Action in relativity

  • I
  • Thread starter Meddusa
  • Start date
  • #1
3
0
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 ?
why there is a minus sign in front of it?
I read that writing (2) is up to a guess (-mc term), is it very interesting, anyone comment on this?

Thanks
 

Answers and Replies

  • #2
haushofer
Science Advisor
Insights Author
2,480
868
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.
 
  • Like
Likes Meddusa
  • #3
vanhees71
Science Advisor
Insights Author
Gold Member
17,073
8,174
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
$$H=T+q A^0.$$
 
  • Like
Likes Meddusa
  • #4
robphy
Science Advisor
Homework Helper
Insights Author
Gold Member
5,779
1,078
In relativistic physics the Lagrangian is not ##T-V##.
Yes. This point is not often highlighted.

The kinetic energy follows from the free-particle Lagrangian in the usual way as, ##\vec{p}=\partial L_0/\partial \dot{\vec{x}}##
$$T=\vec{p} \cdot \vec{x}-L_0=\frac{m c^2}{\sqrt{1-\dot{\vec{x}}^2/c^2}}.$$
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).
 
  • Like
Likes Meddusa, vanhees71 and haushofer
  • #5
haushofer
Science Advisor
Insights Author
2,480
868
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
3
0
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
PeterDonis
Mentor
Insights Author
2020 Award
32,978
11,450
I found that action in relativity is written in two ways:

1) S = ∫ ds

2) S= -mc ∫ ds
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
3
0
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.
I am clearer now.
In the notation ds=cdτ where τ is the proper time, right?

Thanks a lot !
 
  • #9
robphy
Science Advisor
Homework Helper
Insights Author
Gold Member
5,779
1,078
Recall that:
in dynamics, action has units of energy*time=momentum*displacement=Units of Planck's constant.
 

Related Threads on Action in relativity

  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
11
Views
1K
  • Last Post
Replies
6
Views
905
Replies
9
Views
1K
Replies
2
Views
860
Replies
12
Views
713
Replies
1
Views
3K
  • Last Post
Replies
3
Views
955
Replies
3
Views
2K
  • Last Post
Replies
18
Views
2K
Top