Derivation of proper time in acceleration in SR

[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

By your post
http://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?

[Mentz114]

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{1 - v^2} = a0dt/c

when t= 0, v = 0

sin-1 (v) = a0t/c

we can simplify this if we choose c = 1, so v is expressed in terms of c or we can call it \beta

sin-1 (\beta) = a0t

\beta = sin (a0t)

this makes no sense as sin (a0t) is a periodic function and \beta is a monotonic increasing function from 0 to 1.

This disagrees with your post
http://physicsforums.com/showpost.php?p=2864903&postcount=112

HELP!

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

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

[stevmg]

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

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

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.)

[JesseM]

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)}
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 (http://en.wikipedia.org/wiki/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 relativistic rocket page (http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html), the correct velocity function in this case would be v(t) = at / sqrt[1 + (at/c)2], where a is the proper acceleration.

[yuiop]

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

By your post
http://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?

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 and this old PF thread http://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
http://en.wikipedia.org/wiki/Four-acceleration

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

[starthaus]

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


This is not correct, see here (http://www.physicsforums.com/blog.php?b=1911) for a correct derivation.

[Passionflower]

By your post
http://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?
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}

[yuiop]

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

By your post
http://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?

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.

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 [tex]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.

[starthaus]

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.

Start from:

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}

[stevmg]

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.

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.

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.

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}

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?

[DrGreg]

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

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).

[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).

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:


If you're not comfortable with hyperbolic functions, begin with


F = \frac{dp}{dt}
= \frac{d}{dt} \left( \gamma m v \right)
= m \left( \frac{d\gamma}{dv}\, \frac{dv}{dt} \, v + \gamma \frac{dv}{dt} \right)

and calculate d\gamma/dv . A bit more painful, but you'll get there in the end.

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!

[DrGreg]

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)]
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.

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

[stevmg]

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.

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


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

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.

[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).

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.

[Passionflower]

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?

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

[stevmg]

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

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

[Passionflower]

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

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.

[stevmg]

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.

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 "

[yuiop]

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 travelled (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.

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, 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.

[Austin0]

ver.

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'.





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.


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

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

This seems to indicate that the measured coordinate acceleration as measured from instantaneous velocities would fall off from initial (a) at a rate of \gamma3 is this correct or is this factor simply the relationship of internal proper acceleration to (a) ???
Wouldn't it seem logical that based on the increasing mass the coordinate acceleration would fall off at a single factor of \gamma???
I can see a reason for the internal proper acceleration to increase by \gamma2 as it is a function of both time dilation and length contraction but where does the third exponential increase come from???
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.
I agree it is surprising and possibly mistaken
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.
Wouldn't the CMIRF '\Deltat' neccessarily be an integrated sum of the intermediate CMIRF's at velocities from 0 to final???
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)


.

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 :wink:

[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???

[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.

From yuiop post http://www.physicsforums.com/showthread.php?p=2869774#post2869774
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"
(http://physicsforums.com/showpost.php?p=2859246&postcount=48)
that dv/dt = F/(\gamma^3m). 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'/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. 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.

[yuiop]

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

This seems to indicate that the measured coordinate acceleration as measured from instantaneous velocities would fall off from initial (a) at a rate of \gamma^3 is this correct or is this factor simply the relationship of internal proper acceleration to (a) ???
Wouldn't it seem logical that based on the increasing mass the coordinate acceleration would fall off at a single factor of \gamma???
I can see a reason for the internal proper acceleration to increase by
\gamma^2 as it is a function of both time dilation and length contraction but where does the third exponential increase come from???

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.

Coordinate distance Δx' in the CMIRF (S')

From 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.

I agree it is surprising and possibly mistaken
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.


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.

Wouldn't the CMIRF '\Deltat' necessarily be an integrated sum of the intermediate CMIRF's at velocities from 0 to final???

[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 decelerating frame so it makes sense that the course of acceleration would be the same. 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'.

[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.

From yuiop post http://www.physicsforums.com/showthread.php?p=2869774#post2869774
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"
(http://physicsforums.com/showpost.php?p=2859246&postcount=48)
that dv/dt = F/(\gamma^3m). 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'}\f rac{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'/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. 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 ................

[yuiop]

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)


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. :smile:

[stevmg]

This is beautiful but what we have a lot of observers here

stationary inertial observer (S): m, t, a

comoving inertial observer (S'): m', t', a'

Now who is this "accelerating observer" you speak of? m0, t0 and a0

First of all, so I have the notation right (the right mta with the right observer)?

How is he/she different than the CMIRF observer?

But, we are getting somewhere.

As you will see later on down, in oder to calculate \gamma^3 I didn't know what velocity to use - the begiining or the end (t = 0, v = 0 or t = 10, v = ?) It seemed like a recursive equation.

[yuiop]

This is beautiful but what we have a lot of observers here

stationary inertial observer (S): m, t, a

Call this the launch pad. Observer S stays at rest in this frame.

comoving inertial observer (S'): m', t', a'

This is an observer onboard a rocket with its engine off, coasting at 0.99875c all the time and is initially way off to the left of the launch pad heading towards the launch pad.

Now who is this "accelerating observer" you speak of? m0, t0 and a0

This an observer onboard a rocket that takes of at time t=t'=t0 =0 from the launch pad with constant proper acceleration 2c/second. He accelerates to the right and eventually matches the speed of the coasting CMIRF (S') rocket and is briefly alongside the CMIRF rocket and "almost" stationary in the reference frame of the CMIRF rocket for an infinitesimal period. We don't use the observations of this observer much in the calculations, because extended distances and line of simultaneity are very hard to define in an accelerating reference frame, so that is why the concept of the CMIRF is used.

First of all, so I have the notation right (the right mta with the right observer)?

How is he/she different than the CMIRF observer?

The accelerating observer in the accelerating rocket is non inertial. When the speed of the accelerating rocket and the coasting rocket are matched, their clock are ticking at the same rate. This matched speed only happens for a infinitesimal period as the accelerating rocket catches up with and overtakes the coasting rocket. We use the CMIRF to analyse the accelerating frame because the relative velocities for an instant are almost zero and relativistic effects are almost zero and we can analyse the acceleration using Newtonian equations. Once we have done that it just a question of transforming what the CMIRF observer measures to the the frame of the launch pad observer, using Lorentz transformations.

But, we are getting somewhere.

As you will see later on down, in oder to calculate \gamma^3 I didn't know what velocity to use - the begiining or the end (t = 0, v = 0 or t = 10, v = ?) It seemed like a recursive equation.

You can use F=ma\gamma^3 = m_0 a_0 in any inertial RF. In the launch frame S use the instantaneous velocity of the rocket so after 10 seconds use v=0.99875c. In the CMIRF the velocity of the accelerating rocket after 10 seconds is zero so use v=0 in the same equation. When v=0 the gamma factor effectively disappears. Don't forget that the acceleration measured in S is continually getting smaller by a factor of gamma cubed, so overall the force is constant. You can also use a = a_0/ \gamma^3 in any inertial reference frame, just use the instantaneous velocity of the accelerating object as measured in the IRF you are considering at the instant as measured in the same IRF.

[yuiop]

Now I have to figure out where you got
\gamma = \sqrt{1 + (a't/c)^2} = \frac{1}{\sqrt{1 - v^2/c^2}} from


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


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 [COLOR="red"]Again, what numbers gave you this? Well done. You have found another typo. :tongue: 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.


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? 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. :wink:

[starthaus]

:

{\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}{(1-v^2/c^2)^{3/2}}\, \, + \frac{a}{\sqrt{1-v^2/c^2}} \right)



This is incorrect, the units are all wrong because you missed a c^2 in calculating the derivative.

[Passionflower]

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. :wink:
You can render Excel formulas or hand-typed Latex directly in Excel by using:

http://www.codecogs.com/components/excel_render/excel_install.php

That way you can keep the formulas and the Latex close together!

[Austin0]

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.
.

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

[yuiop]

You can render Excel formulas or hand-typed Latex directly in Excel by using:

http://www.codecogs.com/components/excel_render/excel_install.php

That way you can keep the formulas and the Latex close together!

Thanks Passionflower. Nice tip. :smile:

This is incorrect, the units are all wrong because you missed a c^2 in calculating the derivative.

Thanks Starthaus. Your right. I have fixed the original in post #24 and inserted the missing c^2.

[stevmg]

starthaus -

in your post, #29, where did you get the yuiop quote from (your hyperlink leads to post 24 from yuiop) because in post 24, he has a/c2 which, as you correctly pointed out, makes things come out correctly

F = mdv/dt\gamma^3
= \gamma^3ma

[Passionflower]

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')

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.

[starthaus]

starthaus -

in your post, #29, where did you get the yuiop quote from (your hyperlink leads to post 24 from yuiop) because in post 24, he has a/c2 which, as you correctly pointed out, makes things come out correctly

F = mdv/dt\gamma^3
= \gamma^3ma

He fixed it, see post 32.

[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. :tongue: 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. :wink:

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 :wink:

[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 travelling 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 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 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 :wink:

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 :wink:


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, 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.

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 :wink:

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 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
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 "The Relativistic Rocket" (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!

stevmgWell, acceleration in relativity is not a novice subject, but it is fairly straightforward. :wink:

[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 "The Relativistic Rocket" (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:




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.


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:


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.



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 travelling 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 text books :tongue: (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 "The Relativistic Rocket" (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 travelling 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 thats 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 text books :tongue: (it appears to be perfectly standard in serious online references anyway.)
I dont 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:
http://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:
http://www.physicsforums.com/showpost.php?p=2875235&postcount=1[/QUOTE]

Doc
stevmg

[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] :wink:



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! :tongue: 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 thats 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 dont 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 :tongue:. They are handy (to me anyway), because when you carry out the integration of lets 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 travelling 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 travelling 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 thats 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 lets 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 :approve: 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 dont 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 here (http://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, here (http://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 vN, 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 :tongue:). 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 travelling 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 travelling 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 travelling at greater than the speed of light. Are you confused? You should be. I probably confused myself. Told you things would get complicated :tongue: 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 :tongue:). 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 travelling 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 travelling 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 travelling at greater than the speed of light. Are you confused? You should be. I probably confused myself. Told you things would get complicated :tongue: 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 :tongue:). 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 travelling 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 travelling 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 travelling at greater than the speed of light. Are you confused? You should be. I probably confused myself. Told you things would get complicated :tongue: 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 vN, 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 :tongue:). 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 travelling 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 travelling 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 travelling at greater than the speed of light. Are you confused? You should be. I probably confused myself. Told you things would get complicated :tongue: 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.