Time taken to travel to the andromeda galaxy, oserved from earth

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Homework Help Overview

The discussion revolves around a problem in special relativity, specifically regarding the time taken for a traveler moving at a significant fraction of the speed of light (0.82c) to reach the Andromeda galaxy, which is approximately 2.55 million light-years away. The problem involves understanding time dilation and how it affects the perception of time for both the traveler and an observer on Earth.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of time dilation and question how it affects the observer on Earth versus the traveler. There is discussion about the relevance of various parameters in the problem, such as velocity and distance, and whether the intuitive approach of dividing distance by speed is sufficient. Some participants express confusion about the relationship between the time experienced by the traveler and the observer on Earth.

Discussion Status

The discussion is ongoing, with participants sharing their thoughts and seeking clarification on the concepts involved. Some guidance has been offered regarding the application of the time dilation equation, but there remains a lack of consensus on the interpretation of the results and the implications of different observers measuring time differently.

Contextual Notes

Participants note that the complexity of relativity is contributing to confusion, and there is an acknowledgment of the challenge in reconciling intuitive understanding with the mathematical framework of the problem. The original poster and others express a desire for further explanations and clarifications regarding the principles of special relativity.

romanawgarlic
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Homework Statement


A traveller with mass m = 66,5 kg travels with a velocity of 0,82 c to the Andromeda galaxy which is a distance of 2,55·106 lightyears away. Light with the wavelength 459,4 nm is emitted from the galaxy. assume the galaxy doesn't move and use the exact speed of light.

How long time (in million years, year=365,25 days) will the journey take for an observer on earth?

Homework Equations


i don't really know...
possibly:
time dilation T = To / √( 1 - (v2/c2)

The Attempt at a Solution


first of all, id just like to confess that the whole relativity concept is, unsurprisingly, doing my head in a little. i can pretty much grasp the way the speed of light is a constant and that time slows down when things approach the speed of light, and I am usually ok with questions on the doppler effect, red/blue shift and relativistic energy, but I am fairly certain all of the pieces have yet to fall into place.

also, i know a lot of the info in the question isn't relevant (im guessing the velocity and the distance are the ones that actually matter) nonetheless, I am giving the entire question just in case.

Ok, so the part that's confusing me about this specific problem is the matter of what effect time dilation would have on the observer on earth. i understand that the observer on Earth would observe clocks moving slowly on the spaceship, but wouldn't that not affect the observer since he is outside the traveling object?

intuitively, i'd say that the answer is just 2,55·106 lightyears divided by 0.82 c, but this is obviously too simple.

my next instinct is to plug the values into the time dilation equation, but i would love to understand why I am supposed to do that. even then, i imagine I'm missing some aspects of the question...

obviously, its a little amateurish to try to do physics on instinct and intuition, but reading up on the subject has only made me more confused so far. would love any help on the subject, even if its just a link to yet another explanation on special relativity.
 
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romanawgarlic said:

Homework Statement


A traveller with mass m = 66,5 kg travels with a velocity of 0,82 c to the Andromeda galaxy which is a distance of 2,55·106 lightyears away. Light with the wavelength 459,4 nm is emitted from the galaxy. assume the galaxy doesn't move and use the exact speed of light.

How long time (in million years, year=365,25 days) will the journey take for an observer on earth?

Homework Equations


i don't really know...
possibly:
time dilation T = To / √( 1 - (v2/c2)

The Attempt at a Solution


first of all, id just like to confess that the whole relativity concept is, unsurprisingly, doing my head in a little. i can pretty much grasp the way the speed of light is a constant and that time slows down when things approach the speed of light, and I am usually ok with questions on the doppler effect, red/blue shift and relativistic energy, but I am fairly certain all of the pieces have yet to fall into place.

also, i know a lot of the info in the question isn't relevant (im guessing the velocity and the distance are the ones that actually matter) nonetheless, I am giving the entire question just in case.

Ok, so the part that's confusing me about this specific problem is the matter of what effect time dilation would have on the observer on earth. i understand that the observer on Earth would observe clocks moving slowly on the spaceship, but wouldn't that not affect the observer since he is outside the traveling object?

intuitively, i'd say that the answer is just 2,55·106 lightyears divided by 0.82 c, but this is obviously too simple.

my next instinct is to plug the values into the time dilation equation, but i would love to understand why I am supposed to do that. even then, i imagine I'm missing some aspects of the question...

obviously, its a little amateurish to try to do physics on instinct and intuition, but reading up on the subject has only made me more confused so far. would love any help on the subject, even if its just a link to yet another explanation on special relativity.

Your instincts are rather more accurate than you give them credit for. It's not a complicated question. You have the right procedure.
 
Your intuitive answer is not only simple, it's right (congratulations!). To see this, consider the following:
1. The observer on Earth sees the Andromeda galaxy as 2 550 000 light-years away.
2. The observer on Earth sees the traveler moving at a speed 0,82 c towards the Andromeda galaxy.
3. The speed of something, as measured by any observer, is just the distance traveled (as measured by that observer) divided by the time interval (as measured by the same observer).

Different observers measure different distances, time intervals, and speeds (for that you need relativity), but we're only talking about one observer (and one who isn't changing motion by firing rocket engines to turn around or something similar).

Therefore: the observer on Earth measures a time (for the trip out to the Andromeda galaxy, not the round trip) of 2 550 000 c year/0,82 c = 2 550 000/0,82. For a round trip (to Andromeda and back), the Earth observer would measure a time twice as long, assuming the traveler turned around as soon as she or he got to Andromeda and returned at the same speed.

The traveler would measure a different time interval (for the traveler, it's the galaxy that moves toward the traveler, but because of the Lorentz constraction, the traveler measures a shorter distance between Earth and Andromeda than we (on Earth) do.
 
Dick said:
Your instincts are rather more accurate than you give them credit for. It's not a complicated question. You have the right procedure.


well, now that I've typed out my thoughts i realized more specifically what my question is: I'm guessing i'd get the time from the velocity times distance, but which side of the equation would that time go to? would that be the T or the To ?

i imagine the To, bbecause the To in the equation refers to the time one would experience on the spaceship, and that's the way these questions are usually formulated. this would imply that the time experienced on the spaceship would be the time one would expect the journey to take if we were dealing with normal speeds and distances, and the time experienced on Earth would be longer. why is it not the other way round?

oh, and thanks for the help so far. i reckon ill be able to do the question now, however, if you have the time i would love some explanation as to why this is so. :)
 
romanawgarlic said:
well, now that I've typed out my thoughts i realized more specifically what my question is: I'm guessing i'd get the time from the velocity times distance, but which side of the equation would that time go to? would that be the T or the To ?

i imagine the To, bbecause the To in the equation refers to the time one would experience on the spaceship, and that's the way these questions are usually formulated. this would imply that the time experienced on the spaceship would be the time one would expect the journey to take if we were dealing with normal speeds and distances, and the time experienced on Earth would be longer. why is it not the other way round?

oh, and thanks for the help so far. i reckon ill be able to do the question now, however, if you have the time i would love some explanation as to why this is so. :)

to clarify: i'd have thought the time one would expect to experience from Earth would be the distance times velocity, and the time experienced on the spaceship would be shorter. why is this not so (phew, its hard to express oneself accurately in words about this sort of thing!)
 
511keV said:
Your intuitive answer is not only simple, it's right (congratulations!). To see this, consider the following:
1. The observer on Earth sees the Andromeda galaxy as 2 550 000 light-years away.
2. The observer on Earth sees the traveler moving at a speed 0,82 c towards the Andromeda galaxy.
3. The speed of something, as measured by any observer, is just the distance traveled (as measured by that observer) divided by the time interval (as measured by the same observer).

Different observers measure different distances, time intervals, and speeds (for that you need relativity), but we're only talking about one observer (and one who isn't changing motion by firing rocket engines to turn around or something similar).

Therefore: the observer on Earth measures a time (for the trip out to the Andromeda galaxy, not the round trip) of 2 550 000 c year/0,82 c = 2 550 000/0,82. For a round trip (to Andromeda and back), the Earth observer would measure a time twice as long, assuming the traveler turned around as soon as she or he got to Andromeda and returned at the same speed.

The traveler would measure a different time interval (for the traveler, it's the galaxy that moves toward the traveler, but because of the Lorentz constraction, the traveler measures a shorter distance between Earth and Andromeda than we (on Earth) do.

ooops, missed you there, sir/mam! thank you, that confirmed my suspicions and made everything a whole lot clearer! sneaky, sneaky question...
 
romanawgarlic said:
to clarify: i'd have thought the time one would expect to experience from Earth would be the distance times velocity,
umm... divided by
and the time experienced on the spaceship would be shorter. why is this not so
What will be the distance as perceived from the spaceship?
 

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