# Why relativistic Lagrangian doesn't simply equal kinetic minus potential energy $L=T-V$?

As the question above, I wonder why the relativistic Lagrangian is written as:

$$L=-mc² \sqrt{1-\frac{v²}{c²}} - V ~=~-\frac{mc^2}{\gamma} -V~?$$

I know that the kinetic energy of a relativistic particle is defined by

$$T=(\gamma -1)m c².$$

so why we can't define simply the Lagrangian of a relativistic particle as its classical analogous:

$$L=T-V=(\gamma -1)m c² - V~?$$

If so.. What is the definition of a Lagrangian which guide us to the above expression?

If you go all the way back to the derivation of the Lagrangian, I think you can see why the relativistic Lagrangian is what it is. Usually, one starts with d'Alembert's principle $$\sum_i \left(F_i - \dot{p}_i\right) \delta x_i = 0,$$

where $$F_i$$ are the forces acting on particle $$i$$, $$p_i$$ is its momentum, $$\delta x_i$$ is a virtual displacement, and the dot designates differentiation with respect to coordinate time $$t$$. For the relativistic case, we use the proper momentum $$p = \gamma m v$$.

Focusing on the relevant term only:

$$-\dot{p}\, \delta x = -\frac{d(\gamma m v)}{dt}\, \delta x$$

We use the chain rule (or integrate by parts):

$$= -\frac{d}{dt}\left(\gamma m v\, \delta x\right) + \gamma m v \frac{d(\delta x)}{dt}$$

When we do the full variation the first term vanishes because $$\delta x=0$$ on the boundary. Under typical assumptions the ordinary and variational derivatives commute:

$$= \gamma m v\, \delta v$$

We want to end up with a statement like $$\delta(\mathrm{stuff})$$. If you are smarter than me, you might see that:

\begin{align} \delta\left(\frac{1}{\gamma}\right) &= \frac{1}{2} \left(1-\frac{v^2}{c^2}\right)^{-1/2} \frac{-2v}{c^2}\,\delta v = -\frac{1}{c^2}\,\gamma v \, \delta v \\ \delta\left(-\frac{mc^2}{\gamma}\right)&= \gamma m v\, \delta v \end{align}

or you could know the answer and `astutely guess' like I did.

When you go through the whole variational rigmarole you'll end up with something like:

$$\int_{t_i}^{t_f}\sum_i \left(F_i - \dot{p}_i\right) \delta x_i \,\mathrm{d}t = \delta \int_{t_i}^{t_f}\left(-\sum_i\frac{m_i c^2}{\gamma_i} - V \right)\mathrm{d}t = 0$$

So for a single particle $$L = - \frac{m c^2}{\gamma} - V.$$

• The trick to integrating $\int\gamma mvdv=-\frac{mc^2}{\gamma}+C$ is to write $v=c\tanh\phi$, viz. $\int\gamma mvdv=\int m\sinh\phi\cosh^{-2}\phi d\phi=-\frac{m}{\cosh\phi}+C$. – J.G. Feb 18 at 21:51
1. In a nutshell because its EL equation would be different from the correct relativistic Newton's 2nd law $$\frac{d{\bf p}}{dt}~=~-\frac{\partial V}{\partial {\bf r}}, \qquad {\bf p}~=~\gamma m{\bf v}.\tag{1}$$ The correct Lagrangian $$L$$ can instead be found by integrating $${\bf p}~=~\frac{\partial L}{\partial {\bf v}}.\tag{2}$$

2. Another argument is that a relativistic action $$S=\int\! \mathrm{d}t~ L$$ should better be Lorentz invariant, e.g. proportional to proper time, which OP's proposal is not.

# Where does the Lagrangian come from?

So my usual approach to teaching Lagrangian physics comes from the idea that you look at the classical Newton's laws in the context of constraint forces,$$\frac{d\vec p}{dt} = -\nabla U + \vec F_{\text{constraint}}.$$ Constraint forces are these very complicated things to reason about since they always act perpendicular to a surface (which might be a very complicated direction) and never have a fixed magnitude (it's always "however strong it needs to be to enforce the constraint.") But since the allowed motions are always perpendicular to the constraint force, we can say one thing with great certainty, $$\vec F_{\text{c}}\cdot \vec v = 0.$$ The problem is that this only gives us a familiar result: the work-energy theorem $$\frac{d}{dt}\left(\frac12 m \vec v\cdot\vec v + U(\vec r)\right) = 0,$$ but that no longer gives back the full equations of motion: somehow by inserting the "correct" velocity into this dot product we have "lost" information. So instead we consider some path perturbation off of the "true" path $$\mathcal P = \{\vec r(t),~~t_0 which is some function $$\delta \vec r(t)$$. If this path perturbation only takes us within the constraints then $$\vec F_{\text c}\cdot \delta\vec r(t) =0$$ similarly. One can readily interpret again $$U(\vec r + \delta \vec r)\approx U + \nabla U\cdot \delta \vec r$$ hence $$-\nabla U \cdot \delta \vec r = -\delta U$$ in a very natural way. But what's somewhat unnatural is this "kinetic energy" term $$\frac{d\vec p}{dt}\cdot\delta \vec r.$$ It looks somewhat like $$\vec v\cdot\delta\vec v$$ which we could interpret as $$\delta\left(\frac12 v^2\right),$$ the idea being similar to the $$\delta U$$ case, but it only looks like that in an "integration by parts" sort of way. So the Lagrangian approach is to legitimately integrate this thing from $$t_0 \to t_1,$$ producing$$\tag{A} \int_{t_0}^{t_1}dt~\left(\frac{d\vec p}{dt} \cdot \delta \vec r(t) + \delta U\right) = 0.$$And then we can integrate by parts, and if we restrict our choice of $$\delta r$$ only a little bit so that the boundary terms are zero, we find afterwards $$-\delta \left[\int_{t_0}^{t_1} dt~\left(\frac12 m v^2 - U\right) \right] = 0.\tag{B}$$ This in turn has all the information you need to get back those equations of motion again through the "calculus of variations", in other words these steps were entirely reversible. And normally now I would launch into a rant extolling the great value of our resulting freedom to choose whatever coordinates we want to apply those remaining variational calculus arguments: we can enforce the constraints by choosing appropriate coordinates rather than introducing complicated constraint forces.

# How does special relativity change this argument?

Surprisingly little! We have only one niggling little change to equation (A) and it does not change our fundamental approach to finding an analogue to equation (B). The change is that our starting point contains the relativistic momentum $$\int_{t_0}^{t_1}dt~\left(\frac{d}{dt}\left({m_0~v\over\sqrt{1 - (v/c)^2}}\right)\cdot \delta \vec r + \delta U\right) = 0.\tag{A'}$$and thus after integrating by parts we have the middle expression $$-{\vec v\cdot \delta \vec v \over\sqrt{1 - (v/c)^2}}.$$ The problem is that what you want to interpret this as something like $$\delta\left(\frac {c^2}{\sqrt{1 - (v/c)^2}}\right)= \frac{\vec v\cdot \delta \vec v}{(1 - (v/c)^2)^{3/2}},$$ perhaps because you have been seduced by the above $$L = K - U$$ form and you wish to substitute in the relativistic kinetic energy. Unfortunately while that is a good form for remembering Lagrangians it is not a good form for understanding their nature as it is somewhat accidental to the actual derivation that we created above: you can see that you get $$\delta\left(\frac12 m v^2\right)$$ more or less because it is the indefinite integral $$\int dv~m~v$$ and this path-perturbation operator $$\delta$$ in many ways acts like a derivative operator, so the indefinite integral is the right way to figure such things out.

By contrast the indefinite integral $$\int dv\frac{m_0~v}{\sqrt{1 - (v/c)^2}} = m_0 c^2 \sqrt{1 - (v/c)^2},$$ so that is the correct expression in our Lagrangian. Any other expression leads to an incorrect canonical momentum. We can find that $$c^2 \delta \sqrt{1 - (v/c)^2} = \vec v\cdot\delta\vec v/\sqrt{1 - (v/c)^2}$$ and thus reason that $$-\delta\left[\int_{t_0}^{t_1}dt~\left(m_0 c^2 \sqrt{1 - \left(\frac v c\right)^2} - U\right)\right] = 0,$$hence $$L = \gamma^{-1}~m_0~c^2 - U.$$

Of course the proper time famously satisfies $$dt = \gamma~d\tau$$ and thus an integral of $$dt/\gamma$$ is an integral of $$d\tau$$, so it makes the action integral relativistically invariant, which one might have desired as a fundamental aspect of the theory: and by minimizing the proper time between beginning and end, one can understand this as hinting at the general relativistic approach, where this is a sort of geodesic equation. But I don't like to begin the derivation from either of those places since it's not 100% obvious that special or general relativity isn't just going to spit back, "okay so Lagrangians are not the right way to think about me, so what?". I prefer to approach from the other side and thus derive that action principles are still a great way to understand relativistic motions.

For a better understanding of the first equation, simplify by setting V to 0. Then adopt the reference frame of the concerned point particle, that means that $$\gamma = 1$$. The result is the Lagrangian of the point particle in its own reference frame. From the point of view of the particle, its kinetic energy is zero! The particle is always remaining at zero in its own reference frame and does not move, it is moving only upwards through time.

By consequence, the remaining simplified Lagrangian shows that not the kinetic energy but the rest energy of the particle is considered by the Lagrangian.

The rest energy of the particle is transported through time, more precisely, through the proper time of the particle. This is the action of a point particle. Example: In the beginning, the particle (that means its rest energy) has the age 0. After one year, the particle has the age one year. The rest energy has been "transported" through the proper time (the aging) of the particle.

Now we can change our reference frame, by changing $$\gamma (v)$$ which is a function of our relative velocity to the particle. What we get is always the rest energy of the particle, but now it is not parameterized by its proper time $$\tau$$ but by our observer's coordinate time t.

You can find more information in Barton/ Zwieback: A first course of string theory, section 5.1: Action of a relativistic point particle.

Because the Euler-Lagrange equations for such Lagrangian would give incorrect equations of motion. The rule "L=T-V", although it is simple and easy to remember, isn't a universal pattern. It is only valid in Newtonian non-relativistic mechanics.

• So, what is its definition which is valid also in relativistic dynamic – El Mouden Feb 18 at 18:35
• For a single particle in a conservative field of force, you've got it in your question, it is the first formula (I've fixed the missing $V$). – Ján Lalinský Feb 18 at 18:38

If you define special relativistic kinetic energy to be

$$T=\gamma m c^2$$

then in the limit of $$v^2/c^2$$ much less than $$1$$,

$$T \rightarrow m c^2 + m v^2/2$$

which disagrees with the non-relativistic value by a constant factor of $$m c^2$$.

Hence the relativistic kinetic energy is defined to be

$$T=(\gamma -1)m c².$$

Rewriting the equation as

$$\gamma m c^2 = T + m c²$$

then $$\gamma mc^2$$ can be interrupted as the total energy of the particle and $$mc^2$$ is the rest mass when $$T=0.$$

The equation

$$L=-mc² / \gamma - V$$

is a suitable Lagrangian in the sense it produces the correct equations of motion in either the Hamiltonian or Euler-Lagrange variations.

One guesses the Lagrangian - probably using $$T=\gamma m c^2$$ then notices it needs to be negative.

• Thank you but u didn't understand what I mean, this os not the relativistic Lagrangian – El Mouden Feb 19 at 2:36