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The spacetime interval $(\Delta s)^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 - c^2(\Delta t)^2$ is invariant under the Lorentz transformation and this isn't the case for the Galilean transformation. As the title might suggest, I have tried to prove that the spacetime interval is not invariant under Galilean transformations. I will try to explain what I have done and I wonder what's wrong with my proof.

In my attempt to prove that $(\Delta s)^2$ isn't invariant under Galilean transformation I wrote $(\Delta s)^2$ as $(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2 - c^2(t_2 - t_1)^2$ and fully expanded it as

$-c^2 t_1^2+2 c^2 t_1 t_2+c^2 \left(-t_2^2\right)+x_1^2-2 x_1 x_2+x_2^2+y_1^2-2 y_1 y_2+y_2^2+z_1^2-2 z_1 z_2+z_2^2$

I then reasoned that if $(\Delta s)^2$ really were invariant during a Galilean transformation $(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 - c^2(\Delta t) = (\Delta x')^2 + (\Delta y')^2 + (\Delta z')^2 - c^2(\Delta t')$ must hold true.

I applied the same principle as before and wrote $(\Delta x')^2 + (\Delta y')^2 + (\Delta z')^2 - c^2(\Delta t')^2 = (x'_2 - x'_1)^2 + (y'_2 - y'_1)^2 + (z'_2 - z'_1)^2 - c^2(t'_2 - t'_1)^2$

Then, the Galilean transformation

$\begin{cases}t' = t \\ x' = x - v_x t \\ y' = y - v_y t \\ z' = z - v_z t\end{cases}$

was applied and I substituted terms in

$(x'_2 - x'_1)^2 + (y'_2 - y'_1)^2 + (z'_2 - z'_1)^2 - c^2(t'_2 - t'_1)^2$

into

$\left((x_2 - v_x t) - (x_1 - v_x t)\right)^2 + \left((y_2 - v_y t) - (y_1 - v_y t)\right)^2 + \left((z_2 - v_z t) - (z_1 - v_z t)\right)^2 - c^2 (t_2 - t_1)^2$

Again, I expanded the expression

$-c^2 t_1^2 + 2 c^2 t_1 t_2 - c^2 t_2^2 + x_1^2 - 2 x_1 x_2 + x_2^2 + y_1^2 - 2 y_1 y_2 + y_2^2 + z_1^2 - 2 z_1 z_2 + z_2^2$

Remember, I reasoned that $(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 - c^2(\Delta t) = (\Delta x')^2 + (\Delta y')^2 + (\Delta z')^2 - c^2(\Delta t')$ must hold true and now we have an expanded from for both sides, one being the spacetime interval in a "rest frame" $S$ and the other the spacetime interval for the same events in $S'$. If we compare them, we find that they are equal to each other:

$(\Delta s)^2_S = -c^2 t_1^2+2 c^2 t_1 t_2+c^2 \left(-t_2^2\right)+x_1^2-2 x_1 x_2+x_2^2+y_1^2-2 y_1 y_2+y_2^2+z_1^2-2 z_1 z_2+z_2^2$

$(\Delta s)^2_{S'} = -c^2 t_1^2 + 2 c^2 t_1 t_2 - c^2 t_2^2 + x_1^2 - 2 x_1 x_2 + x_2^2 + y_1^2 - 2 y_1 y_2 + y_2^2 + z_1^2 - 2 z_1 z_2 + z_2^2$

Since they are equal, $(\Delta s)^2_S = (\Delta s)^2_{S'}$ and the spacetime interval is invariant under Galilean transformations.

What's wrong with my proof?

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closed as off-topic by Danu, Kyle Kanos, ACuriousMind, HDE 226868, user36790 Oct 15 '15 at 6:28

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    $\begingroup$ I'm voting to close this question as off-topic because check-my-work questions are off-topic on this site. $\endgroup$ – Danu Oct 14 '15 at 20:35
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    $\begingroup$ As a hint... does $x_1 - x_2 = x_1' - x_2'$ make sense if $t_1\neq t_2$? $\endgroup$ – Andrew Oct 14 '15 at 21:03
  • $\begingroup$ @Timaeus I went with how the question was framed (as per the last sentence), rather than with the precise contents. $\endgroup$ – Danu Oct 14 '15 at 21:08
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Recall that to disprove something, you only need one counterexample. So I'd simply advise: don't bother with the symbols: choose a simple example. Think of the events at $X_1=(t,\,x,\,y,\,z)=(0,0,0,0)$ and $X_2=(t,\,x,\,y,\,z)=(1,0,0,0)$. The proper time between them is 1. Now transform to co-ordinates boosted in the $x$ direction by speed $v_x=1/2$. Using the Galilean boost, we have $X_1^\prime = X_1$, $X_2^\prime = (1,\,-\frac{1}{2},\,0\,0)$ and the proper time between them is now $\frac{\sqrt{3}}{2}$. quod erat non demonstrandum

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  • $\begingroup$ Really? I get the spacetime interval $-(-0.5)^2+1^2 = 0.75$. What do you get $\frac{\sqrt{3}}{2}$ from? $\endgroup$ – Markus Klyver Oct 23 '15 at 15:38
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    $\begingroup$ @MarkusKlyver We're talking the same language. You've calculated the metric element $(\Delta s)^2$ and I've calculated the proper time $\tau = \Delta s$. $\endgroup$ – WetSavannaAnimal Oct 23 '15 at 22:28
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You just correctly showed that a Galileian "transformation" where $v_x=v_y=v_z=0$ "preserves" the interval.

You were supposed to find two Galilean frames and two events where the two frames compute different intervals. You didn't even pick two frames since you let $v_x=v_y=v_z=0$.

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  • $\begingroup$ Where do I assume that? $\endgroup$ – Markus Klyver Oct 14 '15 at 21:07
  • $\begingroup$ @MarkusKlyver You introduced $v_x$ and such then used it once and then it disappeared, so you set it equal to zero. Things that aren't zero don't disappear and you never said it was non zero. $\endgroup$ – Timaeus Oct 14 '15 at 21:11
  • $\begingroup$ Well, I put Expand[FullSimplify[((x2 - vx*t) - (x1 - vx*t))^2 + ((y2 - vy*t) - (y1 - vy*t))^2 + ((z2 - vz*t) - (z1 - vz*t))^2 - c^2 (t2 - t1)^2]] into Mathematica and got back -c^2 t1^2+2 c^2 t1 t2-c^2 t2^2+x1^2-2 x1 x2+x2^2+y1^2-2 y1 y2+y2^2+z1^2-2 z1 z2+z2^2. $\endgroup$ – Markus Klyver Oct 14 '15 at 21:14
  • $\begingroup$ @MarkusKlyver But why would you type that into Mathematica? And why wouldn't you be alarmed that the velocity dropped out? None of those things with the v are the primed coordinates. They are just translated unprimed coordinates so you just picked a different origin. $\endgroup$ – Timaeus Oct 14 '15 at 21:18
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    $\begingroup$ @MarkusKlyver: $t$ is the random symbol. $t_1$ and $t_2$ are not random symbols. If you're getting a "naked" $t$ then you have screwed up. Furthermore, if $x_1'$ depends on $t_2$ or $x_2'$ depends on $t_1$ you have screwed up. There exists one way to do $x' = x - v_x~t$ that avoids all of these pitfalls, can you see what it is? (Hint: it is a simple relabeling of the terms in that equation.) Can you then understand why everything else you're doing has been physically wrong, as in it does not perform a Galileian transformation but rather a coordinate translation? $\endgroup$ – CR Drost Oct 15 '15 at 12:56
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Your problem is buried somewhere in "if we expand them then we find out that they are equal to each other." So if we map $(t, \vec r) \mapsto (t, \vec r - \vec v~t)$ we find $\Delta t \mapsto \Delta t,\; \Delta \vec r \mapsto \Delta \vec r - \vec v~\Delta t$ where the operator $\Delta$ takes a symbol $U$ and transforms it into $U_2 - U_1.$

Therefore, $(\Delta \vec r)^2 - c^2 (\Delta t)^2 \mapsto (\Delta \vec r - \vec v~\Delta t)^2 - c^2 (\Delta t)^2,$ and it is not very hard to expand the right hand side to find:$$(\Delta \vec r)^2 - c^2 (\Delta t)^2 - \Delta t \Big[2 \vec v\cdot\Delta\vec r - \vec v\cdot\vec v ~\Delta t\Big],$$which is obviously not equal to $(\Delta \vec r)^2 - c^2 (\Delta t)^2$ unless $\Delta \vec r = \vec v~\Delta t~/2.$

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  • $\begingroup$ No, the problem is not getting an $x_1'$ or an $x_2'$ correctly. $\endgroup$ – Timaeus Oct 14 '15 at 21:56

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