# Can special relativity be derived from the invariance of the interval?

As far as I know, the classical approach to special relativity is to take Einstein's postulates as the starting point of the logical sequence, then to derive the Lorentz transformations from them, and finally to derive the invariance of the interval and other consequences from the latter.

I'm curious to know if it would be sufficient to take the frame-independence of the interval alone as the initial hypothesis to derive the other results (including the Lorentz transformations and the postulates). Is this possible, or do we have to prove that the interval is Lorentz-invariant first? If this approach is consistent, would it offer any technical advantage w/r/t the classical one?

• Most of Einstein's methods are based on the invariance of light-like intervals, no? – dmckee Apr 9 '16 at 23:48
• possible duplicate: physics.stackexchange.com/q/80511 – Wolpertinger Apr 10 '16 at 7:41
• Related: physics.stackexchange.com/q/12664/2451 and links therein. Note that the Lorentz group $O(3,1):= \{ \Lambda\in {\rm Mat}_{4\times 4}(\mathbb{R}) \mid \Lambda^T\eta \Lambda = \eta \}$ is defined as the $4\times 4$ matrices that preserve the Minkowski metric $\eta$. – Qmechanic Apr 10 '16 at 11:01
• @DavidHerreroMartí i guess "related" would express it better, you're right it's not really a duplicate – Wolpertinger Apr 11 '16 at 12:34
• @David Getting the general invariance of the interval takes some work, but the invariance of light-like intervals is the invariance of $c$. Like this: $c = (\Delta x)/(\Delta t)$ so $c \Delta t = \Delta x$ so $(c\Delta t)^2 = (\Delta x)^2$ so $(c\Delta t)^2 - (\Delta x)^2 = 0$. All those games Einstein plays with light clocks and so on rely on the invariance of light-like intervals. – dmckee Apr 11 '16 at 13:47

Yes.

Let's define that the interval needs to be the quantity constructed by adding the square of the spatial interval to the negative of the square of the temporal interval. Now, we postulate that this is frame-invariant.

To be elegant and fast, I define $\eta_{\mu\nu} = +1$ for $\mu=\nu=1,2,3$ and $\eta_{\mu\nu}=-1$ for $\mu=\nu=0$ and $\eta_{\mu\nu}=0$ for the rest of the cases. $\mu, \nu$ run from $0$ to $3$. Now, the interval is $ds^2=\eta_{\mu\nu}dx^\mu dx^\nu$ where the Einstein summation convention is in use.

Now, in a different frame, the interval will be $\eta_{\alpha \beta}dx'^{\alpha}dx'^{\beta}$. Notice that the matrix $\eta$ will remain the same because we have taken it as a definition that the quantity we call the interval needs to be constructed by adding up the square of the spatial interval to the negative of the square of the temporal interval. And now, we invoke our postulate of the invariance of this interval via demanding that $$\eta_{\mu\nu}dx^{\mu}dx^{\nu}=\eta_{\alpha\beta}dx'^{\alpha}dx'^{\beta} \tag{1}$$

Or, $$\eta_{\mu\nu}dx^{\mu}dx^{\nu}=\eta_{\alpha\beta}\dfrac{\partial x'^{\alpha}}{\partial x^\sigma}\dfrac{\partial x'^{\beta}}{\partial x^\rho}dx^\sigma dx^\rho \tag{2}$$

But since the indices $\sigma$ and $\rho$ are dummy indices, we can change them to $\mu$ and $\nu$ without any harm. So, we re-write $(2)$ as

$$\eta_{\mu\nu}dx^{\mu}dx^{\nu}=\eta_{\alpha\beta}\dfrac{\partial x'^{\alpha}}{\partial x^\mu}\dfrac{\partial x'^{\beta}}{\partial x^\nu}dx^\mu dx^\nu \tag{3}$$

The only way $(3)$ can be generically true is if the following equality holds: $$\eta_{\mu\nu}=\eta_{\alpha\beta}\dfrac{\partial x'^{\alpha}}{\partial x^\mu}\dfrac{\partial x'^{\beta}}{\partial x^\nu} \tag{4}$$ Now, as the matrix $\eta$ doesn't change with place, its differentiation with respect to any of the coordinates will be zero. Thus, $$0=\eta_{\alpha\beta}\bigg[\dfrac{\partial^2 x'^\alpha}{\partial x^{\mu}\partial {x^\kappa}}\dfrac{\partial x'^\beta}{\partial x^\nu}+\dfrac{\partial^2 x'^\beta}{\partial x^{\nu}\partial {x^\kappa}}\dfrac{\partial x'^\alpha}{\partial x^\mu}\bigg] \tag{5}$$ Since the right hand side of $(5)$ is identically zero for all the values of the independent indices, we can interchange them. So, we interchange $\mu$ and $\kappa$ and add the resulting expression to $(5)$. Similarly, we interchange $\kappa$ and $\nu$ and subtract the resulting expression from $(5)$.

The resultant is

$$0=\eta_{\alpha\beta}\bigg[\dfrac{\partial^2 x'^\alpha}{\partial x^{\mu}\partial {x^\kappa}}\dfrac{\partial x'^\beta}{\partial x^\nu}+\dfrac{\partial^2 x'^\beta}{\partial x^{\nu}\partial {x^\kappa}}\dfrac{\partial x'^\alpha}{\partial x^\mu} + \dfrac{\partial^2 x'^\alpha}{\partial x^{\kappa}\partial {x^\mu}}\dfrac{\partial x'^\beta}{\partial x^\nu}+\dfrac{\partial^2 x'^\beta}{\partial x^{\nu}\partial {x^\mu}}\dfrac{\partial x'^\alpha}{\partial x^\kappa} - \dfrac{\partial^2 x'^\alpha}{\partial x^{\mu}\partial {x^\nu}}\dfrac{\partial x'^\beta}{\partial x^\kappa}-\dfrac{\partial^2 x'^\beta}{\partial x^{\kappa}\partial {x^\nu}}\dfrac{\partial x'^\alpha}{\partial x^\mu}\bigg] \tag{6}$$

As you can see, the last term cancels the second and the fifth cancels the fourth. Notice that this argument is based on the symmetry of $\eta_{\alpha\beta}$ which is inherent in its definition. The remaining terms simplify to $$0=2\eta_{\alpha\beta} \dfrac{\partial^2x'^\alpha}{\partial x^\mu \partial x^\kappa} \dfrac{\partial x'^\beta}{\partial x^\nu} \tag{7}$$

This can be generically true only if $$\dfrac{\partial^2x'^\alpha}{\partial x^\mu \partial x^\kappa} = 0 \tag{8}$$

The generic solution of this equation would then be $$x'^\alpha = \Lambda^\alpha _\mu x^\mu + c^\alpha \tag{9}$$ Here, $\Lambda$ and $c$ are constants in the sense they do not depend on the position in spacetime.

Therefore, $(4)$ translates to $$\eta_{\mu\nu} = \eta_{\alpha\beta}\Lambda^\alpha _\mu \Lambda^\beta _\nu \tag{10}$$ Or, in the matrix form, $$\eta = \Lambda^{T} \eta \Lambda \tag{11}$$

Thus, the allowed the set of proper homogeneous transformations is isomorphic to $SO(3, 1)$ and the set of all transformations is isomorphic to the Poincare group. All the results of special relativity can now be derived as we know the exact symmetry group of its transformations.

Edit

I gathered the arguments presented above from the wonderful GR book of Steven Weinberg, Gravitation and Cosmology.

I'm curious to know if it would be sufficient to take the frame-independence of the interval alone as the initial hypothesis to derive the other results (including the Lorentz transformations and the postulates). Is this possible, or do we have to prove that the interval is Lorentz-invariant first?

Yes, this is certainly possible. A nice way to develop this is to use a coordinate-free approach, and an exposition along these lines is given in Bertel Laurent, Introduction to spacetime: a first course on relativity. Laurent gives two postulates: (1) there is a metric, and (2) inertial motion gives maximal proper time. Postulate #1 inherently assumes frame-independence, since it refers to an inherent property of vectors, a bilinear inner product, which exists without our ever even having mentioned frames of reference or coordinates.

If this approach is consistent, would it offer any technical advantage w/r/t the classical one?

It typically makes very little difference in the end what axiomatization you use for SR, because you end up with the same theory in the end. But IMO Einstein's 1905 postulates are not the way anyone would express them today based on how we think about physics, e.g., we don't conceive of optics and mechanics as separate sciences, and we don't think of photons as being any more special than any other massless particle.

My understanding (which is based somewhat on Jackson's chapter on SR in Classical Elecrodynamics) is that the invariance of the interval is not enough to derive the Lorentz transformations - you also need the second postulate (that the speed of light is constant in all frames).

The invariance of the interval follows from the fact that spherical light waves obey the equation

$c^2t^2-(x^2+y^2+z^2)=0$

in one frame and

$c'^2t'^2-(x'^2+y'^2+z'^2)=0$

in another. If space is homogeneous and isotropic, as implied by the first postulate (that the laws of physics are invariant wrt the translational motion of frames moving in rectilinear paths), then the above equations can be related by a scale-factor only, which is independent of space and time:

$c^2t^2-(x^2+y^2+z^2)=\lambda(c'^2t'^2-(x'^2+y'^2+z'^2))$

You can show in order for the inverse transforms to be symmetric, $\lambda$ has to $=1$.

Then you need the second postulate ( $c=c'$ ) before you can find the frame transformations.

• This doesn't seem right to me. If you've already established what the metric is, then applying the metric to the energy-momentum four-vector gives $E^2=p^2+m^2$ (in units with c=1). This implies that massless particles have to travel at c, for otherwise you'd have a frame of reference in which the particle was at rest, but there would be no preferred direction for its momentum vector. – Ben Crowell Jun 16 '17 at 21:54