Varying The Relativistic Action

[Luke Tan]

In his book, Landau mentioned varying the relativistic lagrangian

However, I do not understand how he got from varying the integral of ds to varying only the contravariant components.

Would the general procedure not be varying
$$\delta S = -mc\delta\int_a^b\frac{dx_idx^i}{\sqrt{ds}}$$ and then expanding using the quotient & chain rule?

Why is the $\delta$ now appearing only on the $$dx^i$$?

Any help would be appreciated

Thanks!

 

[Orodruin]

The original integral is not a quotient. The quotient appears after the variation.

 

[vanhees71]

Landau-Lifshitz is sometimes very short in explaining things ;-). It's much easier to just write out the integral with help of an independent scalar parameter. Note that this is the parametrization-invariant formulation of the relativistic free-particle action, and thus $\lambda$ can be any independent parameter. Then the action reads
$$A=-m c \int_{\lambda_1}^{\lambda_2} \mathrm{d} \lambda \sqrt{\dot{x}_{\mu} \dot{x}^{\mu}}.$$
Here the dot denotes the derivative wrt. $\lambda$. The variation reads
$$\delta A = -m c \int_{\lambda_1}^{\lambda 2} \mathrm{d} \lambda \delta \dot{x}_{\mu} \dot{x}^{\mu} \frac{1}{\sqrt{\dot{x}_{\mu} \dot{x}^{\mu}}}.$$
Integration by parts and setting the variation to 0 then yields the EoM
$$\frac{\mathrm{d}}{\mathrm{d} \lambda} \left (\frac{\dot{x}^{\mu}}{\sqrt{\dot{x}_{\mu} \dot{x}^{\mu}}} \right)=0.$$

 

[stevendaryl]

To me, it's a little clearer if you put the problem into the Lagrangian format:

Write $ds = \sqrt{dx^i dx_i} = \sqrt{\frac{dx^i}{ds} \frac{dx_i}{ds}} ds = \sqrt{\dot{x}^i \dot{x}_i} ds$, where $\dot{x}^i \equiv \frac{dx^i}{ds}$. I know that seems kind of silly, because the expression $\sqrt{\dot{x}^i \dot{x}_i}$ is just identically equal to 1, but it makes the problem more conceptually similar to the classical Lagrangian formulation of mechanics:

$S = \int \mathcal{L}(x^i, \dot{x}^i) dt$

where the Lagrangian depends on location, $x^j$ as well as velocity, $\dot{x}^j$. In our case, we switch the "time" parameter from $t$ to $s$ to get:

$S = \int \mathcal{L}(x^i, \dot{x}^i) ds$

where the relativistic lagrangian $\mathcal{L}$ is $\sqrt{\dot{x}^i \dot{x}_i}$.

Now, if you vary the lagrangian, you get:
$$\delta \mathcal{L} = \frac{\partial \mathcal{L}}{\partial x^i} \delta x^i + \frac{\partial \mathcal{L}}{\partial \dot{x}^i} \delta \dot{x}^i$$
If you're dealing with Special Relativity, as opposed to General Relativity, then the relativistic Lagrangian doesn't depend on $x^i$, only on $\dot{x}^i$, so you have:
$$\delta \mathcal{L} = \frac{\partial \mathcal{L}}{\partial \dot{x}^i} \delta \dot{x}^i$$

Since $\dot{x}_i = \pm \dot{x}^i$ for Special Relativity (in GR, or when using noninertial coordinates, you have to write: $\dot{x}_i = \sum_{ij} g_{ij} (\dot{x}^j)^2$), we can write: $\mathcal{L} = \sqrt{\dot{x}^i \dot{x}_i} = \sqrt{\sum_i \pm \dot{x}^i}$. (In taking the derivative, there is a factor of 2 from the square-root, and a factor of 2 from $(\dot{x}^j)^2$, and they cancel). So

$$\frac{\partial \mathcal{L}}{\partial \dot{x}^i} = \frac{\pm \dot{x}^i}{\sqrt{\dot{x}^i \dot{x}_i}} = \frac{\dot{x}_i}{\sqrt{\dot{x}^i \dot{x}_i}} $$

(That's slightly misleading: in the numerator, the $i$ is some particular index, while in the denominator, $i$ is a dummy index that is summed over.) So

$$\delta \mathcal{L} = \frac{\dot{x}_i}{\sqrt{\dot{x}^i \dot{x}_i}} \delta \dot{x}^i$$

Since $\delta \dot{x}^i = \dot{\delta x^i}$, we get to the final form:

$$\delta S = \int \delta \mathcal{L} ds = \int \frac{\dot{x}_i}{\sqrt{\dot{x}^i \dot{x}_i}} \dot{\delta x}^i ds$$

Now is when you use partial integration: Write $\frac{\dot{x}_i}{\sqrt{\dot{x}^i \dot{x}_i}} \dot{\delta x}^i = \frac{d}{ds} (\frac{\dot{x}_i}{\sqrt{\dot{x}^i \dot{x}_i}} \delta x^i) - \frac{d}{ds} (\frac{\dot{x}_i}{\sqrt{\dot{x}^i \dot{x}_i}}) \delta x^i$. That's just using $A \frac{d}{ds} B = (\frac{d}{ds} AB) - (\frac{d}{ds} A) B$ with $A = \frac{\dot{x}_i}{\sqrt{\dot{x}^i \dot{x}_i}}$ and $B = \delta x^i$. So we have:

$$\delta S = \int [\frac{d}{ds} (\frac{\dot{x}_i}{\sqrt{\dot{x}^i \dot{x}_i}} \delta x^i) - \frac{d}{ds} (\frac{\dot{x}_i}{\sqrt{\dot{x}^i \dot{x}_i}}) \delta x^i] ds$$

Since integration and taking derivatives are sort of inverses, we can write (putting the limits back in): $\int_a^b \frac{d}{ds} (\frac{\dot{x}_i}{\sqrt{\dot{x}^i \dot{x}_i}} \delta x^i) ds = (\frac{\dot{x}_i}{\sqrt{\dot{x}^i \dot{x}_i}} \delta x^i) |_{a}^b$. So

$$\delta S = [\frac{\dot{x}_i}{\sqrt{\dot{x}^i \dot{x}_i}} \delta x^i]|_a^b - \int_a^b [\frac{d}{ds} (\frac{\dot{x}_i}{\sqrt{\dot{x}^i \dot{x}_i}}) \delta x^i] ds$$

Now, we can remember, after intentionally forgetting temporarily, that $\sqrt{\dot{x}^i \dot{x}_i} = 1$. So this simplifies to:

$$\delta S = [\dot{x}_i \delta x^i]|_a^b - \int_a^b (\frac{d}{ds} \dot{x}_i) \delta x^i ds$$

Your textbook is apparently using $u^i$ to mean $\dot{x}^i$.

I'm a little fuzzy about why this business of forgetting and then remembering that $\sqrt{\dot{x}^i \dot{x}_i} = 1$ is necessary, but it works.

 

[Orodruin]

So

∂L∂˙xi=±˙xi√˙xi˙xi=˙xi√˙xi˙xi∂L∂x˙i=±x˙ix˙ix˙i=x˙ix˙ix˙i​

\frac{\partial \mathcal{L}}{\partial \dot{x}^i} = \frac{\pm \dot{x}^i}{\sqrt{\dot{x}^i \dot{x}_i}} = \frac{\dot{x}_i}{\sqrt{\dot{x}^i \dot{x}_i}}

(That's slightly misleading: in the numerator, the iii is some particular index, while in the denominator, iii is a dummy index that is summed over.)

There is no reason to use i as the index in the denominator and doing things like that is the source of countless student errors. Rename it to j instead.

 

[vanhees71]

Well, this is misleading too! You cannot use $\lambda=\tau$, where $\tau$ is the proper time and then do the variation but you must use an independent world-line parameter. Then everything on this formal level is very simple (see my posting $3$).

It's of course cumbersome to use an arbitrary world-line parameter in practice. You'd rather like to use an affine parameter, for which $\dot{x}_{\mu} \dot{x}^{\mu}=\text{const}$. This can be achieved using this condition on the world-line parameter as a constraint and work through the formalism with a corresponding variational problem with this constraint, using a Lagrange multiplier. At the end it turns out, you can use the simpler expression
$$L_0=-\frac{m}{2} \dot{x}_{\mu} \dot{x}^{\mu}.$$
For the free-particle (kinetic) part of the particle (for a massive particle of course; for massless particles it get's a bit more complicated).

For the interaction piece of the Lagrangian you can just use the parameter-independent expression of the action, which is a homogeneous functions, wrt. $\dot{x}_{\mu}$ of rank one. E.g., for the motion of a charged particle in an external e.m. field you get
$$L_{\text{int}}=-\frac{q}{c} \dot{x}^{\mu} A_{\mu}(x),$$
where $A_{\mu}$ is the four-vector potential of the external field.

Now the total Lagrangian
$$L=L_0+L_{\text{int}} = -\frac{m}{2} \dot{x}_{\mu} \dot{x}^{\mu} - \frac{q}{c} \dot{x}^{\mu}$$
has nice properties. First of all since it depends not explicitly on the world-line parameter $\lambda$, ohne has
$$H=p_{\mu} \dot{x}^{\mu}-L=\text{const}, \quad \text{with} \quad p_{\mu}=\frac{\partial L}{\partial{\dot{x}}^{\mu}}.$$
While in the form with the square-root form of the kinetic term this leads to nothing since it just gives the identity "$0=0$". In the form with the square, it rather gives
$$H=-\frac{m}{2} \dot{x}_{\mu} \dot{x}^{\mu}=\text{const}.$$
This implies that now automatically the arbitrary parameter $\lambda$ is an affine parameter. Since still you can choose any affine parameter you like, this allows you too set
$$\dot{x}_{\mu} \dot{x}^{\mu}=1, \qquad (*)$$
which makes $\lambda$ the proper time of the particle. Note that this is done only after the equations of motion are derived, and thus makes $\tau$ to a valid choice to parametrize the world line, which however is not indepenent anymore but subject to the contraint (*), but this constraint (which is of course equivalent to the on-shell condition $p_{\mu} p^{\mu}=m^2 c^2$ for a classical massive particle) is automatically consistent with the equations of motion and thus a valid model of a relativistic classical particle.

 

[stevendaryl]

Well, this is misleading too! You cannot use $\lambda=\tau$, where $\tau$ is the proper time and then do the variation but you must use an independent world-line parameter. Then everything on this formal level is very simple (see my posting $3$).

That doesn't seem to be what the textbook was doing, though.

 

[vanhees71]

That was what the textbook implicitly did! I'm not sure whether the notation of LL is mathematically correct. I also don't see that it's very helpful for students either ;-)).

 

[stevendaryl]

Personally, it seems to me that writing the action as $S = \int ds$ is pretty useless. There's nothing to vary, there. Of course, that's numerically equivalent to $S = \int \sqrt{g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} d\lambda$, but the latter form gives you something to vary. You're varying the function $x^\mu(\lambda)$, holding $x^\mu(\lambda_1)$ and $x^\mu(\lambda_2)$ constant.

 

[vanhees71]

Here's #9 in clear form (with an important correction in the definition of $S$):

Personally, it seems to me that writing the action as $S = \int ds$ is pretty useless. There's nothing to vary, there. Of course, that's numerically equivalent to $S = \int \sqrt{g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}} d\lambda$, but the latter form gives you something to vary. You're varying the function $x^\mu(\lambda)$, holding $x^\mu(\lambda_1)$ and $x^\mu(\lambda_2)$ constant.

The point is that I introduce an independent parameter $\lambda$ for the worldline and I vary as you say at fixed boundary values as one should in Hamilton's principle in the Lagrange formalism (in the Hamilton formalism the variation is in phase space and only the configuration variables are fixed at the boundaries, but that's another story and not relevant here).

Only after the variation for the Euler-Lagrange equations which defines the actual trajectory of the particle you can choose $\lambda = \tau$, then simplifying the equations somewhat.

The advantage of this "square-root form" of the free-particle Lagrangian is that it is parametrization independent from the very beginning since obviously the so defined action is parametrization independent. Particularly you can also fix an (inertial) reference frame and set $\lambda=t$ (the coordinate time). Then you get the (3+1)-formalism which is not manifestly covariant but an easily interpretable set of equations of motion, i.e., the interpretation for an observer at rest in the chosen computational frame.

 

[TeethWhitener]

I want to point out that I think there is at least one typo in the excerpt of Landau-Lifshitz above. In the line where it says:
$$
\delta S = -mc \int_a^b\frac{dx_i\delta dx^i}{\sqrt{ds}}$$
I think the denominator shouldn't have a square root sign (since $ds = \sqrt{dx_i dx^i}$ already). This comports with $ds$ as proper time (and therefore $\frac{dx_i}{ds}$ as a four-velocity component).

 

[TSny]

@Luke Tan

You are wondering why the textbook varies the contravariant $dx^i$ but apparently does not consider variations in the covariant $dx_i$. Actually, the variations in $dx_i$ are accounted for.

Noting @TeethWhitener 's correction in post #11 that the square root in the denominator should not be there, you would first have
$$
\delta S = -mc \int_a^b\frac{\delta \left(dx_idx^i \right)}{2ds}$$ The factor of 2 in the denominator comes from taking the derivative of a square root. ($ds = \sqrt{dx_idx^i}$)

Now, $\delta \left(dx_idx^i \right) = dx_i \delta dx^i + dx^i \delta dx_i$ where variations in both the contravariant and covariant quantities are included.

But, you can show that the two terms on the right are equal. So, $\delta \left(dx_idx^i \right) =2 dx_i \delta dx^i$. The 2 here cancels the 2 in the denominator and you get the desired result.