# Time and distance at constant C: A sieries of questions for Umbran or other physicists.



## Scott DeWar (Sep 22, 2015)

Ok, As the title says, Questions that has my brain in a twist: A thought experiment on travel at the velocity of C.

in the picture I post here, 

given: The clock on a ship moving at Velocity of C for 100 years

would the star we travel around here on earth move 100 years worth while the shp moves to the end point at the far right, off the page?

and on the return, would it arrive where Sol wold have traveled another 100 years?

According to what I am to be told thus far, the answer should be no. Is that correct?


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## tomBitonti (Sep 22, 2015)

Do you mean 'C' to be the speed of light?  A ship couldn't actually travel at the speed of light, but, in theory, it could travel close to the speed of light.

A ship traveling close to the speed of light will experience time dilation: Clocks on the ship will appear to advance more slowly to an observer which is not in motion (say, someone still on the earth).  A clock on a ship moving close to the speed of light will advance quite slowly, and the closer the ships speed gets to C the more slowly the clock will advance.

An observer on the ship won't notice any slowdown of the clock; they are moving more slowly, too.  However, an observer on the ship will notice other effects: Distances outside of the ship will shrink in the direction of motion of the ship.

If the ship travels 100 light years away from the earth, then travels back to the earth, both at close to C the speed of light, then, to an observer on the earth, the trip will appear to take 200 years, and the earth will circle the sun 200 times during the journey.

On the other hand, the observer on the ship will notice a much reduced passage of time.  How much depends on how long the ship takes to accelerate from a standstill up to C, but in principle the journey can be made arbitrarily short.  _For the observer on the ship._

Thx!

TomB


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## Umbran (Sep 22, 2015)

Scott DeWar said:


> Ok, As the title says, Questions that has my brain in a twist: A thought experiment on travel at the velocity of C.




Travel *at* c is impossible.  How about I answer as if you said "0.9c"?



> given: The clock on a ship moving at Velocity of C for 100 years




100 years as measured by who?  Remember, the clocks are relative.

A ship leaves Earth, and travels at 0.9c for what people on Earth call a century.  The Earth and Sun and all will all do their normal things, moving pretty slowly, just as they have since WWI until now.

Folks on the ship will experience about three and a half years of time passing.  Earth would think the ship had moved 90 light years away.  Let's say they went to check out https://en.wikipedia.org/wiki/HD_70642



> and on the return, would it arrive where Sol wold have traveled another 100 years?




Pretty much, yes.  I mean, Earth is a moving target, so they have to aim where the target *will* be, but yeah, in general.

Overall, about 7 years of time will pass on the ship (not counting times to accelerate or decelerate here), while 200 pass on Earth.  The Earth will have moved much as it has since the War of 1812.  People will have died, new ones born, wars fought, technology developed, and all that jazz.


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## Scott DeWar (Sep 22, 2015)

tomBitonti said:


> Do you mean 'C' to be the speed of light?  A ship couldn't actually travel at the speed of light, but, in theory, it could travel close to the speed of light.
> 
> Thx!
> 
> TomB






Umbran said:


> Travel *at* c is impossible.  How about I answer as if you said "0.9c"?
> 
> 100 years as measured by who?  Remember, the clocks are relative.
> 
> ...




Ok, I should have been more detailed of what I was thinking. Sorry.

I _meant_ to say at .99C, but we will go with the theoretical .9 C

Also, I meant to mention two clocks synced on earth - both atomic clocks for accuracy

Also, let us say, for the sake of argument they can "instantly accelerate to C" for passing of time.

So, the time dilation would be:

a. they wold have traveled a total of 180 light years, round trip

b. it took 7 years to those on the ship

c. Earth has moved through the cosmos the distance it would have after 200 years at the present movement that we are at right now, baring any form of "eternal force"

d. To the earth, It would look like 200 Years had passed

e. (and of course I HAVE to have the humor) it is caused that way because it is a wibbly wobbly timey wimy thing.

Do I have that correct?


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## Morrus (Sep 22, 2015)

Talking of time dilation and relativistic speeds, I was wondering if there were games other than my WOIN system which addressed that?  GURPS, perhaps?  Does Traveller?


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## Scott DeWar (Sep 22, 2015)

I don't remember anything on GURPS, but it has been a while since I played. It was never addressed In any star trek, Firefly or other interstellar travel show/movie except one and that was way inflated in the single season show called Outcasts. And to be honest, Outcasts is a major reason for this question.


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## tomBitonti (Sep 23, 2015)

There is a discussion here:

https://en.m.wikipedia.org/wiki/Time_dilation

The ratio of elapsed times is represented by the lower case gamma, and is (sorry about the representation):

1/sqrt(1-v^2/c^2)

There is a table here:

https://en.m.wikipedia.org/wiki/Lorentz_factor

For 0.9c, the value is 2.3, for 0.99c -- 7.1, for 0.999c -- 22.

A trip which starts with synchronized clocks, one of which goes on a trip while another stays at home is a version of what is called The Twins Paradox, with clocks and twins  exchanged.

Basically, if you put one twin on a rocket and send one on a trip of 11 light years and back (so 22 ly total), at 0.999c, the twin on the rocket will age 1 year while the twin that stays at home will age 22 years.

What happens on the rocket won't affect what happens at home (unless it crashes on return without slowing down!). And other than accelerations, the twin on the rocket won't perceive anything different inside the rocket.

Thx!

TomB


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## Umbran (Sep 23, 2015)

Morrus said:


> Talking of time dilation and relativistic speeds, I was wondering if there were games other than my WOIN system which addressed that?  GURPS, perhaps?  Does Traveller?




The games I know of don't go into it, largely because they either use distance scales in which you'd not travel at such speeds (like Firefly, where all the worlds are within fairly short distances), or they have FTL travel.


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## Mustrum_Ridcully (Sep 23, 2015)

Scott DeWar said:


> I don't remember anything on GURPS, but it has been a while since I played. It was never addressed In any star trek, Firefly or other interstellar travel show/movie except one and that was way inflated in the single season show called Outcasts. And to be honest, Outcasts is a major reason for this question.




The problem migh be that it's not really a rules thing, but a world-building problem. 

If you travel across half the Galaxy with a very close-to-lightspeed speed so it is for you only a 3-day trip, 50,000 years would pass in the mean-time on Earth and any other planet that isn't flying that fast.
How could you possibly build a world - as DM or setting book author - thta could deal with such long timespans?

And even if you stay at closer distances of, say, Alpha Centauri, every trip would take over 4 years. Imagine you had to pace an RPG adventure over several years every time the players decide to travel a bit...


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## Morrus (Sep 23, 2015)

Mustrum_Ridcully said:


> The problem migh be that it's not really a rules thing, but a world-building problem.
> 
> If you travel across half the Galaxy with a very close-to-lightspeed speed so it is for you only a 3-day trip, 50,000 years would pass in the mean-time on Earth and any other planet that isn't flying that fast.
> How could you possibly build a world - as DM or setting book author - thta could deal with such long timespans?
> ...




I've tackled this in various ways.

1) You create settings of a size appropriate to your speeds. No FTL, then you stay in the solar system!  You can have some amazing space adventures in one solar system.

2) If nobody else can move faster than light either, it really doesn't matter, as you'll never know what's happening back at home, and it can't affect you. Freeze the PCs for the time required, wake them up. Have some fund with time dilation.

3) Make time and age part of your advancement system. You can "spend" XP in real-time, or "years" in downtime to purchase character advancements, at the cost of growing older.

4) Say screw it, and allow FTL travel, wormholes, spacegates, what-have-you.


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## Umbran (Sep 23, 2015)

Morrus said:


> I've tackled this in various ways.
> 
> 1) You create settings of a size appropriate to your speeds. No FTL, then you stay in the solar system!  You can have some amazing space adventures in one solar system.




And if our own solar system seems a bit small, make a bigger one, like Firefly.



> 2) If nobody else can move faster than light either, it really doesn't matter, as you'll never know what's happening back at home, and it can't affect you. Freeze the PCs for the time required, wake them up. Have some fund with time dilation.




Well, the "fun" in tie dilation is in the interaction with home.  Otherwise, relativistic travel is just "one-way trip to another universe" really.



> 3) Make time and age part of your advancement system. You can "spend" XP in real-time, or "years" in downtime to purchase character advancements, at the cost of growing older.




It still becomes a worldbuilding issue - say some of the PCs take a relativistic jaunt, and others don't.  The ones at home advance using years, the ones on the trip by XP.  The GM still has to figure out decades worth of world-changes.  A system to handle advancing cultures (not just tech, but *cultures*) over time would be a good part of a relativistic travel system.


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## Scott DeWar (Sep 23, 2015)

_*IF*_ we were to be able to travel at FTL, I recently saw an interesting  video on YouTube where it shows travel through our own solar system at light speed for 45 minutes - you do not reach Saturn, but you make it to Jupiter. 

but enough of that aside.

As for this:

[sblock=Tom B.]


tomBitonti said:


> Do you mean 'C' to be the speed of light? A ship couldn't actually travel at the speed of light, but, in theory, it could travel close to the speed of light.
> 
> A ship traveling close to the speed of light will experience time dilation: Clocks on the ship will appear to advance more slowly to an observer which is not in motion (say, someone still on the earth). A clock on a ship moving close to the speed of light will advance quite slowly, and the closer the ships speed gets to C the more slowly the clock will advance.
> 
> ...






tomBitonti said:


> There is a discussion here:
> 
> https://en.m.wikipedia.org/wiki/Time_dilation
> 
> ...



[/sblock]

I will have to write down stuff to get it to process, otherwise it means a lot of dream time and that takes too long.


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## freyar (Sep 24, 2015)

I'm late to the party, but...



Scott DeWar said:


> Ok, I should have been more detailed of what I was thinking. Sorry.
> 
> I _meant_ to say at .99C, but we will go with the theoretical .9 C
> 
> ...




Just about, but we have to be a bit more careful, actually (I'd have to mark off a bit in my class  ).  What we need to know is what the .9c speed of the rocket is measured with respect to.  People generally seem to be assuming that the rocket is moving at 0.9c with respect to the earth, but then that means the earth wouldn't move IF the earth weren't accelerating.  In other words, all that time passes on the earth, but the earth is sitting still.

But the earth's motion around the sun accelerates, so we don't really want to use that as a benchmark. It's better to say that the rocket moves at 0.9c with respect to the sun, while the earth orbits the sun.  That's a much better approximation, since the sun accelerates very little in its motion around the galaxy. Of course, from the point of view of the earth, that 0.9c rocket travel is modulated just a bit by the motion of the earth around the sun.

Or we could say that the rocket moves at 0.9c with respect to the galaxy.  Then the sun is moving through the galaxy, too, which sounds closest to what you've described.  But then, according to the sun, the rocket isn't moving quite at 0.9c either on the way out or the way back.

Of course, the motion of the sun around the galaxy and the earth around the sun are at very small speeds compared to c, so these are all very minor corrections to the big picture everyone's talked about.  But usually in relativity we need to be pretty clear about the question we're asking as the answer can come out quite differently in slightly different situations.


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## Scott DeWar (Sep 24, 2015)

Frrayer, you already mentioned in another thread that you were going to bee scarce, so no worries.


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## Scott DeWar (Sep 24, 2015)

freyar said:


> I'm late to the party, but... * * * * *stuff * * * * *Of course, the motion of the sun around the galaxy and the earth around the sun are at very small speeds compared to c, so these are all very minor corrections to the big picture everyone's talked about.  But usually in relativity we need to be pretty clear about the question we're asking as the answer can come out quite differently in slightly different situations.



 there you go againd messing up my nice neat little world. 

SiGH . . . I will think on this. I am not sure of what my answer should be here.


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## freyar (Sep 24, 2015)

Scott DeWar said:


> there you go againd messing up my nice neat little world.
> 
> SiGH . . . I will think on this. I am not sure of what my answer should be here.




Well, in this case it doesn't make a practical difference.  In fact, in class, I usually pretend the earth doesn't accelerate.  But suppose you said that the rocket is moving at 0.9c compared to a distant galaxy that's moving at 0.5c compared to our Milky Way galaxy (just as a silly example).  That would make a big difference.


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## Umbran (Sep 25, 2015)

freyar said:


> Well, in this case it doesn't make a practical difference.




There is a phrase often used by physicists (and other hard sciences): "...to first approximation..."  It is a useful concept here.

There's an actual technical definition (see Wikipedia, "Orders of Approximation") , but, for our purposes, it means, "Taking the strongest factor into account, and leaving out the smaller factors".

A *lot* of what I say here is really to first approximation, because the secondary (and tertiary, and so on) factors will often seem like major complications, that will typically get in the way of understanding the base concept.   If Freyar and I have disagreements, it is usually that I think he's jumped on to secondary effects before we've established that folks understand the primary ones. 

So, to first approximation, as compared to the speed of light, the Earth can be considered stationary. 

Then, you can lay upon that the motions of the sun through the galaxy, and the Earth around the Sun.  Note that, on long average (say, over a trip that will take the Earth around the Sun 90 times) the Earth's motion around the Sun won't mean much, as it is pretty much a closed loop - over the long haul, that loop will average out to almost zero.  And yes, the Sun moves a bit, but you already had that in mind.  The way you drew it, the Sun's motion was pretty much perpendicular to the ship's motion, which also decreases the impact of that motion to the main picture.


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## Scott DeWar (Sep 25, 2015)

Re: Orders of approximation

I figured there was some sort of saying to go with this. And yes, I am working at the first order her. Especially since I am on day one of caffeine deprivation. Also, considering my level of knowledge and understanding, I think keeping it at the first order is always a safe bet.

So take that Freyar! :rofl:


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## freyar (Sep 26, 2015)

Umbran said:


> A *lot* of what I say here is really to first approximation, because the secondary (and tertiary, and so on) factors will often seem like major complications, that will typically get in the way of understanding the base concept.   If Freyar and I have disagreements, it is usually that I think he's jumped on to secondary effects before we've established that folks understand the primary ones.




Haha, yes. Either that or we're thinking about slightly different problems with somewhat different distinction between lowest order and higher order effects (see example below).  I also have a pretty low tolerance for introducing small errors to keep the focus on the lowest order stuff, which is partly because I teach upper-division university students.  



Scott DeWar said:


> I figured there was some sort of saying to go with this. And yes, I am working at the first order her. Especially since I am on day one of caffeine deprivation. Also, considering my level of knowledge and understanding, I think keeping it at the first order is always a safe bet.



Caffeine deprivation?  Horrors!  You have my sympathies.

Here's why I brought up that more subtle point.  It's true, if you're looking at time dilation effects, that the speed of the earth around the sun (for example) is a totally trivial consideration.  But suppose we talk about the motion of the earth while the rocket was gone for 100 years (and let's completely ignore the motion of the sun around the galaxy).  There are two cases we might be interested in.  One is that the rocket is moving in a straight line with respect to the rest frame of the sun, which is the same as the earth's average rest frame.  Then the rocket goes out and comes back and gets to the earth 100 earth years later.  But what if the rocket moves in a straight line at 0.9c out and then back as seen from the earth's instantaneous rest frame at the time of launch.  At any given moment, the earth is moving about 30 km/s with respect to the sun.  If the rocket moves along the same direction, everything is the same as the other case.  But if the rocket moves perpendicularly to that, it's path isn't just straight away from the earth and straight back.  It's away from the earth and back to where the earth would have been if the earth left its orbit around the sun and went off at a constant velocity.  That'd be about 90 billion km away from the earth's position.

Anyway, I looked more carefully at your drawing.  Your rocket is moving in straight lines at 0.9c with respect to the sun but as shown from the perspective of the galaxy.


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## Scott DeWar (Sep 26, 2015)

Ok, professors, I am working on paper what you have shown me, but I have a quick question regarding a light year.

given velocity of C is (Apx.) 299,792,458 M/s

how long is one year?

is it:
1. 365 days at 24 hours/ day at 60 mines per hour at 60 seconds/ minute?

2. 365.25 [taking in account leap year to get closer to the astronomical year] 24 hours/ day at 60 mines per hour at 60 seconds/ minute?

or 

3. do we use 365.256363004 days here?

also, I am guessing the ship that traveled for 100 earth years, if given enough fuel and rations to travel 100 years traveling at .9c to get to the star that is 92 light years away would only use a little over 7 years of food and fuel, returning to earth to find a whole new generation of people on earth, right?

also,
I saw in one of the wiki pages that velocity is not the only effect on time, gravity too.
on this, if the ship is moving at .9 c, it will slow time in reference to Terra, but at 0 g it will speed time, but at an amount way less then the velocity effect is
but if there was a way to produce say a gravity of 1 earth grav, there would be no time disparity . . . Right?

on communication, radio or light either one moving at C, would anything happen to the signal en-route, time wise?

if the ship has traveled .9 light year, and earth sent a message aimed to intercept the ship at that point and timed to transmit 0.11...[repeat 1] Earth years down the road using the first order of approximation of 100 earth years to 3.5 ship years. Do I understand this right? I figure the ship will have traveled .035 SY. Am I right?




freyar said:


> Anyway, I looked more carefully at your drawing.  Your rocket is moving in straight lines at 0.9c with respect to the sun but as shown from the perspective of the galaxy.




by the way, that is correct

As for all of the various movements of the celestial bodies being factored in, that is a couple of orders of complications that is way over what I can handle at my early stages of comprehension.


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## Staffan (Sep 27, 2015)

Morrus said:


> Talking of time dilation and relativistic speeds, I was wondering if there were games other than my WOIN system which addressed that?  GURPS, perhaps?  Does Traveller?



There was a sidebar about it in the Spaceships sourcebook for Alternity, but nothing major.

I think most RPGs ignore it because:

1. It's pretty damn complicated.

2. It does not make for good ongoing storytelling. A character going on an expedition somewhere and returning home to a place where decades or centuries have passed? His reaction to that can make for a cool story. So can a relativistic colony ship being caught up to by a FTL ship (that's sort of the backstory to Clarke's Songs of a Distant World, except the ship catching up isn't FTL per se, just a lot faster because of more recent tech). But a bunch of PCs zipping all over the galaxy while centuries or millenia pass at home in between visits? No thanks.


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## freyar (Sep 27, 2015)

Scott DeWar said:


> Ok, professors, I am working on paper what you have shown me, but I have a quick question regarding a light year.
> 
> given velocity of C is (Apx.) 299,792,458 M/s
> 
> ...




Depends on the definition of "year" you want to use.  Astronomers use several different "years" to talk about the orbit of the earth (see wikipedia).  The main ones I know of are the Julian year which is exactly 356.25 days (at 24 hrs of 60 mins of 60 s) and the sidereal year, which is a bit longer and describes how long the earth takes to return to a set position with respect to "fixed stars."

A lightyear is c times a Julian year.

Incidentally, c is defined at a fixed value, not measured.  We measure the length of a second in terms of radiation from a cesium atom and then multiply by c to get the length of a meter.



> also, I am guessing the ship that traveled for 100 earth years, if given enough fuel and rations to travel 100 years traveling at .9c to get to the star that is 92 light years away would only use a little over 7 years of food and fuel, returning to earth to find a whole new generation of people on earth, right?



Well, I think rather than defining an amount of fuel in terms of the time it would last, you'd probably define the amount of fuel by what kind of trip you can take with it.  But for food, you're right.  For the people in the rocket, time really does pass more slowly, so they only experience the shorter length of time and therefore eat less food.  In this example, it would be the amount of food they'd normally eat in 7 years.  And, yes, there'd be a whole new generation of people on earth when they returned.




> also,
> I saw in one of the wiki pages that velocity is not the only effect on time, gravity too.
> on this, if the ship is moving at .9 c, it will slow time in reference to Terra, but at 0 g it will speed time, but at an amount way less then the velocity effect is
> but if there was a way to produce say a gravity of 1 earth grav, there would be no time disparity . . . Right?




Well, you'd remove the gravitational contribution to the time difference.  But that's very small for earth's gravity in comparison to the effect of the speed.

Gravitational time dilation and time dilation due to orbital speed are both about the same size for a satellite around the earth, and both effects have to be included in calculations of position by GPS.  So there are real world applications of all this stuff!



> on communication, radio or light either one moving at C, would anything happen to the signal en-route, time wise?
> 
> if the ship has traveled .9 light year, and earth sent a message aimed to intercept the ship at that point and timed to transmit 0.11...[repeat 1] Earth years down the road using the first order of approximation of 100 earth years to 3.5 ship years. Do I understand this right? I figure the ship will have traveled .035 SY. Am I right?




OK, the ship is moving at 0.9c compared to earth. So each year that passes on earth, the ship gets 0.9 lightyears farther away.  And vice-versa: according to someone on the ship, the ship is sitting still, and the earth is moving away at 0.9c, so the earth gets 0.9 lightyears farther away each year that passes on the ship.  But note that a year on the ship is not a year on earth, and a lightyear measured from the ship is not the same as a lightyear measured from earth either!

Just working from the point of view of earth, if the earth sends out a radio signal 1 year after the ship leaves, the radio signal has to travel both the 0.9 lightyears plus however far the ship travels (according to earth) until the radio signal catches up!  With the ship moving that fast, it actually takes 9 years for the radio signal to reach the ship (meaning, it reaches the ship 10 years after the ship leaves earth, as according to earth).  And that's 9 lightyears away from earth, as measured by earth. This is one of those situations where you have to think carefully about which reference system you're using.


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## Scott DeWar (Sep 27, 2015)

Ok, lets use a less complicated example: Poxima Centari, and theorize an Earth like (M type is it?) at a livable distance of 8:32 light mins away from Centari.

It is 4.3 light years away. We have a thriving colony there and a ship headed there for trade. Sol 3 sends a message via modulated light to the planet to warn them that the ship has an escaped criminal looking to exact revenge on some leader on Centari 3. I do not know how each is moving in reference to each other, so for the sake of argument and simplicity they are close enough to each others movement that no dilation due to velocity occurs, and each planet is nearly the same mass so no gravitational dilation occurs.

Now the ship is traveling at .9 c [chugga chugga chugga]
leaves on 1 October 2115
murderer escaped the week before, and snuck on board

He was discovered to had snuck on board on 2 October 2115, and the message was sent.

message will arrive 4.3 years later, ship will arrive 4.777... years later

those on the ship will have felt aprox .1672 years pass on the trip, or 61 days. 

It is now 1 December to them. the criminal barely had time to get settled in and act like a crew member, or what ever, when he is greeted by the centary prime federal forces.

ok, here is what all of this is getting to:

Do I understand this right? or does the message intelligence get scrambled up in timespace?


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## Staffan (Sep 27, 2015)

You're right in concept, except the numbers are off. For 0.9c the Lorentz factor is about 2.3, so while 4.8 years passed on the "outside", the time on the "inside" is more like 2.1 years than 2 months.


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## Staffan (Sep 27, 2015)

Staffan said:


> You're right in concept, except the numbers are off. For 0.9c the Lorentz factor is about 2.3, so while 4.8 years passed on the "outside", the time on the "inside" is more like 2.1 years than 2 months.




And sending a message between two more-or-less stationary objects does not scramble anything. Sending it to a moving object (or receiving it from one) would lead to blue- or red-shifting (higher/lower frequencies) - basically the same as a moving car sounding more high-pitched when it's moving toward you than when moving away.


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## Scott DeWar (Sep 27, 2015)

of curse, the Doppler effect.

And that can bee remodulated to a corrected factor.


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## Umbran (Sep 27, 2015)

Staffan said:


> You're right in concept, except the numbers are off. For 0.9c the Lorentz factor is about 2.3, so while 4.8 years passed on the "outside", the time on the "inside" is more like 2.1 years than 2 months.




Aha!  Staffan is right.  And, I find that my calculator wasn't functioning as expected, such that I was taking a square when intended to take a square root.

On other models I have used that didn't have a specific square root key, it took "inverse x^2" to be square root.  This one didn't, and just processed it as a square.  Stupid calculator.

This means that my 7 years upthread is off.  Apologies.


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## Scott DeWar (Sep 27, 2015)

Aha, I need to revisit my scribbling and scrawling to se if I had gotten it right after all. I had done it on paper and I coul not come up with 7 SY for 200 EY .


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## freyar (Sep 27, 2015)

What Staffan said. 


Scott DeWar said:


> of curse, the Doppler effect.
> 
> And that can bee remodulated to a corrected factor.



And Proxima Centauri/any of its planets are moving slowly enough compared to earth that the Doppler effect isn't a concern there.

Of course, if the ship has a receiver, it could receive a very red-shifted version of the "arrest that murderer!" message and could very simply translate it as noted.  So the criminal could be arrested long before the ship reaches its destination.


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## Scott DeWar (Sep 27, 2015)

Well, I was thinking of how hard it would be to hit a small target moving at .9c with such a narrow beam like that of laser communications.


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## Staffan (Sep 27, 2015)

Scott DeWar said:


> Well, I was thinking of how hard it would be to hit a small target moving at .9c with such a narrow beam like that of laser communications.




Presumably, the angular speed would be relatively (heh) low if they're moving away from us toward Proxima Centauri.


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## Umbran (Sep 28, 2015)

Scott DeWar said:


> Well, I was thinking of how hard it would be to hit a small target moving at .9c with such a narrow beam like that of laser communications.




At that distance, a planet is a small target too.  Use radio - chew more power, but you're more sure it can be received.


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## Scott DeWar (Sep 28, 2015)

Hrm. There might be a need for some sort of relay system then. That much power even for microwave would need to be immense, on the order of tropospheric scatter level.


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## Umbran (Sep 29, 2015)

Scott DeWar said:


> Hrm. There might be a need for some sort of relay system then. That much power even for microwave would need to be immense, on the order of tropospheric scatter level.




We already have radio transmitters that we could hear 4 light years away.


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## Scott DeWar (Sep 29, 2015)

And how do we know this? How is this possible?


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## Umbran (Sep 29, 2015)

Scott DeWar said:


> And how do we know this? How is this possible?




How do we know this?  Signal strength drops very predictably with distance.  We know the distance, and can calculate the signal strength, and compare that to what signal strengths we already know we can detect.

How is it possible?  We have *really good* receivers.


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## Scott DeWar (Sep 29, 2015)

I guess we can extrapolate the effects of 1 billion miles to the ship that just passed Pluto, huh?

I have a different thought forming in my mind, but I need to dwell on the thought a lot more yet.


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## Umbran (Sep 29, 2015)

Scott DeWar said:


> I guess we can extrapolate the effects of 1 billion miles to the ship that just passed Pluto, huh?




Don't even have to do that.  Radio is just light.  It follows an inverse-square law - the energy received at some distance D from the source drops off as 1/D^2.


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## tomBitonti (Sep 29, 2015)

For those looking for a (somewhat math-y) summary, I found this:

http://www.ece.rutgers.edu/~orfanidi/ewa/ch16.pdf

Thx!

TomB


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## tomBitonti (Sep 29, 2015)

Something neat that I just found: For lasers and other beam type emitters, there is a thing called the "far field", and (speaking very coarsely), emission does not follow the least square law until the far field is reached.

See, for example:

"The intensity (W/m^2) of an electromagnetic wave from an ordinary antenna decreases with the square of the distance from the emitter (in the far field.) Is the same true for a laser beam?"

https://www.physicsforums.com/threads/do-lasers-suffer-r-2-propagation-loss.473549/



> Not just isotropic sources, but any finite source will follow the inverse square law, including lasers. The intensity of the beam is inversely proportional to the square of the beam width. The beam width is proportional to the distance along the beam's axis. So it still works out to be inverse square. However, the beam width also has a constant term and it scales the distance along the axis by another reference value. *So it takes an appreciable distance until the beam width behaves asymptotically as proportional to the distance.*




Bold added by me.

Then:



> In general, the size of a single radiator of an antenna is on the order of a wavelength for efficiency purposes. But we can construct an array of antennas built out of such components that the overall size of the antenna is many wavelengths. The advantage being that an array allows us control over the beam width and angle. Just look at the Very Large Array. Wikipedia gives dimensions of 21 km and a wavelength of 7 mm. *That means that the far field of the VLA is around 10^{11} km. That's only 842 Au so that's perfectly suitable for deep space observation.* We can also note that the size of the dish is much much larger than the wavelength which is practical because the physics of a dish antenna is different from that of a wire antenna.




Thx!
TomB


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## Scott DeWar (Sep 29, 2015)

The distance is measured in what? Meters? and the energy is what? Joules?


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## Umbran (Sep 29, 2015)

Scott DeWar said:


> The distance is measured in what? Meters? and the energy is what? Joules?




Typically.  But really, the units don't matter so long as you are consistent about them.


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## Scott DeWar (Sep 30, 2015)

Aha! Eureka, a simple explaining of the theory of special relativity!!!!

http://imgur.com/MoRhm


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## Sadras (Oct 5, 2015)

This is such an awesome thread, love the topic, thanks @_*Scott DeWar*_. I played a little Eclipse Phase but we never got into this much detail...


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## Scott DeWar (Oct 5, 2015)

It can get really messy if you try and put it to the gaming world.


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## MarkB (Oct 6, 2015)

Scott DeWar said:


> It can get really messy if you try and put it to the gaming world.




Especially if not everyone is on the same page. I've played some Traveller games at cons, and several players (including, in one notable case, the GM) had trouble just grasping basic orbital dynamics. I dread to think what it would've been like if relativity had been a major plot point.


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## tomBitonti (Oct 6, 2015)

Two places where relativistic effects matter, but that can be absorbed in tables, is effects on projectile energies (say, what damage would be done by a 1KG projectile which is at 0.99 C), and on reaction times.  Assuming near instantaneous acceleration and deceleration, an incoming ship might be just behind its wavefront, meaning, there is less time to react to the arrival of the ship than if the ship approached at a lower speed.

Also, energy costs and time taken to accelerate increases would become more and more severe and would have to be taken into account.  Getting from the earth to mars at 0.99 C would cost a lot more than at 0.5 C.  You could make cost tables that factored in the relativistic effects and ignore time dilation effects.

Thx!

TomB


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## Scott DeWar (Oct 6, 2015)

MarkB said:


> Especially if not everyone is on the same page. I've played some Traveller games at cons, and several players (including, in one notable case, the GM) had trouble just grasping basic orbital dynamics. I dread to think what it would've been like if relativity had been a major plot point.




***gets 'messy head'***​


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## Scott DeWar (Oct 16, 2015)

*Bell's theorum on FTL communications*

http://imgs.xkcd.com/comics/bells_theorem.png


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## Mustrum_Ridcully (Oct 19, 2015)

Scott DeWar said:


> http://imgs.xkcd.com/comics/bells_theorem.png




OMG... Bell's Theorem means we have no free will!


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## MarkB (Oct 19, 2015)

Mustrum_Ridcully said:


> OMG... Bell's Theorem means we have no free will!



I already told you that tomorrow.


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## Scott DeWar (Oct 19, 2015)

MarkB, I saw you had ninja'd me the day after tomorrow.


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## Scott DeWar (Oct 20, 2015)

Ok, I am going to attempt some math in this post. Please, no comments until I post -END-<end>.

http://www.disclose.tv/news/nasa_as...l&utm_source=facebook.com&utm_campaign=buffer

Meteor 2015 TB 145 Velocity = 

78,830 MPH = [78,830 * 1.6] = 126,128 KPH; 126,128/3600 = 35,035 M/S
</end>
Time dilation formula

<end>*1/sqrt(1-v^2/c^2)*

[1 /  (1 - 35035^2 / 300,000^2)^ -2] = 
1 / (1,227,451,225 / 90,000,000,000) ^-2 =
1/ (.01363834694)^2 = 
1 / 0.1167833333 = 
8.56286570597

So, as an exorcise of thought, I am going to say that a ship has taken a mining crew to dig for  . . . we will say platinum . . . . yup, that sounds good, to the meteor 2015 TB 145 and it was sent to be timed with the rock's passing by earth at 1.3 lunar distances on 31 Halloween. 

All this in theory at a minor level of depth for approximations.

</end>So, given the above formula on time dilation is the final number I have 8.56286570597,  what is the final answer's unit of measure? Seconds per day  of decrease? 

IE: 100 days on the rock and you will be 856.2865 seconds younger then those people on  earth?


<end></end>-END-


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## Umbran (Oct 20, 2015)

I am wondering what you're trying to calculate, especially considering relevance to the thread topic.


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## Scott DeWar (Oct 20, 2015)

Time dilation. It was mentioned somewhere up on the thread. I am having trouble doing math for some reason. I know how, its just  . . .  .  sometimes I look at it and it stares back at me  tauntingly saying ,"your brain is now scrambled"


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## Staffan (Oct 20, 2015)

Scott DeWar said:


> </end>So, given the above formula on time dilation is the final number I have 8.56286570597,  what is the final answer's unit of measure? Seconds per day  of decrease?
> 
> IE: 100 days on the rock and you will be 856.2865 seconds younger then those people on  earth?




The result is a ratio: for every unit of time that passes in the accelerated frame of reference ("on the ship"), X units pass in the frame at rest ("on Earth"). So for every day on the ship, 8.6 days pass at home.

Except you made a rather big mistake by putting the speed of light at 300,000 - the right value is _300,000,000_ m/s (rounded a bit). You also forgot the *1-*(v^2/c^2) on your first line. The two combined lead to a dilation factor that's pretty much too low to be noticed in your example (1.0000000432, or about one day extra in 70,000 years).

It's probably easier to use _c_ itself as the unit of speed (that is, instead of expressing a speed as 100,000,000 m/s, it's 0.33_c_), because that gets rid of the huge "divide by 90 quadrillions" step (_c_^2 = 90 quadrillion m^2/s^2). That way, the time dilation equation simplifies to:

1 / (1-v^2)^-2

That is: take the speed you're traveling at as a fraction of the speed of light. Square it, and subtract it from 1. Divide 1 by the root of the result.

So let's take the example earlier of 1/3 the speed of light and plug it in:

1 / (1-(1/3)^2)^-2 =
1 / (1-1/9)^-2 =
1 / (8/9)^-2 = 1 / 0.943 = 1.06

In other words, when traveling at 1/3 the speed of light, you experience a time dilation of about 6%. Travel for 100 subjective years, and 106 years will have passed at home.


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## Scott DeWar (Oct 20, 2015)

A BIG mistake from the gittgo. See, I should have knowen that. That is why I tried to do it.


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## Umbran (Oct 20, 2015)

Scott DeWar said:


> 8.56286570597




That's not what I get.  I get Gamma = 1.0000678...

Think of it this way, the thing is moving at 0.011% the speed of light.  You don't start seeing human-noticeable effects until you get up around 10% the speed of light.


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## Scott DeWar (Oct 21, 2015)

I will double check my math and make corrections tomorrow.


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## Umbran (Oct 21, 2015)

Scott DeWar said:


> I will double check my math and make corrections tomorrow.




Note that with those calculations, there's a problem.  Not that you didn't get the math right, but one of experimental error.

The article quotes speeds of 35 km/sec (78,000 mph).  That's reported to what we in the business would call "two significant digits".

It isn't *exactly* 35 km/sec.  Maybe it is 35.4 km/sec.  Or 35.43 km/sec.  We don't know.  The answer is rounded off to 35, the measurement reported with limited precision.  We don't know what comes to the right of the decimal point.

Numbers I calculate from that must also be of limited precision.  I reported a gamma above of 1.0000678.  Most of that is garbage - I don't know it is really 1.0000678.  I can't trust those far out decimal places.  I can only count on the first two places in my answer being right.  Gamma = 1.0  At the level of precision these measurements are taken, the time dilation effect will not be measurable.

In order to see an effect down around the fifth or sixth decimal place, I'd need a *very* precise speed measurement - like 35.01243 km/sec.  I have to know the speed of the thing down to the single centimeter or millimeter per second in order to start to trust my calculation of the effect.


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## Scott DeWar (Oct 21, 2015)

I think it was down to 10 mm to the right of the decimal, but I will have to re work it tomorrow.


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## Umbran (Oct 21, 2015)

Well, we have 35 km/sec.  Let's look at what the stuff to the right of the decimal would be...

35.abcdef

a = 100s of meters
b = 10s of meters
c = meters
d = tenths of meters (decimeters)
e = hundredths of meters (centimeters)
f = thousandths of meters (millimeters)


I gave gamma = 1.0000678.  I need at least 6 significant digits in the speed to get 1.00006, so I need to know the speed down to decimeters per second.
To get 1.0000678 I'd need to know down to the millimeter per second.


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## freyar (Oct 21, 2015)

Another way to work out the spread of values in the "gamma factor" (ie, how many digits your answer is good to) would be to calculate based on the range of input speeds.  So if we know the speed is between 34.5 and 35.5 km/s, we know gamma is between 1.0000000066.... and 1.0000000070... .  What's interesting is that some of the calculator programs I regularly use know enough about significant digits to just say that the answer is "1" unless you really force them to give more digits.


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## Scott DeWar (Oct 21, 2015)

I will get to this when I am done with all of the game threads I am in, be right back professors!


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## Scott DeWar (Oct 21, 2015)

> Ok, I am going to attempt some math in this post. Please, no comments until I post -END-



this is take two on my incorrect math.


[1 /  (1 - 35035^2 / 300,000^2)^ -2] = 
1 / (1,227,451,225 / 90,000,000,000) ^-2 =
1/ (.01363834694)^2 = 
1 / 0.1167833333 = 
8.56286570597

Meteor 2015 TB 145 Velocity = 

78,830 MPH = [78,830 * 1.6] = 126,128 KPH; 126,128/3600 = 35,035 Km/S or 35.035555556 [repeat the 5]

Time dilation formula

*1/sqrt(1-v^2/c^2)*

[1/(1-(35.035555556^2/300,000,000^2)^-2]=
[1/(1-1227.4901531/9*10^16)^ -2]=

I will continue later, but take note that the numbers I came up with for meters/second has a repeating 5 at the mm  level. I also added the 000 to the velocity of C.

-old text-
So, as an exorcise of thought, I am going to say that a ship has taken a mining crew to dig for  . . . we will say platinum . . . . yup, that sounds good, to the meteor 2015 TB 145 and it was sent to be timed with the rock's passing by earth at 1.3 lunar distances on 31 Halloween. 

So, given the above formula on time dilation is the final number I have 8.56286570597,  what is the final answer's unit of measure? Seconds per day  of decrease? 

IE: 100 days on the rock and you will be 856.2865 seconds younger then those people on  earth?


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## MarkB (Oct 21, 2015)

One thing to bear in mind is that the article only provides a simple figure for the asteroid's speed, not an actual velocity. Is it moving at 78,000 mph relative to the solar system in general, or is that its relative speed of approach to Earth? Bear in mind that Earth is travelling around the sun at 70,000 mph.


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## Scott DeWar (Oct 21, 2015)

I am going to take the "relative to earth" choice here just for the sake of a math test. this is all it is. just a test for me.


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## Staffan (Oct 21, 2015)

Scott DeWar said:


> 78,830 MPH = [78,830 * 1.6] = 126,128 KPH; 126,128/3600 = 35,035 Km/S or 35.035555556 [repeat the 5]
> 
> [stuff]
> 
> I will continue later, but take note that the numbers I came up with for meters/second has a repeating 5 at the mm  level. I also added the 000 to the velocity of C.



Except the "source" number isn't the speed in meters per second. It's the speed in MPH, which has four (or possibly five, but zeroes at the end usually doesn't count) significant digits. More properly, you'd say that the speed is 35,040 m/s. Those trailing fives are "false" accuracy.

Hmm, come to think of it, you actually only have two significant digits, because you use 1.6 as the conversion factor between mph and km/h. Plug in 1.609 instead and you get up to four again. That gets you to 35,230 m/s... which actually works out well as an illustration of why significant digits are a thing.


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## Scott DeWar (Oct 22, 2015)

**BANG BANG BANG** That is the sound of my head hitting the table top.


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## Umbran (Oct 22, 2015)

Staffan said:


> Except the "source" number isn't the speed in meters per second. It's the speed in MPH, which has four (or possibly five, but zeroes at the end usually doesn't count) significant digits.




I am going to be picky, because this is science, and we are talking about precision.

The article says it has a speed of "35 kilometers per second (78,000 mph)".  These are, actually, both given with two significant digits.  In the absence of other graphic elements telling us which are significant (a bar above or below the rightmost significant zero is common), you generally assume that all three trailing zeroes are not significant.  The other indicator is that, typically, shifts in units of measure do not introduce or remove significant figures - so the km/sec and mph figures should have the same number of significant figures.  

https://en.wikipedia.org/wiki/Significant_figures
*Concise rules*

All non-zero digits are significant
Zeros between non-zero digits are significant.
Leading zeros are never significant.
In a number with a decimal point, trailing zeros, those to the right of the first non-zero digit, are significant.
In a number without a decimal point, trailing zeros may or may not be significant. More information through additional graphical symbols or explicit information on errors is needed to clarify the significance of trailing zeros.


By the way, if you are going to do a calculation in metric units, and they *hand* you a metric measurement (km/sec), it doesn't make a whole lot of sense to use the Imperial measure, and then convert it to metric.  35 km/s = 35,000 m/s, and off you go!


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## Staffan (Oct 22, 2015)

Umbran said:


> By the way, if you are going to do a calculation in metric units, and they *hand* you a metric measurement (km/sec), it doesn't make a whole lot of sense to use the Imperial measure, and then convert it to metric.  35 km/s = 35,000 m/s, and off you go!



Ah, I didn't see that the original measure was given as 35 km/s and assumed the MPH figure was the original.


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