Assume we have dimensions \((x, y, t)\). We've already shown through thought experiments that all that matters is:
So to exhibit all effects, \(2+1D\) suffices: 2 dimensions of space, one along our direction of motion, one perpendicular. One direction of time.
Assume we send a pulse of light at \((x_0, y_0, t_0)\). Say light travels till time \(t_1\), and reaches the point \((x_1, y_1, t_1)\). By definition, the distance travelled by light is \(c(t_2 - t_1)\) as \(\text{distance} = \text{speed} \times \text{time}\). Also note that the distance is equal to \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Equating the two, we get:
\[ c(t_2 - t_1) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ (x_2 - x_1)^2 + (y_2 - y_1)^2- c^2(t_2 - t_1)^2 = 0 \]
By the assumption of special relativity, the speed of light is constant in every inertial frame of reference. So we know that \(dx/dt = c = dx'/dt'\).
TODO(Siddharth): How do we derive thatthe spacetime interval is constant in all reference frames from the constancy of speed of light? That is, for a new reference frame, where the coordinates of the points are \((x_1', y_1', t_1')\) and \((x_2', y_2', t_2')\), show that we have:
\[ (x_2' - x_1')^2 + (y_2' - y_1')^2- c^2(t_2' - t_1')^2 = 0 \]
Answer (Goose): The exact same formulation as above with all variables replaced by their primed versions. Since you have a 0 on the right hand side of the final quantity, the quantity on the left side (a.k.a the "interval") is conserved in all frames. Alternatively, \((\Delta x,\Delta y,\Delta z,\Delta t)\) are coordinates in a particular inertial frame, but we do not put any additional restrictions on that frame. Hence:
\[
\frac{distance}{time} = speed \\
\frac{\sqrt{(\Delta x)^2 + (\Delta y)^2}}{\Delta t} = c \\
(\Delta x)^2 + (\Delta y)^2 = c^2(\Delta t)^2 \\
c^2(\Delta t)^2 - (\Delta x)^2 - (\Delta y)^2 = 0 \\
\]
in any inertial frame.
Rebuttal(Siddharth)
The argument that because a relationship betwen \(\Delta x, \Delta y, \Delta t = 0\) in one inertial frame implies that it will remain so in all intertial frames assumes that our transformation is tensorial. A mild assumption to be sure, but an assumption nonetheless.
What the principle that says that \(\text{distance}/\text{time} = \text{speed}\) holds in all reference frames? This feels like deep magic to me :P
Re-Rebuttal(Goose):
Re^3 buttal (Siddharth)
Re^4 buttal (Goose):
Better TODO (Goose): Actually prove invariance of the interval (starting from light speed invariance)
Better TODO (Goose): Prove interval invariance along non-null curves in spacetime.Let \(O, O'\) be to overlapping inertial frames. Two events \(E1\) and \(E2\) are observed by both frames and the coordinate separations are measured as \((\Delta x, \Delta y, \Delta t)\) and \((\Delta x', \Delta y', \Delta t')\). Prove
\[ c^2(\Delta t)^2 - (\Delta x)^2 - (\Delta y)^2 = c^2(\Delta t')^2 - (\Delta x')^2 - (\Delta y')^2\] (In general, the interval maybe non-zero)
I have a feeling the proof cannot avoid using light clocks in some sense, but I'm not sure.
Question about invariance of interval (Siddharth)
\(\Delta x^2 + \Delta y^2 - \Delta t^2 = 0 = \Delta x'^2 + \Delta y'^2 - \Delta t'^2\) is only the invariance of "light intervals", and not the invariance of "arbitrary intervals"? I guess that is the difference?
o whose coordinates in the \((x, y, t)\) [unprimed] frame is \(x_o = t_o v\). That is, it moves with a constant velocity \(v\) along the \(x\) axis as measured by the unprimed frame.o started at the origin of the primed frame. Hence, \(x_o' = 0\). This gives us:\[ t_o^2 - t_o^2v^2 = t_o'^2 - 0 \\ t_o^2 (1 - v^2) = t_o'^2 \\ t_o = t_o'/\sqrt{1 - v^2} ~ \text{(when $x'_o = 0$)} \]
We write \(\gamma \equiv 1/\sqrt{1 - v^2}\). This simplifies the equation to \(t_o = \gamma t_o'\). Since \(v < 1\), \(\gamma > 1\), and is thus called the time stretch factor.
Substituting this into \(x_o = t_o v\) gives \(x_o = (\gamma t'_o)v\), when \(x'_o = 0\) .
TODO: fill this in
For contradiction, assume that \(x = f(x', t')\) is non-linear.
\[ \begin{bmatrix} x_o \\ t_o \\ y_o \end{bmatrix} = \begin{bmatrix} A & B & 0 \\ C & D & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x'_o \\ t'_o \\ y_o \end{bmatrix} \]
we need to figure out the values of \(A, B, C, D\).
\[ \begin{bmatrix} x_o \\ t_o \\ y_o \end{bmatrix} = \begin{bmatrix} A & v \gamma & 0 \\ C & \gamma & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x'_o \\ t'_o \\ y_o \end{bmatrix} \]
TODO: Can we now fix \(t'_o = 0\) and repeat the same derivation to get \(A, C\)?
\[x_o^2 - t_o^2 = x'_o^2 - t'_o^2\]
We have two inertial, overlapping, parallel (axes are parallel and origins coincide) frames of reference \(O, O'\). \(O'\) is moving with respect to \(O\) with speed \(v\) along \(x\) axis. For our purposes it is sufficient to consider 2 + 1 dimensions. Hence we use coordinate systems \((t, x, y)\) and \((t', x', y')\). Our goal is to find the following transformation functions. Note that we are not talking about any objects or any events yet.
\[\begin{align}
x' &= f(x, y, t) \\
y' &= h(x, y, t) \\
t' &= g(x, y, t) \\
\end{align}\]
Since the transverse dimension is invariant, the second equation becomes \(y' = y\) and can be safely dropped. Further, it also follows that the functions \(f,g\) cannot depend on the \(y\) coordinate. To illustrate, put a straight rod parallel to the \(y\) axis with one end on the \(x\) axis at some point \(x=k\) in the \(O\) frame. If \(f\) is a function of \(y\), the \(O'\) will no longer observe the rod to be straight. Further, \(O'\) could get information about its own speed w.r.t \(O\) solely by observing the rod, which violates Special Relativity. If \(g\) is a function of \(y\), then \(O'\) will observe different parts of the rod to exist at different points in time. Both are clearly non-sensical predictions for a physical theory. (Yes Siddharth, I know this isn't a very mathematical argument, but we are doing physics. Just as reality is the final verifier of a theory, non-sense is a sure-shot rejector). Our transformations now become:
\[\begin{align}
x' &= f(x, t) \\
t' &= g(x, t) \\
\end{align}\]
We now proceed in the following steps.
TODO: Formalize the proof. Arguments about isometry and isotropy of space. Specifically remove the second case Siddharth and I discussed about.
Since our origins coincide, our transformations now are:
\[\begin{align}
x' &= Ax + Bt \\
t' &= Cx + Dt \\
\end{align}\]
First, we use the fact that the \(O'\) is travelling with velocity \(v\) with respect to \(O\). This means that any point stationary in \(O'\) will be observed by \(O\) to have velocity \(v\) independent of position and time. Let us write the trajectory of the point \(x' = k\). We're still not dealing with objects, just pure math.
\[\begin{align}
x' &= k \\
Ax + Bt &= k \\
x &= \frac{-B}{A} t + \frac{k}{A} \\
\frac{dx}{dt} &= \frac{-B}{A} \\
v &= \frac{-B}{A} \\
B &= -Av \\
\end{align}\]
Our position transform now becomes \(x' = A(x - vt)\). Since we are trying to derive a general transform, the same equation must also apply in the reverse direction. The only change is that if \(O'\) has velocity \(v\) w.r.t \(O\), then \(O\) has velocity \(-v\) w.r.t \(O'\). This allows us to write the following set of equations:
\[\begin{align}
x' &= A(x-vt)\\
x &= A(x'+vt')\\
\end{align}\]
Second, we use the invariance of light speed. Specifically, imagine a light flash from the common spacetime origin. \(O\) observed the trajectory of this flash to be \(x = ct\) and \(O'\) observes the trajectory to be \(x' = ct'\). Here we've used the fact that the velocity of light is the same \(c\) in all frames. As you will note, we haven't used invariance of any intervals here. Substituting that into our equations above:
\[\begin{align}
ct' &= At(c-v)\\
ct &= At'(c+v)\\
\end{align}\]
Multiplying them together:
\[\begin{align}
c^2t't &= A^2tt'(c^2-v^2)\\
\frac{c^2}{c^2-v^2} &= A^2\\
A &= \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\\
\end{align}\]
This factor is the usual gamma \(\gamma = \frac{1}{\sqrt{1-(v/c)^2}}\). Our final position transform now becomes \(x' = \gamma(x-vt)\).
The state of our equations is:
\[\begin{align}
x' &= \gamma(x-vt) \\
t' &= Cx + Dt \\
\end{align}\]
Again, we use the invariance of light speed, but with different substitutions. Again, the equations \(x'=ct'\) and \(x=ct\) must simultaneously be satisfied. We make the substitutions \(x'=ct'\), \(x=ct\), and \(t=x/c\) in the first equation alone.
\[ct' = \gamma(ct-vx/c) \implies t' = \gamma(t-vx/c^2)\]
This gives us the transformation of time along a null trajectory in spacetime. Since we are deriving a general transform, our time transformation equation must also obey this equation for null trajectories. This allows us to compare the above equation with
\[t' = Cx + Dt\]
and we get \(C = -\gamma v / c^2\) and \(D = \gamma\), completing the Lorentz transform.
\[\begin{align} \gamma &= \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \\ x' &= \gamma(x-vt) \\ t' &= \gamma\big(-\frac{vx}{c^2} + t\big) \end{align}\]
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