Derivation of proper time in acceleration in SR

[stevmg]

yuiop,
In post 28 (cited below) you point out that you derived in post 8.
$$v = \frac{a't}{\sqrt{1+(a't/c)^2}}$$

$$a'$$ is the acceleration in S'.

But you actually derived $a$ in that post (the acceleration in S, not S') I have included that post 8 below this post 28 below

Now, based on the original $a$ and final $t$, what did you do?

I am lost.

Post 28, yiuop

In post #8 I derived:

$$v = \frac{a't}{\sqrt{1+(a't/c)^2}}$$
Substitute this for the v in $$\gamma = 1/\sqrt{1-v^2/c^2}$$ and you eventually get after some algebraic manipulation:

$$\gamma = \sqrt{1 + (a't/c)^2}$$

I see you have already done the numerical example for t=10 and v=0.99875c in S and a proper acceleration of 2c per second in your edit. Just for clarity, here is another example for the event t=1 in S. At this event the velocity is:

$$v = \frac{a't}{\sqrt{1+(a't/c)^2}} = \frac{2}{\sqrt{1+(2)^2}} = 0.8944c$$

The gamma factor at time t=1 and v = 0.8944c in S is:

$$\gamma = \sqrt{1 + (a't/c)^2} = \sqrt{1+2^2} = 2.2361$$

or

$$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1-0.8944^2}} =2.2361$$

Well done. You have found another typo. It should have been:

$$v = \frac{a't}{\sqrt{1+(a't/c)^2}} = \frac{a't}{\gamma} = \frac{2*10}{20.025} = 0.99875c$$

where a'=2 and t=10 as given in S and the gamma factor of 20.025 was calculated earlier.

Yes, as in the title of the paragraph.

Sorry about the typos in the equations. I know the numbers are right because I did them in a spreadsheet, but the hassles of entering and previewing TEX equations means some errors sometimes creep in during the translation. At least this this is a two way medium and we can get right between us.

Post 8, yiuop

As I mentioned in post #2, the coordinate acceleration is a factor of gamma cubed smaller than the proper acceleration. In the last two equations above, capital T is the coordinate time and small t is the proper time, but to avoid further confusion I will use the symbols recommended by Passionflower, so that:

$$a = \frac{dv}{dt} = \alpha \gamma^{-3} = \alpha (1-v^2/c^2)^{3/2}$$

This can be rearranged to:

$$\frac{dt}{dv} = \frac{1}{\alpha(1-v^2/c^2)^{3/2}}$$

Integrating both sides with respect to v gives:

$$t= \int \left( \frac{1}{\alpha(1-v^2/c^2)^{3/2}} \right) dv = \frac{v}{\alpha \sqrt{1-v^2/c^2}}$$

When rearranged:

$$v= \alpha t \sqrt{1-v^2/c^2}$$

$$\rightarrow v^2 = (\alpha t)^2 - (\alpha t v/c)^2$$

$$\rightarrow v^2 (1+ (\alpha t /c)^2) = (\alpha t)^2$$

$$\rightarrow v = \frac{\alpha t}{\sqrt{1+(\alpha t/c)^2}}$$
which is the last equation you asked about in your post.

 

[yuiop]

yuiop,
In post 28 (cited below) you point out that you derived in post 8.
$$v = \frac{a't}{\sqrt{1+(a't/c)^2}}$$

$$a'$$ is the acceleration in S'.

But you actually derived $a$ in that post (the acceleration in S, not S') I have included that post 8 below this post 28 below

Now, based on the original $a$ and final $t$, what did you do?

I am lost.

In post 8, I was using the alpha symbol $\alpha$ to represent proper acceleration measured by an accelerometer on the accelerating rocket. (click on the latex symbol to see the underlying code). By post 28 the symbol I was using for proper acceleration was $a_0$ to identify the measurement made by the accelerating observer in a consistent way. The proper acceleration is numerically equal to the (give or take a sign) to the coordinate acceleration measured by the CMIRF observer so $\alpha$, a' and $a_0$ are equivalent and interchangeable. So the coordinate acceleration a is related to the others by $a = a'\gamma^3 = a_0\gamma^3 = \alpha \gamma^3$. Like you said, lots of observers so lots of symbols required.

So given:

$$v = \frac{a_0t}{\sqrt{1+(a_0t/c)^2}}$$

and substitute into $$\frac{1}{\gamma} = \sqrt{1-v^2/c^2}$$

you get:

$$\frac{1}{\gamma} = \sqrt{1-\frac{(a_0t/c)^2}{(1+(a_0t/c)^2)}}= \sqrt{\frac{1+(a_0t/c)^2-(a_0t/c)^2}{(1+(a_0t/c)^2)}}= {\frac{1}{\sqrt{1+(a_0t/c)^2}}$$

$$\rightarrow \gamma = \sqrt{1+(a_0t/c)^2}}$$

 

[yuiop]

Another interesting titbit is the question what is the slowest time to go from Event 1 to Event 2?

It turns out to be around 3.084 Years. In other words the traveler took a 40% shortcut 'through' spacetime.

While checking out your observation, I noticed another interesting relationship.

The slowest time between the two events is the elapsed proper time interval $t_i$ measured by a clock moving inertially between the two events, i.e. $$t_i = \sqrt{\Delta t^2 - \Delta x^2/c^2}$$

Now it seems from playing around with my spreadsheet, that the following (without proof) is true:

$$\Delta x = \frac{1}{2} \alpha t_i^2$$

which has the same form as the Newtonian equation.

It obviously follows that the constant proper acceleration can be obtained from this simple equation:

$$\alpha = 2\frac{\Delta x}{t_i^2}$$

P.S. That should be seconds, not years, to be consistent with the original scenario

 

[yuiop]

Hiyuiop My first reaction to your post was "wonderful", one of those few questions with a neat definitive empirical answer. But having though it over I have more questions.
In the accelerator tests do they actually do a coordinate acceleration profile based on short interval velocities, during the course of acceleration??
I would assume they would have no imperative to maintain constant proper acceleration for the electrons but would just go for the flattest . most constant coordinate acceleration obtainable . yes??

I have no knowledge about the practical design and operation of particle accelerators. Maybe someone like ZapperZ can help you with this.

Although there is as far as know, no imperative to go for constant proper acceleration, it is entirely natural to do so, because that is what happens naturally when you apply a constant force in the accelerator frame. This would seem the simplest option, rather than applying a progressively increasing force trying to maintain a constant coordinate acceleration.

Also, AFAIK there are accelerating sections in the ring and constant speed sections where they carry out the experiments e.t.c. As the particles go around the ring, the energy to the accelerating section has to to be timed to coincide with the arrival of a bunch of particles at the accelerating section of the ring. This requires knowledge of the exact velocity and location of the particles in the ring at all times and this requires using relativistic calculations. If they used Newtonian calculations, the timing would be significantly out and the accelerator simply wouldn't work. The magnitude of relativistic effects in a particle accelerator is far greater than the often quoted example of GPS satelites.

SO does the gamma cubed factor relate to the energy required to maintain maximal acceleration or is it also a declining acceleration curve?

As above, my guess is a declining coordinate acceleration curve, but I am ready to concede to any real experts on the subject.

The above derivation, not surprisingly makes complete sense but...
it is based on acceleration relative to an abstract CMIRF and then this is transformed into rest frame coordinate acceleration at the end. This may of course be absolutely valid but it seems to me that in this circumstance the CIMRFs are somewhat of a bootstrap construct i.e. accelerating relative to one and then there is automatically another one there to accelerate from , with no direct connection to the observation from the reference frame , of either the acceleration of the actual system or the acceleration of the CMIRF.
The increased velocity is just assumed. The coordinate acceleration in the reference frame would actually have to be based on a series of short interval "instantaneous" velocity measurements , no?

Maybe a little more thought on my part.
Thanks for the info

If we ask the question, "What happens if we accelerate an object that is already moving at 0.9c?", the initial analysis might seem daunting. Here is one way of thinking about it that might help. Consider a car that accelerates from 0 to 180 mph in 6 seconds. That is quick for a car, but the relativistic effects are insignificant and can be ignored and we can use Newtonian equations and be correct to a high order of accuracy. To know hat happens if the car was initially traveling at 0.9c and then accelerated with the same force (measured by the car accelerometer) then the relativity principle tells us that all we have to do is do a Lorentz transformation of our car to the point of view of an inertial observer going past us at -0.9c. When we do that, the gamma cubed factor comes out. To build a complete picture then you are right that we need a whole series of observers with velocities varying by an infinitesimal step size. The 180 mph and 5 second figures are arbitrary and I was just trying to demonstrate that the infinitesimal steps assumed by integration, do not actually have to be that small in practice, to get a reasonably accurate result.

 

[starthaus]

While checking out your observation, I noticed another interesting relationship.

The slowest time between the two events is the proper time $t_i$ by a clock moving inertially between the two events, i.e. $t_i = \Delta t^2 - \Delta x^2/c^2$

You must mean

$$d \tau=\sqrt{ \Delta t^2 - \Delta x^2/c^2}$$

Now it seems from playing around with my spreadsheet, that the following (without proof) is true:

$$\Delta x = \frac{1}{2} \alpha t_i^2$$

This cannot be since:

$$x=\frac{c^2}{a}(cosh(a \tau/c)-1)$$

$$dx=c d\tau sinh(a \tau/c)$$

which has the same form as the Newtonian equation.

Maybe as an approximation but not in reality (see above)

It obviously follows that the constant proper acceleration can be obtained from this simple equation:

$$\alpha = 2\frac{\Delta x}{t_i^2}$$

No. See above.

 

[yuiop]

Coordinate distance Δx traveled in frame S

$$\Delta x = (c^2/a)*(\sqrt{(1+(a \Delta t /c)^2)} -1) = (c^2/a)*(\gamma-1) = (1/2)*(20.025-1) = 9.5125$$

I never actually derived the equation for coordinate distance above, so here it is.

The relativistic equation for kinetic energy is:

$$KE = \sqrt{(mc^2)^2 + (mcv\gamma)^2} - mc^2$$

This is the total energy that has been added to the particle that is accelerated from rest to v.

The work done on the particle is:

$$W = F\Delta x = ma\gamma^3 \Delta x = ma_0 \Delta x$$

Since the work is an expression for energy and is equal to the final kinetic energy of the accelerated particle (in a 100% efficient accelerator) then the two above equations can be equated as:

$$ma_0 \Delta x = \sqrt{(mc^2)^2 + (mcv\gamma)^2} - mc^2$$


$$\rightarrow \Delta x = \frac{ \sqrt{(c^2)^2 + (cv\gamma)^2} - c^2}{a_0}$$

$$\rightarrow \Delta x = (c^2/a_0) ( \sqrt{1 + (v\gamma/c)^2} - 1)$$

In an earlier post it was shown that $v\gamma = a_0\Delta t$ and this is substituted into the above equation to obtain:

$$\Delta x = (c^2/a_0) ( \sqrt{1 + (a_\Delta t/c)^2} - 1)}$$

There are probably easier ways to derive it, but I thought it would be nice to introduce the relationships between force, energy and acceleration.

 

[yuiop]

You must mean

$$t_i =\sqrt{ \Delta t^2 - \Delta x^2/c^2}$$

Yes, that is what I meant. (Fixed the original.) Thanks

This cannot be since:

$$x=\frac{c^2}{a}(cosh(a \tau/c)-1)$$

$$dx=c d\tau sinh(a \tau/c)$$

The tau in the equations you have quoted is the proper time of the accelerating object moving between the two events. This is not the same as the proper time $t_i$ that I was using, which is the proper time of an inertially moving object (with constant velocity) between the same two events. To try and avoid confusion, I did not use tau, because we are already using that for the proper time of the accelerating object.

As Passionflower pointed out, the elapsed proper time measured by the accelerating clock moving between the two events is about 40% of the elapsed proper time measured by an inertial clock that is present at the same two events.

Now it seems from playing around with my spreadsheet, that the following (without proof) is true:

$$\Delta x = \frac{1}{2} \alpha t_i^2$$

which has the same form as the Newtonian equation.

Maybe as an approximation but not in reality (see above)

No, it is intended to be exact, when the correct time measurement is used.

 

[starthaus]

Yes, that is what I meant. (Fixed the original.) Thanks

Well, your $$t_i$$ as defined is the proper time of the accelerated observer.

The tau in the equations you have quoted is the proper time of the accelerating object moving between the two events. This is not the same as the proper time $t_i$ that I was using,

The mathematical definition that you are using shows them to be one and the same.

No, it is intended to be exact, when the correct time measurement is used.

Try to do a derivation and let's see what you get.

 

[yuiop]

Well, your $$t_i$$ as defined is the proper time of the accelerated observer.

I defined $$\Delta t_i$$ as $$\sqrt{\Delta t^2- \Delta x^2/c^2}$$ in post #38:

This is directly obtained from the Minkowski metric and it is the elapsed proper time of a clock moving with constant velocity that is present at the two events, (0,0) and (x,t).

This is clearly not the same as the proper elapsed time $$\Delta \tau$$ or $$\Delta t_0$$ of a clock with constant acceleration, that is present at the same two events. The elapsed proper time of the accelerating clock is defined by:

$$\Delta \tau = \Delta t_0 = \frac{c}{a_0}\, arsinh ( \frac{a_0 \Delta t}{c})\, = \, \frac{c}{a_0} \, arcosh({\frac{a_0 \Delta x}{c^2}+1)$$

It is easy to see that $$\Delta t_i \ne \Delta \tau$$ because:$$\sqrt{\Delta t^2- \Delta x^2/c^2} \qquad \ne \qquad \frac{c}{a_0}\, arsinh ( \frac{a_0 \Delta t}{c})$$
and

$$\sqrt{\Delta t^2- \Delta x^2/c^2}\qquad \ne \qquad \frac{c}{a_0} \, arcosh({\frac{a_0 \Delta x}{c^2}+1)$$

The tau in the equations you have quoted is the proper time of the accelerating object moving between the two events. This is not the same as the proper time $t_i$ that I was using, which is the proper time of an inertially moving object (with constant velocity) between the same two events.

The mathematical definition that you are using shows them to be one and the same.

I am afraid you are seriously mistaken about the maths and physics of this issue.

 

[starthaus]

I clearly defined $$t_i$$ as $$\sqrt{\Delta t^2- \Delta x^2/c^2}$$ in post #38

Yes. And this is not a very good definition to begin with because the RHS is a differential while the LHS is not.

This is directly obtained from the Minkowski metric and it is the elapsed proper time of a clock moving with constant velocity that is present at the two events, (0,0) and (x,t).

The standard interpretationfollows correctly from the INVARIANCE of the Minkowski metric as follows:

$$(c d \tau)^2=(cdt)^2-dx^2$$

I am sure I have shown you this before.

This is clearly not the same as the proper elapsed time $$\tau$$ or $$t_0$$ of a clock with constant acceleration, that is present at the same two events. The elapsed proper time of the accelerating clock is defined by:

$$\tau = t_0 = \frac{c}{a_0}\, arcosh ( \frac{a_0 t}{c})\,$$

Yes, I just showed you this in the previous post.

$$=\frac{c}{a_0} \, arsinh{\frac{a_0 \Delta x}{c^2})$$

Definitely wrong since you are effectively writing $$t=\frac{\Delta x}{c}$$

It is easy to see that $$t_i \ne \tau$$ because:$$\sqrt{\Delta t^2- \Delta x^2/c^2} \qquad \ne \qquad \frac{c}{a_0} arcosh ( \frac{a_0 t}{c})$$

Sure, if you use wrong derivations, it is easy to show anything. Do you understand the difference between $$\tau$$ and $$d \tau$$? One is a variable, the other one is a differential, you are mixing them freely and incorrectly.
Speaking of derivations, where is the derivation for your claim that started this debate?

 

[stevmg]

I am going to introduce some new notation because of the complication that arise when considering the measurements according to the inertial observer with velocity relative to the accelerating observer, the accelerating observer and the co moving inertial observer.

Quantities that are proper measurements made by the accelerating observer will be denoted by zero subscript such as m₀, t₀ and a₀.

Quantities measured by the inertial observer at rest in frame S with velocity relative to the accelerating observer do not have a subscript or a superscript, e.g. m, t and a.

Quantities measured in the Co-Moving Inertial Reference Frame (CMIIRF or S') will be denoted by a prime symbol, eg m', t' and a'.

m = m' = m₀

a₀ = a'

In earlier posts we have established with help from DrGreg that a = dv/dt = a'/γ³.

We have also established that

F = dp/dt = d(mvγ)/dt' = m d(vγ)/dt = may³ = ma'

and it is also true that:

F = may³ = m(dv/dt)y³ = m(a'/γ³)γ³ = ma'

From the above we can conclude that F = F' = F₀ because in the co-moving frame where the v=0, F' = ma'γ³ = ma'*(sqrt(1-v²))³ = ma'*(sqrt(1-0²))³ = ma'.

Unfortunately the values you have chosen for acceleration and the time period, means that relativistic effects are extremely small and barely distinguishable from Newtonian calculations.

Staying with the car metaphor, let's fit a performance exhaust, go-faster-stripes and nitro injection and boost the acceleration up to 2c per second and use units of c=1. Yes, surprisingly 2c per second is allowed, because 2c is the hypothetical terminal velocity that reached if it was possible to maintain a constant coordinate acceleration for a full second, which is of course impossible, but that sort of acceleration is in principle possible for an infinitesimal time period.

Problem statement:

Proper acceleration = 2c /s.

The x axes of S and S' are parallel to each other and the relative velocity of the frames and the acceleration and the velocity of the accelerated object is parallel to the x axes.

The velocity of accelerating object is v=0 at time t=0 and the object accelerates for 10 seconds as measured in frame S. We will consider two events, one at the start and at one at the end of the acceleration period.

Event 1:
(x1,t1) = (0,0)
(x1',t2') = (0,0)

Event 2:
(x2,t2) = (Δx,10)
(x2',t2') = (Δx', Δt')

$$\Delta x = (x_2-x_1) = x_2 \quad , \quad \Delta t = (t_2-t_1) = t_2$$
$$\Delta x' = (x_2'-x_1') = x_2' \quad , \quad \Delta t' = (t_2'-t_1') = t_2'$$

v is the final velocity of the accelerating object in frame S and is also the relative velocity of frame S' to frame S.

The instantaneous gamma factor at the terminal velocity in frame S

$$\gamma = \sqrt{1+(a't/c)^2} = \frac{1}{\sqrt{1-v^2/c^2}} = 20.025$$

Final coordinate velocity in S

$$v = \frac{a't}{\sqrt{1+(a't/c)^2}} = a't/\gamma = \frac{2*10}{20.025} = 0.99875c$$

Final coordinate acceleration in S

The initial coordinate acceleration is equal to the proper acceleration, but while the proper acceleration remains constant the coordinate acceleration does not and the final coordinate acceleration is:

$$a = a_0/\gamma^3 = 2/20.025^3 = 0.00025c/s$$

Final coordinate velocity in the CMIRF (S')

This is zero by definition. Note that the initial velocity in S' was -0.99875c.

Coordinate distance Δx traveled in frame S

$$\Delta x = (c^2/a)*(\sqrt{(1+(a \Delta t /c)^2)} -1) = (c^2/a)*(\gamma-1) = (1/2)*(20.025-1) = 9.5125$$

Coordinate distance Δx' in the CMIRF (S')

Fram the Lorentz transformation:

$$\Delta x' = \frac{\Delta x-v \Delta t}{\sqrt{1-v^2/c^2}} = \gamma(\Delta x-v \Delta t) = 20.025*(9.5125-0.99875*10) = -9.5125$$

You might find it surprising and unintuitive that the distance between the two events is the same in frame S and frame S', except for the sign. I know I did and maybe I made a mistake somewhere.

Proper distance:

Distance is poorly defined in an accelerating reference frame. Proper distance in inertial RFs is normally the distance measured by a ruler between two simultaneous events. Since the two events are not simultaneous in any inertial reference frame in this example it is hard to define a proper distance even in the inertial reference frames. We can however invoke a notion of the invariant interval which can be thought of as the proper distance between events 1 and 2.

$$(c \Delta t)^2 - \Delta x^2 = (10)^2 - 9.5125^2 = 9.5125$$

This is invariant for any inertial reference frame (moving parallel to the x axis) but I am not sure if it applies to accelerating frames.

Coordinate elapsed time Δt' in the CMIRF

Using the Lorentz transformation:

$$\Delta t' = \frac{ \Delta t-v \Delta x/c^2}{\sqrt{1-v^2/c^2}} = \gamma(\Delta t-v\Delta x/c^2) = 20.025*(10-0.99875*9.5125) = 10 s$$

Again this is a counter-intuitive result.

Proper elapsed time:

The proper elapsed time for the accelerating object is clearly defined because it measured by a single clock between the two events. It is derived like this. The total elapsed proper time is the integral of the instantaneous proper time at any instant which is a function of the instantaneous velocity u at any instant, so:

$$\frac{dt_0}{dt} = \frac{1}{\gamma^2} = \sqrt{1-u^2/c^2} = \frac{1}{\sqrt{1+(a_0\Delta t/c)^2}}$$

Integrating both sides with respect to t:

$$\Delta t_0 = \int \left( \frac{1}{\sqrt{1+(a_0t/c)^2}} \right) dt = (c/a_0)\, arsinh(a_0 \Delta t/c)$$

Now I know you wanted no hyperbolic functions, but it is almost unavoidable in the proper time calculation. However, there is an alternative in the form of the natural log Ln which is:

$$t_0 = (c/a)\, arsinh(a_0t/c) = (c/a_0) Ln((at/c)+\gamma) = 1/2*Ln( 20+ 20.025) = 1.845 s$$

Hyperbolic curves have the form x^2-y^2 = Constant and the Minkowski metric has the same form (dx/dt_0)^2 - (cdt/dt_0)^2 = c^2, so hyperbolic functions will keep popping up.

Some hyperbolic functions can only be expressed in terms of exponential functions and logarithms, but these are transcendental functions too and cannot be expressed in simple algebraic terms.

Proper velocity in S

I will leave this for another post as I am still thinking about it and this post is long enough.

This long post is me gathering my thoughts on the topic and may contain typos/ mistakes/ misunderstandings/ misconceptions so any corrections are welcome.

Having explained your derivations of the equations used, my original question: For a given mass m at rest, what would be the final velocity if a constant force (in the resting FR or S) were to be applied to it, has been answered as well as the general approach to delineating this.

Any texts that cover this as you contributors did? I haven't seen any. Maybe those of Wolfgang Rindler, which starthaus uses, but I don't know which one as he has written several. I am cheap and I don't need the latest text so a cheaper earlier edition such as the first edition of Taylor/Wheeler Spacetime Physics would be ideal.

When I started the original thread, I thought this would be a "wham-bam" quickie amswer but this has resulted in a thread of some 40+ posts with thoughtful responses in these posts (other than mine which were still at the novice level.)

I will start a new thread on hyperbolic functions in relation to the Lorentz transformations. starthaus and DrGreg have used the hyperbolic functions numerous times in proving their points and I must get a working knowledge of them, at least to the level you have gone with times, distances and velocities in accelerated frames.

For starters, given x, t, in S and x', t' in S' then:

x² - t² = x'² - t'² (if spacelike)
t² - x² = t'² - x'² (if timelike)

what do you normally use - which hyperbola?

 

[stevmg]

Yes. And this is not a very good definition to begin with because the RHS is a differential while the LHS is not.

I take it...

"RHS" = "Right Hand Side"
"LHS" = "Left Hand Side"

The standard interpretation follows correctly from the INVARIANCE of the Minkowski metric as follows:

$$(c d \tau)^2=(cdt)^2-dx^2$$

I am sure I have shown this before.

I love this one as it leads into my next thread on hyperbolic functions and the Lorentz transformations or Minkowski metric.

 

[stevmg]

While checking out your observation, I noticed another interesting relationship.

The slowest time between the two events is the elapsed proper time interval $t_i$ measured by a clock moving inertially between the two events, i.e. $$t_i = \sqrt{\Delta t^2 - \Delta x^2/c^2}$$
Now it seems from playing around with my spreadsheet, that the following (without proof) is true:

$$\Delta x = \frac{1}{2} \alpha t_i^2$$

which has the same form as the Newtonian equation.

It obviously follows that the constant proper acceleration can be obtained from this simple equation:

$$\alpha = 2\frac{\Delta x}{t_i^2}$$

P.S. That should be seconds, not years, to be consistent with the original scenario

To starthaus - isn't this claim a basic tenet of SR - that (in the timelike area) for inertially moving objects, $\tau$ = SQRT[(delta x)² - (delta t)²]? I am sure that I have the wrong claim that you speak of.

 

[starthaus]

To starthaus - isn't this claim a basic tenet of SR - that (in the timelike area) for inertially moving objects, $\tau$ = SQRT[(delta x)² - (delta t)²]? I am sure that I have the wrong claim that you speak of.

The correct derivation starts with:

$$(c d \tau)^2=(cdt)^2-dx^2$$

It is important that you understand the differential form. The LHS and RHS are BOTH total differentials.

 

[yuiop]

I clearly defined $$t_i$$ as $$\sqrt{\Delta t^2- \Delta x^2/c^2}$$ in post #38: [/tex]

Yes. And this is not a very good definition to begin with because the RHS is a differential while the LHS is not.

Yes, I am guilty of sometimes using just x to mean $\Delta x$ and vice versa, but you do know what I mean from the context. In post 40 you said:

You must mean

$$d \tau=\sqrt{ \Delta t^2 - \Delta x^2/c^2}$$

You are guilty of using a similar mixing of symbols here. $d \tau$ and $\Delta \tau$ do not necessarily mean the same thing. The equation should be either:

$$d \tau=\sqrt{ dt^2 - dx^2/c^2}$$

or:

$$\Delta \tau=\sqrt{ \Delta t^2 - \Delta x^2/c^2}$$

I hope you understand that for example, $$\frac{\Delta x}{\Delta t}$$ is not always the same as $$\frac{dx}{dt}$$

In the worked acceleration example given in post we had $${\Delta x} = 9.5125$$ and $${\Delta t} = 10$$ so the average velocity of the accelerating object between the two events:

$$\frac{\Delta x}{\Delta t} = 0.95125c$$

while the final instantaneous coordinate velocity was found to be:

$$\frac{dx}{dt} = 0.99875c$$

Do you see they are not the same thing? One is the average velocity over a non-infinitesimal period while the other is an instantaneous velocity over an infinitesimal period. The distinction is important in an accelerating context.

The standard interpretation follows correctly from the INVARIANCE of the Minkowski metric as follows:

$$(c d \tau)^2=(cdt)^2-dx^2$$

I am sure I have shown you this before.

Please stop saying insulting things like this trying to put me down. That is against the rules of PF. I had used that equation hundreds of time in this forum before you even joined it. You did not "show" me this.

Definitely wrong since you are effectively writing $$t=\frac{\Delta x}{c}$$

Yes, I misquoted some formulas from http://www.xs4all.nl/~johanw/PhysFAQ/Relativity/SR/rocket.html and I have now corrected them in the original post (#38). You are nit-picking about typos and minutia and completely missing the whole point and substance of the physics, which is that the elapsed time on a clock accelerating between two events is not the same as the elapsed time of a clock moving inertially between the same two events. Do you now agree that the after editing, that post #38 is now correct and that you are wrong about the substance and physics of the issue?

Sure, if you use wrong derivations, it is easy to show anything.

I did not do any derivations in post #38. I was just quoting the standard Minkowski metric and the standard relativistic rocket equations from http://www.xs4all.nl/~johanw/PhysFAQ/Relativity/SR/rocket.html
Admittedly I made several typos in the transcription of the equations, which I hope are now fixed in the original post.

 

[starthaus]

I hope you understand that for example, $$\frac{\Delta x}{\Delta t}$$ is not always the same as $$\frac{dx}{dt}$$

You can't be serious.

I did not do any derivations in post #38. I

You mean, you guessed

$$\Delta x=\frac{a t_i^2}{2}$$?

For good reason this is not part of the webpage you are citing ("The Relativistic Rocket"), it is total nonsense as I have showed . If you just guessed it without any derivation, try deriving it. This is all I was asking.

 

[yuiop]

I hope you understand that for example, $$\frac{\Delta x}{\Delta t}$$ is not always the same as $$\frac{dx}{dt}$$

You can't be serious.

I am serious. Do you disagree?

 

[Austin0]

Problem statement:

Proper acceleration = 2c /s.

The x axes of S and S' are parallel to each other and the relative velocity of the frames and the acceleration and the velocity of the accelerated object is parallel to the x axes.

The velocity of accelerating object is v=0 at time t=0 and the object accelerates for 10 seconds as measured in frame S. We will consider two events, one at the start and at one at the end of the acceleration period.

Event 1:
(x1,t1) = (0,0)
(x1',t2') = (0,0)

Event 2:
(x2,t2) = (Δx,10)
(x2',t2') = (Δx', Δt')

$$\Delta x = (x_2-x_1) = x_2 \quad , \quad \Delta t = (t_2-t_1) = t_2$$
$$\Delta x' = (x_2'-x_1') = x_2' \quad , \quad \Delta t' = (t_2'-t_1') = t_2'$$

v is the final velocity of the accelerating object in frame S and is also the relative velocity of frame S' to frame S.

The instantaneous gamma factor at the terminal velocity in frame S

$$\gamma = \sqrt{1+(a't/c)^2} = \frac{1}{\sqrt{1-v^2/c^2}} = 20.025$$

Final coordinate velocity in S

$$v = \frac{a't}{\sqrt{1+(a't/c)^2}} = a't/\gamma = \frac{2*10}{20.025} = 0.99875c$$

Coordinate distance Δx traveled in frame S

$$\Delta x = (c^2/a)*(\sqrt{(1+(a \Delta t /c)^2)} -1) = (c^2/a)*(\gamma-1) = (1/2)*(20.025-1) = 9.5125$$

The origen of S' at (t=10) will be colocated with (t=10,x=9.9875) yes?
Considering that the accelerating system started out at v=0 while the S' system was moving at .99875c and the accl. Frame only attains that velocity at the end. how could the accl system end up having traveled almost an equal distance?
After 10 secs S' (.4994,0) will be at S (10,9.9875)
S(0.4994,0) will be located at S' (10,-9.9875) correct?
If this is correct it would seem to follow that event 2 must lie somewhere in the middle between these two events both spatially and temporally.

Coordinate distance Δx' in the CMIRF (S')

Fram the Lorentz transformation:

$$\Delta x' = \frac{\Delta x-v \Delta t}{\sqrt{1-v^2/c^2}} = \gamma(\Delta x-v \Delta t) = 20.025*(9.5125-0.99875*10) = -9.5125$$

You might find it surprising and unintuitive that the distance between the two events is the same in frame S and frame S', except for the sign. I know I did and maybe I made a mistake somewhere. .

A frame traveling 0.9521 relative to both S and S' would end up in the middle.
I.e. v= 0.9512 relative to S ,,,,and v= -0.9512 relative to S' but it could not have traveled 10 secs in either frame.

I may be totally wrong about all this but I think there may be a problem with using the CMIRF as I said last post.

Proper distance:

Distance is poorly defined in an accelerating reference frame. Proper distance in inertial RFs is normally the distance measured by a ruler between two simultaneous events. Since the two events are not simultaneous in any inertial reference frame in this example it is hard to define a proper distance even in the inertial reference frames. We can however invoke a notion of the invariant interval which can be thought of as the proper distance between events 1 and 2.

$$(c \Delta t)^2 - \Delta x^2 = (10)^2 - 9.5125^2 = 9.5125$$ .

Coordinate elapsed time Δt' in the CMIRF

Using the Lorentz transformation:

$$\Delta t' = \frac{ \Delta t-v \Delta x/c^2}{\sqrt{1-v^2/c^2}} = \gamma(\Delta t-v\Delta x/c^2) = 20.025*(10-0.99875*9.5125) = 10 s$$

Again this is a counter-intuitive result. .

Proper elapsed time:

As above ,,I think there may be a problem here

The proper elapsed time for the accelerating object is clearly defined because it measured by a single clock between the two events. It is derived like this. The total elapsed proper time is the integral of the instantaneous proper time at any instant which is a function of the instantaneous velocity u at any instant, so:

$$\frac{dt_0}{dt} = \frac{1}{\gamma^2} = \sqrt{1-u^2/c^2} = \frac{1}{\sqrt{1+(a_0\Delta t/c)^2}}$$

Integrating both sides with respect to t:

$$\Delta t_0 = \int \left( \frac{1}{\sqrt{1+(a_0t/c)^2}} \right) dt = (c/a_0)\, arsinh(a_0 \Delta t/c)$$

.

In this situation the elapsed time in S and S' do not really represent elapsed proper time but time as applied at different locations, so the clock desynchronization accounts for most of the difference between local clocks at event 2 and the accelerated systems elapsed proper time , yes?

 

[stevmg]

With all this discussion going on from my original thread I do NOT feel that I am such a novice after all!

stevmg

 

[yuiop]

Now it seems from playing around with my spreadsheet, that the following (without proof) is true:

$$\Delta x = \frac{1}{2} \alpha t_i^2$$

which has the same form as the Newtonian equation.

It obviously follows that the constant proper acceleration can be obtained from this simple equation:

$$\alpha = 2\frac{\Delta x}{t_i^2}$$

This cannot be since:

$$x=\frac{c^2}{a}(cosh(a \tau/c)-1)$$

$$dx=c d\tau sinh(a \tau/c)$$

You mean, you guessed

$$\Delta x=\frac{a t_i^2}{2}$$?

For good reason this is not part of the webpage you are citing ("The Relativistic Rocket") . If you just guessed it without any derivation, try deriving it. This is all I was asking.

When I first noticed the relationship, it "jumped out" at me when looking at the numbers on a spreadsheet and I did not not have a derivation, so yes in that sense I "guessed it". I now have the derivation as shown below:

Using notation defined thus:

Coordinate distance interval in S = $$\Delta x$$ = d
Coordinate time interval in S = $$\Delta t$$ = t
Proper acceleration = $$\alpha$$

.. the equation for coordinate distance from http://www.xs4all.nl/~johanw/PhysFAQ/Relativity/SR/rocket.html" is given as:

$$d = (c^2/\alpha)*(\sqrt{(1+(\alpha t /c)^2)} -1)$$

$$\rightarrow \alpha d = \sqrt{c^4+(\alpha t c)^2} - c^2$$

$$\rightarrow (\alpha d +c^2)^2 = c^4+(\alpha t c)^2$$

$$\rightarrow \alpha^2d^2 + c^4 + 2c^2\alpha d = c^4+\alpha^2 t^2 c^2$$

$$\rightarrow \alpha d^2 + 2c^2 d = \alpha t^2 c^2$$

$$\rightarrow 2c^2d =\alpha (t^2 c^2 - d^2)$$

$$\rightarrow \alpha = \frac{2d}{(t^2 - d^2/c^2)}$$


Since I defined [itex]t_i[/tex] as $$\sqrt{(t^2 - d^2/c^2)$$ the following statents are true:

$$\alpha = \frac{2d}{t_i^2}$$

$$t_i = \sqrt{\frac{2d}{\alpha}}$$

$$d = \frac{1}{2} \, \alpha t_i^2$$

It is also an exact solution because I derived it directly from from an exact formula.

Q.E.D.

 

[yuiop]

With all this discussion going on from my original thread I do NOT feel that I am such a novice after all!

stevmg

Well, acceleration in relativity is not a novice subject, but it is fairly straightforward.

 

[stevmg]

When I first noticed the relationship, it "jumped out" at me when looking at the numbers on a spreadsheet and I did not not have a derivation, so yes in that sense I "guessed it". I now have the derivation as shown below:

Using notation defined thus:

Coordinate distance interval in S = $$\Delta x$$ = d
Coordinate time interval in S = $$\Delta t$$ = t
Proper acceleration = $$\alpha$$

.. the equation for coordinate distance from http://www.xs4all.nl/~johanw/PhysFAQ/Relativity/SR/rocket.html" is given as:

$$d = (c^2/\alpha)*(\sqrt{(1+(\alpha t /c)^2)} -1)$$

$$\rightarrow \alpha d = \sqrt{c^4+(\alpha t c)^2} - c^2$$

$$\rightarrow (\alpha d +c^2)^2 = c^4+(\alpha t c)^2$$

$$\rightarrow \alpha^2d^2 + c^4 + 2c^2\alpha d = c^4+\alpha^2 t^2 c^2$$

$$\rightarrow \alpha d^2 + 2c^2 d = \alpha t^2 c^2$$

$$\rightarrow 2c^2d =\alpha (t^2 c^2 - d^2)$$

$$\rightarrow \alpha = \frac{2d}{(t^2 - d^2/c^2)}$$


Since I defined [itex]t_i[/tex] as $$\sqrt{(t^2 - d^2/c^2)$$ the following statents are true:

$$\alpha = \frac{2d}{t_i^2}$$

$$t_i = \sqrt{\frac{2d}{\alpha}}$$

$$d = \frac{1}{2} \, \alpha t_i^2$$

It is also an exact solution because I derived it directly from from an exact formula.

Q.E.D.

Beautiful!

How did you get the "quote within a quote" as you did with this (your quoted post was quoted within starthaus's quote.)

 

[yuiop]

Beautiful!

How did you get the "quote within a quote" as you did with this (your quoted post was quoted within starthaus's quote.)

You can do it like this:

[quote="starthaus, post: 2873083"]

[quote="yuiop, post: 2873062"]
I hope you understand that for example,
  [tex]\frac{\Delta x}{\Delta t}[/tex] 
is not always the same as
 [tex]\frac{dx}{dt}[/tex] 
[/QUOTE]

You can't be serious.
[/QUOTE]

To get:

I hope you understand that for example, $$\frac{\Delta x}{\Delta t}$$
is not always the same as $$\frac{dx}{dt}$$

You can't be serious.

or:

[quote="yuiop, post: 2873062"]
I hope you understand that for example,
  [tex]\\frac{\Delta x}{\Delta t}[/tex] 
is not always the same as
 [tex]\frac{dx}{dt}[/tex] 

[quote="starthaus, post: 2873083"]
You can't be serious.
[/QUOTE]
[/QUOTE]

to get:

I hope you understand that for example, $$\frac{\Delta x}{\Delta t}$$
is not always the same as $$\frac{dx}{dt}$$

You can't be serious.

I wrapped it in PHP tags so that you can actually see the quote tags. Just use the multi quote feature and copy and paste to nest the quotes inside each other.

 

[stevmg]

You can do it like this:

To get:



or:

to get:





I wrapped it in PHP tags so that you can actually see the quote tags. Just use the multi quote feature and copy and paste to nest the quotes inside each other.

Thanks...

stevmg

 

[yuiop]

The origen of S' at (t=10) will be colocated with (t=10,x=9.9875) yes?

Yep, that makes sense if frame S' is traveling at 0.99875c and if the origins of S and S' were initially collocated.

Considering that the accelerating system started out at v=0 while the S' system was moving at .99875c and the accl. Frame only attains that velocity at the end. how could the accl system end up having traveled almost an equal distance?

The acceleration is extreme. In post #28 , I calculated that the accelerating object gets to nearly 0.8944c in the first second.

The accelerating object travels a distance of 9.5125 in 10 seconds so its average velocity in S is 0.95125c so it is not that much slower than the 0.99875c velocity of the CMIRF.

After 10 secs S' (.4994,0) will be at S (10,9.9875)
S(0.4994,0) will be located at S' (10,-9.9875) correct?
If this is correct it would seem to follow that event 2 must lie somewhere in the middle between these two events both spatially and temporally.

(t,x) = (10,9.5125) is event 2 in frame S and (t',x') = (10,-9.5125) is the same event 2 in frame S'. Where is this event 2 that lies somewhere between those events that you speak of?

You can't simply apply the gamma factor to the time in S to get the time in S' as 0.4994 because the initial vent and the final event are not at the same place in either reference frame. It is basically a simultaneity issue.

A frame traveling 0.9521 relative to both S and S' would end up in the middle.
I.e. v= 0.9512 relative to S ,,,,and v= -0.9512 relative to S' but it could not have traveled 10 secs in either frame.

The figure you have calculated for a reference frame that sees the origins of S and S' going away at equal speeds in opposite directions appears to be correct. This new frame would not appear to be in the middle in either frame S or S'. I am not sure why you think that is significant.

I may be totally wrong about all this but I think there may be a problem with using the CMIRF as I said last post.

All I can say is that using the CMIRF concept is a perfectly standard method in textbooks. Well at least I think it would be if I had any textbooks (it appears to be perfectly standard in serious online references anyway.)

In this situation the elapsed time in S and S' do not really represent elapsed proper time but time as applied at different locations, so the clock desynchronization accounts for most of the difference between local clocks at event 2 and the accelerated systems elapsed proper time , yes?

Correct again. That is why I have always referred to the elapsed times in S and S' as coordinate times rather than as proper times.

 

[starthaus]

When I first noticed the relationship, it "jumped out" at me when looking at the numbers on a spreadsheet and I did not not have a derivation, so yes in that sense I "guessed it".

Ok, I thought so.

$$\Delta x$$ = d
$$\Delta t$$ = t

Interesting choice of notation, so, for you:

$$\Delta x = x(t)-x(0)=x(t)$$

.. the equation for coordinate distance from http://www.xs4all.nl/~johanw/PhysFAQ/Relativity/SR/rocket.html" is given as:

$$d = (c^2/\alpha)*(\sqrt{(1+(\alpha t /c)^2)} -1)$$

$$\rightarrow \alpha d = \sqrt{c^4+(\alpha t c)^2} - c^2$$

$$\rightarrow (\alpha d +c^2)^2 = c^4+(\alpha t c)^2$$

$$\rightarrow \alpha^2d^2 + c^4 + 2c^2\alpha d = c^4+\alpha^2 t^2 c^2$$

$$\rightarrow \alpha d^2 + 2c^2 d = \alpha t^2 c^2$$

$$\rightarrow 2c^2d =\alpha (t^2 c^2 - d^2)$$

$$\rightarrow \alpha = \frac{2d}{(t^2 - d^2/c^2)}$$

So the above is nothing but

$$d = (c^2/\alpha)*(\sqrt{(1+(\alpha t /c)^2)} -1)$$

rearranged to express $$\alpha$$ as a function of $$d$$ with the notation $$t_i=\sqrt{(t^2 - d^2/c^2)$$ . There is no physics in this, just an elementary algebraic manipulation.

 

[yuiop]

Coordinate distance interval in S = $$\Delta x$$ = d
Coordinate time interval in S = $$\Delta t$$ = t

Interesting choice of notation, so, for you:

$$\Delta x = x(t)-x(0)=x(t)$$

Yes it is a form of shorthand that is often used if the initial event is (0,0).
If event 1 is (0,0) and event 2 is (x,t) then $$\Delta x = x-0 = x$$ and $$\Delta t = t-0 = t$$. Perfectly valid.

So the above is nothing but

$$d = (c^2/\alpha)*(\sqrt{(1+(\alpha t /c)^2)} -1)$$

rearranged to express $$\alpha$$ as a function of $$d$$ with the notation $$t_i=\sqrt{(t^2 - d^2/c^2)$$ . There is no physics in this, just an elementary algebraic manipulation.

Yep, elementary and yet you could not see it and kept insisting it was wrong, until I showed you the proof. In the end all physics equations are an algebraic manipulation of something else.

The interesting physics aspect is that the proper time of an accelerating clock moving between two events is not the same as proper time of a clock with inertial motion between the same two events. Something you did not seem to realize, judging from your earlier postings. You seemed to think that the invariant proper time between two events, applied to accelerating clocks too.

 

[starthaus]

Yes it is a form of shorthand that is often used if the initial event is (0,0).
If event 1 is (0,0) and event 2 is (x,t) then $$\Delta x = x-0 = x$$ and $$\Delta t = t-0 = t$$. Perfectly valid.

I understand now why you believe that $$dx$$ and $$\Delta x$$ are two different things.

Yep, elementary and yet you could not see it and kept insisting it was wrong, .

You are right, I couldn't see any physics , just numerology.

 

[Passionflower]

The interesting physics aspect is that the proper time of an accelerating clock moving between two events is not the same as proper time of a clock with inertial motion between the same two events.

And while we are at it, a traveler could have a special clock linked to an accelerometer that is calibrated at:

$$d \tau_{twin} = \frac{1}{2} \frac{\alpha}{c} \, \sinh \left(\frac{1}{2} \eta\right)$$

This clock would at each instant record the elapsed time of a hypothetical twin traveling between the same events but reaching him on a geodesic.

While another clock that shows the coordinate time has to be calibrated at:

$$d \tau_{coordinate} = \frac{c}{\alpha} \, \sinh \left(\eta)$$With Rapidity: $$\eta = \frac {\alpha d \tau}{c}$$

A more interesting question would be what would the above two formulas be for a positive or negative jerk (no slight intended). Anyone willing to take a stab at that one?

 

[Austin0]

The origen of S' at (t=10) will be colocated with (t=10,x=9.9875) yes?
Considering that the accelerating system started out at v=0 while the S' system was moving at .99875c and the accl. Frame only attains that velocity at the end. how could the accl system end up having traveled almost an equal distance?

Yep, that makes sense if frame S' is traveling at 0.99875c and if the origins of S and S' were initially collocated.
The acceleration is extreme. In post #28 , I calculated that the accelerating object gets to nearly 0.8944c in the first second.

The accelerating object travels a distance of 9.5125 in 10 seconds so its average velocity in S is 0.95125c so it is not that much slower than the 0.99875c velocity of the CMIRF.

The difference in total distance traveled is 0.4750 = [9.9875- 9.8512 ]
If the accl F reaches v=0.8944 in 1 sec
S' has traveld 0.99875 after 1 sec while acclF has only moved a percentage of ,8944 somewhat over 50% making a lead for S' that is already a large part of the final difference in dx',,, with a 0.10435 c velocity differential that will be reduced at a cubic falloff rate for acclF for 9 more secs.
It seems hard for the difference at the end to be so slight.

After 10 secs S' (.4994,0) will be at S (10,9.9875)
S(0.4994,0) will be located at S' (10,-9.9875) correct?
If this is correct it would seem to follow that event 2 must lie somewhere in the middle between these two events both spatially and temporally.

(t,x) = (10,9.9875) is event 2 in frame S and (t',x') = (10,-0.9875) is the same event 2 in frame S'. Where is this event 2 that lies somewhere between those events that you speak of?

The events I noted above are not two views of your event 2 at all.

They are event #3
Colocation S(10, 9.9875), S'(0.4994, 0)

And event #4
Colocation S'( 10, -9.9875) , S(0.4994, 0)

Unless I am mistaken your event #2 is:
The colocation of acclF at reaching v=0.99875 with S( 10, 9.512),S'( 10, -9.512)
Is this not correct?
DO you disagree that this event #2 lies between events 3 and 4 in both frames??

You can't simply apply the gamma factor to the time in S to get the time in S' as 0.4994 because the initial vent and the final event are not at the same place in either reference frame. It is basically a simultaneity issue.

Colocation S(10, 9.9875), S'(0.4994, 0),,,,OK you are telling me I am wrong here
,so please tell me what you think is the correct time for S' x'=0 cojacent with S t=10 ,x=9.9875?
ANd likewise for S t=0.4994 colocated with S' t'=10, x'= -9.9875.

A frame traveling 0.9521 relative to both S and S' would end up in the middle.
I.e. v= 0.9512 relative to S ,,,,and v= -0.9512 relative to S' but it could not have traveled 10 secs in either frame.

The figure you have calculated for a reference frame that sees the origins of S and S' going away at equal speeds in opposite directions appears to be correct. This new frame would not appear to be in the middle in either frame S or S'. I am not sure why you think that is significant.

You could say I found it an interesting coincidence that this v was exactly the average velocity you have calculated for asslF. And it may be significant.
You have the final distance from the origen and elapsed time the same in both frames for event 2.
Given that the origens clocks both read t,t'=0 and an understanding of simultaneity how do you think it is possible for clocks from both frames, colocated at a later point ,could agree on the proper time?
If your figures are correct and the end point of acceleration would be 9.512 in S because the acceleration was so rapid then this would seem to neccessarily imply an equally rapid deceleration relative to S'
This being the case how then could acclF end up traveling so far in S' i.e. -9.512 if the velocity differential dropped off so radically. From -0.99875 to -0.10435 in the first sec. Yes??
If on the other hand you assume that the deceleration in S' is the inverse of S
I.e. Starting out very slowly with a long term cubic increase in acceleration then that's fine but I think it would open a whole new can o' wormholes physicswise ,no?
Having frame agreement on profile between S and S' would seem problematic at best ,,for one , yeh??.
I suggest you may want to look at a drawing as far as colocating the event 2 spatial points.

Originally Posted by Austin0
it is based on acceleration relative to an abstract CMIRF and then this is transformed into rest frame coordinate acceleration at the end. This may of course be absolutely valid but it seems to me that in this circumstance the CIMRFs are somewhat of a bootstrap construct i.e. accelerating relative to one and then there is automatically another one there to accelerate from , with no direct connection to the observation from the reference frame , of either the acceleration of the actual system or the acceleration of the CMIRF.
The increased velocity is just assumed. The coordinate acceleration in the reference frame would actually have to be based on a series of short interval "instantaneous" velocity measurements , no?

I may be totally wrong about all this but I think there may be a problem with using the CMIRF as I said last post.

All I can say is that using the CMIRF concept is a perfectly standard method in textbooks. Well at least I think it would be if I had any textbooks (it appears to be perfectly standard in serious online references anyway.)

I don't doubt that and I am equally bereft of books, but what I am talking about is not a quantitative one , not about having infinitesimal measurements etc. Sporadic measurements would be fine. Its the fact that CMUFOs are not a matter of measurement in an inertial frame whatever. They are an ad hoc abstract creation without history or physics , simply a handy tool for the a priori assumtion of constant proper acceleration. It seems to me possible that they have the same problem to be found with accelerated lines of simultaneity which are simply another mainifestation of CMIRFs
I.e. they may have a questionable relation to the real world and its physics but that just MHO

SO at best they appear to be a superfluous addition ,easily eliminated simply by applying differential calculus directly to the accelerated frame or is there something I am missing?

In this situation the elapsed time in S and S' do not really represent elapsed proper time but time as applied at different locations, so the clock desynchronization accounts for most of the difference between local clocks at event 2 and the accelerated systems elapsed proper time , yes?

Correct again. That is why I have always referred to the elapsed times in S and S' as coordinate times rather than as proper times.

OK ,,,,,well then what is the completely inertial clock that shows greater elapsed proper time ?

 

[stevmg]

starthaus and yuiop:

There IS a difference between a "WAG" (wild a-- guess) and a "SWAG" (Scientific wild a-- guess.) This was an example of a SWAG.

SWAGs have more credibility than WAGs.

If we keep this straight, we're in business.

Doc
(stevmg)

 

[yuiop]

I understand now why you believe that $$dx$$ and $$\Delta x$$ are two different things.

Do you agree with the following statements:

When describing the motion of a particle with constant velocity $$dx = \Delta x$$ .

When describing the motion of an accelerating particle $$dx \ne \Delta x$$ .

I stated that $$dx$$ and $$\Delta x$$ are not necessarily the same thing, which is consistent with the above two statements.

You are right, I couldn't see any physics , just numerology.

This is just offensive You asked for a proof and I provided it for you and yet you still call it numerology.

When Kepler came up with his laws of orbital motion he analysed the data and discerned some patterns and came up with relationships that are now called his laws. Even though Kepler did not know the physics behind his laws (Newton did that), Kepler contribution is not thought of as insignificant and not many people would be so uncharitable as to describe Kepler as just a numerologist. I noticed a relationship and eventually provided a proof as well, so calling numerology is uncharitable. Analyzing data and discerning patterns is part of the scientific process.

In this thread I have provided a derivation for relativistic force and acceleration, a derivation for distance traveled by an accelerating object based on the relationship between force acceleration, momentum and energy and finally the equation for acceleration from the distance equation. I have also provided derivations for the terminal velocity achieved by an accelerating object and the proper time that elapses for the accelerating object. Saying that all you see is "numerology" (when you can not find any technical faults in my derivations) is just designed to be confrontational and against the rules of this forum. I suspect it is just sour grapes on your part, because you probably believed I would not be able to come up with a proof for $$\alpha = \frac{2d}{(t^2 - d^2/c^2)}$$. (I now have a much simpler proof btw)

 

[stevmg]

"Doc" = stevmg = WAG
Starthaus and yiuop = SWAG
Definitive proof (now by yiuop & by starthaus) = true science

Guys, keep it cool! I'm the dummy, not you. If you want to criticize anyone for being unknowledgeable, MAKE IT ME. I can handle it. If that's as bad as I've been called in my life, trust me, I can handle it. When someone calls me an a--hole, my response is, "Is that the best you can do?" If it is worse, I take it as a compliment.

But, now, sports fans, to my new topic, "Derivation of Hyperbolic Equations from the Lorentz transformations." I bet you this gets interesting and heated, too.

Doc
stevmg

Go to this new thread:
https://www.physicsforums.com/showpost.php?p=2875235&postcount=1

 

[starthaus]

Do you agree with the following statements:

When describing the motion of a particle with constant velocity $$dx = \Delta x$$ .

When describing the motion of an accelerating particle $$dx \ne \Delta x$$ .

I stated that $$dx$$ and $$\Delta x$$ are not necessarily the same thing, which is consistent with the above two statements.

You are trying to make new inventions in differential calculus. In mainstream science, both dx and $$\Delta x$$ mean $$x_i-x_{i-1}$$

 

[stevmg]

Go to this new thread:
https://www.physicsforums.com/showpost.php?p=2875235&postcount=1[/QUOTE]

Doc
stevmg