In his lectures, Feynman compares the transformations for rotations of coordinate space – $x'= x\cos\alpha + y\sin\alpha$, $y' = y\cos\alpha – x\sin\alpha$ and z' = z – with the Lorentz transformations, both being effective 'rotations' in space and spacetime respectively. He then writes that the difference in sign between the two sets of transformations and the fact that the rotation in space is written in terms of $\cos(\alpha)$ and $\sin(\alpha)$, rather than algebraic quantities, leads to a difference in geometry. I am struggling to get an intuition for why this is the case, though it makes sense mathematically. I would appreciate an intuition for this.
Special Relativity – Why Does Spacetime Exhibit a Different Geometry to Space?
coordinate systemsinertial-framesmetric-tensorspacetimespecial-relativity
Related Solutions
Recall that the Lorentz group $O(1,3)$ consists of all $4\times 4$ matrices $\Lambda$ satisfying $\Lambda^T\eta\Lambda=\eta$. This gives, in particular, $\det(\Lambda)=\pm1$. The subgroup $SO(1,3)$ consists of all transformations with determinant $+1$.
$\vec{E}\cdot\vec{B}$ is invariant under $SO(1,3)$ but not under all of $O(1,3)$ - it's a pseudoscalar. This is because it's a product of the vector $\vec{E}$ and the pseudovector $\vec{B}$. Here's an easy way of seeing that $\vec{B}$ transforms as a pseudovector:
Consider the Lorentz force law $\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$. Now apply a reflection: \begin{equation}R:(t,\vec{x})\mapsto(t,-\vec{x}).\end{equation} We have: \begin{equation}\vec{F}\mapsto-\vec{F} \qquad \text{and} \qquad\vec{E}\mapsto-\vec{E},\end{equation} so that $\vec{v}\times\vec{B}\mapsto-\vec{v}\times\vec{B}$.
But the velocity transforms as $\vec{v}\mapsto-\vec{v}$ so we have the transformation of $\vec{B}$ under $R$:\begin{equation}\vec{B}\mapsto\vec{B}=(-1)(-\vec{B})=\det(R)(R\cdot\vec{B}).\end{equation}
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.
Best Answer
The difference of the geometries is due to certain sign differences and the use of hyperbolic-trig functions vs circular-trig functions --- not the use of algebraic variables vs trig variables.
Note that rotations can written in terms of slope as $$x'=x\frac{1}{\sqrt{1+m^2}}+y\frac{m}{\sqrt{1+m^2}}$$ $$y'=x\frac{-m}{\sqrt{1+m^2}}+y\frac{1}{\sqrt{1+m^2}}$$ where $m=\tan\alpha$,
and Lorentz boosts can be written trigonometrically as $$t'=t\cosh\theta+\frac{y}{c}\sinh\theta$$ $$\frac{y}{c}'=t\sinh\theta+\frac{y}{c}\cosh\theta$$ where $\frac{v}{c}=\tanh\theta$.
Both are examples of affine Cayley-Klein geometry (see also https://en.wikipedia.org/wiki/Cayley%E2%80%93Klein_metric#Special_relativity ) (which involves projective geometry).
UPDATE:
I think it is enlightening to point out that this "difference of geometry" already occurs with Galilean relativity. (Special relativity made us more aware of this situation... and we see this when we look back to Galilean physics.)
It is known (but not well known) that the position-vs-time graph (regarded as a Galilean spacetime) does not have a Euclidean geometry. In general, two inertial worldline segments with equal elapsed times do not necessarily have equal Euclidean lengths. [Take one inertial worldline to be at rest, and the other at traveling at 3 m/s. Suppose they met at event O. Stop the diagram 1-second after event O. Although their elapsed times are equal to 1, the drawn-length of their worldline segments are unequal.] (The position-vs-time graph also has a different metric and, thus, a different "circle" and a different notion of perpendicularity from that of Euclidean geometry.)
Thus, the position-vs-time graph (from PHY 101) has a non-Euclidean geometry (with zero curvature).
Aspects of this can be seen in my visualization https://www.desmos.com/calculator/kv8szi3ic8
(Tune the E-slider from 1(Minkowski) to -1(Euclidean), and observe the case E=0 (Galilean).)
