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Travel at a fraction of light speed. Subjective time question.


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#1 owl   Banned   -  Reputation: 364

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Posted 09 May 2011 - 06:01 PM

To keep the calculations simple, say that a traveler had a spaceship that could instantly accelerate to 1/2 of light speed (without killing him) and he started a journey to a point in space 10 light years away to finally instantly decelerate at the moment of arrival.

How many time would it pass from the point point of view of the traveler inside the spaceship?

Intuitive thinking may make one think it'd be 20 years but I know relativistic speeds don't work that way. What would be the equations to calculate that "t"?
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#2 Emergent   Members   -  Reputation: 967

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Posted 09 May 2011 - 06:11 PM

See this article. This is the equation you're after (explained in the article):

Posted Image.

(The "gamma" in the above equation is known as the Lorentz factor.)

#3 owl   Banned   -  Reputation: 364

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Posted 09 May 2011 - 06:26 PM

For the given values it would be:

10 / ( Root( 1-( (1/2)^2 / 1) ) ) = 11,5470

11 years and a half?
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#4 White Dwarf   Members   -  Reputation: 96

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Posted 09 May 2011 - 06:33 PM

To keep the calculations simple, say that a traveler had a spaceship that could instantly accelerate to 1/2 of light speed (without killing him) and he started a journey to a point in space 10 light years away to finally instantly decelerate at the moment of arrival.

How many time would it pass from the point point of view of the traveler inside the spaceship?

Intuitive thinking may make one think it'd be 20 years but I know relativistic speeds don't work that way. What would be the equations to calculate that "t"?


Correct me if I'm wrong, but I'm pretty sure that for the the guy in the spaceship it would be twenty years.

However, see this video for a simple explanation of time dilation. Skip to about 6:40 to avoid the junk.

Hope this helps!

Also a book called 'The Elegant Universe' explains this and many other related issues very well.

#5 Postie   Members   -  Reputation: 867

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Posted 09 May 2011 - 08:05 PM

From the traveler's perspective, the calculation is simple... Time = Distance / Speed... which is 10 / 0.5 = 20 years.

The time dilation effect occurs when you compare what an external observer sees versus what the traveler sees. In this example, more than 20 years will have passed for the rest of the universe. The reason for this is that the non-moving reference frame sees the traveler's clock running slow (the actual amount is given by the Lorentz factor as stated by Emergent). So 20 years for a slow running clock means more than 20 years will have passed for the rest of us.

However, no matter what the speed of travel is, the time taken for the journey from the traveler's point of view is the simple formula : Time = Distance / Speed.

Edit: The value of the Lorentz factor for the given setup is about 1.1547, so about 23.094 years will have passed. In your calculation you used the distance (10 light years) not the time (20 years).
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#6 owl   Banned   -  Reputation: 364

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Posted 10 May 2011 - 12:33 AM

no, wait I meant 10 years traveling at speed c (which still makes no sense I realise). The resulting value is the elapsed time for an observer outside the ship then

No I got that right.


Not sure about the formula Emergent provided though. As the traveler approaches the speed of light, the flow of time in his local frame decreases by which the object in which he is traveling, to him, appears to be moving faster than what it really is (covering more external distance in less time). So, his perception of the time passed since he started traveling will be inferior than the time that passes for an external observer.

I double checked this in Isaac Asimov's "Extraterrestrial Civilizations" non-fiction book (Chapter "The speed of light"). He mentions that at 293,800 km/s or 98% of the speed of light, time inside a spaceship will flow at 1/5 of the rate it would if it was at rest. He states then than for a distance of 10 light years at that speed, from the point of view of an external observer (at rest in relation to the ship, say the Earth) the travel would seem to last a little longer than 10 years but for the traveler on board the ship the travel would seem to have lasted just one week.

So, again, what's the formula with which I can calculate those numbers?
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#7 szecs   Members   -  Reputation: 2099

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Posted 10 May 2011 - 01:04 AM

Well, if he travels at 0.5*c, then 10 lightyears means 20 years from the observers POV. But the equation should be dt'/gamma because the distance is measured in the observer's coordinate system (which is considered standing: we defined distances in Space like that).

You can verify it, because the time from the traveler's CS should be less than the time in the observers CS, because we know (from sci-fies....) if the traveler would travel with c his time would be zero. (if you substitute v = 0, dt'/gamma will be zero as expected.)

so 17.32 years is the correct answer.

Another verification:

v/c = 0.98, dt' = 1 ->

dt = 0.198997 -> 1/5, see Asimov's

EDIT: edited dt' dt confusion. Emergent's formula is perfect, it's just we are not calculating dt', but dt.

#8 owl   Banned   -  Reputation: 364

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Posted 10 May 2011 - 01:11 AM

Thanks, szecs. Could you write the formula for me in pseudo-code so I can see it better? Just pretend I'm scientific-notation blind :)
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#9 szecs   Members   -  Reputation: 2099

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Posted 10 May 2011 - 01:16 AM

20 * ( Root( 1-( (1/2)^2 / 1) ) ) = 17.32

or

time_for_traveler = time_for_observer * sqroot(1-(v/c)^2)

just pretending I'm not having sense of humor (which I'm really not having..)

#10 owl   Banned   -  Reputation: 364

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Posted 10 May 2011 - 01:22 AM

20 * ( Root( 1-( (1/2)^2 / 1) ) ) = 17.32

or

time_for_traveler = time_for_observer * sqroot(1-(v/c)^2)

just pretending I'm not having sense of humor (which I'm really not having..)


But, but, that doesn't very for the numbers provided by Asimov:

10*(Root(1-(293800^2/299792^2))) = 1,9893 years (he states 1 week)

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#11 szecs   Members   -  Reputation: 2099

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Posted 10 May 2011 - 01:24 AM

A more interesting question:
How would we relativistically simulate a space-fighting game with all these time dilatations in a multiplayer game? All controls/the behavior of the spaceship would slow down compared to the outside scene? That is an interesting game design question....

#12 szecs   Members   -  Reputation: 2099

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Posted 10 May 2011 - 01:25 AM


20 * ( Root( 1-( (1/2)^2 / 1) ) ) = 17.32

or

time_for_traveler = time_for_observer * sqroot(1-(v/c)^2)

just pretending I'm not having sense of humor (which I'm really not having..)


But, but, that doesn't very for the numbers provided by Asimov:

10*(Root(1-(293800^2/299792^2))) = 1,9893 years (he states 1 week)


Well, you said 1/5 time at first then the week thing. It's obvious, that 1 week is not 1/5 of 10 years .... What is the correct context in Asimov's?

10 ly -> dt' = 10/0.98 = 10.2 years (Earth's CS).
dt = 2,0298 years (traveler's CS).

#13 owl   Banned   -  Reputation: 364

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Posted 10 May 2011 - 01:32 AM



20 * ( Root( 1-( (1/2)^2 / 1) ) ) = 17.32

or

time_for_traveler = time_for_observer * sqroot(1-(v/c)^2)

just pretending I'm not having sense of humor (which I'm really not having..)


But, but, that doesn't very for the numbers provided by Asimov:

10*(Root(1-(293800^2/299792^2))) = 1,9893 years (he states 1 week)


Well, you said 1/5 time at first.... What is the context in Asimov's?


Wait, I missread, the velocity he is talking about is 299791. One kilometer per second less than the speed of light. It does verify then.

Thanks szecs. Thanks Emergent!
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#14 szecs   Members   -  Reputation: 2099

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Posted 10 May 2011 - 01:35 AM

299792/299791*10*(Root(1-( 299791^2/299792^2))) = 0,02583 years
which is 1.3 weeks...

EDIT: I'm working, so..... anyway, fixed it: fucked up the first two numbers.

#15 owl   Banned   -  Reputation: 364

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Posted 10 May 2011 - 01:42 AM

299791/299792*10*(Root(1-( 299791^2/299792^2))) = 0,02583 years
which is 0.02583 1.3 weeks...


Yeah, Emergent provided the right formula. It was just that I didn't get it right. Not a surprise tought :)

EDIT: Then Asimov adds:

In a lapse that would appear of 60 years to them (the travelers) they'd reach the Galaxy of Andromeda, which is located at 2,300,000 light years from us.



Then he states that if the energy obtained from fusion of hydrogen was used as a propeller, 3,500 tons of hydrogen would have to used in the fusion to accelerate (at a constant 1g) 1 ton of matter to 98% of the speed of light

Kind of hard. Yet SO COOL! :)
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#16 owl   Banned   -  Reputation: 364

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Posted 10 May 2011 - 12:47 PM

A more interesting question:
How would we relativistically simulate a space-fighting game with all these time dilatations in a multiplayer game? All controls/the behavior of the spaceship would slow down compared to the outside scene? That is an interesting game design question....


Look at that, I managed to miss this post and still rate it up. I think I was tired.

Last night I kept reading a little more and learned that for a spaceship traveling almost at the speed of light, all the radiations that arrives to the spaceship would be so shifted to the extremes (X-rays, ultra-infra-red) of the spectrum that nothing from the outside would be visible from the inside. All the free non dense elements floating in space, would be (relatively) so accelerated and condensed that they would behave like an enormous flood of cosmic rays going through the spaceship messing everything up inside. Some scientist stated that even at a 10nth of the speed of light, such levels of radiation would be unstoppable, making traveling faster than that, virtually impossible.

To make something realistic, there should exist a realistic way of solving all of the above.
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#17 szecs   Members   -  Reputation: 2099

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Posted 11 May 2011 - 01:15 AM

You can ask someone to rate that post down for you...

Anyway, take a look at Carl Sagan's Cosmos, it's awesome

#18 forsandifs   Members   -  Reputation: 154

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Posted 11 May 2011 - 03:31 AM

[EDIT: This post was based on a misreading of "years" as "light years" in the OP and therefore doesn't answer the OP properly. My post after the next is the correct one form me because it calculates for a journey that is 10 light years long as opposed to 10 years long]

The observer on earth will have one clock, named cE, and the traveller, will have another clock named cT. These are the only clocks we are interested in and thus they are the only way we have of measuring the time passed.

The question is, what do these clocks say after the journey is over? But the question is ambiguous as to how long the journey actually takes and which clock we are interested in looking at. The journey takes 10 years, but according to which clock?

We now use the formula quoted by Emergent:

Assuming that we mean it takes 10 years on cE then cT reads ~8.66 years. In other words the traveller would be 8.66 years older whilst people on earth would be 10 years older.

Assuming that we mean it takes 10 years on cT then cE reads ~11.55 years. In other words the traveller would be 10 years older and people on earth would be 11.55 years older.

EDIT: owl, in other words your first calculation was correct.

#19 szecs   Members   -  Reputation: 2099

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Posted 11 May 2011 - 03:43 AM

in other words your first calculation was correct.


Well, no, because lightyears as distance is defined in the Earth's coordinate system (not exactly, but you know what I mean). So 10 years means 10 years in cE not cT.

Plus I he calculated with 10 years and not with 20.


x = distance

dt' = x/v.

if x is in lightyears that means

dt' = c/v * x [years]

dt = dt' * sqrt(1-(v/c)^2)

so

dt = x/v * sqrt(1-(v/c)^2)

if x is in lightyears:

dt = c/v * x * sqrt(1-(v/c)^2) [years]

#20 forsandifs   Members   -  Reputation: 154

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Posted 11 May 2011 - 03:49 AM

in other words your first calculation was correct.


Well, no, because lightyears as distance is defined in the Earth's coordinate system (not exactly, but you know what I mean). So 10 years means 10 years in cE not cT.


Oh I miss-read years instead of light years. My bad. I'll repost when I've taken that into account.




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