Is Time's Speed Just Wild Speculation?

[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?

 

[Dale]

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. [nitpick]First, the velocity profile does not represent a mass being pushed by a spring[/nitpick] 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:

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

 

[Dale]

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 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]

There are two problems here.

1. you must start and end your clocks together.
2. you need to use the 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?

 

[Dale]

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]

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

 

[Dale]

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]

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:

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.

 

[Dale]

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.

 

[Dale]

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 into 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]

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.]

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.

 

[Dale]

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]

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 into 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]

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.

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.

 

[Dale]

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.