Relativistic rocket equation

• B
Gold Member
Hi guys.

I am studying the relativistic implementation of the standard Tsiolkosvky rocket equation, but ran into some doubts.

Doesn't special relativity apply only to inertial frames of reference(non-accelerated)? Then, how could it be used for the rocket equation, which speaks of a ##\Delta v##, of course caused by an acceleration?

Moreover, could someone give me a shallow and brief(since I haven't actually studied it, nor have the mathematical knowledge to fully get it) explanation of the hyperbolic motion equation, which takes to this ##t'=\frac{c}{a}sinh(\frac{at}{c}## relation between proper and dilated time?

Staff Emeritus
Doesn't special relativity apply only to inertial frames of reference(non-accelerated)?
No.

Mentor
Doesn't special relativity apply only to inertial frames of reference(non-accelerated)?
As @Vanadium 50 has already commented, no, it doesn't. It applies to flat spacetime, which is the underlying assumption used to derive the relativistic rocket equation. It is in no way restricted to inertial frames. People thought it was at the time Einstein first published it, but that was 116 years ago and we've learned a lot since then.

could someone give me a shallow and brief(since I haven't actually studied it, nor have the mathematical knowledge to fully get it) explanation of the hyperbolic motion equation, which takes to this ##t'=\frac{c}{a}sinh(\frac{at}{c}## relation between proper and dilated time?

https://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html

greg_rack
2022 Award
Doesn't special relativity apply only to inertial frames of reference(non-accelerated)?
As Peter says, SR is not restricted to inertial frames. But even if it were, you can describe accelerating objects in inertial frames. You do it all the time in Newtonian physics - that's what the SUVAT equations do.

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greg_rack
Mentor
Doesn't special relativity apply only to inertial frames of reference(non-accelerated)?
As everyone else has aready said... No, SR works just fine as long as the spacetime is flat, meaning no significant tidal gravitational effects. Indeed, it's called "special" relativity because it only works in the special case of flat spacetime, while general relativity works for all spacetimes whether flat or not.

Your misunderstanding is common though, because the math involved is appreciably more complicated - when working with non-inertial frames there's no way of avoiding calculus and the hyperbolic functions you asked about. Many introductory texts don't want to assume that the student has the necessary mathematical background so don't go there. This leaves the student with the idea that SR only works with intertial motion, when what's really going on is that that's all their text covered.

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SolarisOne, greg_rack and berkeman
Gold Member
Your misunderstanding is common though, because the math involved is appreciably more complicated - when working with non-inertial frames there's no way of avoiding calculus and the hyperbolic functions you asked about. Many introductory texts don't want to assume that the student has the necessary mathematical background so don't go there. This leaves the student with the idea that SR only works with intertial motion, when what's really going on is that that's all their text covered.
That's exactly the case! So excited about studying university physics... tired of high-school's shallowness :)
Thank you so much

Gold Member
Mentor
how does it switch from hyperbolic functions definition to their implementation into the rocket equations?
Hm, I thought I remembered that being briefly covered in the article, but I see it isn't. There are two ways to do it:

(1) You can apply conservation of momentum in the momentarily comoving inertial frame. When you do this in Newtonian physics, you get the Newtonian rocket equation; when you do it in relativity, you get the relativistic rocket equation.

(2) You can compute the worldline of an object with constant proper acceleration (i.e., path curvature) ##a## in Minkowski spacetime. The ##t## and ##d## in the article are just the ##t## and ##x## coordinates of the worldline, in the frame in which the rocket is initially at rest, and ##v## in the article is just ##dx / dt## in that frame. The ##T## in the article is the proper time along the worldline. The fact that the path curvature is constant is what makes the worldline a hyperbola (it's the Minkowski analogue to a circle, which is the curve of constant curvature in Euclidean space), which in turn is what brings in the hyperbolic trig functions (for a circle the functions would be the corresponding ordinary trig functions).

vanhees71
Staff Emeritus
I honestly didn't fully get it... how does it switch from hyperbolic functions definition to their implementation into the rocket equations?
A detailed explanation is not necessary, it would be enough to roughly get the concept behind it
See the section entitled "How much fuel is needed". I think that it should answer your question, if not I probably don't understand the question :(.

Gold Member
[Moderator's note: This was posted as a separate thread, but it is actually the same question as the relativistic rocket equation, so the threads have been combined.]

Hi guys.

Assume we want to calculate the variation in velocity of an accelerated frame with respect to one at rest, using Lorentz transformations, as a function of the rest frame time ##t'## and of the acceleration.

This should be the resulting equation:
$$\Delta v=\frac{at'}{\sqrt{1+(\frac{at'}{c})^2}}$$
but how is it derived, making use only of ##x'=\gamma(x-Vt)## and ##t'=\gamma(t-\frac{V}{c^2}x)##?

Gold Member
Are you familiar with 4-velocity and then 4-acceleration ? Then it comes easily.

greg_rack
Gold Member
Are you familiar with 4-velocity and then 4-acceleration ? Then it comes easily.
Hm no, I don't think I've heard it before

You can think of the four velocity as the rate of change of the position of the particle in ##\mathbf{R}^4## with respect to proper time along the worldline ##U = \dfrac{1}{c} \dfrac{dX}{d\tau}##; it satisfies ##U \cdot U = -1##. Similarly, the four acceleration is ##A = \dfrac{1}{c} \dfrac{dU}{d\tau}##; we will write ##A \cdot A = a^2##.

Consider a particle accelerating along the ##x##-axis of some inertial reference system from rest (i.e. ##u^x(0) = 0##), such that ##U = u^t \mathbf{e}_t + u^x \mathbf{e}_x## and ##A = a^t \mathbf{e}_t + a^x \mathbf{e}_x##. Then\begin{align*}
U \cdot U &= -(u^t)^2 + (u^x)^2 = -1 \\
A \cdot A &= -(a^t)^2 + (a^x)^2 = a^2 \\ \\

\implies a^x &= \sqrt{a^2 + (a^t)^2}
\end{align*}We also have that ##u^t = \sqrt{1+(u^x)^2}## hence
\begin{align*}
a^t = \frac{1}{c} \frac{du^t}{d\tau} = \frac{1}{c} \frac{u^x}{\sqrt{1+(u^x)^2}} \frac{du^x}{d\tau}
\end{align*}Finally write\begin{align*}
A \cdot A = -\left( \frac{1}{c} \frac{u^x}{\sqrt{1+(u^x)^2}} \frac{du^x}{d\tau}\right)^2 + \left( \frac{1}{c} \frac{du^x}{d\tau} \right)^2 &= a^2 \\ \\

\implies \int \frac{1}{\sqrt{1+(u^x)^2}} \frac{du^x}{d\tau} d\tau &= \int ca d\tau \\ \\

\implies u^x &= \sinh{(ac\tau)}

\end{align*}and it follows that ##u^t = \sqrt{(u^x)^2 + 1} = \cosh{(ac\tau)}##. By differentiating you can determine ##A##, and we have finally the results\begin{align*}
U &= \cosh{(ac\tau)} \mathbf{e}_t + \sinh{(ac\tau)} \mathbf{e}_x \\
A &= a\sinh{(ac\tau)} \mathbf{e}_t + a\cosh{(ac\tau)} \mathbf{e}_x \\

\end{align*}Remember that ##U = \dfrac{1}{c} \dfrac{dX}{d\tau}## so we can integrate the components of ##U## to find that\begin{align*}
X^t &= \dfrac{1}{a} \sinh{(ac\tau)} \\

X^x &= \dfrac{1}{a}\left( \cosh{(ac\tau)} - 1 \right)
\end{align*}and you may check with the standard hyperbolic identity that\begin{align*}
(aX^x + 1)^2 - (aX^t)^2 = 1
\end{align*}This you can re-write as\begin{align*}
X^x = \dfrac{1}{a} \left( \sqrt{1+(aX^t)^2} -1\right)
\end{align*}which is then easily differentiated for\begin{align*}

\frac{dX^x}{dX^t} = \frac{aX^t}{\sqrt{1+(aX^t)^2}}

\end{align*}It's more conventional to use ##X^t = ct## and ##X^x = x##, so that e.g. ##\dfrac{dx}{dt} = \dfrac{ac^2t}{\sqrt{1+(act)^2}}##.

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JD_PM, anuttarasammyak, vanhees71 and 1 other person
Gold Member
@PeroK the point of asking such a question is that I'm doing just really basic, shallow SR in school, but I'd wish to know more in order to study on my own some concepts of astrodynamics, which profoundly fascinates me and on which I'd like to provide a presentation to my class, before the final high-school exam.

It may be textbook stuff, but of textbooks I do not have nor know, surely not thought for my level, and which I wouldn't be able to understand properly with my current mathematical skillset.
I don't see why you point that out as being incorrect... moreover, here, I can establish a direct dialogue with the answerer in case some passages aren't clear.
Your kindness, reading all your answers, getting the different ways of thinking behind them, is helping me a whole lot to mature under many points of view; isn't it part of the power of PF, for newbies like me, who still strive to get a bit more advanced concepts?
Why shouldn't I get the most out of this?

Mentor
unless you are studying SR seriously what's the point of posting a question like this for someone else to post what must be either in your textbook or set as an exercise?
It looks like the OP is studying SR seriously, but doesn't have easy access in school to helpful resources. This thread seems to me to be perfectly within bounds for PF discussion.

The only potential issue I can see would be whether this thread should be in the homework forum, but I don't think that's necessary for this particular question.

greg_rack and vanhees71
Mentor
It may be textbook stuff, but of textbooks I do not have nor know
If you can get hold of a copy of Taylor & Wheeler's Spacetime Physics, I would recommend doing so; it is the best introductory textbook on SR that I know of, but unfortunately is not used as an actual textbook nearly as often as it should be. It looks like an electronic version is now available for free download:

https://www.eftaylor.com/spacetimephysics/

greg_rack and vanhees71
Mentor
Why shouldn't I get the most out of this?
You definitely should.

vanhees71
Mentor
how is it derived, making use only of ##x'=\gamma(x-Vt)## and ##t'=\gamma(t-\frac{V}{c^2}x)##?
You might want to consider this: in the Lorentz transformations, ##V## is assumed to be known, but in the scenario you describe, you know the proper acceleration ##a##, but ##V## is what you're trying to figure out, not something you know to begin with. So the Lorentz transformations might not be the best tool for this particular problem.

This should be the resulting equation:
$$\Delta v=\frac{at'}{\sqrt{1+(\frac{at'}{c})^2}}$$
I'm not sure why you are using ##t'## here, since the ##t## in this formula is the coordinate time in the original rest frame (the frame in which the accelerating object starts out at rest).

greg_rack
Mentor
I thought this was the third or fourth post on SR
I do see that there was a previous thread by the OP on the relativistic rocket equation, which is the same problem; so I have now combined the two threads into one.

@greg_rack please note that your previous question on the relativistic rocket equation is really the same question as your more recent one on the velocity of an accelerating object; the rocket is the accelerating object. Accordingly, as noted just above, your two threads on that topic have been combined under the "relativistic rocket equation" title.

greg_rack
Staff Emeritus
Good news - there's a treatment of the relativistic rocket in Misner, Thorne, Wheeler, "Gravitation", page 166.

Bad news. The treatment there uses 4-velocities and 4-accelerations. If the concept is totally unfamiliar, you'd have to learn it to use MTW's derivation. The book as a whole is graduate level, the tensor notation used may be confusing.

The derivation is pretty simple, using three key facts that give three key equations. These are that the magnitude of a 4 velocity is always c, the 4-acceleration is always perpendicular to the 4-velocity, and that the magnitude of the 4-acceleration is just the magnitude of the acceleration in the rocket's rest frame. These could be reframed without the need for the tensor notation, I imagine.

To utilize this approach you'd have to be willing to learn about 4-velocities and 4-accelrations. Or you'd need to find another treatment if you don't want to or can't learn about them. If you want to look for another treatment, google for "hyperbolic motion". Wiki has some writing about this, but it doesn't derive key facts, such as the ##\gamma^3## relationship between coordinate acceleration and proper acceleration, or a detailed description of proper acceleratiion. (##\gamma## is just ##1/\sqrt{1 - v^2/ c^2}##).

Some posters have already posted the basics of 4-velocities and 4-accelrations in this thread, but it may be intimidating without the support of a textbook. Taylor & Wheeler's "Spacetime Physics" will provide some of the background needed, but I don't think they actually work through the relativistic rocket.

PeroK
Homework Helper
Gold Member
2022 Award
Good news - there's a treatment of the relativistic rocket in Misner, Thorne, Wheeler, "Gravitation", page 166.

Bad news. The treatment there uses 4-velocities and 4-accelerations. If the concept is totally unfamiliar, you'd have to learn it to use MTW's derivation. The book as a whole is graduate level, the tensor notation used may be confusing.
The relevant 1D relativistic kinematics is covered in Helliwell's book, in appendices F and G. That the material is in the appendices suggests to me that the OP is diving in at the deep end. In any case, Helliwell would be my recommendation for a first textbook in SR:

greg_rack
Gold Member
I'm not sure why you are using ##t'## here, since the ##t## in this formula is the coordinate time in the original rest frame (the frame in which the accelerating object starts out at rest).
Doesn't that ##t'## indicate time in the rest frame, which is the dilated one?
With ##t## I indicate proper time(measured in the rocket frame).

Gold Member
2022 Award
Motion with constant proper acceleration is equivalent to the motion of a point charge in a homogeneous static electric field (neglecting radiation damping). That calculation you find in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

p. 30. There it's solved using one inertial reference frame and the 3D (non-covariant) formalism with the (coordinate) time as the independent variable.

The manifestly covariant calculation with proper time as the independent variable is done in #13.

Homework Helper
Gold Member
2022 Award
Doesn't that ##t'## indicate time in the rest frame, which is the dilated one?
With ##t## I indicate proper time(measured in the rocket frame).
Usually ##t## would indicate time in the original rest frame and ##\tau## would be used for the proper time in the rocket frame.

If you use ##t'## for the rest frame, then to be consistent you should have ##\Delta v'## for the change in speed in that frame.

Mentor
Doesn't that ##t'## indicate time in the rest frame, which is the dilated one?
With ##t## I indicate proper time(measured in the rocket frame).
Normally the primed frame would be the rocket frame. Also, normally the rocket frame would be the one described as "dilated".

Staff Emeritus
In case it’s not clear already, here’s the route that I would use in hindsight.

Let ##\tau## be the proper time for the rocket, related to coordinate time ##t## through
##\frac{d\tau}{dt} = \sqrt{1-\frac{v^2}{c^2}}##.

Then define a bunch of spacetime vectors:

##X = (ct, x)##
##V = \dfrac{dX}{d\tau}##
##A = \dfrac{dV}{d\tau}##

In the launch frame, we have, initially,
##V = (c,0)##
##A = (0,a)##

Even though that’s only true, initially, we can come up with invariants that are always true.

##(V^0)^2 - (V^1)^2 = c^2##
##(A^0)^2 - (A^1)^2 = - a^2##

Now, anytime you have a difference of squares equal to a constant, you can write it as hyperbolic trig functions:

##V^0 = c cosh(\theta)##
##V^1 = c sinh(\theta)##

This implicitly defines the rapidity ##\theta## in terms of velocity.

Taking derivatives gives:

##A^0 = c sinh(\theta) \frac{d\theta}{d\tau}##
##A^1 = c cosh(\theta) \frac{d\theta}{d\tau}##.

Then the invariant condition on ##A## tells us that

##c^2 (\frac{d\theta}{d\tau})^2 = a^2##

So ##\theta =\frac{a\tau}{c}##

Therefore, ##A^1 = a cosh(\frac{a\tau}{c})##.

Integrating twice gives:

##X^1 = \frac{c^2}{a} cosh(\frac{a\tau}{c})## + a constant

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Homework Helper
Gold Member
2022 Award
By contrast, the most elementary approach would be:

1) Derive the (inverse) relativistic velocity transformation formula.

2) Derive the (inverse) relativistic acceleration transformation.

3) Use 2) in the case where the acclerating rocket is instantaneously at rest and has constant proper acceleration ##a## (in the primed frame), to give the specific result: $$\gamma^3 \frac{dv}{dt} = a$$

4) Integrate!

SiennaTheGr8
I'll use ##c = 1##. Overdots mean proper-time derivatives.

Four-position: ## \mathbf{R} = (t, \mathbf{r}) ##

Four-velocity: ##\mathbf{V} = \dfrac{d \mathbf{R}}{d \tau} = (\gamma, \gamma \mathbf{v}) ##

Four-acceleration (requires some effort): ## \mathbf{A} = \dfrac{d \mathbf{V}}{d \tau} = (\gamma^4 (\mathbf{v} \cdot \mathbf{a}), \gamma^2 \mathbf{a} + \gamma^4 (\mathbf{v} \cdot \mathbf{a} ) \mathbf{v}) ##

In the case of rectilinear motion, ## \mathbf{v} \cdot \mathbf{a} = \pm va## and ##\mathbf{a} = \pm a \hat{\mathbf v}##, and the four-acceleration's components simplify to:

$$\mathbf{A} = (\pm \gamma^4 v a, \gamma^4 \mathbf{a}) .$$
In this special case (that is, for all inertial observers for whom the accelerating object's motion is rectilinear), the magnitude of the four-acceleration can be reckoned from the components like so:

$$\sqrt{ \mathbf{A} \cdot \mathbf{A} } = \sqrt{(\pm \gamma^4 v a)^2 - (\gamma^4 \mathbf{a})^2} = \gamma^3 a \, \mathrm{i} .$$
In the accelerating object's instantaneous rest frame, ##\gamma^3 = 1##, and the four-acceleration's magnitude (divided by ## \mathrm{i} ##) is just ##a##, which is the magnitude of acceleration that the object "feels." We'll call this the "proper acceleration" ##\alpha##; as the magnitude of a four-vector, its value is invariant, and any observer for whom the motion is rectilinear can calculate it like ##\alpha = \gamma^3 a##.

We'll come back to this result.

Because ##\gamma^2 - \gamma^2 v^2 = 1##, we can appeal to the identity ##\cosh^2 \phi - \sinh^2 \phi = 1## and introduce the "rapidity" ##\phi##:

## \cosh \phi = \gamma ,##
## \sinh \phi = \gamma v ,##
## \tanh \phi = \sinh \phi / \cosh \phi = v .##

This is especially useful for doing calculus, since, e.g., ##\sinh## and ##\cosh## are derivatives of each other.

Note in passing that ##\sinh## and ##\tanh## (and their inverses) are odd functions. That means we're free to define signed rapidities, too, which we can pair with (signed) velocity-components. I'll make some simplifications below to avoid dealing with them directly, but allowing for them isn't exactly rocket science. (Sorry.)

The four-velocity ##\mathbf{V} = (\gamma, \gamma \mathbf{v}) ## can now be rewritten:

$$\mathbf{V} = (\cosh \phi, \hat{\mathbf v} \sinh \phi ) ,$$

and four-acceleration (where the overdot means a proper-time derivative):

$$\mathbf{A} = \dfrac{d \mathbf{V}}{d \tau} = (\dot{\phi} \sinh \phi, \hat{\mathbf v} \dot{\phi} \cosh \phi + \dot{\hat{\mathbf v}} \sinh \phi) .$$

Its magnitude is:

$$A = \sqrt{ \mathbf{A} \cdot \mathbf{A} } = \sqrt{ (\dot{\phi} \sinh \phi)^2 - (\hat{\mathbf v} \dot{\phi} \cosh \phi + \dot{\hat{\mathbf v}} \sinh \phi)^2 } .$$

For rectilinear motion, ##\dot{\hat{\mathbf v}} = \mathbf 0##, and:

$$A = \sqrt{ \dot{\phi}^2 ( \sinh^2 \phi - \cosh^2 \phi ) } = | \dot \phi | \, \mathrm{i} .$$

Put that together with our previous result for ##A## in the rectilinear case, and we have:

$$\dfrac{d \phi}{d \tau} = \pm \alpha .$$

So for rectilinear motion, the rapidity's proper-time derivative is the proper acceleration, up to a sign (we defined ##\alpha## as positive).

Now we can integrate with respect to ##\tau## to get the rapidity as a function of proper time. (We'll actually get a signed rapidity in the general case, but we'll narrow things down in a moment.) If the proper acceleration is constant, that's:

$$\int_0^\tau \dfrac{d \phi}{d \tau} d \tau = \pm \alpha \int_0^\tau d \tau ,$$

giving ## \phi (\tau) = \phi_i \pm \alpha \tau ##, where ##\phi_i## is the initial rapidity (its value at ##\tau = 0##).

For convenience, let's now set the initial rapidity to zero and interest ourselves only in the "plus" case, so that our function is ## \phi (\tau) = \alpha \tau ##. By the definition of ##\phi##, we immediately also get:

##\gamma (\tau) = \cosh (\alpha \tau), ##
## (\gamma v)(\tau) = \sinh (\alpha \tau) ,##
## v(\tau) = \tanh (\alpha \tau ).##

How about position as a function of proper-time? Say the object accelerates in the positive ##x##-direction. Then we have ##\dot x = \gamma v_x = \gamma v##. To get ##x(\tau)##, integrate ##\gamma v = \sinh (\alpha \tau)##:

$$\int_0^\tau \dfrac{d x}{d \tau} d \tau = \int_0^\tau \sinh(\alpha \tau) d \tau ,$$

giving ## x(\tau) = x_i + \cosh(\alpha \tau) / \alpha ##.

If you want functions of coordinate time ##t## instead, note that ##\alpha = \gamma^3 a = \frac{d (\gamma v)}{d t}## in the rectilinear case, integrate with respect to ##t## to get a function for ##\gamma v##, and use ##\phi = \sinh^{-1} (\gamma v)## to get ##\phi(t)##, etc.

Hope I didn't make any mistakes.

vanhees71
Staff Emeritus
In case it’s not clear already, here’s the route that I would use in hindsight.

Integrating twice gives:

##X^1 = \frac{c^2}{a} cosh(\frac{a\tau}{c})## + a constant

A good derivation with a classic textbook result. I have a feeling the original poster might want to find x(t), not yet having any intuition for proper time tau.

One more integration gives ##t(\tau)## from
$$V^0 = \frac{dt}{d\tau} = \cosh{\frac{a \tau}{c}} \quad dt = \cosh{\frac{a \tau}{c}} d\tau$$

yielding

$$t = \frac{a}{c} \sinh{ \frac{a \tau}{c}} \quad \tau = \frac{c}{a} arcsinh{\frac{a\,t}{c}}$$

Then x(t) is ##x(t(\tau))##, though it might be simpler to note
$$x^2 - c t^2 = \frac{c^2}{a} \left( cosh^2 - \sinh^2 \right) = c^2/a$$

which can be simply solved for x(t), giving rise to a hyperbola.

The asymptotes of the hyperbola reveal the interesting fact that an observer with a headstart and a constant proper acceleration can stay slightly ahead of a light ray, approaching the light ray asymptotically. This also makes it clear why the motion is called hyperbolic motion.

Staff Emeritus
By contrast, the most elementary approach would be:

1) Derive the (inverse) relativistic velocity transformation formula.

2) Derive the (inverse) relativistic acceleration transformation.

3) Use 2) in the case where the acclerating rocket is instantaneously at rest and has constant proper acceleration ##a## (in the primed frame), to give the specific result: $$\gamma^3 \frac{dv}{dt} = a$$

4) Integrate!

I have a feeling this might be the sort of approach the OP is looking for, though it's hard to be sure.

If the OP wants to go this route, the first thing to note is that the velocity composition formula in special relativity can be derived by a pair of boosts.

We can write
$$\begin{bmatrix} {ct' \, \\ x'} \end{bmatrix} = \begin{bmatrix} \Lambda \end{bmatrix} \begin{bmatrix} {ct \,\\ x} \end{bmatrix}$$

(note - I think there is a latex bug here, those should be column matrices ).

where ##\Lambda## is the matrix form of the Lorentz boost.

Then velocity composition consists of noting that a pair of boosts in the same direction is equivalent to a single, larger boost, with the usual velocity composition formula ##u \oplus v = \frac{u+v}{1-\frac{u\,v}{c^2}}##

Alternatively, one might already know the velocity composition formula, but the OP did want to derive everything from the Lorentz transform, and this is how one would go about doing that.

The effects of velocity compositon should give a factor of ##\gamma^2##. The remaining factor of gamma comes from the relationship between coordinate time t and and proper time ##\tau## as experienced on the rocket. The final result that ##\alpha = \gamma^3 a##, ##\alpha## being the proper acceleration as measured by clocks and rods on the rocketship, a being the coordinate acceleration ##\frac{d^2 x}{dt^2}## in terms of the (x,t) coordinates of an inertial frame in which the rocket is initially at rest.

A sidenote - replacing a pair of boosts with a single boost only works if the boosts are in the same direction. A pair of boosts in different directions also generates a rotation. This isn't relevant to the problem at hand, but it becomes interesting later on.

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2022 Award
Here is my try. There are two ways. One is to use coordinate time and the other is to work with the manifestly covariant equations of motion. For the latter way we use the normalized four-velocity and assume that the constant acceleration is in ##x^1##-direction. So we can work in (1+1)-dimensional Minkowski space.

$$(u^{\mu})=\begin{pmatrix} \cosh y \\ \sinh y \end{pmatrix},$$
where ##y## is the (momentum-space) rapidity, which parametrization works in the constraint ##u_{\mu} u^{\mu}=1##.

The equations of motion are
$$\mathrm{d}_{\tau} u^0=\alpha/c u^1, \quad \mathrm{d}_{\tau} u^1=c \alpha/c u^0.$$
The two equations are not independent because of the constraint. So we use the 2nd equation with the parametrization using the rapidity,
$$\cosh y \mathrm{d}_{\tau} y =\alpha/c \cosh y \; \Rightarrow \; y=\frac{\alpha \tau}{c}+y_0.$$
A direct check shows that indeed also the first eom is fulfilled so we have
$$(u^{\mu})=\begin{pmatrix} \cosh(\alpha \tau/c+y_0) \\ \sinh(\alpha \tau/c+y_0) \end{pmatrix}.$$
The spacetime vector is then defined by
$$\mathrm{d}_{\tau} x^{\mu}=c u^{\mu}$$
from which
$$x^{\mu}=\begin{pmatrix} \frac{c^2}{\alpha} \sinh(\alpha \tau/c+y_0) \\ x_0^{1} + \frac{c^2}{\alpha} [\cosh(\alpha \tau/c + y_0)-1]. \end{pmatrix}$$
I've chosen the initial time such that ##t_0=t(0)=0##.

In terms of coordinate time we have
$$x^1(t)=x_0^1 + \frac{c^2}{\alpha} \left [\sqrt{\alpha^2 t^2/c^2+1}-1 \right]$$
and for the velocity
$$v(t)=\mathrm{d}_t x^1=\frac{\alpha t}{\sqrt{\alpha^2 t^2/c^2+1}}.$$

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JD_PM and PeroK