Problem with dimensionless quantities in an equation requiring M^2 (Super Hamiltonian Formulation for Geodesic Motion)

[JimWhoKnew]

$\mathcal H = -\frac {1}{2} = \frac {1}{2}p_\mu p^\mu$

$p_\mu p^\mu = -1 = m^2 \textbf u^2$

But $\textbf u^2 = -1$

So $p_\mu p^\mu = -1 = -m^2$

So $m^2 = 1, \quad \textbf p = \textbf u, \quad \frac {d}{d\lambda} = \frac {d}{d\tau} , \quad \lambda = \tau$
...

What does the $~m~$ in $~m^2~$ stand for?

 

[TerryW]

m is the rest mass. Should have stuck to $\mu$.

 

[JimWhoKnew]

m is the rest mass. Should have stuck to $\mu$.

So how can you get $~m^2=-1~$ as in post #30 ?

Take for example the trivial case $~\lambda=\tau~$ in Minkowski SR spacetime, where the geodesic is the straight line $~x^\mu(\tau)=(\tau,0,0,0)~$ . You get $~p^\mu=\frac{dx^\mu}{d\lambda}=u^\mu=\delta^\mu_0~$ and $~\mathcal H=-\frac 1 2~$ . But how can you infer a mass from it?

Edit: Addition
$p^\mu=\frac{dx^\mu}{d\lambda}~$ is the vector field on the curve of tangent vectors to it, whose components (in a given chart) are scaled by the parametrization $~\lambda~$. It is proportional to $~u^\mu~$ , but the proportionality factor is not $~\mu~$ (except for the case where $~\mathcal H=-\frac 1 2 \mu^2~$ ), so in general it is not equal to the "mechanical 4-momentum" $~\mu \mathbf u~$ . Moreover, what sense does it make to talk of masses in the case of spacelike geodesics?

 

[TerryW]

I know that $m^2 = -1$ is problematical! It's just that the maths lead to that result.

I looked on the internet for some kind of discussion of what is happening in the spacelike realm of spacetime and didn't find anything useful. I have hazy memories of people postulating tachyons which accelerate when a retarding force is applied and decelerate when a positive force is applied (negative mass but not imaginary mass?).

 

[JimWhoKnew]

Hi TerryW

After 3 months on this thread, I give up.
I apologize for accepting the challenge (which turned out to be beyond my capabilities), since by doing so I possibly deprived you from getting a better help.

Thanks for your pleasant company

Regards,

JimWhoKnew

 

[TerryW]

I actually feel that you’ve helped me a great deal with this problem and the exchange has improved my understanding of what parameterisation is all about and that $\frac{d}{d\lambda}$ is something more abstract than classical momentum.
So you’ve most certainly helped me more than you think you have!
Regards

TerryW