Derivation of the geodesic equations

Question

Pg 79 of "Tensors, Relativity and Cosmology"

In order to construct the geodesic equations which define the curve with a stationary arc length, we may choose the arc length itself as the action integral with zero variation. Using $$s(t)=\int_{t_o} ^t\sqrt{g_{mn}\frac{dx^m}{dt}\frac{dx^n}{dt}}dt \tag{1}$$

we write $$I=\int_{S_A}^{S_B}\sqrt{g_{mn}\frac{dx^m}{ds}\frac{dx^n}{ds}}ds \tag{2}$$

where $I$ is the action integral and is equal to unity along the geodesic line C, where $$ds^2=g_{mn}dx^mdx^n \tag{3}$$

But isn't (2) just equal to 1 since $\frac{ds}{ds}=1$ and can this still be an action integral (functional) if the LHS of (2) is just a constant?

Or was it the author's intention to introduce to a scalar parameter such as $d\lambda$ but accidentally used ds instead?

(The author actually made tons of mistakes in the previous chapters, so I'm just making sure that this is not another error but I could be wrong...)

Or to put it another way, under what circumstances is (3) true? — puppetsock, Jun 14 '19 at 14:34

Qmechanic · Accepted Answer · 2019-06-14T14:39:35.920

2

Yes, OP is right. One cannot consistently formulate a variational principle for geodesics with arc length/affine parameter as the independent curve parameter for all virtual curves. Although the square root action is indeed independent of parametrization $\lambda$, the point is that the parameter interval $[\lambda_i,\lambda_f]$ must be common for all virtual curves, so that if $\lambda$ happens to be the arc length/affine parameter for one particular curve, this will not be the case for all neighboring curves. For more details, see e.g. my Phys.SE answer here.

edited Jun 14 '19 at 14:39

answered Jun 14 '19 at 10:28

Qmechanic

201,751

1

The author might (admittedly confusingly) claim that the notation $s$ in eq. (1) denotes arc length while the notation $s$ in eq. (2) denotes an arbitrary parameter. – Qmechanic Jun 14 '19 at 11:49
And this arbitrary parameter cannot be/is not necessary an affine parameter/arc length parameter? – Chern-Simons Jun 14 '19 at 13:14
Not for all virtual curves. – Qmechanic Jun 14 '19 at 19:08

score 2 · Answer 2 · answered Jun 14 '19 at 14:20

One can choose any parameter one likes to parameterize a curve and the standard trick is to use $s$ itself (sometimes called "proper time"). So the length of a curve is defined to be $$ L = \int_{t_0}^{t_1}\sqrt{g_{ab}\frac{dx^a}{dt}\frac{dx^b}{dt}}\,dt $$ but if we choose $t=s$ we have $$ L = \int_{s_0}^{s_1} ds = s_1-s_0\,. $$ This is called "proper time" or sometimes "unit speed". This does not help compute the length of the curve however, since we still need to evaluate $s_1-s_0$.

The geodesic equations arise from minimizing an action. There are two equivalent choices, one can minimize the length $L$ or one can also minimize the energy functional $$ I = \int_{t_0}^{t_1}g_{ab}\frac{dx^a}{dt}\frac{dx^b}{dt}\,dt $$ The actions $L$ and $I$ are different but their variations both lead to the geodesic equations.

Not sure I answered exactly the question, the text your using sounds a little imprecise (maybe try Sean Carroll's book Spacetime and Geometry), but I hope it helps.

Derivation of the geodesic equations

2 Answers2