3
$\begingroup$

If yes:

  • Then why are coordinate transformations such as Cartesian $\to$ spherical not included in this set, despite the fact that the spacetime interval / proper time is unchanged (since the space coordinates $x$, $y$, $z$ are simply being re-labeled, while time stays the same)?

If no:

  • Then what is the name of the full group of transformation that leaves invariant the spacetime interval (including Lorentz transformations and Cartesian $\to$ spherical coordinate transformations) ?
  • Then what makes the Lorentz transformations different/special with respect to this group (outlined in the previous question), meaning, why do we only talk about them in SR, disregarding the others members of the group?
$\endgroup$
1

3 Answers 3

3
$\begingroup$

Spacetime interval, defined as $g_{\mu\nu}dx^\mu dx^\nu$ (where $g_{\mu\nu}$ are the components of the metric tensor in a given set of coordinates, and $dx^\mu$ are the components of the infinitesimal displacement in a given set of coordinates), is, by construction, invariant under general coordinate transformations -- assuming that you are transforming the $g_{\mu\nu}$ and $dx^\mu$ as tensor components. This follows simply from the mathematical fact about tensors that transformations acting on upper indices "cancel out" the transformations acting on lower indices. So, the spacetime interval is invariant under any general coordinate transformation.

What is special about Lorentz transformations is that they leave the Minkowski metric invariant! In particular, under a general coordinate transformation, the metric components $g_{\mu\nu}$ will change. For example, you see various terms such as $r^2$ or $\sin^2\theta$ in the flat-space metric when you write it down in polar coordinates. On the contrary, under a Lorentz transformation, the flat-space metric remains Minkowskian, i.e., $\mathrm{diag}(1,-1,-1,-1)$. Lorentz transformations are not the only transformations that have this property, the exhaustive group of transformations that leave the Minkowskian metric invariant is called the Poincare group. The Lorentz group is the homogeneous part of the Poincare group (i.e., such transformations that map the origin to the origin).


PS: It should also be noted that if you define the spacetime interval to be $\eta_{\mu\nu}dx^\mu dx^\nu$ where you take $\eta_{\mu\nu}$ to be a constant matrix given by $\mathrm{diag}(1,-1,-1,-1)$ then the Lorentz transformations (or, Poincare transformations if you include non-homogeneous transformations) are the only transformations that leave the spacetime interval invariant. See this old answer of mine for worked-out details of the proof.

$\endgroup$
7
  • $\begingroup$ focusing only on special relativity, and in minkowski spacetime, how exactly does the Cartesian->spherical coordinate transformation alter the value of the minkowski spacetime interval ($\eta_{\mu\nu}dx^\mu dx^\nu$)? we are not altering the time component, and this spacetime interval correspond to the proper time (i.e. time measured by an observer passing through both events)... with this coordinate transformation (which does not involve time) how are we allowed to "object" to what that proper time observer is measuring on their clock? $\endgroup$
    – TrentKent6
    Feb 12 at 11:36
  • $\begingroup$ @TrentKent6 You're altering the length. If you calculate $\eta_{\mu\nu}dx^\mu dx^\nu$ in spherical coordinates, you'd be calculating $dt^2-dr^2-d\phi^2-d\theta^2$. $\endgroup$
    – ACat
    Feb 13 at 0:55
  • $\begingroup$ Hi, I was asking from the FoR of the proper time observer, who is observing both events at the same place (so no length is involved); in cartesian->spherical coordinate transformations time is not affected, so how can his proper time (which is spacetime interval in SR) be altered? also this comment by G. Smith: "if two events are time-like separated by two seconds, this is true whether the coordinates are cartesian or polar" $\endgroup$
    – TrentKent6
    Feb 13 at 12:12
  • $\begingroup$ @TrentKent6 Proper time can be evaluated by any observer, you can't restrict it to be evaluated by only the proper observer. About G.Smith's comment, that is correct. The point is that proper time is a physical quantity that is frame-invariant. The point is that if you don't transform your metric/coordinates properly then you will calculate it to be something that it actually isn't. [...] $\endgroup$
    – ACat
    Feb 13 at 12:25
  • 1
    $\begingroup$ @TrentKent6 Yes, nobody is claiming otherwise :) What I am (or others are) emphasizing is that this is not enough because we need all observers to be able to calculate the proper time correctly. As I said, in order to do this, you either have to transform the metric appropriately or you have to stick to Lorentz/Poincare transformations. $\endgroup$
    – ACat
    Feb 13 at 12:44
1
$\begingroup$

We use the special coordinate system and the (indefinite) quadratic form (Minkowski metric) to define Minkowski space (as a manifold, if it suits you). Then we can attach to it any other coordinate system (including local or coordinate charts) to express the same manifold in any other coordinate system ("atlas"). Then, of course, we can say that the tensor of the Minkowski metric is the same in all coordinates, therefore the spacetime interval (proper time) remains the same in any coordinate system.

It is important, however, to not forget that the original choice of coordinates -- and the quadratic form expressed in those coordinates -- has a very privileged role because that's how we defined the manifold (Minkowski space in this case). It could have been done in any other coordinate system, to begin with, but this appears simplest and easiest to correlate with observations.

With this, the inhomogeneous Lorentz transformations (the Poincaré group) should be the only ones that preserve the Minkowski metric (I am not sure if there are some pathological transformations, mathematical artifacts without physical meaning). Any such transformation can be expressed in any coordinate system. The form ("how it is written") will, in general, differ from one coordinate system to another, so the representation in the spherical coordinates will differ from that in the Cartesian.

$\endgroup$
1
$\begingroup$

(Note I am considering every transformation here as a passive transformation, i.e. as transformations of globally defined coordinates rather than transformations of the actual spacetime. This is very important, because the reasoning here applies to passive transformations only!)

If you're asking what passive transformations preserve what the spacetime interval actually is, then all possible coordinate transformations preserve it, because the spacetime distance between any two events is a coordinate-independent notion.

However, if you're asking about what passive transformations preserve the way the spacetime interval is written, then the answer is that it is the Poincare group (reinterpreted not as group of isometries but as a group of passive transformations). The transformations from the Poincare group preserve the form of the spacetime interval as $$ ds^{2} = -c^{2}dt^{2} + dx^{2} + dy^{2} + dz^{2} \qquad (*) $$ in Cartesian inertial coordinates.

The Poincare group is the group of all (1) Lorentz boosts, (2) rotations, (3) translations, and (4) reflections, and these are the only passive transformations that preserve the way the spacetime interval is written.

Again, I emphasize that there is a difference in asking what changes the spacetime metric vs asking what changes the way the spacetime metric is written. It is extremely important to understand what is coordinate-independent and what is not.

This gets us to another question you posed.

Then why are coordinate transformations such as Cartesian → spherical not included in this set, despite the fact that the spacetime interval / proper time is unchanged (since the space coordinates x, y, z are simply being re-labeled, while time stays the same)?

The Cartesian → spherical transformations are not included, because they change the way the spacetime metric is written. In spherical coordinates, the spacetime interval is $$ ds^{2} = -c^{2}dt^{2} + dr^{2} + r^{2}d\theta^{2} + r^{2}\sin^{2}\theta d\phi^{2}. $$ Clearly this is not the same form as in $(*)$, so this transformation is not in the Poincare group.

So to summarize,

  • All coordinate transformations of any kind (no matter how crazy and weird) preserve the spacetime distance between two events, because the spacetime distance between two events is coordinate-independent.
  • Only transformations of the Poincare group preserve the way the spacetime distance between two events is written and calculated.
$\endgroup$
9
  • $\begingroup$ Nit: Rotations and reflections are subsets of Lorentz transformations. What you probably mean to single out is boosts. $\endgroup$
    – ACat
    Feb 11 at 20:48
  • 1
    $\begingroup$ @TrentKent6 I should note that I am thinking of all transformations as passive transformations here, i.e. actions on the coordinate system rather than the spacetime. Maybe this makes things more confusing, I apologize, but in my defense the Cartesian -> spherical transformation can only be interpreted as an action on coordinates (hence why I talk about everything as a passive transformation). $\endgroup$ Feb 13 at 16:59
  • 1
    $\begingroup$ @TrentKent6 If we were thinking about active transformations (actions on the spacetime itself), then some actions would be able to actually change the metric (and fail to preserve it). The story is a lot different there. I'll edit my post to clarify that I am talking about passive transformations only. $\endgroup$ Feb 13 at 17:06
  • 1
    $\begingroup$ @TrentKent6 I guess I did make somewhat of a mistake here. The usual definition of the Poincare group is that it is a group of active transformations. If that's the case, I am using a different definition of the Poincare group (as a group of transformations on globally defined coordinate systems). In my case, I define it as the group of passive transformations that leave the way the metric is written unchanged. This ends up being isomorphic to the usual Poincare group nonetheless if you reinterpret each action as an active transformation. $\endgroup$ Feb 13 at 17:17
  • 1
    $\begingroup$ @TrentKent6 To answer your original question, there are two ways we can fix the reasoning in (1)-(5). One way is to keep the Poincare group as a group of active transformations and then recognize that Cartesian -> spherical is a passive transformation, so then (4) and (5) are wrong (just because something is distance preserving, doesn't mean it is an isometry... an isometry must be an active transformation that preserves distance). Alternatively, we can redefine the Poincare group in the way I suggested, but then (1) wouldn't be true. $\endgroup$ Feb 13 at 17:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.