Does everything that has to do with velocity

[epkid08]

Given that you have a constant force applied to any mass, what is the rate at which the mass' acceleration decreases as its velocity approaches c?

Does everything that has to do with velocity in some way, have a relativistic transformation i.e. t'=f(t)?

 

[Fredrik]

Given that you have a constant force applied to any mass, what is the rate at which the mass' acceleration decreases as its velocity approaches c?

I would define a constant force as something that causes constant proper acceleration. The world line of an object that has constant proper acceleration is a hyperbola (-t²+x²=constant). I have to go to bed, so I'm not going to work out the details now, but maybe this thread will be helpful. (See the first question I asked and DrGreg's answer).

Does everything that has to do with velocity in some way, have a relativistic transformation i.e. t'=f(t)?

I'm not sure how to interpret that question. There is a coordinate system called an inertial frame associated with each velocity, and everything that happens in spacetime can be described using the coordinates of any inertial frame. A "relativistic transformation" (a Lorentz transformation) is a coordinate change from one inertial frame to another, i.e. if you know how something looks in one inertial frame, you would use a Lorentz transformation to find out how it looks in another.

 

[MeJennifer]

Since force obviously decreases with distance could you explain what you mean when you say you are applying a constant force to a mass that is rapidly moving away from you?

 

[epkid08]

Since force obviously decreases with distance could you explain what you mean when you say you are applying a constant force to a mass that is rapidly moving away from you?

Constant force applied to a mass causes the mass to have a constant acceleration.

 

[DaveC426913]

Since force obviously decreases with distance could you explain what you mean when you say you are applying a constant force to a mass that is rapidly moving away from you?

Yah, I think you're misunderstanding the question. Pretend he's asking about a rocket with constant thrust.

 

[DaveC426913]

Given that you have a constant force applied to any mass, what is the rate at which the mass' acceleration decreases as its velocity approaches c?

Does everything that has to do with velocity in some way, have a relativistic transformation i.e. t'=f(t)?

Here is a http://www.1728.com/reltivty.htm" you can play with and draw your own conclusions.

 

[MeJennifer]

Yah, I think you're misunderstanding the question. Pretend he's asking about a rocket with constant thrust.

In that case the rocket acceleration will never descrease and the speed relative to something else will never be c.

 

[epkid08]

I would define a constant force as something that causes constant proper acceleration. The world line of an object that has constant proper acceleration is a hyperbola (-t²+x²=constant). I have to go to bed, so I'm not going to work out the details now, but maybe this thread will be helpful. (See the first question I asked and DrGreg's answer).I'm not sure how to interpret that question. There is a coordinate system called an inertial frame associated with each velocity, and everything that happens in spacetime can be described using the coordinates of any inertial frame. A "relativistic transformation" (a Lorentz transformation) is a coordinate change from one inertial frame to another, i.e. if you know how something looks in one inertial frame, you would use a Lorentz transformation to find out how it looks in another.

To further my first question, if we have a relativistic transformation of t, t', as well as of x, x', why can't we take \frac{x'}{t'^2}, and have that be a relativistic transformation of a, which then simplifies to

a'=a\sqrt{1-(at/c)^2}?Does the above equation correctly represent the boundary of a constant acceleration as velocity approaches c?

 

[epkid08]

In that case the rocket acceleration will never descrease and the speed relative to something else will never be c.

The acceleration has to decrease as the velocity approaches c.

 

[MeJennifer]

The acceleration has to decrease as the velocity approaches c.

No that is not true at all.

 

[Al68]

Given that you have a constant force applied to any mass, what is the rate at which the mass' acceleration decreases as its velocity approaches c?

by a factor of 1/sqrt[1 +(at/c)^2]. So delta v = at / sqrt[1 + (at/c)^2] where a is proper acceleration (ie force applied/proper mass), t is time in the rest frame in which the mass was initially at rest, and v is relative velocity between that rest frame and the accelerated mass.

Al

 

[russ_watters]

The acceleration has to decrease as the velocity approaches c.

Acceleration is the force you feel when pushed against your chair in an accelerating rocket. This force is always present when the engines are firing. That's one of the main points of special relativity.

 

[DaveC426913]

No that is not true at all.

Explain.

The OP is asking why - if the first hour of his trip gets him from .98c to .99c (delta v=+.01c), but the second hour of his trip gets him from .99c to only .999c (delta v=+.001c) - why that doesn't constitute a decease in acceleration.

 

[epkid08]

Explain.

The OP is asking why - if the first hour of his trip gets him from .98c to .99c (delta v=+.01c), but the second hour of his trip gets him from .99c to only .999c (delta v=+.001c) - why that doesn't constitute a decease in acceleration.

It might not constitute a decrease in acceleration in an objects inertial frame, but I think it obviously constitutes a decrease in observed acceleration of that object, if not, then I'm definitely missing something.

 

[jtbell]

This is the difference between proper acceleration (loosely speaking, what the object itself "feels") and coordinate acceleration in some inertial reference frame.

 

[epkid08]

This is the difference between proper acceleration (loosely speaking, what the object itself "feels") and coordinate acceleration in some inertial reference frame.

Heh, please explain

 

[Fredrik]

The proper acceleration (what the object "feels", what it measures with an accelerometer) is the coordinate acceleration (d^2x/dt^2) in a co-moving inertial frame. So "constant proper acceleration" doesn't mean that d^2x/dt^2 is constant in one particular inertial frame. It means that if we calculate d^2x/dt^2 in the inertial frame that's moving with the same velocity as the object we get the same result at every point on the object's world line. (Note that this means that we're using a different inertial frame at each point).

When the proper acceleration is constant, the coordinate acceleration (in one specific inertial frame) is decreasing.

 

[epkid08]

The proper acceleration (what the object "feels", what it measures with an accelerometer) is the coordinate acceleration (d^2x/dt^2) in a co-moving inertial frame. So "constant proper acceleration" doesn't mean that d^2x/dt^2 is constant in one particular inertial frame. It means that if we calculate d^2x/dt^2 in the inertial frame that's moving with the same velocity as the object we get the same result at every point on the object's world line. (Note that this means that we're using a different inertial frame at each point).

When the proper acceleration is constant, the coordinate acceleration (in one specific inertial frame) is decreasing.

That's what I thought coming into this topic, the acceleration of an object as viewed by an observer decreases as the object's velocity approaches c, but my real question was, at what rate does it decrease? Is this the correct way to show the change from proper acceleration to coordinate acceleration?

a_c=a_p\sqrt{1-(\frac{a_pt}{c})^2}

Or is it more complicated than this? I got that equation using lorentz transformations, solving for x'/(t'^2).

 

[DrGreg]

Acceleration, like almost everything else in relativity, is a relative concept. Although everyone agrees what zero acceleration is, they disagree on the magnitude of a non-zero acceleration.

Thus an acceleration that is constant according the object feeling the acceleration (constant "proper acceleration") would, according to an inertial observer, be a decreasing acceleration.

It turns out, surprisingly, that, for linear motion of a constant mass, the "force" is invariant, i.e. all inertial observers agree on its value. (The appropriate relativistic equation for force is \textbf{f} = d\textbf{p}/dt, not \textbf{f} = m \, d^2\textbf{x}/dt^2.)

Thus a constant force applied to a constant mass in the same direction as its velocity causes constant proper acceleration but decreasing "coordinate acceleration" relative to an inertial observer.

The two accelerations are linked by

a_{proper} = \gamma^3 \, a_{coord}
\gamma = 1/\sqrt{1-v^2/c^2}​

This is proved in this thread, posts #13,#14,#15 equation (12) (with a correction in post #28).

For an example of a force acting on a non-constant mass, consider friction. Friction may cause a body to heat up, and that gain in energy is accounted for as an increase in mass (i.e. rest mass). Another, more obvious, example is a rocket ejecting burnt fuel and therefore losing mass.

_______________

Proof that, for linear motion of a constant mass, force is invariant:

In the notation of the posts I referred to above

p = m \, dx/d\tau = mc \, \sinh\phi
\frac {dp}{d\tau} = mc \, \frac{d\phi}{d\tau} \, \cosh\phi
f = \frac {dp}{dt} = \frac {dp/d\tau}{dt/d\tau} = mc \, \frac{d\phi}{d\tau} = m \, a_{proper}​

 

[epkid08]

I read your proof and understood it, but my question was more directed at the at the equation a_c=f(a_p), not a_p=f(a_c).

Before I go further let me ask you a simple question, in your equation, a_{p} = \gamma^3 \, a_{c}, isn't the velocity equal to \frac{x}{\tau}? So in that equation, you will need to know both proper time and coordinate time in order to execute the function, which doesn't really show \frac{da_c}{da_p}.Back to what I was saying before, here are the relativistic transformations for x and \tau:

x'=x\gamma
t'=\tau\gamma

My question is why can't we use this to calculate coordinate acceleration, in terms of proper velocity/acceleration i.e. solve for x'/t'^2?

a_c = a' = \frac{x'}{t'^2} = \frac{x\gamma}{\tau^2\gamma^2} = \frac{x}{\tau^2\gamma} = a_p\sqrt{1-(\frac{a_p\tau}{c})^2}

 

[DrGreg]

I read your proof and understood it, but my question was more directed at the at the equation a_c=f(a_p), not a_p=f(a_c).

Before I go further let me ask you a simple question, in your equation, a_{p} = \gamma^3 \, a_{c}, isn't the velocity equal to \frac{x}{\tau}? So in that equation, you will need to know both proper time and coordinate time in order to execute the function, which doesn't really show \frac{da_c}{da_p}.


Back to what I was saying before, here are the relativistic transformations for x and \tau:

x'=x\gamma
t'=\tau\gamma

My question is why can't we use this to calculate coordinate acceleration, in terms of proper velocity/acceleration i.e. solve for x'/t'^2?

a_c = a' = \frac{x'}{t'^2} = \frac{x\gamma}{\tau^2\gamma^2} = \frac{x}{\tau^2\gamma} = a_p\sqrt{1-(\frac{a_p\tau}{c})^2}

In my posts, x and t are coordinates of the "stationary" inertial observer, \tau is the proper time of the accelerating particle (and I used x' and t' as coordinates of a co-moving inertial observer moving with speed v_0 = c \tanh \phi_0 relative to the "stationary" observer.) I'm not sure that your notation is the same as mine.

The equations

x'=x\gamma
t'=\tau\gamma​

can't be true whatever your symbols mean.

Maybe you meant

x'=\gamma_0(x - v_0t)
t'=\gamma_0(t - v_0x/c^2)​

in my notation?

 

[DrGreg]

isn't the velocity equal to \frac{x}{\tau}?

No, it's dx/dt (which turns out be be c^2 t / (x + c^2/\alpha)). Do you understand calculus? dx/dt means something very different to x/t.

If you want equations in terms of particle time \tau, you've already got them in the posts in the other thread I referred to earlier. If you want equations in terms of coordinate time t, they are:

x = \frac {c^2}{\alpha} \left[ \sqrt{1 + \left( \frac{\alpha t}{c} \right)^2 } - 1 \right] ...(A)
v = \frac{\alpha t} {\sqrt{1 + \left( \frac{\alpha t}{c} \right)^2 }} ...(B)
\gamma = \sqrt{1 + \left( \frac{\alpha t}{c} \right)^2 } ...(C)
a = \frac{\alpha} {\gamma^3} ...(D)​

where \alpha = a_p is constant proper accceleration, a = a_c is coordinate acceleration.

(A) follows from equations (16) and (17) in my original proof in the other thread (with a change-of-origin so x=0 when t=0), and the rest follows by differentiation.

And if the particle's mass is m,

p = m \alpha t ...(E)
E = \gamma m c^2 ...(F)
f = dp/dt = m \alpha ...(G)​

 

[epkid08]

In my posts, x and t are coordinates of the "stationary" inertial observer, \tau is the proper time of the accelerating particle (and I used x' and t' as coordinates of a co-moving inertial observer moving with speed v_0 = c \tanh \phi_0 relative to the "stationary" observer.) I'm not sure that your notation is the same as mine.

The equations

x'=x\gamma
t'=\tau\gamma​

can't be true whatever your symbols mean.

Maybe you meant

x'=\gamma_0(x - v_0t)
t'=\gamma_0(t - v_0x/c^2)​

in my notation?

I think we had the same notation, I'm just going off about a different way of solving for coordinate acceleration of an object. My idea was to use the Lorentz transforms of (x,t), (x',t'), to find a', the coordinate acceleration. I would do this by taking x', and dividing it by t'^2. I was wondering why this leads to a wrong solving of coordinate acceleration, as it's different than the equation you had posted.

 

[Fredrik]

epkid, you are mistaken about the form of Lorentz transformations. You can't transform the time and space coordinates separately. A Lorentz transformation always mixes them up, unless we're talking about a rotation in space (which doesn't change the time coordinate at all). The general form in 1+1 dimensions (excluding translations, reflections and time-reversal, and using units such that c=1) is

\begin{pmatrix}t' \\ x'\end{pmatrix}=\gamma\begin{pmatrix}1 && -v\\ -v && 1\end{pmatrix}\begin{pmatrix}t \\ x\end{pmatrix}

If you don't like matrices, then look at DrGreg's post again. He said the same thing, but without using matrices.

 

[DrGreg]

I think we had the same notation, I'm just going off about a different way of solving for coordinate acceleration of an object. My idea was to use the Lorentz transforms of (x,t), (x',t'), to find a', the coordinate acceleration. I would do this by taking x', and dividing it by t'^2. I was wondering why this leads to a wrong solving of coordinate acceleration, as it's different than the equation you had posted.

a' = x'/t'^2 would be wrong even for constant coordinate acceleration. You ought to know the equation x' = u't' + a't'^2/2 which has an extra factor of 1/2. But the coordinate acceleration a' is not constant, so you have to use calculus techniques like I did.

Please also remember that the accelerating object is not an inertial observer, so there are no inertial coordinates associated with the object. The best you can do is consider a co-moving inertial observer (x',t') who is momentarily traveling at the same speed as the object. But to examine the whole motion you need an infinite number of different co-moving inertial observer all at different points along the trajectory of the accelerating object. That's where the calculus comes in.

 

[epkid08]

Please also remember that the accelerating object is not an inertial observer

Ahh, this is what I was doing wrong.

 

[JANm]

Most of the answers are about question 2 aren't they. The first question has to do with the mass velocity relation. Mass increases with velocity: m=gamma(v)m_r
In relativity theory gamma = 1/sqrt(1-v^2/c^2), and mass and impulse are increasing with this factor and Kinetic energy = (m-m_r)*c^2.

 

[HallsofIvy]

Since force obviously decreases with distance could you explain what you mean when you say you are applying a constant force to a mass that is rapidly moving away from you?

Where did you get the idea that "force" decreases with distance?

Gravitional force between two masses decreases with distance and electrical force between two charges decreases with distance but those are not all forces.

 

[HallsofIvy]

Force= rate of change of momentum or F= d(p)/dt. In special relativity p= m₀v(1- v²/c²)^-1/2 so dp/dt= (1- v²/c²)^1/2 (dv/dt)- (v²/c²)(1- v²/c²)^-3/2= F. Solve that for dv/dt.

 

[JANm]

dp/dt= (1- v²/c²)^1/2 (dv/dt)- (v²/c²)(1- v²/c²)^-3/2= F. Solve that for dv/dt.

so that is fast! in the first term I miss the mass and in the second dv/dt

 

[JANm]

But H is right you get:

F= dp/dt = gamma^3*m_0*dv/dt

So with increasing velocity gamma is increasing and with constant force dv/dt must decrease!

 

[MeJennifer]

Where did you get the idea that "force" decreases with distance?

Gravitional force between two masses decreases with distance and electrical force between two charges decreases with distance but those are not all forces.

I agree with the gravitational force but the electro-magnetic force, the weak nuclear force and the strong nuclear force are all forces. Hint: why do you think they are called forces in the fist place?

I think you are confused by Newtonian dynamics. In fact Newtonian dynamics by itself does not really deal with forces as macro objects "collide" due to EM forces not real collisions. The difference is obviously irrelevant at the macro level but at the micro level two objects coming very close together do not need to collide to feel a force. And fact is, that force depends on distance.

If I use one magnet to accelerate another magnet the force on the pushed magnet does decrease with distance.

 

[JANm]

The difference is obviously irrelevant at the macro level but at the micro level two objects coming very close together do not need to collide to feel a force. And fact is, that force depends on distance.

If I use one magnet to accelerate another magnet the force on the pushed magnet does decrease with distance.

I'am not so familliar with forces on the microlevel, so how is this dependence on distance there?
secondly
If the magnet repulsion remains constant then Mejennifer let us take a very long tube; place a magnet at the one end and put in a magnetic canon-ball.

 

[MeJennifer]

I'am not so familliar with forces on the microlevel, so how is this dependence on distance there?
secondly
If the magnet repulsion remains constant then Mejennifer let us take a very long tube; place a magnet at the one end and put in a magnetic canon-ball.

The point is that it does not remain constant over distance.

 

[JANm]

Dear MeJ
You are answering none of my two questions and discussing the starter of this thread.
Suppose F constant: WHAT IF.
This question has resulted in the explaining of Impulse as a function of velocity and acceleration at high velocities. Surely a micro-physics knowledger as you must know that it is an important question whether velocity has an upper limit.

 

[Austin0]

Hi I havr followed this thread with interest and have a question.
Isnt it true that as v => c
Mass => infinite
Processes/time =>0
Force/energy required for acceleration => infinite ?

This would seem to mean that in this universe both proper accceleration and coordinate acceleration would slow to a halt. Otherwise you have Zenos Rocket perpetually accelerationg off down the rabbithole.
Or am I just taking this discussion to literally?

 

[JesseM]

The point is that it does not remain constant over distance.

Imagine you are driving an object forward by aiming a constant stream of pellets at its backside, and imagine taking the limit as the spacing between the pellets (and their individual mass) goes to zero so the stream is applying a continuous force rather than a discrete series of impulses. Using this method it would certainly be possible in principle to apply a force to accelerate the object which is constant in our own rest frame, although it would not appear constant in the instantaneous inertial rest frame of the object from one moment to another.

 

[HallsofIvy]

IF it were possible to apply a constant force to an object then, since dp/dt= d(mv)dt= m(dv/dt)+ (dm/dt)v= F, you would have m(dv/dt)= F- (dm/dt)v so dv/dt= (F- dm/dt)/m. And since "m" increases without bound as v would increase toward c as a limit.

 

[MeJennifer]

Imagine you are driving an object forward by aiming a constant stream of pellets at its backside, and imagine taking the limit as the spacing between the pellets (and their individual mass) goes to zero so the stream is applying a continuous force rather than a discrete series of impulses. Using this method it would certainly be possible in principle to apply a force to accelerate the object which is constant in our own rest frame, although it would not appear constant in the instantaneous inertial rest frame of the object from one moment to another.

I assume it is clear that I would have to accelerate as well to achieve that. But think of the impossibility of Born rigidy in this context, since what you are describing is nothing else than the acceleration of a rod from the back.

 

[JesseM]

I assume it is clear that I would have to accelerate as well to achieve that.

Why would you have to accelerate? The gun firing the pellets can have a single fixed rest frame (though it might have to steadily increase the velocity which with it fires successive pellets in order to achieve a constant force on the object, if that's what you mean).

But think of the impossibility of Born rigidy in this context, since what you are describing is nothing else than the acceleration of a rod from the back.

If we're just discussing the fact that the force needed to accelerate an object to c in finite time would be infinite, I don't see why it's necessary to impose the condition of Born rigidity at all.

Anyway, Born rigidity is an idealization, but I would guess that approximate Born rigidity is achieved "naturally" as an equilibrium of all the forces between atoms in a solid object when you push the object from the back (since each atom is only directly being accelerated by the force from nearby atoms, when you push an object's back you aren't directly applying a force to atoms far from the point you're pushing, only indirectly since the atoms next to the point being pushed then apply a force to farther atoms, which apply their own force to farther atoms, etc.--so while the initial push will at first cause a compression wave which distorts the shape of the object, if the external force is constant eventually an equilibrium should be reached where each atom is being accelerated at a constant rate by nearby atoms.) My reasoning is that the equivalence principle says that a small object sitting on a platform on the Earth should behave like the same object sitting on a platform which is accelerating at 1G in empty space, with gravitational time dilation between clocks at the top and bottom of the object being equivalent to the time dilation of clocks at front and back of an accelerating object that satisfies the Born rigidity condition (see this post). If the object being pushed from the back at 1G did not "naturally" achieve an equilibrium distribution of forces such that the front was being accelerated at a lesser rate, then you wouldn't have the same time dilation between clocks at "top" and "bottom" of the object as you do with clocks at the top and bottom of an object resting on Earth, and an observer standing on the ground next to the object could determine whether he was on a gravitating body or an accelerating platform in deep space, which would seem to violate the equivalence principle.

 

[MeJennifer]

If we're just discussing the fact that the force needed to accelerate an object to c in finite time would be infinite, I don't see why it's necessary to impose the condition of Born rigidity at all.

That is not a fact at all, you are mistaken. The Einstein velocity addition has no limit. I suggest you read up on limits and hyperbolic functions.

 

[JesseM]

That is not a fact at all, you are mistaken. The Einstein velocity addition has no limit.

I'm not sure what "the Einstein velocity addition has no limit" means in this context. No limit to what equation as what variable approaches what value? And what would this have to do with my statement "the force needed to accelerate an object to c in finite time would be infinite"? It's certainly true that if we imagine applying a constant proper acceleration to an object for some finite time (time as measured in the inertial frame where the object is at rest initially), then in the limit as the acceleration goes to infinity, the object's final speed at the end of the time interval approaches c. You can also draw an "impossible" worldline which reaches c in some finite time, like v(t) = (c/1 year)*t (where t and v are defined in terms of a particular inertial frame, so the coordinate velocity reaches c at t=1 year), and then calculate the proper acceleration at any point on this worldline--the thing that makes it impossible is that in the limit as t approaches 1 year, the proper acceleration approaches infinity.

 

[MeJennifer]

It's certainly true that if we imagine applying a constant proper acceleration to an object for some finite time (time as measured in the inertial frame where the object is at rest initially), then in the limit as the acceleration goes to infinity, the object's final speed at the end of the time interval approaches c.

Indeed it approaches c but never reaches c and that is what you claimed. The limit is not c as the limit does not exist.

 

[JesseM]

Indeed it approaches c but never reaches c and that is what you claimed. The limit is not c as the limit does not exist.

It is standard when talking about limits to say that the function "approaches" a certain value in the limit as some variable "approaches" another value (if a mathematician wants to verbally describe the symbols \lim_{x\rightarrow a} f(x), they'll use words like 'the limit of f(x) as x approaches a'). c satisfies the formal definition of a limit in this case:

A function f(z) is said to have a limit \lim_{z\rightarrow a} f(z) = c if, for all \epsilon > 0, there exists a \delta > 0 such that | f(z) - c | < \epsilon whenever 0 < | z - a | < \delta.

If in this case we replace z with t so the function f(t) is the instantaneous velocity of the object accelerating with velocity given by f(t) = (c/1 year)*t, then it's true that for any velocity you choose which is less than c by some tiny amount epsilon, it's possible to find a delta such that at the time (1 year - delta) the velocity given by f(t) becomes larger than (c - epsilon). Just name any epsilon, I'll give you a delta! For example, say you chose epsilon = 0.0001c, so that c - epsilon was equal to 0.9999c. Then I could pick any delta smaller than 0.0001 years, say delta=0.00009 years, and f(t) would be closer to c than 0.9999c at time f(1 year - delta). This is all that's required for f(t) to have the limit c as t approaches 1 year, that the function f(t) gets arbitrarily close to the value of c as t gets arbitrarily close to 1 year--if you disagree, you need to review your calculus.

 

[MeJennifer]

JesseM, it is clear to me that your teacup is full, just keep believing that massive objects can reach c in the limit.

Apparently you do not want to learn that some functions do not have limits.

 

[JesseM]

JesseM, it is clear to me that your teacup is full, just keep believing that massive objects can reach c in the limit.

I don't believe it is physically possible, since it would require infinite energy and infinite proper acceleration. But as a purely mathematical matter it is certainly true that the function v(t) = (c/1 year)*t approaches c in the limit as t approaches 1 year, and it is also true that the only reason this function cannot describe a physical worldline is because certain physical quantities (like energy and proper acceleration) go to infinity as t approaches 1 year. This is why I said that if we imagine that infinite energy or proper acceleration were physically possible, then objects could reach c in finite time (hence my statement 'the force needed to accelerate an object to c in finite time would be infinite' which you objected to). But it isn't physically possible, so in the real universe they can't.

Apparently you do not want to learn that some functions do not have limits.

I am not sure what bizarre misreading of my posts could lead you to think that I am denying the statement "some functions do not have limits". Of course some functions don't have limits. The velocity function v(t) = (c/1 year)*t does have a well-defined limit of c in the limit as t approaches 1 year, however (but the different function that describes proper acceleration as a function of time for a worldline with that velocity function would not have any finite limit as t approaches 1 year, since proper acceleration goes to infinity in the limit as t approaches 1 year).

It would really help if you would give some mathematically precise statement of what you are arguing instead of speaking in nebulous generalities. What specific function that doesn't have a limit do you think is relevant to showing I am wrong when I say "the force needed to accelerate an object to c in finite time would be infinite"? Or do you not have any clear idea of what you are arguing yourself?

 

[JANm]

IF it were possible to apply a constant force to an object then, since dp/dt= d(mv)dt= m(dv/dt)+ (dm/dt)v= F, you would have m(dv/dt)= F- (dm/dt)v so dv/dt= (F- dm/dt)/m. And since "m" increases without bound as v would increase toward c as a limit.

As I claimed earlier HallsofIvy: the mass is asumed to have relation to velocity, so dm/dt has a dv/dt in it and it is not fair to isolate this dv/dt this way!