I'm trying to describe projectile motion wrt a reference frame $S'$ that moves with velocity $v=0.6c$ in the $x$ direction wrt to the reference frame $S$ where the projectile was launched. I know projectile motion equations and Lorentz transformations but I'm kind of struggling to understand what is it that $S'$ sees. In $S$ I know $$x(t) = x_0 + v_o\cos(\alpha)t$$ $$y(t) = y_0 + v_o\sin(\alpha)t-\frac{g}{2}t^2$$ And Lorentz transformations are $$ \left[ {\begin{array}{c} x' \\ y' \\ t' \end{array} } \right] = \underbrace{\left[ {\begin{array}{ccc} \gamma & 0 &-v\gamma\\ 0 & 1 & 0 \\ -v\gamma/c^2 & 0 & \gamma \\ \end{array} } \right]}_{L_{xv}} \left[ {\begin{array}{c} x(t) \\ y(t) \\ t \end{array} } \right]$$ Now here are my questions,

  • Is the last equation correct? (Correct in the sense that I simply have to substitute the expressions for $x(t)$ and $y(t)$) If yes, then $x',y'$ would depend on $t$ and shouldn't they depend on $t'$ instead?
  • I am quite intrigued by the angle $\alpha$. Shouldn't I also apply some sort of transformation $\alpha\rightarrow\alpha'$? because $S'$ won't see the same angle that $S$ does, right?
  • Also should I make a velocity transformation for the initial velocity and somehow account for it in $S'$? If that's the case how would I need to approach and add this to the equations of $S'$?
  • Lastly, if I were to calculate the range and maximum height that $S'$ sees (which in $S$ are $R = x_0+\frac{v_0^2}{g}\sin(2\alpha)$ and $h_{max} = y_0 + v_0^2\frac{\sin(\alpha)^2}{2g}$ respectively) since $y' = y$ then $h_{max}':$ maximum height seen from $S'$, should be the same right? And for $R':$ range seen from $S'$, would I need to find $t'$ (or $t$ not even sure) such that $y' = 0$ and substitute this value of $t'$ (or $t$) in $x'$ to get $R'$? Or would it be possible I use some transformation for $\alpha$, $v_0$ and $x_0$ and get $R'$ that way?

Any information or help would be greatly appreciated.

  • 1
    $\begingroup$ You are going to need to work out (or look up) how velocity and accelertion transform. Try to find an explanation of motion under a constant acceleration. It's supposed to be called the Bondi space traveler, but searches on that seem to be overpowered by a brand name of luggage. Sigh. Maybe start with en.wikipedia.org/wiki/Bondi_k-calculus $\endgroup$
    – Dan
    Dec 24, 2021 at 4:50
  • $\begingroup$ @Dan Thanks for the info! I mean I kind of get the k-calculus but how would transforming the velocity and acceleration be necessary in this case if I'm only interested in the position of the projectile in the $S'$ frame? $\endgroup$
    – Kevin
    Dec 24, 2021 at 15:15

1 Answer 1


Yes, your matrix equation is correct, although we normally put the time component at the top, not the bottom.

You don't actually need to worry about transforming $g, v_o, \alpha$. Once you have equations for $x', t'$ in terms of $x, t$, you can substitute in $$x = x_0 + v_o\cos(\alpha)t$$ and then eliminate $t$, which will let you write $x'$ as a function of $t'$.

For small $v_o$, the projectile's motion in the $S'$ frame is almost a straight line going backwards at 0.6c. If $v_o$ is large, then your basic projectile motion equations for $x, y$ in $S$ are probably inadequate, unless you happen to have a gigantic plane with uniform gravity. ;)

I won't do the substitution I mentioned above, but here's a simpler example to give you the general idea. Instead of parabolic motion, the projectile has simple uniform motion: $$x = ut$$

with $y=0$, so we can ignore $y$ and $y'$.

The Lorentz transformation gives us $$x' = \gamma(x-vt)$$ $$ct' = \gamma(ct-vx/c)$$

We have $v=\frac35 c$, therefore $\gamma=\frac54$. So $$x'=\frac54 x - \frac34 ct$$ $$ct'=\frac54 ct - \frac34 x$$ Substituting $x = ut$, $$x'=\left(\frac54 u - \frac34 c\right)t$$ $$ct'=\left(\frac54 c - \frac34 u\right)t$$ Thus $$t = \frac{ct'}{\frac54 c - \frac34 u}$$ Substituting that into our last equation for $x'$ we get $$x'=\left(\frac{\frac54 u - \frac34 c}{\frac54 c - \frac34 u}\right)ct'$$ or $$x'=\left(\frac{5u - 3c}{5c - 3u}\right)ct'$$

We can write that as $$x'=u't'$$ where $$u'=\left(\frac{5u - 3c}{5c - 3u}\right)c$$ Note that for small $u$, $u'\approx u-v$

The Lorentz transformation is symmetrical, and it can often be useful to work with the inverse form,

$$x = \gamma(x'+vt')$$ $$ct = \gamma(ct'+vx'/c)$$

Here's a more general way to find $u'$.

$$x' = \gamma(x-vt)$$ $$t' = \gamma(t-vx/c^2)$$

Now $u'=x'/t'$, thus $$u'=\frac{x-vt}{t-vx/c^2}$$

Substituting $x=ut$ $$u'=\frac{ut-vt}{t-uvt/c^2}$$


which can also be written in this form: $$\frac{u'}c=\frac{\frac uc-\frac vc}{1-\frac uc \frac vc}$$

And due to symmetry, $$u=\frac{u'+v}{1+uv/c^2}$$

This is the well-known law of addition of relativistic velocities.

  • $\begingroup$ Yup I am working with non-relativistic initial velocities. Nice! Thanks for the info, so in that sense I'm actually right. Now, would it be possible you help me with the range and maximum height? For the range in $S'$ ($R'$) would it be enoough to find $t'$ such that $y'=0$ and then substitute it in the equation for $x'$? $h'_{max}$ should be the same right? $\endgroup$
    – Kevin
    Dec 24, 2021 at 15:20
  • 1
    $\begingroup$ @Kevin Sure, you can do that. And since $y'=y$, you can just work out the relevant coords of the range & max height in $S$ then transform them to $S'$. The key point is to always work with spacetime events. Get the $(t,x,y,z)$ coords of event $A$ in $S$ and use the transformation to see what the $(t',x',y',z')$ coords of $A$ are in $S'$. $\endgroup$
    – PM 2Ring
    Dec 24, 2021 at 23:23

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