Reconciling Minkowski and 3+1 view of special relativity

Question

I am having some trouble reconciling the Minkowski (4-dimensional) and the pre-Minkowski (3+1-dimensional) approach to special relativity. Let me describe (how I interpret) the Lorentz transformations in these two approaches.

3+1 view

There exist two reference frames, $S(t,x,y,z)$ and $S'(t',x',y',z')$ so that $S'$ moves with velocity $\vec{u} = u \vec{e}_x$ with respect to the frame $S$. We can take the axes of the two frames to be parallel $\vec{e}_x = \vec{e}_{x'}$, $\vec{e}_y = \vec{e}_{y'}$ and $\vec{e}_z = \vec{e}_{z'}$. Here, it is assumed that the basis vectors of both frames live in the same vector space so that we can compare and equate them.

Now, let us relate the description of motion in these two frames. If a particle moves with velocity $\vec{v} = v^x \vec{e}_x + v^y \vec{e}_y + v^z \vec{e}_z$ in $S$ and with the corresponding velocity $\vec{v}' = v^{x'} \vec{e}_{x'} + v^{y'} \vec{e}_{y'} + v^{z'} \vec{e}_{z'}$ in $S'$, then the Lorentz transformations imply the following transformation law for the components of the velocities $$v^{x'} = \frac{v^x - u}{1 - \frac{u v^x}{c^2}}, \quad v^{y'} = \frac{1}{\gamma}\frac{v^y}{1 - \frac{u v^x}{c^2}}, \quad \quad v^{z'} = \frac{1}{\gamma}\frac{v^z}{1 - \frac{u v^x}{c^2}},$$ where $\gamma = (1-u^2/c^2)^{-1/2}$ is the standard Lorentz factor. Now, we can multiply these transformation laws with the appropriate basis vectors and sum to obtain the transformation law for the velocity vector (and not just the components) $$\vec{v}'_\parallel = \frac{\vec{v}_\parallel - \vec{u}}{1 - \frac{\vec{u} \cdot \vec{v}}{c^2}}, \quad \vec{v}'_\perp = \frac{1}{\gamma} \frac{\vec{v}_\perp}{1 - \frac{\vec{u} \cdot \vec{v}}{c^2}}.$$ Here, parallel and perpendicular is meant with respect to $\vec{u}$. The two last equations make perfect sense since all involved vectors are elements of the same vector space.

Minkowski view

In Minkowskian view, the two reference frames are described by two tetrads $\{\vec{e}_t, \vec{e}_x, \vec{e}_y,\vec{e}_z\}$ and $\{\vec{e}_{t'}, \vec{e}_{x'}, \vec{e}_{y'},\vec{e}_{z'}\}$ which are related via hyperbolic rotation $$\vec{e}_{t'} = \vec{e}_{t} \cosh \chi + \vec{e}_{x} \sinh \chi, \quad \vec{e}_{x'} = \vec{e}_{x} \cosh \chi + \vec{e}_{t} \sinh \chi,$$ so that, while $\vec{e}_y = \vec{e}_{y'}$ and $\vec{e}_z = \vec{e}_{z'}$, the $x$ and $x'$ basis vectors do no longer coincide $\vec{e}_x \neq \vec{e}_{x'}$.

So, the first puzzling differences between the two approaches is that they do not agree on the equality of the basis vectors.

Let's see what this implies for the velocity transformation. A particle with four-velocity $\vec{V}$ is equally well described in both frames, $$\vec{V} = V^t \vec{e}_t + V^x \vec{e}_x + V^y \vec{e}_y + V^z \vec{e}_z = V^{t'} \vec{e}_{t'} + V^{x'} \vec{e}_{x'} + V^{y'} \vec{e}_{y'} + V^{z'} \vec{e}_{z'},$$ where the transformation laws for the components is inherited from the transformation law for the basis vectors $$V^{t'} = V^{t} \cosh \chi - V^{x} \sinh \chi, \quad V^{x'} = V^{x} \cosh \chi - V^{t} \sinh \chi,$$ and $V^{y'} = V^y, V^{z'} = V^z$. The standard relation between the three- and four-velocity is taken to be $$\vec{V} = V^t (\vec{e}_t + \vec{v}/c) = V^{t'} (\vec{e}_{t'} + \vec{v}'/c).$$ If we write the 3-velocities in terms of the components, we have $v^k =c V^k/V^t$, with $k \in \{x,y,z\}$ and similarly for the primed vector. From the known transformation laws of the components of the 4-velocity, we can deduce the transformation law for the components of the 3-velocity $$v^{x'} = \frac{v^x - c \tanh \chi}{1 - \frac{v^x c \tanh \chi}{c^2}}, \quad v^{y'} = \frac{1}{\cosh \chi}\frac{v^y}{1 - \frac{v^x c \tanh \chi}{c^2}}, \quad v^{z'} = \frac{1}{\cosh \chi}\frac{v^z}{1 - \frac{v^x c \tanh \chi}{c^2}},$$ which agrees with the 3+1 transformation law for the components of 3-velocity if we identify $u = c \tanh \chi$ and $\gamma = \cosh \chi$. However, in contrast to the 3+1 approach, we can no turn the above transformations to vector transformations because $x$ and $x'$ basis vectors no longer point in the same direction.

My question: What is the correct way to interpret this "paradox"? Also, what is the correct interpretation of the velocity-addition formula? Does is hold only component-wise, or at the level of 3-vectors?

Answer 1

I think the fact that $\vec{e}_{x'} = \vec{e}_x \cosh\chi + \vec{e}_t \sinh\chi \neq \vec{e}_x$ can be interpreted as a direct expression of relativity of simultaneity.

Based on our non-relativistic experience, when we say that $S$ observes $S'$ moving in the x direction, while $S'$ observes $S$ moving in the x' direction, we automatically conclude that the x-axis in $S$ coincides with the x'-axis in $S'$. There is no reason not to do so as long as we consider strictly 3D-geometry in 3D space, without reference to the time coordinate.

However, when we bring in time and relativity, this is no longer correct, although we commonly ignore the change because it lies outside the domain of our standard experience. To see why, imagine the $x$ and $x'$ axes as very thin straight rods stretching from $-\infty$ to $\infty$ and at rest in their respective frames. As seen in either frame, one rod is stationary, while the other slides past at uniform velocity. In $S$ the points of the x-rod described by $\vec{e}_{x}$ correspond to the same time $t$. But $S$ observes the $x'$-rod as being in motion. If we examine the latter at the same given $t$, according to the Lorentz transformations we find that each of its points corresponds to a different time $t'$ in $S'$. This is simply relativity of simultaneity. From this point of view, the fact that $\vec{e}_{x'} \neq \vec{e}_x$ simply tells us that what is simultaneous in $S$ is no longer simultaneous in $S'$ and conversely.

As for the velocity formulas, if you refer to $\vec{V}$ and $\vec{V'}$, note that the time-like component of a 4-velocity is not independent of the space-like components. So addition of two 4-velocities does not yield a 4-velocity, although a 4-velocity is a contravariant vector. In order to have four independent components amenable to a true addition of 4-vectors you must use the 4-momentum, which for a particle is the 4-velocity multiplied by the (scalar) rest mass, $\vec{P} = m_0 \vec{V}$. But 4-momentum has a completely different significance: there is no "relative momentum" of two inertial frames.

If you refer to $v^k$ and $v^{k'}$, the addition formula obviously concerns 3-velocities as observed by $S$ and $S'$ in their respective views of in 3D-space.

Answer 2

I disagree that in the 3+1 view you have $\mathbf{e}_x = \mathbf{e}_{x'}$. This is because these vectors do not in fact belong to the same space. To make sense of the notion of three-dimensional space you need simultaneity; you need to be able to form a 3d space of vectors at the same time, where you can put $(x,y,z)$ coordinates. This is implicit in the Galiliean view of spacetime. The thing is, you can construct such a simultaneity, but it is observer dependent; I'm sure you knew this already.

The easiest way to understand what this implies is to consider 2+1 spacetime, with one observer moving in the x-direction with respect to the other. In the $S$ frame you can draw mutually perpendicular $t$-, $x$-, and $y$-axes. What are the corresponding axes (i.e., the basis vectors) of $S'$? The $y'$-axis is the same, but the $t'$- and $x'$-axes are rotated according to the Lorentz transformation. Again, this is nothing new; your Minkowski view says that $\mathbf{e}_x \neq \mathbf{e}_{x'}$. What I think the conclusion should be is that the 3+1 view is invalid; the 3d basis vectors are only "equal" if you pretend that simultaneity is absolute. When you write the transformation law for 3-velocity, you are contradicting yourself. You say that you get it from the Lorentz transformation, but the Lorentz transformation is telling you that the basis vectors change!

I think this gets to the core of your question. You have assumed that it is possible to have the basis vectors not change, and your reasoning is telling you that you have assumed something false! In other words, the 3+1 view is wrong; at least I cannot see how to make sense of it.

Answer 3

In the 3+1 case the spatial basis vectors live in the Euclidean space and the transformation law you derived, which is a Lorentz boost and does NOT include spatial rotations, does not affect the spatial basis vectors. In other words, you have found how the two observers view the velocity using the SAME spatial basis.

In the Minkowski space a Lorentz boost is a hyperbolic rotation of the spatial-temporal axes so, even in the absence of spatial rotations, the basis "rotate" hyperbolically (e.g. in the 1+1 space the transformed temporal and spatial axes are no longer orthogonal but they converge symmetrically to the light cone).

Concluding, in the 3+1 case, a Lorentz boost does not rotate the spatial basis but in the Minkowski space it rotates the spatial and temporal axes.

Answer 4

Suppose you were comparing two approaches to keeping track of orientation in three dimensions - quaternions and Euler angles, for example. You might worry about the fact that they have completely different behavior under rotation. You might even see it as a "paradox", and try to find a physical "interpretation" that relates quaternions to Euler angles. But the physically observable results of, say, rotating a broomstick from the z axis by 90 degrees about the y axis are the same for each system. The rest is just your choice of mathematical abstraction to get to that answer.

In the same way, you can worry about the different ways Lorentz and Minkowski axes transform under boosts and try to find some physical intuition about it. But you shouldn't. The only connection to reality for either system is through their predictions of observable results. The physically observable results are the same in both systems. The rest is just abstraction.

Answer 5

This is merely the equivalence between passive and active transformations.

In your "pre-Minkowskian" case, you can consider the $S$ and $S'$ frames to actually be the same in every way, and it is the velocity vector that is being transformed. Since all the basis vectors are the same, there is no way to say that $S$ and $S'$ are actually different. The differing velocities measured between the two is therefore an active transformation.

In the "Minkowskian" case, there are two clearly distinct frames, and we evaluate the components of a single geometric quantity in two different bases. That's a passive transformation.

But we know from mathematics that passive and active transformations are the same, and physically you can't distinguish the two. So both of these describe the same physics.

Answer 6

You could view the pre-Minkowski approach as a projection on the Euclidean part of Minkowski space, i.e.

$\left( \begin{array}{ccc} 0 \\ \mathbf{e_x} \\ \mathbf{e_y} \\ \mathbf{e_z} \end{array}\right) = \left( \begin{array}{ccc} 0 & 0 & 0& 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) \left( \begin{array}{ccc} \mathbf{e_0} \\ \mathbf{e_1} \\ \mathbf{e_2} \\ \mathbf{e_3}\end{array}\right) $

When you have a Lorentz transformation, then you have:

$\left( \begin{array}{ccc} 0 \\ \mathbf{e_{x'}} \\ \mathbf{e_y} \\ \mathbf{e_z} \end{array}\right) = \left( \begin{array}{ccc} 0 & 0 & 0& 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) \left( \begin{array}{ccc} \mathbf{e_0}\cosh\chi + \mathbf{e_1} \sinh \chi\\ \mathbf{e_0} \sinh \chi + \mathbf{e_1} \cosh \chi \\ \mathbf{e_2} \\ \mathbf{e_3}\end{array}\right) $

It's clear that $\mathbf{e_{x'}} \neq \mathbf{e_x}$ in this projective space. You might think to equate the two conceptually but I think that fails too. The following example I rotate the coordinate system before boosting.

$\left( \begin{array}{ccc} 0 \\ \mathbf{e_{x'}} \\ \mathbf{e_y} \\ \mathbf{e_z} \end{array}\right) = \left( \begin{array}{ccc} 0 & 0 & 0& 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) \left( \begin{array}{ccc} \cosh \chi & \sinh \chi & 0& 0 \\ \sinh \chi & \cosh \chi & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)\left( \begin{array}{ccc} 1 & 0 & 0& 0 \\ 0 & \cos \theta & \sin \theta & 0 \\ 0 & - \sin \theta & \cos \theta & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)\left( \begin{array}{ccc} \mathbf{e_0} \\\mathbf{e_1} \\ \mathbf{e_2} \\ \mathbf{e_3}\end{array}\right) $

$\left( \begin{array}{ccc} 0 \\ \mathbf{e_{x'}} \\ \mathbf{e_{y'}} \\ \mathbf{e_z} \end{array}\right) = \left( \begin{array}{ccc} 0 & 0 & 0& 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) \left( \begin{array}{ccc} \mathbf{e_0}\cosh\chi + (\mathbf{e_1} \cos \theta + \mathbf{e_2} \sin \theta) \sinh \chi\\ \mathbf{e_0} \sinh \chi + (\mathbf{e_1} \cos \theta + \mathbf{e_2} \sin \theta) \cosh \chi \\ -\mathbf{e_1} \sin \theta + \mathbf{e_2} \cos \theta \\ \mathbf{e_3}\end{array}\right) $

In the standard pre-Minkowski vector space you'd have $\mathbf{e_{x'}} = \mathbf{e_x} \cos \theta + \mathbf{e_y} \sin \theta$ and $\mathbf{e_{y'}} = -\mathbf{e_x} \sin \theta + \mathbf{e_y} \cos \theta$ even after the boost. To recover the original unprimed coordinate system you would reverse the rotation, however if you rotate the above vector back, you get the following:

$\left( \begin{array}{ccc} \mathbf{e_{x}} \\ \mathbf{e_{y}} \\ \mathbf{e_z} \end{array}\right) = \left( \begin{array}{ccc} \cos \theta & - \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array}\right) \left( \begin{array}{ccc} \mathbf{e_0} \sinh \chi + (\mathbf{e_1} \cos \theta + \mathbf{e_2} \sin \theta) \cosh \chi \\ -\mathbf{e_1} \sin \theta + \mathbf{e_2} \cos \theta \\ \mathbf{e_3}\end{array}\right) $

$ = \left( \begin{array}{ccc} \mathbf{e_0} \cos \theta \sinh \chi + \mathbf{e_1} (\cos^2 \theta \cosh \chi + \sin^2 \theta) + \mathbf{e_2} (\cos \theta \sin \theta \cosh \chi - \cos \theta \sin \theta) \\ \mathbf{e_0} \sin \theta \sinh \chi +\mathbf{e_1} (\cos \theta \sin \theta \cosh \chi - \cos \theta \sin \theta) + \mathbf{e_2}(\sin^2 \theta \cosh \chi + \cos^2 \theta) \\ \mathbf{e_3}\end{array}\right) $

This is not the coordinate system we started with. Even if we were to boost back, the y direction would still contain contribution from the original $\mathbf{e_x}$. You would have to boost back before reversing the rotation, but the requirement that $\mathbf{e_{x'}} = \mathbf{e_x}$ hides the fact that rotations and boosts do not generally commute. I believe that $\mathbf{e_{x'}} = \mathbf{e_x}$ is not a good concept to be working with, as it lends itself to easy oversights like this. If you do not require that $\mathbf{e_{x'}} = \mathbf{e_x}$ and explicitly keep track that $\mathbf{e_{x'}}$ is a different basis vector than $\mathbf{e_{x}}$ , I think you're far less likely to make a mistake like that. It should now be clear the 3D vector formulae for the velocity are recoverable because they are derivable even if $\mathbf{e_{x'}} \neq \mathbf{e_x}$ if you treat the 3D pre-Minkowski approach as a projection.