# Force-free motion from a busted Lagrangian

1. Aug 27, 2008

### Mentz114

Maybe this should be in the classical physics section, but there is a connection with GR so I'm posting here.

I've been looking at Lagrangians for oscillating systems because I'm interested in gravitational time dilation and stumbled on this. The Lagrangian for a free particle is just the kinetic energy, and I wondered what would happen if I allowed the inertial mass term to vary locally so that

$$L = \frac{1}{2}m(x)\dot{x}^2$$

$$\frac{\partial L}{\partial x} = 0, \frac{\partial L}{\partial\dot{x}} = m(x)\dot{x}$$

and

$$\frac{d}{dt}\left(m(x)\dot{x}\right) = m'(x)\dot{x}^2 + \ddot{x}m(x)$$

where the ' indicates differentiation wrt x. So for force free motion this is zero which gives the EOM

$$\ddot{x} = \frac{-m'(x)\dot{x}^2}{m(x)}$$

This is interesting, because we have motion here without a force, ie coordinate acceleration.

Now, making the assertion that

$$m(x) = m_0f(x)$$ where m0 is a constant, we can say

$$\ddot{x} = \frac{-f'(x)\dot{x}^2}{f(x)}$$

and the acceleration is independent of m0. This is a strong clue that the motion is being caused by gravity. The EOM also looks like the GR equivalent, where the coefficients of the velocities are the affine connections.

In order to conform to the known gravitational time dilation effect which is proportional to

$$(g_{00})^{-1}$$

we have to say

$$m(x) = (g_{00})^{-1}$$.

Doing the calculation for the Schwarzschild spacetime one finds

$$\frac{-m'(x)}{m(x)} = -\left(1-\frac{2m}{r}\right)^{-1}\frac{m}{r^2} = \Gamma^x_{xx}$$.

We get exactly the spatial part of the GR EOM.

What's happening is that translational symmetry of the Lagrangian is broken, giving rise to a gauge field that looks a lot like gravity, and seems to reproduce to some degree the EOM of GR. This is known already (Teleparallel gravity), but I haven't seen it done by allowing the inertial mass to be local.

My questions are

1. Have I made a mistake in my calculation ?

2. can anyone point me to related work ?

3. Did you spot the pun on 'connection' ?

M

Last edited: Aug 27, 2008
2. Aug 27, 2008

### Peeter

If m = m(x) then how can you say this derivative is zero? Example, m = x :

$$\frac{\partial L}{\partial x} = \frac{1}{2}\dot{x}^2$$

3. Aug 27, 2008

### Mentz114

Thank you for pointing that out, Peeter.

Surprisingly, my major bungle makes little difference, just adding

$$\frac{f'(x)}{2f(x)}\dot{x}^2$$ to the RHS of the EOM giving

$$\ddot{x} = \frac{-f'(x)\dot{x}^2}{2f(x)}$$

Phew !

(Peeter, I'd be much obliged if you could check the sums again !).

If this is correct, we have a factor of 1/2 that isn't in the GR version.

I've had more time to think about the idea in general and I will have to do a lot more work in 4-D. It may be necessary to add spin as well to get the whole picture. I'm not sure if this is a gauge situation and I'm having trouble coming up with covariant derivatives that restore energy conservation.

Last edited: Aug 27, 2008
4. Aug 28, 2008

### Peeter

Hi Mentz, yes I get the same result:

$$\ddot{x} = -\frac{m_x \dot{x}^2}{2 m}$$

However, I don't understand anything GR related that you did after that;)

Fwiw, I was able to do the same derivatives with proper time/velocity and I get:

$$m\dot{v} + v \sum \dot{x}^{\beta} \frac{\partial m}{\partial x^{\beta}} = \frac{1}{2} v^2 \nabla m$$

where:
$$v = \sum \gamma_{\mu} \dot{x}^{\mu}$$
$$\nabla = \sum \gamma^{\mu} \frac{\partial}{\partial x^{\mu}}$$
$$\gamma^{\mu} \cdot \gamma_{\nu} = \delta^{\mu}_{\nu}$$

I don't know if this is what you were also getting (and for GR I have the suspision that you let the frame vectors vary with position/time)?

5. Aug 28, 2008

### Mentz114

Peeter,

I'm glad we agree. In fact, with the extra term I can see the possibility of getting covariant derivative which will restore enrgy conservation.

I think I understand what you're doing, but is this what you mean,

$$m\dot{v} + v \sum \dot{x}^{\beta} \frac{\partial m}{\partial x^{\beta}} = m\dot{v} + \frac{1}{2} v^2 \nabla m$$

in the second equation ? You seem to be doing the Lagrangian in curved space.

I'm not very practised in the diff. geom. approach, being a tensor person by training. If the $\gamma_{\mu}$ are frames and $\gamma^ {\mu}$ are co-frames, we can substitute (?)

$$\gamma_{1} = \gamma^{0} = (1-2M/r)^{-\frac{1}{2}}, \gamma^{1} = \gamma_{0} = (1-2M/r)^{\frac{1}{2}}$$

for the Schwarzschild case and try it.

Anyhow, thanks for the input so far. I have to think about this. I need to get the proper time from my gauge field.

M

6. Aug 28, 2008

### Peeter

Hi Mentz,

I was using flat space, but allowing for a non-orthonormal frame if desired.

Now, it's worth pointing out here that I'm not a phyisist, nor even a physics student. I have an undergrad Engineering degree, am a computer programmer, and have been self studying (special) relativity lately. My way of doing things is probably weird to anybody who actually knows what they are doing. I've typed up my derivation of the above, explaining the approach and notation. I've probably got too much detail, but it should be sufficient that somebody that isn't me can understand what I did and my assumptions:

http://www.geocities.com/peeter_joot/geometric_algebra/mass_vary_lagrangian.pdf

(despite taking only half a page on paper, it turned out too big in latex with explainations to conviently post inline).

7. Aug 28, 2008

### Mentz114

Thanks, Peeter, I'll have a look at that a little later. Have you read the paper I attached in post#3 in this thread ?

Join the club.

M

8. Aug 28, 2008

### Peeter

No I hadn't, but it is an interesting looking paper. I'd say I have a considerable way to go before I'm capable of digesting much of it though;)

I do have a text of one of the authors referenced in that paper (Doran), that I've been working my way through very slowly. Its skims over a lot of details so you have to figure a lot out for yourself. Near the end I think it has some details on the GR approach in this paper. I was hoping to eventually get to that too, and after doing so I'll probably be in better shape to read that paper.

9. Aug 28, 2008

### Mentz114

Peeter,
I've had quick look at your paper - it all looks right, you have deduced the correct Lagrange principle and

$$L = \frac{1}{2}m(x)g_{\mu\nu}d\dot{x}^{\mu}d\dot{x}^{\nu}$$

(you can drop the Sigma's since we assume summation over repeated indexes).

is the relativistic expression I'm looking for. Nice work. Sadly, I'm out of time for today, but you'll hear from me again on this.

Finally - how did you produce such beautiful PDF so quickly ?

M

Last edited: Aug 28, 2008
10. Aug 28, 2008

### Peeter

Wasn't that quick, though I thought it would be;) I'd say it probably took me an hour to write it up, vs. 5 min on paper this morning while I ate my breakfast (and I was at work, so I'll have to work late tonight to make up for it)

11. Aug 28, 2008

### Peeter

Hi Mentz, I've updated the pdf. I realized that the results can be algebraically reduced significantly, and when this is done the final result is nothing more than:

$$\frac{d (mv)}{d\tau} = \nabla \frac{1}{2} m v^2$$

(in the pdf writeup I also included a non-velocity dependant potential term).

When written like so:

$$m \ddot{x} = \frac{1}{2} v^2 \nabla m - \dot{m} v$$

(using $v^2 = c^2$ to pull it out of the gradient), you can see the similarity in form to the corrected form of your original calculation.

12. Aug 31, 2008

### Mentz114

Peeter,
I read the update, thanks. I have found an attempt to make a gravity theory with variable mass here

Unfortunately the author does not give a Lagrangian so I can't follow his logic.

I now don't think that the Lagrangian I started with can give a realistic theory because the field is a scalar and that never works for gravity except in weak fields.

M

13. Aug 31, 2008

### Jonathan Scott

To get a semi-classical view of GR, you can use a Lagrangian in a flat background space using isotropic coordinates (relying on the spherically symmetrical case), treating the rest mass and c as variables affected by a scalar potential which is -Gm/rc^2 to first order. With that approximation, one can calculate orbits and for example determine the effect on the perihelion precession of different second-order terms in the potential.

All you have to do is assume the action is the usual integral of $mc^2 d\tau$, giving

$$L = -mc^2 \sqrt{1-v^2/c^2}$$

(where the sign is a matter of convention).

Let $\Phi$ represent the timelike scalar factor in the metric, which is to the first order equal to $(1 - Gm/rc^2)$. The value of $m$ in the above equation is then equal to $\Phi^{-3} m_0$ and the value of $c$ is equal to $\Phi^2 c_0$ where $m_0$ and $c_0$ are the standard fixed values of the mass and speed of light as seen locally.

A bit of manipulation of the Euler-Lagrange equations then gives rise to the following equation of motion:

$$d\mathbf{p}/dt = \frac{E}{c^2} \mathbf{g} ( 1 + v^2/c^2)$$

where $E$ is the total energy and the Newtonian-style acceleration vector $g$ is given by

$$\mathbf{g} = \frac{c^2}{\Phi} \nabla \Phi$$

Note that if you only take into account the effect of the potential on time and not on space, you get a Newtonian-style equation of motion instead, changing the sign of the $v^2/c^2$ term, which does not give the correct deflection of fast-moving objects:

$$d\mathbf{p}/dt = \frac{E}{c^2} \mathbf{g} ( 1 - v^2/c^2)$$

or, dividing both sides by $\sqrt{1-v^2/c^2}$,

$$d\mathbf{p}/d\tau = m \mathbf{g}$$

which is the obvious (but incorrect) special relativity Newtonian gravitational equation of motion.

14. Aug 31, 2008

### Mentz114

Jonathan, I can see how coupling $\Phi$ to the mass and the speed of light gives those equations. I think in my original Lagrangian, I only coupled to the mass. Anyhow, I think the lesson is that a scalar field is only going to give weak field gravity at best. Interesting, though.

The GR Lagrangian, whose 4-D volume integral is an action

$$S = \int dx^4 L = m\int dx^4 g_{ab}\dot{x}^a\dot{x}^b$$

may be extremised to get the geodesic equations ( the $g_{ab}$ being functions of the coordinates now)

If the m term is dropped, then what is being extremised is the proper length of the curve. So in curved space the proper length is proportional to the action (?).

M

15. Sep 2, 2008

### Mentz114

I completed the calculations in Minkowski space-time for the mass varying Lagrangian,

$$L = m(x_{\mu})\eta_{ab}\dot{x}^a\dot{x}^b$$

from which we get by standard Euler-Lagrange procedure,

$$mg_{a\mu}\ddot{x}^a = \frac{1}{2}(\partial_{\mu}m)g_{ab}\dot{x}^a\dot{x}^b - (\partial_{\beta}m)\dot{x}^{\beta}g_{\alpha\mu}\dot{x}^{\alpha}= A_{\mu} - B_{\mu}$$
( see peeter's paper, above ) and so

$$\ddot{x}^a = \frac{1}{m}g^{a\mu}(A_{\mu} - B_{\mu})$$

In order to compare the EOM with those of the Rindler space-time, we set

$$\phi_{\mu} \equiv \frac{\partial_{\mu}m}{m} = 0 , \mu \ne 1$$

which gives,

$$\ddot{x}^0 = \phi_1\dot{x}^0\dot{x}^1$$
$$\ddot{x}^1 = \frac{\phi_1}{2}((\dot{x}^0)^2+(\dot{x}^1)^2)$$

Compare these to the proper time EOM for the Rindler metric, which is the nearest thing in GR to the scenario above,

$$\ddot{x}^0 = -\frac{2a}{(1+ax)}\dot{x}^0\dot{x}^1$$
$$\ddot{x}^1 = -a(1+ax)(\dot{x}^0)^2$$

Which look incompatible to me.

16. Sep 2, 2008

### Mentz114

There are error in the above. The EOMs are

$$\ddot{x}^0 = \phi_1\dot{x}^0\dot{x}^1$$
$$\ddot{x}^1 = \frac{\phi_1}{2}((\dot{x}^0)^2+(\dot{x}^1)^2)$$

and
$$\ddot{x}^0 = -\frac{2a}{(1+ax)}\dot{x}^0\dot{x}^1$$
$$\ddot{x}^1 = -a(1+ax)(\dot{x}^0)^2$$