Co-ordinate vs. true acceleration

  • Thread starter masudr
  • Start date
  • #1
934
0
In G.R. we do physics in any kind of frame, inertial or not. To compensate for that, we get nice little Christoffel symbols in all our derivatives. And straight lines become geodesics.

But is there truly no way to distinguish between co-ordinate acceleration and "real" acceleration? Or is there no such thing as real acceleration, since we can always transform it away? Or is is that in a reference frame where we did transform the acceleration of a particle away, geodesics would no longer be straight lines? But if the curvature tensor's components all vanished, then we know that the space is Minkowski, and hence we can distinguish between these two types of acceleration.

Hmm, I'm not sure if I've made sense here. Is every question [posed above] answerable by a resounding "yes" or have I made a mistake somewhere?
 

Answers and Replies

  • #2
Garth
Science Advisor
Gold Member
3,577
105
You have answered your own question I think:
But if the curvature tensor's components all vanished, then we know that the space is Minkowski, and hence we can distinguish between these two types of acceleration.
It's all in the Riemannian.

Garth
 
  • #3
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,482
1,206
But is there truly no way to distinguish between co-ordinate acceleration and "real" acceleration? Or is there no such thing as real acceleration, since we can always transform it away?

No coordinate transformation will tranform away the real 4-acceleration of an observer. Let [itex]\left\{ x^\mu \left( \tau \right) \right\}[/itex] be the coordinate description of an observer's worldline parametrized by proper time. Then the observer's 4-velocity is

[tex]u = \frac{dx^\mu}{d\tau} \frac{\partial}{\partial x^\mu},[/tex]

and 4-acceleration is

[tex]
\begin{equation*}
\begin{split}
a &= \nabla_u u \\
&= u^\mu \nabla_\mu \left( u^\nu \frac{\partial}{\partial x^\nu} \right) \\
&= u^\mu \left[ \left( \nabla_\mu u^\nu \right) \frac{\partial}{\partial x^\nu} + u^\nu \nabla_\mu \frac{\partial}{\partial x^\nu} \right] \\
&= u^\mu \left[ \frac{\partial u^\nu}{\partial x^\mu} \frac{\partial}{\partial x^\nu} + u^\nu \Gamma^{\alpha}_{\nu \mu} \frac{\partial}{\partial x^\alpha} \right] \\
&= u^\mu \left[ \frac{\partial u^\alpha}{\partial x^\mu} + u^\nu \Gamma^{\alpha}_{\nu \mu} \right] \frac{\partial}{\partial x^\alpha}.
\end{split}
\end{equation*}
[/tex]

If an observer's 4-acceleration is zero in one coordinate system (freely falling obsrever), it is zero in all coordinate systems; if an observer's 4-acceleration is non-zero in one coordinate system, it is non-zero in all coordinate systems.
 
Last edited:
  • #4
Garth
Science Advisor
Gold Member
3,577
105
The operative word here of course is "4-acceleration".

In the OP it sounded as if 3-acceleration was being talked about.

Hmm, I'm not sure if I've made sense here.
Was that the cause of your confusion masdur?

Garth
 
  • #5
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,482
1,206
The operative word here of course is "4-acceleration".

In the OP it sounded as if 3-acceleration was being talked about.

But, [itex]1 = g \left( u , u \right)[/itex] gives that

[tex]0 = \nabla_u g \left( u , u \right) = g \left( \nabla_u u , u \right) + g \left( u , \nabla_u u \right),[/tex]

so

[tex]0 = g \left( a , u \right).[/tex]

Thus, since the 4-velocity [itex]u[/itex] is timelike, the 4-acceleration [itex]a[/itex] is spacelike. Therefore, if the 4-acceleration is non-zero, the three "spatial" (assuming that three of the coordinates are spacelike and one is timelike) coordinates of the 4-acceleration cannot be simultaneously transformed away.
 
Last edited:
  • #6
934
0
Thanks for your timely responses.

Was that the cause of your confusion masdur?

This probably demonstrates my lack of experience in thinking in GR terms, but I was just concerned that there should be something physically different in the situation of using spherical polars and detecting acceleration, and using Cartesians and detecting acceleration. But in G.R. neither system has an advantage: both types of acceleration should be deemed equivalent. This seems especially troublesome to me, especially when considering non-gravitational motion, i.e. in a flat space.

[tex]
\begin{equation*}
\begin{split}
a &= \nabla_u u \\
&= u^\mu \nabla_\mu \left( u^\nu \frac{\partial}{\partial x^\nu} \right) \\
&= u^\mu \left[ \left( \nabla_\mu u^\nu \right) \frac{\partial}{\partial x^\nu} + u^\nu \nabla_\mu \frac{\partial}{\partial x^\nu} \right] \\
&= u^\mu \left[ \frac{\partial u^\nu}{\partial x^\mu} \frac{\partial}{\partial x^\nu} + u^\nu \Gamma^{\alpha}_{\nu \mu} \frac{\partial}{\partial x^\alpha} \right] \\
&= u^\mu \left[ \frac{\partial u^\alpha}{\partial x^\mu} + u^\nu \Gamma^{\alpha}_{\nu \mu} \right] \frac{\partial}{\partial x^\alpha}.
\end{split}
\end{equation*}
[/tex]

If an observer's 4-acceleration is zero in one coordinate system (freely falling obsrever), it is zero in all coordinate systems; if an observer's 4-acceleration is non-zero in one coordinate system, it is non-zero in all coordinate systems.

Again this probably demonstrates my lack of knowledge, but I fail to see how your conclusion follows directly from your expression for the 4-acceleration.
 

Related Threads on Co-ordinate vs. true acceleration

  • Last Post
Replies
12
Views
3K
  • Last Post
Replies
6
Views
4K
Replies
11
Views
3K
Replies
37
Views
2K
  • Last Post
Replies
18
Views
7K
  • Last Post
Replies
2
Views
1K
  • Last Post
2
Replies
44
Views
3K
  • Last Post
Replies
13
Views
1K
Replies
2
Views
1K
Replies
1
Views
2K
Top