Why ct² - x² is Invariant under Lorentz Transformation

The spacetime interval s^2 = ct^2 - x^2 is important in special relativity because it stays the same in all inertial reference frames. Here we use some linear algebra to give a proof of exactly why this interval is invariant!

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Music: C418 - Pr Department

[tomkerruish]

Nicely done! Do you plan to cover rapidity? Looking forward to your series on e&m!

[magneticocampo]

Awesome. I Just found that lorentz transformation determinant is 1 and I was looking for geometric interpretation. And I found this video

[kleinpca]

Let T be the Lorentz matrix. Let J = [1 0
0 -1].
Evidently x^2-t^2 = [x t]J[x
t].
Thus it suffices to show that T'JT = TJT = J. This follows from straightforward algebra.

[sirknightartorias]

Wow 😍 👏 the most amazing video on the subject. Kudos bruh.

[amit.o]

5:21 but γ=1/√[1-(v²/c²)] so does it mean the β=v/c

[depressedguy]

Plz go through tensor as well it will be grate help

[NH-zhmp]

Let U = {u1, u2, u3, u4} be an ordered basis of R^4.
Let [x]_U = {x1, x2, x3, x4) for all x in R^4.
We define a pseudo inner product
⟨, ⟩ : R^4 × R^4 → R
⟨x, y⟩
= x1y1 + x2y2 + x3y3 - x4y4
= ([x]_U)^T . diag(1, 1, 1, -1). [y]_U

Clearly, we have
⟨ui, uj⟩
= 0 when i ≠ j,
= 1 when i = j = 1, 2, 3,
=-1 when i = j = 4.

So then U is called a pseudo orthonormal basis of R^4.

Let V = {v1, v2, v3, v4} is another ordered pseudo orthornormal basis of R^4, such that
⟨vi, vj⟩ = ⟨ui, uj⟩
and
[x]_V = [x5, x6, x7, x8]

We will show that
⟨x, y⟩ = x5y5 + x6y6 + x7y7 - x8y8.

Let (U→V) be the change basis matrix. Let
f : R^4 → R^4,
f(x) = (U→V).x
so, f is linear operator and
[f(x)]_U = (U→V).[x]_U = [x]_V
and f(ui) = vi, i = 1, 2, 3, 4.


We have
f(x) = f(x1u1 + … + x4u4)
= x1f(u1) + … + x4f(u4)
= x1v1 + … + x4v4

Hence
⟨f(x), f(y)⟩ = ⟨x1v1 + … + x4v4, y1v1 + … + y4v4⟩

Since ⟨, ⟩ is bilinear, and ⟨vi, vj⟩ = 0 when i ≠ j, we imply
⟨f(x), f(y)⟩
= x1y1⟨v1, v1⟩ + … + x4y4⟨v4, v4⟩
= x1y1⟨u1, u1⟩ + … + x4y4⟨u4, u4⟩
= x1y1 + x2y2 + x3y3 - x4y4
= ⟨x, y⟩

Moreover, we have
⟨f(x), f(y)⟩
= [f(x)]_U^T . diag(1, 1, 1-1). [f(y)]_U
= [x]_V^T . diag(1, 1, 1, -1). [y]_V
= x5y5 + x6y6 + x7y7 - x8y8

Therefore, we have
⟨x, y⟩
= x1y1 + x2y2 + x3y3 - x4y4
= x5y5 + x6y6 + x7y7 - x8y8

So ⟨x, x⟩ is invariant in any compatibly ordered pseudo orthornomal basis of R^4.