Accelerated Motion In Special Relativity

Question

I have a number of questions regarding particles which accelerate in special relativity.

1) What parameter do you use to describe the particle's world line? Can you use proper time as a parameter? How does one make sense of proper-time? The definition I have been introduced to says that , proper time is the time in an inertial frame in which the particle is at rest,but if the particle is accelerating then there is no inertial frame in which it is at rest.

2) If proper time is used then will velocity transform as a four-vector?

As far as my understanding goes,

${ y }^{ \mu }={ x }^{ \nu }{ { L }_{ \nu }^{ \mu } }\\ \\ $

Where $y^{\mu}$ is the coordinate in the frame of the particle and $x^{\nu}$ is the coordinate in laboratory frame.(I understand here that I am comparing a non-inertial frame with an inertial frame, I don't know if I can do this.) If we parameterise the curve using $\lambda$ as the parameter then,

$$\frac{d{y^{\mu}}}{d{\lambda}}= \frac{d{x^{\nu}}}{d{\lambda}}L^{\mu}_{\nu} + x^{\nu}\frac{dL^{\mu}_{\nu}}{d\lambda}$$

If, velocity must transform like a four vector then, $$\frac{dL^{\mu}_{\nu}}{d\lambda}=0$$

How do we know such a parametrisation exists?

3) I have learnt that in S-R, the Lorentz transformation matrix must be constant, is this true?

Answer 1

The definition I have been introduced to says that , proper time is the time in an inertial frame in which the particle is at rest,but if the particle is accelerating then there is no inertial frame in which it is at rest.

At any given moment, there is still an inertial frame in which the particle is at rest. You can also simply define proper time using the metric, $d\tau^2=t^2-x^2-y^2-z^2$ (in natural units, where $c=1$).

If, velocity must transform like a four vector then $\frac{dL^{\mu}_{\nu}}{d\lambda}=0$. How do we know such a parametrisation exists?

What you are requiring is something much stronger than requiring that velocity transform as a four-vector. By requiring that ${ y }^{ \mu }={ x }^{ \nu }{ { L }_{ \nu }^{ \mu } }$ and then taking a time derivative, you're requiring that the particle's entire motion, at all times, be derivable by applying a Lorentz transformation to an inertial motion in some other inertial frame. That isn't possible. A Lorentz transformation takes inertial motion to inertial motion.

Note that all the issues you're talking about are identical to the corresponding issues in Galilean relativity, if you just substitute "Galilean transformation" for "Lorentz transformation."

I have learnt that in S-R, the Lorentz transformation matrix must be constant, is this true?

Yes. It is also possible to do SR in an arbitrary coordinate system, e.g., spherical coordinates, as long as the spacetime itself is intrinsically flat. But the transformation from, e.g., Minkowski coordinates to spherical coordinates is not a Lorentz transformation. If you want to do this kind of generalization, you have to deal with Christoffel symbols. (People don't usually encounter Christoffel symbols until they study GR, but they don't inherently have anything to do with curvature. They basically describe fictitious forces.)