Does the unit hypercube in Minkowski space always have the 4-volume of 1?
Suppose we have a unit hypercube in Minkowski space defined by the column vectors in the identity matrix $$ \mathbf I = \begin{bmatrix} 1 & 0 & 0 & 0 \\[0.3em] 0 & 1 & 0 & 0 \\[0.3em] 0 & 0 & 1 & 0 \\[0.3em] 0 & 0 & 0 & 1 \end{bmatrix}$$
Now; the length of one edge would have units of time, but this is solved by multiplying the time interval with the speed of light $c = 1.$
Obviously, this hypercube would have the 4-volume of 1, as seen by its determinant:
$$\det \left(\begin{bmatrix} 1 & 0 & 0 & 0 \\[0.3em] 0 & 1 & 0 & 0 \\[0.3em] 0 & 0 & 1 & 0 \\[0.3em] 0 & 0 & 0 & 1 \end{bmatrix}\right) = \det \mathbf I = 1$$
Now, I have performed some numerical testing on the using the Lorentz transformation written as a matrix,
$$ \left[ \begin{array}{c} t' \\x'\\ y' \\ z' \\\end{array}\right] = \\ \left[ \begin{array}{c} t \\x\\ y \\ z \\\end{array}\right]\left[\begin{array}{cccc} \frac{1}{\sqrt{-v_x^2-v_y^2-v_z^2+1}} & -\frac{v_x}{\sqrt{-v_x^2-v_y^2-v_z^2+1}} & -\frac{v_y}{\sqrt{-v_x^2-v_y^2-v_z^2+1}} & -\frac{v_z}{\sqrt{-v_x^2-v_y^2-v_z^2+1}} \\ -\frac{v_x}{\sqrt{-v_x^2-v_y^2-v_z^2+1}} & \frac{\left(\frac{1}{\sqrt{-v_x^2-v_y^2-v_z^2+1}}-1\right) v_x^2}{v_x^2+v_y^2+v_z^2}+1 & \frac{v_x v_y \left(\frac{1}{\sqrt{-v_x^2-v_y^2-v_z^2+1}}-1\right)}{v_x^2+v_y^2+v_z^2} & \frac{v_x v_z \left(\frac{1}{\sqrt{-v_x^2-v_y^2-v_z^2+1}}-1\right)}{v_x^2+v_y^2+v_z^2} \\ -\frac{v_y}{\sqrt{-v_x^2-v_y^2-v_z^2+1}} & \frac{v_x v_y \left(\frac{1}{\sqrt{-v_x^2-v_y^2-v_z^2+1}}-1\right)}{v_x^2+v_y^2+v_z^2} & \frac{\left(\frac{1}{\sqrt{-v_x^2-v_y^2-v_z^2+1}}-1\right) v_y^2}{v_x^2+v_y^2+v_z^2}+1 & \frac{v_y v_z \left(\frac{1}{\sqrt{-v_x^2-v_y^2-v_z^2+1}}-1\right)}{v_x^2+v_y^2+v_z^2} \\ -\frac{v_z}{\sqrt{-v_x^2-v_y^2-v_z^2+1}} & \frac{v_x v_z \left(\frac{1}{\sqrt{-v_x^2-v_y^2-v_z^2+1}}-1\right)}{v_x^2+v_y^2+v_z^2} & \frac{v_y v_z \left(\frac{1}{\sqrt{-v_x^2-v_y^2-v_z^2+1}}-1\right)}{v_x^2+v_y^2+v_z^2} & \frac{\left(\frac{1}{\sqrt{-v_x^2-v_y^2-v_z^2+1}}-1\right) v_z^2}{v_x^2+v_y^2+v_z^2}+1 \\ \end{array}\right]$$
and the determinant of the resulting matrix always seem to be $1 \forall \{v_x, v_y, v_z\}$, even when $\sqrt{v_x^2 + v_y^2 + v_z^2} >1$, indicating that this "cube" will always have the same 4-volume, regardless of the inertial frame of reference (including tachyonic ones).
It seems that if the 4-volume of an arbitrary "hypercube" is 1 in one intertial reference frame, it must also have the 4-volume equal to 1 in every other inertial reference frame. Is this really true? How would one prove a such proposition?
special-relativity inertial-frames lorentz-symmetry
Qmechanic♦
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Markus KlyverMarkus Klyver
$\begingroup$ A Lorentz transformation could also have determinant $-1$. $\endgroup$ – Qmechanic♦ Sep 29 '16 at 21:37
$\begingroup$ You are absolutely right! The transformations with determinant $-1$ mentioned in the comments belong to a disconnected component of the Lorentz group which does not have the identity in it, so it probably is irrelevant to your question. $\endgroup$ – Prof. Legolasov Sep 30 '16 at 7:19
In terms of matrix components, Lorentz transformations have matrices that satisfy $$\eta = \Lambda^T \eta \Lambda$$ where $\eta$ is the Minkowski metric. Taking determinants of each side, we have $$|\eta| = |\Lambda^T| |\eta| |\Lambda| = |\Lambda|^2 |\eta|$$ which implies that $|\Lambda| = \pm 1$. Since a transformation by $A$ changes volume by $|A|$, this implies that Lorentz transformations preserve the (absolute) volume.
knzhouknzhou
$\begingroup$ Does this apply to all 4-volumes in Minkowski space? $\endgroup$ – Markus Klyver Sep 29 '16 at 21:40
$\begingroup$ @MarkusKlyver Yes, because every $4$-volume can be built out of little cubes. $\endgroup$ – knzhou Sep 29 '16 at 21:55
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