Clarifications on Susskind's argument for relativistic laws of motion

Question

I am reading Special Relativity and Classical Field Theory by Susskind. Lecture 3 is about relativistic laws of motion, and I had a few questions about Susskind's arguments here.

1. He says that, if we take the case of a particle, then we can write it's classical lagrangian (in a region of no external forces) as $$\mathcal{L = \frac{1}{2}mv^2}$$ and, therefore, its action as $$S = \mathcal{L = m \int^b_a \frac{1}{2}v^2} dt$$ Then, he makes the claim that, if we are to develop a relativistic lagrangian, we know that it has to reduce to the classical lagrangian in the nonrelativistic limit. Therefore, we can expect it to be proportional to mass and $$S = -m \int^b_a d\tau$$ My question is why this does not undermine the entire argument. If the goal is to come to conclusions about invariant quantities, conversation, etc., why is mass here assumed to be constant? I would expect that we put some $m(\tau)$ into the integral and for this to reduce to $m$ in the nonrelativistic limit.
2. One of the formulas that Susskind derives in this lecture is that $E^2 - P^2 = m^2$. To my understanding, $m$ is an invariant quantity. On the other hand, $E^2 - P^2$ is a conserved quantity. What I understand from this is that $m$ can transform into other things (i.e. into energy), but it cannot change under different reference frames. However, my understanding about $E^2 - P^2$ based on the fact that it is a four-vector, which is associated with a conserved quantity via the fact that it's line element engrains an inner product, which is conserved. However, there is no mention to my knowledge of this quantity being invariant. Therefore, I am confused because if, for example, we have Susskind's positronium example. Here, we have a positron and an electron orbiting with some collective mass. If the mass is all transformed into some kind of energy, this would say that $E^2 - P^2$ is now equal to zero for the system. However, it was just nonzero (back when there was mass), so how can I reconcile this problem?
3. If we have the formula (that Susskind gives) $$d\tau = \sqrt{dt^2 - d\vec{x}^2}$$ then, we can get $$\frac{d\tau}{dt} = \sqrt{1-\vec{v}^2}$$ So my understanding is there, there is an implicit $c^2$ below the $v^2$ and that we can write or not write this because $c=1$. My question is if this is entirely independent of the speed of light being c? Is it just a constant that we can take or leave as we please because it is equal to one, or is it here because of its connection to the speed of light?

Answer 1

Let's try to understand what a relativistic invariant entity even is. The formal definition of invariant scalars is based on taking contraction between two four vectors (or more precisely between a vector and a one form), so that under general coordinate transformation you get the two vectors transformation cancelling the metric transformation. Assuming mostly positive convention, an example of an invariant scalar can then be,

$$\eta_{\mu\nu}p^{\mu}p^{\nu}=-m^2$$

You can derive this relation by noting that if the particle has no other form of energy in its rest frame,

$$E=mc^2$$

Now taking $c=1$, if we boost this along a radial $\mathbf{p}$, we will have,

$$\begin{pmatrix}E\\\mathbf{p}\end{pmatrix}=\begin{pmatrix}\cosh\eta & -\sinh\eta\\-\sinh\eta & \cosh\eta\end{pmatrix}\begin{pmatrix}m\\0\end{pmatrix}=\begin{pmatrix}m\cosh\eta\\-m\sinh\eta\end{pmatrix}$$

with some rapidity $\eta$. Taking the contraction in mostly positive convention,

$$-E^2+\mathbf{p}^2=m^2(-\cosh^2\eta+\sinh^2\eta)=-m^2$$

Which should tell you that the point particle's momentum norm has to be the rest mass. You can check here for the derivation of $E=m$.

With this we come to your questions.

1. Notice first that we have two invariant scalar in the definition of the relativistic point particle Lagrangian. One comes from the norm of momentum and another comes from the norm of infinitesimal displacement. These are $m$ and $d\tau$. This makes sure that the action is a relativistic invariant. You can take a point source to have a mass varying with the affine parameter $\tau$, but that's a needless complication for the simplest Lagrangian that can be formed for a point particle. Also note that the mass here is the rest mass so setting it dependent on the affine parameter would would necessarily violate conservation of energy.

2. Again mass is invariant because it is a scalar coming from vector contractions. We additionally assume that nothing is supplying our little particle with additional rest mass, so that the entire system has a conserved energy. Note also that mass doesn't transform to energy. In the language of relativity mass is energy. Additionally $m=0$ is also perfectly fine to consider. In that case one needs to change the affine parameter for a photon and the conservation relation becomes $E=|\mathbf{p}|$. For such a scenario however the action would have to be formulated using an einbein (or an auxiliary field). Check here for example.

3. In that equation $c=1$ is assumed. When you set the constant $c$ to 1 you're essentially scaling the value of velocity for all massive particles to be less than 1. Such a rescaling is perfectly fine and would give you the same physics.

Note: There seem to be a misconception about invariants and conserved quantities. $E^2-\mathbf{p}^2$ is an invariant quantity, on the other hand $E$ is a conserved quantity for a given frame. The frame invariant quantities will never change unless externally tinkered with, on the other hand conserved quantity can be frame dependent. For example if a particle is moving with a given momentum with respect to you, when you look at its rest frame, it has zero 3-momentum. However both in their respective frames are conserved.