# LOG#028. Rockets and relativity.

The second post in this special thread of 3 devoted to Neil Armstrong memory has to do with rocketry.

Firstly, for completion, we are going to study the motion of a rocket in “vacuum” according to classical physics. Then, we will deduce the relatistic rocket equation and its main properties.

CLASSICAL NON-RELATIVISTIC ROCKETS

The fundamental law of Dynamics, following Sir Isaac Newton, reads:

$\mathbf{F}=\dfrac{d\mathbf{p}}{dt}$

Suppose a rocket with initial mass $M_i$ and initial velocity $u_i=0$. It ejects mass of propellant “gas” with “gas speed” (particles of gas have a relative velocity or speed with respect to the rest observer when the rocket move at speed $\mathbf{v}$) equals to $u_0$ (note that the relative speed will be $u_{rel} and the propellant mass is$latex m_0\$. Generally, this speed is also called “exhaust velocity” by engineers. The motion of a variable mass or rocket is given by the so-called Metcherski’s equation:

$\boxed{M\dfrac{d\mathbf{v}}{dt}=-\mathbf{u_0}\dfrac{dM}{dt}+\mathbf{F}}$

where $-\mathbf{u_0}=\mathbf{v_{gas}}-\mathbf{v}$. The Metcherski’s equation can be derived as follows: the rocket changes its mass and velocity so $M'=M+dM$ and $V'=V+dV$, so the change in momentum is equal to $M'V'=(M+dM)(V+dV)$, plus an additional term $v_{gas}dv_{gas}$ and $-mV$. Therefore, the total change in momentum:

$dP=Fdt=(M+dM)(V+dV)+v_{gas}dv_{gas}-mV$

Neglecting second order differentials, and setting the conservation of mass (we are in the non-relativistic case)

$dM+dm_{gas}=0$

we recover

$MdV=v_{rel}dM+Fdt$

that represents (with the care of sign in relative speed) the Metcherski equation we have written above.

Generally speaking, the “force” due to the change in “mass” is called thrust.  With no external force, from the remaining equation of the thrust and velocity, and it can be easily integrated

$u_f=-u_0\int_{M_i}^{M_f}dM$

and thus we get the Tsiolkowski’s rocket equation:

$\boxed{\mathbf{u_f}=\mathbf{u_0}\ln \dfrac{M_i}{M_f}}$

Engineers use to speak about the so-called mass ratio $R=\dfrac{M_f}{M_i}$, although sometimes the reciprocal definition is also used for such a ratio so be aware, and in terms of this the Tsiokolski’s equation reads:

$\boxed{\mathbf{u_f}=\mathbf{u_0}\ln \dfrac{1}{R}}$

We can invert this equation as well, in order to get

$\boxed{R=\dfrac{M_f}{M_i}=\exp\left(-\dfrac{u_f}{u_0}\right)}$

Example: Calculate the fraction of mass of a one-stage rocket to reach the Earth’s orbit. Typical values for $u_f=8km/s$ and $u_0=4km/s$ show that the mass ratio is equal to $R=0.14$. Then, only the $14\%$ of the initial mass reaches the orbit, and the remaining mass is fuel.

Multistate rockets offer a good example of how engineer minds work. They have discovered that a multistage rocket is more effective than the one-stage rocket in terms of maximum attainable speed and mass ratios. The final n-stage lauch system for rocketry states that the final velocity is the sum of the different gains in the velocity after the n-th stage, so we can obtain

$\displaystyle{u_f=\sum_{i=1}^{n}u_i^f=u_1^f+\cdots+u_n^f}$

After the n-th step, the change in velocity reads

$u_i^f=c_i\ln \dfrac{1}{R_i}$

where the i-th mass ratios are definen recursively as the final mass in the n-th step and the initial mass in that step, so we have

$\displaystyle{u_f=\sum_i c_i\ln \dfrac{1}{R_i}}$

and we define the total mass ratio:

$\displaystyle{R_T=\prod_i R_i}$

If the average effective rocket exhaust velocity is the same in every step/stage, e.g. $c_i=c$, we get

$\displaystyle{u_f=c\ln \left( \prod_{i=1}^{n} R_i^{-1}\right)}$

or

$\displaystyle{u_f=c \ln \left[ \left(\dfrac{M_0}{M_f}\right)_1\left(\dfrac{M_0}{M_f}\right)_2\cdots \left(\dfrac{M_0}{M_f}\right)_n\right]=c\ln \left[\left(\dfrac{M_0}{M_f}\right)_T\right]}$

The influence of the number of steps, for a given exhaust velocity, in the final attainable velocity can be observed in the next plots:

RELATIVISTIC ROCKETS

We proceed now to the relativistic generalization of the previous rocketry. An observer in the laboratory frame observes that total momentum is conserved, of course, and so:

$M'du'=-u'_0dM'$

where $du'$ is the velocity increase in the rocket with a rest mass M’ in the instantaneous reference frame of the moving rocket S’. It is NOT equal to its velocity increase measured in the unprimed reference frame, du. Due to the addition theorem of velocities in SR, we have

$u+du=\dfrac{u+du'}{1+\dfrac{udu'}{c^2}}$

where u is the instantenous velocity of the rocket with respect to the laboratory frame S. We can perform a Taylor expansion of the denominator in the last equation, in order to obtain:

$u+du=(u+du')\left(1-\dfrac{udu'}{c^2}\right)$

and then

$u+du=u+du'\left(1-\dfrac{u^2}{c^2}\right)$

and finally, we get

$du'=\dfrac{du}{1-\dfrac{u^2}{c^2}}=\gamma^2_u du$

Plugging this equation into the above equation for mass (momentum), and integrating

$\displaystyle{\int_{0}^{u_f}\dfrac{du}{1-\dfrac{u^2}{c^2}}=-u'_0\int_{M'_0}^{M'_f}dM}$

we deduce that the relativistic version of the Tsiolkovski’s rocket equation, the so-called relativistic rocket equation, can be written as:

$\dfrac{c}{2}\ln \dfrac{1+\dfrac{u_f}{c}}{1-\dfrac{u_f}{c}}=u'_0\ln\dfrac{M'_i}{M'_f}$

We can suppress the primes if we remember that every data is in the S’-frame (instantaneously), and rewrite the whole equation in the more familiar way:

$\boxed{u_f=c\dfrac{1-\left(\dfrac{M_f}{M_0}\right)^{\frac{2u_0}{c}}}{1+\left(\dfrac{M_f}{M_0}\right)^{\frac{2u_0}{c}}}=c\dfrac{1-R^{\frac{2u_0}{c}}}{1+R^{\frac{2u_0}{c}}}}$

where the mass ratio is defined as before $R=\dfrac{M_f}{M_i}$. Now, comparing the above equation with the rapidity/maximum velocity in the uniformly accelerated motion:

$u_f=c\tanh \left(\dfrac{g\tau}{c}\right)$

we get that relativistic rocket equation can be also written in the next manner:

$u_f=c\tanh \left[ -\dfrac{u_0}{c}\ln \left(\dfrac{1}{R}\right)\right]$

or equivalently

$u_f=c\tanh \left[ \dfrac{u_0}{c}\ln R\right]$

since we have in this case

$\dfrac{g\tau}{c}=-\dfrac{u_0}{c}\ln \left(\dfrac{1}{R}\right)=\dfrac{u_0}{c}\ln R$

and thus

$R^{\frac{u_0}{c}}=\left(\dfrac{M_f}{M_i}\right)^{\frac{u_0}{c}}=\exp \left(-\dfrac{g\tau}{c}\right)$

If the propellant particles move at speed of light, e.g., they are “photons” or ultra-relativistic particles that move close to the speed of light we have the celebrated “photon rocket”. In that case, setting $u_0=c$, we would obtain that:

$\boxed{u_f=c\dfrac{1-\left(\dfrac{M_f}{M_0}\right)^{2}}{1+\left(\dfrac{M_f}{M_0}\right)^{2}}=c\dfrac{1-R^{2}}{1+R^{2}}=c\tanh \ln R}$

and where for the photon rocket (or the ultra-relativistic rocket) we have as well

$\dfrac{g\tau}{c}=-\ln \left(\dfrac{1}{R}\right)=\ln R$

Final remark: Instead of the mass ratio, sometimes is more useful to study the ratio fuel mass/payload. In that case, we set $M_f=m$ and $M_0=m+M$, where M is the fuel mass and m is the payload. So, we would write

$R=\dfrac{m}{m+M}$

so then the ratio fuel mass/payload will be

$\dfrac{M}{m}=R^{-1}-1=\exp \left(\dfrac{g\tau}{c}\right)-1$

We are ready to study the interstellar trip with our current knowledge of Special Relativity and Rocketry. We will study the problem in the next and final post of this fascinating thread. Stay tuned!

# LOG#027. Accelerated motion in SR.

Hi, everyone! This is the first article in a thread of 3 discussing accelerations in the background of special relativity (SR). They are dedicated to Neil Armstrong, first man on the Moon! Indeed,  accelerated motion in relativity has some interesting and sometimes counterintuitive results, in particular those concerning the interstellar journeys whenever their velocities are close to the speed of light(i.e. they “are approaching” c).

Special relativity is a theory considering the equivalence of every  inertial frame ( reference frames moving with constant relative velocity are said to be inertial frames) , as it should be clear from now, after my relativistic posts! So, in principle, there is nothing said about relativity of accelerations, since accelerations are not relative in special relativity ( they are not relative even in newtonian physics/galilean relativity). However, this fact does not mean that we can not study accelerated motion in SR. The own kinematical framework of SR allows us to solve that problem. Therefore, we are going to study uniform (a.k.a. constant) accelerating particles in SR in this post!

First question: What does “constant acceleration” mean in SR?   A constant acceleration in the S-frame would give to any particle/object a superluminal speed after a finite time in non-relativistic physics! So, of course, it can not be the case in SR. And it is not, since we studied how accelerations transform according to SR! They transform in a non trivial way! Moreover, a force growing beyond the limits would be required for a “massive” particle ( rest mass $m\neq 0$). Suppose this massive particle (e.g. a rocket, an astronaut, a vehicle,…) is at rest in the initial time $t=t'=0$, and it accelerates in the x-direction (to be simple with the analysis and the equations!). In addition, suppose there is an observer left behind on Earth(S-frame), so Earth is at rest with respect to the moving particle (S’-frame). The main answer of SR to our first question is that we can only have a constant acceleration in the so-called instantaneous rest frame of the particle.  We will call that acceleration “proper acceleration”, and we will denote it by the letter $\alpha$. In fact, in many practical problems, specially those studying rocket-ships, the acceleration is generally given the same magnitude as the gravitational acceleration on Earth ($\alpha=g\approx 9.8ms^{-2}\approx 10 ms^{-2}$).

Second question: What are the observed acceleration in the different frames? If the instantaneous rest frame S’ is an inertial reference frame in some tiny time $dt'$, at the initial moment, it has the same velocity as the particle (rocket,…) in the S-frame, but it is not accelerated, so the velocity in the S’-frame vanishes at that time:

$\mathbf{u}'=(0,0,0)$

Since the acceleration of the particle is, in the S’-frame, the proper acceleration, we get:

$\mathbf{a}'=(a'_x,0,0)=(\alpha,0,0)=(g,0,0)=\mbox{constant}$

Using the transformation rules for accelerations in SR we have studied, we get that the instantaneous acceleration in the S-frame is given by

$\mathbf{a}=(a_x,0,0)=\left(\dfrac{g}{\gamma^3},0,0\right)$

Since the relative velocity between S and S’ is always the same to the moving particle velocity in the S-frame, the following equation holds

$v=u_x$

We do know that

$a_x=\dfrac{du_x}{dt}=\left(1-\dfrac{u_x^2}{c^2}\right)^{3/2}g$

Due to time dilation

$dt'=dt/\gamma$

so in the S-frame, the particle moves with the velocity

$du_x=\left(1-\dfrac{u_x^2}{c^2}\right)^{3/2}g dt$

We can now integrate this equation

$\int_0^{u_x}\dfrac{1}{(c^2-u_x^2)^{3/2}}du_x=\dfrac{g}{c^3}\int_0^t dt$

The final result is:

$\boxed{u_x=\dfrac{g t}{\sqrt{1+\left(\dfrac{g t}{c}\right)^2}}}$

We can check some limit cases from this relativistic result for uniformly accelerated motion in SR.

1st. Short time limit: $gt<< c\longrightarrow u_x\approx gt=\alpha t$. This is the celebrated nonrelativistic result, with initial speed equal to zero (we required that hypothesis in our discussion above).

2nd. Long time limit: $t\rightarrow \infty$. In this case, the number one inside the root is very tiny compared with the term depending on acceleration, so it can be neglected to get $u_x\approx \dfrac{gt}{gt/c}=c$. So, we see that you can not get a velocity higher than the speed of light with the SR framework at constant acceleration!

Furthermore, we can use the definition of relativistic velocity in order to integrate the associated differential equation, and to obtain the travelled distance as a function of $t$, i.e. $x(t)$, as follows

$u_x=\dfrac{dx}{dt}=\dfrac{gt}{\sqrt{1+\left(\dfrac{g t}{c}\right)^2}}$

$\int_0^x dx=\int_0^t\dfrac{gt dt}{\sqrt{1+\left(\dfrac{g t}{c}\right)^2}}=\int_0^t\dfrac{ctdt}{\sqrt{\dfrac{c^2}{g^2}+t^2}}$

We can perform the integral with the aid of the following known result ( see,e.g., a mathematical table or use a symbolic calculator or calculate the integral by yourself):

$\int \dfrac{ctdt}{\sqrt{\left(\dfrac{c}{g}\right)^2+t^2}}=c\sqrt{\left(\dfrac{c}{g}\right)^2+t^2}+\mbox{constant}=c\sqrt{\left(\dfrac{c}{g}\right)^2+t^2}+C$

From this result, and the previous equation, we get the so-called relativistic path-time law for uniformly accelerated motion in SR:

$x=c\sqrt{\left(\dfrac{c}{g}\right)^2+t^2}-\dfrac{c^2}{g}$

or equivalently

$\boxed{x=x(t)=\dfrac{c^2}{g}\left(\sqrt{1+\left(\dfrac{gt}{c}\right)^2}-1\right)}$

For consistency, we observe that in the limit of short times, the terms in the big brackets approach $1+\frac{1}{2}\left(\frac{gt}{c}\right)^2$, in order to get $x\approx \frac{1}{2}gt^2$, so we obtain the nonrelativistic path-time relationship $x=\frac{1}{2}gt^2$ with $g=a_x$. In the limit of long times, the terms inside the brackets can be approximated to $gt/c$, and then, the final result becomes $x\approx ct$. Note that the velocity is not equal to the speed of light, this result is a good approximation whenever the time is “big enough”, i.e., it only works for “long times” asymptotically!

And finally, we can write out the transformations of accelaration between the two frames in a explicit way:

$a_x=\left[1-\dfrac{\left(\dfrac{gt}{c}\right)^2}{1+\left(\dfrac{gt}{c}\right)^2}\right]^{3/2}g$

that is

$\boxed{a_x=\dfrac{1}{\left[1+\left(\dfrac{gt}{c}\right)^2\right]^{3/2}}g}$

Check 1: For short times, $a_x\approx g=\mbox{constant}$, i.e., the non-relativistic result, as we expected!

Check 2: For long times, $a_x\approx \dfrac{c^3} {g^2t^3}\rightarrow 0$. As we could expect, the velocity increases in such a way that “saturates” its own increasing rate and the speed of light is not surpassed. The fact that the speed of light can not be surpassed or exceeded is the unifying “theme” through special relativity, and it rest in the “noncompact” nature of the Lorentz group due to the $\gamma$ factor, since it would become infinity at v=c for massive particles.

It is inevitable: as time passes, a relativistic treatment is indispensable, as the next figures show

The next table is also remarkable (it can be easily built with the formulae we have seen till now with any available software):

Let us review the 3 main formulae until this moment

$\boxed{a_x=\dfrac{1}{\left[1+\left(\dfrac{gt}{c}\right)^2\right]^{3/2}}g}$ $\boxed{u_x=\dfrac{\alpha t}{\sqrt{1+\left(\dfrac{g t}{c}\right)^2}}}$ $\boxed{x=x(t)=\dfrac{c^2}{g}\left(\sqrt{1+\left(\dfrac{gt}{c}\right)^2}-1\right)}$

We have calculated these results in the S-frame, it is also important and interesting to calculate the same stuff in the S’-frame of the moving particle. The proper time $\tau=t'$ is defined as:

$\boxed{d\tau=dt\sqrt{1-\left(\dfrac{u_x}{c}\right)^2}}$

Therefore,

$d\tau=dt\left[1-\dfrac{\left(\dfrac{gt}{c}\right)^2}{1+\left(\dfrac{gt}{c}\right)^2}\right]^{1/2}$

We can perform the integral as before

$\displaystyle{\int_0^\tau d\tau=\int_0^t\dfrac{dt}{\sqrt{1+\left(\dfrac{gt}{c}\right)^2}}}$

and thus

$\tau=\dfrac{c}{g}\int_0^\tau\dfrac{dt}{\sqrt{\left(\dfrac{c}{g}\right)^2+t^2}}=\dfrac{c}{g}\ln \left(\dfrac{gt}{c}+\sqrt{\left(\dfrac{gt}{c}\right)^2+1}\right)\bigg|_0^t$

Finally, the proper time(time measured in the S’-frame) as a function of the elapsed time on Earth (S-frame) and the acceleration is given by the very important formula:

$\boxed{\tau=\dfrac{c}{g}\ln \left(\dfrac{gt}{c}+\sqrt{1+\left(\dfrac{gt}{c}\right)^2}\right)}$

And now, let us set $z=gt/c$, therefore we can write the above equation in the following way:

$\dfrac{g\tau}{c}=\ln \left( z+\sqrt{1+z^2}\right)$

Remember now, from our previous math survey, that $\sinh^{-1}z=\ln \left( z+\sqrt{1+z^2}\right)$, so we can invert the equation in order to obtain t as function of the proper time since:

$\boxed{\tau=\dfrac{c}{g}\sinh^{-1}\left(\dfrac{gt}{c}\right)}$

$\boxed{t=\dfrac{c}{g}\sinh \left(\dfrac{g\tau}{c}\right)}$

Inserting this last equation in the relativistic equation path-time for the uniformly accelerated body in SR, we obtain:

$x=x(\tau)=\dfrac{c^2}{g}\left(\sqrt{1+\sinh^2\left(\dfrac{g\tau}{c}\right)}-1\right)$

i.e.,

$\boxed{x=x(\tau)=\dfrac{c^2}{g}\left[\cosh \left(\dfrac{g\tau}{c}\right)-1\right]}$

Similarly, we can calculate the velocity-proper time law. Previous equations yield

$u_x=\dfrac{c\sinh\left(\dfrac{g\tau}{c}\right)}{\sqrt{1+\sinh^2\left(\dfrac{g\tau}{c}\right)}}=\dfrac{c\sinh \left(\dfrac{g\tau}{c}\right)}{\cosh \left(\dfrac{g\tau}{c}\right)}$

and thus the velocity-proper time law becomes

$\boxed{u_x=c\tanh \left(\dfrac{g\tau}{c}\right)}$

Remark: this last result is compatible with a rapidity factor $\varphi= \left(\dfrac{g\tau}{c}\right)$.

Remark(II): $a_x=\dfrac{du_x}{dt}=\left(1-\dfrac{u_x^2}{c^2}\right)^{3/2}g=\left(1-\tanh^2\left(\dfrac{g\tau}{c}\right)\right)^{3/2}g=\dfrac{1}{\cosh^3\left(\dfrac{g\tau}{c}\right)}g$. From this, we can read the reason why we said before that constant acceleration is “meaningless” unless we mean or fix certain proper time in the S’-frame since whenever we select a proper time, and this last relationship gives us the “constant” acceleration observed from the S-frame after the transformation. Of course, from the S-frame, as this function shows, acceleration is not “constant”, it is only “instantaneously” constant. We have to take care in relativity with the meaning of the words. Mathematics is easy and clear and generally speaking it is more precise than “words”, common language is generally fuzzy unless we can explain what we are meaning!

As the final part of this log entry, let us summarize the time-proper time, velocity-proper time, acceleration-proper time-proper acceleration and distance- proper time laws for the S’-frame:

$\boxed{t=\dfrac{c}{g}\sinh \left(\dfrac{g\tau}{c}\right)}$ $\boxed{u_x=c\tanh \left(\dfrac{g\tau}{c}\right)}$ $\boxed{a_x=\dfrac{1}{\cosh^3\left(\dfrac{g\tau}{c}\right)}g}$ $\boxed{x=x(\tau)=\dfrac{c^2}{g}\left[\cosh \left(\dfrac{g\tau}{c}\right)-1\right]}$

My last paragraph in this post is related to express the acceleration $g\approx 10ms^{-2}$ in a system of units where space is measured in lightyears (we take c=300000km/s) and time in years (we take 1yr=365 days). It will be useful in the next 2 posts:

$g=10\dfrac{m}{s^2}\dfrac{1ly}{9.46\cdot 10^{15}m}\dfrac{9.95\cdot 10^{14}s^2}{1yr^2}=1.05\dfrac{lyr}{yr^2}\approx 1\dfrac{lyr}{yr^2}$

Another election you can choose is

$g=9.8\dfrac{m}{s^2}=1.03\dfrac{lyr}{yr^2}\approx 1\dfrac{lyr}{yr^2}$

so there is no a big difference between these two cases with terrestrial-like gravity/acceleration.