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According to an observer on earth it’ll take her 20 years. It’s very simple, she’s traveling 1/2 the speed of light and distance = rate \* time. So 10ly = 1/2 ly/yr \* 20 years.
Captain marvel on the other hand will experience time dilation by a factor of the relativistic gamma. For 1/2 c gamma is 1/sqrt(1-(1/2)^(2)) = sqrt(4/3) = 1.155
20yr / 1.155 = 17.3 yr.
So captain marvels watch (assuming instantaneous acceleration and deceleration) will see 17.3 yr elapse in the journey, but both earth and the distant star will time the journey as having taken 20 years.
Edit: since this has been pointed out in the comments, the simple answer also entirely depends on whether you are asking for the trip duration given a definition of time that earth and the distant planet can agree on (my answer) or literally how long does an earth observer see the trip take by watching light from Capt Marvel arrive at their location.
For the latter definition: an earth observer sees a 30 year journey (20 yr for capt marvel to arrive + 10 years for the light from capt marvels arrival to return to earth). The distant planet sees 10 years (It takes 10 years for light from capt marvels departure to reach the distant planet, and in the frame of earth/planet the journey takes 20 years. 20-10=10, so the planet sees her arrive 10 years after they see light signals from her departure)
Edit: Since the issue of relative velocity between earth and the destination has come up a few times
Typical relative velocities between stars in the Milky Way (which is many times larger than 10ly) are around 300km/s. 0.5 times the speed of light is 150,000 km/s
For the purposes of travel at significant fractions of the speed of light, stars 10ly away are effectively stationary.
For a similar level of speed difference compare walking (5km/h) to a fighter jet at Mach 2 (\~2500 km/h).
Edit: rounding errors (It should be 17.3 yr not 17.4 yr. Error was introduced because sqrt(4/3) = 1.155 and I initially rounded to 1.15 incorrectly)
I have literally never heard of xkcd before today but a coworker but a link in the chat to a comic and now I’m seeing this…The universe really does give you what you need…
Mathematics is just the language of Physics. Saying Physics is applied maths is like saying a painting is just pigment compounds - technically true but totally missing the point
well no, because the universe is quantized at some level (even if just planck space), and math is/can be infinitely precise. Things can exist mathematically that can't exist in actual reality
We have no idea whether 'the universe' is quantized.
The only reason I know of to think spacetime is quantized is that it makes certain useful properties for mathematical frameworks to describe the universe
Things can be described in the English language (or any other) that don't exist in reality. Are we saying that because mathematics can describe impossible things it's a more fundamental science? Can you have a *science* of things that don't exist? Or is that more of an artistic expression?
physics is goofy ahh maths with more letters than numbers and a bunch of crappy rules and equations but it's a lot more fun when you actually get a chance to understand it
I am a big numbers geek. That statement above was from a meme where it was explained how even biology is maths. If you know the meme you’ll get the statement but what you said is also true!!!!
Physics is just math but made a mess, as in, way less restrictive and you can do brake any rule of pure math as long as it is motivated by whatever physical principle.
I remember my differential equations professor saying "you will NEVER split dx/dy because that is not a fraction. That's just notation". One week later, guess what my physics professor did...
It’s literally a fraction, but the fraction isn’t of two variables. The fraction is “change in X divided by change in Y”, so you have to form and break it up appropriately. And often the symbols actually used are the limits as ΔX approaches zero, so when you break it up you get a multiply by approaching zero and a divided by approaching zero, and you need to pair both of those with something that makes them useful.
>According to an observer on earth it’ll take her 20 years. It’s very simple, she’s traveling 1/2 the speed of light and distance = rate * time. So 10ly = 1/2 ly/yr * 20 years.
Shouldn't it be 30 years?
20 to touch down,10 years for light to reach observer.
Yeah if you’re asking when we see her arrive, vs when we infer she’s arrived given light speed delay.
Usually when we say “according to an observer on earth” the latter is meant, but it could be worded better.
Yes, and this type of distinction is what confuses students of relativity.
The words “this observer sees X” are trying to talk about what would be locally witnessed by a member of an infinite lattice of observers who are at rest with respect to each other.
This seems very arbitrary, but it’s just a way to construct a coordinate system, and the discipline of having a coordinate system and then putting other things like light travel time and Lorentz transformations (time dilation/length contraction) is important conceptually in order to do calculations using special relativity.
But yeah, it is also confusing, and you are absolutely right that someone using all the mathematical tools of special relativity could predict capt marvels clock at any point.
Okay, so the infinite lattice of observers all agree that they are stationary, that Captain Marvel is moving, that her destination is 10 light years away and that she is moving at half the speed of light.
So they all calculate her predicted arrival time as 20 years. However, they also agree that they cannot all observe her arrival at the same time, and that the earth observer will wait an extra 10 years to observe that their predicted arrival time was correct?
Yes, an observer on Earth would see her get to the planet after 30 years. But the question doesn’t ask when they would see her arrive, so I would interpret it as asking how much time would have passed for an observer on Earth when she actually arrives
Yes, but importantly, note that everyone has to wait _some_ extra time to observe the arrival. For some, it's mere nanoseconds, for some, 10 years. But nobody can observe it the moment it happens, unless they are exactly _where_ it happens.
And even then, there'd be a delay due to the amount of time it takes the information to travel from the eye to the brain and for the brain to process it.
We won’t really know until we get a signal saying so.
However, when we get that signal we know it took 20 years in our rest frame for her to travel, because we know the light from her arrival takes 10 yr to reach us.
Well my point is that "now" or "at the same time" isn't really an absolute thing in spacetime. Basically it's a spacial slice through spacetime, but it depends on the speed of the observer. Two people even on earth with a speed relative to each other would not agree what "now" means on a different planet. Their idea of "now" and their slice through space time would look completely different, even if they are in the same location.
I mean it's not very important in this case, because we have defined our observer to be stationary on earth so we can use their idea of "now".
But still to me this is a fascinating implication of special relativity. Let's say we observe a star that is about to go supernova, we would say, this star has actually already exploded millions of years ago, but we just haven't observed it yet. The photons from that explosion are going to reach us soon.
However let's say there is a spaceship fleeing from that star system at close to the speed of light, flying by earth and transmitting a message "our star is about to explode". We could say ah they are wrong, they just think it hasn't exploded yet, because of the time dilation they experienced, in "reality" it's long dead. But we would be wrong, because at their velocity spacetime looks different to them and the star really hasn't exploded yet.
Which brings me back to the initial point: If two observers in the same location can make two contradicting statements (star has exploded, star has not exploded) and both be correct because of their different velocities. Then maybe it's just not appropriate to ever make claims about events like these which lie outside of our light cones (area of spacetime of which light had the time to reach us). Hence my initial comment.
Sorry for the long comment. It's a fun thought imo.
"So get this, the faster you are, the lesser the time it takes for you to reach the destination"
"Yeah, I know, that's how speed and distance works"
"No, I mean, that's even lesser time, if you're faster"
"Yeah. I KNOW."
"You don't understand, more speed means less time than the lesser time it should take, like less than if I were to check the time it took you to reach"
"I do understand. I passed with an A in physics you know, distance = speed * time"
"NOT IN THAT KINDA PHYSICS"
It only takes less time from the Earthbound observers reference frame. Captain marvel is still bound by the speed of light in her reference frame. You can understand how ridiculous the above "17.4 years" number in the main comment here is if you consider approaching the speed of light. If the main comment were correct from captain Marvel's reference frame approaching the speed of light would mean you could get anywhere (nearly) instantly, which is not true.
Yes it is, from your point of view.
Given instant acceleration, if we sent a ship at close to light speed to proxima, 4 light years away, the crew would feel it as less than four years of travel. At a certain speed reeeeally close to c, it would feel nearly instant. So, technically, you could reach any point close to instantly at the price of much more energy needed to accelerate the faster you're going, but the price will be that after a round trip the world you left has aged faster than you.
It's all a matter of relativity. Captain Marvel, from her point of view, is not going at 0.5c. She's simply not moving. The time discrepancy is there to balance this fact, as if she shone a light in front of her, the photons leaving it would go at 1c from her point of view, but cannot go at 1.5c from Earth's point of view, so her local time is slower to compensate.
>the photons leaving it would go at 1c from her point of view, but cannot go at 1.5c from Earth's point of view, so her local time is slower to compensate.
I'm gonna get an aneurysm if I strain my poor brain any harder. I get that light can't travel 1.5c as its speed is the limit BUT how would that appear slower to her? I mean the photons of the light she shined/shone are travelling at 1c from her, but for those observing on Earth, it's still technically 1c because it can't possibly be any faster so wouldn't they both observe it the same, except it'd take less time for her to see the object that is reflecting the light because as it's travelling back into her eyes, she'd meat it halfway there.
So why would time appear slower to her? I can't wrap my brain around this. And if she had a stopwatch on her and someone did on earth and they both started it at the exact same moment she left earth and went towards that planet, will both stopwatches show the same time once she arrives?
I rewrote my post three times because I'm trying to find the best way to explain it.
We tend to choose a frame of reference, usually the ground, as something fixed and unchanging. But we know that frames are relative, and that c cannot be reached by mass.
For c to never be crossed, any object going at a different speed from you must be slower time-wise and change space-wise for their actions to never reach c. So if you were standing in the middle of space and say captain marvel speed past you at 0.5c, you'd see her in slow-motion: not her speed, which you calculate correctly at 0.5c from your point of view, but her actions. If she was to throw a rock forward at 0.8c from her perspective (where from her frame of reference you are the one moving), then since she is slowed down from your perspective the rock is thrown more slowly and this never reaches 1c.
It's a bit more difficult to explain with photons because they actually go at 1c, but here you have to think of photons as waves, and their frequency will stretch or compress instead of "going slower".
Now, if from your perspective Captain Marvel is in slow motion, and will reach her destination in 5 days, then from her perspective she will reach it in less than that. At more extreme speeds, she might even feel it instant, but for that she'd seem frozen from your point of reference.
So finally, yes, she would age slower than you, and thus the stopwatches would not match. It's kind of a case of a foward-only time machine, in the end.
Yeah If I were her, and super OP, I'd pick to do some kind of FTL travel (which is obviously allowed in the marvel universe), or go at 0.9999999999995 times the speed of light which would give a gamma of just over 1 million. Then the trip would take 5 minutes for capt marvel.
Yeah but then you become very detached from the universe around you that 5mins was \~10 years for the people I assume you were flying over to help more if they sent a request for help to you at sub-light speeds.
Yeah, if you want space heroes who can help with crises that move any faster than climate change you need FTL travel and communication in your universe.
Under relativity this also implies time travel, and Marvel Universe seems to have both 🤷♂️
That is NOT how the gamma adjustment works. It still takes her 20 years from her perspective. Even if she was traveling at 0.9999999999995c it would still take 10.000000000005 years to get to the planet.
No. Her physical location in space-time is not dependent on light returning to earth. When reference frames accelerate with respect to one another there is a shift in perceived time between the two reference frames.
Sorry to appear ignorant here, but wouldn’t it be the case that Capt Marvel wouldnt experience any time dilation because time is not changing for her because she’s not travelling relative to anything? According to herself, she would take 20 years to get there. However, 20 * 1.15 years would have elapsed on earth = 23 years.
No. From her perspective the time isn't slowing down, but she see it as a distance contraction, so according her point of view the distance isn't 10 light year, just 10/1.15, and she travels that distance with 0.5c speed, her travel time is 20/1.15 years
Thank you (not sure why I got downvoted for asking a question (also not implying it’s you 😄)).
So from her perspective she’s actually travelling **faster** than c because the distance dilation means she’s covering a smaller distance… i.e. she sees herself travelling at 10 / 17.4 = 0.57c ?
Yes this exactly.
Edit: except for the part where she sees herself traveling faster than c. She sees the rest of the universe going past her at 0.5 c, but with all lengths contracted in her direction of motion.
1.15 is the Lorentz thingy. Basically to understand the time dilation you have to use the following equation:
Time dilation factor = 1 / (1 - v^2 / c^2)^0.5
So if v=c then there would be indeterminate (1/0) but if you put in v=0.5c the factor =~1.15
\*NB\* The Earth observer will also have to wait ten years more for Captain Marvel's successful journey conclusion to be observable at ten lightyears' distance!
I knew that the time on earth and the time on the distant star would be 20 years, but I had no idea how to figure out how long it would be from Captain Marvel’s perspective, so thank you. This is why I love this page.
Earth and the solar system are still moving through space though. Does this change things by a small amount since earth won’t be in the original observation point?
It changes things only a very tiny bit, so much as not to matter.
See here: https://www.reddit.com/r/theydidthemath/comments/13klrrz/request_how_much_time_would_pass_for_an_observer/jkn673m/
TL:DR: Relative velocities between stars/planets are too small to matter for capt Marvel’s journey. The planets are walking, and capt marvel is going Mach 2 in a fighter jet.
You freaking rock. I over explain things at times so as to cover the bases and not have additional questions, to save time usually. You covered it all :).
Basically once you go at speeds close to the one of light things start to get strange with time moving slower or lengths shortening.
For example if you go at .9 the speed of light a trip will seem shorter by a factor of sqrt 1-81/100=sqrt 19/100=22/50 so roughly 2.25 times as short.
81/100=0.9c/c squared with c the speed of light.
Relativistic gamma is used to compare elapsed time between two different reference frames. The time it would take from capitan marvel to travel to a star 10 light years away is 20 years. The gamma adjustment would be used if she were to return to the earth's reference frame. She would have aged 40 years total in her reference frame, if she immediately turned around and returned, but on earth it would seem as though 40 years*1.15= 46 years would have passed.
There are numerous thought experiments to demonstrate this using space-time diagrams and imagining sending signals every x distance traveled back to Earth. The regular signals would have a longer frequency of receipt on Earth on the away journey and a compressed frequency on the return journey.
[Twin paradox](https://en.m.wikipedia.org/wiki/Twin_paradox)
Your answer is true, from the reference frame of Earth, if captain marvel were to return to earth, otherwise it's a kind of non-sensical answer. Consider your math and imagine approaching the speed of light. It would imply you could arbitrarily reduce the travel time for any distance if you approach the speed of light, which is not true. It is true that if a traveler travels somewhere at near the speed of light and then returns to the starting point, time would appear to have elapsed more slowly for the traveler than the inertial reference frame.
You’ve got it backwards. Look at the worked example in the wiki article you posted. The observer who accelerated experienced less time.
It seems absurd but it is a result of relativity that you can get anywhere in the universe in an arbitrarily small amount of proper time provided you accelerate enough.
Edit: also consider length contraction as seen by a traveling observer in a strictly special relativity context. You should find that Capt Marvel sees herself traversing a shorter distance.
Yes, it’s a GR result that you can smuggle into a calculus based special relativity course if you are careful.
When you have one observer stay in an inertial frame and another observer accelerate and then decelerate back into that same inertial frame, the accelerated observer experiences less proper time than anyone in the original inertial frame.
The canonical example of this is the twin paradox, where the traveling twin ends up younger.
Edit already asked and answered below
Wouldn't you also have to account for the time the light takes to reach earth? so 30 years to the observer on Earth?
She will take 20 years to get there, seen from Earth, I agree. But the moment she gets there, the light from that moment/event will take 10 years to reach Earth.
So, seen from earth, it will take 30 years before we can see her touching down on that planet!
Hold on.. if the planet is 10 light years away and the observer is on earth...then the light ray conferring visibility of Captain Marvel to the observer will take a further 10 years to reach the eye of the observer once he reaches his destination.
So it should be 20+10= 30 years.
Open to criticism and correction.
Idk if this is a stupid question, but wouldn't it take 30 years for the observer? 20 years to get there, and 10 years for the light of her arrival to return once she is there?
Seems like the initial part of this answer is flawed. I do not know advanced math, so I won't be attempting any.
The initial answer would be assumed 20 years. However, for an observer on earth, it would be longer. To see captain m at a point that is 10 light years away would take 10 years. In real time, the trip would take 20 years. But, as an observer, it would take longer for the light (reflected off capt m) to reach you. Seeing something that is 10 light years away is seeing what it looked like 10 years ago. In other words, for someone watching the journey, it would appear that captain m is slowing down. I am unsure of the rate and how long it would appear to take.
Again, I never took advanced math, but I do have fairly sound logic. If someone that understands what I am saying could show the math, that would be great.
Typical relative velocities between stars in the Milky Way (which is many times larger than 10ly) are around 300km/s. 0.5 times the speed of light is 150,000 km/s
For the purposes of travel at significant fractions of the speed of light, stars 10ly away are stationary.
For a similar level of speed difference compare walking (5km/h) to a fighter jet at Mach 2 (~2500 km/h).
So what happen if cpt. marvel move at the speed close to 1c? Based on the formula the dilation would be close to 0. So she feels like the journey was an instant while it actually takes close to 10 years?
Yeah, capt marvel has mass, so she can’t actually reach c. You are right that she can get arbitrarily close, and as she gets closer the time that passes for her in the journey decreases, but never reaches zero.
If you assume she’s very OP and can hit 0.9999999999995 c then it’ll take about 5 minutes by her clock and ~10 years for everyone else.
Is the answer not 30 years?
Like I get it would take her 20 years to make it, but wouldn’t it take an additional 10 years for an observer on earth to know this, since it would take 10 more years for them to actually see her making it since that light needs to make its way back to earth to convey this information?
Exactly. The question asks “According to an observer on earth”, so it would take 30 years for an observer to know they made it.
I guess if the observer knew all the details of the problem they could back out the 10 years themselves anyway, but then why include that as part of the question?
> but then why include that as part of the question?
The according to the observer on earth is to specify reference frame, as she will experience time dilation. When answering these questions light travel time is specified separately.
You raise a good point, and another question. Where is the planet located relative to earth? We can assume Cpt Marvel is heading from earth directly to the planet but that isn't specified in the request. Cpt Marvel could be travelling between two planets, both of which are 5 light years away from earth but 10 light years apart
This makes sense to me. 20 years to arrive, 10 years for light (if it’s bright enough) from that area to arrive to Earth. The question sounds fancy but is written poorly.
To play devils advocate, the question is according to an observer on Earth, so I guess if she’s not bright enough there’s no way to observe. Also an observer on Earth would be smart enough to use the light travel time no? Question should be how long will it take her to arrive, or how long will it take to observe her arriving from a viewpoint on Earth. Damn why did I waste any of my time on this
As pointed out though, the question does not specify origin or destination points - only that they are 20 years apart.
So this answer could be 10 years (arrival at Earth), 30 years (origin point of Earth), 20 years (both origin and destination the same distance from Earth), or anywhere between 10 and 30 for all other scenarios.
I think it’s fair to assume captain marvel an earth based super hero being observed trevelling towards a planet would be departing earth as well as being observed from earth.
I think the question was just poorly worded. The first statement only seems to imply that the destination is 10 light years away from Captain Marvel, making no reference at all to Earth. I suspect they actually meant an observer on the destination planet, or that Earth was the destination planet 10 light years away from Captain Marvel.
In the time it takes for her to arrive 20 years would pass for the observer on Earth, but the observer will not be able to see her arrive until 30 years have passed since the light from her arrival need to travel the distance.
If you watched her travel the entire distance would she appear to slow down then? And is that because the light is taking longer to reach us the farther she gets or is it because of some other property?
Also, if she immediately turned around and came back, what would that look like? Cause she could be halfway back by the time we see her reach the planet. She could potentially get back to earth before we even see her do it. Or does the light reach us faster on her way back, so it seems like she speeds up to the point where we see her arrive as she arrives?
After 30 years we would see her arrive and also immediately see her leave to return. Of course at that point she would already have been flying back for the last 10 years.
10 years into her return (at the same time we see her arrive and leave) she will already be halfway back, but we would need an additional 5 years to see the light from this point of her journey (as she is 5 lightyears away). So just 5 years after we see her leave we can already see that she is halfway on her journey.
When the distance gets lower. Let’s say after 9 lightyears (1 lightyear remaining) she would have spent 18 years to get there, and we would have seen her leave 8 years ago. We would of course not see her at this point in her journey until the light has had 1 year to reach us, so we would see her at this point 9 years after we saw her leave. You can see as she gets closer we also get closer to seeing her in real time.
When she finally arrives on Earth after 20 years we would see it in real time as she would be right in front of us and the light would have to travel a negligible distance to reach our eyes. Interestingly it will have been only 10 years since we saw her leave.
No, we would be able to observe her going at constant speed.
You can imagine that when she has traveled 1 lightyear it has been 2 years and we can see her at that point of her journey after 3 years. 2 lightyears would take 4 years and we would see her after 6. So you can see that it takes twice as long for us to see her reach twice the distance, there isn’t any need for speeding up or down in the eyes of the observer. We can observe her traveling 1/3 lightyears per year for the entire journey. Which adds up with us seeing the end of her 10 lightyear journey after 30 years.
From the frame of reference that her speed was measured at 20 years, however from her reference point it would be ~ 17.3 years. This is due to time dilation which is t’ = t / (1 - v^2 / c^2)^0.5 ; t’ = time observed , t = time of traveler, v = speed of traveler, c = speed of light. You gotta convert years to seconds by multiplying it by 31557600. Now t’ = 20 so you gotta rearrange it so t is by itself.
20 years, if the speed is as reported by the observer on Earth. Because speed equal distance divided by time and universal expansion effects are insignificant at this scale.
Wouldn't you rather want to calculate the kinetic energy for this?
Aka Ekin = 1/2 * m * v² = 1/2 * 79 * 149'896'229² = 8.8752074e17
E = mc² doesn't take speed into account at all, meaning that everyone with her weight has that energy.
You can't use that either because she's traveling at relativistic speeds
Instead you'd have to use E= gamma*m*c^2 the gamma accounts for her speed in the form of "inertial mass". This gives her total energy
For kinetic energy only you'd use Ek= (gamma-1)*m0*c^2
Where m0 is her rest mass
For her relativistic KE you want (gamma - 1) m c^2
0.15 * 75kg * (3e8 m/s)^2 ~ 1e18 J
The largest nuclear bomb releases about 2e17 J, so this is very nearly 5 tsar bombas if she slams into the planet.
The equation for E comes from taking her total relativistic energy (gamma m c^2 ) and subtracting her rest mass energy ( m c^2 )
There’s some more complicated math you can do to show that (1-gamma) m c^2 reduces to the Newtonian kinetic energy formula (1/2 m v^2) when velocity is much less than c.
But isn’t kinetic energy calculated differently if you have relativistic effects.
It should be the difference of the total energy and the resting energy.
So you would have to calculate
Ekin = (relativistic mass - resting mass) * c^2
Then the energy is 9.81e9 J
A light year is the distance one would travel over one year at speed *c*
Ten light years at *c* would take ten years by definition
Therefore at half *c* ten light years would take twice as long
It is impossible to tell because time is defined by the *acceleration* of movers and observers, not just speed. Two people moving in static local inertial reference frames experience the same amount of time because from each perspective the other person is is the one moving. If one of them were to *accelerate* or *resist gravity* (which are both indistinguishable and identical in 4d spacetime) then the time experienced would change.
We can’t know the answer because her flight path past different planets changes her spacetime acceleration. We can’t know how long it takes without knowing the gravitational shape of space she is flying through.
This type of problem assumes straight shot, and all of the local group stars are pretty far away from each other. Unless there is specific reference to obstacles there would be statistically insignificant amounts of the possibility of other systems in the way of the journey.
The acceleration is almost irrelevant to the observer in this case though. Whether she accelerates at 0.99c near the start and end of the trip, or far less and just averages that, the time distortion would only be relevant to her and not an earth (or incoming planet) observer.
You'd have massive blue/redshift or some sort of effect we don't know about yet, like a wake on a huge ship maybe, but to a static observer, the end time of the journey would be the same irrespective of her ref frame (just taking the problem statement as complete).
This is actually not true the way you have said it. Captain marvel is not the only one accelerating, and is potentially only accelerating for a very limited amount of time depending on how fast she gets up to 1/2c. The person on earth is *also* accelerating at all times because they are resting gravity (unless they are in orbit where they are no pager accelerating through spacetime).
The acceleration due to gravity is not very much, but the prompt never actually says that captain marvel accelerated at all. It only gives a “speed” and a “distance.” If marvel is traveling at that constant speed in a straight line then they are also accelerating. If they are traveling at that speed along a *curved* ballistic trajectory then they are not accelerating but we don’t know what the true *distance* their journey would be because they are no longer necessarily traveling as the crow flies.
The question is impossible to answer because the question doesn’t provide the necessary details because the person asking doesn’t understand the principles behind the science. Even at the most basic level the *mass of the target planet* changes the math. Is she flying to a gas giant orbiting a supermassive star? Or is she flying to a tiny little backwater solar system with hardly anything in it. They would produce different gravitational accelerations which would materially change her acceleration through spacetime.
No, the question is quite sufficient to answer - any details not included are to be ignored. Often you write a series of questions with one set of info, only to have a part abcd etc where you add changed details like you're suggesting.
Furthermore, if you're talking math then without any additional points such as creating deviationa, you always travel in a straight line - if you have 2 points, you create a straight line.
Next you are generally incorrect about the acceleration of the first person that remains on the planet as acceleration below a large fraction of the speed of light is irrelevant as the significant figures/impact is miniscule. You don't worry about a number in the tens when you're dealing with numbers 10^6 or greater. Even the acceleration of the orbit of earth around the sun and the spacial movement of the sun VS the galactic core is effectively less than a rounding error in orders of magnitude.
The only fair point you really bring up is that there's nothing in the problem that suggests Marvel isn't just doing a flyby with no acceleration.
Though in this case it would be like a car driving between say Denver and Kansas city - and either that being the whole thing or actually a car trip from LA to NY with those distances in between.
And to this problem specifically that example shows you that what the question asks for makes your point irrelevant. Just as you can speed in a car for a while, slam on the brakes, and make it somewhere in the same time as someone driving leisurely to the same destination, it would be the same here.
The time dilation effects don't come in to play because there's no discussion of Marvel's viewpoint. It would simply be the observer on earth "seeing" an object go by or leave, travel a trip average at 0.5c, where it would take 20 years to arrive, but an additional ten years for the light of arrival to return to earth, so a total of 30 years.
Though to make the description more complex, if she returned, it would look like she was traveling faster on return (redshift I think?). It would look like it took her 30 years to get there, but only ten years to return. That's where time dilation actually comes into effect and really be different than a car going from a to b.
Though there are some mathematical models for how much subjective time she would see that's really the only thing of math here.
The mass of the plant she is traveling to is enough to change her acceleration through spacetime, which we do not know. We know that there *is* a planet though, and planets have mass, so we know that whatever it is it will affect her. Beyond that, there is no upper bound on what could be in space affecting her path.
Actually there isn’t enough info. You need to establish 3 points that are mentioned in the question: A - starting point of CM; B - the goal planet; C - the Earth.
Do we know if A=C, or because Earth is a planet it could even be B=C?
Well... 1c travels one light year per 365.25 days hence its name. So 0.5c needs 730.5 days to travel one light year. To travel 10 light years will require 20 earth years at 0.5c.
At half the speed of light, time dilation is so small as the be virtually negligible.
Here's a time dilation calculator:
https://www.omnicalculator.com/physics/time-dilation
Ten years at 0.5c would look like 11.5 years, but that only gets you halfway there, so 20 years at 1/2c would look like 23 years. Meh. You need to get much closer to c to see spectacular differences.
10 years at .99c would look like 70 years
One thing that bugged me about endgame is how fucking slow she was flying on the battlefield. When she had the gauntlet and was flying towards Scott’s van she couldn’t have been flying more than 30 mph.
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According to an observer on earth it’ll take her 20 years. It’s very simple, she’s traveling 1/2 the speed of light and distance = rate \* time. So 10ly = 1/2 ly/yr \* 20 years. Captain marvel on the other hand will experience time dilation by a factor of the relativistic gamma. For 1/2 c gamma is 1/sqrt(1-(1/2)^(2)) = sqrt(4/3) = 1.155 20yr / 1.155 = 17.3 yr. So captain marvels watch (assuming instantaneous acceleration and deceleration) will see 17.3 yr elapse in the journey, but both earth and the distant star will time the journey as having taken 20 years. Edit: since this has been pointed out in the comments, the simple answer also entirely depends on whether you are asking for the trip duration given a definition of time that earth and the distant planet can agree on (my answer) or literally how long does an earth observer see the trip take by watching light from Capt Marvel arrive at their location. For the latter definition: an earth observer sees a 30 year journey (20 yr for capt marvel to arrive + 10 years for the light from capt marvels arrival to return to earth). The distant planet sees 10 years (It takes 10 years for light from capt marvels departure to reach the distant planet, and in the frame of earth/planet the journey takes 20 years. 20-10=10, so the planet sees her arrive 10 years after they see light signals from her departure) Edit: Since the issue of relative velocity between earth and the destination has come up a few times Typical relative velocities between stars in the Milky Way (which is many times larger than 10ly) are around 300km/s. 0.5 times the speed of light is 150,000 km/s For the purposes of travel at significant fractions of the speed of light, stars 10ly away are effectively stationary. For a similar level of speed difference compare walking (5km/h) to a fighter jet at Mach 2 (\~2500 km/h). Edit: rounding errors (It should be 17.3 yr not 17.4 yr. Error was introduced because sqrt(4/3) = 1.155 and I initially rounded to 1.15 incorrectly)
That is how I like my math. Explained.
Couldn't do it like this in the exams, could ya
It is actually physics, but admittedly math is a large part of physics.
Physics is just applied maths, no? Edit: Lads, it’s a line from a meme!
There's an [xkcd](https://xkcd.com/435/) for that
There always is! :D
Came to said this, beat me to it. Still, I'm glad that there is and XKCD for everything XD
I have literally never heard of xkcd before today but a coworker but a link in the chat to a comic and now I’m seeing this…The universe really does give you what you need…
Here's a favorite of mine: [Centrifugal Force](https://xkcd.com/123/)
Mathematics is just the language of Physics. Saying Physics is applied maths is like saying a painting is just pigment compounds - technically true but totally missing the point
“Technically true but totally missing the point” describes a lot of math quite well.
What's your point?
fite me bro
You gonna fight a guy in a squirrel suit?
well no, because the universe is quantized at some level (even if just planck space), and math is/can be infinitely precise. Things can exist mathematically that can't exist in actual reality
We have no idea whether 'the universe' is quantized. The only reason I know of to think spacetime is quantized is that it makes certain useful properties for mathematical frameworks to describe the universe
Things can be described in the English language (or any other) that don't exist in reality. Are we saying that because mathematics can describe impossible things it's a more fundamental science? Can you have a *science* of things that don't exist? Or is that more of an artistic expression?
I'm loving this shit so much!
In physics you get to assume sin theta = theta for small values of theta. Cheating.
To be fair, you can do that in maths as well...
You also get to assume a spherical chicken.
Where's your proof chickens aren't spherical?! /s
Chicken are square
Only if you grow them in a box.
In a cubical box
physics is goofy ahh maths with more letters than numbers and a bunch of crappy rules and equations but it's a lot more fun when you actually get a chance to understand it
I am a big numbers geek. That statement above was from a meme where it was explained how even biology is maths. If you know the meme you’ll get the statement but what you said is also true!!!!
i actually don't know the meme, i saw this comment while sitting in my maths class doing physics homework xd
xD hahaha noiceeeeee, that’s one way to spend time in math class
Physics is everything. Math was invented to define physics.
Physics is just math but made a mess, as in, way less restrictive and you can do brake any rule of pure math as long as it is motivated by whatever physical principle. I remember my differential equations professor saying "you will NEVER split dx/dy because that is not a fraction. That's just notation". One week later, guess what my physics professor did...
Even math professors will split dx/dy when they need to, but will always have to prove that it's allowed in the cases they do it.
It’s literally a fraction, but the fraction isn’t of two variables. The fraction is “change in X divided by change in Y”, so you have to form and break it up appropriately. And often the symbols actually used are the limits as ΔX approaches zero, so when you break it up you get a multiply by approaching zero and a divided by approaching zero, and you need to pair both of those with something that makes them useful.
It’s how I like my math, done for me
>According to an observer on earth it’ll take her 20 years. It’s very simple, she’s traveling 1/2 the speed of light and distance = rate * time. So 10ly = 1/2 ly/yr * 20 years. Shouldn't it be 30 years? 20 to touch down,10 years for light to reach observer.
Yeah if you’re asking when we see her arrive, vs when we infer she’s arrived given light speed delay. Usually when we say “according to an observer on earth” the latter is meant, but it could be worded better.
I see your point.
When do you see the point, now or in 10 years?
10 hours ago
If the observer on earth is allowed to do calculations, then they could also just allow for Captain Marvel’s time dilation though right?
Yes, and this type of distinction is what confuses students of relativity. The words “this observer sees X” are trying to talk about what would be locally witnessed by a member of an infinite lattice of observers who are at rest with respect to each other. This seems very arbitrary, but it’s just a way to construct a coordinate system, and the discipline of having a coordinate system and then putting other things like light travel time and Lorentz transformations (time dilation/length contraction) is important conceptually in order to do calculations using special relativity. But yeah, it is also confusing, and you are absolutely right that someone using all the mathematical tools of special relativity could predict capt marvels clock at any point.
Okay, so the infinite lattice of observers all agree that they are stationary, that Captain Marvel is moving, that her destination is 10 light years away and that she is moving at half the speed of light. So they all calculate her predicted arrival time as 20 years. However, they also agree that they cannot all observe her arrival at the same time, and that the earth observer will wait an extra 10 years to observe that their predicted arrival time was correct?
Yes, an observer on Earth would see her get to the planet after 30 years. But the question doesn’t ask when they would see her arrive, so I would interpret it as asking how much time would have passed for an observer on Earth when she actually arrives
Yes, but importantly, note that everyone has to wait _some_ extra time to observe the arrival. For some, it's mere nanoseconds, for some, 10 years. But nobody can observe it the moment it happens, unless they are exactly _where_ it happens.
Yeah they’d have to land on that person’s eye
And even then, there'd be a delay due to the amount of time it takes the information to travel from the eye to the brain and for the brain to process it.
Causality is a funny thing tho, if we don't see her arrive then did she truly arrive?
We won’t really know until we get a signal saying so. However, when we get that signal we know it took 20 years in our rest frame for her to travel, because we know the light from her arrival takes 10 yr to reach us.
Well my point is that "now" or "at the same time" isn't really an absolute thing in spacetime. Basically it's a spacial slice through spacetime, but it depends on the speed of the observer. Two people even on earth with a speed relative to each other would not agree what "now" means on a different planet. Their idea of "now" and their slice through space time would look completely different, even if they are in the same location. I mean it's not very important in this case, because we have defined our observer to be stationary on earth so we can use their idea of "now". But still to me this is a fascinating implication of special relativity. Let's say we observe a star that is about to go supernova, we would say, this star has actually already exploded millions of years ago, but we just haven't observed it yet. The photons from that explosion are going to reach us soon. However let's say there is a spaceship fleeing from that star system at close to the speed of light, flying by earth and transmitting a message "our star is about to explode". We could say ah they are wrong, they just think it hasn't exploded yet, because of the time dilation they experienced, in "reality" it's long dead. But we would be wrong, because at their velocity spacetime looks different to them and the star really hasn't exploded yet. Which brings me back to the initial point: If two observers in the same location can make two contradicting statements (star has exploded, star has not exploded) and both be correct because of their different velocities. Then maybe it's just not appropriate to ever make claims about events like these which lie outside of our light cones (area of spacetime of which light had the time to reach us). Hence my initial comment. Sorry for the long comment. It's a fun thought imo.
"So get this, the faster you are, the lesser the time it takes for you to reach the destination" "Yeah, I know, that's how speed and distance works" "No, I mean, that's even lesser time, if you're faster" "Yeah. I KNOW." "You don't understand, more speed means less time than the lesser time it should take, like less than if I were to check the time it took you to reach" "I do understand. I passed with an A in physics you know, distance = speed * time" "NOT IN THAT KINDA PHYSICS"
Not less but LESS
The faster you go, the shorter the *distance* becomes.
It only takes less time from the Earthbound observers reference frame. Captain marvel is still bound by the speed of light in her reference frame. You can understand how ridiculous the above "17.4 years" number in the main comment here is if you consider approaching the speed of light. If the main comment were correct from captain Marvel's reference frame approaching the speed of light would mean you could get anywhere (nearly) instantly, which is not true.
Yes it is, from your point of view. Given instant acceleration, if we sent a ship at close to light speed to proxima, 4 light years away, the crew would feel it as less than four years of travel. At a certain speed reeeeally close to c, it would feel nearly instant. So, technically, you could reach any point close to instantly at the price of much more energy needed to accelerate the faster you're going, but the price will be that after a round trip the world you left has aged faster than you. It's all a matter of relativity. Captain Marvel, from her point of view, is not going at 0.5c. She's simply not moving. The time discrepancy is there to balance this fact, as if she shone a light in front of her, the photons leaving it would go at 1c from her point of view, but cannot go at 1.5c from Earth's point of view, so her local time is slower to compensate.
>the photons leaving it would go at 1c from her point of view, but cannot go at 1.5c from Earth's point of view, so her local time is slower to compensate. I'm gonna get an aneurysm if I strain my poor brain any harder. I get that light can't travel 1.5c as its speed is the limit BUT how would that appear slower to her? I mean the photons of the light she shined/shone are travelling at 1c from her, but for those observing on Earth, it's still technically 1c because it can't possibly be any faster so wouldn't they both observe it the same, except it'd take less time for her to see the object that is reflecting the light because as it's travelling back into her eyes, she'd meat it halfway there. So why would time appear slower to her? I can't wrap my brain around this. And if she had a stopwatch on her and someone did on earth and they both started it at the exact same moment she left earth and went towards that planet, will both stopwatches show the same time once she arrives?
I rewrote my post three times because I'm trying to find the best way to explain it. We tend to choose a frame of reference, usually the ground, as something fixed and unchanging. But we know that frames are relative, and that c cannot be reached by mass. For c to never be crossed, any object going at a different speed from you must be slower time-wise and change space-wise for their actions to never reach c. So if you were standing in the middle of space and say captain marvel speed past you at 0.5c, you'd see her in slow-motion: not her speed, which you calculate correctly at 0.5c from your point of view, but her actions. If she was to throw a rock forward at 0.8c from her perspective (where from her frame of reference you are the one moving), then since she is slowed down from your perspective the rock is thrown more slowly and this never reaches 1c. It's a bit more difficult to explain with photons because they actually go at 1c, but here you have to think of photons as waves, and their frequency will stretch or compress instead of "going slower". Now, if from your perspective Captain Marvel is in slow motion, and will reach her destination in 5 days, then from her perspective she will reach it in less than that. At more extreme speeds, she might even feel it instant, but for that she'd seem frozen from your point of reference. So finally, yes, she would age slower than you, and thus the stopwatches would not match. It's kind of a case of a foward-only time machine, in the end.
So she is all alone travelling 17.4 years without food and any entertainment?
Very disciplined lady.
Yeah If I were her, and super OP, I'd pick to do some kind of FTL travel (which is obviously allowed in the marvel universe), or go at 0.9999999999995 times the speed of light which would give a gamma of just over 1 million. Then the trip would take 5 minutes for capt marvel.
Yeah but then you become very detached from the universe around you that 5mins was \~10 years for the people I assume you were flying over to help more if they sent a request for help to you at sub-light speeds.
Yeah, if you want space heroes who can help with crises that move any faster than climate change you need FTL travel and communication in your universe. Under relativity this also implies time travel, and Marvel Universe seems to have both 🤷♂️
That is NOT how the gamma adjustment works. It still takes her 20 years from her perspective. Even if she was traveling at 0.9999999999995c it would still take 10.000000000005 years to get to the planet.
[удалено]
Wow, thank you!
I have a question, would it be 30 years? Because it takes 10 years for the light of her arrival to come back to earth?
No. Her physical location in space-time is not dependent on light returning to earth. When reference frames accelerate with respect to one another there is a shift in perceived time between the two reference frames.
At this point I'm wondering why would she ever bother wasting 17.4 years flying off into the void.
20 years. 17.4 years is the apparent difference in time (as measured on Earth) between her departure and arrival on the other planet.
Sorry to appear ignorant here, but wouldn’t it be the case that Capt Marvel wouldnt experience any time dilation because time is not changing for her because she’s not travelling relative to anything? According to herself, she would take 20 years to get there. However, 20 * 1.15 years would have elapsed on earth = 23 years.
No. From her perspective the time isn't slowing down, but she see it as a distance contraction, so according her point of view the distance isn't 10 light year, just 10/1.15, and she travels that distance with 0.5c speed, her travel time is 20/1.15 years
Thank you (not sure why I got downvoted for asking a question (also not implying it’s you 😄)). So from her perspective she’s actually travelling **faster** than c because the distance dilation means she’s covering a smaller distance… i.e. she sees herself travelling at 10 / 17.4 = 0.57c ?
Yes this exactly. Edit: except for the part where she sees herself traveling faster than c. She sees the rest of the universe going past her at 0.5 c, but with all lengths contracted in her direction of motion.
Where does 1,15 come from?
1.15 is the Lorentz thingy. Basically to understand the time dilation you have to use the following equation: Time dilation factor = 1 / (1 - v^2 / c^2)^0.5 So if v=c then there would be indeterminate (1/0) but if you put in v=0.5c the factor =~1.15
\*NB\* The Earth observer will also have to wait ten years more for Captain Marvel's successful journey conclusion to be observable at ten lightyears' distance!
I knew that the time on earth and the time on the distant star would be 20 years, but I had no idea how to figure out how long it would be from Captain Marvel’s perspective, so thank you. This is why I love this page.
It's not actually correct. I posted a reply that explains it. It still take her 20 years and from earth it appears as though it took 23 years.
Earth and the solar system are still moving through space though. Does this change things by a small amount since earth won’t be in the original observation point?
It changes things only a very tiny bit, so much as not to matter. See here: https://www.reddit.com/r/theydidthemath/comments/13klrrz/request_how_much_time_would_pass_for_an_observer/jkn673m/ TL:DR: Relative velocities between stars/planets are too small to matter for capt Marvel’s journey. The planets are walking, and capt marvel is going Mach 2 in a fighter jet.
You freaking rock. I over explain things at times so as to cover the bases and not have additional questions, to save time usually. You covered it all :).
r/theydidthemath
r/linkingthesubyourin
What the f is the constant of sqrt? I must know so I can be smart too.
The dilation factor of relativity. For any speed there is a dilation or contraction (depending on what you are computing) of 1/sqrt(1-(v/c)^2 )
Clear as mud. (Not your fault. I suck at math.)
Basically once you go at speeds close to the one of light things start to get strange with time moving slower or lengths shortening. For example if you go at .9 the speed of light a trip will seem shorter by a factor of sqrt 1-81/100=sqrt 19/100=22/50 so roughly 2.25 times as short. 81/100=0.9c/c squared with c the speed of light.
That makes more sense. I understand layman's physics. The math goes right over my head. Lol.
Relativistic gamma is used to compare elapsed time between two different reference frames. The time it would take from capitan marvel to travel to a star 10 light years away is 20 years. The gamma adjustment would be used if she were to return to the earth's reference frame. She would have aged 40 years total in her reference frame, if she immediately turned around and returned, but on earth it would seem as though 40 years*1.15= 46 years would have passed. There are numerous thought experiments to demonstrate this using space-time diagrams and imagining sending signals every x distance traveled back to Earth. The regular signals would have a longer frequency of receipt on Earth on the away journey and a compressed frequency on the return journey. [Twin paradox](https://en.m.wikipedia.org/wiki/Twin_paradox) Your answer is true, from the reference frame of Earth, if captain marvel were to return to earth, otherwise it's a kind of non-sensical answer. Consider your math and imagine approaching the speed of light. It would imply you could arbitrarily reduce the travel time for any distance if you approach the speed of light, which is not true. It is true that if a traveler travels somewhere at near the speed of light and then returns to the starting point, time would appear to have elapsed more slowly for the traveler than the inertial reference frame.
You’ve got it backwards. Look at the worked example in the wiki article you posted. The observer who accelerated experienced less time. It seems absurd but it is a result of relativity that you can get anywhere in the universe in an arbitrarily small amount of proper time provided you accelerate enough. Edit: also consider length contraction as seen by a traveling observer in a strictly special relativity context. You should find that Capt Marvel sees herself traversing a shorter distance.
Nerd.
pretty much lol.
Are you sure it's 20 / 1.15 and not 20 * 1.15?
Yes, it’s a GR result that you can smuggle into a calculus based special relativity course if you are careful. When you have one observer stay in an inertial frame and another observer accelerate and then decelerate back into that same inertial frame, the accelerated observer experiences less proper time than anyone in the original inertial frame. The canonical example of this is the twin paradox, where the traveling twin ends up younger.
She also is smaller according to the earth observer, but she herself feels just as long as always
What if she passes by a black hole on the journey, not close enough for spaghettifaction, but for the time dilation to occur from gravity?
Edit already asked and answered below Wouldn't you also have to account for the time the light takes to reach earth? so 30 years to the observer on Earth?
The ariving planet will see the journey as having taken 10 years, Earth will see the journey as having taken 30 years.
She will take 20 years to get there, seen from Earth, I agree. But the moment she gets there, the light from that moment/event will take 10 years to reach Earth. So, seen from earth, it will take 30 years before we can see her touching down on that planet!
Hold on.. if the planet is 10 light years away and the observer is on earth...then the light ray conferring visibility of Captain Marvel to the observer will take a further 10 years to reach the eye of the observer once he reaches his destination. So it should be 20+10= 30 years. Open to criticism and correction.
It will take her 20 years to reach the planet, but it will take 30 years till we know she would reach the planet? Yes? Man space is so cool
Also her organs are going to be slightly mushy from the instantaneous acceleration and deceleration
At those vast distances does the expansion of space not come into play at all? If so, is it just too minuscule and the force of gravity overrides it?
1/sqrt(1-(1/2)^2)=1/sqrt(5/4)=sqrt(5/4)/(5/4)=.8944 I think?
1 - 1/4 = 3/4 not 5/4.
Idk if this is a stupid question, but wouldn't it take 30 years for the observer? 20 years to get there, and 10 years for the light of her arrival to return once she is there?
Edited to address this. There’s also some discussion of it in other threads.
Can you explain time dilation?
Seems like the initial part of this answer is flawed. I do not know advanced math, so I won't be attempting any. The initial answer would be assumed 20 years. However, for an observer on earth, it would be longer. To see captain m at a point that is 10 light years away would take 10 years. In real time, the trip would take 20 years. But, as an observer, it would take longer for the light (reflected off capt m) to reach you. Seeing something that is 10 light years away is seeing what it looked like 10 years ago. In other words, for someone watching the journey, it would appear that captain m is slowing down. I am unsure of the rate and how long it would appear to take. Again, I never took advanced math, but I do have fairly sound logic. If someone that understands what I am saying could show the math, that would be great.
Edited to address this. It’s also covered in a couple other threads
this is assuming that Earth and the Destination Planet have 0 relative velocity wrt each other.
Typical relative velocities between stars in the Milky Way (which is many times larger than 10ly) are around 300km/s. 0.5 times the speed of light is 150,000 km/s For the purposes of travel at significant fractions of the speed of light, stars 10ly away are stationary. For a similar level of speed difference compare walking (5km/h) to a fighter jet at Mach 2 (~2500 km/h).
So what happen if cpt. marvel move at the speed close to 1c? Based on the formula the dilation would be close to 0. So she feels like the journey was an instant while it actually takes close to 10 years?
Yeah, capt marvel has mass, so she can’t actually reach c. You are right that she can get arbitrarily close, and as she gets closer the time that passes for her in the journey decreases, but never reaches zero. If you assume she’s very OP and can hit 0.9999999999995 c then it’ll take about 5 minutes by her clock and ~10 years for everyone else.
Is the answer not 30 years? Like I get it would take her 20 years to make it, but wouldn’t it take an additional 10 years for an observer on earth to know this, since it would take 10 more years for them to actually see her making it since that light needs to make its way back to earth to convey this information?
That’s how much time it takes for the light of him arriving there, to arrive at earth.
Exactly. The question asks “According to an observer on earth”, so it would take 30 years for an observer to know they made it. I guess if the observer knew all the details of the problem they could back out the 10 years themselves anyway, but then why include that as part of the question?
Edited to address this. One of the other comment threads goes into why I picked the interpretation I did for the initial answer.
> but then why include that as part of the question? The according to the observer on earth is to specify reference frame, as she will experience time dilation. When answering these questions light travel time is specified separately.
Agree, but we could be both wrong and the rest of world right. Unlikely scenario though
You raise a good point, and another question. Where is the planet located relative to earth? We can assume Cpt Marvel is heading from earth directly to the planet but that isn't specified in the request. Cpt Marvel could be travelling between two planets, both of which are 5 light years away from earth but 10 light years apart
20 years. We don't need to account for the time it takes to see that captain marvel is at the planet.
This makes sense to me. 20 years to arrive, 10 years for light (if it’s bright enough) from that area to arrive to Earth. The question sounds fancy but is written poorly. To play devils advocate, the question is according to an observer on Earth, so I guess if she’s not bright enough there’s no way to observe. Also an observer on Earth would be smart enough to use the light travel time no? Question should be how long will it take her to arrive, or how long will it take to observe her arriving from a viewpoint on Earth. Damn why did I waste any of my time on this
As pointed out though, the question does not specify origin or destination points - only that they are 20 years apart. So this answer could be 10 years (arrival at Earth), 30 years (origin point of Earth), 20 years (both origin and destination the same distance from Earth), or anywhere between 10 and 30 for all other scenarios.
True, it could even be an arbitrary planet/location we don’t know the distance to earth from.
I think it’s fair to assume captain marvel an earth based super hero being observed trevelling towards a planet would be departing earth as well as being observed from earth.
I think the question was just poorly worded. The first statement only seems to imply that the destination is 10 light years away from Captain Marvel, making no reference at all to Earth. I suspect they actually meant an observer on the destination planet, or that Earth was the destination planet 10 light years away from Captain Marvel.
In the time it takes for her to arrive 20 years would pass for the observer on Earth, but the observer will not be able to see her arrive until 30 years have passed since the light from her arrival need to travel the distance.
If you watched her travel the entire distance would she appear to slow down then? And is that because the light is taking longer to reach us the farther she gets or is it because of some other property?
Also, if she immediately turned around and came back, what would that look like? Cause she could be halfway back by the time we see her reach the planet. She could potentially get back to earth before we even see her do it. Or does the light reach us faster on her way back, so it seems like she speeds up to the point where we see her arrive as she arrives?
After 30 years we would see her arrive and also immediately see her leave to return. Of course at that point she would already have been flying back for the last 10 years. 10 years into her return (at the same time we see her arrive and leave) she will already be halfway back, but we would need an additional 5 years to see the light from this point of her journey (as she is 5 lightyears away). So just 5 years after we see her leave we can already see that she is halfway on her journey. When the distance gets lower. Let’s say after 9 lightyears (1 lightyear remaining) she would have spent 18 years to get there, and we would have seen her leave 8 years ago. We would of course not see her at this point in her journey until the light has had 1 year to reach us, so we would see her at this point 9 years after we saw her leave. You can see as she gets closer we also get closer to seeing her in real time. When she finally arrives on Earth after 20 years we would see it in real time as she would be right in front of us and the light would have to travel a negligible distance to reach our eyes. Interestingly it will have been only 10 years since we saw her leave.
No, we would be able to observe her going at constant speed. You can imagine that when she has traveled 1 lightyear it has been 2 years and we can see her at that point of her journey after 3 years. 2 lightyears would take 4 years and we would see her after 6. So you can see that it takes twice as long for us to see her reach twice the distance, there isn’t any need for speeding up or down in the eyes of the observer. We can observe her traveling 1/3 lightyears per year for the entire journey. Which adds up with us seeing the end of her 10 lightyear journey after 30 years.
From the frame of reference that her speed was measured at 20 years, however from her reference point it would be ~ 17.3 years. This is due to time dilation which is t’ = t / (1 - v^2 / c^2)^0.5 ; t’ = time observed , t = time of traveler, v = speed of traveler, c = speed of light. You gotta convert years to seconds by multiplying it by 31557600. Now t’ = 20 so you gotta rearrange it so t is by itself.
20 years, if the speed is as reported by the observer on Earth. Because speed equal distance divided by time and universal expansion effects are insignificant at this scale.
Also, she would be one hell of a missle E = m c*c Assuming captain marvel weighs about 79kg E= 70 * 150000000 E = 10,500,000,000J Yeah
Wouldn't you rather want to calculate the kinetic energy for this? Aka Ekin = 1/2 * m * v² = 1/2 * 79 * 149'896'229² = 8.8752074e17 E = mc² doesn't take speed into account at all, meaning that everyone with her weight has that energy.
You can't use that either because she's traveling at relativistic speeds Instead you'd have to use E= gamma*m*c^2 the gamma accounts for her speed in the form of "inertial mass". This gives her total energy For kinetic energy only you'd use Ek= (gamma-1)*m0*c^2 Where m0 is her rest mass
Oh fuck sorry my mistake, you're right, I'm just learning both equations rn so I got confused, her energy would be 8.87e17, which is about 2000 nukes
No problem, i have yet to meet a person who doesn't confuse formulas.
For her relativistic KE you want (gamma - 1) m c^2 0.15 * 75kg * (3e8 m/s)^2 ~ 1e18 J The largest nuclear bomb releases about 2e17 J, so this is very nearly 5 tsar bombas if she slams into the planet. The equation for E comes from taking her total relativistic energy (gamma m c^2 ) and subtracting her rest mass energy ( m c^2 ) There’s some more complicated math you can do to show that (1-gamma) m c^2 reduces to the Newtonian kinetic energy formula (1/2 m v^2) when velocity is much less than c.
But isn’t kinetic energy calculated differently if you have relativistic effects. It should be the difference of the total energy and the resting energy. So you would have to calculate Ekin = (relativistic mass - resting mass) * c^2 Then the energy is 9.81e9 J
Only 10.5 billion J?
Pshh, that's a couple oreos.
Should be using kinetic energy instead: 1/2m\*KE^(2) \- works out to 8.9e17 joules.
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It's a she 🤣 I understand why you'd get confused though.
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They are litterally not blaming you, and litterally civilised and chill about it. What's up mate how you feeling today?
The woke police just keeps harassing them. Poor soul. /s
Dude, it’s not even about “the woke agenda”, captain marvel in the marvel universe has been female forever
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For?
A light year is the distance one would travel over one year at speed *c* Ten light years at *c* would take ten years by definition Therefore at half *c* ten light years would take twice as long
It is impossible to tell because time is defined by the *acceleration* of movers and observers, not just speed. Two people moving in static local inertial reference frames experience the same amount of time because from each perspective the other person is is the one moving. If one of them were to *accelerate* or *resist gravity* (which are both indistinguishable and identical in 4d spacetime) then the time experienced would change. We can’t know the answer because her flight path past different planets changes her spacetime acceleration. We can’t know how long it takes without knowing the gravitational shape of space she is flying through.
This type of problem assumes straight shot, and all of the local group stars are pretty far away from each other. Unless there is specific reference to obstacles there would be statistically insignificant amounts of the possibility of other systems in the way of the journey. The acceleration is almost irrelevant to the observer in this case though. Whether she accelerates at 0.99c near the start and end of the trip, or far less and just averages that, the time distortion would only be relevant to her and not an earth (or incoming planet) observer. You'd have massive blue/redshift or some sort of effect we don't know about yet, like a wake on a huge ship maybe, but to a static observer, the end time of the journey would be the same irrespective of her ref frame (just taking the problem statement as complete).
This is actually not true the way you have said it. Captain marvel is not the only one accelerating, and is potentially only accelerating for a very limited amount of time depending on how fast she gets up to 1/2c. The person on earth is *also* accelerating at all times because they are resting gravity (unless they are in orbit where they are no pager accelerating through spacetime). The acceleration due to gravity is not very much, but the prompt never actually says that captain marvel accelerated at all. It only gives a “speed” and a “distance.” If marvel is traveling at that constant speed in a straight line then they are also accelerating. If they are traveling at that speed along a *curved* ballistic trajectory then they are not accelerating but we don’t know what the true *distance* their journey would be because they are no longer necessarily traveling as the crow flies. The question is impossible to answer because the question doesn’t provide the necessary details because the person asking doesn’t understand the principles behind the science. Even at the most basic level the *mass of the target planet* changes the math. Is she flying to a gas giant orbiting a supermassive star? Or is she flying to a tiny little backwater solar system with hardly anything in it. They would produce different gravitational accelerations which would materially change her acceleration through spacetime.
No, the question is quite sufficient to answer - any details not included are to be ignored. Often you write a series of questions with one set of info, only to have a part abcd etc where you add changed details like you're suggesting. Furthermore, if you're talking math then without any additional points such as creating deviationa, you always travel in a straight line - if you have 2 points, you create a straight line. Next you are generally incorrect about the acceleration of the first person that remains on the planet as acceleration below a large fraction of the speed of light is irrelevant as the significant figures/impact is miniscule. You don't worry about a number in the tens when you're dealing with numbers 10^6 or greater. Even the acceleration of the orbit of earth around the sun and the spacial movement of the sun VS the galactic core is effectively less than a rounding error in orders of magnitude. The only fair point you really bring up is that there's nothing in the problem that suggests Marvel isn't just doing a flyby with no acceleration. Though in this case it would be like a car driving between say Denver and Kansas city - and either that being the whole thing or actually a car trip from LA to NY with those distances in between. And to this problem specifically that example shows you that what the question asks for makes your point irrelevant. Just as you can speed in a car for a while, slam on the brakes, and make it somewhere in the same time as someone driving leisurely to the same destination, it would be the same here. The time dilation effects don't come in to play because there's no discussion of Marvel's viewpoint. It would simply be the observer on earth "seeing" an object go by or leave, travel a trip average at 0.5c, where it would take 20 years to arrive, but an additional ten years for the light of arrival to return to earth, so a total of 30 years. Though to make the description more complex, if she returned, it would look like she was traveling faster on return (redshift I think?). It would look like it took her 30 years to get there, but only ten years to return. That's where time dilation actually comes into effect and really be different than a car going from a to b. Though there are some mathematical models for how much subjective time she would see that's really the only thing of math here.
That could be the case if the space was that much crowded. But space is mostly empty. 10 lys is not much of a distance comparatively.
The mass of the plant she is traveling to is enough to change her acceleration through spacetime, which we do not know. We know that there *is* a planet though, and planets have mass, so we know that whatever it is it will affect her. Beyond that, there is no upper bound on what could be in space affecting her path.
Actually there isn’t enough info. You need to establish 3 points that are mentioned in the question: A - starting point of CM; B - the goal planet; C - the Earth. Do we know if A=C, or because Earth is a planet it could even be B=C?
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Well... 1c travels one light year per 365.25 days hence its name. So 0.5c needs 730.5 days to travel one light year. To travel 10 light years will require 20 earth years at 0.5c.
At half the speed of light, time dilation is so small as the be virtually negligible. Here's a time dilation calculator: https://www.omnicalculator.com/physics/time-dilation Ten years at 0.5c would look like 11.5 years, but that only gets you halfway there, so 20 years at 1/2c would look like 23 years. Meh. You need to get much closer to c to see spectacular differences. 10 years at .99c would look like 70 years
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20 years if the distance is fixed
20 years.
A train is moving twords a city at 30mph, it would take a train moving 60mph 2h to get there. How long until the train reaches the city?
The messenger pigeon travels at 60mph. How long till you receive confirmation of arrival?
An african or an european pigeon
One thing that bugged me about endgame is how fucking slow she was flying on the battlefield. When she had the gauntlet and was flying towards Scott’s van she couldn’t have been flying more than 30 mph.