speed of time? [Archive] - Physics Forums

Naty1

Jun20-09, 02:37 PM

Time appears to pass at a constant rate and that was a cornerstone of Newton's theories. But Einstein showed that space and time are NOT fixed and immutable; they vary according to one's relative motion....space contracts and time slows as relative speed between observers increases. Identical events are observed to occur at different times in different frames. Time also slows down near massive gravitational bodies where gravitational potential is greater.

Check Wikipedia at http://en.wikipedia.org/wiki/Time#Modern_physics

for some additional description and further references.

DaleSpam

Jun20-09, 02:49 PM

Time is the thing measured by an ideal clock, and an ideal clock ticks at a rate of 1 second/light-second.

Max™

Jun20-09, 04:32 PM

What is the speed of Left?

negitron

Jun20-09, 04:47 PM

What is the speed of Left?

More accurately reflecting the OP: what is the speed of height? If you consider time as a dimension, it doesn't move at all; instead, objects move through the time dimension in the same way that they can move through the spatial dimensions.

fluidistic

Jun20-09, 04:51 PM

According to wikipedia Speed is the rate of motion, or equivalently the rate of change of distance.. I think we all agree on any similar definitions.
Hence the question what is the speed of time? has no sense.

HallsofIvy

Jun20-09, 04:57 PM

what is the speed of time?

One second per second!

Naty1

Jun20-09, 05:51 PM

Another related interpretation has been discussed in physics forums: our speed through spacetime...in that analogy we pass thru time at "c" when stationary and our passage slows as our speed through space increases....at speed "c" thru space, our passage thru time would slow to zero....

Here is one such excerpt: (Brian Greene, THE ELEGANT UNIVERSE) :

"...Einstein found that precisely this idea - the sharing of motion between different dimensions - underlies all of the remarkable physics of special relativity...
...Einstein proclaimed that all objects in the universe are always traveling through space-time at one fixed speed - that of light...
...If an object dose not move through space all of the objects motion is used to travel through time...
...Something traveling at light speed through space will have no speed left for motion through time. Thus light does not get old; a photon that emerged from the big bang is the same age today as it was then. There is no passage of time at the speed of light."

(quoted from another thread which I just lost....)
Some pooh pooh this concept; I find it at least a very useful analogy; others are unable to conceptualize speed through time.

MeJennifer

Jun20-09, 06:25 PM

Another related interpretation has been discussed in physics forums: our speed through spacetime...
Point objects are lines in spacetime and spatially extended objects are like tubes. Nothing passes through spacetime with any speed as spacetime is frozen.

DaleSpam

Jun20-09, 07:11 PM

True, but you can still talk of speed in a geometric sense. Geometrically coordinate speed is the slope of the worldline, and geometrically Naty1's speed (the four-velocity) is a tangent to the worldline.

Max™

Jun21-09, 09:09 AM

More accurately reflecting the OP: what is the speed of height? If you consider time as a dimension, it doesn't move at all; instead, objects move through the time dimension in the same way that they can move through the spatial dimensions.

Left is not a direction/dimension?

It is not part of Width?

matheinste

Jun21-09, 09:54 AM

Hello all.

Speed in the directions or dimensions x,y and z is distance paramaterized by time and so we can, if we wish have speed measured in meters per second, one meter being the unit basis vector in each of the spatial directions. I suppose that in spacetime, taking time as a dimension on an equal footing with the other three dimensions, and taking one second as the unit basis vector in the time direction, we should, for consistency, also paramatrerize it by time and so this would give the speed of time, as much as it can "mean" anything, in units of seconds per second. Thus we could give, for a clock at rest in an inertial frame a rate of 1 second per second as measured by itself, or by an observer at rest in the same reference frame.

So as has been said many times before that as much as any meaning can be attached to the question "what is the speed of time?", one second per second seems, to me, to be the most appropriate.

My apologies to any mathematicians for lack of rigor.

Matheinste

DaveC426913

Jun21-09, 10:00 AM

Left is not a direction/dimension?
It is not part of Width?
Left is a direction in a dimension. If negatron were going to be perfectly literal in correcting you, he would have said "More accurately, what is the speed of width?"

Naty1

Jun21-09, 11:14 AM

"Point objects are lines in spacetime and spatially extended objects are like tubes."

Isn't this like plotting the course progress of a plane on a map and proclaiming "A plane is line on a map?"

Any attempts to visualize physical phenomena, interpret and record our visualizations and represent them as mathematical formulations are designed more to represent the physical world in ways our limited senses can comprehend than they are objective descriptions. It's only we humans that seem to have trouble understanding natures rules; an atom,for example, always seems to know how its supposed to respond to forces of all types.

Nothing passes through spacetime with any speed as spacetime is frozen.

What does "spacetime is frozen" mean? Is this akin to "all history is fixed"?

And while some on this forum object to this concept of speed thru spacetime, it seems good enough for Brian Greene to utilize. Of course, he might be entirely wrong!

Fredrik

Jun21-09, 11:15 AM

The riddler

Jun21-09, 12:43 PM

its one second per second

A.T.

Jun21-09, 12:44 PM

"Point objects are lines in spacetime and spatially extended objects are like tubes."

Isn't this like plotting the course progress of a plane on a map and proclaiming "A plane is line on a map?"
I second that. No one objects if a plane icon is shown advancing along a path on a map. Visualizing objects as advancing in space time is no different.

Sure, you can see space time as frozen. That's the point of diagrams with time as dimensions: they describe movement with a static graph. But nothing stops you from imaging the time flow as an animated graph.

And while some on this forum object to this concept of speed thru spacetime, it seems good enough for Brian Greene to utilize. Of course, he might be entirely wrong!
It's not question of right or wrong, but if you like this dynamic visualzation or not.

negitron

Jun21-09, 12:46 PM

One second per second is not a satisfactory answer because it's not a rate, it's a scale. It's no more a speed than one meter per meter is.

A.T.

Jun21-09, 01:09 PM

One second per second is not a satisfactory answer because it's not a rate, it's a scale. It's no more a speed than one meter per meter is.
In the Geometrized unit system (http://en.wikipedia.org/wiki/Geometrized_unit_system) time has the dimensin of length, so meter per meter can be a speed. It basically means that a clock at rest advances trough time with 1 lightsecond per second.

matheinste

Jun21-09, 01:15 PM

One second per second is not a satisfactory answer because it's not a rate, it's a scale. It's no more a speed than one meter per meter is.

I agree. It is not a satisfactory answer because it is a pretty much meaningless or at least ill defined question.

Matheinste

Phrak

Jun21-09, 01:17 PM

In the Geometrized unit system (http://en.wikipedia.org/wiki/Geometrized_unit_system) time has the dimensin of length, so meter per meter can be a speed. It basically means that a clock at rest advances trough time with 1 lightsecond per second.

Using this logic we can ask what is the speed of distance, and the answer will be one meter per meter.

But why stop here. Can we cast frogs in generalized units too?

The riddler

Jun21-09, 02:21 PM

O.K Listen, the speed of time cannot be compared to anything. Basic Speed is different because speeds vary, for example you can have a car moving at 6m/s and another car moving at 9m/s. Imagine a universe where everything must move at 1m/s constantly all the time so nothing is going at a speed different to 1m/s, now lets say someone in this universe asks "What is our speed?", but because this universe has nothing to compare its speed to (as everything is going at 1m/s) then their speed is a universal constant and the speed is always 1m/s per 1m/s.

This is the same with time there are no other forms of time you can compare with time, meaning the speed of time can only be compared to itself.

Thus the speed of time is one second per second, please amuse me with more insufficent reasons why you may think this is not so. :biggrin:

P.S: Good anwser Phrak. :rofl:

negitron

Jun21-09, 02:44 PM

In the Geometrized unit system (http://en.wikipedia.org/wiki/Geometrized_unit_system) time has the dimensin of length, so meter per meter can be a speed. It basically means that a clock at rest advances trough time with 1 lightsecond per second.

According to that page, velocity has the dimension of 1, so time is expressed in units of 1/L.

A.T.

Jun21-09, 02:59 PM

Using this logic we can ask what is the speed of distance, and the answer will be one meter per meter.

I agree that "speed of time" doesn't really make sense. But I have no problem talking about "speed in time" of an object, when referring to the ratio: objects_proper_time / coordinte_time

There is no deeper meaning in this. It is rather trivial.

Sumo

Jun21-09, 03:36 PM

I think Lewis Epstein puts it best in his book "Relativity Visualized".

Inquiring minds will insist in wondering: Why, why, why?...There is no way to explain [it], all that are known is empirical facts. But the inquiring mind is yet unsatisfied...What the inquiring mind yearns for is a little story or myth that will give the feeling of being back home on logical ground.... The turtle myth is not good ("turtles all the way down") because it predicts that turtles are at the South Pole. On the other hand the Quantum Field Theory myth is bad because virtually all physicists spend their energy advancing the subject rather than making it easy to understand, so psychic mediums fill the void.

He then goes on to describe the myth that is the one being discussed: That time runs slower in objects moving faster because objects are always moving through time at the speed of light, and must devert that speed to move through space.
As we can see, many people have been deluded by this myth, so that they say things like the wikipedia quote, Green's quote, or the idea that objects "move through time".

I think this question points out the error of the myth. But I also think the myths (or analogies) are important, because thats largely how people think, conceptualize, and thus form new ideas. What we need to do is investigate what time is, and come up with a better myth.

Phrak

Jun21-09, 06:07 PM

I agree that "speed of time" doesn't really make sense. But I have no problem talking about "speed in time" of an object, when referring to the ratio: objects_proper_time / coordinte_time

There is no deeper meaning in this. It is rather trivial.

I can't see this an being any more than content-free physics. You make the fraction dt/dT, where T is proper time and give it the name 'speed of time'. I take the fraction dx/dX, where X is proper length and call it the 'speed of distance'. The naming is equally content-free.

A.T.

Jun21-09, 06:33 PM

I agree that "speed of time" doesn't really make sense. But I have no problem talking about "speed in time" of an object, when referring to the ratio: objects_proper_time / coordinte_time

You make the fraction dt/dT, where T is proper time and give it the name 'speed of time'.
No, I don't.
1) I call it "speed in time" not "speed of time"
2) It is dT/dt not dt/dT, (if T is proper time and t is coordinate time).
This is equivalent to calling dx/dt the "speed in space".
I take the fraction dx/dX, where X is proper length and call it the 'speed of distance'. The naming is equally content-free.
Usually a "speed" or "rate" is defined as change_of_something / coordinate_time. This fits my definition above, but not quite yours.

jefswat

Jun21-09, 10:17 PM

diazona

Jun22-09, 02:18 AM

I don't like the label seconds per second as it is misleading and very confusing. If one is inquireing how fast time passes in a different frame relative to another frame I think it linguistically makes more sense to make a distinction between the two seconds(for each different reference frame). Perhaps give it the label "prime seconds per second or something". I would think there is a convention for the label. I don't think thats where the original question was directed but it seems thats where conversation is headed
Confusing to you, but clear to me - on the other hand, I would find "prime seconds per second" quite bewildering. It's a matter of opinion which you prefer, of course, but I think most people would get the point if you say "seconds per second." (In fact, now that I think about it, isn't this "speed in time" related to the rapidity? That's a dimensionless angle, so "seconds per second" makes sense in the sense that the units cancel out...)

monty37

Jun22-09, 06:23 AM

now,time is a dimension,to us on earth we say its speed is "1 sec per sec"??
may i know the speed in m/sec. its us passing through time,this what differentiates past
from present.so,how to express speed of time in m/sec??

monty37

Jun22-09, 06:36 AM

i think the speed of time is the distance ticked every instant by the needle
of the atomic clock per every second. what say?atomic clock is supposed to
be the most accurate timepiece.

A.T.

Jun22-09, 06:39 AM

now,time is a dimension,to us on earth we say its speed is "1 sec per sec"??
may i know the speed in m/sec.
It depends on which conversion factor between meters and seconds you choose.Usually the speed of light is used, so the speed trough time of an object at rest would be the speed of light (299 792 458 m / s). But this is just a convenient convention.

Max™

Jun22-09, 07:38 AM

Left is a direction in a dimension. If negatron were going to be perfectly literal in correcting you, he would have said "More accurately, what is the speed of width?"

I said it is a part of width, which of course implies that you can change it such that height or depth are labeled as such depending on your chosen set of coordinates, x/y/z, etc.

There's no need to "correct" my smart-aleck response, replace "what is the speed of time" with "what is the speed of future" and the quip "what is the speed of left" with "what is the speed of width" would fit just as well.

:P

DaveC426913

Jun22-09, 05:57 PM

There's no need to "correct" my smart-aleck response, replace "what is the speed of time" with "what is the speed of future" and the quip "what is the speed of left" with "what is the speed of width" would fit just as well.
:PYou're suggesting the original question should be changed to better suit your answer? :confused:

Nisse

Jun22-09, 06:29 PM

petm1

Jun22-09, 07:25 PM

I vote for 9.8 meters per second as the speed of time relative to Earth.:tongue2:

Phrak

Jun22-09, 07:31 PM

Max™

Jun23-09, 08:02 AM

You're suggesting the original question should be changed to better suit your answer? :confused:

I'm suggesting the answer was pretty well given already, but I have a compulsion to make dumb jokes, and "the speed of left" is one of them.

DaleSpam

Jun23-09, 08:26 AM

You're suggesting the original question should be changed to better suit your answer? :confused:Hey, that's a great idea. A professor could determine what percentage of the question would need to be changed in order to make a student's answer right, and that is the student's score on the question!

Naty1

Jun23-09, 09:47 AM

I very much hope that in a few hundred years time some physicist finds this archived thread and experiences much mirth at all the confusion surrounding something that, to her, will seem obvious.

This is not to criticise any of the contributers; it is the nature of science.

To the contrary, I was about to post a similar thought.

My only critique about this wonderful forum is that conventional scientific answers are given with such authority many would be led to think they are a final, complete and accurate answer. I do believe conventional scientific understanding should form the basis of replies but it is also important to foster new ways of thinking about problems or progress will be stifled. Clearly there is more we do not understand than we do.

I am beginning to wonder if the recent lack of major theoretical scientific breakthroughs to some degree results from an overreliance on conventional approaches, ridicule of those who propose revolutionary ideas, and a general lack of a willingness to listen to opposing views.
Seems like experimentalists have found the most interesting new science lately.

If government grants go only to conventional thinkers, mankind may be shooting itself in the foot. That's not the type thinking that led people like Copernicus, Einstein, Witten, Wheeler and others to new scientific insights.

matheinste

Jun23-09, 09:59 AM

Hello all.

It is interesting to note that the phrase "shooting oneself in the foot" originally had a very different meaning from its current one of accidently doing harm to one's cause. It was a deliberate act by a soldier on himself to render himself no longer fit for service to escape the horrors of trench warfare.. I believe it originated during the First World War in the trenches in France.

Matheinste.

DaleSpam

Jun23-09, 01:36 PM

I am beginning to wonder if the recent lack of major theoretical scientific breakthroughs to some degree results from an overreliance on conventional approaches, ridicule of those who propose revolutionary ideas, and a general lack of a willingness to listen to opposing views. I suspect that it is probably more due to the general unwillingness of people who propose "revolutionary" ideas to actually put even minimal effort into making sure that their idea is logically sound and consistent with existing experiments. The scientific community is certainly willing to listen good ideas, but simply being "revolutionary" is not the same as being good.

Also, most "revolutionary" ideas are actually "counter-revolutionary". In other words they seek to reject modern physics when they should instead be demonstrating that their theory reduces to relativity and quantum mechanics in the appropriate limits.

DaveC426913

Jun23-09, 02:01 PM

Hey, that's a great idea. A professor could determine what percentage of the question would need to be changed in order to make a student's answer right, and that is the student's score on the question!You'd want the score to be 100 - the percentage. :wink:

negitron

Jun23-09, 02:48 PM

It is interesting to note that the phrase "shooting oneself in the foot" originally had a very different meaning from its current one of accidently doing harm to one's cause. It was a deliberate act by a soldier on himself to render himself no longer fit for service to escape the horrors of trench warfare.. I believe it originated during the First World War in the trenches in France.

Worldwide Words (http://www.worldwidewords.org/qa/qa-sho4.htm) (a source which has consistently proven itself to me) disagrees with you:

In the sense of a minor self-inflicted injury for the reasons you give, it is certainly older [than the 1980s]. My erratic memory suggests it was a well-known tactic in the First World War, rather too well known to officers and medics even then to be easily carried off. I found a reference in a 1933 book, Death in the Woods and Other Stories by Sherwood Anderson.

...

As a literal expression describing an accidental injury it is earlier still, from the middle of the nineteenth century....A search of US newspapers found 187 items between 1960 and 1965 reporting that a man had accidentally shot himself in the foot; it’s no doubt a common injury down to the present day (it’s difficult to search for, as most examples are now figurative).

I’m sure the expression shoot oneself in the foot derives from such accidents, usually the result of incompetence, and has led to our current meaning of making an embarrassing error of judgement or inadvertently making one’s own situation worse. That men did it deliberately as a way to avoid combat is only a side meaning.

matheinste

Jun23-09, 04:44 PM

Worldwide Words (http://www.worldwidewords.org/qa/qa-sho4.htm) (a source which has consistently proven itself to me) disagrees with you:

I stand corrected.

Matheinste.

Phrak

Jun24-09, 12:06 AM

I stand corrected.

Matheinste.

I don't buy the 'corrected' version. So there :)

pawan_ctn

Jun26-09, 05:39 PM

time is a dimention...... it has no speed......

m.starkov

Jun27-09, 07:40 PM

I believe the time is just a human-made term.
The term is made for forecasting needs.
If you have a ball and two bullets fired in directions to its different sides then
if left bullet will come first it will move ball to the right and the other bullet will miss.
if right bullet will come first it will move ball to the left and the other bullet will miss.
So to understand the final position of the ball we need to know bullets Speeds.
And with using the term "Speed" we come to use also the term "Time".
So Time is a measure of "what will come earlier".

DaleSpam

Jun27-09, 07:59 PM

Hi m.starkov, welcome to PF,

By "time is just a human-made term" what do you mean? Do you just mean that the word itself is human-made, or that there is no physical reality to time itself (the thing represented by the term)?

m.starkov

Jun27-09, 08:20 PM

Hi DaleSpam,

I believe the physical reality of the thing represented by the term Time is just a cause-effect chains of this world.

DaleSpam

Jun27-09, 08:23 PM

OK, that sounds like you do believe that time is physically real. So then I don't really understand your original comment.

m.starkov

Jun27-09, 08:28 PM

I believe the following question is related to the topic.

I have one simple (i believe) question to people who are familiar with relativity theory.
We have following cases in which we need to compare elapsed time on different clocks.
At the beginning of each experiment clocks will be syncronized.
We have measured marks on the ground: A 250m B 250m C 250m D 250m E
Clocks starts from a distance of 1 kilometer at points A and E.
At the end of each experiment elapsed times will be t1 and t2 accordingly.
So the Clock which started from A, at the end will show t1.
And the Clock which started from E, at the end will show t2.

The question for all cases is the same: t1 ? t2.
I mean:
Answer "1": t1 = t2
Answer "2": t1 < t2
Answer "3": t1 > t2

1. Both clocks are in inertial frames. At the end they meet in point C. (I hope the answer is "1")
2. Both clocks are in inertial frames. At the end they meet in point B.
3. Only the Right clocks (which come from E) is in inertial frame.
The Left clocks come to point C first and stops (smoothly).
After a second the Right clock comes to point C (and this is the end).

Could please somebody give answers for these all cases?

m.starkov

Jun27-09, 08:30 PM

How the time can be unreal if we can observe it?
So we have to separate the term Time used for measuring and the world Time flow.
My original post was about the term Time which people use for their measuring needs.
And if the Topic question regards to world time flow then the answer will be the following.
We have cause-effect chains observed in this world.
We take any one cause-effect chain with known size and name it as Etalon.
Then we can compare other observed cause-effect chains with the Etalon chain and
determine which ends earlier.
This is how I understand the Time.

DaleSpam

Jun27-09, 08:54 PM

So the Clock which started from A, at the end will show t1.
And the Clock which started from E, at the end will show t2.

The question for all cases is the same: t1 ? t2.The general formula for the time elapsed on a clock is called the proper time (http://en.wikipedia.org/wiki/Proper_time) and is given by:
dτ² = dt² - dx²/c² - dy²/c² - dz²/c²

I will leave it to you to plug in the numbers for your specific problem. Note that the proper time, dτ, is an invariant quantity, meaning that it has the same value in all coordinate systems. I.e. it is absolute, not relative. This is in contrast with the coordinate time, dt, which is relative or frame dependent.

m.starkov

Jun27-09, 09:02 PM

Dear DaleSpam,

Actually I expect only 3 bytes of information for my Clocks question.
For example: 111 or 133 or 132, ok?

Thank you

DaleSpam

Jun27-09, 09:05 PM

You are welcome. Please calculate the 3 bytes yourself, I gave you the formula. I have already passed the class and have no desire to do someone else's homework.

m.starkov

Jun27-09, 09:14 PM

There is nothing to calculate if you know the principle.
I don't need exact value of difference.
The conditions are trivial!
And I just need a clear answer < , > or =.
If you are not sure about the answer you can skip it.
And this is not a homework..

DaleSpam

Jun27-09, 09:53 PM

And this is not a homework..Excellent, if it is not homework then an answer which is true in general is always preferable to an answer which is only true for a special case. As I mentioned above, the general answer is:
dτ² = dt² - dx²/c² - dy²/c² - dz²/c²

Simply plug in the coordinate time for dt and the coordinate distances for dx, dy, and dz, and you can get your special-case answers and the special case answer for any other possible scenario that you can invent.

m.starkov

Jun28-09, 07:25 AM

Simply plug in the coordinate time for dt and the coordinate distances for dx, dy, and dz

Do you mean there is not enough arguments in the question?
Why I need to plug something?
Is it possible to say what is greater just looking at the conditions?

DaleSpam

Jun28-09, 09:54 AM

Look, I am too busy and not interested enough in the problem to work through the details. I have given you the formula. If you are really interested in the answer then use the equation and find it. If you don't care enough about your own question to do that then why shoud you expect anyone else to?

Besides, if you don't work through it yourself then you won't learn. It should be a good exercise for you in finding out which details are important and how they affect the end result.

DaveC426913

Jun28-09, 11:05 AM

Do you mean there is not enough arguments in the question?
Why I need to plug something?
Is it possible to say what is greater just looking at the conditions?
Starkov, Dalespam has given you the formula that will allow you to answer your question. You say you don't need high accuracy so simply put in simple numbers for the clocks in your question such as 1m in each direction. That'll give you your >, < and =.

monty37

Jun28-09, 11:10 AM

coming back to the point,do i conclude since time is a dimension it has
no speed

DaleSpam

Jun28-09, 11:35 AM

That is certainly a reasonable conclusion.

TVP45

Jun28-09, 11:52 AM

I have seen this phrase "speed of time" used in so many confused discussions that I wonder whether we might be better off to simply ban it. Any Physics graduate is not going to be confused by what might seem sloppy language, but others seem to be.

Should we simply talk about stationary clocks or moving clocks or say that a time interval is measured to be different in different frames? Should we insist on always saying "proper time", "coordinate time", or "spacetime"?

Or, should we insist on using expressions like d\tau/dt or dt/ds or whatever we might be thinking of? (I don't really think this is a good idea, but it beats the current state of confusion.)

DaleSpam

Jun28-09, 12:11 PM

Should we insist on always saying "proper time", "coordinate time", or "spacetime"?That would be nice! I think the term "speed of time" is an inevitable by product of phrases like "time slows down" or even "clocks slow down".

Vanadium 50

Jun28-09, 12:57 PM

If you don't care enough about your own question to do that then why shoud you expect anyone else to?

Um...because my time is important? :uhh:

Joans

Jun28-09, 03:10 PM

sqrt(c*c-v*v) - That is speed of time as I call. Because time is dimension, and max speed is c!
Firstly I found it myself at 16 and I was very interested in it, spend dozens of sleepless nights. (:
When we stand speed is c and when we move faster it goes slower: sqrt(c*c-v*v)

in fact it is in main equations of special relativity, example m=m0c/sqrt(c*c-v*v)=m0/sqrt(1-(v*v/c*c))

OR m/m0=c/ts
ts- speed of time or whatever we call it, it have dimensions and it is meters per sec.

E/E0=m/m0=t/t0=l0/l=c/ts

gamma is just a ratio between speed of time to moving object and speed of light.

In fact it is very interesting thing to think on.
We always travel at speed of time! Our speed is constant, never we slower never we faster. Whats like a vectors sum of speed in all dimensions and it is always equal to c. Judge me, but do not copy.

In fact good reply:
Another related interpretation has been discussed in physics forums: our speed through spacetime...in that analogy we pass thru time at "c" when stationary and our passage slows as our speed through space increases....at speed "c" thru space, our passage thru time would slow to zero....

Here is one such excerpt: (Brian Greene, THE ELEGANT UNIVERSE) :

"...Einstein found that precisely this idea - the sharing of motion between different dimensions - underlies all of the remarkable physics of special relativity...
...Einstein proclaimed that all objects in the universe are always traveling through space-time at one fixed speed - that of light...
...If an object dose not move through space all of the objects motion is used to travel through time...
...Something traveling at light speed through space will have no speed left for motion through time. Thus light does not get old; a photon that emerged from the big bang is the same age today as it was then. There is no passage of time at the speed of light."

(quoted from another thread which I just lost....)
Some pooh pooh this concept; I find it at least a very useful analogy; others are unable to conceptualize speed through time.

m.starkov

Jun29-09, 07:08 PM

Ok let's try to use given formula: dτ² = dt² - dx²/c² - dy²/c² - dz²/c² for my "homework"

Let's take the first one: "Both clocks are in inertial frames. At the end they meet in point C."

We can use the following formula since we have only inertial movement:
Δτ = sqrt( (Δt)² - (Δx)²/c² - (Δy)²/c² - (Δz)²/c² ), ok?

Δy = 0
Δz = 0
So we will use Δτ = sqrt( (Δt)² - (Δx)²/c² )
Δx = 300000 km
Δt = 2 seconds

Calculation 1 - for inertial reference frame associated with A:
A: Δτ = sqrt( (Δt)² - (Δx)²/c² ) = sqrt( (2)² - ( 0 )² / c² ) = sqrt( (2)² ) = 2
B: Δτ = sqrt( (Δt)² - (Δx)²/c² ) = sqrt( (2)² - ( 300000 )² / c² ) = sqrt( 4 - 1 ) = sqrt( 3 ) = 1.732
So the time on clock A is 2 sec and the time on clock B is 1.732 sec.

Calculation 2 - for inertial reference frame associated with B:
A: Δτ = sqrt( (Δt)² - (Δx)²/c² ) = sqrt( (2)² - ( 300000 )² / c² ) = sqrt( 4 - 1 ) = sqrt( 3 ) = 1.732
B: Δτ = sqrt( (Δt)² - (Δx)²/c² ) = sqrt( (2)² - ( 0 )² / c² ) = sqrt( (2)² ) = 2
So the time on clock A is 1.732 sec and the time on clock B is 2 sec.

So which calculation is correct? Obviously neither. But why?
How to use this formula in a correct way?

DaleSpam

Jun29-09, 08:18 PM

Excellent effort using the formula! You haven't been here long, but you will soon realize how that simple effort sets you apart from the "cranks".

The nice thing about this formula is that it is frame-invariant. This means that you do not need to perform it in any particular reference frame since you will get the same answer in any inertial frame. Specifically, you can do it in the "ground" frame (the frame in which the clocks are initially synchronized). You should use that frame because that is the only frame where the coordinate time difference (Δt) is the same for both clocks.

Remember that simultaneity is relative so the fact that they are initially synchronized is only true in the ground frame. Therefore, if you transform into either clock's frame then you will need to find the start time separately for each clock which will result in a different Δt for each clock.

m.starkov

Jun30-09, 02:59 PM

Please take a look

19524

The coordinate time difference (Δt) is the same for both clocks in both cases.
So why not use the clock inertial frame for calculations? Can we?

DaleSpam

Jun30-09, 06:58 PM

Your bottom drawing is wrong. You are forgetting the relativity of simultaneity. Your two clocks are synchronized at t1 in the ground frame, so they cannot be synchronized at t1 in any other frame. Don't feel bad about it, the relativity of simultaneity is the most difficult concept to master in SR.

Privalov

Jun30-09, 10:32 PM

Hi M.Starkov!

Nice to see you here as well. Your English is not bad. And especially high achievement is you somehow managed DaleSpam to read your posts. I was not able to do that yet. Perhaps I'm too cranky.

the relativity of simultaneity is the most difficult concept to master in SR.

Unless you, M.Starkov, will start using Minkowski diagram, as I advised you long time ago.

So which calculation is correct? Obviously neither. But why?

In fact, both calculations are correct (from the sounds of it - sorry, I did not check the math).

The only problem is you see a problem here, while there are none. Time is relative. Coordinate time, that is.

Clock A will measure time intervals, as shown in Calculation 1. From the point of view of clock B, Calculation 2 applies. Therefore, both clocks will meet at point C, showing 1.732 sec as duration of the trip from point A/E to point C.

Please keep in mind I do not know proper English terms. You should pick the terminology from someone else's posts.

DaleSpam

Jul1-09, 05:05 AM

And especially high achievement is you somehow managed DaleSpam to read your posts. I was not able to do that yet.This is a rather strange comment. I did read and answer your posts. If you feel that the answer was insufficient then you should say so in that thread rather than making strange side comments in another thread.

m.starkov

Jul1-09, 12:31 PM

Your bottom drawing is wrong.
Your two clocks are synchronized at t1 in the ground frame, so they cannot be synchronized at t1 in any other frame.

Ok, let me modify the case 1 conditions to try to fit this requirement.
Now clocks are connected with a spring.
At the beginning they stay together and they are in sync at the moment.
Then they get equal impulses in different directions (left and right) and they fly away from each other (but they are still connected with a spring).
After a time spring will pull them back to be together.
In this case they are not in inertial frames but the major question is:
can we use the formula for the clock frame only since we have clocks synchronized in it?

m.starkov

Jul1-09, 12:38 PM

Hi Alexander! (Privalov)

Unfortunately I don't have enough knowledge how to use Minkowski diagram yet.
I would like to go step by step with understanding Relativity.
May be I will come to it in near future.
Now the major task for me is to understand Relativity twin paradox official explanation.

DaleSpam

Jul1-09, 01:44 PM

In this case they are not in inertial frames but the major question is:
can we use the formula for the clock frame only since we have clocks synchronized in it?The standard "textbook" formulas only apply in inertial frames. (That is essentially the definition of an inertial frame.) You can do the work in non-inertial frames only if you either use the mathematical framework of General Relativity or if you carefully modify the formulas for the specific frame (eg by including fictitious forces).

m.starkov

Jul1-09, 03:17 PM

Wait wait wait...
The
http://en.wikipedia.org/wiki/Proper_time
says that
"To make things even easier, inertial motion in special relativity is where the spatial coordinates change at a constant rate with respect to the temporal coordinate. This further simplifies the proper time equation to..."
So I believe the main formula is for non-inertial frames, isn't it?

DaleSpam

Jul1-09, 05:37 PM

That is a very astute observation. The formula that I posted is called the Minkowski metric. It applies to any form of an object's motion, inertial or non-inertial, as analyzed in an inertial reference frame. There are other metric equations for non-inertial reference frames, and they also can be used to analyze any form of motion in that reference frame, but their form is different from what I posted.

Do you understand the difference between an inertial object and an inertial reference frame?

m.starkov

Jul1-09, 07:21 PM

I can imagine inertial object:
no gravity, no force fields, object moves on straight line without rotation with a constant speed.
Inertial Reference Frame is an abstraction which can be associated with inertial object in order to simplify some calculations for Relativity.

But what about my case with clocks on spring?
Which formula can I use?

And let me know you my following question please.
Once again we have two clocks connected with a spring.
But one of the clocks is a little bit heavier than another one.
If clocks are identical then obviously time will be the same at the end of experiment.
But here we have no more symmetry - we have heavy clock and lite clock.
So one of them will have a little bit more velocity than another one.
They will meet in the same point where they start.
So what going to happen in this case?
Will heavier clock show more time than the lite one?
I believe Relativity's answer will be "yes".
But once again which formula can explain it?

DaleSpam

Jul1-09, 08:37 PM

I can imagine inertial object (without gravity and force fields).Exactly, an object that is not acted on by any external forces is inertial.Inertial Reference Frame is an abstraction which can be associated with inertial object in order to simplify some calculations for Relativity.Pretty much. The concept of an inertial reference frame is not just for relativity, there are also inertial frames in Newtonian physics. In both Relativity and Newtonian Physics an inertial frame can be (but is not required to be) associated with an inertially moving object. The basic definition of an inertial reference frame is simply any coordinate system in which the laws of physics have their "standard" form. This definition applies from Newton on.

But what about my case with clocks on spring?
Which formula can I use?The same formula as I posted above, provided you are doing the analysis in any inertial frame. (Note that you will have to integrate over the path since it is not a series of inertial segments). In a non-inertial frame you will have to derive the correct expression for the metric.

Will heavier clock show more time than the lite one?
I believe the Relativity answer is yes.
But once again which formula can explain it?Again, the same formula for the metric as above. It is a very fundamental formula.

DaveC426913

Jul1-09, 10:20 PM

Exactly, an object that is not acted on by any external forces is inertial.Maybe I'm taking this out of context but...

A rocket under thrust is not being acted upon by an external force, yet it is not inertial.

DaleSpam

Jul1-09, 10:25 PM

DaveC426913

Jul1-09, 10:31 PM

A rocket under thrust is being acted on by an external force, the pressure of the exhaust. Now, if you take the object to be the rocket + exhaust then that overall object is not acted on by an external force and is inertial.

But the exhaust is not an external force. OK, I guess it could be, from a certain point of view*.

*Damn you Obi Wan. You sullied the purity of truth with subjectivity!

DaleSpam

Jul1-09, 10:36 PM

*Damn you Obi Wan. You sullied the purity of truth with subjectivity!Use the force Dave, it is equal to mass times acceleration! :smile:

m.starkov

Jul2-09, 05:27 AM

Exactly, an object that is not acted on by any external forces is inertial.

Ok, just one offtopic question: Gravity is a force?

A.T.

Jul2-09, 06:00 AM

Ok, just one offtopic question: Gravity is a force?
In Newtons theory gravity is a force based on interaction between two bodies, fullfiling his 3rd law. In GR a free falling object is forcefree, so there is no "force of gravity". Howerver if you switch to a noninertial frame of reference, like the earths surface, you get intertial forces, like the one we know as gravity.

Compared to Newton GR redefines who is inertial and who is not in a gravtational field.

m.starkov

Jul2-09, 09:34 AM

Hi again,

I have made some calculations for the case with two clocks on spring.
Calculations are made for Inertial Reference Frame moving with speed "s" relative to start center point.
I have got the same proper time for both clocks (Clocks will show the same time).
So I'm happy with this.
But can somebody please take a look at my calculations below and tell if may be something is wrong?
Because I'm going to calculate for my second case using the same formulas.

19552

m.starkov

Jul2-09, 09:38 AM

In Compared to Newton GR redefines who is inertial and who is not in a gravtational field.

So if we have a planet and free-falling object then can I say the object is in Inertial Reference Frame?

m.starkov

Jul2-09, 02:26 PM

Now I have calculations for the Case 2: clocks with different mass on the spring.
Mass affects velocity acceleration of each clock.

19558

At the bottom of the image below you can find two results:
1. for Inertial Reference Frame associated with center start point.
2. for Inertial Reference Frame moving with speed "p".
The results are different.
But the formula I use is frame invariant.
So I would like to know why am I getting different results?

19557

P.S. I believe we will use the same formula to calculate twins age difference?
So now I'm a little bit confused - which IRF I should take to get the real difference?

DaleSpam

Jul2-09, 08:36 PM

But can somebody please take a look at my calculations below and tell if may be something is wrong?I calculated the first one given the velocity profile you specified. It was all correct. I didn't bother to calculate the second one using the time-reversed velocity profile.

There are a couple of minor things wrong. First, the velocity profile does not represent a mass being pushed by a spring but that really doesn't matter since you are free to specify any force you wish. More importantly you have specified a velocity profile that reaches c, and it should always be less than c. However, you only reached c instantaneously so it is not horrendously wrong.

By the way, in case you didn't know, you can loosely derive the formula you used from the formula I gave as follows:

\begin{array}{1}
\text{d$\tau$}^2=\text{dt}^2-\frac{\text{dx}^2}{c^2}-\frac{\text{dy}^2}{c^2}-\frac{\text{dz}^2}{c^2}\\
\text{d$\tau$}^2=\text{dt}^2-\frac{\text{dx}^2}{c^2}\\
\text{d$\tau$}=\sqrt{\text{dt}^2-\frac{\text{dx}^2}{c^2}}\\
\int \text{d$\tau$}=\int \sqrt{\text{dt}^2-\frac{\text{dx}^2}{c^2}} \\
\tau =\int \frac{\sqrt{\text{dt}^2-\frac{\text{dx}^2}{c^2}}}{\text{dt}} \, dt\\
\tau =\int \sqrt{\frac{\text{dt}^2}{\text{dt}^2}-\frac{\text{dx}^2}{\text{dt}^2 c^2}} \, dt\\
\tau =\int \sqrt{1-\frac{v^2}{c^2}} \, dt
\end{array}

DaleSpam

Jul2-09, 08:48 PM

The results are different.
But the formula I use is frame invariant.There are two problems here.

The first is that if you want to avoid relativity of simultaneity issues then you must start and end your clocks together. If they are not reunited at the end then if you stop the clocks simultaneously in one frame they will not be stopped simultaneously in any other frame.

The second issue is that you need to use the relativistic velocity addition (http://en.wikipedia.org/wiki/Velocity-addition_formula) formula if you want to find what the velocity profile looks like in another frame. Note that simple addition can easily result in velocities > c whereas the correct relativistic formula ensures that never happens.

m.starkov

Jul3-09, 03:28 AM

There are two problems here.

1. you must start and end your clocks together.
2. you need to use the relativistic velocity addition (http://en.wikipedia.org/wiki/Velocity-addition_formula) formula if you want to find what the velocity profile looks like in another frame.

1. My clocks start and end together! Do we still have an issue here? I'm not sure why you're telling me again about it.

2. Ok, but can you please be more specific: what exactly I have to do with the formula?
Look, we have 3 speed values here:
a) left clock speed relative to clocks mass center (is it still I guess?)
b) right clock speed relative to clocks mass center
c) calculations IRF speed relative to clocks mass center
So what do I have to do with these speeds and given formula?

Ok, I guess I have to do the following.
Before passing "v(t)" (velocity of time function) to the formula I have to acquire a "proper speed" from relativistic velocity addition formula, ok? And I have to do this for both clocks.
So I take "calculations IRF" speed as U and clocks speed as V and put it into the relativistic velocity addition formula, correct?

DaleSpam

Jul3-09, 07:09 AM

1. My clocks start and end together! Do we still have an issue here? I'm not sure why you're telling me again about it.Then I must be misunderstanding your graphs. In both of your plots of position it looks like the red clock and the dotted-blue clock start at the same point but end at different points. Could you please explain what each graph represents?

So I take "calculations IRF" speed as U and clocks speed as V and put it into the relativistic velocity addition formula, correct?Yes. U will be constant (the relative velocity of the two frames), and V will be your v(t) function from the first reference frame. That will give you a new V'(t) which you can then put into f and integrate to get the time.

m.starkov

Jul3-09, 07:50 AM

The red graphs means calculation for IRF associated with center of mass.
The blue graphs means calculation for IRF moving with speed "s".
Upper calculations - is for heavy clock.
Bottom calculations - is for lite clock.

P.S. Sorry, I'll put a legend next time

DaleSpam

Jul3-09, 08:13 AM

Ah, ok, I was misunderstanding. Now I see that the horizontal dashed green line is the x position where the two clocks meet at the end in the moving frame. So it looks like the velocity addition is the only problem.

m.starkov

Jul3-09, 08:21 PM

Now my calculations are made with regards to Velocity-addition formula.

But once again results are different!

May be I'm wrong with calculations?
Or just forgot one another thing?

Appreciate if you could look at my calculations below:

19568

I suggest that may be my clocks will not meet in the "Moving IRF" according to new proper velocities?
But I have no idea how can I manage it. I'm really confused with this.

DaleSpam

Jul4-09, 02:06 PM

I am sorry about the confusion, it is my fault. The problem is that my most recent advice (use the velocity addition formula) contradicts some of my previous advice (make sure to account for the relativity of simultaneity). Because I made a mistake I will work this problem out completely and post it, probably tomorrow.

Basically, all of the usual relativity formulas (time dilation, length contraction, relativistic velocity addition, etc.) are special cases of the Lorentz transform. Because they are special cases if they are accidentally misused you can get contradictions, which is what happened here. It is always safer to use the Lorentz transform rather than the special-case formulas whenever possible, which I will do in the write-up.

DaleSpam

Jul4-09, 04:47 PM

OK, here the "center of mass" frame will be the unprimed frame and the "moving" frame will be the primed frame. The two frames are moving wrt each other with a velocity u.

From the equation I gave above if we have in any arbitrary reference frame the time and position (as a function of some parameter r) for any clock undergoing any arbitrary motion then we can find the accumulated proper time by:

\tau = \int
\sqrt{\left(\frac{dt}{dr}\right)^2-\left(\frac{dx}{c \, dr}\right)^
2-\left(\frac{dy}{c \, dr}\right)^
2-\left(\frac{dz}{c \, dr}\right)^
2} \, dr

So the problem is simply to express the time and position of each clock in each reference frame as a function of some parameter and then evaluate the above integral. Since we are ignoring motion in y and z those terms immediately drop out and I will not refer to them again.

The velocity of a mass in the unprimed frame is given by:

v(t)=\begin{cases}
k t & t\leq 2 \\
-k (t-4) & 2<t\leq 6 \\
k (t-8) & \text{otherwise}
\end{cases}

Integrating this gives the position and time in the unprimed frame:

t=t

x(t)=\int v(t) \, dt=\begin{cases}
\frac{k t^2}{2} & t\leq 2 \\
-\frac{k t^2}{2}+4 k t-4 k & 2<t\leq 6 \\
\frac{k t^2}{2}-8 k t+32 k & \text{otherwise}
\end{cases}

Taking the Lorentz transform gives the time and position in the primed frame:

t'=\frac{c^2 t-u \, x(t)}{c^2 \sqrt{1-\frac{u^2}{c^2}}}

x'=\frac{-t u + x(t)}{\sqrt{1-\frac{u^2}{c^2}}}

Now, we have the time and position of each clock expressed as a function of the parameter t. Taking the derivatives we get:

\frac{dt}{dt}=1

\frac{dx}{c \, dt}=\frac{v(t)
}{c}

\frac{dt'}{dt}=\frac{c^2-u \, v(t)}{c^2 \sqrt{1-\frac{u^2}{c^2}}}

\frac{dx'}{c \, dt}=\frac{-u + v(t)
}{c \sqrt{1-\frac{u^2}{c^2}}}

Substituting in to the above integral gives

\tau=\int_0^8 \sqrt{1-\frac{v(t)^2}{c^2}} \, dt=\frac{2 \left(2 \sqrt{\frac{c^2-4 k^2}{c^2}} k+c \sin
^{-1}\left(\frac{2 k}{c}\right)\right)}{k}

\tau'=\int_0^8 \sqrt{1-\frac{v(t)^2}{c^2}} \, dt=\frac{2 \left(2 \sqrt{\frac{c^2-4 k^2}{c^2}} k+c \sin
^{-1}\left(\frac{2 k}{c}\right)\right)}{k}

which are clearly equal to each other. Evaluating for the specific conditions you gave above gives:

\tau_{heavy \, clock} = \tau'_{heavy \, clock} = 7.849
\tau_{lite \, clock} = \tau'_{lite \, clock} = 7.360

Askalany

Jul5-09, 05:30 AM

dt/dt=1

but as far as I understand that speed is how much distance you travel in how much time, how can time travel distance!

A.T.

Jul5-09, 09:02 AM

So if we have a planet and free-falling object then can I say the object is in Inertial Reference Frame?
Not quite the right wording. The free-falling object exists in all kinds of frames. But a frame attached to the object is inertial.

Given it is a small point like object. The earth for example can have different kinds of frames attached to it. A frame attached to the center is inertial. A frame attached to the surface is not.

DaleSpam

Jul5-09, 09:23 AM

dt/dt=1

but as far as I understand that speed is how much distance you travel in how much time, how can time travel distance!Don't think of those derivatives as speeds. If you look at the units you will see that they are all unitless. Instead, think of them as a conversion factor between the parameter and proper time.

m.starkov

Jul10-09, 10:44 AM

The time and position in the primed frame:
t'=\frac{c^2 t-u \, x(t)}{c^2 \sqrt{1-\frac{u^2}{c^2}}}
x'=\frac{-t u + x(t)}{\sqrt{1-\frac{u^2}{c^2}}}
Now, we have the time and position of each clock expressed as a function of the parameter t. Taking the derivatives we get:

\frac{dt}{dt}=1
\frac{dx}{c \, dt}=\frac{v(t)
}{c}
\frac{dt'}{dt}=\frac{c^2-u \, v(t)}{c^2 \sqrt{1-\frac{u^2}{c^2}}}
\frac{dx'}{c \, dt}=\frac{-u + v(t)
}{c \sqrt{1-\frac{u^2}{c^2}}}

Substituting in to the above integral gives
\tau=\int_0^8 \sqrt{1-\frac{v(t)^2}{c^2}} \, dt=\frac{2 \left(2 \sqrt{\frac{c^2-4 k^2}{c^2}} k+c \sin
^{-1}\left(\frac{2 k}{c}\right)\right)}{k}
\tau'=\int_0^8 \sqrt{1-\frac{v(t)^2}{c^2}} \, dt=\frac{2 \left(2 \sqrt{\frac{c^2-4 k^2}{c^2}} k+c \sin
^{-1}\left(\frac{2 k}{c}\right)\right)}{k}
which are clearly equal to each other. Evaluating for the specific conditions you gave above gives:
\tau_{heavy \, clock} = \tau'_{heavy \, clock} = 7.849
\tau_{lite \, clock} = \tau'_{lite \, clock} = 7.360

Actually the above is not so explicit for me.
May be I'm not so strong in Math but I believe that explanation can be more detailed.

We have v(t) = k sin(t) and x(t) = - k cos(t) + k (actually these two are more realistic I believe)

So now to caclulate the proper time I'm still have to use the function
\tau=\int_0^8 \sqrt{1-\frac{v(t)^2}{c^2}} \, dt
and for this I have to put into it not my v(t) = k sin(t) velocity but
I have to put a velocity which will be observed from the Primed Frame ("Moving IRF"), correct?

So the question is how can I find this "observed velocity"?
Here your manipulations seems not be explicit enough.

But anyway I will refer to it in my further investigation.

Thank you.

m.starkov

Jul10-09, 11:12 AM

Dalespam, Here I have used the formula you gave me to see the dependency of elapsed time from "k".
Once again, may be i'm wrong in my calculations but I getting very strange results.
And I can't get the results you have got in your calculations.

19668

Anyway your formula gets me into bewilderment.
How can it possible: the amount of time elapsed depends on "k"!
The "k" is relative coefficient! It will be another in different IRFs! (and so the elapsed time will not be the same)

For sure it needs a "little bit" more explanation.
Because it absolutely not explicit for usual person at least.

DaleSpam

Jul11-09, 12:31 PM

So now to caclulate the proper time I'm still have to use the function
\tau=\int_0^8 \sqrt{1-\frac{v(t)^2}{c^2}} \, dt
and for this I have to put into it not my v(t) = k sin(t) velocity but
I have to put a velocity which will be observed from the Primed Frame ("Moving IRF"), correct?No, this is a "special case" formula that I derived as shown above specifically for the scenario you proposed. The v(t) in that formula is the specific function given above, not an arbitrary velocity in any frame.

Please use only the general equation I showed in post 55 (or the integral version of the same which I showed as the first integral in post 99). Once you have become proficient using the general formulas then you will know when you can apply "special case" formulas.

m.starkov

Jul11-09, 07:58 PM

Ok, got it. But what can you say about the second point?

DaleSpam

Jul11-09, 09:19 PM

Ok, got it. But what can you say about the second point?I assume that by "second point" you mean: So the question is how can I find this "observed velocity"?Use the Lorentz transform. Follow my example in post 99.

m.starkov

Jul12-09, 06:04 PM

No! )
I mean my post from Jul10-09, 03:12 PM.
I can't understand how amount of time elapsed can depend on "k"?!

DaleSpam

Jul12-09, 08:00 PM

How can it possible: the amount of time elapsed depends on "k"!The time must depend on k since the speed is, at all times, proportional to k and therefore the time dilation is at all times a function of k. The "k" is relative coefficient!k is not relative, it is proportional to the acceleration, which is absolute.

Go ahead and do the derivation using the general formula above. You will get the right dependence on k if you do it carefully. Be aware, not all values of k will be possible, so don't be surprised if you get strange (eg imaginary) results for large k.

m.starkov

Jul14-09, 01:39 AM

Ok, but what about values?
I have put my values (50000 and -100000) as you see on the picture, but I getting values different from yours ones. Is that ok?

And there are some more points there about this.
You say acceleration is absolute. Ok.
If I have acceleration caused by gravity then can I use this formula too?

So if I have a rotating alpha particle then if it will get into some electric field then the rotation will get slower due to time dilation effect, correct?

DaleSpam

Jul14-09, 04:58 PM

Ok, but what about values?
I have put my values (50000 and -100000) as you see on the picture, but I getting values different from yours ones. Is that ok?
What are you refering to? The only two values I showed agreed with your values.
If I have acceleration caused by gravity then can I use this formula too?I would rather not get into general relativity while you are still shaky on special relativity. But in essence, yes, a generalization of this equation is called the metric and is of critical importance for gravity and GR.

m.starkov

Jul15-09, 08:33 AM

Ok,
So if I will sit every day on a carousel I will be yonger than others?
Acceleration will be always with me, so my time should go slower, yes?

vin300

Jul15-09, 08:47 AM

,
So if I will sit every day on a carousel I will be yonger than others?
If you stay on the fifth floor you be 3 minutes older in 1000 years than groundfloor

DaleSpam

Jul15-09, 12:37 PM

So if I will sit every day on a carousel I will be yonger than others?Yes, I think it would be a worthwhile exercise for you to calculate this. It is a useful problem for its own sake since rotation is so common, and it will help build your confidence and proficiency.

Use the spacetime interval (as shown above) to determine the difference between two clocks, one that is on the carousel and the other that is on the ground that is right next to the carousel. The only real difference from what I showed above is that you will need to use two spatial dimensions to describe the worldline.

m.starkov

Jul19-09, 05:54 AM

Yes I can try. But first I would like to clarify some points.
Here we have no changes in velocity, only movement direction changes.
Does it cause the time dilation effect?
To make conditions simple: we are in open space and not on an orbit of any planet.

A.T.

Jul19-09, 03:51 PM

Yes I can try. But first I would like to clarify some points.
Here we have no changes in velocity, only movement direction changes.
That's still a change in velocity (vector) just no change in speed (scalar). But it still means proper acceleration.
Does it cause the time dilation effect?
In flat space time the proper accelerated will experiences less proper time between meetings that the inertial one.

It would be interesting to compute this also in the non-inertial frame of the carousel guy. There are metrics that describe space-time for a uniformly accelerated observer. Can they be used here?

m.starkov

Jul19-09, 05:24 PM

So I really can't understand now why these guys who performed time dilation approvement experiment use Aircrafts?
It's enough to use carousel in a garden or just to take a rope and spin the clocks by hands!
Can I reproduce that experiment by myself?
I mean can I take two electronic watches, ensure its sync during a month and then put one of them on a carousel and see what will happen in the following month?
I believe we can compensate china's watch precision with a long time of experiment. We will know watches precisions.
Am I right in my conclusions?

DaveC426913

Jul19-09, 07:42 PM

m.starkov

Jul19-09, 11:59 PM

Wait a minute. For aircraft way for several thousands of kilometers the discrepancy was NOT so tiny tiny tiny.
And here I'm putting my clocks on the end of a rope and the other end of a rope I bind to ventilator making clocks to rotate at high speed.
So the acceleration is high here and it suffer acceleration for ALL the time (for aircraft case acceleration is only at the take off and landing).
Thus I expect here not so tini discrepancy.
If you are strong in formulas then I would like you to give approx. value for the following conditions.
Radius = 0.5 meter
Ventilator Speed = 2000 rpm
EarthTime = 1 month

Gigan

Jul20-09, 01:06 AM

The linear speed of a clock on that rig would be .5m x 2000rpm = 1000 m/min = ~ 17m/sec (I know I'm rounding up)

The equation for time dilation is:
\triangle t \prime = \frac{\triangle t}{ \sqrt{1-v^2/C^2}}

Put in your data (assuming your month has 31 days of exactly 24 hours) and you get a difference of 5.3 \times 10^{-9} seconds.
The clock that has been spinning will be 5.3 nanoseconds behind one that is stationary.

(I'm fairly certain that math is right even if this is my first post :) )

matheinste

Jul20-09, 02:13 AM

m.starkov

Jul20-09, 07:14 AM

The linear speed of a clock on that rig would be .5m x 2000rpm = 1000 m/min = ~ 17m/sec (I know I'm rounding up)

The equation for time dilation is:
\triangle t \prime = \frac{\triangle t}{ \sqrt{1-v^2/C^2}}

Put in your data ...
5.3 nanoseconds behind one that is stationary.

The Keyword is "linear".
I'm not so sure we can use this formula in this experiment.
Because the movement is not linear.
The major sense of this experiment is that clocks is always under acceleration.
Your formula doesn't "feel" that acceleration.
Dalespam recommends to use always Lorentz transformation formula.
But I'm not strong enough in Math to handle it.
May be someone else can do it.
It will be very good if we can reproduce time dilation effect just at home using ventilator and china's watches.
Yes, temperature will be different and also may be some electrostatic charge can affect results.
But we can reduce these conditions as we can and... since this is not for public this is just for ourselves. If calculated value will be just aprox. as actual - it will be very nice!

m.starkov

Jul20-09, 07:18 AM

Aircraft circling the earth are continually accelerating.
Matheinste.

During the flight it suffers only centripetal acceleration which is not so high.

matheinste

Jul20-09, 07:34 AM

Just FYI.
Aircraft accelerates only at takoff and landing.
During the flight it suffers only centripetal acceleration which is not so high.

Circular motion, with or without gravity is accelerated motion. An object following a curved path is accelerating towards the instantaneous center of curvature of that path. In the case of circular motion the acceleration is at all times towards the center of the circle. The acceleration is real.
What would be the point of the aircraft circling the earth if the only effective accelerations were at landing and take off?

Matheinste.

Ich

Jul20-09, 07:46 AM

I'm not so sure we can use this formula in this experiment.
Of course you can. It is completely general (in flat spacetime). Have a look at Einstein §4 (http://www.fourmilab.ch/etexts/einstein/specrel/www/).

DaleSpam

Jul20-09, 08:23 AM

Dalespam recommends to use always Lorentz transformation formula.
But I'm not strong enough in Math to handle it.I think you can handle it. You did very well previously, especially considering that I gave some bad advice. Plus, you have access to some math software (Maple?). Just follow the steps:

1) Pick some inertial frame.

2) Write down the parametric form of the path ( t(r), x(r), y(r), z(r) ) in terms of some arbitrary parameter r.

3) Optionally, (not necessary, will not affect the results, but you seem interested in doing it) use the Lorentz transform to find the path, or worldline, in any other inertial frame. The parameter, r, is unchanged.

4) Find the derivatives: dt/dr, dx/dr, dy/dr, dz/dr.

5) Evaluate the integral form of the metric:
\tau = \int
\sqrt{\left(\frac{dt}{dr}\right)^2-\left(\frac{dx}{c \, dr}\right)^
2-\left(\frac{dy}{c \, dr}\right)^
2-\left(\frac{dz}{c \, dr}\right)^
2} \, dr

ibcnunabit

Jul20-09, 05:07 PM

what is the speed of time?

I understand the sense of the question, but within a given frame of reference, the question is meaningless. If time "sped up" or "slowed down", it would do so for everything in that frame, and would be indetectible. Also, speed is a concept that's derived from time, so to try to build it into time somehow would be faulty reasoning.

AND YET:

Physicists talk all the time about time speeding up or slowing down in one frame relative to another. You need the 2nd frame of reference to describe the other frame--but even then, that's only comparing to your own frame--it's not an absolute measure. It can be described as a proportion between the two, but that proportion would be unitless. It would be a constant like 1.2 or 0.7. The same would be true if a strong gravitational field were applied. Time would be "slower," but that change in the time stream would be indetectible to the entity upon which it was applied. Others, however, could conceivably detect it and measure relative to their own frame.

Cheers,
Mike from Shreveport

m.starkov

Aug4-09, 11:25 AM

Basically, all of the usual relativity formulas (time dilation, length contraction, relativistic velocity addition, etc.) are special cases of the Lorentz transform. Because they are special cases if they are accidentally misused you can get contradictions, which is what happened here. It is always safer to use the Lorentz transform rather than the special-case formulas whenever possible, which I will do in the write-up.

Well, finally I have managed to bring my brains together.

The major question for me for the moment: how Lorentz transformation formulas can help here?
AFAIK Lorentz transformation formulas can be used only for calculating Position and Time in another INERTIAL frame.
But my clocks frames are NOT inertial!

So I really can't understand how we can find TIME elapsed in not-inertial frame using only inertial-to-inertial transformations?

It is not in Math, I just can't get the principle...

Can somebody help me?

Dalespam?

DaleSpam

Aug4-09, 05:33 PM

The major question for me for the moment: how Lorentz transformation formulas can help here?
AFAIK Lorentz transformation formulas can be used only for calculating Position and Time in another INERTIAL frame.
But my clocks frames are NOT inertial!

So I really can't understand how we can find TIME elapsed in not-inertial frame using only inertial-to-inertial transformations?

It is not in Math, I just can't get the principle...

Can somebody help me?Don't worry, this is a very common misconception (see Baez's Physics FAQ (http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html)). Special relativity can easily handle accelerating objects, as long as you do the analysis in an inertial reference frame.

In other words, don't forget that reference frames are nothing more than coordinate systems. There is never any requirement that you attach a coordinate system to every object, nor is there even a requirment that every coordinate system be attached to some object. In fact, the most generally useful reference frame, the center of mass frame, usually does not correspond to the rest frame of any single particle. The whole point of SR is that you can use any inertial reference frame to do your analysis, not that you need to attach a coordinate system to each object and re-do your analysis multiple times.

RLS.Jr

Aug4-09, 06:06 PM

The speed of time is dependant on our speed of travel. Tell me how fast we are going and I will tell you the speed of our time.

DaveC426913

Aug5-09, 11:14 PM

The speed of time is dependant on our speed of travel. Tell me how fast we are going and I will tell you the speed of our time.
OK, I'll bite.

We're travelling at 93,141 mi/s relative to Milky Way galactic centre. What is the speed of our time?

GoodPR

Aug5-09, 11:15 PM

rls.jr your going to have to caculate the effect gravity has on this speed of time your speaking of also

Austin0

Aug6-09, 08:21 AM

what is the speed of time?

I propose that the speed of Time is c.

Count Iblis

Aug6-09, 09:20 AM

What is the speed of the x-coordinate? :confused:

RLS.Jr

Aug6-09, 05:46 PM

OK, I'll bite.

We're travelling at 93,141 mi/s relative to Milky Way galactic centre. What is the speed of our time?

Our speed is determined by adding all the vectors of travel in respect to the center of the Universe. Picking one vector (93,141 mi/s relative to Milky Way galactic centre) is as incomplete as determining Lance Armstrong's average speed of a 8 hour race by measuring distance in a random 60 second interval.

However, if you wanted the speed of our time based on 'Milky Way Galaxy Time', and added all vectors of travel (earth's rotation: ~900 mph; earth's orbit: ~66,600 mph; our solar system's motion within the galaxy: ~41,666 mph) and came up with an instant vector (much less than 93,141 mi/s) then it could be done. Let's hypothetically say that at this second it is 84,000 mph relative to Milky Way galactic centre. Now, based on our relative speed compared to C, what is the speed of our time (in Milky Way terms)?

PaulRacer

Aug6-09, 06:48 PM

OK, I'll bite.

We're travelling at 93,141 mi/s relative to Milky Way galactic centre. What is the speed of our time?

Half the speed of light? Now were talkin'. How do we know this? From my understanding time spans from infinity to zero. But 1 second will always be 1 second to the person experiencing it firsthand. Love this site.

DaveC426913

Aug6-09, 10:49 PM

Our speed is determined by adding all the vectors of travel in respect to the center of the Universe. Picking one vector (93,141 mi/s relative to Milky Way galactic centre) is as incomplete as determining Lance Armstrong's average speed of a 8 hour race by measuring distance in a random 60 second interval.

However, if you wanted the speed of our time based on 'Milky Way Galaxy Time', and added all vectors of travel (earth's rotation: ~900 mph; earth's orbit: ~66,600 mph; our solar system's motion within the galaxy: ~41,666 mph) and came up with an instant vector (much less than 93,141 mi/s) then it could be done. Let's hypothetically say that at this second it is 84,000 mph relative to Milky Way galactic centre. Now, based on our relative speed compared to C, what is the speed of our time (in Milky Way terms)?
Well, it was your question; presumably you had an answer.

I was just interested if your were going to propose an answer that was anything other than 'one second per second' - to which I would have cried foul.

RLS.Jr

Aug7-09, 01:20 PM

Well, it was your question; presumably you had an answer.

I was just interested if your were going to propose an answer that was anything other than 'one second per second' - to which I would have cried foul.

Fine, here is your answer:

Assuming we are traveling at 84,000 mph or 23.333333333 mps relative to the center of the Milky Way Galaxy...the speed of our time is 1.000000007844765 MGT (Milky Way Galaxy Time)

Of course, this is just a hypothetical 'instant'. If we put a vector model together tracking our constantly changing speed relative to the center of the Galaxy we would accurately know the speed of our time (relative to MGT). As it is, we can assume a relative time flux of 1-1.000000013249398 MGT.

This is derived by adding all known speed vectors between the earth and the center of the Milky Way Galaxy, ~30.323888 mps, which illustrates the highest potential speed if all vectors aligned. With 0 mps being the lowest potential speed relative to the center of the Galaxy, the most accurate assessment of the speed of our time is:
1 - 1.000000013249398 MGT

DaveC426913

Aug7-09, 09:40 PM

Fine, here is your answer:

Assuming we are traveling at 84,000 mph or 23.333333333 mps relative to the center of the Milky Way Galaxy...the speed of our time is 1.000000007844765 MGT (Milky Way Galaxy Time)

All that calculating for nothing...

I picked 93,141 mi/s deliberately because it is half of c.

Of course, this is just a hypothetical 'instant'. If we put a vector model together tracking our constantly changing speed relative to the center of the Galaxy we would accurately know the speed of our time (relative to MGT). As it is, we can assume a relative time flux of 1-1.000000013249398 MGT.

This is derived by adding all known speed vectors between the earth and the center of the Milky Way Galaxy, ~30.323888 mps, which illustrates the highest potential speed if all vectors aligned. With 0 mps being the lowest potential speed relative to the center of the Galaxy, the most accurate assessment of the speed of our time is:
1 - 1.000000013249398 MGT
OK, I kind of thought that's what you were going to say.

I don't know why you think our time would change because I've specified our speed relative to some arbitrary object. Note that I can give you my speed relative to multiple objects simultaneously. How are you going to calculate our speed of time if I give you speeds relative to six different galaxies?

Our time is 1s per s.

RLS.Jr

Aug8-09, 02:12 AM

All that calculating for nothing...

I picked 93,141 mi/s deliberately because it is half of c.

OK, I kind of thought that's what you were going to say.

I don't know why you think our time would change because I've specified our speed relative to some arbitrary object. Note that I can give you my speed relative to multiple objects simultaneously. How are you going to calculate our speed of time if I give you speeds relative to six different galaxies?

Our time is 1s per s.

Wasted math? Truth is never a waist.

Had you just posted the speed I would have given you a simple answer. However, using the center of our Galaxy as the reference point makes much more sense. That's why I modified your number to one that fits that reference point. Of course you can see that our speed in relation to the center of the Galaxy is in constant flux. Thus our 'speed of time' is in constant flux.

Thank you for bringing up the idea of multiple galaxies. The best reference point for determining our speed of time is the center of the universe. When we can measure all the speed vectors (actually only three more than I have already added..one of which is known)
then we can accurately determine the speed of our time and any galaxy. It would be in the unit of UT (Universal Time).

Finally, time is 1s/s only at a point of reference. If you consider yourself as the only real point of reference, then yes you are right. That does come across a little egocentric though.

matheinste

Aug8-09, 03:47 AM

DaveC426913

Aug8-09, 10:07 AM

Wasted math? Truth is never a waist.
There was no truth in it. Your logic is flawed. See below:

Had you just posted the speed I would have given you a simple answer.
Had I just posted the speed it would have been nonsensical. Speed is meaningless without providing a reference point by which to measure it.

However, using the center of our Galaxy as the reference point makes much more sense. That's why I modified your number to one that fits that reference point. Of course you can see that our speed in relation to the center of the Galaxy is in constant flux. Thus our 'speed of time' is in constant flux.
The speed of our time does not change because of a reference to some far off point. As I point out, the reference point is arbitrary. I can observe a hundred or a thousand of them in less time than it takes to name them. Do you think my speed of time changes everytime I cast my eyes upon a different star?

Thank you for bringing up the idea of multiple galaxies. The best reference point for determining our speed of time is the center of the universe. When we can measure all the speed vectors (actually only three more than I have already added..one of which is known)
then we can accurately determine the speed of our time and any galaxy. It would be in the unit of UT (Universal Time).
There is no such thing as the centre of the universe.

Finally, time is 1s/s only at a point of reference. If you consider yourself as the only real point of reference, then yes you are right. That does come across a little egocentric though.
You really need to read up on relativity. It would clear up many of the misconceptions you've stated above. All time and velocity is relative; there is no absolute.

DaveC426913

Aug8-09, 10:09 AM

If our clocks, that is the clocks in the frame in which we are at restJust a clarification: By definition we are at rest in our own frame, as are all clocks in our frame.

matheinste

Aug8-09, 01:40 PM

Just a clarification: By definition we are at rest in our own frame, as are all clocks in our frame.

Yes. What I really should have said was a clock carried with us, which amounts to the same thing.

Matheinste.

scott haig

Aug8-09, 06:09 PM

I can see getting a bit confused because of our language: we say "time is passing" etc.--but in the most basic (newtonian) sense time is just a ratio of distance (a thing we measure) to speed. Nothing is moving when "time moves" and therefore there is no speed of time.

BUT

I've always wondered--along similar lines-- about the putative "speed" of gravity. (Have assumed it's c but don't remember/never knew why). I'm guessing that its something like this:as a distant mass moves the change in the gravitational force vector's direction does transmit information--which has to move at subluminal speeds to prevent problems with causation.

So the question I'd like to add here is this: why is the vector particle of gravitation so hard to sense? Aren't we "sensing" them when we observe tidal forces of, say, the moon? And couldn't a pretty easy experiment be done with large masses undergoing rapid accelleration (like a a satellite between earth and moon) to time nearby gravitational effects relative to a light signal?

(sorry to branch off here but it seems the crowd on this thread would know)
svh

RLS.Jr

Aug8-09, 06:34 PM

Just some simple thoughts regarding clock rates.I am not sure if that is what is meant by the speed of time but I suppose it must be.

If our clocks, that is the clocks in the frame in which we are at rest,run at a rate other than 1 second per second then

1-- How would we know.

2-- If we knew to, what did we compare them with to arrive at this knowledge.

3-- If we know they are running at the wrong rate perhaps it would be best to set them at the correct rate, whatever that is.

4-- If a rate with which we compare our clocks, and everybody else compares theirs exists, should we call it universal time or absolute time.

Matheinste.

First, remember that as a primary reference point your time always runs at 1s/s. It's only when using an object traveling a different speed as the primary reference point that your time is either faster or slower than 1s/s (1s/s is always the clock speed of the primary reference point).

1-- We know that everything traveling at different speeds have different speeds of time (rates of clock speed). Since it makes more sense to have a standardized point of reference as a primary reference point than everyone using themselves. We can conclude that our clock speed is only 1s/s when our speed equals that of the primary reference point.

2-- When comparing time speeds in and around Earth (satellites, planes, the moon) , it would be best to use the earth as the primary reference point and have units ET (Earth Time). Progressively more significant primary reference points would be the sun (SST-Solar System Time); the center of the Milky Way Galaxy (MGT); with the center of the universe as the best (UT) when all vectors of speed can be calculated.

3-- There is no wrong or right rate. Just more significant primary reference points when talking about two or more objects that have different velocities.

4-- UT or Universal Time (when we are able to calculate all the speed vectors) would be most significant when dealing with referencing objects outside our galaxy. For most practical purposes ET is adequate. Absolute time would only be a valid unit if we were able to determine that space had limits and what all the speed vectors from the center were.

Lastly, an example showing that it is less valid for each of us to use ourselves as primary points of reference:
If you are sitting in a seat. Are you motionless? What if that seat is on a train traveling
80 mph? You could still consider yourself as the primary point of reference and think that the earth and everyone on it was traveling, but you can see the inherent problems with this point of view. It is more valid to use the earth as the primary reference point and describe yourself as traveling 80 mph along it's surface.

RLS.Jr

Aug8-09, 07:30 PM

There was no truth in it. Your logic is flawed. See below:

Had I just posted the speed it would have been nonsensical. Speed is meaningless without providing a reference point by which to measure it.

The speed of our time does not change because of a reference to some far off point. As I point out, the reference point is arbitrary. I can observe a hundred or a thousand of them in less time than it takes to name them. Do you think my speed of time changes everytime I cast my eyes upon a different star?

There is no such thing as the centre of the universe.

You really need to read up on relativity. It would clear up many of the misconceptions you've stated above. All time and velocity is relative; there is no absolute.

Ok, my logic is flawed? Let's see:

- If only a speed is given then the 'speed of time' difference can be calculated using 0 mps as the primary reference point (which is inherent in the statement). Is it nonsensical to say that you are in a rocket ship traveling 17580 mph?

- If your speed is constantly changing in relation to a primary reference point (e.g. center of the Milky Way Galaxy) then your speed of time is constantly changing in direct proportion. It is easier to see using the center of our solar system as the primary reference point. When the revolution of the earth (~900 mph) aligns (is the same direction) with it's orbit around the sun, then you are traveling ~1800 mph faster than when it's traveling in the opposite direction of the orbit.

- Your speed of time is always experienced as 1s/s as you can only experience reality with yourself as the primary reference point. But in relation to every star you see in the sky, your speed of time is different. You said it yourself "All time and velocity is relative; there is no absolute".

- There is no center of the universe? Is there a center of your room? If there was a 'Big Bang' and mater/energy continues to spread outward in all directions, then there is a center. If fact, there is most likely a massive black hole at the center of the universe as there is at the centers of Galaxies.

Hyrage

Aug8-09, 08:44 PM

Ok, my logic is flawed? Let's see:

- If only a speed is given then the 'speed of time' difference can be calculated using 0 mps as the primary reference point (which is inherent in the statement). Is it nonsensical to say that you are in a rocket ship traveling 17580 mph?

- If your speed is constantly changing in relation to a primary reference point (e.g. center of the Milky Way Galaxy) then your speed of time is constantly changing in direct proportion. It is easier to see using the center of our solar system as the primary reference point. When the revolution of the earth (~900 mph) aligns (is the same direction) with it's orbit around the sun, then you are traveling ~1800 mph faster than when it's traveling in the opposite direction of the orbit.

- Your speed of time is always experienced as 1s/s as you can only experience reality with yourself as the primary reference point. But in relation to every star you see in the sky, your speed of time is different. You said it yourself "All time and velocity is relative; there is no absolute".

- There is no center of the universe? Is there a center of your room? If there was a 'Big Bang' and mater/energy continues to spread outward in all directions, then there is a center. If fact, there is most likely a massive black hole at the center of the universe as there is at the centers of Galaxies.
There should be a center to everything that exists.
Time is created by existing things & doesn't exist; it's a perception and you can act and think faster than it.

DrGreg

Aug8-09, 09:24 PM

There is no center of the universe?No, in the same way that there is no point on the Earth's surface that is the centre of the Earth's surface.

RLS.Jr

Aug8-09, 10:44 PM

No, in the same way that there is no point on the Earth's surface that is the centre of the Earth's surface.

Are you then saying that our Universe is two dimensional? Because there is most certainly a center of our three dimensional Earth.

DaveC426913

Aug8-09, 11:05 PM

Are you then saying that our Universe is two dimensional? Because there is most certainly a center of our three dimensional Earth.
Really? What latitude and longitude is it at?

The surface of the Earth has no centre, just as the volume of the universe has no centre.

Your statements are based upon incorrect preconceptions. You'll have to get the basics right before you can argue sensically about things built upon them.

- There is no center of the universe? Is there a center of your room? If there was a 'Big Bang' and mater/energy continues to spread outward in all directions, then there is a center. If fact, there is most likely a massive black hole at the center of the universe as there is at the centers of Galaxies.

More misconceptions. How can you possibly deign to put forth a hypothesis about cosmology when you are being tripped up by the basics? There is no way to be more gentle. You need to read more. Or at least, do more asking of questions and less stating of claims. At PF, our goal is to educate.

RLS.Jr

Aug9-09, 12:22 AM

Really? What latitude and longitude is it at?

The surface of the Earth has no centre, just as the volume of the universe has no centre.

Your statements are based upon incorrect preconceptions. You'll have to get the basics right before you can argue sensically about things built upon them.

More misconceptions. How can you possibly deign to put forth a hypothesis about cosmology when you are being tripped up by the basics? There is no way to be more gentle. You need to read more. Or at least, do more asking of questions and less stating of claims. At PF, our goal is to educate.

LOL
You talk about basics, but you don't even concede that there is a center to the Earth. Let me give you a hint, it's hot. You offer that there is no center to the outside of a sphere as proof that there is no center to the universe. Nice logic.

It's easy to act condescending saying that I need to read more without giving reasoning as to how my position is wrong. Well educate us at PF with evidence or logic on how my assessment on the speed of our time is wrong. (We can get back to your two dimensional universe later)

DaveC426913

Aug9-09, 12:54 AM

You offer that there is no center to the outside of a sphere as proof that there is no center to the universe. Nice logic.
It is not a proof; it is an analogy, to help you grasp the idea.

And it is not my idea. I take credit for none of it. You will find this in any basic book or website on the subject.
Here are the first couple of Google hits for "where is the centre of universe":
http://math.ucr.edu/home/baez/physics/Relativity/GR/centre.html
http://www.physlink.com/Education/AskExperts/ae614.cfm
Feel free to find answers for yourself.

Or, let's let anyone else here on PF weigh in.

It's easy to act condescending saying that I need to read more I have no wish to be condescending. You are making claims that you have no business making. That puts you in an awkward position.

If I added 2 and 2 and got 5, you'd correct me before letting me proceed with a discussion. If I insisted, you wouldn't try to refute my logic, you'd tell me to go read a math book - it's not your job to teach me arithmetic.

I say again: ask more questions; make fewer claims.

RLS.Jr

Aug9-09, 02:19 AM

It is not a proof; it is an analogy, to help you grasp the idea.

And it is not my idea. I take credit for none of it. You will find this in any basic book or website on the subject.
Here are the first couple of Google hits for "where is the centre of universe":
http://math.ucr.edu/home/baez/physics/Relativity/GR/centre.html
http://www.physlink.com/Education/AskExperts/ae614.cfm
Feel free to find answers for yourself.

Or, let's let anyone else here on PF weigh in.

I have no wish to be condescending. You are making claims that you have no business making. That puts you in an awkward position.

If I added 2 and 2 and got 5, you'd correct me before letting me proceed with a discussion. If I insisted, you wouldn't try to refute my logic, you'd tell me to go read a math book - it's not your job to teach me arithmetic.

I say again: ask more questions; make fewer claims.

Stick with the subject.
I have no issue with you saying that my use of Einstein's theory is inaccurate. However, you give no logic or support. I am stating a logical extension of Einstein's time dilation. If you disagree, indicate why. If it's because you read that it's wrong, let us know what the author said and we will give him credit for the argument.

I maintain that my use of primary reference points is valid. If you can't or are unwilling to state the error in my reasoning or calculations on the speed of our time, stop telling me I'm wrong. (i.e. I can explain to you why your 2+2=5 is wrong. Can you explain why you feel my 'speed of our time' is wrong?)

And I encourage others to join in..on either side. (Just please add more to the discussion than "I've read more books")

Nabeshin

Aug9-09, 03:22 AM

- There is no center of the universe? Is there a center of your room? If there was a 'Big Bang' and mater/energy continues to spread outward in all directions, then there is a center. If fact, there is most likely a massive black hole at the center of the universe as there is at the centers of Galaxies.

I must reiterate DaveC's grievances. This statement is, to put it bluntly, uneducated. You are essentially attempting to apply common sense and analogies to a problem which inherently defies common sense. Analogies such as the surface of the earth one DaveC brings up (Note the word SURFACE) are very carefully crafted by people who already understand the mathematics behind the theory attempting to explain it in a visual manner without giving too many misconceptions. I am sorry, but when I read this claim that you so boldly state without any uncertainty, you lose all credibility. To use another of DaveC's analogies, if a man wants to lecture me in mathematics and begins by saying 2+2=5, so clearly ... I will walk away immediately.

The point here is that the assertion that there exists a central point of the universe seems to be a critical point of whatever... theory or whatever you have been discussing. Since this is not the case, any further discussion is moot. End of story.

DaveC426913

Aug9-09, 10:56 AM

I maintain that my use of primary reference points is valid. If you can't or are unwilling to state the error in my reasoning or calculations on the speed of our time, stop telling me I'm wrong. (i.e. I can explain to you why your 2+2=5 is wrong. Can you explain why you feel my 'speed of our time' is wrong?)
You yourself followed you own logic of "absolute reference points" to reach this conclusion: The best reference point for determining our speed of time is the center of the universe. ... It would be in the unit of UT (Universal Time).
There is no such thing. This is basic modern cosmology. You need to know this in order to have a meaningful discussion.

The onus is not upon me to refute your arguments, the onus is upon you to first know the facts before basing claims upon them.

Hyrage

Aug9-09, 11:02 AM

So there is no center to our universe, really? If there is such thing as gravity, there must be a center, isn't it?

DaveC426913

Aug9-09, 11:06 AM

So there is no center to our universe, really?Really. (http://physicsforums.com/poll.php?do=showresults&pollid=1643)

DaleSpam

Aug9-09, 11:08 AM

Are you then saying that our Universe is two dimensional? Because there is most certainly a center of our three dimensional Earth.The geometric point is that when you are dealing with curved geometry it is easily possible for the "center" of a space to lie outside that space. I.e. it exists only in some higher-dimensional flat space in which the curved space is embedded.

When talking about the geometry of the universe any such higher dimensional space is entirely non-physical. Thus the universe has no center in any physical sense.

The surface of a sphere is really a good analogy, you should try to learn from it.

Hyrage

Aug9-09, 11:17 AM

Even if the universe really was a torus, including that everything moves inside and outside of it, it would be possible to find it's center.

Everything that exist is geometrical... and I guess our universe too.

http://www.soulsofdistortion.nl/images/scalar_torus.gif
http://www.smeddum.net/images/articles/torus.gif
http://www.benpadiah.com/otherstuff/1torus.gif

Hootenanny

Aug9-09, 11:20 AM

Even if the universe really was a torus, including that everything moves inside and outside of it, it would be possible to find it's center.
What do you mean inside and outside our universe?

Everything that exist is geometrical... and I guess our universe too.

Nonsense, many things have a geometrical interpretation, but that doesn't necessarily mean that are geometrical, what ever 'being geometrical' means.

Hyrage

Aug9-09, 11:24 AM

What do you mean inside and outside our universe?

Nonsense, many things have a geometrical interpretation, but that doesn't necessarily mean that are geometrical, what ever 'being geometrical' means.
If you look at the .gifs... you'll find your answer.

And from that following image, it strangely looks like a part of the torus...

http://www.scienceimage.csiro.au/mediarelease/images/expanding_universe_nasa.jpg

Hootenanny

Aug9-09, 11:27 AM

If you look at the .gifs... you'll find your answer.
I don't need to look at the images, I know precisely what a torus looks like and they don't answer my question.

What do you mean a torus? Do you mean the surface of a torus, or do you mean the volume enclosed by the surface?

Why don't you stop being cryptic and just directly answer the direct questions?

Hyrage

Aug9-09, 11:31 AM

I meant torus like a toroidal field.

DaveC426913

Aug9-09, 11:33 AM

Hyrage

Aug9-09, 11:59 AM

Hyrage, when you are using a torus to represent the universe, you cannot talk about the "centre of the torus". The representation involves only the surface of the object. You cannot fly to- point to- or have gravity emanating from- any point that is not on the surface.

And I totally disagree with that. A center is a fictive point that you create (an object itself), that goes beyond any surfaces...

By using math, you can find the center of an object. If you take the Tetrahedron by example, it is mostly easy to find it out... You can find the middle point of any polygon and link the centers to their opposite vertex to get the center of the geometrical form and it's not on the surface. You could do that for anything, even a sphere or a torus...

http://www.unifiedworlds.com/tetrahedron.jpg

However, maybe you could give me a deeper explanation, because for now... I can't agree.

Hootenanny

Aug9-09, 12:10 PM

By using math, you can find the center of an object. If you take the Tetrahedron by example, it is mostly easy to find it out... You can find the middle point of any polygon and link the centers to their opposite vertex to get the center of the geometrical form and it's not on the surface. You could do that for anything, even a sphere or a torus...
Of course one can find the geometrical centre of an object, but the question is how is that interpreted physically. If the torus representation pertains only to the surface of the torus, then although one can find a point that is geometrically at the centre of a torus, this point lies outside the representation and therefore cannot be interpreted.

Hyrage

Aug9-09, 12:27 PM

Of course one can find the geometrical centre of an object, but the question is how is that interpreted physically. If the torus representation pertains only to the surface of the torus, then although one can find a point that is geometrically at the centre of a torus, this point lies outside the representation and therefore cannot be interpreted.However, what if the center of the is unified instead of being made of a hole like a donut. A "hole" so thin that it doesn't even look like a hole, but the geometrical shape really looks like a torus or a toroidal field. So if everything moves as a toroidal field, everything will one day pass by there.

http://www.geocities.com/human_energy_fields/index_files/image004.gif
In other terms, maybe there is really a center afterall.