# I Action in relativity

1. Jun 24, 2016

### Meddusa

Hi,
I found that action in relativity is written in two ways:

1) S = ∫ ds

2) S= -mc ∫ ds

is (1) for non-relativistic particles and (2) for the relativistic particle ?
I read that writing (2) is up to a guess (-mc term), is it very interesting, anyone comment on this?

Thanks

2. Jun 25, 2016

### haushofer

No. Both are for relativistic particles. The minus sign is because the Lagrangian L is the difference between kinetic T and potential energy U, so in this case one has L=T. The mc in front of it is to have the units of energy right. See e.g. Zwiebach's book on String Theory.

3. Jun 25, 2016

### vanhees71

In relativistic physics the Lagrangian is not $T-V$.

The argument goes as follows. For a free particle you should have a Poincare invariant variation of the action. Because of the symmetry under space-time translations the Lagrangian must be a function of $\dot{\vec{x}}$ only. The only scalar action you can build is
$$S_0=-m c^2 \int \mathrm{d} t \sqrt{1-\dot{\vec{x}}^2/c^2}.$$
Interactions with external fields can be easily built in by assuming reparametrization invariance, i.e., to keep the Lagrangian fist-order homogeneous in $\dot{x}^{\mu}=(c,\dot{\vec{x}})$. The most simple example is the motion in an electromagnetic field represented by the four-vector potential $A^{\mu}$:
$$L_{\text{em}}=-mc^2 \sqrt{1-\dot{\vec{x}}^2/c^2}-\frac{q}{c} \dot{x}^{\mu} A_{\mu}.$$
The kinetic energy follows from the free-particle Lagrangian in the usual way as, $\vec{p}=\partial L_0/\partial \dot{\vec{x}}$
$$T=\vec{p} \cdot \vec{x}-L_0=\frac{m c^2}{\sqrt{1-\dot{\vec{x}}^2/c^2}}.$$
Note that in the case of the em. field the momenta are not the mechanical but the canonical momenta, which differ in this case
$$\vec{p}=\frac{\partial L}{\partial \dot{\vec{x}}} = \frac{m \dot{\vec{x}}}{\sqrt{1-\dot{\vec{x}}^2/c^2}}+\frac{q}{c} \vec{A}.$$
The Hamiltonian in this case thus is
$$H=T+q A^0.$$

4. Jun 25, 2016

### robphy

Yes. This point is not often highlighted.

The Legendre transformation of the free-particle Lagrangian gives the free-particle Hamiltonian.
In special relativity, the expression $\vec{p} \cdot \vec{x}-L_0=\frac{m c^2}{\sqrt{1-\dot{\vec{x}}^2/c^2}}$ is equal to the "relativistic energy" (the time-component of the particle's 4-momentum).

However, curiously, that is not equal to the "relativistic kinetic energy", which is
$m c^2\left(\frac{1}{\sqrt{1-\dot{\vec{x}}^2/c^2}}-1\right)$, as one gets from the work-energy theorem.
In fact, it appears that the relativistic kinetic energy is not a term in either the Lagrangian or Hamiltonian of a free relativistic particle.

I would argue that the minus sign in the action (and in the Lagrangian) is needed to get the canonical momentum of the free-particle equal to the "relativistic momentum" (the spatial-components of the particle's 4-momentum).

5. Jun 25, 2016

### haushofer

Yes, you're right, I was confused. Something similar confused me some time ago concerning the Hilbert action, regarding it as a kinetic energy term for the metric.

6. Jun 25, 2016

### Meddusa

Thank you all for detailed answers but still I do not feel I got the point.
Writing action without (mc) term means what?

On wikipedia, derivation of the geodesic equation is starting with S = ∫ ds
where ds is specified as "line element"

7. Jun 25, 2016

### Staff: Mentor

The first isn't really an action, it's a time (the elapsed proper time on the particle's clock between two fixed events). The second multiplies that time by a constant with units of energy (the particle's rest energy) in order to make it an action. The sign flip is because the equation of motion is derived by minimizing the action, but that corresponds to maximizing the elapsed proper time.

8. Jun 25, 2016

### Meddusa

I am clearer now.
In the notation ds=cdτ where τ is the proper time, right?

Thanks a lot !

9. Jun 25, 2016

### robphy

Recall that:
in dynamics, action has units of energy*time=momentum*displacement=Units of Planck's constant.