We can verify that the spacetime interval is invariant through brute force computation. Is there a deeper reason why the interval is invariant under Lorentz transform?

  • $\begingroup$ How do you define a "Lorentz transformation"? Some might define it as "linear transformation that leaves the spacetime interval invariant", thus dissolving this question. $\endgroup$ – ACuriousMind Jun 3 '17 at 14:05

The invariance of the line element is one of the fundamental principles of relativity.

Suppose we start with regular Newtonian mechanics, and suppose you move by a small vector $(dx, dy, dz)$. Then the total distance moved is just given by Pythagoras' theorem:

$$ ds^2 = dx^2 + dy^2 + dz^2 $$

Different observers might be using different coordinate systems, for example my coordinates might be rotated compared to yours so my $x$ axis might be your $y$ axis. That means you and I may disagree about the components of the vector $(dx, dy, dz)$, but we will both agree on the value of $ds^2$ because, well, it's just the length of that vector and that doesn't change when we rotate or displace the vector. The key thing is that:

$ds^2$ is an invariant

What special relativity does is change the definition of $ds^2$ to:

$$ ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 $$

where $c$ is a constant. But $ds^2$ is still an invariant i.e. all observers will agree on its value. This equation is called the metric, and it is pretty much all you need to define special relativity. It is the metric that defines special relativity, not the Lorentz transformations. So when you say:

Is there a deeper reason why the interval is invariant under Lorentz transform?

you are putting this the wrong way round. The Lorentz transformations work because they preserve $ds^2$.

General relativity extends the definition of the metric to:

$$ ds^2 = \sum_i \sum_j g_{ij} dx^i dx^ j $$

where $g$ is the metric tensor. Einstein's equations tell us how to calculate the metric tensor from the mass/energy distribution. Special relativity is the special case where the metric tensor takes the simple form:

$$ g = \left( \begin{matrix} -1 && 0 && 0 && 0 \\ 0 && 1 && 0 && 0 \\ 0 && 0 && 1 && 0 \\ 0 && 0 && 0 && 1 \end{matrix} \right) $$

There are lots of questions explaining how to use the metric to calculate stuff in relativity. For example in my answer to How do I derive the Lorentz contraction from the invariant interval? I explain how to calculate time dilation and Lorentz contraction. You can also work with accelerated motion for example in Can a ultracentrifuge be used to test general relativity? I use the metric to calculate time dilation for rotational motion. And indeed my answer to Is gravitational time dilation different from other forms of time dilation? explains how this extends in a simple way to gravitational time dilation.

The point of all this is that understanding the importance of the metric in special relativity is a key step in achieving a deep understanding of the theory, and it puts you on the path to understanding general relativity as well.

  • $\begingroup$ Such a brilliantly written answer, I learnt so much... Thanks so much for taking the time to compose such coherent answers. $\endgroup$ – user929304 Jun 14 '17 at 8:52

Consider the Euclidean analogue of your question.
"Why is the radius of a circle [with center at the origin] invariant under a Rotation?
It's because a rotation maps points on such circles to points on those same circles... thus, preserving the distance of points from the origin.
If you accept that answer, I offer the analogue for the spacetime interval.

So, from a Minkowski spacetime-geometrical viewpoint,
where the interval is the Minkowski-radius of a Minkowski-circle (a hyperbola) and the Lorentz Transformation--which maps events on a hyperbola [centered at the origin] to other events on that hyperbola--is analogous to a Rotation. Thus, the Minkowski-radius (the interval) is preserved.

(The Lorentz transformation also preserves the asymptotes of these hyperbolas, which is physically interpreted as leaving the speed of light unchanged. In fact, this is related to the eigenvectors of the Lorentz transformation.)


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