Local acceleration experienced by observer

In summary, the conversation discusses the computation of acceleration in the context of special relativity and how to transform it into different inertial frames. It is explained that the local acceleration experienced by a particle can be determined using the 4-vector treatment, where the magnitude of the 4-acceleration is a Lorentz invariant and can be computed by any observer. The proper acceleration is given by the Lorentz transform that maps from the particle's instantaneous co-moving frame to the inertial frame. However, it is noted that in general, the Lorentz transform from one frame to another is not equal to the product of the individual Lorentz transforms from each frame to the particle's instantaneous co-moving frame.
  • #1
gordonblur
6
0
I have read a number of threads on acceleration and special relativity, but can't find what I'm looking for. I would like to know, in the context of special relativity, how to compute the acceleration "felt" by an observer, and how to transform this acceleration into different inertial frames.

Given a particle's world line as a 4-vector [tex] x(\tau) [/tex] parameterised by proper time and relative to some inertial frame, then it's 4-velocity and 4-acceleration are

[tex] \dot{x}=\frac{dx}{d\tau}=\gamma(e_0+V) [/tex]

[tex] \ddot{x}=\dot{\gamma}(e_0+V)+\gamma^2 A [/tex]

where [tex] V\cdot e_0=A\cdot e_0=0, \gamma^{-2}=1+V\cdot V [/tex] and [tex] \dot{\gamma}=-\gamma^4 A\cdot V [/tex].

I appologise for the slightly informal notation that follows. [tex] V=(0,V_1,V_2,V_3) [/tex] and [tex] A [/tex] are the instantaneous 3-velocity and 3-acceleration.

Let the Lorentz tranform that maps into the particle's instantaneous co-moving frame frame be [tex] \Lambda_V [/tex], s.t. [tex] \Lambda_V \dot{x}=e_0 [/tex] and [tex] \Lambda_V e_0=\gamma(e_0-V) [/tex]. My initial thought was that the "local" acceleration experienced by the particle was simply [tex] \Lambda_V \ddot{x} [/tex].

However, consider the same world line expressed in a different inertial frame [tex] y=\Lambda_U x [/tex], [tex] \dot{y}=\Lambda_{U} \dot{x}=\delta(e_0+W) [/tex] and [tex] \ddot{y}=\Lambda_{U} \ddot{x} [/tex]. Then the local acceleration is either [tex] \Lambda_W \ddot{y} [/tex] or [tex] \Lambda_V \Lambda_{-U} \ddot{y} [/tex], but these are not equal since in general [tex] \Lambda_W \neq \Lambda_V \Lambda_{-U} [/tex] (although I think [tex] \Lambda_W e_0 = \Lambda_V \Lambda_{-U}e_0 [/tex]).

I guess the solution lies in splitting the velocity into two components, one parallel and one perpendicular to [tex] A [/tex], but can't find a mathematical justification. Any help would be greatly appreciated.

Thanks, Gordon.
 
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  • #2
gordonblur said:
I have read a number of threads on acceleration and special relativity, but can't find what I'm looking for. I would like to know, in the context of special relativity, how to compute the acceleration "felt" by an observer, and how to transform this acceleration into different inertial frames.

Given a particle's world line as a 4-vector [tex] x(\tau) [/tex] parameterised by proper time and relative to some inertial frame, then it's 4-velocity and 4-acceleration are

[tex] \dot{x}=\frac{dx}{d\tau}=\gamma(e_0+V) [/tex]

[tex] \ddot{x}=\dot{\gamma}(e_0+V)+\gamma^2 A [/tex]

where [tex] V\cdot e_0=A\cdot e_0=0, \gamma^{-2}=1+V\cdot V [/tex] and [tex] \dot{\gamma}=-\gamma^4 A\cdot V [/tex].

I appologise for the slightly informal notation that follows. [tex] V=(0,V_1,V_2,V_3) [/tex] and [tex] A [/tex] are the instantaneous 3-velocity and 3-acceleration.

Let the Lorentz tranform that maps into the particle's instantaneous co-moving frame frame be [tex] \Lambda_V [/tex], s.t. [tex] \Lambda_V \gamma(e_0+V)=e_0 [/tex] and [tex] \Lambda_V e_0=\gamma(e_0-V) [/tex]. My initial thought was that the "local" acceleration experienced by the particle was simply [tex] \Lambda_V \ddot{x} [/tex].

However, consider the same world line expressed in a different inertial frame [tex] y=\Lambda_U x [/tex], [tex] \dot{y}=\Lambda_{U} \dot{x}=\delta(e_0+W) [/tex] and [tex] \ddot{y}=\Lambda_{U} \ddot{x} [/tex]. Then the local acceleration is either [tex] \Lambda_W \ddot{y} [/tex] or [tex] \Lambda_V \Lambda_{-U} \ddot{y} [/tex], but these are not equal since in general [tex] \Lambda_W \neq \Lambda_V \Lambda_{-U} [/tex] (although I think [tex] \Lambda_W e_0 = \Lambda_V \Lambda_{-U}e_0 [/tex]).

According to the "Principle of General Covariance" which states that the laws of physics are valid in all frames of reference and in all valid coordinate systems I would expect the local acceleration experienced by the observer to be the same in all frames.

I guess the solution lies in splitting the velocity into two components, one parallel and one perpendicular to [tex] A [/tex], but can't find a mathematical justification. Any help would be greatly appreciated.

Thanks, Gordon.
In the momentarily co-moving inertial reference frame the http://en.wikipedia.org/wiki/Proper_acceleration" is equal to (0,A), which is obviously the same as the proper acceleration A. So you can always determine the magnitude of the proper acceleration as the norm of the four-acceleration.
 
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  • #3
I would suggest using the 4-vector treatment. If you are not familiar with 4-vectors, read about them online at:

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html
http://www.physics.ohio-state.edu/~cleo/hep/explain/4vector_1.html
http://en.wikipedia.org/w/index.php?title=Four-vector&oldid=171495581

Now, to apply 4-vectors to accelerations. Let's start with 3-vectors.

[tex]\frac{d^2 x}{d t^2}, \frac{d^2 y}{d t^2}, \frac{d^2 z}{d t^2} [/tex]

would be a 3-acceleration, or coordinate acceleration, where (t,x,y,z) are coordinates in some specific inertial frame, and we have functions

x(t), y(t), z(t)

to describe the path the particle takes.

If you use [itex]\tau[/itex] as the time variable, where [itex]\tau[/itex] is the proper time of the accelerating particle, then we describe the path the particle takes by 4 functions

t([itex]\tau[/itex]), x([itex]\tau[/itex]), y([itex]\tau[/itex]), z([itex]\tau[/itex])

and the 4-acceleration is given by

[tex]
\left(\frac{d^2 t}{d \tau^2}, \frac{d^2 x}{d \tau^2}, \frac{d^2 y}{d \tau^2}, \frac{d^2 z}{d \tau^2} \right) [/tex]

It transforms via the Lorentz transform, just like any other 4-vector.

The magnitude of the 4-acceleration is the "proper" acceleration that's "felt" by the observer, and is a Lorentz invariant of the 4-vector so it can be computed via any observer.
 
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  • #4
Thanks for the links guys, that all makes sense. My understanding of SR comes from Malcolm Ludvigsen's book, General Relativity a Geometric Approach, and searching the web for help. Unfortunately I'm not comfortable with using tensor notation yet, so forgive my linear algebra type style.

Am I correct that given a world line [tex] x(\tau) [/tex] (a 4-vec relative to some inertial frame [tex] X [/tex]) that the proper acceleration

[tex] a=(0,a_1,a_2,a_3)=\Lambda \frac{d^2 x}{d_2 \tau} [/tex],

where [tex] \Lambda [/tex] is the Lorentz transform that maps from frame [tex] X [/tex] to the particle's instantaneous co-moving frame?
 
  • #6
Great, so here is the bit I don't understand.

Suppose [tex] x(\tau) [/tex] and [tex] y(\tau) [/tex] represent the same world line, but w.r.t. two different inertial frames [tex] X [/tex] and [tex] Y [/tex] respectively. Then

[tex] y=\Lambda \, x [/tex]

where [tex] \Lambda [/tex] is the Lorentz transform that maps from [tex] X [/tex] to [tex] Y [/tex] and the proper acceleration is

[tex] a=(0,a_1,a_2,a_3)=\Lambda_X \ddot{x}=\Lambda_Y \ddot{y} [/tex]

where [tex] \Lambda_X [/tex] is the Lorentz transform that maps from [tex] X [/tex] into the instantaneous co-moving inertial frame of [tex] x [/tex] and similarly [tex] \Lambda_Y [/tex] from [tex] Y [/tex] to [tex] y [/tex].

However, [tex] \ddot{y}=\Lambda \, \ddot{x} \Rightarrow a=\Lambda_Y \Lambda \, \ddot{x} [/tex], but in general [tex] \Lambda_X \neq \Lambda_Y \Lambda [/tex].

Where have I gone wrong?
 
  • #7
So, here is a simple example that should demonstrate the problem I'm having. I'll use only 2 spatial dimensions for brevity.

A particle under constant acceleration [tex] g [/tex] has world line:

[tex] x=\frac{1}{g}(sinh(g\tau),cosh(g\tau),0), \; \dot{x}=(cosh(g\tau),sinh(g\tau),0),
\; \ddot{x}=g^2x [/tex]

Using the formula from post #4 the proper acceleration is:

[tex] a_x=\left(\begin{array}{ccc}
cosh(g\tau)&-sinh(g\tau)&0 \\
-sinh(g\tau)&cosh(g\tau)&0 \\
0&0&1 \\ \end{array}\right)\ddot{x}=(0,g,0) [/tex]

Everything fine so far, but change frame into one moving at speed [tex] 1/\sqrt 2 [/tex] perpendicular to the acceleration. The same world line in this frame is:

[tex] y=\left(\begin{array}{ccc}
\sqrt 2&0&1 \\
0&1&0 \\
1&0&\sqrt 2 \\ \end{array}\right) x, \; \dot{y}=(\sqrt 2 \, cosh(g\tau),sinh(g\tau),cosh(g\tau)), \; \ddot{y}=g(\sqrt 2 \, sinh(g\tau),cosh(g\tau),sinh(g\tau)) [/tex]

A quick sanity check shows [tex] \dot{y}\cdot\dot{y}=1, \dot{y}\cdot\ddot{y}=0 [/tex] and [tex] \ddot{y}\cdot\ddot{y}=-g^2 [/tex].

At [tex] \tau=asinh(1)/g [/tex], [tex] \dot{y}=(2,1,\sqrt 2) [/tex], [tex] \ddot{y}=g(\sqrt 2,\sqrt 2,1) [/tex] and the proper acceleration is:

[tex] a_y=\left(\begin{array}{ccc}
2&-1&-\sqrt 2 \\
-1&\frac{4}{3}&\frac{\sqrt 2}{3} \\
-\sqrt 2&\frac{\sqrt 2}{3}&\frac{5}{3} \\ \end{array}\right) \ddot{y}=\frac{g}{3}(0,2 \sqrt 2,1) [/tex]

So [tex] a_x\cdot a_x=a_y\cdot a_y=-g^2 [/tex] but [tex] a_x\neq a_y [/tex].

This is causing a simulation I'm running a problem. I want to specify the proper acceleration of a particle, then transform this into an arbitrary "universal" reference frame in which I'm keeping track of the particle's world line and numerically integrate the path of the particle in this frame. The problem is that as this example shows, using the above formula (from post #4), the proper acceleration of the particle depends on the choice of "universal" frame, but the particle shouldn't care which frame its world line is being integrated in. This is borne out in the simulations in the form of inconsistent results depending on the particle's initial velocity (which is equivalent to changing universal frame).

I hope this makes sense, what have I missed?
 
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  • #8
gordonblur said:
However, [tex] \ddot{y}=\Lambda \, \ddot{x} \Rightarrow a=\Lambda_Y \Lambda \, \ddot{x} [/tex], but in general [tex] \Lambda_X \neq \Lambda_Y \Lambda [/tex].

Where have I gone wrong?
Be careful here. If I am reading correctly [tex] \ddot{y}[/tex] denotes differentiation wrt coordinate time in Y and [tex] \ddot{x}[/tex] denotes differentiation wrt coordinate time in X, so I am not 100% sure that [tex] \ddot{y}=\Lambda \, \ddot{x}[/tex] is true. You may be correct, but I have never tried to transform derivatives wrt coordinate time before so I am unsure.
 
  • #9
[tex] \ddot{y}=\frac{d^2 y(\tau)}{d \tau^2} [/tex]

That is to say the second derivative of a 4-vec w.r.t. proper time. I'm using the notation from Malcolm Ludvigsen's book.
 
  • #10
The example I gave demonstrates that the problem lies in this line from post #6.

gordonblur said:
[tex] a=(0,a_1,a_2,a_3)=\Lambda_X \ddot{x}=\Lambda_Y \ddot{y} [/tex]

In general [tex] \Lambda_X \ddot{x} \neq \Lambda_Y \ddot{y} [/tex].
 
  • #11
gordonblur said:
[tex] \ddot{y}=\frac{d^2 y(\tau)}{d \tau^2} [/tex]

That is to say the second derivative of a 4-vec w.r.t. proper time.
That should transform correctly then. Let me look into it in detail.
 

Related to Local acceleration experienced by observer

1. What is local acceleration experienced by an observer?

The local acceleration experienced by an observer refers to the rate at which an observer's velocity changes with respect to time in a specific location or frame of reference. It is a measure of how quickly the observer is speeding up or slowing down in their immediate surroundings.

2. What factors can affect the local acceleration experienced by an observer?

The local acceleration experienced by an observer can be affected by various factors, such as the gravitational pull of nearby objects, the rotation of the Earth, and the observer's own movement or position relative to their surroundings.

3. How is local acceleration different from global acceleration?

Local acceleration and global acceleration are different concepts. Local acceleration refers to the acceleration experienced by an observer in a specific location, while global acceleration refers to the overall acceleration of an object in a larger scale, such as the acceleration due to the Earth's gravity.

4. How is local acceleration measured?

Local acceleration can be measured using various instruments, such as accelerometers, which can detect changes in an object's velocity over time. It can also be calculated using mathematical equations based on the observer's position, velocity, and time.

5. Why is understanding local acceleration important in scientific research?

Understanding local acceleration is important in scientific research because it allows us to accurately describe and predict the movements of objects and observers in different locations and frames of reference. It also plays a crucial role in many fields of study, such as physics, astronomy, and engineering.

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