Jump to content

Is there a common moment of now throughout the Universe?


1x0

Recommended Posts

OK.

What I say is that the points that this line is made of are connected by simultaneity, which is unphysical.

False, two events can be simultaneous, has nothing to do with "force"

 

 

 

Because all forces need time. IOW this line is unphysical.

 

While forces need time to propagate this does not mean that two (or multiple ) events cannot be simultaneous. So, your claim has no basis. It is perhaps best if you opened a separate thread on your misconceptions about physics.

Link to comment
Share on other sites

OK.

What I say is that the points that this line is made of are connected by simultaneity, which is unphysical. Because all forces need time. IOW this line is unphysical.

'Unphysical' isn't the right word, because it has a physical meaning, and it can be measured. Eg. send a light signal to two receivers the same distance from you, and they'll receive them simultaneously, even though the two reception events are not directly causally connected to each other.

 

The best you could say is that simultaneity is 'conventional'. http://plato.stanford.edu/entries/spacetime-convensimul/gives an overview.

It is widely accepted (though not settled) that simultaneity is indeed conventional.

 

 

To relate that back to this thread... simultaneity is meaningfully defined in flat spacetime, using Einstein's definition, whether or not that is merely a convention. It would seem there's no such meaningful definition for curved spacetime, even though you could arbitrarily make one (eg with a foliation of Cauchy surfaces) if the universe fits the requirements (if it's globally hyperbolic or whatever).

 

You could for example take all events in the universe, and construct a Cauchy surface that means "this is all the events that I consider to be happening 'now'". That too would be a convention, but I think it would be physically meaningless, because you would probably have no way of unambiguously measuring the simultaneity of events no matter how you choose to define it. In flat spacetime Einstein's definition of simultaneity defines an unambiguous measurement of simultaneity.

Link to comment
Share on other sites

To relate that back to this thread... simultaneity is meaningfully defined in flat spacetime, using Einstein's definition, whether or not that is merely a convention. It would seem there's no such meaningful definition for curved spacetime, even though you could arbitrarily make one (eg with a foliation of Cauchy surfaces) if the universe fits the requirements (if it's globally hyperbolic or whatever).

 

 

You could do that only locally. You can never do it globally.

 

 

 

You could for example take all events in the universe, and construct a Cauchy surface that means "this is all the events that I consider to be happening 'now'".

 

You can only get local Cauchy patches, you cannot construct a full Cauchy surface out of those patches.

Edited by xyzt
Link to comment
Share on other sites

'Unphysical' isn't the right word, because it has a physical meaning, and it can be measured. Eg. send a light signal to two receivers the same distance from you, and they'll receive them simultaneously, even though the two reception events are not directly causally connected to each other.

 

The best you could say is that simultaneity is 'conventional'. http://plato.stanford.edu/entries/spacetime-convensimul/gives an overview.

It is widely accepted (though not settled) that simultaneity is indeed conventional.

 

 

To relate that back to this thread... simultaneity is meaningfully defined in flat spacetime, using Einstein's definition, whether or not that is merely a convention. It would seem there's no such meaningful definition for curved spacetime, even though you could arbitrarily make one (eg with a foliation of Cauchy surfaces) if the universe fits the requirements (if it's globally hyperbolic or whatever).

 

You could for example take all events in the universe, and construct a Cauchy surface that means "this is all the events that I consider to be happening 'now'". That too would be a convention, but I think it would be physically meaningless, because you would probably have no way of unambiguously measuring the simultaneity of events no matter how you choose to define it. In flat spacetime Einstein's definition of simultaneity defines an unambiguous measurement of simultaneity.

Yes that is what i mean. all simultaneous events that form a line parallel to the x axis are unrelated to each other. That is what i mean by unphysicalThe line has no "structure".. An observer upon this line cannot observe it. If you want to observe this line, you cannot do that at an instant T, you need a duration., you need to wait.

Link to comment
Share on other sites

An observer upon this line cannot observe it. If you want to observe this line,

This is false. Any observer situated at the midpoint between the two simultaneous events observes them without any trouble. And, to boot, the observer in cause observes the two events as being....simultaneous. You are hijacking this thread with your misconceptions.

Link to comment
Share on other sites

This is false. Any observer situated at the midpoint between the two simultaneous events observes them without any trouble. And, to boot, the observer in cause observes the two events as being....simultaneous. You are hijacking this thread with your misconceptions.

At midpoint yes, but not on the same x line.

i mean, if the observer is upon the same x line as the 2 events, he cannot observe the 2 events.

Link to comment
Share on other sites

At midpoint yes, but not on the same x line.

i mean, if the observer is upon the same x line as the 2 events, he cannot observe the 2 events.

Looks like you do not understand what "midpoint" and or "simultaneous" means. There are two (simultaneous) events :

 

[math](x_1,t)[/math]

[math](x_2,t)[/math]

 

The observer located at midpoint has the coordinate [math]\frac{x_1+x_2}{2}[/math]. This observer is located on the x-axis, contrary to your claim. He or she will observe the two events as being simultaneous, contrary to your claim.

Edited by xyzt
Link to comment
Share on other sites

Looks like you do not understand what "midpoint" and or "simultaneous" means. There are two (simultaneous) events :

 

[math](x_1,t)[/math]

[math](x_2,t)[/math]

 

The observer located at midpoint has the coordinate [math]\frac{x_1+x_2}{2}[/math]. This observer is located on the x-axis, contrary to your claim. He or she will observe the two events as being simultaneous, contrary to your claim.

Oh, we are not dicussing about the same thing.

The x axis is part of a foliation in a spacetime diagram.

Link to comment
Share on other sites

Oh, we are not dicussing about the same thing.

The x axis is part of a foliation in a spacetime diagram.

Using buzzwords is not going to cover your lack of understanding.

Link to comment
Share on other sites

i mean, if the observer is upon the same x line as the 2 events, he cannot observe the 2 events.

 

I agree with xyzt. I think you've asked the same questions in various threads over the years, and now you're adding a misuse of "foliation" into your arguments. A point on a spacetime diagram represents an event. An observer is better represented by its worldline, a more-vertical-than-not path on the diagrams, and the events on the x-axis will eventually be observed if the observer sticks around long enough. A short-lived observer could be represented by a point, and it wouldn't observe those events you're talking about, but so what? Is something not real if you can contrive a particular observer that can't observe it? I've seen this argument before and it doesn't seem to have progressed. Forget foliations, look up world lines and space-like intervals, and why not devote one thread (not this) to figuring out your idea?

Link to comment
Share on other sites

 

I agree with xyzt. I think you've asked the same questions in various threads over the years, and now you're adding a misuse of "foliation" into your arguments. A point on a spacetime diagram represents an event. An observer is better represented by its worldline, a more-vertical-than-not path on the diagrams, and the events on the x-axis will eventually be observed if the observer sticks around long enough. A short-lived observer could be represented by a point, and it wouldn't observe those events you're talking about, but so what? Is something not real if you can contrive a particular observer that can't observe it? I've seen this argument before and it doesn't seem to have progressed. Forget foliations, look up world lines and space-like intervals, and why not devote one thread (not this) to figuring out your idea?

You are correct that I struggle for years.

If you accept just for a while that the observer is "moving" though time, i.e. changing coordinates, then maybe you could catch a glimpse of what I am struggling for.

Say that the observer cannot observe what is happening "really NOW" but only what "was simultanate". i.e. ONLY what lies upon the diagonals of the spacetime diagram, then...

For example:

_we suppose that there is a galaxy "NOW" 1 billion LY away from us. We cannot see this galaxy NOW, so it is a supposition.

In this galaxy, we suppose there exist an earthlike planet with an alien looking at us (that we cannot observe because it is NOW.

We suppose that this alien NOW is observing the Earth as it was a billion years ago. That is also a supposition. All suppositions. There is no physical way to communicate with this earthlike planet, because it is NOW. That makes a lot of suppositions.

Because if we are indeed "moving" through time, then this alien, if existing, is not looking at the Earth a billion years ago. He is looking at something else.

That we have no clue about.

Wecome to my universe.

Link to comment
Share on other sites

If you accept just for a while that the observer is "moving" though time, i.e. changing coordinates, then maybe you could catch a glimpse of what I am struggling for.

 

That doesn't make much sense. If the observer is moving through time, then you can define a speed at which it is moving. Which means you must need a second sort of time to measure that rate of "movement".

 

This is why the 4D space-time view is static: nothing "moves" through time. The object has a "world line" which defines all the places-times at which it exists.

 

 

We suppose that this alien NOW is observing the Earth as it was a billion years ago. That is also a supposition.

 

The only supposition is your invented alien. The fact that light takes a billion years to travel one billion light years is not supposition.

 

 

Because if we are indeed "moving" through time, then this alien, if existing, is not looking at the Earth a billion years ago. He is looking at something else.

That we have no clue about.

 

Perhaps you should take that conclusion as an indication that your mental model doesn't correspond to reality in any way.

Link to comment
Share on other sites

 

This is why the 4D space-time view is static: nothing "moves" through time. The object has a "world line" which defines all the places-times at which it exists.

 

That is an interpretation that we get from of a 2D graph on a sheet of paper.

 

I am not the only one who has ever argued that time does not flow but that all objects "move" in time. IOW that Time (exactly as Space) is a receptacle, a container. It is not something that makes us stretch. And we are not 4D objects that extend in a 4D space, we are 3D objects that "move" into a 4D substract.

Link to comment
Share on other sites

The t coordinate in the Robertson-Walker metric represents cosmic time. This is the time measured by a fundamental observer. A fundamental observer is one whose only motion is a result of the expansion or contraction of homogenous, isotropic spacetime. At any particular value of cosmic time, all fundamental observers will be measuring the same slice of three dimensional space.

 

http://en.m.wikipedia.org/wiki/Cosmic_time

Link to comment
Share on other sites

Exactly david345, which is just the same as my statement that these space-times are globally hyperbolic. However, remember that when you write a metric in local coordinates you have made some choices.

Link to comment
Share on other sites

The t coordinate in the Robertson-Walker metric represents cosmic time. This is the time measured by a fundamental observer. A fundamental observer is one whose only motion is a result of the expansion or contraction of homogenous, isotropic spacetime. At any particular value of cosmic time, all fundamental observers will be measuring the same slice of three dimensional space.

 

http://en.m.wikipedia.org/wiki/Cosmic_time

I don't understand.

Where is the "slice" that you are "measuring"?

 

From your link

 

Cosmic time (also known as time since the big bang) is the time coordinate commonly used in the Big Bang models of physical cosmology. It is defined for homogeneous, expanding universes as follows: Choose a time coordinate so that the universe has the same density everywhere at each moment in time (the fact that this is possible means that the universe is, by definition, homogeneous). Measure the passage of time using clocks moving with the Hubble flow. Choose the big bang singularity as the origin of the time coordinate.

Cosmic time is the standard time coordinate for specifying the Friedmann–Lemaître–Robertson–Walker solutions of Einstein's equations.

If instead the present is chosen as the origin, the scale is called the lookback time (or look back time).

 

 

"Choose a time coordinate so that the universe has the same density everywhere at each moment in time (the fact that this is possible means that the universe is, by definition, homogeneous)"

 

The "at each moment in time" means NOW on the horizontal X axis of a spacetime diagram.

We have NO access to any direct information from this axis. The indirect info we can have is the sum of information gathred from Galileo and after, that is 500 years, which is minuscule at universal scale. What we actually have is info from a few years at best. That is equal to nothing.

The real info we get from the universe comes from its past.

 

So, I am asking (again) where is the "slice"?

Is it on the horizontal? NOW?

Or on the diagonal? in the past?

post-19758-0-27013100-1431153601_thumb.jpg

Edited by michel123456
Link to comment
Share on other sites

I don't understand.

Where is the "slice" that you are "measuring"?

From some general arguments you know the local topology of the Universe is [math]R\times \Sigma[/math]. You then pick coordinates adapted to this and view each [math]\Sigma[/math] for a given t as 'now'.

Link to comment
Share on other sites

From some general arguments you know the local topology of the Universe is [math]R\times \Sigma[/math]. You then pick coordinates adapted to this and view each [math]\Sigma[/math] for a given t as 'now'.

So it is on X axis.

 

(edit)

Or not necessarily?

Edited by michel123456
Link to comment
Share on other sites

I don't quite follow. Anyway, because you know it looks like time + space globally you can pick a global coordinate for time. This choice of coordinates is still a choice.

 

More generally you can only locally make such a cut into space and time. Again, this corresponds to a choice of local coordinates.

Link to comment
Share on other sites

From some general arguments you know the local topology of the Universe is [math]R\times \Sigma[/math]. You then pick coordinates adapted to this and view each [math]\Sigma[/math] for a given t as 'now'.

Do the general arguments include these assumptions?

1. The universe cannot be closed (must be globally hyperbolic or flat).

2. It can't have local singularities like those at the center of black holes.

 

I think these are accepted mainstream assumptions, but they're not proven. Black hole singularities are not expected to be real, but it is probably not answerable without a quantum gravity theory. I'm guessing they would not permit a Cauchy surface???

 

 

 

 

Edit: I googled and came up with this: "if you look at the Penrose diagram for an astrophysical black hole, there clearly are Cauchy surfaces. The singularity is spacelike, so Cauchy surfaces don't have to intersect it. – Ben Crowell"

So it would seem I'm wrong... black hole singularities don't interfere with the ability to define a "now" everywhere.

 

If I understand that right, it's kind of interesting... You can define a "now" for the universe, but it won't include the centers of black holes! At first I think "that's not the whole universe then", but it is... the singularities are completely in the future of any such "now" that we're talking about. They could exist in 4d spacetime, just not "now". (Not so weird if you think of spacelike intervals to an outside observer being timelike inside a BH horizon.)

 

 

 

Edit: Just thinking some more... we can speak of a BH's singularity at a specific time in our coordinates. We can define from here its existence in a moment of time "now" (here), but there or near enough, our defined moment isn't even a moment. The usual concept of "now" that humans use, essentially assigns a local (ie. Earth) time to all events in the universe, which is not a common moment throughout the universe, and is not a Cauchy surface. Thinking about black holes and how spacelike curves can be timelike elsewhere, it becomes pretty obvious pretty quick why we don't try to define a common now in the universe, and instead work in local coordinates.

 

If we chose to define a "now" using a Cauchy surface, everyone would agree that the surface is a moment, with a consistent clear separation of past and future, but no one (not even us who get to choose how to define the surface) would say that far distant points on that surface agree with local coordinates. I think.

Edited by md65536
Link to comment
Share on other sites

The cosmological principle states that on a sufficiently large scale the universe is homogeneous and isotropic. Homogeneous means the number of stars per unit volume is roughly the same. Isotropic means uniform in all directions. In a sphere with a radius of 100 Mpc the average density is the same as any other sphere the same size. 1 pc=3.26 light-years. Cosmological time is the proper time measured by those who are at rest with respect to the local matter distribution. Their only motion is that which results from the expansion of the spacetime. These are known as fundamental observers. Cosmological time is useful globally rather then locally. Fundamental observers can agree to synchronize their clocks to a standard time when the universal homogeneous density reaches a certain value such as the big bang singularity. The age of the universe can be approximated using the velocity distance law. The velocity distance law is v=H(t)d. v is the proper radial velocity. v equates the rate of change of proper distance with respect to cosmic time. d is the proper distance. H(t) is the Hubble parameter. This states that in RW spacetime fundamental observers are moving away from each other with a proper radial velocity that is proportional to the proper distance between them.

 

t0=d/v=1/H0

 

t0=13.9 billion years.

 

t0 is time since the big bang. H0 is the Hubble constant. A more accurate approximation involves the density of radiation, matter, and dark energy along with the Hubble constant.

post-107966-0-49029200-1431232059_thumb.jpg

Link to comment
Share on other sites

The assumptions for the FRW cosmologies are homogeneity and isotropy of space and that the spatial component of the metric can be time-dependent. This gives possible metrics for the universe quite independent of the specifics of the theory of gravity you are using. The only thing is that it should be a metric theory.

Edited by ajb
Link to comment
Share on other sites

I see that the shwarzchild metric uses the gravitational constant and the frw does not. I just finished my first book on gr. You are a few books ahead of me. You say frw is time dependent. It appears that density of mass, radiation, dark energy and the Hubble constant can be used to come up with a global cosmic time which all fundamental observers could determine based off the density.

Edited by david345
Link to comment
Share on other sites

Guest
This topic is now closed to further replies.
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.