Co-ordinate vs. true acceleration

In summary: The 4-acceleration cannot be zero if it exists in one coordinate system, but it can be zero if it exists in all coordinate systems.
  • #1
masudr
933
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?
 
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  • #2
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
masudr said:
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
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
Garth said:
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
Thanks for your timely responses.

Garth said:
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.

George Jones said:
[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 to Co-ordinate vs. true acceleration

What is the difference between co-ordinate and true acceleration?

Co-ordinate acceleration refers to the rate of change of velocity with respect to a specific chosen frame of reference, while true acceleration refers to the rate of change of velocity with respect to an absolute frame of reference, such as the Earth.

Why is it important to distinguish between co-ordinate and true acceleration?

It is important to distinguish between the two because the choice of frame of reference can greatly affect the measured acceleration. Co-ordinate acceleration is relative to a chosen frame of reference, while true acceleration is an objective measure that is not affected by the choice of frame of reference.

How do you calculate co-ordinate and true acceleration?

Co-ordinate acceleration can be calculated using the equation a = Δv/Δt, where a is the acceleration, Δv is the change in velocity, and Δt is the change in time. True acceleration can be calculated using the equation a = Δv/Δt + ar, where ar is the acceleration due to the chosen frame of reference.

What are some examples of co-ordinate and true acceleration?

An example of co-ordinate acceleration would be a car accelerating on a straight road. The acceleration would be measured relative to the road. An example of true acceleration would be the acceleration of a rollercoaster as it moves through loops and curves. The true acceleration would be measured relative to the Earth's frame of reference.

How does understanding co-ordinate and true acceleration relate to real-world applications?

Understanding co-ordinate and true acceleration is crucial in fields such as physics, engineering, and navigation. Knowing the difference between the two can help accurately measure and predict the movement of objects and vehicles in various frames of reference. This knowledge is also important in developing technologies such as GPS systems, which rely on accurate measurements of true acceleration to determine location and movement.

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