# Special Relativity and the Lagrangian

1. Sep 19, 2008

### dx

Hi,

The invariant quantity in special relativity is $$t^2 - x^2$$ and the form of the lagrangian is $$T - U$$. The kinetic energy T has to do with time while the potential energy U has to do with space. I was just wondering if there's some known connection here.

2. Sep 19, 2008

### dx

Rereading my post, I think maybe I didn't explain my question properly. In classical mechanics the Lagrangian has the form $$T - U$$. In special relativity, the interval $$t^2 - x^2$$ is invariant. I've always found the specific form T - U of the Lagrangian pretty baffling, and have wondered whether there's any deeper way of looking at it. Recently I got the idea that maybe there's some connection between the relativistic interval and the Lagrangian, since the kinetic energy is associated with time and the potential energy is associated with space, like t and x in the expression for the interval. So is there any known connection between the relativistic interval and the Lagrangian?

3. Sep 19, 2008

### clem

Not as you put it. The form T-U is unconnected to the form t^2-x^2.

4. Sep 19, 2008

### Mentz114

Because we consider potential energy to be negative, T - U is the total energy. The Lagrangian will give the same dynamics if it is rewritten,

L = T + (K-U(x))

where K is a very large constant representing all the potential energy in the system.

I think the similarity in form to x2-t2 is coincidental.

5. Sep 19, 2008

### dx

Mentz,

T - U is not the total energy, I don't understand why you think that. Also, consider the Lagrangian of a string, where the particles are vibrating transversely. In appropriate units, it has the form

$$L = \frac{1}{2} \int (\frac{dy}{dt})^2 - (\frac{dy}{dx})^2 dx$$.

where the integrand is of the form $$(time derivate)^2 - (space derivative)^2$$, which also suggests an analogy with $$t^2 - x^2$$.

And yes I know $$L = T - U + K$$ where K is a constant will give the same dynamics, but notice that $$t^2 - x^2 + K$$ is also an invariant.

Last edited: Sep 19, 2008
6. Sep 19, 2008

### Mentz114

Dx,

yes, I was thinking of the Hamiltonian H = T + U for the SHO. Whoops.

To get back to your question. The invariance of x2-t2 under Lorentz transformations tells us about the symmetry of SR. If the form T - U is significant it will be related to a symmetry.

So, what transformations leave the dynamics of

$$L = \frac{1}{2} \int (\frac{dy}{dt})^2 - (\frac{dy}{dx})^2 dx$$

unchanged ?

M

Last edited: Sep 19, 2008
7. Sep 20, 2008

### atyy

I don't know exactly, but there is some sort of connection.

From the Lagrangian we get the Euler-Lagrange equations, which describe geodesic motion. In relativity, free particles and free photons move on geodesics, trajectories of extremal arc length, with arc length given by ds2.

See the discussion beginning at 3.4.3:
John Baez, Blair Smith and Derek Wise, Lectures on Classical Mechanics.
http://math.ucr.edu/home/baez/classical/texfiles/2005/book/classical.pdf

Last edited: Sep 20, 2008
8. Feb 4, 2009

### dx

I started this thread a long time ago, and the issue wasn't resolved to my satisfaction at that time. I have since found the relationship I was looking for. It is essentially an elaboration of what atyy said. I thought I would post it in case any one is interested. The question was whether there was any relationship between the Lagrangian expression $L = T - U$ and the proper time $s^2 = t^2 - x^2$, since they look formally similar and also the relationship between T and t, and U and x, is analogous. It turns out that there is indeed a very deep connection between the two, involving general relativity!

The law of motion in gravitational fields in general relativity says that the proper time along the correct path $\int ds$ is maximum. For a particle moving vertically on the surface of the earth, choose an arbitrary path $$h(t)$$. General relativity says that the relative rate of a clock is higher if the clock is higher in the gravitational field by an amount $gh/c^2$. If the clock is moving, then there is an additional change in the relative rate due to special relativity which is approximately $-v^2/2c^2$. Therefore, the condition that the proper time is maximum will be

$$\int ( gh - v^2/2 )dt = max.$$​

If you multiply this by $-m$, you get

$$\int ( mv^2/2 - mgh )dt = min,$$​

which is exactly the condition that the the action $\int (T - U) dt$ is a minimum. I found this in "The Feynman Lectures" (vol. II), if you want more details.

Last edited: Feb 4, 2009