November 14th, 2009, 12:00 PM
Time relative to whom/what? Besides the fact that your OP was short on details, the question in your thread title reads like, "What's the difference between a duck?"
oh im just trying to solve this apparent contradiction of weather you are in a or b calculated mass of the star comes out differently
certainly this is a posible case and i dont see why such contraption couldnt orbit a star
ah and time relative to earth if you please
oh and yes its a problem such limited space for the title
If the two planets aren't connected then they will orbit at different speeds, and Fc and Fg will be equal in both cases.Originally Posted by luxtpm
If you connect the two planets such that they have the same angular velocity, then they will orbit at a speed equal to that of a mass located at a point halfway between them (the point of the center of Mass of the system.) This less than the angular needed to maintain an orbit for B and more than needed to maintain an orbit for A.
IOW, "centrifugal force" is smaller than gravitational force at body B, and greater at body A. The system is pulled in opposing directions by this differential in force. This same differential acts across the diameter of the Earth from the side facing the Sun to the opposite side and is what causes the Solar tides.
So, if you solve for the Sun's mass with the planet's unconnected, you solve separately for each planet's orbital speed and distance and you will get the same answer, and if you solve for the connected system you have to solve for the connected system's orbital speed and center of mass distance, and again you will get the same answer.
There is no contradiction.
A word of advice: Every time that you think you've found a contradiction in basic Physics or uncovered a situation where the laws of conservation have been violated, you've made a mistake or applied a misconception somewhere.
*Cough*
*Cough*
So, just out of curiosity, what is the relationship between distance from a massive body and amount of time dilation? Does anyone know the formula? (Is there a simple formula, or is it really complicated?)
I mean, I kind of know how you'd go about determining it, but I'd really rather save myself the headache and just ask.
i did an experiment that has me puzzled with the outcome:
i took two plastiline balls as planets, i united them by a pencil and atached a string to the midel of the pencil and rotate the string
amazingly the balls pencil system locks in phase to the string rotation and yet more amazingly the pencil remains perpendicular to the radius of the orbit
so its neither you can simplify the thing into analizing the cog cause if so it would have rotated flacidily not in phase
neither you can consider two separated weights and apply centrifugal force formula cause if so the pencil would have adopted a radial position not perpendicular to the radius
where T' is the time for an observer far removed from mass M and r is the distance from M.
So for instance, at the surface of the Earth the time dilation is about 1.000000001. Meaning that for every 100 years that passes on Earth, 100 yrs and 2 sec pass for our far removed observer.
No contradictions here. While behavior at the quantum level might differ from what we would expect from what common experience tells us, it does not contradict it. IOW, there is nothing in quantum mechanics that says objects cannot behave the way we see them behave on the macroscopic level and still behave as they do at the quantum level.*Cough*
*Cough*
Now while we have not yet been able to completely reconcile Relativity with QM, this dos not mean that it can't be done. Besides that, such reconcilation is far beyond basic physics.
Thanks Janus. :-)
where T' is the time for an observer far removed from mass M and r is the distance from M.
So for instance, at the surface of the Earth the time dilation is about 1.000000001. Meaning that for every 100 years that passes on Earth, 100 yrs and 2 sec pass for our far removed observer.
Now another question: Doesn't that equation imply that you could have negative values for T?
It looks to me like, if there's a high enough M, or a low enough r, then 2GM/RC^2 could become larger than 1. I guess time would slow all the way to a stop if the density of a black hole ever began to approach that limit, however, which would prevent it from surpassing the crucial point.
[quote="kojax"][quote="Janus"]If you look closely enough at that equation you will notice that it is the same as the standard time dilation formula with the escape velocity inserted for "v". Or put another way, the only way that 2GM/rc² becomes larger than 1 it where the escape velocity is greater than c, and this only occurs within the event horizon of a black hole.
In addition, this still would not give you negative value for T. 1-2GM/rc² is inside a radical sign, which would give the square root of a negative number, which is imaginary.
[quote="Janus"][quote="kojax"]So, if time stops at the event horizon of a black hole, wouldn't that imply that nothing ever falls in? All the matter that has ever reached the horizon must still be there, then.If you look closely enough at that equation you will notice that it is the same as the standard time dilation formula with the escape velocity inserted for "v". Or put another way, the only way that 2GM/rc² becomes larger than 1 it where the escape velocity is greater than c, and this only occurs within the event horizon of a black hole.
Good point. I can't believe I missed that.
In addition, this still would not give you negative value for T. 1-2GM/rc² is inside a radical sign, which would give the square root of a negative number, which is imaginary.
From an observer's perspective, yes, but not from the perspective of the object crossing the event horizon. Also, AFAIK, in practice you would stop being able to see the object before you'd see it stop short of the event horizon, as the light leaving it would be redshifted beyond detectable levels.So, if time stops at the event horizon of a black hole, wouldn't that imply that nothing ever falls in? All the matter that has ever reached the horizon must still be there, then.
True, though we would both agree as to its present location. Suppose an asteroid fell into a black hole 2 billion years ago.
Our perspective: 2 billion years have passed, and it's still at the event horizon.
Its perspective: 0.00000 seconds have passed, and it's still at the event horizon.
I like how you describe it as extreme red shifting. That's kind of the way I see it too, because it's not like the light is ever going to turn around and go back in. Or would it in some cases?Also, AFAIK, in practice you would stop being able to see the object before you'd see it stop short of the event horizon, as the light leaving it would be redshifted beyond detectable levels.
I don't think it works like this. As I understand it you would be able to fly a space ship right past the event horizon without any problems (barring physical effects). Time doesn't stop from the asteroid's perspective and if time doesn't stop, movement also doesn't and it crosses the event horizon.True, though we would both agree as to its present location. Suppose an asteroid fell into a black hole 2 billion years ago.
Our perspective: 2 billion years have passed, and it's still at the event horizon.
Its perspective: 0.00000 seconds have passed, and it's still at the event horizon.
I don't think so. Inside the event horizon all paths lead parallel or away from the event horizon, but outside of it, it would escape.Or would it in some cases?
Kalster is correct. The asteroid would carry on about its way as if nothing had changed... It would just sail right past the horizon. [EDIT: This next section is not a very accurate view... it's about being infinitely red-shifted and the light never reaching the outside observer... not appearing "frozen" to them] However, to an outside observer, it would appear to be frozen right at the horizon. This is why the Russians used to refer to blackholes as "frozen stars." They appeared frozen from our frame of reference. [/EDIT - See clarifying remarks above]. However, the object crossing the EH sails right past as if nothing has changed... and really wouldn't notice anything until the tidal effects ripped it apart.
if time on earth is normal and as you fall into the black hole your time with respect to earth becomes zero wouldnt you reach transfinity after one second as you fall into the black hole?
Yeah. That's kind of what I'm thinking here too. If we see the asteroid stand still from our vantage point, then an observer standing on the asteroid must see time move very very fast when he's looking back at us.
He could barely have time to blink, and the Earth's sun would have burned out before he could open his eyes again. Or... actually that's if he were near the event horizon. If he were in it, then infinity time would pass during that blink, instead of just a few billion years.
The asteroid's perspective of its immediate vicinity would be unchanged, but what would it see when it looks back from the singularity toward us? What do we look like to it? Do we look frozen, or do we look like we're racing around at supersonic speed?
well if light slows down at the same rate i think he should see things unchanged
If the asteroid looked back at us after crossing the event horizon, from its perspective time would appear to be passing very quickly for those of us outside.
Light, by definition, does not accelerate (or decelerate... aka, negatively accelerate). It is always moving at constant speed if it exists at all. There is no speeding or slowing.
True, but since light is the great clock that all time is measured relative to, saying that time slows down, and saying the clock slows down are basically the same thing. It's slightly less precise, but still accurate.Light, by definition, does not accelerate (or decelerate... aka, negatively accelerate). It is always moving at constant speed if it exists at all. There is no speeding or slowing.
... Just so long as luxtpm knows all the implications of what he's saying. We have to make sure and be consistent. We can't talk about C and time as identical things in one part of the conversation, and then treat them as being different in another.
Light does not slow down. Instead, its wavelength gets longer.
BUT IMAGINE YOU ARE ALMOST AT Light speed and you look back
you are frozen in time so you should see things happening very fast
but this would mean that the info transmitted by light is reaching you much faster than you are going
No, from your own frame of reference, you are not frozen. You are moving along as if everything were perfectly normal. Others would appear to be moving rather fast, but you would not sense that you are "still." You are moving along as per usual, it's just that the relative velocity of those in other frames of reference seems to have increased (note... from their own perspective, they too are moving along as per usual... no change... just a normal day at the park).
Also, from other frames of reference, you are still not frozen when they look back at you... You are just red-shifted. Eventually, as you get more and more red-shifted, the people outside can no longer see you (as you get closer to infinite red-shift), but you are not "frozen" ... even from their perspective (I really should have made this more clear in my post above... sorry about not doing so with my first contribution and for being a bit lazy and somewhat inaccurate).
my point:
if you fall into a blackhole you reach transfinity since after 1 normal second for you more than infinite time has happened on earth
yet as time dilation has a gradient everithing must be syncronized
i mean what problem should you have if your upper body goes faster than your botton as it happens minuscally
i dont understand it of course but i see that without sync things wouldnt work
the only reason cause you cant comunicate beyond the event horizont is cause at that point scape vel for g transcends c
yet you could pseudocomunicate with quantum enatnglement
in the end thats not perfect info but with just one bit and quantum entanglenmnet is enough to send any message
so in the end the question could be reduce to are you in the mirror world or am i
so rry since everybody says nonsenses i say mine, for what is known i could be as right as anybody
An interesting thing to add is that the frequency of light is a time-dependent observation. If you perceive time to be moving more slowly, then you will also perceive the frequency of all incoming light to be higher. (blue shifted)
I'm not sure if that would fully counter the redshifting of the light coming from behind you. I'd have work through some equations to be sure, but it would definitely lessen the effect.
If I understand correctly, the what happens to light in a gravitational field is that its path becomes curved, which means that it is taking an indirect course when it travels between objects, rather than moving in a straight line. This should also give it the appearance of taking longer to arrive.
No.
Light follows a geodesic path in spacetime. There is a no such thing as a Euclidean "straight line" because we are talking about non-Eudlidean geometry. A geodesic is as straight as it gets.
No. Time slows down, but never that much. I can't provide many details on this point though.
The problem is that near enough to a black hole, the difference isn't miniscule anymore. Look up the word "spaghettification".
True. This is why even light can't get out.
False. You cannot communicate at all with any number of entangled qubits. While something is going on between those bits, there is no way to extract information from that. Also, it's very likely that the entanglement would be destroyed rapidly for any number of reasons.
Sorry, but just because some people say nonsense doesn't make your nonsense any more right.
I think that's a slightly clearer version of what I was trying to say. Gravity creates non-euclidean geometry, forcing the light to follow a course that isn't a straight line, which in turn causes it to take longer to arrive at its destination.No.
Light follows a geodesic path in spacetime. There is a no such thing as a Euclidean "straight line" because we are talking about non-Eudlidean geometry. A geodesic is as straight as it gets.
I mean that, from the perspective of an observer far away from the gravitational field, it looks like it's not a straight line, but if you're inside the field, I guess the geodesic would look like a straight line to you. So, there's no absolute perspective that says any line would be "straight" but there are relative perspectives, where one perspective would say the other was curved. Each perspective thinks its geodesic is the straight one.
Or ... I might be misunderstanding your version too.
I think the tricky bit is that far away observers wouldn't see it as curved either, but there are indirect ways to tell that it must be curved (such as seing the same stars on two sides of a massive body). That's just my understanding though.
A geodesic is the shortest path between two points, so the geodesic is the fastest route light can take, by anyone's perspective.
For example. If you walk due west to a destination, from your perspective you are walking a straight line. Another person walks the geodesic between the two points at the same speed. From your perspective, he walks a curve, starting off walking in a direction just a bit North of West and ending up walking in a direction just a little South of West. But by both your perspectives, he takes less time to make the trip.
if light curves respect space which really curves?
whats the shortest path what you feel percieve as curve or as straight
therefore the shortest path can be what we percieve as a curve?
therefore a staright line can be curved?
edit:
so you can see one thing 1 m away and the shortest way is going trillion km to the opposite side?
oh come on if space was even curved with radius of million km we would tell, i dont swallow gravity curves space
That's the wrong path to your brain. Maybe you want to study this:
http://en.wikipedia.org/wiki/Tests_o...ght_by_the_Sun
http://en.wikipedia.org/wiki/Gravitational_lens
So it sounds like the path light takes in a gravitational field appears curved even to a nearby observer, but if the Earth were really flat, the light wouldn't curve at all. There would be no geodesic.A geodesic is the shortest path between two points, so the geodesic is the fastest route light can take, by anyone's perspective.
For example. If you walk due west to a destination, from your perspective you are walking a straight line. Another person walks the geodesic between the two points at the same speed. From your perspective, he walks a curve, starting off walking in a direction just a bit North of West and ending up walking in a direction just a little South of West. But by both your perspectives, he takes less time to make the trip.
So, far away from a gravitational body, the light pretty much travels a straight line.
Yes, far away from massive objects, space is nearly Euclidean, which is what we think of when we thing of straight lines and such. All though at the largest scales it may not actually be Euclidean.
Near massive objects, the question is much more complicated. I can't say for sure, but it may look to any observer like light always travels in straight lines, since the view is curved just as much as the light is.
Light always follows a geodesic path in space-time.
A geodesic is a close to a straight line as it gets. In fact a straight line is nothing more than a geodesic in Euclidean space.
So, the closer you are to being in Euclidean space, the closer you are to having light move in a straight line. Because Euclidean space is the place where a straight line and a geodesic become identical to each other.Light always follows a geodesic path in space-time.
A geodesic is a close to a straight line as it gets. In fact a straight line is nothing more than a geodesic in Euclidean space.
I think what magimaster might be suggesting is that an observer is always presented with the illusion that the area of space where they are located is fully euclidean, even though it is not.
No,
Light always moves in a "straight line". A geodesic is the equivalent of a straight line in an arbitrary geometry. It is as straight as it gets.
A line is nothing but another word for a geodesic, which is applicable only in a strictly Euclidean geometry.
Sorry, I just meant that we might not be able to see the non-Euclidean geodesics as non-Euclidean, since we're in the same space. That particular thought is kind of hard to put into everyday words.Light always follows a geodesic path in space-time.
A geodesic is a close to a straight line as it gets. In fact a straight line is nothing more than a geodesic in Euclidean space.
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