- #1

kmarinas86

- 979

- 1

[tex]\gamma \equiv \frac{c}{\sqrt{c^2 - v^2}} = \frac{1}{\sqrt{1 - \beta^2}} = \frac{\mathrm{d}t}{\mathrm{d}\tau}[/tex]

I can derive:

[tex]\left(\frac{\mathrm{d}\tau}{\mathrm{d}t}\right)^2 + \left(\frac{v}{c}\right)^2 = 1[/tex]

The term [itex]\left(\frac{v}{c}\right)^2[/itex] can be expanded to three terms:

[tex]\left(\frac{\mathrm{d}\tau}{\mathrm{d}t}\right)^2 + \left(\frac{v_x}{c}\right)^2 + \left(\frac{v_y}{c}\right)^2 + \left(\frac{v_z}{c}\right)^2 = 1[/tex]

[tex]\mathrm{d}\tau^2 +\left(\frac{\mathrm{d}x}{c}\right)^2 + \left(\frac{\mathrm{d}y}{c}\right)^2 + \left(\frac{\mathrm{d}z}{c}\right)^2 = \mathrm{d} t[/tex]

[tex]\left(c\mathrm{d}\tau\right)^2 +\left(\mathrm{d}x\right)^2 + \left(\mathrm{d}y\right)^2 + \left(\mathrm{d}z\right)^2 = \left(c\mathrm{d}t\right)^2[/tex]

Using slightly different notation, we have:

[tex]\left(c \Delta\tau \right)^2 +\left(\Delta x\right)^2 + \left(\Delta y\right)^2 + \left(\Delta z\right)^2 = \left(c \Delta t \right)^2[/tex]

This would give us a metric signature of (+,+,+,+). All components are space-like, with no time-like components.

Mathematically, the spacetime interval would be nothing other than:

[tex]s^2 = - \left(c \Delta \tau\right)^2[/tex]

Which would mean that either [itex]s[/itex] or [itex]\Delta \tau[/itex] is imaginary.

Is the facility of hyperbolic trigonometric functions the primary motivation of formulating relativity on the basis of the metric signatures (-,+,+,+) and (+,-,-,-)?

It would seem that if relativity were formulated on the basis of non-hyperbolic trigonometric functions, the metric signature (+,+,+,+) would work as well.