What is the Meaning of Lagrangian in Special Relativity?

[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.

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.

 

[Ibix]

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.

 

[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.

 

[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.

 

[vanhees71]

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]

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 $\tau$.

 

[vanhees71]

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).