# GR analog to F=ma

1. Oct 14, 2011

### nickyrtr

I am interested to know what Newton's second law (F=ma) looks like in general relativity. Looking at the geodesic equation, it appears to have some similarity. Multiply the mass of a particle by both sides and it looks like this:

$$m\frac{d^2x^{\mu}}{ds^2}+m\Gamma^{\mu}_{\nu\lambda}\frac{dx^{\nu}}{ds}\frac{dx^{\lambda}}{ds} = 0$$

The left side resembles 'ma' in F=ma. The above describes a particle influenced only by gravity, so now I wonder how the equation changes when other forces are present, electromagnetism for example. From web searching and looking in textbooks I do not find any modified version of the geodesic equation that would fit, but there are some that look similar. This is my best guess, where q is the particle's charge and A is a potential due to some field:

$$m\frac{d^2x^{\mu}}{ds^2}+m\Gamma^{\mu}_{\nu\lambda}\frac{dx^{\nu}}{ds}\frac{dx^{\lambda}}{ds} = q\frac{dx^{\nu}}{ds}\frac{dA^{\mu}}{dx^{\nu}}$$

Is this equation correct? It seems the right hand side is the GR analog to 'F' in F=ma. If anyone can refer me to a textbook or publication with this kind of equation, I would appreciate it very much.

2. Oct 14, 2011

### Bill_K

Close. For a charged particle the equation of motion is actually

m d2xμ/ds2 + m Γμνσ dxν/ds dxσ/ds = q Fμν dxν/ds

3. Oct 14, 2011

### pervect

Staff Emeritus
The left-hand side is indeed the four-fource in special relativity. I'm not sure if the four-force is still a tensor in the context of GR though - I'm pretty sure it transforms in a more complex manner, but I'm not positive.

This is important because you are assured covariance (i.e. independence of your predictions from the specific coordinate system you use) when everything you use transforms as a tensor, and you have no such guarantee if you use entities that do not transform as tensors.

The right-hand side of the expression should be:

$$m\frac{d^2x^{\mu}}{ds^2}+m\Gamma^{\mu}_{\nu\lambda}\frac{dx^{\nu}}{ds}\frac{dx^{\lambda}}{ds} = qF^{\mu}{}_{\delta} \frac{d\,x^{\delta}}{ds}$$

where F is the Faraday tensor.

In flat space time
$$F_{\mu\nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$$

in curved space-time, one replaces the partial derivatives with covariant derivatives.

$$F_{\mu\nu} = \nabla_{\mu}A_{\nu} -\nabla_{\nu}A_{\mu}$$

I'm not quite sure what your background is - I hope I didn't underexplain or overexplain, but I'm sort of guessing where you're coming from.

4. Oct 14, 2011

### nickyrtr

Thanks for the replies, especially the definition of the Faraday tensor. That's what I was looking for.

5. Oct 15, 2011

### juanrga

Newton equation is, in reality, the equation

$$\frac{dp}{dt} = F$$

The GR analog is

$$\frac{Dp_i}{d\tau} = F_i$$

where $\frac{D}{d\tau}$ is the absolute (covariant) derivative, $p_i$ the four-momentum and $F_i$ a non-gravitational four-force.

6. Oct 15, 2011

### jfy4

Bill I have a question. Is what you wrote there
$$\nabla_{\mu}u^{\nu}=\frac{q}{m} F_{\;\mu}^{\,\nu}?$$
In other words, for geodesic motion the right hand side is zero. But in the presence of an E&M field the motion is not geodesic, but instead, covariantly we can say elegantly that the parallel transport of the tangent vector is the same as the charge to mass ratio times the field strength tensor, does that sound right?

If so, that's one of the prettiest equations I have ever seen...

7. Oct 17, 2011

### Phrak

An insightful question, nickyrtr. Very cool.

Amu is a one-form, thus an antisymmetric tensor. The exceptional beauty of antisymmetric tensors with lower indices is that

$$\nabla_{\mu}A_{\nu} -\nabla_{\nu}A_{\mu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu} \ .$$
The connections cancel. More generally, for a (completely) antisymmetric tensor T,

$$U_{\mu \nu_1...\nu_k} = \partial_{[\mu}T_{\nu_1...\nu_k]} \ .$$
U is also a tensor on curved manifolds, and also antisymmetric.

The bracket notation, [abc...z] says you add elements with even permutations of [abc...z] and subtract elements with odd permutations. For example,

[rst] = rst + stu + tus - srt - rts - tsr.

This isn't standard fair in general relativity so isn't well known.

Last edited: Oct 17, 2011
8. Oct 17, 2011

### nickyrtr

Do the connections also cancel for this?:

$$F^{\mu}_{\nu} = \nabla^{\mu}A_{\nu} - \nabla^{\nu}A_{\mu}$$

And also, I am not sure about the definition of ∂μ. Is this right?:

$$\partial_{\mu} = \frac{\partial}{\partial x^{\mu}}$$

I am unsure because ∂μ has a lower index, but ∂/∂xμ has an upper index. Thanks much for the information.

9. Oct 17, 2011

### DrGreg

Yes, it's correct. It turns out that when you differentiate with respect to a contravariant vector you get a covariant (not contravariant) operator.

10. Oct 17, 2011

### pervect

Staff Emeritus
Yes, as Dr. Greg said, $$\partial_{\mu} = \frac{\partial}{\partial x^{\mu}}$$

In the notation I was using, $\mu$ is a placeholder or index. This means that if $u^0=t, u^1=x, u^2=y, u^3=z$ then $\partial_u = (\partial / \partial_t, \, \partial / \partial_x, \, \partial / \partial_y, \, \partial / \partial_z)$.

If you have a vector u with components $u^a$ and you want to find the directional derivative in that direction, in the index notation I was using you need to write

$u^a \partial_a$

for instance, if u had the components (0,1,0,0) then you'd get $\partial / \partial_x$

Sometimes, people write the directional derivative as $\partial_u$, and I think a few people did it in this thread, but that's not index notation. It can be confusing when the two styles are mixed, especially at first, though usually context tells you what's meant.

11. Oct 20, 2011

### Phrak

$\partial_\mu$ can have more than one meaning.

To be precise, it is $\frac{\partial}{\partial x^\mu}$, as Pervect and and Dr. Greg pointed out.

However sometimes it is shorthand for an entire dual vector, $\partial_\mu = \frac{\partial}{\partial x^\mu} dx^\mu$, where it is convenient to drop the basis $dx^\mu$.

This is not an uncommon practice, at all. It's convenient, and the basis don't usually lend addiitional information.