Asymptotic velocity under constant force?

1. Feb 20, 2015

Timothy S

[Moderator's note: thread title edited for greater clarity]

If you apply a constant force to an object, will it eventually begin to asymptote towards a specific velocity? My reasoning is that with special relativity we can re-write f = ma as F =(m0/(1 - v2/c2)1/2)a.

Because of this equation, as the mass is accelerated due to the force, the object's mass increases. This makes the acceleration due to the force applied smaller.

Am i correct in my thinking?

Last edited by a moderator: Feb 21, 2015
2. Feb 20, 2015

Staff: Mentor

Yes; that velocity is $c$, as should be evident if you analyze the formula you gave.

3. Feb 20, 2015

bcrowell

Staff Emeritus
Yes, as the object's velocity approaches c, its acceleration approaches zero. Therefore there is a limiting velocity, and the limiting velocity is c.

Note that the modern convention is to treat mass as invariant, and absorb the factor of gamma into Newton's second law rather than considering it part of the mass. See http://physics.stackexchange.com/a/133395/4552 .

4. Feb 20, 2015

pervect

Staff Emeritus
Your conclusion is correct, the details of the argument are a bit suspect.

force can be defined as the rate of change of momentum with respect to time, and the relativistic momentum is equal to $m_0 \, v \, / \sqrt{1 - v^2/c^2}$

It's unclear to me if you are thinking a constant "proper force" (which would be the derivaitve of the relativistic momentum with respect to proper time), or a constant force as measured in some particular inertial frame (which would be the derivative of the relativistic momentum with respect to the coordinate time of the chosen frame). If you meant the former, there's a brief discussion of the resulting equations of motion, known as hyperbolic motion, in
http://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity)

In the later case (constant force), you take the derivative with respect to coordinate time. What is correct is to say that f = dp/dt and p = $m_0 \, v / \sqrt{1-v^2/c^2}$. You need to apply the rules of calculus as to how to take the derivative of the quotient $v / \sqrt{1-v^2/c^2}$ correctly. This is called the "quotient rule", see for instance https://www.math.hmc.edu/calculus/tutorials/quotient_rule/. If you are familiar with calculus this should get you on the right track. (You'd also need the chain rule, to write dp/dt as (dp/dv) (dv/dt).) If you're not familiar with calculus (I don't know your background), all I can really say is that the result of this process is NOT $m_0 / \sqrt{1-v^2/c^2}$.

5. Feb 21, 2015

DrStupid

At least not for a constant force. There are special cases where Timothy S' formula applies but they don't work with constant force and cannot lead to an asymptotic velocity.

6. Feb 21, 2015

Staff: Mentor

More precisely, they don't work with constant force in the original rest frame. Yes, that's true. I had interpreted the OP as meaning constant proper force, i.e., constant force felt by the object, but as pervect pointed out, that is not the only possible interpretation.

This part is incorrect. A constant force in the original rest frame will still lead to a velocity that asymptotes to $c$; what will be different is the specific functional form of the velocity vs. time.

7. Feb 21, 2015

pervect

Staff Emeritus
I don't see why you say that. Most (though by no means all) situations with a continuously applied force which accelerates an object are going to make the object's asymptotic velocity equal to c. The velocity will be constantly increasing, and it can never get above c. The only thing that remains is demonstrate that the asymptotic velocity is not lower than c - it can never be above c!

Letting t be coordinate time and tau be proper time, when dp/dt is constant, dp/dtau will be (dp/dt) (dt/dtau) = (dp/dt) * gamma = some constant * gamma, which will be constantly increasing as gamma increases with time. So the "constant force" case (dp/dt = constant) should have an increasing proper acceleration (dp/dtau, which is also the "felt" acceleration, the acceleration measured by an onboard accelerometer). This means the constant force case will always be accelerating harder than the constant proper force case, and implies that there is "enough force" to make the asymptotic velocity equal to c.

Note that the textbook case of constant proper acceleration has the property of an asymptotic worldline (this is different than an asymptotic velocity). So with a constant proper acceleration, you'll asumptotically approach some particular worldline as you accelerate, this worldline happens to be light-like. There will be some specific light-beam that will never reach you, but that will get closer and closer.

With the increasing acceleration of a "constant force", I'm unclear about the existence of an asymptotic worldline - rather than guess I'll leave it to someone motivated enough to work it out, if there is sufficient interest (I'm not sure there is).

8. Feb 21, 2015

Staff: Mentor

I'm not sure I understand the difference, unless you're just saying that the term "asymptotic velocity" is awkward. If the worldline asymptotes to a lightlike worldline, then the velocity asymptotes to $c$. (If the worldline asymptotes to some timelike worldline, then the velocity will asymptote to some value less than $c$.)

9. Feb 21, 2015

maline

Timothy, you are assuming that F=ma, using the new "mass". But that law assumes a constant mass. Force means rate of change of momentum, which is the product m0v(gamma) so at high speeds most of the force goes into increasing the gamma factor.
Pervect, isn't it true that a constant force (defined using coordinate time) causes a constant proper acceleration, leading to the textbook hyperbolic motion?

Last edited: Feb 21, 2015
10. Feb 21, 2015

DrStupid

Show me a case with asymptotic velocity where the formula applies.

then Timothy S' formula doesn't apply (except for dp/dt=0). If it applies than pd/dt is not constant or zero. The assumptions of the OP are mutually exclusive.

11. Feb 21, 2015

Staff: Mentor

More precisely, they're ambiguous. It's not made clear what "constant force" means (constant proper force, or constant force in the original rest frame?), and it's not made clear whether $a$ in his formula is supposed to be a constant.

I think I was misunderstanding your statement. I thought you were saying that a constant force in the original rest frame could not lead to an asymptotic velocity. It does. But you're right that the OP's formulas don't apply to that case; as above, they can't be applied at all without clarifying what "constant force" means, and if it means "constant force in the original rest frame", then the OP's formulas are not correct.

12. Feb 21, 2015

maline

Isn't it true that constant force as measured in an IRF leads to constant proper acceleration and hyperbolic motion?

13. Feb 21, 2015

Staff: Mentor

Not if by "measured in an IRF" you mean "measured in a fixed IRF". Constant force as measured in the object's momentarily comoving inertial frame leads to constant proper acceleration and hyperbolic motion. But the object's MCIF changes as it accelerates.

14. Feb 21, 2015

Staff: Mentor

I got curious enough to work it out, so here it is.

[Edit: The below turned out to be incorrect; see the follow-up post.]

The key thing is that, if we write the 4-acceleration of the object for this case, it must have constant components in the original rest frame of the object. This is because 4-acceleration is 4-force divided by rest mass (i.e., in the 4-vector formalism we don't have to worry about "relativistic mass"), so a constant 4-force means a constant 4-acceleration. (I'll do the constant proper acceleration case in a follow-up post for comparison; for that case, the 4-force is not constant in the original rest frame.)

What are those constant components of the 4-acceleration? The other key thing that tells us that is that, in the original rest frame, the 4-acceleration should be purely spatial--i.e., it should have a zero "time" component. A frame-independent way of stating this is that the 4-acceleration is orthogonal to the worldlines of observers at rest in the original rest frame (whereas for the case of constant proper acceleration, the 4-acceleration is orthogonal to the worldline of the accelerated object).

So we now know what form the 4-acceleration takes; it is simply $a^a = (0, a)$ (where I am considering only a single spatial dimension, and vector components are listed in $(t, x)$ order, "time" then "space"), where $a$ is a constant. All we have to now is integrate; the 4-acceleration is $d u^a / d \tau$, where $u^a$ is the 4-velocity and $\tau$ is proper time (we'll get to coordinate time shortly, don't worry), so the 4-velocity is given by

$$u^x = \int a^x d\tau = a \tau$$

$$u^t = \sqrt{1 + ( u^x )^2}$$

where the second equation just uses the fact that the 4-velocity is a unit timelike vector. So we have $u^a = (\sqrt{1 + a^2 \tau^2}, a \tau)$.

The 3-velocity of the object, in the original rest frame, is then just $u^x / u^t$, or

$$v = \frac{a \tau}{\sqrt{1 + a^2 \tau^2}}$$

It is obvious from the above that $v \rightarrow 1$ as $\tau \rightarrow \infty$. As we'll see in a moment, $\tau \rightarrow \infty$ is equivalent to $t \rightarrow \infty$, so this proves that $v$ asymptotes to $c$ (in conventional units).

We can integrate once more to get the spacetime position $x^a$:

$$x^a = \left( \frac{1}{2a} \left[ a \tau \sqrt{1 + a^2 \tau^2} + \ln | a \tau + \sqrt{1 + a^2 \tau^2} | \right], \frac{1}{2} a \tau^2 \right)$$

The first component is $t$, and it goes to $\infty$ as $\tau \rightarrow \infty$, as I said above. The second component is $x$. I'll defer finding an expression for $x(t)$ and assessing whether there is an asymptotic worldline to a follow-up post.

Last edited: Feb 23, 2015
15. Feb 21, 2015

Staff: Mentor

Following up first with a brief treatment of the constant proper acceleration case, for comparison. Of course we already know how the solution looks for this case, but it's interesting to work through it using the same method I used in the previous post for the case of constant force in the original rest frame.

Here it is easiest to start with the 4-velocity $u^a$, which we know must have components $(\gamma, \gamma v)$ in the original rest frame. The key thing about the 4-acceleration for this case is that, as I mentioned in the previous post, it must be orthogonal to $u^a$. So we must have $\eta_{ab} u^a a^b = 0 = \gamma a^t - \gamma v a^x$. This reduces to $a^t = v a^x$.

The other key thing is that the magnitude of the 4-acceleration must be constant, because if $a^a$ is orthogonal to $u^a$, then the magnitude of $a^a$ is the proper acceleration felt by the object. (Note carefully that this is only true if the two vectors are orthogonal; in the case in the previous post, the 4-acceleration also has constant magnitude, but since it's not orthogonal to the 4-velocity of the object except at the instant $\tau = 0$, the proper acceleration felt by the object is not constant.) So we must have $\left( a^t \right)^2 - \left( a^x \right)^2 = - a^2$, where $a$ is the constant proper acceleration and the squared magnitude is negative because $a^a$ is a spacelike vector. Substituting from above, this gives $a^2 = \left( a^x \right)^2 \left( 1 - v^2 \right)$, or $a^x = \gamma a$. So we have

$$a^a = a \left( \gamma v, \gamma \right)$$

So now, since we know that $a^a = d u^a / d \tau$, we can write:

$$\frac{d \gamma}{d \tau} = a \gamma v$$

$$\frac{d \left( \gamma v \right)}{d \tau} = v \frac{d \gamma}{d \tau} + \gamma \frac{dv}{d\tau} = a \gamma$$

Combining these gives $dv / d\tau = a \left( 1 - v^2 \right)$, or

$$\int \frac{dv}{1 - v^2} = \tanh^{-1} v = a \tau$$

This gives the familiar result $v = \tanh \left( a \tau \right)$, which we know describes a hyperbolic worldline with proper acceleration $a$. It is easy to confirm that $\gamma = \cosh \left( a \tau \right)$ and $\gamma v = \sinh \left( a \tau \right)$. So we can write

$$a^a = a \left( \sinh \left( a \tau \right), \cosh \left( a \tau \right) \right)$$

$$u^a = \left( \cosh \left( a \tau \right), \sinh \left( a \tau \right) \right)$$

Integrating $u^a$ with respect to $\tau$ gives the position $(t, x)$:

$$x^a = \frac{1}{a} \left( \sinh \left( a \tau \right), \cosh \left( a \tau \right) \right)$$

We can easily confirm that, as expected, $x^2 - t^2 = 1 / a^2$, so that we do indeed have a hyperbolic worldline. That also allows us to write

$$x(t) = \frac{1}{a} \sqrt{1 + a^2 t^2}$$

Computing $dx / dt$ gives an interesting expression for $v$ in terms of $t$:

$$v(t) = \frac{a t}{\sqrt{1 + a^2 t^2}} = \frac{t}{x}$$

Notice that this expression is identical in form to the expression for $v$ as a function of $\tau$ in the previous example.

We can also compute $dv / dt$ to get an expression for the coordinate acceleration:

$$a(t) = \frac{a}{\left( \sqrt{1 + a^2 t^2} \right)^3}$$

Finally, we note that comparing the expression for $x(t)$ above with the expressions for $x^a$ and $u^a$ gives $\gamma = \cosh \left( a \tau \right) = \sqrt{1 + a^2 t^2}$. So we can write our three functions above compactly as $x(t) = \gamma / a$, $v(t) = a t / \gamma$, and $a(t) = a / \gamma^3$.

Last edited: Feb 22, 2015
16. Feb 22, 2015

maline

The equation for hyperbolic motion is x2=c2t2+c42, where x and t are the coordinate position & time, α is the constant proper acceleration, and x0=c2/α.
This gives v(t)=dx/dt=ct/(t2+c22)1/2

γ(t)=(α/c)(t2+c22)1/2

p(t)=mv(t)γ(t)=mαt

F=dp/dt=mα=constant force in original IRF
Am I mistaken here?
Note that I'm not discussing 4-force, just dp/dt. This was what Pervect mentioned in the post I originally commented on.

Last edited: Feb 22, 2015
17. Feb 22, 2015

Staff: Mentor

We're using different definitions of "force". See below.

The notation here is ambiguous; does $p$ mean the 4-momentum, or just the ordinary 3-momentum? The context of pervect's post makes it clear that he means 4-momentum, which means $dp / d\tau$ is the 4-force. It should more correctly be written as $d p^a / d \tau$ to make it clear that $p$ is a 4-vector. In the case of constant proper force, i.e., constant force felt by the object, we have $d p^a / d \tau$ constant (more precisely, its magnitude is constant, as I discuss in post #15); but $d p^a / dt = \left( 1 / \gamma \right) d p^a / d\tau$ is not constant.

What you calculated in your post was the 3-momentum $\vec{p}$ (we are only considering one spatial dimension so the vector symbol can be left off), which is indeed $p = m a t$, so $dp / dt$ is indeed constant. But $dp / dt$ is just the spatial component of $dp^a / dt$. The time component of $dp^a / dt$ is the rate of change of the object's energy, which is $E = m \gamma = m \sqrt{1 + a^2 t^2}$; the time derivative of this is $dE / dt = m a v$, which increases as the object accelerates.

Note also that if we use the OP's definition of "force" as $\vec{F} = m \gamma \vec{a} = E \vec{a}$, then this is not constant either. If we interpret $\vec{a}$ as the proper acceleration, then this "force" increases as the object accelerates; if we interpret $\vec{a}$ as the coordinate acceleration, i.e., as $a / \gamma^3$ where $a$ is the proper acceleration (see post #15), then we have $F = m a / \gamma^2$, which decreases as the object accelerates.

18. Feb 23, 2015

Staff: Mentor

On thinking it over, I realized that this won't work. The 4-acceleration has to be orthogonal to the 4-velocity, because the 4-velocity's length can't change; it has to be a unit vector. Only its direction in spacetime can change. I'll need to rethink this analysis.

19. Feb 23, 2015

PAllen

I have some questions about Peter's #14 and #15. Much of this is likely terminology. I normally use 4-force as derivative of 4-momentum by proper time, 4 acceleration as derivative of 4-velocity by proper time. If this 4-force has constant in magnitude, and mass (rest) doesn't change, then proper acceleration = magnitude of 4-acceleration is constant. This all describes the case of constant force/acceleration as measured in momentary rest frame of the accelerating object (whatever coordinates you use for computation). This is what #15 is about.

However, to distinguish #14, I would think the difference would be that for constant force as measured in a starting rest frame, the force is derivative of 4-momentum by coordinate time rather than proper time. It obviously does not follow that coordinate acceleration is constant. Further, 4-acceleration, as I use it, is not constant in magnitude. Then, you have a choice what to make constant here. You could posit constancy of this coordinate 4-force magnitude. I do not think that is a physically meaningful quantity. Constancy of the spatial part is meaningful: it is constancy of force as measured in the starting rest frame. From this, I get a different result than Peter. I also don't get his argument about the time component of 4-acceleration being zero (except at the start). Once the 4-velocity of the object has a nonzero spatial component, its 4-acceleration must have a nonzero time component. And 4-acceleration of an object must always be orthogonal to its 4-velocity, no exceptions as far as I know.

20. Feb 23, 2015

PAllen

I believe maline's #16 is correct for showing that constant force as measured in the original rest frame produces constant proper accleration (I got this result by a much more laborious, direct, route of setting the general expression for relatvistic 3-force (dp/dt constant, with p being relativistic 3-momentum). Further, one can see that this is expected from analysis in Peter's #15:

a(γv,γ)

is derived as 4 acceleration with 'a' being the proper acceleration. Then, 4 force is m times this. The derivative by t of instead of tau divides this by gamma, leading to ma as the 3 force (with 'a' as proper acceleration not coordinate acceleration!). Note, coordinate acceleration (in this collinear case) is simply proper acceleration divided by gamma cubed. [ Also, note that the magnitude of 4-force is ma as expected, while the magnitude of coordinate 4-force is ma/γ.]

Last edited: Feb 23, 2015