Aging When Travelling

relativity

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#1 terryble

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Posted 26 April 2017 - 03:48 AM

I had an exam today and in the exam a question was this, "A 50 year old tortoise jumps onto a rocket and goes to a distant planet 370 lightyears away, the rockets speed was 0.7c . If a tortoise lives 450 years in avarage, would he live till reaching the destination or not?" cruel question I know, but i calculated like this that relativity won't affect the tortoise and it'd take him (370)©/(0.7c)=528 years, so he won't survive. Is my process correct? because as i understand it, for the tortoise everything is normal. All my friends have answered using time dilation. So I am really confused. Please help.

#2 spartan45

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Posted 26 April 2017 - 08:47 AM

The tortoise rocket clock would record an elapse time of 377.47 years so the tortoise would only be 427.5 (rounded) years old, giving it another 22.5 years to explore the planet. 528.57years is the period of time an Earth based observer would record for the 370 light year journey of the craft travelling at 0.7c. The tortoise travelling in the rocket would experience time dilation of that figure, giving 377.47 years.

#3 A-wal

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Posted 26 April 2017 - 11:54 AM

The tortoise rocket clock would record an elapse time of 377.47 years...

How did you get that?

I had an exam today and in the exam a question was this, "A 50 year old tortoise jumps onto a rocket and goes to a distant planet 370 lightyears away, the rockets speed was 0.7c . If a tortoise lives 450 years in avarage, would he live till reaching the destination or not?" cruel question I know, but i calculated like this that relativity won't affect the tortoise and it'd take him (370)©/(0.7c)=528 years, so he won't survive. Is my process correct? because as i understand it, for the tortoise everything is normal. All my friends have answered using time dilation. So I am really confused. Please help.

They got it wrong as well then. The distance traveled in space is shortened by the same amount traveled in time. Apply whatever length contraction is at 0.7c and then work work how long it would take to travel that distance then apply time dilation to the journey time. The tortoise doesn't feel time dilated so it that sense everything is normal. The tortoise should have plenty of time.

Edited by A-wal, 26 April 2017 - 04:20 PM.

#4 spartan45

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Posted 26 April 2017 - 03:34 PM

How did you get that?

Special relativity means moving clocks run slow. The time dilation equation is:

∆t0 = ∆t√ 1 – v2/c2

∆t0 = spacecraft time interval (in light years for this example)

∆t = Earth based time interval in light years (528.57 in this case)

v = spacecraft velocity 0.7c

c = speed of light (2.99792458x108 m/s)    (3.0x108) rounded    (Cancels out in this example)

∆t0 = ∆t√1 – v2/c2     ∆t0 = 528.57√1 – (0.7)2    ∆ t0 = 528.57√1 – 0.49         ∆ t0 = 528.57√0.51

∆ t0 = 528.57 x 0.7141428 = 377.47 years

#5 A-wal

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Posted 26 April 2017 - 04:20 PM

The tortoise does feel time dilated so it that sense everything is normal.

I meant: The tortoise DOESN'T feel time dilated so it that sense everything is normal.

#6 Darky

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Posted 04 May 2017 - 05:54 AM

Hi,

Mathimatically, the tortoise would have quite some time to explore the planet. However, time dilation is based on moving at fast speeds which can be caused by gravity. 0.7c is hardly fast enough to escape gravity of a black hole and if you're going to a planet 370 light years away, you are bound to run into a 100 different large black holes. In technical terms, he would be dead before he reached even 30% of his journey. To really go into depth, you'll need to know the route, the gravitational field Emmitt; only then will your answer me 100% correct.

#7 spartan45

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Posted 04 May 2017 - 09:46 AM

if you're going to a planet 370 light years away, you are bound to run into a 100 different large black holes.

I’m not so sure, the nearest black hole is 2,800 light years away in the faint Monoceros  (Unicorn) constellation, (viewed in the night sky between Sirius and Procyon), according to the reference below.

https://en.wikipedia...est_black_holes

#8 sluggo

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Posted 04 May 2017 - 11:15 AM

sparten45 gave the correct answer per SR and the initial conditions, which did not include black holes.

#9 A-wal

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Posted 18 May 2017 - 09:34 AM

sparten45 gave the correct answer per SR

No he didn't, he left out length contraction. To get the difference in the amount of proper time it would take to make the journey you need to work out out the difference in coordinate time over the distance in space.

Length contracts by the same amount as time dilates so it should take 270 years proper time.

#10 billvon

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Posted 18 May 2017 - 09:40 AM

Hi,

Mathimatically, the tortoise would have quite some time to explore the planet. However, time dilation is based on moving at fast speeds which can be caused by gravity. 0.7c is hardly fast enough to escape gravity of a black hole and if you're going to a planet 370 light years away, you are bound to run into a 100 different large black holes. In technical terms, he would be dead before he reached even 30% of his journey. To really go into depth, you'll need to know the route, the gravitational field Emmitt; only then will your answer me 100% correct.

No, there are no black holes that close to the Earth.

#11 spartan45

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Posted 18 May 2017 - 03:09 PM

No he didn't, he left out length contraction. To get the difference in the amount of proper time it would take to make the journey you need to work out out the difference in coordinate time over the distance in space.

Length contracts by the same amount as time dilates

This is interesting, as you have tackled the problem from a different perspective. Here is my reasoning looking at the problem from the point of view of length contraction.

L= L0√ 1 – v2/c2

L0 = Earth based distance in light years (370 in this case)

L = spacecraft measured distance (in light years for this example)

v = spacecraft velocity 0.7c

c = speed of light (2.99792458x108 m/s)    (3.0x108) rounded    (Cancels out in this example)

L= L0√1 – v2/c2     L= 370√1 – (0.7)2    L= 370√1 – 0.49         L= 370√0.51

L= 370 x 0.7141428 = 264.23 light years.

Using time=distance/speed   t=264.23/0.7 years

t = 377.47 years.

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#12 OceanBreeze

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Posted 18 May 2017 - 03:45 PM

This is interesting, as you have tackled the problem from a different perspective. Here is my reasoning looking at the problem from the point of view of length contraction.

L= L0√ 1 – v2/c2

L0 = Earth based distance in light years (370 in this case)

L = spacecraft measured distance (in light years for this example)

v = spacecraft velocity 0.7c

c = speed of light (2.99792458x108 m/s)    (3.0x108) rounded    (Cancels out in this example)

L= L0√1 – v2/c2     L= 370√1 – (0.7)2    L= 370√1 – 0.49         L= 370√0.51

L= 370 x 0.7141428 = 264.23 light years.

Using time=distance/speed   t=264.23/0.7 years

t = 377.47 years.

You are much too kind! A-wal didn’t tackle the problem at all, he just posted a number of 270 years, that is wrong without showing any calculations at all.

You, on the other hand have done your calculations properly, and your solutions of 264.23 LY distance due to length contraction, and 377.47 years time, due to time dilation, are correct.

One thing to keep in mind about this sort of problem is both observers (the one on earth and the one in the rocket) will not agree on the time or the distance, but they always agree on the velocity.

That gives you a quick and easy way to check your answer.

Since velocity = distance / time

For the earth observer: 0.7c = 370 LY / 528 Y

For the space farer: 0.7C = 264.23 LY / 377.47 Y

The velocities agree, as they must, so that is a good indication your answers are correct, and of course they are.

But don’t expect A-wal “Mr special relativity” to admit he is wrong. He never does and he never learns

#13 A-wal

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Posted 18 May 2017 - 05:06 PM

Right, length contracts by the same amount as time dilates so you get the same result. So when both are taken into account you get a proper time of 270 years for the trip, right?

#14 OceanBreeze

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Posted 19 May 2017 - 03:38 AM

Right, length contracts by the same amount as time dilates so you get the same result.

You get the same result for velocity, yes.

So when both are taken into account you get a proper time of 270 years for the trip, right?

No, Wrong. See here

For an object in a SR spacetime traveling with a velocity of v for a time Δ T  the proper time interval experienced is the same as in the SR time dilation formula, 377.47 years in this case.

Why do you think the proper time is different from the dilated time?

#15 A-wal

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Posted 21 May 2017 - 11:52 AM

The velocities agree, as they must, so that is a good indication your answers are correct, and of course they are.

But don’t expect A-wal “Mr special relativity” to admit he is wrong. He never does and he never learns

You get the same result for velocity, yes.

No, Wrong. See here

For an object in a SR spacetime traveling with a velocity of v for a time Δ T  the proper time interval experienced is the same as in the SR time dilation formula, 377.47 years in this case.

It shouldn't be, that doesn't take into account traveling the shorter distance caused by length contraction after accelerating.

Why do you think the proper time is different from the dilated time?

Time dilation is the coordinate difference between two inertial frames, proper time is the amount of time (dilated) that it takes to moves across length contracted space. This...

Special relativity means moving clocks run slow. The time dilation equation is:

∆t0 = ∆t√ 1 – v2/c2

∆t0 = spacecraft time interval (in light years for this example)

∆t = Earth based time interval in light years (528.57 in this case)

v = spacecraft velocity 0.7c

c = speed of light (2.99792458x108 m/s)    (3.0x108) rounded    (Cancels out in this example)

∆t0 = ∆t√1 – v2/c2     ∆t0 = 528.57√1 – (0.7)2    ∆ t0 = 528.57√1 – 0.49         ∆ t0 = 528.57√0.51

∆ t0 = 528.57 x 0.7141428 = 377.47 years

...is time dilation. This...

This is interesting, as you have tackled the problem from a different perspective. Here is my reasoning looking at the problem from the point of view of length contraction.

L= L0√ 1 – v2/c2

L0 = Earth based distance in light years (370 in this case)

L = spacecraft measured distance (in light years for this example)

v = spacecraft velocity 0.7c

c = speed of light (2.99792458x108 m/s)    (3.0x108) rounded    (Cancels out in this example)

L= L0√1 – v2/c2     L= 370√1 – (0.7)2    L= 370√1 – 0.49         L= 370√0.51

L= 370 x 0.7141428 = 264.23 light years.

Using time=distance/speed   t=264.23/0.7 years

t = 377.47 years.

...is length contraction. To get the proper time you need both. When the tortoise accelerates the distance between the starting point and the destination decreases and so does the coordinate time it takes to cover that distance. The amount of proper time is how long it takes on the watch of the of the tortoise.

Maybe the equations are taking both into account in both examples but that's not how it reads.

Edited by A-wal, 21 May 2017 - 11:54 AM.

#16 OceanBreeze

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Posted 21 May 2017 - 12:33 PM

I gave you the link to follow. If that doesn't convince you, I won't waste any more of my time on you.

$\Delta \tau = {\sqrt {\Delta T^{2}-(v_{x}\Delta T/c)^{2}-(v_{y}\Delta T/c)^{2}-(v_{z}\Delta T/c)^{2}}}=\Delta T{\sqrt {1-v^{2}/c^{2}}}$

As you can see, the last part of the equation for proper time, is exactly the same as the expression for dilated time in SR.

Edited by OceanBreeze, 21 May 2017 - 12:46 PM.

#17 A-wal

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Posted 21 May 2017 - 01:30 PM

But this...

This is interesting, as you have tackled the problem from a different perspective. Here is my reasoning looking at the problem from the point of view of length contraction.

L= L0√ 1 – v2/c2

L0 = Earth based distance in light years (370 in this case)

L = spacecraft measured distance (in light years for this example)

v = spacecraft velocity 0.7c

c = speed of light (2.99792458x108 m/s)    (3.0x108) rounded    (Cancels out in this example)

L= L0√1 – v2/c2     L= 370√1 – (0.7)2    L= 370√1 – 0.49         L= 370√0.51

L= 370 x 0.7141428 = 264.23 light years.

Using time=distance/speed   t=264.23/0.7 years

t = 377.47 years.

... is time dilation alone, meaning that the journey in proper time would be 377.47 years if length in space weren't contracted, and this...

This is interesting, as you have tackled the problem from a different perspective. Here is my reasoning looking at the problem from the point of view of length contraction.

L= L0√ 1 – v2/c2

L0 = Earth based distance in light years (370 in this case)

L = spacecraft measured distance (in light years for this example)

v = spacecraft velocity 0.7c

c = speed of light (2.99792458x108 m/s)    (3.0x108) rounded    (Cancels out in this example)

L= L0√1 – v2/c2     L= 370√1 – (0.7)2    L= 370√1 – 0.49         L= 370√0.51

L= 370 x 0.7141428 = 264.23 light years.

Using time=distance/speed   t=264.23/0.7 years

t = 377.47 years.

... is length contraction alone, meaning that the journey in proper time would be 377.47 years if length in time weren't dilated.

Are you saying that both ways of doing it take both time dilation and length contraction into account? It doesn't look like it.