# What is the Meaning of Lagrangian in Special Relativity?

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• jbergman
In summary, the Lagrangian in special relativity is a mathematical expression that describes the dynamics of a system in terms of its position, momentum, and time. It plays a crucial role in Einstein's theory of special relativity, allowing for the prediction of the behavior of particles traveling at high speeds. The Lagrangian approach provides a more elegant and intuitive way to understand the laws of motion in special relativity, making it a fundamental concept in modern physics.
jbergman
TL;DR Summary
I have a question about choosing proper time for the parametrization of the Lagrangian in special relativity.
According to @vanhees71 and his notes at https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf under certain conditions one can choose ##\tau## as the parameter to parametrize the Lagrangian in special relativity.

For instance if we have,

$$A[x^{\mu}]=\int d\lambda \left[-mc\sqrt{\eta_{\mu\nu}\dot{x}^{\mu} \dot{x}^{\nu}} - \frac{q}{c}\eta_{\mu\nu}\dot{x}A^{\nu}(x) \right]$$

then we can choose ##\lambda=\tau##.

I am trying to follow the proof in the above mentioned notes and I get hung up on the following line of reasoning.

vanhees71 said:
Since ##\dot{x}^{\mu}\frac{d}{d\lambda}\frac{\partial L}{\partial \dot{x}^{\mu}} = \dot{x}^{\mu}\frac{\partial L}{\partial x^{\mu}}## holds for any word line, only three of the four space-time variables, ##x^{\mu}## are independent.

I am not seeing how the above equation implies that.

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jbergman said:
I am not seeing how the above equation implies that.
I think he means that with ##\lambda=\tau## you've chosen that ##\dot{x}^\mu\dot{x}_\mu=\pm 1##, so once you've chosen three components the fourth is implied by the normalisation.

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vanhees71
Ibix said:
I think he means that with ##\lambda=\tau## you've chosen that ##\dot{x}^\mu\dot{x}_\mu=\pm 1##, so once you've chosen three components the fourth is implied by the normalisation.
How does that statement relate to the equation that was cited. I don't see the connection. I agree that if you a priori choose ##\lambda=\tau## ehat you say is true, but I think he's trying to prove the other direction.

The point is that the Lagrangian is a homogeneous function i.e. ##L(x, \alpha \dot{x}) = \alpha L(x,\dot{x})## so Euler's theorem implies the identity ##\dot{x}^a (\partial L / \partial \dot{x}^a) = L##. As per the manuscript you cited, one can use this identity to show that ##\dot{x}^a[ \partial L/\partial x^a - (d/d\lambda)(\partial L/\partial \dot{x}^a)] = 0## without having to use the Euler-Lagrange equations - i.e. this relationship holds everywhere, not just on the solution to the EoM. The constraint so-obtained reduces the number of independent Euler-Lagrange equations by one.

You can only re-parameterise ##\lambda \rightarrow \tau## to proper time after having already solved the Euler-Lagrange equations, otherwise you excessively restrict the possible variations of the Lagrangian.

PeroK and vanhees71
ergospherical said:
As per the manuscript you cited, one can use this identity to show that ##\dot{x}^a[ \partial L/\partial x^a - (d/d\lambda)(\partial L/\partial \dot{x}^a)] = 0## without having to use the Euler-Lagrange equations - i.e. this relationship holds everywhere, not just on the solution to the EoM. The constraint so-obtained reduces the number of independent Euler-Lagrange equations by one.
Can you clarify how this only reduce "the independent Euler-Lagrange equations by one"? It looks like we have all of the Euler-Lagrange equations except for the extra factor of ##\dot{x}^a## from that identity.

Note that an index appearing twice in an equation (one as an upper, one as a lower index) means that you have to sum over this index. The equation thus says that independent of the worldline of the particle the functional derivative of the action,
$$\frac{\delta S}{\delta x^{\mu}} = \frac{\partial L}{\partial x^{\mu}} -\frac{\mathrm{d}}{\mathrm{d} \lambda} \frac{\partial L}{\partial \dot{x}^{\mu}},$$
is Minkowski-orthogonal to ##\dot{x}^{\mu}##, which means that there is a constraint reducing the four Euler-Lagrange equations of motion to three independent ones, as it must be for a single particle's motion.

jbergman
vanhees71 said:
Note that an index appearing twice in an equation (one as an upper, one as a lower index) means that you have to sum over this index. The equation thus says that independent of the worldline of the particle the functional derivative of the action,
$$\frac{\delta S}{\delta x^{\mu}} = \frac{\partial L}{\partial x^{\mu}} -\frac{\mathrm{d}}{\mathrm{d} \lambda} \frac{\partial L}{\partial \dot{x}^{\mu}},$$
is Minkowski-orthogonal to ##\dot{x}^{\mu}##, which means that there is a constraint reducing the four Euler-Lagrange equations of motion to three independent ones, as it must be for a single particle's motion.
Thanks for that explanation. I missed the Einstein summation part.

I'm still fuzzy on the final step. We've reduced our equations of motions from 3 to 4 and we can take one of our space-time variables to be a function of the other 3. But, I still don't see how that let's us re-parametrize with ##\tau##.

Maybe it becomes clearer if you read a bit further in this section of my above quoted SRT manuscript starting from Eq. (2.4.33) ff.

More elegant is the alternative Lagrangian discussed in the following Sect. 2.4.3, where the parameter becomes automatically an affine parameter along the trajectory of the particle (i.e., the solution of the equations of motion).

jbergman

## 1. What is the Lagrangian in Special Relativity?

The Lagrangian in Special Relativity is a mathematical function that describes the dynamics of a system in terms of its position, velocity, and time. It is used to calculate the equations of motion for a system, and is a fundamental concept in the field of physics.

## 2. How is the Lagrangian related to Special Relativity?

In Special Relativity, the Lagrangian is used to describe the motion of particles in a four-dimensional space-time. It takes into account the effects of time dilation and length contraction on the dynamics of a system, and is a crucial component in understanding the behavior of objects moving at high speeds.

## 3. What is the significance of the Lagrangian in Special Relativity?

The Lagrangian is significant because it allows us to describe the motion of particles in a way that is independent of the reference frame. This means that the laws of physics remain the same, regardless of the observer's perspective. It also allows us to calculate the equations of motion for a system without having to consider external forces.

## 4. How is the Lagrangian calculated in Special Relativity?

The Lagrangian is calculated by subtracting the kinetic energy from the potential energy of a system. In Special Relativity, the kinetic energy is given by the relativistic formula, which takes into account the effects of mass and velocity on an object's energy. The potential energy is determined by the forces acting on the system, such as gravity or electromagnetic forces.

## 5. What are some practical applications of the Lagrangian in Special Relativity?

The Lagrangian has many practical applications, such as in particle physics, cosmology, and astrophysics. It is used to study the behavior of particles in high-energy accelerators, to calculate the trajectories of objects in space, and to understand the dynamics of black holes. It is also used in the development of technologies such as GPS, which relies on the principles of Special Relativity to function accurately.

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