NASA's Kepler Mission Announces Largest Collection of Planets Ever Discovered

[Gadion]

It's not like I understood everything the first time I heard about it. It IS difficult and easy to forget that I wasn't born with the knowledge. I certainly wasn't.
Don't worry, after a while it will feel relatively natural. You can view spacetime as two glasses. If you fill the space one with speed you need to take away speed from the time glass. You have only a specific amount of speed that you switch between the glasses. If it's almost empty you are "frozen" in space or time depending on the glass and very quick in the other glass. Quick space travelling means slow time travelling. It's not linear though. At our human speeds it doesn't practically feel any difference.

Okay... if I understand this any at all, using your glass analogy:
There is a relationship between time and speed, where the speed you travel at can alter the time required to cover a distance, but this relationship is nonlinear. So, if you increase the speed enough, the time required to cover any distance, becomes infinitesimal?
Say, you could somehow travel at the speed of light, then your speed from your vantage and dimension would be such that infinitesimal time passes over a vast distance.
While if this is viewed from a stationary point relative to this traveler, it takes whatever distance in light-years the object being traveled to is away. And this is because the speed being traveled is perceived in linear terms, while it is actually being performed on nonlinear scales?

Excuse the pseudoscience babble :s

[Ripster42]

Okay... if I understand this any at all, using your glass analogy:
There is a relationship between time and speed, where the speed you travel at can alter the time required to cover a distance, but this relationship is nonlinear. So, if you increase the speed enough, the time required to cover any distance, becomes infinitesimal?
Say, you could somehow travel at the speed of light, then your speed from your vantage and dimension would be such that infinitesimal time passes over a vast distance.
While if this is viewed from a stationary point relative to this traveler, it takes whatever distance in light-years the object being traveled to is away. And this is because the speed being traveled is perceived in linear terms, while it is actually being performed on nonlinear scales?

Excuse the pseudoscience babble :s

I think you're getting it if I understand you correctly as long as you mean the amount of time can alter from within the reference frame of the thing travelling and not due to just covering the ground quicker. It's posited that from a photon's frame of reference no time passes between its emission and absorption.

[Gadion]

as long as you mean the amount of time can alter from within the reference frame of the thing travelling and not due to just covering the ground quicker

I'm afraid then I don't understand.
I was under the impression that the frame of reference for whatever the travelling object is, is quite dependent on the covering the ground quicker.

[Ripster42]

I'm afraid then I don't understand.
I was under the impression that the frame of reference for whatever the travelling object is, is quite dependent on the covering the ground quicker.

I'll just pose this scenario then: both the runner and the stationary guy are actually stationary within their own reference frames. It has nothing to do with covering ground quicker, but with reconciling the fact that the speed of light is observably constant no matter how fast different reference frames are moving with relation to each other.

Edit: I'll use the train example instead of the runner example.

Lets say there is a train on a track with an observer on it. There is another observer off of the track. The guy on the train releases a photon and in his frame of reference it travels from the photon emitter to a mirror and then to a detector right next to the emitter. This whole apparatus of emitter/mirror/detector is on the train. Lets say the distance from detector->mirror or mirror->emitter is D. So the distance traveled by the photon is 2*D. Since the speed of light is constant this took 2*D/c=t. For the observer off the (I forgot to mention, the train is moving) train the distance the light travels isn't just 2*D, but actually travels further. It isn't ACTUALLY 2*D+distance train traveled d because the light would be taking the straight line path there and not up+down then over but for this exercise we'll just say it is. So for the guy off the train the distance traveled by the photon is 2*D+d. Since the speed of light is constant the time(t) it took to travel that distance was (2*D+d)/c=t. So it took longer for the stationary observer to see the photon emit+bounce+absorb than it took for the guy on the train because the light actually travels farther.


(The reason I'm just using 2*D+d is because I'm trying to keep the math simple and for the actual expression it incorporates the velocity of the train to figure out distance the train traveled and then using that to get an isosceles right triangle with both same sides being the D for the stationary observer to get you a new 2*D/c=t)

[Gadion]

I'll just pose this scenario then: both the runner and the stationary guy are actually stationary within their own reference frames. It has nothing to do with covering ground quicker, but with reconciling the fact that the speed of light is observably constant no matter how fast different reference frames are moving with relation to each other.

Edit: I'll use the train example instead of the runner example.

Lets say there is a train on a track with an observer on it. There is another observer off of the track. The guy on the train releases a photon and in his frame of reference it travels from the photon emitter to a mirror and then to a detector right next to the emitter. This whole apparatus of emitter/mirror/detector is on the train. Lets say the distance from detector->mirror or mirror->emitter is D. So the distance traveled by the photon is 2*D. Since the speed of light is constant this took 2*D/c=t. For the observer off the (I forgot to mention, the train is moving) train the distance the light travels isn't just 2*D, but actually travels further. It isn't ACTUALLY 2*D+distance train traveled d because the light would be taking the straight line path there and not up+down then over but for this exercise we'll just say it is. So for the guy off the train the distance traveled by the photon is 2*D+d. Since the speed of light is constant the time(t) it took to travel that distance was (2*D+d)/c=t. So it took longer for the stationary observer to see the photon emit+bounce+absorb than it took for the guy on the train because the light actually travels farther.


(The reason I'm just using 2*D+d is because I'm trying to keep the math simple and for the actual expression it incorporates the velocity of the train to figure out distance the train traveled and then using that to get an isosceles right triangle with both same sides being the D for the stationary observer to get you a new 2*D/c=t)

I'm guessing it doesn't really matter in which direction the train is moving, but to my mind it makes this entire scenario difficult (lol) My thinking is that the train is moving in the direction of the light and so, the distance the light travels becomes ever smaller. I do understand that this would make the distance for the light to travel to the observer larger, but this is probably not what you are trying to tell me.

Off-topic, I'm a little pressed for time, and I won't be able to consider this any further just yet, but I promise to return to it later. And I thank you for taking the time to teach me and your patience.

[Ripster42]

I'm guessing it doesn't really matter in which direction the train is moving, but to my mind it makes this entire scenario difficult (lol) My thinking is that the train is moving in the direction of the light and so, the distance the light travels becomes ever smaller. I do understand that this would make the distance for the light to travel to the observer larger, but this is probably not what you are trying to tell me.

Off-topic, I'm a little pressed for time, and I won't be able to consider this any further just yet, but I promise to return to it later. And I thank you for taking the time to teach me and your patience.

Hmm, I think I didn't explain it well enough. The photon is moving perpendicular to the direction of motion of the train.



That's for a ball, but the concept of the ball travelling different distances is the same as for a photon. Well, close enough.

[Uzkin]

I'm afraid then I don't understand.
I was under the impression that the frame of reference for whatever the travelling object is, is quite dependent on the covering the ground quicker.

The geometry of Minkowski space-time is such that the INTERVAL s between two given events is the same for all (inertial) observers. The interval is defined as



Here, Δr is the spatial distance between the events and Δt the temporal distance between them. Thus, if upon a change of frame the spatial distance decreases, the temporal distance must decrease as well to keep the interval invariant.

In my previous post, there were two relevant events: the launch of the rocket from Earth and its arrival at the distant star. For the Earth-based observer, Δr = 4 light years and Δt ≈ 4 years (approx), giving s ≈ 0. For the passengers, those are 0 light years and ∼0 years, respectively, so again s = 0. Space-time interval is invariant.

[Yvaelle]

You can't travel faster than light (probably) but you don't need to. All you need is to travel close to light speed and then you could reach any spot in the universe in a matter of seconds.

Actually, this is exactly, perfectly wrong in every way.

You very likely can travel faster than light, but you can't travel 'close to lightspeed'. Even if you could travel 99.999999999999999% (or ~100%) the speed of light, you certainly could fucking not get anywhere in the universe in a matter of seconds.

The common unit of measure in space is a "Lightyear" - that is the distance light travels if you just let it go, for a year. If you could get anywhere in the universe in a matter of seconds by travelling "close to the speed of light" - the entire universe would have to be about the size of Sun, which is 4.6 lightseconds across the diameter. It takes minutes (8) just for light from the sun to get to Earth.

The observable universe is 93 Billion lightyears across: even if you travelled at ~100% the speed of light, it would take you 93 billion years to cross it, and in those years the continued expansion would make it far, far, far bigger still.

Ok so - going back to the starting bit - accelerating mass close to the speed of light requires a literally infinite amount of energy - more energy than all the energy in the universe: the speed of light is an asymptote for matter - so no - it's not easy - it's literally impossible.

Travelling faster than light is almost certainly easier than travelling close to the speed of light, because to travel faster than light we need to think differently. An alcubierre warp drive works by building a ship which never moves, objects which don't move, don't approach the speed of light - and therefore don't require infinite energy (Sloth ships are lazy as f--k). Instead, they would compel the shadows of the universe to shift around the ship, creating a bubble of non-existence with the ship inside. The ship never moves, never accelerates, requires no engines - it instead bends existence around itself until it reappears elsewhere.

Shadows move faster than light, we just need to become the shadows

[Ripster42]

You very likely can travel faster than light, but you can't travel 'close to lightspeed'. Even if you could travel 99.999999999999999% (or ~100%) the speed of light, you certainly could fucking not get anywhere in the universe in a matter of seconds.
)

We've already talked about this in the thread. It depends on your frame of reference.

[Uzkin]

Actually, this is exactly, perfectly wrong in every way.

No it's not. Me, Holas, Ripster42 and so on have already explained this. It is how Special Relativity works, and it seems to describe the nature very well within its domain of application.

[TheLimonTree]

That's nice but uh... how do we get there.

Obviously wait for the warlock to summon us.

[Gadion]

Hmm, I think I didn't explain it well enough. The photon is moving perpendicular to the direction of motion of the train.



That's for a ball, but the concept of the ball travelling different distances is the same as for a photon. Well, close enough.

I'm afraid that ball graph helps me very little, but is it anything like this Wikipedia section on time dilation?
This section at least seems to make mathematical sense, although it does seem counter-intuitive. To me, it does explain though that the time that passes for a moving object is lower than that for a stationary object (although, if the movement of the object is greater than the speed of light, the function is undefined, so the speed of light is set as a maximum velocity).

- - - Updated - - -

The geometry of Minkowski space-time is such that the INTERVAL s between two given events is the same for all (inertial) observers. The interval is defined as



Here, Δr is the spatial distance between the events and Δt the temporal distance between them. Thus, if upon a change of frame the spatial distance decreases, the temporal distance must decrease as well to keep the interval invariant.

In my previous post, there were two relevant events: the launch of the rocket from Earth and its arrival at the distant star.

For the Earth-based observer, Δr = 4 light years and Δt ≈ 4 years (approx), giving s ≈ 0.

Okay... I'm trying to understand, but given that s is not only a function of r, and t, but also of c, won't s= √(16-16c^2)?

For the passengers, those are 0 light years and ∼0 years, respectively, so again s = 0. Space-time interval is invariant.

Over here, I can see that the solution to this equation would be zero, given that both Δr and Δt equal zero. I can also understand that the time required to travel this distance approaches infinitesimal if your speed is incredibly high, so I am willing to accept that Δt is just about zero. However, I do not understand why Δr would be zero. Can you please explain?

[Uzkin]

Okay... I'm trying to understand, but given that s is not only a function of r, and t, but also of c, won't s= √(16-16c^2)?

Let y = year and ly = light year. Now in the Earth's frame, Δr = 4 ly, c = 1 ly/y, Δt = 4 y. Therefore s^2 = (4 ly)^2 - (1 ly/y)^2 * (4 y)^2 = 16 ly^2 - 16 ly^2 = 0, that is s = 0.

Over here, I can see that the solution to this equation would be zero, given that both Δr and Δt equal zero. I can also understand that the time required to travel this distance approaches infinitesimal if your speed is incredibly high, so I am willing to accept that Δt is just about zero. However, I do not understand why Δr would be zero. Can you please explain?

In the rocket's frame the position of the rocket is always r = 0. For example, in the rockets frame, the launch happens at r(launch) = 0. Likewise, the rocket encounters the planet at r(arrival) = 0. The spatial distance between these two events (launch and arrival) in the rocket's frame is therefore Δr = r(arrival) - r(launch) = 0.

[Gadion]

Let y = year and ly = light year. Now in the Earth's frame, Δr = 4 ly, c = 1 ly/y, Δt = 4 y. Therefore s^2 = (4 ly)^2 - (1 ly/y)^2 * (4 y)^2 = 16 ly^2 - 16 ly^2 = 0, that is s = 0.

Okay, if c is measured in light-years, then c is one, and the equation makes perfect sense.

In the rocket's frame the position of the rocket is always r = 0. For example, in the rockets frame, the launch happens at r(launch) = 0. Likewise, the rocket encounters the planet at r(arrival) = 0. The spatial distance between these two events (launch and arrival) in the rocket's frame is therefore Δr = r(end) - r(start) = 0.

So, r isn't the distance the rocket has traveled, but the distance that any object is currently from the rocket?

[Ripster42]

I'm afraid that ball graph helps me very little, but is it anything like this Wikipedia section on time dilation?
This section at least seems to make mathematical sense, although it does seem counter-intuitive. To me, it does explain though that the time that passes for a moving object is lower than that for a stationary object (although, if the movement of the object is greater than the speed of light, the function is undefined, so the speed of light is set as a maximum velocity).

Yes, that's exactly what I'm trying to convey. The "non-moving" observer sees the photon move a longer path than the observer on the train perceives. They both measure light moving at the same speed. Since c is a constant and d is different, it means time also has to be different since d=c*t.

- - - Updated - - -

So, r isn't the distance the rocket has traveled, but the distance that any object is currently from the rocket?

From the frame of reference ya if I remember right.

[Uzkin]

So, r isn't the distance the rocket has traveled, but the distance that any object is currently from the rocket?

A spacetime EVENT has a certain time and a certain place associated with it. In the rocket's frame the launch event takes place at t = 0 and r = 0, and the arrival at t = (something very small) and r = 0. The spatial distance between these events is 0, because both take place where the rocket is at that time. The temporal distance is "something very small", assuming that the rocket is traveling "almost" at c.

[Voidism]

Actually, this is exactly, perfectly wrong in every way.

You very likely can travel faster than light, but you can't travel 'close to lightspeed'. Even if you could travel 99.999999999999999% (or ~100%) the speed of light, you certainly could fucking not get anywhere in the universe in a matter of seconds.

The traveller would feel like just a few seconds would've passed if they flew close to the speed of light, even if the distance was to the Andromeda galaxy - because relative to earth, the travellers time is close to frozen. People on the earth would feel like 3 million years have passed when the traveller arrives at the Andromeda galaxy. The whole thing is that there's no universal time - everything is relative.

- - - Updated - - -

Okay... if I understand this any at all, using your glass analogy:

You pretty much got the basics yes.

[Gadion]

Yes, that's exactly what I'm trying to convey. The "non-moving" observer sees the photon move a longer path than the observer on the train perceives. They both measure light moving at the same speed. Since c is a constant and d is different, it means time also has to be different since d=c*t.

A spacetime EVENT has a certain time and a certain place associated with it. In the rocket's frame the launch event takes place at t = 0 and r = 0, and the arrival at t = (something very small) and r = 0. The spatial distance between these events is 0, because both take place where the rocket is at that time. The temporal distance is "something very small", assuming that the rocket is traveling "almost" at c.

You pretty much got the basics yes.

I think I sort of understand now, and I would like to thank you all for your effort in helping me to get this
At this point, I'd also wish to say thanks to @Holas for starting this branch of the discussion, and issue him that apology I owe him.

Thanks guys

[Yvaelle]

The traveller would feel like just a few seconds would've passed if they flew close to the speed of light, even if the distance was to the Andromeda galaxy - because relative to earth, the travellers time is close to frozen. People on the earth would feel like 3 million years have passed when the traveller arrives at the Andromeda galaxy. The whole thing is that there's no universal time - everything is relative.

Which is fine for the traveller, but a pretty shitty form of transportation if we're building a space empire. By the time you finally get there, our entire civilization has risen, fallen to the inexorable conquering of the Militant Hyper-Intelligent Space Dolphin Empire, fallen to dust, and then the intergalactic dolphin empire too fell to dust and ruin.

Even if we could travel close to the speed of light for millions of years (again, more than all the energy in the universe) - it's at best a last ditch effort by a soon-to-be extinct civilization.

Saying that time is relative is useful for knowing that you will not age during the trip, but it's useless for virtually any other discussion - because outside your spaceship, once you stop moving at lightspeed, time still really, really, really matters: time moves on for literally everything else in the universe, even if you don't perceive it.

[Noradin]

We've already talked about this in the thread. It depends on your frame of reference.

You wouldn't survive the acceleration needed to reach the speed you speak of in seconds, let alone both the acceleration and the deceleration, thus even if you could pass any two points in the universe in a few seconds the need to start and stop the journey makes it impossible.
You wouldn't be able to get enough energy to do it compressed enough either.

And that still leaves you with the problem that when you get there nothing will be like it was before, so unless you can predict the future a few milion years in advance in just a few seconds your proposal is simply the most expensive suizide ever dreamed up.