Derivation of proper time in acceleration in SR

[yuiop]

The difference in total distance traveled is 0.4750 = [9.9875- 9.8512 ]

I think that should be 0.4750 = [9.9875- 9.5125]

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'

After 1 sec the accelerating object will have traveled 0.618 units so it has only fallen behind by a distance of 0.99875-0.618 = 0.3807 units. After that the speed differential is very small and getting smaller so it does not lose much more distance and only loses 0.475 units after 10 seconds.

,,, 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.

Seems OK to me.

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??

O.K. you threw me off balance by introducing two new events to an already complicated situation! Yes, your definition of my event #2 is correct for the location and time that the accelerating object reaches it terminal velocity after 10 seconds in S and I have corrected the original post to reflect that. Your Lorentz transformations of the other events are also correct, but I am not sure what you are trying to demonstrate here.

Your event #3 = S(10, 9.9875) = S'(0.4994, 0), is where the CMIRF origin is after 10 seconds in S.

Your event #4 = S'( 10, -9.9875) = S(0.4994, 0) is where the origin of S is after 10 seconds in S'.

I wouldn't say that event 2 is half way between events 3 and 4 in either S or S' spatially or temporally. 0.4994 is approximately 1/20 of the time measured in the other frame. Events 2 and 3 but not 4 are simultaneous in S and events 2 and 4 but not 3 are simultaneous in S'.

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.

I agree.

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?

That is what comes out of the Lorentz transformations and I showed the calculations in post #20 of this thread. I also said it is surprising and unintuitive, but that is what comes out of the maths and it unique to constant acceleration.

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?

The second case is correct (the can of wormholes).

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.

A drawing might help us both. Unfortunately, all the velocities are near 0.9c in the chosen example and it is hard to see anything interesting clearly in the diagram. If you are really interested I can run all the numbers for an acceleration of 0.1c per second which results in a final CMIRF velocity of about 0.7c and a average velocity of about 0.4c for the accelerating object after 10 seconds. The lines on the diagram would be more spread out then.

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?

O.K. I take it you don't like CMIRFs . They are handy (to me anyway), because when you carry out the integration of let's say the instantaneous gamma factor to obtain the total elapsed proper time, it the CMIRF concept that allows you to justify the assumption that there is no additional time dilation due to acceleration per se.

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

It is a clock traveling at a constant velocity equal to the average velocity of the accelerating clock. If this inertial clock starts out at the same time as the accelerating clock from location d1=0 in S they both end up at location d2 = 9.5125 in S at the same time.

 

[Passionflower]

I think that should be 0.4750 = [9.9875- 9.5125]
It is a clock traveling at a constant velocity equal to the average velocity of the accelerating clock. If this inertial clock starts out at the same time as the accelerating clock from location d1=0 in S they both end up at location d2 = 9.5125 in S at the same time.

Indeed, a loop, a twin experiment showing differential aging.

 

[Austin0]

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??

Yes, your definition of my event #2 is correct for the location and time that the accelerating object reaches it terminal velocity after 10 seconds in S
Your Lorentz transformations of the other events are also correct, but I am not sure what you are trying to demonstrate here.
Not trying to demonstrate simply trying provide a context and boundaries to see how these, as you said , surprising results work out.
Your event #3 = S(10, 9.9875) = S'(0.4994, 0), is where the CMIRF origin is after 10 seconds in S.

Your event #4 = S'( 10, -9.9875) = S(0.4994, 0) is where the origin of S is after 10 seconds in S'.

I wouldn't say that event 2 is half way between events 3 and 4 in either S or S' spatially or temporally. 0.4994 is approximately 1/20 of the time measured in the other frame. Events 2 and 3 but not 4 are simultaneous in S and events 2 and 4 but not 3 are simultaneous in S'.

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.

I agree.

If the events are viewed from the frame M traveling 0.9512c wrt each, it turns out that event #2 does occur at the midpoint between S origen and S' origen where each are 9.512 away from that event In that frame they are obviously separated by 19.24 at that time , or so it appears , According to their owns metrics. In M that distance would be reduced by a gamma of 3.2407.
I am still looking at this picture to see if it makes things any less surprising

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?

That is what comes out of the Lorentz transformations and I showed the calculations in post #20 of this thread. I also said it is surprising and unintuitive, but that is what comes out of the maths and it unique to constant acceleration.

Last night I remembered and realized I was mistaken about simultaneity and the clocks agreeing.
IF two frames are synched at the coincidence of the origens then all other locations are out of synch between them with one exception. There is a singularity. A static point equidistant between the origens where the colocated clocks from both frames will continue to agree on proper time as they pass each other.
In this case, that point coincides with the origen of frame M so the passing clocks of S and S' will always have the same time reading there.
In this frame the accelFrame will first be decelerating and then accelerating but reaching +0.9512c at the origen of M at S t=10,S' t'=10

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?

The second case is correct (the can of wormholes).

Well this presents an interesting study. In one frame a rapid increase in velocity that falls off over increasing distance and time is measured in another frame as a slow decrease that augments over a decreasing distance and time ,over the same course of acceleration. Yes?
Given that both measuring frames are inertial and uniform doesn't it seem hard to picture how they could agree on the short interval changes of velocity ?
The reciprocal acceleration profiles this requires

A drawing might help us both. Unfortunately, all the velocities are near 0.9c in the chosen example and it is hard to see anything interesting clearly in the diagram. If you are really interested I can run all the numbers for an acceleration of 0.1c per second which results in a final CMIRF velocity of about 0.7c and a average velocity of about 0.4c for the accelerating object after 10 seconds. The lines on the diagram would be more spread out then.

What would be interesting would be the acceleration profiles for an acceleration to 0.7 in one frame and the reciprocal deceleration in the 0.7 frame.
To be comprehensible it would seem to need spot values for each frame , maybe every 0.05 c and the appropriate locations and times for both frames. I worked on something like that previously but was working pn a wrong profile. I figured a single gamma dropoff.

O.K. I take it you don't like CMIRFs . They are handy (to me anyway), because when you carry out the integration of let's say the instantaneous gamma factor to obtain the total elapsed proper time, it the CMIRF concept that allows you to justify the assumption that there is no additional time dilation due to acceleration per se.

Yep , see a handy concept for justification but no actual use
Isn't the clock hypotheses enough to justify your assumption.
If that assumption is wrong there won't be any CMIRF around to apologize. Besides I thought that question had been fairly well , empirically answered??
It's not really that I don't like the critters as I haven't actually seem enough of em around to have an opinion.

 

[Passionflower]

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?

Oops, I just realized I made a mistake entering the first formula, here is the complete posting but now hopefully correct.
Sorry for that.

With Rapidity: $$\eta = \frac {\alpha \tau}{c}$$

Hypothetical inertial twin:

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

Coordinate time:

$$d \tau_{coordinate} = \frac{c}{a} \sinh(d \eta) = d \tau_{twin} \cosh \left(\frac{1}{2} d\eta\right)$$

 

[Anamitra]

To yuiop:

Consider an object of mass m₀ which, subjected to a constant force, accelerates at a₀ initially. Initially, the velocity of this mass is zero but then picks up as this force is applied.

By the relativistic momentum equation,

a = dv/dt = a₀$\sqrt{(1 - v^2/c^2)}$

dv/$\sqrt{(c^2 - v^2)}$ = a₀dt/c

sin^-1 (v/c) = a₀t/c

v/c = sin (a₀t/c)

This is a periodic function (up and down depending on t) which is impossible

$${a_{0}}$$ has been assumed to be a constant[wrt time] in the process of integration . But this is raising a contradiction in the relation

$${\frac{dv}{dt}{=}{a_{0}}{\sqrt{(1-v^2/c^2)}}$$

If the left side is positive increasing the right hand side should also increase.But velocity should increase due to the acceleration and hence $${\sqrt{(1-v^2/c^2)}}$$decreases. This disturbs the balance of the above equation.
Again if the left side is positive constant the right side decreases due to increase of v.The balance gets disturbed.

Should the left side be negative and increasing in magnitude there is a chance of compatibility since v would decrease on the right side[if $${a_{0}}$$ is a negative constant in this case].But if initial speed is zero then this won't work.Incidentally v=0 at t=0 has been assumed in the calculations.

 

[stevmg]

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

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

starthaus

The above is a hyperbolic relationship between $$(cdt)^2 \text {and } dx^2$$

This is not the same as the hyperbolic relationship between
$$(ct)^2 \text {and } x^2$$
which also does exist.
What is the relationship between the two equations? One is not the derivative of the other. I am trying to figure out what
$$(ct)^2 - x^2 = a^2$$
means. These are a family of hyperbolas with a parameter
$$a$$
but what does each curve represent? What does
$$a$$
represent?

 

[starthaus]

starthaus

The above is a hyperbolic relationship between $$(cdt)^2 \text {and } dx^2$$

This is not the same as the hyperbolic relationship between
$$(ct)^2 \text {and } x^2$$
which also does exist.
What is the relationship between the two equations? One is not the derivative of the other. I am trying to figure out what
$$(ct)^2 - x^2 = a^2$$
means. These are a family of hyperbolas with a parameter
$$a$$
but what does each curve represent? What does
$$a$$
represent?

You can find all the answers https://www.physicsforums.com/blog.php?b=1911 . $$a=\frac{c^2}{a_p}$$ where $$a_p$$ represents the proper acceleration.

 

[stevmg]

Referring to post 77, I have downloaded the second .pdf (Lorentz...& Force in SR) as I had already downloaded the first months ago before I could make any attempt at comprehending it. Thanks for the clarification about $a$.

Now this leads to a "correlative" question about the answer in which $a$ is the proper acceleration in that equation $(ct)^2 - x^2 = a^2$. Your answer would imply that any FR with any $v$ and $a = 0$ would would be on the hyperbola $(ct)^2 - x^2 = 0$ which maps out to be the two diagonal asymptotes $x = \pm (ct)$ for the general equation of the hyperbola $(ct)^2 - x^2 = a^2$.

In other words, just working with the positives, for given $v$, then, if $x$ = $vt$ then $a = 0$.

If this last paragraph is true, I am going to have to "digest" that to understand what that means.

Also, with proper times themselves, in non-accelerating FRs proper time represents the time that two events are apart given no displacement of $x$ over $t$ (i.e, forcing $v = 0$) and using Lorentz to show that $\tau = \Delta t$ which is obvious (used to oblivious to me.)

But given that non-trivial acceleration $\Rightarrow$ displacement of $x$ then I don't know what the hell proper time means in that case.

 

[starthaus]

Referring to post 77, I have downloaded the second .pdf (Lorentz...& Force in SR) as I had already downloaded the first months ago before I could make any attempt at comprehending it. Thanks for the clarification about $a$.

Now this leads to a "correlative" question about the answer in which $a$ is the proper acceleration in that equation $(ct)^2 - x^2 = a^2$. Your answer would imply that any FR with any $v$ and $a = 0$ would would be on the hyperbola $(ct)^2 - x^2 = 0$ which maps out to be the two diagonal asymptotes $x = \pm (ct)$ for the general equation of the hyperbola $(ct)^2 - x^2 = a^2$.

In other words, just working with the positives, for given $v$, then, if $x$ = $vt$ then $a = 0$.

If this last paragraph is true, I am going to have to "digest" that to understand what that means.

Also, with proper times themselves, in non-accelerating FRs proper time represents the time that two events are apart given no displacement of $x$ over $t$ (i.e, forcing $v = 0$) and using Lorentz to show that $\tau = \Delta t$ which is obvious (used to oblivious to me.)

But given that non-trivial acceleration $\Rightarrow$ displacement of $x$ then I don't know what the hell proper time means in that case.

Let's look together at the first file, https://www.physicsforums.com/blog.php?b=1911 . Eq (15) tells you that :

$$x=\frac{c^2}{a_p}cosh\frac{a_p \tau}{c}$$
$$t=\frac{c}{a_p}sinh\frac{a_p \tau}{c}$$

so $$-(ct)^2+x^2=(\frac{c^2}{a_p})^2$$

What does this tell you?

Further, differentiating eq (15):

$$dx=c*sinh\frac{a_p \tau}{c} *d \tau$$
$$dt=cosh \frac{a_p \tau}{c}* d \tau$$

meaning that :

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

What does this mean?

 

[stevmg]

Tried to answer but window locked up too many times. Will try again tomorrow, 9-22.

 

[stevmg]

starthaus -

I have broken your post up into smaller segments and there will be sequential posts by me. This will prevent my computer from locking up. I am assuming that $S$ and S' are constructed such that $x = x' = 0$ and $t = t' = 0$ at the beginning.

$$x=\frac{c^2}{a_p}cosh\frac{a_p \tau}{c}$$
$$t=\frac{c}{a_p}sinh\frac{a_p \tau}{c}$$

so $$-(ct)^2+x^2=(\frac{c^2}{a_p})^2$$

What does this tell you?

These equations (15) according to your article are the equations of hyperbolic motion. This is proven by the last equation in your quote which are in the form of a hyperbola with variables $x \text { and } ct$.

 

[stevmg]

Moving right along,

Further, differentiating eq (15):

$$dx=c*sinh\frac{a_p \tau}{c} *d \tau$$
$$dt=cosh \frac{a_p \tau}{c}* d \tau$$

meaning that :

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

What does this mean?

This establishes a hyperbolic equation on the differentials
$(cdt)$ text [ and ] dx [/iex]

The parameter here is $(c* d \tau)^2$

 

[starthaus]

starthaus -

I have broken your post up into smaller segments and there will be sequential posts by me. This will prevent my computer from locking up. I am assuming that $S$ and S' are constructed such that $x = x' = 0$ and $t = t' = 0$ at the beginning.



These equations (15) according to your article are the equations of hyperbolic motion. This is proven by the last equation in your quote which are in the form of a hyperbola with variables $x \text { and } ct$.

Good. What happens when the proper acceleration $$a_p$$ increases?

 

[starthaus]

Moving right along,
This establishes a hyperbolic equation on the differentials
$(cdt)$ text [ and ] dx [/iex]

The parameter here is $(c* d \tau)^2$

No, this gives you the relationship between proper ($\tau$) and coordinate (t) time:

$$d \tau =\sqrt{1-(\frac{dx}{dt})^2}*dt$$

used in the twins paradox

 

[stevmg]

Good. What happens when the proper acceleration $$a_p$$ increases?

as $a_p$ increases, $(c/a_p)^2$ gets smaller ($\Rightarrow 0$.) There is no upper limit on $a_p$ as there is with $v\Rightarrow c$ provided the $v$ achieved is $<c$.

 

[stevmg]

No, this gives you the relationship between proper ($\tau$) and coordinate (t) time:

$$d \tau =\sqrt{1-(\frac{dx}{dt})^2}*dt$$

used in the twins paradox

Of course. What the hell am I thinking.

The equation $(cdt)^2-dx^2=(c* d \tau)^2$ is not a hyperbola as none of those variables $cdt, dx \text { or } c*d \tau$ is a constant. Where is my damn brain?

Now, I have seen $c \Delta t \text { and/or } \Delta x$ used in the past when your equation above suggests $cdt \text { or } dx$. When is that appropriate?

 

[stevmg]

Since $a_p = c*a$ (Equation 12), then the "constant" side of the hyperbolic equation is $(\frac{c^2}{a_p})^2 \text { or } (\frac{c}{a})^2$.

As $a \Rightarrow \infty$ that constant portion $\Rightarrow 0$

As $a \Rightarrow 0$ that constant portion $\Rightarrow \infty$

That makes no sense.

By the Minkowski equations, $(ct)^2 - x^2 = (\text {some constant})^2$ even at a constant velocity. I am missing something.

 

[starthaus]

Since $a_p = c*a$ (Equation 12), then the "constant" side of the hyperbolic equation is $(\frac{c^2}{a_p})^2 \text { or } (\frac{c}{a})^2$.

As $a \Rightarrow \infty$ that constant portion $\Rightarrow 0$

As $a \Rightarrow 0$ that constant portion $\Rightarrow \infty$

That makes no sense.

By the Minkowski equations, $(ct)^2 - x^2 = (\text {some constant})^2$ even at a constant velocity. I am missing something.

Think again, what happens for $$a_p->\infty$$?

 

[JesseM]

But given that non-trivial acceleration $\Rightarrow$ displacement of $x$ then I don't know what the hell proper time means in that case.

Physically the "proper time" along a given worldline just means the time elapsed on a clock that travels along that worldline. If you want to know the proper time between two events on the worldline of a clock moving at constant velocity along the x-axis of some frame, it's just $$\Delta \tau = \sqrt{\Delta t^2 - (1/c^2) \Delta x^2 }$$, where $$\Delta t$$ is the difference in coordinate time between the two events and $$\Delta x$$ is the difference in coordinate position. This can be rewritten as $$\Delta \tau = \Delta t \sqrt{1 - (1/c^2) (\Delta x / \Delta t )^2 } = \Delta t \sqrt{1 - v^2 /c^2 }$$ which is just the standard time dilation formula. For a worldline where the velocity is varying in a continuous way, you can approximate its worldline by a "polygonal" worldline made up of a series of constant-velocity segments each lasting a time interval of $$\Delta t$$ with instantaneous accelerations between them, then if the velocity during the first time interval v₁ (which could just be the average velocity during the same interval for the path with continuously changing velocity that you are trying to approximate), the second has velocity v₂, and the final segment has velocity v_N, then the total elapsed time would just be the sum of the elapsed time on each segment, or $$\Delta t \sqrt{1 - v_1^2 /c^2} + \Delta t \sqrt{1 - v_2^2 / c^2} + ... + \Delta t \sqrt{1 - v_N^2 /c^2}$$. And in the limit as as the time interval of each segment goes to zero (so the number of segments approaches infinity), this approximation should approach perfect agreement with the proper time on the original path with continuously-changing velocity. Since an integral is just the limiting case of a sum with smaller and smaller intervals (or an infinite series of intervals which each have an 'infinitesimal' time dt), that means the proper time along a worldline where the velocity as a function of time is given by v(t) can always be computed according to the integral $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2 /c^2} \, dt$$

 

[yuiop]

Of course. What the hell am I thinking.

The equation $(cdt)^2-dx^2=(c* d \tau)^2$ is not a hyperbola as none of those variables $cdt, dx \text { or } c*d \tau$ is a constant. Where is my damn brain?

Now, I have seen $c \Delta t \text { and/or } \Delta x$ used in the past when your equation above suggests $cdt \text { or } dx$. When is that appropriate?

You can make any of those variables into a constant. If you draw a line from the origin that intercepts the hyperbola $\Delta t = \sqrt{(3^2 + \Delta x^2)}$ then for ANY velocity $\Delta x/ \Delta t$ in the range (-c<v<c) the elapsed proper time is a constant equal to 3 at the point where the line intercepts the hyperbola. This is for the "horizontal hyperbola". For the vertical hyperbola, $\Delta x = \sqrt{(3^2 + \Delta t^2)}$ the 3 represents the constant proper distance, but imaginary quantities are involved and things get complex (pun intended ). Start with the Minkowski relationship $(c\tau)^2 = (ct^2) - x^2$ and multiply by both sides by minus one to obtain $-(c\tau)^2 = -(ct^2) + x^2$ then we get $ic\tau = \sqrt{(x^2-(ct)^2)}$. For real velocities, where dt>dx the quantity on the left is imaginary, but it is easy to see that for imaginary velocities greater than the speed of light, dt<dx, the proper time is imaginary and the quantity on the left becomes real and is equal to the proper distance, $dx_o = ic\tau$. Now if we plot lines from the origin, with slopes of dt/dx rather than the usual dx/dt, they intercept the vertical hyperbola at points that all have the same constant proper distance measured by a particle traveling with arbitary velocity v = dx/dt. The reason the imaginary quantities appear is that a proper time interval is the interval between two timelike events measured by a real clock traveling inertially between the two events at less than the speed of light, while the proper distance is measurement of the spacelike interval that is measured by an imaginary clock traveling at greater than the speed of light. Are you confused? You should be. I probably confused myself. Told you things would get complicated Anyway, for the vertical hyperbola, the constant parameter is the real constant proper distance.

 

[stevmg]

You can make any of those variables into a constant. If you draw a line from the origin that intercepts the hyperbola $\Delta t = \sqrt{(3^2 + \Delta x^2)}$ then for ANY velocity $\Delta x/ \Delta t$ in the range (-c<v<c) the elapsed proper time is a constant equal to 3 at the point where the line intercepts the hyperbola. This is for the "horizontal hyperbola". For the vertical hyperbola, $\Delta x = \sqrt{(3^2 + \Delta t^2)}$ the 3 represents the constant proper distance, but imaginary quantities are involved and things get complex (pun intended ). Start with the Minkowski relationship $(c\tau)^2 = (ct^2) - x^2$ and multiply by both sides by minus one to obtain $-(c\tau)^2 = -(ct^2) + x^2$ then we get $ic\tau = \sqrt{(x^2-(ct)^2)}$. For real velocities, where dt>dx the quantity on the left is imaginary, but it is easy to see that for imaginary velocities greater than the speed of light, dt<dx, the proper time is imaginary and the quantity on the left becomes real and is equal to the proper distance, $dx_o = ic\tau$. Now if we plot lines from the origin, with slopes of dt/dx rather than the usual dx/dt, they intercept the vertical hyperbola at points that all have the same constant proper distance measured by a particle traveling with arbitary velocity v = dx/dt. The reason the imaginary quantities appear is that a proper time interval is the interval between two timelike events measured by a real clock traveling inertially between the two events at less than the speed of light, while the proper distance is measurement of the spacelike interval that is measured by an imaginary clock traveling at greater than the speed of light. Are you confused? You should be. I probably confused myself. Told you things would get complicated Anyway, for the vertical hyperbola, the constant parameter is the real constant proper distance.

You can make any of those variables into a constant. If you draw a line from the origin that intercepts the hyperbola $\Delta t = \sqrt{(3^2 + \Delta x^2)}$ then for ANY velocity $\Delta x/ \Delta t$ in the range (-c<v<c) the elapsed proper time is a constant equal to 3 at the point where the line intercepts the hyperbola.

yiuop, why the "3?"

For the vertical hyperbola, $\Delta x = \sqrt{(3^2 + \Delta t^2)}$ the 3 represents the constant proper distance, but imaginary quantities are involved and things get complex (pun intended ). Start with the Minkowski relationship $(c\tau)^2 = (ct^2) - x^2$ and multiply by both sides by minus one to obtain $-(c\tau)^2 = -(ct^2) + x^2$ then we get $ic\tau = \sqrt{(x^2-(ct)^2)}$. For real velocities, where dt>dx the quantity on the left is imaginary, but it is easy to see that for imaginary velocities greater than the speed of light, dt<dx, the proper time is imaginary and the quantity on the left becomes real and is equal to the proper distance, $dx_o = ic\tau$. Now if we plot lines from the origin, with slopes of dt/dx rather than the usual dx/dt, they intercept the vertical hyperbola at points that all have the same constant proper distance measured by a particle traveling with arbitary velocity v = dx/dt. The reason the imaginary quantities appear is that a proper time interval is the interval between two timelike events measured by a real clock traveling inertially between the two events at less than the speed of light, while the proper distance is measurement of the spacelike interval that is measured by an imaginary clock traveling at greater than the speed of light. Are you confused? You should be. I probably confused myself. Told you things would get complicated Anyway, for the vertical hyperbola, the constant parameter is the real constant proper distance.

I thought of imaginary or complex numbers and remembered their additive qualities but that pertained to mulitplying. To wit:

Take the value "-1." In complex numbers that's $(1)*(sin \pi + i cos \pi) = -1$

The $\sqrt {-1} = 1*(sin (\pi/2) + i cos (\pi/2) = 0 + i = i$

Adding the two angles is for multiplication rather than addition.

 

[stevmg]

Physically the "proper time" along a given worldline just means the time elapsed on a clock that travels along that worldline. If you want to know the proper time between two events on the worldline of a clock moving at constant velocity along the x-axis of some frame, it's just $$\Delta \tau = \sqrt{\Delta t^2 - (1/c^2) \Delta x^2 }$$, where $$\Delta t$$ is the difference in coordinate time between the two events and $$\Delta x$$ is the difference in coordinate position. This can be rewritten as $$\Delta \tau = \Delta t \sqrt{1 - (1/c^2) (\Delta x / \Delta t )^2 } = \Delta t \sqrt{1 - v^2 /c^2 }$$ which is just the standard time dilation formula. For a worldline where the velocity is varying in a continuous way, you can approximate its worldline by a "polygonal" worldline made up of a series of constant-velocity segments each lasting a time interval of $$\Delta t$$ with instantaneous accelerations between them, then if the velocity during the first time interval v₁ (which could just be the average velocity during the same interval for the path with continuously changing velocity that you are trying to approximate), the second has velocity v₂, and the final segment has velocity v_N, then the total elapsed time would just be the sum of the elapsed time on each segment, or $$\Delta t \sqrt{1 - v_1^2 /c^2} + \Delta t \sqrt{1 - v_2^2 / c^2} + ... + \Delta t \sqrt{1 - v_N^2 /c^2}$$. And in the limit as as the time interval of each segment goes to zero (so the number of segments approaches infinity), this approximation should approach perfect agreement with the proper time on the original path with continuously-changing velocity. Since an integral is just the limiting case of a sum with smaller and smaller intervals (or an infinite series of intervals which each have an 'infinitesimal' time dt), that means the proper time along a worldline where the velocity as a function of time is given by v(t) can always be computed according to the integral $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2 /c^2} \, dt$$

JesseM -
That is succinct explanation of proper time. Thank you. Now
a) what is a similar explanation for proper velocity?
b) for proper acceleration?

 

[stevmg]

starthaus -

Good. What happens when the proper acceleration $$a_p$$ increases?

$$-(ct)^2+x^2=(\frac{c^2}{a_p})^2$$
should be added to your blog of "Acceleration in SR - II.pdf" as equation 16 as you do have much occasion to refer to it in subsequent discussions.

Also, I relooked at this same .pdf and realized that you were just dealing with an acceleration mode, i.e., where one has an inertial time frame S and a (constant) accelerating time frame S'. Thus your equations pertain to only that and not to two time frames (say, O and O') in which there is constant velocity of O' from O.

S-o-r-r-y... Brain dead again. (I'll rename myself "Fred" so I may be referred to as "Brain Dead Fred.")

stevmg

 

[starthaus]

starthaus -



$$-(ct)^2+x^2=(\frac{c^2}{a_p})^2$$
should be added to your blog of "Acceleration in SR - II.pdf" as equation 16 as you do have much occasion to refer to it in subsequent discussions.

Also, I relooked at this same .pdf and realized that you were just dealing with an acceleration mode, i.e., where one has an inertial time frame S and a (constant) accelerating time frame S'. Thus your equations pertain to only that and not to two time frames (say, O and O') in which there is constant velocity of O' from O.

S-o-r-r-y... Brain dead again. (I'll rename myself "Fred" so I may be referred to as "Brain Dead Fred.")

stevmg

Hence the name of the blog file, "Accelerated Motion in SR" :-)

 

[stevmg]

proper time = time elapsed by the inertial observer (must use time-dilation formula or $\Delta \tau^2 = \Delta t^2 - \Delta x^2/c^2) = \Delta t^2/\gamma)$

What's proper velocity?

What is proper acceleration? (?acceleration as measured by a co-moving inertial ('steady state") observer going at the same instantaneous velocity as the accelerating particle?)

 

[DrGreg]

What's proper velocity?

What is proper acceleration? (?acceleration as measured by a co-moving inertial ('steady state") observer going at the same instantaneous velocity as the accelerating particle?)

See post #11

 

[starthaus]

proper time = time elapsed by the inertial observer (must use time-dilation formula or $\Delta \tau^2 = \Delta t^2 - \Delta x^2/c^2) = \Delta t^2/\gamma)$

What's proper velocity?

What is proper acceleration? (?acceleration as measured by a co-moving inertial ('steady state") observer going at the same instantaneous velocity as the accelerating particle?)

From the blog:1. Proper speed

$$v_p=\frac{dx}{d\tau}=\gamma \frac{dx}{dt}=\gamma v$$2. Proper acceleration

$$a_p=c \frac{d\phi}{d\tau}=\gamma^3*\frac{dv}{dt}=\gamma^3*\frac{d^2 x}{dt^2}=\gamma^3 a$$

t=coordinate time
$\tau$=proper time
v=coordinate speed
a=coordinate acceleration

 

[yuiop]

yiuop, why the "3?"

3 is just a random number that I chose for the constant, but I could have picked any other number. If I had put a letter such as k or n or represent a constant, some people get confused and think I mean a variable. There are various sorts of constants. Some are physical constants are a particular number. Others are sometimes called constants because they are invariant under a transformation and some are constants with respect to time. I meant the last kind. You can choose any number to be the constant (so in that sense it is a variable or a parameter) but once you have chosen it, it remains fixed over time for a given equation. Believe it or not, we have had a long argument on this forum, where someone used the fact that I had used a letter to represent a constant, as "proof" that the constant I using was in fact a variable. LOL

 

[stevmg]

If you are working in a (t,x) coordinate system, and $\tau$ is proper time:

coordinate acceleration = d²x/dt²

proper acceleration = acceleration measured in the coordinate system of a comoving inertial observer = what an accelerometer measures

rapidity = $$\tanh^{-1} \frac {dx/dt}{c}$$

coordinate time = t

coordinate velocity = dx/dt

proper velocity = $$dx/d\tau$$ although I prefer to call it "celerity" because of possible confusions that can occur (especially over proper acceleration).

Thanks Dr Greg, starthaus.

One more thing:

What is proper distance? (I assume this would only apply to events that are spacelike in relationship.)

stevmg

 

[stevmg]

If proper speed (or velocity) is $\tau = v*\gamma$ then it is conceivable that $\tau$ exceed the speed of light. Is that true? If so, what physical meaning does that have? Is there a limit on $v_p < c?$. Seems like there should be in the sense that there should be NO FR for a given observation in which $v \geq c$

Likewise, proper acceleration or $a_p = a\gamma^3$ also can be $> c$ but I have been told that there is no limit on acceleration other than that the coordinate velocity be $< c$. Is that valid?

starthaus -

How do you create your blog with the .pdf s? To wit, your 3-page MMX.pdf. You obvously create the original in some program or do you use the PF "reply - advanced" box and then delete that box? You can print the LATEX formatted "reply" as a .pdf and then delete (or navigate away from an unposted reply.) Is there another platform which you can use to see LATEX formatted characters?

Don't worry, I am not going to start a blog and if I do, it would be for brain-dead dummies like me (you know, "advanced physics for idiots.pdfs.") There are things I would like to ".pdf-alize" and would like to know how.

stevmg

 

[stevmg]

You can make any of those variables into a constant. If you draw a line from the origin that intercepts the hyperbola $\Delta t = \sqrt{(3^2 + \Delta x^2)}$ then for ANY velocity $\Delta x/ \Delta t$ in the range (-c<v<c) the elapsed proper time is a constant equal to 3 at the point where the line intercepts the hyperbola. This is for the "horizontal hyperbola". For the vertical hyperbola, $\Delta x = \sqrt{(3^2 + \Delta t^2)}$ the 3 represents the constant proper distance, but imaginary quantities are involved and things get complex (pun intended ). Start with the Minkowski relationship $(c\tau)^2 = (ct^2) - x^2$ and multiply by both sides by minus one to obtain $-(c\tau)^2 = -(ct^2) + x^2$ then we get $ic\tau = \sqrt{(x^2-(ct)^2)}$. For real velocities, where dt>dx the quantity on the left is imaginary, but it is easy to see that for imaginary velocities greater than the speed of light, dt<dx, the proper time is imaginary and the quantity on the left becomes real and is equal to the proper distance, $dx_o = ic\tau$. Now if we plot lines from the origin, with slopes of dt/dx rather than the usual dx/dt, they intercept the vertical hyperbola at points that all have the same constant proper distance measured by a particle traveling with arbitary velocity v = dx/dt. The reason the imaginary quantities appear is that a proper time interval is the interval between two timelike events measured by a real clock traveling inertially between the two events at less than the speed of light, while the proper distance is measurement of the spacelike interval that is measured by an imaginary clock traveling at greater than the speed of light. Are you confused? You should be. I probably confused myself. Told you things would get complicated Anyway, for the vertical hyperbola, the constant parameter is the real constant proper distance.

I'm going to have to hold off on this one as I don't have a bloody idea what or why this is being done. I will say "enough for now" on this but the other questions I have in the last two posts still stand.