Is Time's Speed Just Wild Speculation?

[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.

 

[m.starkov]

Please take a look

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?

 

[Dale]

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]

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.

 

[Dale]

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]

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]

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.

 

[Dale]

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]

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?

 

[Dale]

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]

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?

 

[Dale]

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]

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.

 

[Dale]

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.

 

[DaveC426913]

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!

 

[Dale]

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

Use the force Dave, it is equal to mass times acceleration!

 

[m.starkov]

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.]

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 Earth's 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]

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.

 

[m.starkov]

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]

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

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?

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?