# Acceleration in special relativity

Required math: calculus, hyperbolic functions

Required physics: relativity basics

A common misconception is that special relativity cannot handle accelerations. In fact, acceleration follows on naturally from the treatment of velocity that we’ve already seen.

The four-velocity is defined as the vector with unit magnitude along a particle’s world line, so in the particle’s rest frame ${\mathcal{O}}$ it is

$\displaystyle \vec{U}\stackrel{\mathcal{O}}{\rightarrow}(1,0,0,0) \ \ \ \ \ (1)$

The magnitude is therefore

$\displaystyle \vec{U}\cdot\vec{U}=-1 \ \ \ \ \ (2)$

which is an invariant.

Now suppose that a particle does accelerate. In that case, we can have an inertial frame at any event in the particle’s life by defining the momentarily comoving reference frame or MCRF for short. This is a reference frame that, at a given event, has the same velocity as the particle. If the particle is accelerating, then the MCRF will change from one event to the next, but at each point it is always an inertial frame. If the particle makes an infinitesimal displacement ${d\vec{x}}$ then in its MCRF, the separation between the two events that define either end of this displacement is just an infinitestimal change in the proper time, ${(d\tau,0,0,0)}$. Therefore, the magnitude of the displacement is

$\displaystyle d\vec{x}\cdot d\vec{x}=-(d\tau)^{2} \ \ \ \ \ (3)$

If we take mathematical liberties with infinitesimals, we can now consider a vector ${d\vec{x}/d\tau}$ which is the displacement vector divided by the increment in the proper time. In particular we can divide the magnitude equation by ${(d\tau)^{2}}$ and get

 $\displaystyle \frac{d\vec{x}}{d\tau}\cdot\frac{d\vec{x}}{d\tau}$ $\displaystyle =$ $\displaystyle -1\ \ \ \ \ (4)$ $\displaystyle \frac{d\vec{x}}{d\tau}$ $\displaystyle \stackrel{MCRF}{\rightarrow}$ $\displaystyle (1,0,0,0) \ \ \ \ \ (5)$

which is the same as the definition of the four-velocity. Despite the dodgy mathematics involved in its definition, we arrive at an expression of the four-velocity as a derivative of a displacement with respect to proper time:

$\displaystyle \vec{U}=\frac{d\vec{x}}{d\tau} \ \ \ \ \ (6)$

To introduce acceleration, we start with 2 and take its derivative with respect to ${\tau}$:

 $\displaystyle \frac{d}{d\tau}(\vec{U}\cdot\vec{U})$ $\displaystyle =$ $\displaystyle 2\vec{U}\cdot\frac{d\vec{U}}{d\tau}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (8)$

where the last line arises because ${\vec{U}\cdot\vec{U}=-1}$ which is a constant. We can define the four-acceleration as the derivative of the four-velocity with respect to proper time:

$\displaystyle \vec{A}\equiv\frac{d\vec{U}}{d\tau} \ \ \ \ \ (9)$

Note that the four-acceleration is always orthogonal to the four-velocity. This does not mean that the direction of acceleration is always perpendicular to the velocity (although it can be, as in the case of circular motion). Remember that orthogonality with four-vectors is a more general concept that it is with three-vectors.

What this relation does mean, though, is that in the MCRF, we can write a specific form for ${\vec{A}}$. In the MCRF, we know that ${\vec{U}\stackrel{MCRF}{\rightarrow}(1,0,0,0)}$, so the orthogonality requires that ${A^{0}=0}$, and we get

$\displaystyle \vec{A}\stackrel{MCRF}{\rightarrow}(0,A^{1},A^{2},A^{3}) \ \ \ \ \ (10)$

To see the relation between the four-acceleration and the more familiar Galilean acceleration, we can consider the case where a particle has a constant four-acceleration along the ${\vec{x}}$ axis, with magnitude ${\vec{A}\cdot\vec{A}=\alpha^{2}}$. In the MCRF, the four-vector ${\vec{U}\stackrel{MCRF}{\rightarrow}(1,0,0,0)}$. After an increment ${\Delta\tau}$ of proper time, the velocity of the particle will have changed to ${\Delta v}$ in the ${x}$ direction, so

 $\displaystyle \vec{U}(\Delta\tau)$ $\displaystyle \stackrel{MCRF}{\rightarrow}$ $\displaystyle (1,\Delta v,0,0)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \vec{U}(0)+\frac{d\vec{U}(0)}{d\tau}\Delta\tau+\mathcal{O}(\Delta\tau)^{2}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle (1,0,0,0)+\Delta\tau(0,\alpha,0,0)+\mathcal{O}(\Delta\tau)^{2} \ \ \ \ \ (13)$

The relation between proper time ${\tau}$ and the time ${t}$ in another inertial frame that is attached momentarily to the particle after the increment is given by the time dilation factor ${\gamma}$ so that

 $\displaystyle \frac{\Delta t}{\Delta\tau}$ $\displaystyle =$ $\displaystyle \gamma\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{1-(\Delta v)^{2}}} \ \ \ \ \ (15)$

In this case, to first order in ${\Delta v}$, ${\gamma=1}$ and ${\Delta\tau=\Delta t}$, so again to first order we can write

$\displaystyle (1,\Delta v,0,0)=(1,\alpha\Delta t,0,0) \ \ \ \ \ (16)$

which means that the constant acceleration is

$\displaystyle \alpha=\frac{\Delta v}{\Delta t} \ \ \ \ \ (17)$

and so is in the limit equivalent to the Galilean acceleration.

Now suppose we have a particle moving along the ${x}$ axis at a momentary speed ${v}$. The four-velocity is found by applying the Lorentz transformation to the four-velocity in the MCRF, which is ${\vec{U}\stackrel{MCRF}{\rightarrow}(1,0,0,0)}$. We get upon transforming to the frame ${\bar{\mathcal{O}}}$ in which the particle is seen to have speed ${v}$:

$\displaystyle \vec{U}\stackrel{\bar{\mathcal{O}}}{\rightarrow}(\gamma,\gamma v,0,0) \ \ \ \ \ (18)$

The four-acceleration is the derivative of this with respect to proper time ${\tau}$, so we get

 $\displaystyle \vec{A}$ $\displaystyle \stackrel{\bar{\mathcal{O}}}{\rightarrow}$ $\displaystyle \frac{d}{d\tau}(\gamma,\gamma v,0,0)\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{v}{(1-v^{2})^{3/2}},\frac{v^{2}}{(1-v^{2})^{3/2}}+\frac{1}{\sqrt{1-v^{2}}},0,0\right)\frac{dv}{d\tau}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{(1-v^{2})^{3/2}}\frac{dv}{d\tau}(v,1,0,0) \ \ \ \ \ (21)$

Since the acceleration in the MCRF is ${\vec{A}\stackrel{MCRF}{\rightarrow}(0,\alpha,0,0)}$, in the frame moving at speed ${v}$, we apply the Lorentz transformation as with the four-velocity, and get

$\displaystyle \vec{A}\stackrel{\bar{\mathcal{O}}}{\rightarrow}(\gamma v\alpha,\gamma\alpha,0,0) \ \ \ \ \ (22)$

Comparing this with the derivation above, we use the second component of the four-acceleration in each case to get

 $\displaystyle \alpha$ $\displaystyle =$ $\displaystyle \frac{1}{1-v^{2}}\frac{dv}{d\tau}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{1-v^{2}}\frac{dv}{dt}\frac{dt}{d\tau}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{(1-v^{2})^{3/2}}\frac{dv}{dt} \ \ \ \ \ (25)$

Assuming ${\alpha}$ is a constant, we can integrate this equation to get (assuming ${v(0)=0}$)

 $\displaystyle \alpha\int dt$ $\displaystyle =$ $\displaystyle \int\frac{dv}{(1-v^{2})^{3/2}}\ \ \ \ \ (26)$ $\displaystyle \alpha t$ $\displaystyle =$ $\displaystyle \frac{v}{\sqrt{1-v^{2}}}\ \ \ \ \ (27)$ $\displaystyle v$ $\displaystyle =$ $\displaystyle \frac{\alpha t}{\sqrt{\alpha^{2}t^{2}+1}} \ \ \ \ \ (28)$

Thus under constant acceleration, as ${t\rightarrow\infty}$, ${v\rightarrow1}$. Unlike the Galilean case where the velocity would increase without limit under constant acceleration, in relativity, the speed tends to the speed of light and never surpasses it, as must be the case.

The distance travelled after a given time ${t}$ can be found by another integration of the ${v}$ relation, since ${v=dx/dt}$:

$\displaystyle x=\frac{1}{\alpha}\sqrt{\alpha^{2}t^{2}+1}-\frac{1}{\alpha} \ \ \ \ \ (29)$

where we’ve taken ${x=0}$ at ${t=0}$.

As an example, we can find how long it takes to reach a speed of ${v=0.999}$. We can invert the ${v}$ relation above to give ${t}$ in terms of ${v}$:

 $\displaystyle t$ $\displaystyle =$ $\displaystyle \frac{v}{\alpha\sqrt{1-v^{2}}}\ \ \ \ \ (30)$ $\displaystyle t(0.999)$ $\displaystyle =$ $\displaystyle \frac{22.34}{\alpha} \ \ \ \ \ (31)$

If the acceleration is ${\alpha=10m\: s^{-2}}$ (about 1g), we must first convert this to relativistic units. Since 1 sec = ${3\times10^{8}}$ m, we get

 $\displaystyle \alpha$ $\displaystyle =$ $\displaystyle \frac{10}{9\times10^{16}}m^{-1}\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.11\times10^{-16}m^{-1}\ \ \ \ \ (33)$ $\displaystyle t(0.999)$ $\displaystyle =$ $\displaystyle \frac{22.34}{1.11\times10^{-16}}\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\times10^{17}m \ \ \ \ \ (35)$

To convert this back to ordinary units, we get

 $\displaystyle t(0.999)$ $\displaystyle =$ $\displaystyle \frac{2\times10^{17}m}{3\times10^{8}m\: s^{-1}}\ \ \ \ \ (36)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 6.7\times10^{8}s\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 21.2\; years \ \ \ \ \ (38)$

$\displaystyle \alpha=\frac{1}{1-v^{2}}\frac{dv}{d\tau} \ \ \ \ \ (39)$

Integrating this, we get

 $\displaystyle \alpha\tau$ $\displaystyle =$ $\displaystyle \tanh^{-1}v\ \ \ \ \ (40)$ $\displaystyle \tau$ $\displaystyle =$ $\displaystyle \frac{1}{\alpha}\tanh^{-1}v\ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\alpha}\tanh^{-1}\left(\frac{\alpha t}{\sqrt{\alpha^{2}t^{2}+1}}\right) \ \ \ \ \ (42)$

In relativistic units, we find

 $\displaystyle \tau(v$ $\displaystyle =$ $\displaystyle 0.999)=\frac{1}{1.11\times10^{-16}}\tanh^{-1}\left(0.999\right)\ \ \ \ \ (43)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3.42\times10^{16}\; m\ \ \ \ \ (44)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3.42\times10^{16}\; m}{3\times10^{8}m\: s^{-1}}\ \ \ \ \ (45)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.14\times10^{8}\; s\ \ \ \ \ (46)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3.61\; years \ \ \ \ \ (47)$

For a trip to the centre of the galaxy, with ${x=2\times10^{20}m}$ we use 29 to work out the time, which comes out to essentially the same value since most of the journey the speed is so close to ${v=1}$ it makes little difference. We can then plug in the numbers and find that

 $\displaystyle \tau$ $\displaystyle =$ $\displaystyle \frac{1}{\alpha}\tanh^{-1}\left(\frac{\alpha t}{\sqrt{\alpha^{2}t^{2}+1}}\right)\ \ \ \ \ (48)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 10.2\; years \ \ \ \ \ (49)$

In the inertial frame of the galaxy, the travel time is found by inverting the relation between ${\tau}$ and ${t}$ above:

$\displaystyle t=\frac{\tanh(\alpha\tau)}{\alpha\sqrt{1-\tanh^{2}(\alpha\tau)}} \ \ \ \ \ (50)$

 $\displaystyle$ $\displaystyle =$ $\displaystyle 7.07\times10^{11}\; s\ \ \ \ \ (51)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 22407\; years \ \ \ \ \ (52)$

This is time dilation in the extreme!

## 13 thoughts on “Acceleration in special relativity”

1. Matthew Kim

This was one of my college HW. I answered it for full credit but it did not occurred to me util now that there seems to be a discrepancy. x, v, and t of the spaceship in the outside inertial frame is given by hyperbolic function of proper time. If the starting position is shifted by 10m to the right, then new x = old x + 10 while v and t are same as before. So, if two objects do the identical acceleration in their rest frame just 10m apart at the start, they will alway remain 10m apart seen from the outside.

Now, consider what happen to 10m long spaceship with two engines with two separate clocks, one in front and one at the back so that both ends so that both ends start acceleration at t=0 exactly. At the start, the spaceship is at rest and
its length is 10m. Let’s say each engine turns off when it’s own ideal clock reached the proper time corresponding to gamma = 2, it turns off acceleration and
coast. According to Lorentz contraction, if the outside measurement is still 10m, then the rest length of the object must be gamma times bigger which is 20m. When did the spaceship get stretched so?

I have used the fact that the clock in the front appears to tick faster than the one in the back seen inside the spaceship to estimate the delay of the rear engine turn off from front engine turn off and it comes out even bigger than the delay behind the cause of Lorentz contraction. So,the outside measurement should have been even less than 5m.

I don’t know how to treat the exact time difference between the two astronaut at the start and at the turn off where accelating frame and inertial frame transitions occur and suspect where the key for the discrepancy lie. Can anyone show qunatitative answer to my quesion?

2. Sarang

This is a treasure trove. Came across it while trying to find a good explanation of how acceleration is treated in special relativity, but now I am very excited with the prospect of going through the other pages as well.

3. Dean

One quick observation of an error above (I believe):
Following the seemingly correct equation:
v = at [(a^2)(t^2) + 1]^(-1/2), is the (incorrect?) equation
x = (1/a) [(a^2)(t^2) + 1]^(1/2).

A little algebra on the first of the two above implies =>
v{(1/a) [(a^2)(t^2) + 1]^(1/2) = t.

Substituting the second equation in implies =>
vx = t,

which looks wrong.

Thank you.

1. growescience

If you mean equation 29, I had missed off a term ${-\frac{1}{\alpha}}$ which sets ${x=0}$ at ${t=0}$. Otherwise I think the equation is right. If you’re confused by the units not seeming to work out properly, remember that I’m taking ${c=1}$ so that velocity is dimensionless, time has the units of length and acceleration has the units of 1/time = 1/length. So ${\alpha^{2}t^{2}}$ in equation 29 is dimensionless and ${x}$ has the units of 1/acceleration which gives units of length, so all is well.

4. John

What would you see if you were in a rocket traveling at v= ½ c (half the speed of light), and a rocket comes at you from the opposite direction at v= -½ c and then passes you? My guess: Since the rocket would be coming toward you at the same speed as light, you wouldn’t see it until it passes you, then you would see a long tube of the rocket, where it has been, but only for an instant, as all the light hits you at the same time.

5. Peter Szilas

Have a problem with equation (11) where U(arrow on top)(∆tau) is -> MCRF given as U(1,∆v,0,0).
Can someone elaborate?
Note: ∆v = dx/dt is in the original frame, say , O.
So one or two words would be in place to explain how it appears in U (MCRF) a moment ∆tau later.

Thanks, Peter

1. gwrowe Post author

Possibly it’s the notation that’s a bit confusing. ${\vec{U}\left(\Delta\tau\right)}$ is the four-velocity at proper time ${\Delta\tau}$ as measured in the MCRF that existed at proper time 0, not the four-velocity as measured in a MCRF that exists at proper time ${\Delta\tau}$ (the latter would, of course, be ${\left(1,0,0,0\right)}$ as ${\vec{U}}$ always has this value in a MCRF at the current proper time).

1. Peter Szilas

Thanks, agree we are considering the MCRF at ∆tau=0. Equation 12 is OK, U is expanded in this MCRF. In eq. 16 and 17 this is compared to eq 11 where U(∆tau) is given as (1,v,o,0).
Agree, U(∆tau) in eq.11 is a vector in a MCRF( at ∆tau=o). But again, how does suddenly (1,v,0,0) appear? Yes, in ∆tau (sets gamma equal to 1 when comparing 11 and 12 later) the particle has moved and acquired speed ∆v.
So U(∆tau=0)= (1,0,0,0) ->U(∆tau) =(1,v,0,0)????
Sorry, but I don’t see the problem. As viewed in the MCRF at ${\tau=0}$, the speed in the ${x}$ direction increases from 0 to ${\Delta v}$ in time ${\Delta\tau}$, so ${U\left(\Delta\tau\right)=\left(1,\Delta v,0,0\right)}$.