{c}^2 ] = 1.97s
= 2.0s

{c} 1.97s ---- 1 beat
? ? -----70 beats

70/1 * 1.97 = 137.9
= 138 beats

{d} Can i say : Increasing the speed of the spacecraft will increase the astronaut's pulse both as measured by the astronaut and by the Earth observer because the speed of the spacecraft is directly proprtional to the relativistic and proper time.

** Is there any other reason/explanation??

** Am i right??

Plz reply..

Enthalpy

13th October 2008 - 04:22 PM

Hi Fahad!
Fairy's post is regrettable as are all comments he posts here. Please consider he isn't representative of all the forum.

As for the relativistic correction to time (the one in v^2/C^2), it makes events at the astronaut slower to us.

fahad

13th October 2008 - 05:06 PM

I'm kinda confused abt that Mr. Enthalpy,are you talking about part d of the question or...??
Are my answers of part a,b,c and d right??

Plz reply..

bm1957

13th October 2008 - 07:04 PM

a) looks good.

b), you've got the right answer but notice that the equation you quoted below and in your last post is not the same as the equation you actually used. The one you used is correct.

c), you've got yourself a bit confused. 1.97s --- 1 beat is good. Now, how many beats in 60s? 1s --- 1/1.97 beats, 60s --- 60*(1/1.97) beats

d) you now know that when the spaceship is travelling at 0.9c, the astronaut measures his pulse as normal, and the observer measures the pulse as the figure above (60*(1/1.97)). What happen if the spaceship was travelling at 0.99c? Would the astronaut measure his own pulse to be any faster or slower? Would the observer measure his pulse to have changed at all? (You will need to re-visit c) before you can answer this question correctly).

fahad

13th October 2008 - 07:36 PM

Thnx for the reply,i gave it a try again..

{c} 1.97s ---- 1 beat
60s ----- ? ?

60 / 1.97 * 1 = 30.46
= 31 beats {is the rounding off correct?}

{d} Can i say : Increasing the speed of the spacecraft will increase the astronaut's pulse measured by the Earth observer and will decrease the astronaut's pulse measured by the astronaut because the speed of the spacecraft is directly proprtional to the proper time and indirectly proportional to the relativistic time.

**Are my answers right and is there any beter explantion for part d??

Thanking you in advance..

Ron

13th October 2008 - 08:42 PM

QUOTE (fahad+Oct 13 2008, 07:36 PM)

Thnx for the reply,i gave it a try again..

{c} 1.97s ---- 1 beat
60s ----- ? ?

60 / 1.97 * 1 = 30.46
= 31 beats {is the rounding off correct?}

{d} Can i say : Increasing the speed of the spacecraft will increase the astronaut's pulse measured by the Earth observer and will decrease the astronaut's pulse measured by the astronaut because the speed of the spacecraft is directly proprtional to the proper time and indirectly proportional to the relativistic time.

**Are my answers right and is there any beter explantion for part d??

Thanking you in advance..

Hi Fahad,
Just to help clarify part 'd', you are correct in the Earth observer's measurement differing by gamma, but the astronaut can never know that he is moving, no matter at what speed, as long as his movement is constant (ie; in an inertial reference frame).
Peace,
Ron

fahad

13th October 2008 - 08:57 PM

But we were not told the speed was going to be constant,so can we assume that it wasnt constant thus Increasing the speed of the spacecraft will decrease the astronaut's pulse measured by the astronaut??

bm1957

14th October 2008 - 07:41 AM

QUOTE (fahad+Oct 13 2008, 08:36 PM)

Thnx for the reply,i gave it a try again..

{c} 1.97s ---- 1 beat
60s ----- ? ?

60 / 1.97 * 1 = 30.46
= 31 beats {is the rounding off correct?}

{d} Can i say : Increasing the speed of the spacecraft will increase the astronaut's pulse measured by the Earth observer and will decrease the astronaut's pulse measured by the astronaut because the speed of the spacecraft is directly proprtional to the proper time and indirectly proportional to the relativistic time.

**Are my answers right and is there any beter explantion for part d??

Thanking you in advance..

That looks better. The rounding isn't quite right though.

If you have 30.46, you can either round to 3sf, which means looking at the 4th no. which is 6, and rounding the 4 up to a 5 (since 6>5), giving 30.5. But that would be too many figures.

Better would be to 2sf. Look at the 3rd no., 4. 4 is less than 5, so you round down to 30 beats.

d). Let's think about it again. The astronaut will measure his pulse to be 70bpm, wherever he is. If he's on Earth, it will be 70bpm, at 0.90c, it's still 70bpm. So does getting faster change it?

Now, when he's moving at 0.90c, the Earthbound observer measures his pulse to be 30bpm, which you calculated in c). So, increasing the speed of the spacecraft has changed his pulse from 70bpm to 30bpm as measured by the Earthbound observer.

Those are the two changes you need to be considering.

QUOTE

But we were not told the speed was going to be constant,so can we assume that it wasnt constant thus Increasing the speed of the spacecraft will decrease the astronaut's pulse measured by the astronaut??

Stating that the astronaut is travelling at 0.90c implies zero acceleration. Any acceleration will make things much more complicated, I recommend forgetting about relativistic acceleration for now!!!

fahad

14th October 2008 - 07:44 PM

I do understand about the earth-bound observer bt im kinda confused about the astronaut's view point.

I didnt understand this statement: ["The astronaut will measure his pulse to be 70bpm, wherever he is. If he's on Earth, it will be 70bpm, at 0.90c, it's still 70bpm."]

** Cant i answer this question using the equation of dialated and proper time??Coz according to this equation:
<>t{proper time} = <>t{dialated time} * sq.root[1-v^2/c^2
whereby proper time is directly proportional to the speed {v} thus Increasing the speed of the spacecraft will decrease the astronaut's pulse measured by the astronaut?? ??

I do apologize for the inconvineance...

fahad

14th October 2008 - 07:52 PM

sry i made a mistake in my previous post,its:

<>t{proper time} = <>t{dialated time} * sq.root[1-v^2/c^2
whereby proper time is directly proportional to the speed {v} thus Increasing the speed of the spacecraft will increase the astronaut's pulse measured by the astronaut?? ??

and:

<>t{dialated time} = <>t{proper time} / sq.root[1-v^2/c^2
whereby dialated time is inversely proportional to the speed {v} thus Increasing the speed of the spacecraft will decrease the astronaut's pulse measured by the Earth observer.

Did i misunderstand anything??

bm1957

15th October 2008 - 08:27 AM

You're tying yourself in knots a little.

When you worked out dilated time in the original question, you had to use a value for proper time that you knew. You were happy that proper time was the time that the astronaut measured for one beat. That is still the proper time, so you don't need to use an equation to find it.

The thing to realise is that the astronaut doesn't know whether he is stationary or moving at 0.9c, in fact, it is just as valid to say that he is stationary and the earthbound observer is travelling at 0.9c, that's what relativity is; all speeds are relative.

The proper time is the time measured for an event by a clock that is near to that event, and the proper time is the same whatever speed the event is moving at. The astronaut will measure his pulse to be the same whatever speed he si moving at. Any observer moving at a different speed will measure a dilated time for any event.

Look here for info on proper time.

Hopefully that all makes sense, post back if you want any clarifications!

[EDIT]
If you're interested in algebra, you might like to try this out:

For the equation:
<>t{proper time} = <>t{dialated time} * sq.root[1-v^2/c^2

<>{dilated time} is itself a function of v, according to your other equation:
<>t{dialated time} = <>t{proper time} / sq.root[1-v^2/c^2

Try substituting your formula for dilated time into you formula for proper time and you'll see why proper time is not proportional to velocity.
[/EDIT]

fahad

16th October 2008 - 01:22 AM

So i guess thats the rule,"the proper time is the same whatever speed the event is moving at"..Thanks for your help..

I tried to solve this other question,plz check it out and let me know if i am right..

A 30-year-old female astronaut leaves her newborn child on Earth {hypothetically, of course}and goes on a roud-trip voyage to a star that is 40 light-years away in a spaceshiptravelling at 0.90c.What will be the ages of the astronaut and her son when she returns?

Ans: son : {0.2}40 + 40 = 48 yrs

Astronaut: <>t{proper time} = <>t{dialated time} * sq.root[1-v^2/c^2]
= 48 * sq.root[1-{0.9c}^2/c^2] = 20.9232

Therefore, the child will have aged 48 years while the astronaut will have aged 21 years

Am i right??

bm1957

16th October 2008 - 06:51 AM

QUOTE (fahad+Oct 16 2008, 02:22 AM)

So i guess thats the rule,"the proper time is the same whatever speed the event is moving at"..Thanks for your help..

I think that would be a fair interpretation of what proper time is. It would be helpful to remember that this is the local time; that is, that any clock near to the event (and at rest with respect to the event) will measure the proper time of the event.

QUOTE

Ans: son : {0.2}40 + 40 = 48 yrs

I'm not sure where you got this equation from. What is 0.2? What exactly are you doing?

I would try

CODE (->

QUOTE

Ans: son : {0.2}40 + 40 = 48 yrs

I'm not sure where you got this equation from. What is 0.2? What exactly are you doing?

I would try
Astronaut: <>t{proper time} = <>t{dialated time} * sq.root[1-v^2/c^2]
= 48 * sq.root[1-{0.9c}^2/c^2] = 20.9232

For the astronaut, your method looks correct, I would just check whether 48 years is correct.

QUOTE

Therefore, the child will have aged 48 years while the astronaut will have aged 21 years

I think they are both wrong, but once you correct the initial mistake you should get both answers right.

There is a useful example here along with some explanations and background.

fahad

16th October 2008 - 06:10 PM

I tried to solve the question using an example in my text book,this is what i did:

According to the stationary new born child,since the spaceship is travelling 10% slower than the speed of light,it should take 10% longer to reach the star and 10% longer to return.
Therefore,the entire trip should take 20% longer than 40 years.
The total time required is 40years + {0.2}40years = 48 yrs.
Therefore,the child will have aged 48 years while the astronaut was en route.

However,the time measured by earth observers is dialated time.The amount of time required for the trip as measured by the asttronaut is:
<>t{proper time} = <>t{dialated time} * sq.root[1-v^2/c^2]
= 48 * sq.root[1-{0.9c}^2/c^2] = 20.9232

Therefore,the child will have aged 48 years while the atronaut will only have aged 21 years.

Where am i wrong??

Ron

16th October 2008 - 07:18 PM

QUOTE (fahad+Oct 16 2008, 06:10 PM)

I tried to solve the question using an example in my text book,this is what i did:

According to the stationary new born child,since the spaceship is traveling 10% slower than the speed of light,it should take 10% longer to reach the star and 10% longer to return.
Therefore,the entire trip should take 20% longer than 40 years.
The total time required is 40years + {0.2}40years = 48 yrs.
Therefore,the child will have aged 48 years while the astronaut was en route.

However,the time measured by earth observers is dialated time.The amount of time required for the trip as measured by the asttronaut is:
<>t{proper time} = <>t{dialated time} * sq.root[1-v^2/c^2]
= 48 * sq.root[1-{0.9c}^2/c^2] = 20.9232

Therefore,the child will have aged 48 years while the atronaut will only have aged 21 years.

Where am i wrong??

Hi Fahad, bm1957,
Fahad,
Bm is correct in that it will take the astronaut 88.9 yrs to travel 80 light years at .9c (80/.9), so she will be gone on her travel for that long. Hence that will be the age of her newborn when she returns(the newborn ages normally). Because of time dilation, gamma for .9c = 2.294 so she will have only aged 38.75 years(88.9/2.294), so she will be 68.75 years old. Her newborn will be about 10.25 years older than his/her mother!
Do you follow?
Peace,
Ron

fahad

16th October 2008 - 07:35 PM

I uderstand Ron's bm1957's argument but my question is why is my solution wrong?Because according to my text book,they solved a similar question using the same approach i used..

Plz reply..

Ron

16th October 2008 - 08:17 PM

Hi Fahad,
The trip takes 10% longer, not 20% longer, so you have 80years +8years (round trip), but you have to realize that the trip is 10% longer than the time it would take to travel the same distance at 100%c. This small difference is where the 88.9 years comes from. Look at the formula that bm used. By dividing the distance traveled by the speed, you get the time. If you use 10%, you would have to divide the distance by 90% to get the true time. If you multiply 40 by 10%, you would have to multiply that 4 extra years by 10% again (then the .4 by 10% ...), hence the repeating 44.444... (x2)= 88.888... or 88.9years.
Still not making sense? Maybe BM can jump in again!
Peace,
Ron

fahad

16th October 2008 - 08:37 PM

sry bt i still dnt get why its not 20% because its a round trip.it takes 10% longer to reach the star and 10% longer to return,thus adding up to 20%.
Unless my textbook is wrong,then i'm kinda stuck..

bm1957

17th October 2008 - 08:25 AM

QUOTE (fahad+Oct 16 2008, 08:35 PM)

I uderstand Ron's bm1957's argument but my question is why is my solution wrong?Because according to my text book,they solved a similar question using the same approach i used..

Plz reply..

I am sure that the example in your textbook is slightly different.

Your solution assumes that if speed is reduced by 10% then time is increased by 10%. This is wrong:

100 miles @ 100mph takes 60min
100 miles @ 90mph does not take (60*1.1)=66min

The correct way to approach the problem is to say that speed has been mutiplied by a factor of 0.9, so time must be divided by a factor of 0.9:

100 miles @ 90mph takes (60/0.9)=66.667min

(Also note that the round trip @ c would be 80 years, not 40 years as you assumed in your calculation)

Hopefully that makes sense, if not, keep asking questions; that's the only way to learn! If you post the example from your textbook I might be able to show you better why it is different.

Ron

17th October 2008 - 11:52 AM

QUOTE (fahad+Oct 16 2008, 08:37 PM)

sry bt i still dnt get why its not 20% because its a round trip.it takes 10% longer to reach the star and 10% longer to return,thus adding up to 20%.
Unless my textbook is wrong,then i'm kinda stuck..

Hi again,
Fahad, to answer this question alone, you have to realize that a percentage would be taken of the entire trip, not one way then the other. 20%, using your argument, would imply that you are traveling at .8c.
You take 10% of the one way, then 10% of the return. It is still 10% of the entire round trip. This seems to be half of your sticking point. The other half is in the approach that Bm has mentioned. It's probably a good idea to write out the entire text book example, then take it from there.
Peace,
Ron

fahad

17th October 2008 - 04:36 PM

The question below is the example inmy textbook,plz explain to me how its different frm my original question..

Ed and Fred are identical 30-year-old twins. Fred is about to go to proxima Centauri,the nearest star to Earth after our sun, aboard a spacecraft travelling at 0.99c. Proxima Centauri is 4.0 * 10^13 km,or 4.2 light-years away. How old will the twins be when Fred returns?

Ans:
According to Ed, the stationary twin, since the spaceship is traveling 1% slower than the speed of light,it should take 1% longer to reach Proxima Centauri and 1% longer to return.
Therefore,the entire trip should take 2% longer than 4.2 years.
The total time required is 8.4years + {0.02}4.2years = 8.484 yrs.
Therefore, Ed will have aged 8.484 years while Fred was en route.

However,the time measured by earth observers is dialated time.The amount of time required for the trip as measured by Fred is:
<>t{proper time} = <>t{dialated time} * sq.root[1-v^2/c^2]
= 8.484 * sq.root[1-{0.99c}^2/c^2] = 1.197 years

Therefore, Ed will have aged 8.484 years while Fred will only have aged 1.197 years.

N/B: I copied everything exactly from the text book

**I am now totally confused,pls help me out..

Ron

17th October 2008 - 06:14 PM

I see where you're at now, Fahad.
firstly, notice in the textbook equation, they use the total round trip of 8.4 years + (.02)*4.2years. You used only the one way in your equation of 40 + (.2)*40=48years when it should have read 80 + (.2)*40 = 88years.
That's the big difference.

That will get you much closer to the real answer. I still maintain that 88 years is an approximation, though. , because the 88 years does not account for the true definition of V=D/T (as Bm said). It is a very subtle difference, and your answer will only be off by a small percentage, but when you use that whole 'it takes 10% longer' approach, you are going to have to take into account that if it 44 years is 10% + 100%, you have to repetitively add 10% of 10% + 100%, add infinity, which gives the more accurate value of the repeating decimal 88.888...
Peace,
Ron

fahad

18th October 2008 - 05:50 AM

I think i now get it.It jst happens that i tried to solve the example in the text book using the formula {t=d/v} and i got the same answer in terms of its accuracy but that doesnt happen in my original question {i.e 88yrs and 88.9yrs are alittle bit different}.

I tried to solve another question,this time i hope i got it right..

A spaceship travels past a planet at a speed of 0.80 c as measured from the planet's frame of reference. An observer on the planet measures the length of a moving spaceship to be 40m
{a} How long is the spaceship,according the astroanut?
{b} At what speed would the spaceship have to travel for its relativistic length to be half its "proper" length?

Ans:
{a} <>l{relativistic length} = <>l{proper length} * sq.root[1-v^2/c^2]
= 40 * sq.root[1-{0.8c}^2/c^2] = 24m

{b} 0.5 * 40 = 20m
<>l{relativistic length} = <>l{proper length} * sq.root[1-v^2/c^2]
20 = 40 * sq.root[1-v^2/c^2]
v = 0.866c or 2.6*10^8 m/s

Plz check it out and reply..

bm1957

18th October 2008 - 09:26 AM

QUOTE (fahad+Oct 17 2008, 05:36 PM)

Ans:
According to Ed, the stationary twin, since the spaceship is traveling 1% slower than the speed of light,it should take 1% longer to reach Proxima Centauri and 1% longer to return.

At 0.99c, this approximation is very close. It is also a very poor method to be teaching a student. I would bring this up with your tutor, or if you are self-learning, perhaps try to find a better text book.

Unfortunately this method is just plain wrong and is not very close below about 0.95c.

QUOTE

{a} <>l{relativistic length} = <>l{proper length} * sq.root[1-v^2/c^2]
= 40 * sq.root[1-{0.8c}^2/c^2] = 24m

The question says that 40m is the length that the observer on the planet measures. This means that 40m is not the proper length, but the relativistic length. You should try using:

<>|{proper length} = <>l{relativistic length} / sq.root[1-v^2/c^2]

and see what you get. A quick check would have been to notice that for your answer, the astronaut was measuring the spaceship to be shorter than the observer was; but an observer should see length contraction.

QUOTE (->

QUOTE

{a} <>l{relativistic length} = <>l{proper length} * sq.root[1-v^2/c^2]
= 40 * sq.root[1-{0.8c}^2/c^2] = 24m

The question says that 40m is the length that the observer on the planet measures. This means that 40m is not the proper length, but the relativistic length. You should try using:

<>|{proper length} = <>l{relativistic length} / sq.root[1-v^2/c^2]

and see what you get. A quick check would have been to notice that for your answer, the astronaut was measuring the spaceship to be shorter than the observer was; but an observer should see length contraction.

{b} 0.5 * 40 = 20m
<>l{relativistic length} = <>l{proper length} * sq.root[1-v^2/c^2]
20 = 40 * sq.root[1-v^2/c^2]
v = 0.866c or 2.6*10^8 m/s

Yep, this is correct. Any values such that the relativistic length was half of the proper length, (1 and 2, 5 and 10, etc) would have given you the correct speed.

fahad

18th October 2008 - 06:52 PM

Thanks for your help..

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