#### Mentz114

Gold Member

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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

[tex]L = \frac{1}{2}m(x)\dot{x}^2[/tex]

which leads to

[tex]\frac{\partial L}{\partial x} = 0, \frac{\partial L}{\partial\dot{x}} = m(x)\dot{x} [/tex]

and

[tex]\frac{d}{dt}\left(m(x)\dot{x}\right) = m'(x)\dot{x}^2 + \ddot{x}m(x) [/tex]

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

[tex]\ddot{x} = \frac{-m'(x)\dot{x}^2}{m(x)}[/tex]

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

Now, making the assertion that

[tex]m(x) = m_0f(x)[/tex] where m

[tex]\ddot{x} = \frac{-f'(x)\dot{x}^2}{f(x)}[/tex]

and the acceleration is independent of m

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

[tex](g_{00})^{-1}[/tex]

we have to say

[tex]m(x) = (g_{00})^{-1}[/tex].

Doing the calculation for the Schwarzschild spacetime one finds

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

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

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

[tex]L = \frac{1}{2}m(x)\dot{x}^2[/tex]

which leads to

[tex]\frac{\partial L}{\partial x} = 0, \frac{\partial L}{\partial\dot{x}} = m(x)\dot{x} [/tex]

and

[tex]\frac{d}{dt}\left(m(x)\dot{x}\right) = m'(x)\dot{x}^2 + \ddot{x}m(x) [/tex]

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

[tex]\ddot{x} = \frac{-m'(x)\dot{x}^2}{m(x)}[/tex]

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

Now, making the assertion that

[tex]m(x) = m_0f(x)[/tex] where m

_{0}is a constant, we can say[tex]\ddot{x} = \frac{-f'(x)\dot{x}^2}{f(x)}[/tex]

and the acceleration is independent of m

_{0}. 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

[tex](g_{00})^{-1}[/tex]

we have to say

[tex]m(x) = (g_{00})^{-1}[/tex].

Doing the calculation for the Schwarzschild spacetime one finds

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

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

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