Superluminal moving points 4-velocity

[PhDeezNutz]

Homework Statement

We can define the 4-velocity of a superluminally moving points in analogy to that of normal particles:

$$\tilde{U} = c \frac{d\tilde{R}}{ds}$$.

Prove that

$$\tilde{U} = \left( \frac{u^2}{c^2} \right)^{-\frac{1}{2}} \left( c , \tilde{u} \right)$$

and

$$U^2 = - c^2$$

But $\tilde{U}$ is still tangent to the worldline.

Relevant Equations

$$\tilde{R} = \left( ct,x,y,z \right)$$

$$\tilde{U} = \frac{d\tilde{R}}{d \tau}$$

$$d \tau^2 = \frac{ds^2}{c^2} \Rightarrow d \tau = \frac{ds}{c}$$

I think for sub-luminal velocities we use signature $(+,-,-,-)$ and for super luminal velocities we use $(-,+,+,+)$. I initially thought we could use either signature in either case but I'm only able to get

$$\tilde{U} = \left( \frac{u^2}{c^2} \right)^{-\frac{1}{2}} \left( c , \tilde{u} \right)$$

using $(-,+,+,+)$

To that end

$$\eta = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}$$

The first thing I want to do is reconcile the two definitions of 4-velocities $\tilde{U} = \frac{d \tilde{R}}{d \tau} = c \frac{ d \tilde{R}}{ds}$Start off with the usual definition of 4-velocity$\tilde{U} = \frac{d \tilde{R}}{d \tau}$we know that d \tau^2 = \frac{ds^2}{c^2} \Rightarrow d \tau = \frac{ds}{c}So $\tilde{U} = \frac{d \tilde{R}}{d \tau} = c \frac{ d \tilde{R}}{ds}$Now going back to the more tractable definition of 4-velocity $\tilde{U} = \frac{d \tilde{R}}{d \tau} = \frac{d}{d \tau} \left( ct,x,y,z \right) = \left( c \frac{dt}{d\tau}, \dot{x} \frac{dt}{d\tau}, \dot{y} \frac{dt}{d\tau}, \dot{z} \frac{dt}{d\tau} \right)$What is $\frac{dt}{d\tau}$ ?$ds^2 = \left( c dt, dx, dy, dz \right) \eta \left( c dt, dx, dy, dz \right) = - c^2 dt^2 + \left( dx^2 + dy^2 + dz^2 \right)$$d \tau^2 = \frac{ds^2}{c^2} = - dt^2 + \frac{1}{c^2} \left(dx^2 + dy^2 + dz^2 \right)$$\frac{d\tau^2}{dt^2} = -1 + \frac{1}{c^2} \left( \dot{x}^2 + \dot{y}^2 + \dot{z}^2 \right) = \left(\frac{u^2}{c^2} - 1 \right)$$\frac{d \tau}{dt} = \sqrt{\frac{u^2}{c^2} - 1} \Rightarrow \frac{dt}{d \tau} = \frac{1}{\sqrt{\frac{u^2}{c^2} - 1}}$Therefore the first result we wanted (correct)$\tilde{U} = \left( \frac{u^2}{c^2} - 1\right)^{-\frac{1}{2}} \left(c, \tilde{u} \right)$However I am having trouble proving $U^2 = - c^2$Let $\Gamma = \left( \frac{u^2}{c^2} - 1\right)^{-\frac{1}{2}}$$U^2 = \tilde{U} \cdot \tilde{U} = \left( \Gamma c, \Gamma u_1, \Gamma u_2 , \Gamma u_3\right) \eta \left( \Gamma c, \Gamma u_1, \Gamma u_2 , \Gamma u_3\right) = \left( \Gamma c, \Gamma u_1, \Gamma u_2 , \Gamma u_3\right) \cdot \left( -\Gamma c, \Gamma u_1, \Gamma u_2 , \Gamma u_3\right) = -\Gamma^2 c^2 + \Gamma u^2 = \Gamma^2 \left( u^2 - c^2 \right) = \Gamma c^2 \left( \frac{u^2}{c^2} - 1\right) = c^2$we wanted $-c^2$ ( Before moving on to prove that the 4-velocity is tangent to the worldline I'd like to reconcile this result first)

 

[PhDeezNutz]

I believe I have reconciled the first two results that I was supposed to get, but in doing so I believe I employed maneuvers that mathematicians would find egregious. Namely treating derivatives as quotients of differentials. There has to be some justification for this, unfortunately I do not know what that is because I am far from a mathematician.

$\tilde{R} = \left(ct,x,y,z \right)$

Using the signature $(-,+,+,+)$

$ds^2 = - c^2 dt^2 + \left(dx^2 + dy^2 + dz^2 \right)$

$ds = \sqrt{-c^2 dt^2 + \left(dx^2 + dy^2 + dz^2 \right)} = sqrt{dt^2 \left( - c^2 + \frac{dx^2}{dt^2} + \frac{dy^2}{dt^2} + \frac{dz^2}{dt^2}\right)}$

$ds = \sqrt{u^2 - c^2} dt$

$d\tilde{R} = \left(cdt, dx, dy, dz \right)$

$\tilde{U} = c \frac{d\tilde{R}}{ds} = \frac{c \left(cdt, dx, dy, dz \right)}{\sqrt{u^2 - c^2} dt}= \left(\frac{c^2}{\sqrt{u^2 - c^2}} , \frac{c\dot{x}}{\sqrt{u^2 - c^2}}, \frac{c\dot{y}}{\sqrt{u^2 - c^2}}, \frac{c\dot{z}}{\sqrt{u^2 - c^2}}\right)$

$\tilde{U} = \left( \frac{c}{\sqrt{\frac{u^2}{c^2} - 1}}, \frac{\dot{x}}{\sqrt{\frac{u^2}{c^2} - 1}}, \frac{\dot{y}}{\sqrt{\frac{u^2}{c^2} - 1}}, \frac{\dot{z}}{\sqrt{\frac{u^2}{c^2} - 1}}\right)$

Therefore,

$\tilde{U} = \left( \frac{u^2}{c^2} - 1\right)^{-\frac{1}{2}} \left( c, \tilde{u} \right)$

Now for $U^2$

$U^2 = \tilde{U} \eta \tilde{U} = \left(\frac{1}{\sqrt{\frac{u^2}{c^2} - 1}}\right)^2 \left( c, u_1, u_2, u_3\right) \cdot \left(c, - u_1, - u_2, - u_3 \right)$

$= \frac{1}{\frac{u^2}{c^2} - 1} \left( c^2 - u^2 \right) = -c^2$

 

[PhDeezNutz]

Now that I think I got the answers to the first two parts I can move onto proving U is tangent to the world line. I don’t know how to prove this. I thought it was the definition.