I Free Particles & Geodesics: Do They Go Hand-in-Hand?

[kent davidge]

Is it correct to say that free particles always follow geodesics?

 

[Dale]

Yes (geodesics in spacetime, not space)

 

[kent davidge]

(geodesics in spacetime, not space)

Can we just use this same concept for Euclidean space of Newton? Say, in 2 or 3 dimensions.

 

[Dale]

Can we just use this same concept for Euclidean space of Newton? Say, in 2 or 3 dimensions.

There is a theory called Newton Cartan gravity which uses curved spacetime geometry to describe Newtonian gravity and pre-relativistic physics. If there is tidal gravity then it is not Euclidean, but free particles still travel on geodesics and gravity is not a real force, just like in GR.

 

[stevendaryl]

Can we just use this same concept for Euclidean space of Newton? Say, in 2 or 3 dimensions.

Not really. A geodesic is a curve through spacetime, which is a one-dimensional set of points in spacetime. If you're talking about "space", rather than "spacetime", then a curve through space is not a particle trajectory, because you lose the information about how quickly the particle traveled along that curve.

You can, as @Dale says, talk about geodesics in Newtonian spacetime, though. Without going all the way to Special Relativity, you can consider space plus time unified into a 4-dimensional spacetime. This is equivalent to the usual Newtonian theory of gravity.

 

[cosmik debris]

Can we just use this same concept for Euclidean space of Newton? Say, in 2 or 3 dimensions.

Isn't this just Newton's first law?

Cheers

 

[stevendaryl]

Isn't this just Newton's first law?

Not quite. As I said, a geodesic is a curve through a space (or spacetime). In contrast, the Newtonian notion of the path of a particle gives the position (in 3-space) as a function of time. So a Newtonian path is not a curve through space--it has additional information, which is the rate at which it travels along that curve.

If you combine 3D space with 1D time to get 4D spacetime, then you can talk about the path of a particle being a geodesic.

 

[kent davidge]

@stevendaryl and @Dale I was thinking about time as the parameter along the path and the three spatial coordinates as the coordinates. So as @cosmik debris mentioned, the particle path in space would be according to the first law, so that a geodesic would mean the particle is free.

 

[Nugatory]

@stevendaryl and @Dale I was thinking about time as the parameter along the path and the three spatial coordinates as the coordinates. So as @cosmik debris mentioned, the particle path in space would be according to the first law, so that a geodesic would mean the particle is free.

Time as a parameter just means that you're doing ordinary classical mechanics, in which the the path of the particle is described by the three functions $x(t)$, $y(t)$, and $z(t)$ (or coordinate transforms of these). Whether the resulting path is a geodesic (that is, a straight line in space) or not is unrelated to whether the particle is free or not.

 

[kent davidge]

Whether the resulting path is a geodesic (that is, a straight line in space) or not is unrelated to whether the particle is free or not.

but why? by the first law I would expect a free particle to follow a straight line.

 

[Nugatory]

but why? by the first law I would expect a free particle to follow a straight line.

Or no line at all, if it happens to be at rest ($\dot{x}(t)=\dot{y}(t)=\dot{z}(t)=0$).
Consider also that the path of a dropped particle is a straight line although it is subject to gravitational force.
(This is all in the context of your #3 where you specified that the manifold in question is the Euclidean space of Newtonian physics. If you're talking about four-dimensional spacetime, Dale's #2 and #4 are your answers).

 

[pervect]

Is it correct to say that free particles always follow geodesics?

If particle implies "test particle", I believe this is correct. If the particle has a significant mass though, it may emit gravitational waves and not follow a geodesic. It would be distinctly unusual to have a situation where the difference between a particle and a test particle would matter, though.

 

[Dale]

@stevendaryl and @Dale I was thinking about time as the parameter along the path and the three spatial coordinates as the coordinates.

So just space and not spacetime then? In that case, no, the geodesic-ness of a spatial path is not directly related to being a free particle as mentioned by @Nugatory. Also, it becomes fairly tricky to identify space and even define what path is taken.

 

[PeterDonis]

I was thinking about time as the parameter along the path and the three spatial coordinates as the coordinates.

In a general curved spacetime you can't do this, since there is no invariant notion of "space". You can parameterize a particle's path through spacetime by its proper time (as long as the particle has positive invariant mass), but it's a path through spacetime, not space.

 

[kent davidge]

Thank you to everyone. Can we go a little bit into the math? I tried this:

First, consider a 2 dimensional Euclidean space of Newtonian theory and polar coordinates. Suppose we know a particle is following a radial trajectory and call its velocity $u$ and the time $\tau$ . Then $\dot \theta = 0$ for all $\tau$. The trajectory is a geodesic because $\nabla_r e_r = 0$ and the equations of motion are $$\frac{F^r}{m} = \frac{D u^r}{D \tau} = \ddot r \\ \frac{F^\theta}{m} = \frac{D u^\theta}{D \tau} = \ddot \theta = 0$$ The particle may or may not be free depending on whether $\ddot r$ is or not equal to zero.

Now, consider the Minkowsky spacetime. Since the definition of the four-force is $F^\alpha = m \frac{D u^\alpha}{D \tau}$ we immediatly conclude that a particle following a geodesic is free. Furthermore, it will be accelerating when observed from a non-inertial frame (ficticious force) and not accelerating in a inertial, or Lorentz, frame.

Is this analysis correct?

 

[pervect]

Here is an example that might illustrate some of the subtle issues.

Suppose we have a massless particle moving at "c", as all massless particles do.

Then, assuming it is low energy enough to qualify as a test particle, it follows a geodesic through space-time, what we would call a null geodesic. But in this case, we can't parameterize the path of the particle through space-time by proper time, as the proper time doesn't exist along a null worldline. We can parameterize it's path via an affine parameter, though.

Now, suppose we want to find the path of this particle through a Schwarzschild space-time. It becomes clear that if we think of the particle as existing in space, it is not following a straight line through space, informally we might say that "gravity makes the path curve". More on this below. So, if we split space-time into space+time, by using the Schwarzschild coordinate "t" to describe what we mean by time and r, $\theta$, and $\phi$ coordinates to represent space, the particle in this example is not traveling in a geodesic of the 3-d spatial hypersurface of space-time that we've defined as "space".

We might question whether or not the particle in the Schwarzschild space-time was a free particle, I suppose. It becomes a bit uninteresting if, by definition, we don't allow the question to apply to anything but a perfectly flat, empty, space-time.

When we talk about what causes the total deflection of the particle, we start to see some of the problems in describing it unambiguously in English words without math. If we talk about the "extra" deflection of light, for instance, we might say that some of the deflection is due to "spatial curvature", and some of the deflection is due to aspects of gravity that are not "spatial curvature". But the words start to get rather vague.

The simplest math treatment is to calculate the geodesic through space-time via the geodesic equation. Given a split of space-time into space + time, we can define what we mean by space, and compare our space-time geodesic to a geodesic in the subspace of space-time we've decided to call "space", and find they are different paths.

 

[stevendaryl]

@stevendaryl and @Dale I was thinking about time as the parameter along the path and the three spatial coordinates as the coordinates.

But that would not be a geodesic, because the parameterization of a geodesic is arbitrary. Some parametrizations (the affine ones) are more convenient than others, but it's the same curve regardless of how you parameterize it. In contrast, considering 3-position as a function of 1-D time, the parameterization makes a difference.

 

[haushofer]

There is a theory called Newton Cartan gravity which uses curved spacetime geometry to describe Newtonian gravity and pre-relativistic physics. If there is tidal gravity then it is not Euclidean, but free particles still travel on geodesics and gravity is not a real force, just like in GR.

And we so happen to have a brilliantly written Insight about it:

https://www.physicsforums.com/insights/revival-Newton-cartan-theory/

:P

 

[stevendaryl]

And we so happen to have a brilliantly written Insight about it:

https://www.physicsforums.com/insights/revival-Newton-cartan-theory/

:P

I'll tell him you said so.

Just a quick point about Newton-Cartan theory. It's a theory of gravity in 4D spacetime. But you can get a feel for it in the gravity-free case.

Newton's equations of motion are actually weird from the point of view of spacetime, because they are written in terms of derivatives with respect to time. It's unusual to take the derivative of one coordinate with respect to another coordinate. But that's easily fixed by introducing a path parameter $s$. In cartesian coordinates, Newton's equations of motion for a point-mass are:

$m \frac{d^2 x^j}{dt^2} = F^j$

To make it a spacetime theory, let's introduce a path parameter, $s$ and a fourth coordinate $t = x^0$. Then we transform these equations into 4 equivalent equations:

$m \frac{d^2 x^j}{ds^2} = F^j$
$m \frac{d^2 x^0}{ds^2} = F^0$

where we let $F^0 = 0$. This is equivalent to the original 3 equations if we choose $\frac{dx^0}{ds} = 1$ (which is really a choice of units for $s$).

Now, these equations as they stand are only valid for inertial cartesian coordinates. But it's actually simple to rewrite them in a way that is valid for any coordinate system:

Let $x^j$ be our inertial cartesian coordinates (where now $j$ ranges from 0 to 3 instead of 1 to 3, with $x^0 = t$). Let $x^\alpha$ be a new set of coordinates (not necessarily inertial or cartesian).

Let $L^\alpha_j = \frac{\partial x^\alpha}{\partial x^j}$ and the inverse matrix $L^j_\alpha = \frac{\partial x^j}{\partial x^\alpha}$

Then the equations $m \frac{d^2 x^j}{ds^2} = F^j$ becomes

$m \ddot{x^\alpha} + L^\alpha_j (\partial_\beta L^j_{\gamma}) \dot{x^\beta} \dot{x^\gamma} = L^\alpha_j F^j$

which we can write as:

$m \ddot{x^\alpha} + \Gamma^\alpha_{\beta \gamma} \dot{x^\beta} \dot{x^\gamma} = F^\alpha$

where the connection coefficients $\Gamma^\alpha_{\beta \gamma}$ are given by:

$\Gamma^\alpha_{\beta \gamma} = L^\alpha_j (\partial_\beta L^j_{\gamma})$

and the force in the new coordinates is given by: $F^\alpha = L^\alpha_j F^j$.

What I think is nice about this equation, which as far as I know is never used in classical physics, even though it's perfectly valid, is that all the "fictitious" forces that come up in classical physics---g-forces, coriolis forces, centrifugal forces---are all just special cases of connection coefficients. The fictitious force in direction $\alpha$ is just given by:

$F^\alpha_{fict} = - m \Gamma^\alpha_{\beta \gamma} \dot{x^\beta} \dot{x^\gamma}$

 

[kent davidge]

@stevendaryl wow thanks, that's almost all of what I was thinking about, written out in equations.

you can get a feel for it in the gravity-free case

so those equations only work for inertial and non inertial frames as long as gravity is absent?

 

[stevendaryl]

@stevendaryl wow thanks, that's almost all of what I was thinking about, written out in equations.

so those equations only work for inertial and non inertial frames as long as gravity is absent?

Actually, they work in the presence of gravity, too. Galileo's discovery that all objects fall at the same acceleration through vacuum means that you can incorporate Newtonian gravity as a "fictious" force, just as General Relativity does. That's Newton-Cartan theory, as @haushofer says.