# Confused by variational principle

1. Apr 22, 2010

My notes give the variational principle for a geodesic in GR:

$$c\tau_{AB} = c\int_A^B d\tau = c\int_A^B \frac{d\tau}{dp}dp = \int_A^B Ldp$$

and then apply the Euler-Lagrange equations. By choosing p to be an "affine parameter" where $$\frac{d^2 p}{d\tau^2}$$ the Euler-Lagrange equations are then expressed as:

$$\frac{dL^2}{dx^\mu} + \frac{d}{dp}\left( \frac{dL^2}{d\dot{x}^\mu} \right) = 0$$

where the dot is wrt p. Apparently the affine parameter is usually chosen to be tau, but then L is just 1 and the equation holds trivially! In fact each term in the equation is zero individually. What am I missing here?

2. Apr 22, 2010

### Mentz114

In my notes, L is

$$\sqrt{g_{\alpha\beta}{x'}^{\alpha}{x'}^{\beta}$$
where the primes are differentiation wrt the affine parameter. Solving the E-L equations for L gives the geodesic equation in terms of the affine param. After that's done, we can substitute $\tau$ for the affine parameter.

3. Apr 22, 2010

Yes that's the same thing, the space-time interval is:

$$ds^2=c^2d\tau^2=g_{\mu\nu}dx^\mu dx^\nu$$

which relates our equations. I don't like substituting for tau at the end, since if we had tau at the beginning then L would be constant and each term in the differential equation would be zero individually. It seems like dividing by x then afterwards setting x=0.

4. Apr 22, 2010

### Mentz114

I don't see the problem. It doesn't matter what the affine parameter is, the Lagrangian does not disappear.

5. Apr 22, 2010

The problem is that L is defined as:

$$L=c\frac{d\tau}{dp}$$

and if p = tau then L=c, a constant. And the each term in the Euler-Lagrange equation is zero individually, which it shouldn't be. In this case any path would satisfy the equations.

6. Apr 22, 2010

### Mentz114

So

$$L=c\frac{d\tau}{dp} = \sqrt{g_{\alpha\beta}{x'}^{\alpha}{x'}^{\beta}$$ ?

Sorry you've lost me. I can't see what you're talking about.

7. Apr 22, 2010

$$ds^2=c^2d\tau^2=g_{\mu\nu}dx^\mu dx^\nu$$

divide both sides by (dp)^2 (not rigourous but that's what my notes say):

$$c^2\left(\frac{d\tau}{dp}\right)^2=g_{\mu\nu}\frac{dx^\mu}{dp}\frac{dx^\nu}{dp}$$

denote differentiation wrt p by a dot and take the square root:

$$c\frac{d\tau}{dp}=\sqrt{g_{\mu\nu}\dot{x}^\mu \dot{x}^\nu}$$

8. Apr 23, 2010

### George Jones

Staff Emeritus
Are you sure that total derivatives are used?
No.

Consider

$$f\left( x, y \right) = x^2 + y^2 .$$

Set

$$1 = f\left( x, y \right).$$

Does this mean that

$$\frac{\partial f}{\partial x} = 2x = 0?$$

9. Apr 23, 2010

No sorry the total derivatives were a typo. You gave a good example, but now I feel like I don't even understand partial differentiation. Setting f(x,y)=1 means x and y are no longer independent, so partial differentiation wrt x and not y doesn't make sense. Is that right? So is partial differentiation always assumed to be applied to the unconstrained function?

10. Apr 23, 2010

### DrGreg

Yes. In this context, $\partial f / \partial x$ means "differentiate with respect to x while holding y constant". You can't vary x and hold y constant if you are keeping f(x,y) = 1.

The chain rule

$$\frac{df}{dp} = \frac{\partial f}{\partial x}\frac{dx}{dp} + \frac{\partial f}{\partial y}\frac{dy}{dp}$$​

relates three derivatives of f, the first along the curve (f=1 in this case), the second if you were to hold y constant, the third if you were to hold x constant. Of course in the case where f is constantly 1 along the curve in question, df/dp would be zero.

11. Apr 23, 2010

### George Jones

Staff Emeritus
Yes, as DrGreg explained.

Have you taken Hamiltonians and Hamilton's equations in a mechanics course? If so, what is the Hamiltonian, and what are Hamilton's equation for a harmonic oscillator? This is a similar application of partial derivatives.

12. Apr 24, 2010