Derivation of proper time in acceleration in SR

[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 v2. 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 &lt; 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 &gt; c but I have been told that there is no limit on acceleration other than that the coordinate velocity be &lt; 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.