Derivation of proper time in acceleration in SR

1. Sep 3, 2010

stevmg

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

https://www.physicsforums.com/showpost.php?p=2864903&postcount=112

$T_{a1} =\frac{c}{a_1}\, sinh(a_{a1} t_{a1}/c)$

$v = \frac{a_1 T_{a1}}{\sqrt{1+(a_1 T_{a1}/c)^2}}$

How do I get my equation to equal your equations, or where is my mistake?

Last edited: Sep 3, 2010
2. Sep 3, 2010

Mentz114

I think those 'sines' should be hyperbolic sine, 'sinh'.

see http://en.wikipedia.org/wiki/Hyperbolic_function

3. Sep 3, 2010

stevmg

They should be, but by using a table of integrals

dv/$\sqrt{(c^2 - v^2)}$ = sin-1(v/c) which messes this up.

I have checked this integral in several different books and it is correct. There is something else I am doing wrong and I think it has something to do with not taking time dilation into consideration (I thought I was with the original equation I posted.)

4. Sep 3, 2010

JesseM

Where does that equation come from? The equation suggests that v(t) = a0*t*sqrt[1 - v2/c2], but that isn't correct for a rocket undergoing constant proper acceleration (which is different from the coordinate acceleration in the inertial frame where the rocket is initially at rest)--as you can see from the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken], the correct velocity function in this case would be v(t) = at / sqrt[1 + (at/c)2], where a is the proper acceleration.

Last edited by a moderator: May 4, 2017
5. Sep 3, 2010

yuiop

Hi, I am a little short on time at the moment, so I can not give your question the full attentian it deserves and when I am in a hurry I tend to make (more) mistakes.

First thing I notice is that the acceleration transformation should probably be a function of gamma cubed. Look at it like this. In the Co-Moving Inertial Reference Frame CMIRF the object increases its distance displacement by x every time interval t squared. In the frame of a inertial observer with velocity relative to the CMIRF, the distance is shorter by 1/gamma and the time interval is longer by gamma^2 so this gives an acceleration transformation that is slower by the order 1/gamma^3. See equation 29 of this document http://www.physics.princeton.edu/~mcdonald/examples/mechanics/matthews_ajp_73_45_05.pdf [Broken] and this old PF thread https://www.physicsforums.com/showthread.php?t=233264

These references might be handy too:

http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/acceleration.html
http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/rocket.html [Broken]
http://en.wikipedia.org/wiki/Four-acceleration

I will have to come back to this when I have more time.

Last edited by a moderator: May 4, 2017
6. Sep 3, 2010

starthaus

This is not correct, see https://www.physicsforums.com/blog.php?b=1911 [Broken] for a correct derivation.

Last edited by a moderator: May 4, 2017
7. Sep 4, 2010

Passionflower

You might be confused stevmg, the title of your topic is "Derivation of proper time in acceleration in SR" however the first formula does not derive the proper time, in fact it does the opposite it reconstructs the coordinate time from the proper time.

I highly suggest everybody to use 'standard' symbols when discussing these problems to avoid problems with misinterpretation of formulas for instance:

$\eta$ = rapidity
$a$ = coordinate acceleration
$\alpha$ = proper acceleration
$t$ = coordinate time
$\tau$ = proper time
$v$ = coordinate velocity
$w$ = proper velocity
$x$ = distance

These are three important equations for constant proper acceleration:

$$\alpha = \frac{\Delta w}{\Delta t} = \frac{\Delta \eta}{\Delta \tau} = \frac{\Delta \gamma}{\Delta x}$$

$$\eta = \alpha \tau = sinh^{-1} w = tanh^{-1} v = cosh^{-1} \gamma$$

$$a = \frac{\alpha}{\gamma^3}$$

Last edited: Sep 4, 2010
8. Sep 4, 2010

yuiop

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:

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

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

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

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

In Einstein's original paper he concluded from the force acceleration relationship for force parallel to the motion $F = m \gamma^3 a$ that the relativistic longitudinal mass is $$m/(\sqrt{1-v^2/c^2})^3[/itex], (See http://www.fourmilab.ch/etexts/einstein/specrel/www/) but it can be seen from the above discussion that it is the proper acceleration that is greater than the coordinate acceleration by a factor of gamma cubed rather than the mass changing. Einstein later withdrew from the concept of relativistic mass. I am still not sure how we get from the $p = mv/\sqrt{1-v^2/c^2}$ and $F=dp/dt$ to $a = \alpha \gamma^3$. I will have to think about that some more. Anyone any ideas? P.S. I will have to come back to how we get to the hyperbolic functions when time allows. 9. Sep 5, 2010 starthaus Start from: [tex]v=\frac{\alpha t}{\sqrt{1+(\alpha t/c)^2}}$$

On the other hand:

$$\alpha=c \frac {d \eta}{d \tau}$$

$$\eta=arccosh(\gamma)$$

so:

$$\alpha=c \frac {d \eta}{dt} \frac{dt}{d \tau}=\frac{c}{\sqrt{\gamma^2-1}} \frac{d \gamma}{dt} \frac{dt}{d \tau}=....=\frac{a}{\gamma^3}$$

10. Sep 5, 2010

stevmg

A) Thank you very much for your derivation for $v$ in terms of $\alpha$ and $t$ when under a constant force over $t$:
$v = \frac{\alpha t}{\sqrt{1+(\alpha t/c)^2}}$

B) My original differential equation was a = a0(1-v^2/c^2)(1/2)
This proof supposes that this equation should be a = a0(1-v^2/c^2)(3/2). As you mention in the turquoise highlighted text Einstein changed his own thinking as presented in A) above. I agree with you as it would be nice to know how this came about. starthaus below uses the hyperbolic functions and the introduction of rapidity ($\eta$) to explain this. Obviously this is valid however the algebraic proof is more "intuitive" and an algebraic derivation of a = a0(1-v^2/c^2)(3/2) would be desirable.

Again, starthaus, I have found all your posts on a wide range of subjects and your .pdf on acceleration - II most insightful. I just don't have a "feel" for hyperbolic function and inroducing rapidity is not intuitive to me as yet. I think there is an algebraic answer to my question.

To all:

Passionflower brought up the symbols and concepts necessary for interpretation of what is going on which I have listed here:

rapidity
coordinate acceleration
proper acceleration
coordinate time
proper time
coordinate velocity
proper velocity
distance

The question of the hour is what are the definitions of each of those terms? I know what proper time and distance are but what about the rest?

What are the definitions of "coordinate acceleration?," "proper acceleration," "rapidity," "coordinate time," "coordinate velocity," and "proper velocity?" Understanding definitions is 2/3 the battle.

Again, any textbooks or websites that have all this and are not too esoteric?

11. Sep 5, 2010

DrGreg

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).

12. Sep 5, 2010

stevmg

I see from your post here in "Why Is It Impossible To Reach The Speed of Light" you have, in essence, algebraically derived my a0$\gamma$-3 at this post:

Now, I hate to be stupid, but how do I get dx/d$\tau$? Just get me started. I know how to calculate proper time:
$\tau$ = SQRT[(t2 - t1)2 - (x2 - x1)2)]
I apologize as I cannot get my LATEX symbols to work out.

Again, I have no textbook that shows this and I am after a good but understandible text which covers these matters.

I did read Einstein's 1905 paper on electromagnetic forces but - no clue. I am not stupid as I DO understand the derivation of the Lorentz transformations.

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

In other words, "proper acceleration = the coordinate acceleration of a comoving inertial observer?"

Thanks for the definitions as they are, as I stated earlier, 2/3 the battle!

Last edited: Sep 5, 2010
13. Sep 5, 2010

DrGreg

That's true for the proper time of an inertial object, but in general for a non-inertial object you have to use

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

from which you get

$$\frac{d\tau}{dt} = \sqrt{1 - \frac{1}{c^2}\,\left(\frac {dx}{dt}\right)^2} = \frac{1}{\gamma}$$​

i.e.

$$\frac{dt}{d\tau} = \gamma$$​

hence

$$\frac{dx}{d\tau} = \frac{dx}{dt} \, \frac{dt}{d\tau} = \gamma \frac{dx}{dt} = \gamma v$$​

in the usual notation.

proper acceleration = the coordinate acceleration measured by a comoving inertial observer

14. Sep 5, 2010

stevmg

Great. I didn't think of differentiating the integral to get back to the differential.

Yes, I see the difference between what I wrote and what you stated.

yuiop and you as well as the other contributors have just cleared up a stumbling block for me and I never would have figured it out for myself if left alone to do it.

15. Sep 6, 2010

stevmg

Now, to have a specific example:

Using the initial acceleration of 9.8 m/sec2 and using the stuff from above

What would be the velocity of an object starting at 0 m/sec and having the acceleration at v = 0 of 9.8 m/sec2 for 10 seconds?
What would be its proper velocity?
What would be its proper acceleration?
What would be the distance this object traveled (coordinate distance?)
What would be the proper distance?
What would be it proper time if coordinate time was 10 seconds?

The above is where the rubber meets the road and demonstrates how to apply what has been stated above. Do not use hyperbolic functions, please. Stick with algebra and calculus.

16. Sep 6, 2010

Passionflower

A constant coordinate or proper acceleration of 9.8 m/s2?
10 seconds of coordinate time or proper time?

17. Sep 6, 2010

stevmg

That's exactly what I am talking about. Don't have anyone here to bounce thoughts off nor any examples to work from.

In other words, if FR S is inertial and we place at (0, 0) a object which accelerates at 9.8 m/sec2 to the right, under classical physics, a = 9.8 m/sec2

OK, what is the coordinate acceleration in S?

Is there a proper acceleration in S and if so, what is it?

If we apply the F necessary to give the initial a = 9.8 m/sec2 when v = 0 [at (0, 0)] what would be the velocity at 10 seconds. In classical physics it would be 98 m/sec. x = 490 m in classical physics.

Do you see where I am going? I need some basic problems worked out as you would have in a basic text so that I would understand what these terms actually mean in the real world.

If there is a site or textbook that does this please direct me to it.

Thanks,

stevmg

18. Sep 6, 2010

Passionflower

The inertial observer will measure a different acceleration than the accelerating observer.

So if we take a coordinate acceleration of a = 9.8 m/sec2 then the accelerating observer will measure an increasing proper acceleration over time. Note that we cannot accelerate forever because we linearly approach the speed of light with a constant coordinate acceleration.

However if we take a proper acceleration of $\alpha$ = 9.8 m/sec2 (notice the Greek letter!) then the coordinate acceleration will decrease in time while the proper acceleration will remain fixed.

The same with time, 10 seconds for the inertial observer is not going to have the same calculation as 10 seconds for the traveler. So distinguish between t and $\tau$ when you state the problem.

19. Sep 6, 2010

stevmg

Oooh, please, please please point me to a text or site where these things are worked out so I get working knowledge of what is what. I understand the English of what I highlighted above. I don't understand how to apply it until I see it applied. That is the way I learn. I have nobody here to learn it from and no text or synopsis that does just that.

When I first was told that "force is directly proportional to mass and directly proportional to acceleration" I didn't have a clue what that meant until I was given the example of, say, "98 newtons = 10 kg $\times$ 9.8 m/sec2 "

20. Sep 7, 2010

yuiop

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'/γ33 = 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, lets 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 travelled 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.

Last edited: Sep 7, 2010
21. Sep 7, 2010

Austin0

Hi yuiop I have been wondering how you arrived at this alias and what was the possible meaning until I just typed it in , the light dawned

[EDIT] later. Wrt the CMIRF elapsed time. Having realized you were looking at the final CMIRF as a continuous frame from the beginning of acceleration it then makes sense that the elapsed time would be the same as the rest frame S . According to this S' the accelerating frame was a decelrating frmae so it makes sense that the course of acceleration would be the same.
Regarding the gamma factor cubed : perhaps it is because although the mass increase is not relevant to the coordinate acceleration or proper acceleration within the accelerating frame it is relevant in the observing inertial frame so the overall difference in between the two frames would be a cubed factor, is this relevant at all???

Last edited: Sep 7, 2010
22. Sep 7, 2010

stevmg

yuiop - I am not at Austin0's level but do have questions about what you wrote which I will highlight in red in your post which is below.

I took out the quote brackets so that the symbols and coloring would appear as they did in your posting.

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 when t = 0 and v = 0

a0 = a' when t= 0 and v = 0, both t and v as measured in the stationary inertial frame or S.

In earlier posts we have established with help from DrGreg that a = dv/dt = a'/γ3.
DrGreg did show that post 48, "Why is impossible to reach the speed of light"
(https://www.physicsforums.com/showpost.php?p=2859246&postcount=48)
that dv/dt = F/($\gamma^3$m). I presume that's where this paragraph comes from.

This does make intuitive sense because as one continually apples the same force to the object in the CMIRF (hence the same acceleration and mass in the CMIRF) and as time is dilated, acceleration will remain constant in the CMIRF (S') but will decrease when looked at from the original S or resting frame because 1 second of dilated time (S') is > 1 sec of original stationary reference time (S) when looked at from the point of view of original stationary reference time (S.) That's what DrGreg proved both with hyperbolic functions as well as with "algebraic" calculus, right?.

Also, a = dv/dt is the same as classical physics, true?

We have also established that

F = dp/dt = d(mvγ)/dt' = m d(vγ)/dt = may3 = ma' ["y" = "$\gamma$," right?]

and it is also true that:

F = may3 = m(dv/dt)y3 = m(a'/γ33 = 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. Yes, sir. I realized that after I submitted the question.

Staying with the car metaphor, lets 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. Is any acceleration, no matter how large, possible for an infinitesimal time period provided we are at < c in linear velocity or speed?

Problem statement:

Proper acceleration = 2c/sec.

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

$$\gamma = (1+(a't/c)^2) = \frac{1}{\sqrt{1-v^2/c^2}} = 20.025$$ where does this come from? What numbers are you putting where in the equation?

Final coordinate velocity in S

$$v = \frac{a't}{\sqrt{1+(a't/c)^2}} = a't\gamma = \frac{2*10}{20.025} = 0.99875c$$ Again, what numbers gave you this?

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: Is the coordinate acceleration the acceleration in S?

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

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 travelled 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'm stopping here until I get my questions above answered

What you have done is yeoman's effort to bust into my head these concepts. Your efforts are most appreciated. I still have to get these points clarified or going on will be fruitless.

Thanks a lot...

EDIT:

I got the 20.25 = SQRT [1 + (10*2)2]= SQRT (401) = 20.25
a' = 2 (given), t = 2 (given) hence the above. However, the noatation in the post should be:
$\gamma = SQRT [1 + (a't/c)^2] = 1/SQRT (1 - v^2/c^2)$
Now I have to figure out where you got
$\gamma = SQRT [1 + (a't/c)^2] = 1/SQRT (1 - v^2/c^2)$ from

DrGreg proved the below:
$\ a = a_0/\gamma^3 = 2/20.20.025^3 = 0.00025c/s$

The rest, later. Still need the answers in red.

Last edited: Sep 7, 2010
23. Sep 7, 2010

yuiop

In the CMIRF, the acceleration is a' = dv'/dt = dx'/dt'2. Now dx = dx'/$\gamma$ and dt2 = dt'2*$\gamma^2$, so after substitutions a = dv/dt = dx/dt2 = (dx'/dt'2)/$/\gamma^3$

The force required to accelerate a particle parallel to its velocity increases significantly as the velocity increases and this gamma cubed factor is routinely observed in particle accelerators. This leads to a diminishing law of returns for particle accelerators. To attain terminal velocities that are 1% higher than the terminal velocity of the CERN accelerator particles, would require an accelerator that is orders of magnitude more expensive to build and operate. However, it is still worth their while investing in larger accelerators because the energy levels of the collisions are significantly higher, for what looks like a small increase in velocity.

I have checked it again and it seems to be OK, but this property is not necessarily true for Lorentz transformations of inertial movement, so this property is unique to constant proper acceleration. See also my comments below.

Yes you are right that I was looking at the CMIRF as a continuous frame from start to finish so it is valid to just do a straightforward Lorentz transformation of the two events and you are also right that in the CMIRF the "accelerating" object decelerates from an initial speed in S' equal to the final speed in S to a final velocity of zero in S'.

24. Sep 7, 2010

yuiop

yuiop - I am not at Austin0's level but do have questions about what you wrote which I will highlight in red in your post which is below.

I took out the quote brackets so that the symbols and coloring would appear as they did in your posting. Thanks, that makes things easier. my comments are in magenta.

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 when t = 0 and v = 0

Not only at t=0 and v=0.

m = m' = m0 all the time in all reference frames. I am not using the concept of relativistic mass, which is an outdated concept.

a0 = a' when t= 0 and v = 0, both t and v as measured in the stationary inertial frame or S.

No, a0 = a' at the final event, when the velocity of the CMIRF in S coincides with the final velocity of the accelerating particle as measured in S. I should have made that clearer. The proper acceleration is always equal to the coordinate acceleration measured in the instantaneously co-moving inertial frame. I have only considered the CMIRF that was co-moving at the final event. At an earlier time, when the velocity was lower, just pick another CMIRF with a matching lower velocity and the same is true. There is a whole sequence of instantaneous co-moving inertial frames. It is however true that initially when t=0 and v=0, that proper acceleration a0 is equal to the coordinate acceleration a in S

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

DrGreg did show that post 48, "Why is impossible to reach the speed of light"
(https://www.physicsforums.com/showpost.php?p=2859246&postcount=48)
that dv/dt = F/($\gamma^3$m). I presume that's where this paragraph comes from.

Yes, DrGreg basically demonstrated:

$${\color{magenta} F = \frac{dp}{dt} = \frac{d}{dt} \left( \gamma m v \right) = m \frac{d}{dt}(\gamma v) = m \left( \frac{d\gamma}{dv}\, \frac{dv}{dt} \, v + \gamma \frac{dv}{dt} \right) = m \left( \frac{av^2/c^2}{(1-v^2/c^2)^{3/2}}\, \, + \frac{a}{\sqrt{1-v^2/c^2}} \right) = m \left( \frac{a}{(1-v^2/c^2)^{3/2}} \right) = ma\gamma^3 }$$

Now by definition;

$${\color{magenta} a =\frac{dv}{dt} = \frac{d}{dt}\left(\frac{dx}{dt}\right) = \frac{d}{dt'}\frac{dt'}{dt}\left(\frac{dx'}{dt'}\frac{dt'}{dt}\frac{dx}{dx'}\right) = \frac{d}{dt'}\frac{1}{\gamma}\left(\frac{dx'}{dt'}\frac{1}{\gamma}\frac{1}{\gamma}\right) = \frac{d}{dt'}\left(\frac{dx'}{dt'} \right) \frac{1}{\gamma^3} = \frac{a'}{\gamma^3} }$$

So:

$${\color{magenta} F=ma' }$$

This does make intuitive sense because as one continually apples the same force to the object in the CMIRF (hence the same acceleration and mass in the CMIRF) and as time is dilated, acceleration will remain constant in the CMIRF (S') but will decrease when looked at from the original S or resting frame because 1 second of dilated time (S') is > 1 sec of original stationary reference time (S) when looked at from the point of view of original stationary reference time (S.) That's what DrGreg proved both with hyperbolic functions as well as with "algebraic" calculus, right?.

Yes

Also, a = dv/dt is the same as classical physics, true?

Yep

We have also established that

F = dp/dt = d(mvγ)/dt' = m d(vγ)/dt = may3 = ma' ["y" = "$\gamma$," right?]

Yep

and it is also true that:

F = may3 = m(dv/dt)y3 = m(a'/γ33 = 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. Yes, sir. I realized that after I submitted the question.

Staying with the car metaphor, lets 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. Is any acceleration, no matter how large, possible for an infinitesimal time period provided we are at < c in linear velocity or speed?

Yes, I believe so, but there there might be QM concerns if the infinitesimal time period is less than the Planck time interval. Not sure about that. Although it might be possible to show that infinite acceleration (i.e. acceleration that occurs in a zero time interval) only results in reaching the speed of light, I personally believe nothing happens at zero time intervals and you certainly cannot measure velocities and accelerations in a zero time interval. Whether there is such a thing as a quantum of time is a subject for another forum. Peter Lynds argued in a published paper that there is no such thing as an "instant of time" in relation to Zeno's paradoxes, but I don't think he went so far as to define a minimum time interval and his ideas are not generally accepted even though they were published.

To be continued ................

Last edited: Sep 8, 2010
25. Sep 7, 2010

yuiop

Oops, well spotted. Thanks for finding that typo quickly so that I was still able to edit and correct the original. Fortunately it does not make a mess of the rest of the post.