Is Time's Speed Just Wild Speculation?

[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}<br /> \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}\\<br /> \text{d$\tau$}^2=\text{dt}^2-\frac{\text{dx}^2}{c^2}\\<br /> \text{d$\tau$}=\sqrt{\text{dt}^2-\frac{\text{dx}^2}{c^2}}\\<br /> \int \text{d$\tau$}=\int \sqrt{\text{dt}^2-\frac{\text{dx}^2}{c^2}} \\<br /> \tau =\int \frac{\sqrt{\text{dt}^2-\frac{\text{dx}^2}{c^2}}}{\text{dt}} \, dt\\<br /> \tau =\int \sqrt{\frac{\text{dt}^2}{\text{dt}^2}-\frac{\text{dx}^2}{\text{dt}^2 c^2}} \, dt\\<br /> \tau =\int \sqrt{1-\frac{v^2}{c^2}} \, dt<br /> \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<br /> \sqrt{\left(\frac{dt}{dr}\right)^2-\left(\frac{dx}{c \, dr}\right)^<br /> 2-\left(\frac{dy}{c \, dr}\right)^<br /> 2-\left(\frac{dz}{c \, dr}\right)^<br /> 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:

<br /> v(t)=\begin{cases}<br /> k t &amp; t\leq 2 \\<br /> -k (t-4) &amp; 2&lt;t\leq 6 \\<br /> k (t-8) &amp; \text{otherwise}<br /> \end{cases}<br />

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

t=t

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

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

t&#039;=\frac{c^2 t-u \, x(t)}{c^2 \sqrt{1-\frac{u^2}{c^2}}}

x&#039;=\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)<br /> }{c}

\frac{dt&#039;}{dt}=\frac{c^2-u \, v(t)}{c^2 \sqrt{1-\frac{u^2}{c^2}}}

\frac{dx&#039;}{c \, dt}=\frac{-u + v(t)<br /> }{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<br /> ^{-1}\left(\frac{2 k}{c}\right)\right)}{k}

\tau&#039;=\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<br /> ^{-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&#039;_{heavy \, clock} = 7.849
\tau_{lite \, clock} = \tau&#039;_{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&#039;=\frac{c^2 t-u \, x(t)}{c^2 \sqrt{1-\frac{u^2}{c^2}}}
x&#039;=\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)<br /> }{c}
\frac{dt&#039;}{dt}=\frac{c^2-u \, v(t)}{c^2 \sqrt{1-\frac{u^2}{c^2}}}
\frac{dx&#039;}{c \, dt}=\frac{-u + v(t)<br /> }{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<br /> ^{-1}\left(\frac{2 k}{c}\right)\right)}{k}
\tau&#039;=\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<br /> ^{-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&#039;_{heavy \, clock} = 7.849
\tau_{lite \, clock} = \tau&#039;_{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.

 

[m.starkov]

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

 

[Dale]

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]

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

 

[Dale]

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]

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?

 

[Dale]

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 referring 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]

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]

,
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

 

[Dale]

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]

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.]

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]

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]

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?

Because the discrepancy is tiny tiny tiny. Too small for watches on carousels.

 

[m.starkov]

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]

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 :) )