# Lagrangian in Special Relativity

• I
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.

Last edited:

2022 Award
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.

Last edited:
vanhees71
jbergman
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.

ergospherical
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
jbergman
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.

Gold Member
2022 Award
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
jbergman
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##.