Is Time's Speed Just Wild Speculation?

[Nisse]

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 criticize any of the contributers; it is the nature of science.

And I also recognise that this comment adds nothing positive to the debate.

 

[petm1]

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

 

[Phrak]

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 criticize any of the contributers; it is the nature of science.

And I also recognise that this comment adds nothing positive to the debate.

They'll also be wondering about your comment, and why most physicists where female for a brief span of history within western culture. :/

 

[Max™]

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

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.

 

[Dale]

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

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]

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 criticize 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]

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.

 

[Dale]

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]

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.

 

[negitron]

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 (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]

Worldwide Words (a source which has consistently proven itself to me) disagrees with you:

I stand corrected.

Matheinste.

 

[Phrak]

I stand corrected.

Matheinste.

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

 

[pawan_ctn]

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

 

[m.starkov]

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".

 

[Dale]

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]

Hi DaleSpam,

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

 

[Dale]

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

 

[m.starkov]

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]

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 anyone 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.

 

[Dale]

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 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]

Dear DaleSpam,

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

Thank you

 

[Dale]

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]

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..

 

[Dale]

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]

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?

 

[Dale]

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]

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]

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

 

[Dale]

That is certainly a reasonable conclusion.

 

[TVP45]

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.)

 

[Dale]

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]

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?

 

[Joans]

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]

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?

 

[Dale]

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.