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.
Best Answer
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 $x=ut$ $$u'=\frac{ut-vt}{t-uvt/c^2}$$
$$u'=\frac{u-v}{1-uv/c^2}$$
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.