Help understanding proof of ds=ds' in Classical Theory of Fields

[tx_kurt]

TL;DR

I am unable to follow the argument used in L&L's Classical Theory of Fields to show that an infinitesimal interval is invariant. Help appreciated.

I just decided to look at Landau & Lifshitz' Classical Theory of Fields (English version, 4th ed), and I am a bit embarrassed to be confused already on page 4&5 of this book. The book can be viewed on archive.org.
The goal of this section of the book is to show $s = s'$ starting from only the principle of relativity and the existence of a maximum signal speed c. The interval s is defined and since c is the same in all frames if follows that $s = 0$ implies $s' = 0$. Next, an infinitesimal interval is considered:
$ds^2=c^2 dt^2 - dx^2 - dy^2 - dz^2$ and similar in a second inertial frame with primed coordinates. It is argued that $ds^2$ and ${ds'}^2$ must be proportional to each other, and that the proportionality constant can only depend on the absolute value of relative velocity between the two inertial systems: $ds^2=a(V){ds'}^2$. (Note that although the interval considered is infinitesimal, there is no restriction on the relative velocity between the two systems.) Once this last equation is established, it follows that $a(V)=1$ so that the differential intervals are equal (and finite intervals as well).
But it's this argument that the coefficient $a(V)$ can only depend on the absolute relative velocity between the systems that I don't understand. The text contains the argument "[the coefficient] cannot depend on the direction of the relative velocity, since that would contradict the isotropy of space." But why can't the proportionalty depend on the angle between the relative velocity and $(dx,dy,dz)$ for example?

(Be patient, this is my first post.)

 

[anuttarasammyak]

Imagine you have full set of dashed systems which have all the directions but same magnitude of relative velocity. If you observe some coefficient a are larger than the other, you have succeeded to find special direction or anisotropy of the universe.

 

[PeterDonis]

why can't the proportionalty depend on the angle between the relative velocity and $(dx,dy,dz)$ for example?

Because such a dependency would not be isotropic. It would mean $a(V)$ would change if the direction of the relative velocity changed.

 

[PeroK]

But it's this argument that the coefficient $a(V)$ can only depend on the absolute relative velocity between the systems that I don't understand. The text contains the argument "[the coefficient] cannot depend on the direction of the relative velocity, since that would contradict the isotropy of space." But why can't the proportionalty depend on the angle between the relative velocity and $(dx,dy,dz)$ for example?

The homogeneity of spacetime allows you to choose a common origin for any two inertial reference frames. And the isotropy allows you to choose the direction of relative motion of the origins as the x-axis, for example. There is, therefore, no physical difference between an arbitrary configuration of coordinates and the simplified configuration with coincident origins and relative motion along the x-axis.

This emphasises the point that there are no absolute angles in space. For example, if an object A was moving at speed $V$ along the positive x-axis and another object $B$ was moving at the same speed $V$ at an angle $\theta$ to the x-axis, then you could change your coordinates so that object $B$ was moving along the x-axis and object A was moving at an angle $\theta$ to the x-axis. And, by the homogeneity and isotropy of space, these two scenarios would be physically identical.

 

[Ibix]

But why can't the proportionalty depend on the angle between the relative velocity and $(dx,dy,dz)$ for example?

Say it depends on the angle away from the x axis. Rotate your coordinate system so that the old x axis points along some arbitrary vector (A, B, C) in the new system. The scale factor can't depend on angle away from the new x axis, since that would be inconsistent with the result in the old coordinate system. It must depend on angle away from that arbitrary vector. The conclusion is that we've picked that direction (the one we first called the x direction and now call the (A, B, C) direction) out as a special direction in space, violating the assumption of isotropy.

 

[tx_kurt]

Imagine you have full set of dashed systems which have all the directions but same magnitude of relative velocity. If you observe some coefficient a are larger than the other, you have succeeded to find special direction or anisotropy of the universe.

I would agree if we already knew that the same a applied for all infinitesimal intervals. However, at this point in the argument, the coefficient a applies only to one infinitesimal interval (dt, dx, dy, dz). This infinitesimal interval has a direction.

 

[PeterDonis]

at this point in the argument, the coefficient a applies only to one infinitesimal interval

No, it doesn't. The argument applies to any infinitesimal interval, not just one.

 

[tx_kurt]

Thank you for the answers so far. Let me try restating my problem. (Btw, I'm used to seeing the invariance of the interval proved using the Lorentz transform. Landau is taking a different approach, going straight for interval invariance armed only with the relativity principle and a maximum signal speed c.)

At this stage, we are considering two events with infinitesimal separation $(dt, dx, dy, dz)$, and asking how the interval $ds^2$ can appear in other inertial systems. As Ibex commented, we have a freedom of choice for our coordinate axes. Let's choose them so that the infinitesimal considered is along the x direction: $(dt, dx, 0, 0)$. Now consider how $ds^2$ appears in a system with relative velocity $(v_x, v_y, v_z)$.

Because $ds'=0$ when $ds=0$, and $ds'$ and $ds$ must be infinitesimals of the same order, they must be proportional with a coefficient depending only on the relative velocity of the second frame: $ds^2 = a {ds'}^2$. (This is the step at the bottom of page 4.)

Taking the most general dependence on velocity, we have $ds^2 = a(v_x, v_y, v_z)\ {ds'}^2$, with $a$ a function of all three components of the velocity. Now, because the infinitesimal separation $(dt, dx, 0, 0)$ is symmetric to rotations around the x axis, $a$ can only depend on $v_y$ and $v_z$ through the total transverse velocity. In other words we must have $a(v_x, v_y^2 + v_z^2)$. I would like to extend that same argument to $v_x$ also, but this is the direction of the infinitesimal separation we are considering. That's essentially my problem: The universe is isotropic, but the infinitesimal separation we are considering is not.

 

[anuttarasammyak]

As Ibex commented, we have a freedom of choice for our coordinate axes. Let's choose them so that the infinitesimal considered is along the x direction: (dt,dx,0,0). Now consider how ds2 appears in a system with relative velocity (vx,vy,vz).

dx^2+dy^2+dz^2:=dl^2
is invariant under translation and rotation invariance in 3D homogeneous and isotropic space as Euclid says. With traslation invariance of time.
c^2dt^2-dl^2=ds^2
is invariant in 3D transformations which does not change dt.　dt^2 and dl^2 are invariant separately.

Now, because the infinitesimal separation (dt,dx,0,0) is symmetric to rotations around the x axis, a can only depend on vy and vz through the total transverse velocity. In other words we must have a(vx,vy2+vz2). I would like to extend that same argument to vx also, but this is the direction of the infinitesimal separation we are considering.

With no regard and in addition to the translation and rotation of 3D coordinates above said, you have freedom to choose direction and magnitute of relative velocity, which is interpreted as rotation in time-space plane. dt^2 and dl^2 are not invariant anymore but c^2dt^2-dl^2 is.

 

[PeterDonis]

At this stage, we are considering two events with infinitesimal separation

No, we aren't. We are considering any possible pair of events with infinitesimal separation, not just one. So this...

The universe is isotropic, but the infinitesimal separation we are considering is not.

...is wrong, because we are not considering just one particular infinitesimal separation. We are considering the entire set of possible infinitesimal separations, and that set, as a whole, is isotropic.

In other words, the factor $a$ in the formula has to be the same for any infinitesimal separation we choose, so $a$ cannot depend on the direction of any particular infinitesimal separation, because then it would be different if we switched to another infinitesimal separation, but it can't be.

 

[PeroK]

... in a homogeneous and isotropic spacetime, there is no way to distinguish one differential spacetime interval from another. Whatever holds for one, holds for all.

The 2D surface of a sphere is homogeneous and isotropic. All points and all differentials at any point are geometrically equivalent.

Saying "this interval might be special" is equivalent to non-homogeneity or non-isotropy.