Local acceleration experienced by observer

1. Feb 21, 2008

gordonblur

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 $$x(\tau)$$ parameterised by proper time and relative to some inertial frame, then it's 4-velocity and 4-acceleration are

$$\dot{x}=\frac{dx}{d\tau}=\gamma(e_0+V)$$

$$\ddot{x}=\dot{\gamma}(e_0+V)+\gamma^2 A$$

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

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

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

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

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

Thanks, Gordon.

Last edited: Feb 21, 2008
2. Feb 21, 2008

Staff: Mentor

In the momentarily co-moving inertial reference frame the four-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.

3. Feb 21, 2008

pervect

Staff Emeritus
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

$$\frac{d^2 x}{d t^2}, \frac{d^2 y}{d t^2}, \frac{d^2 z}{d t^2}$$

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 $\tau$ as the time variable, where $\tau$ is the proper time of the accelerating particle, then we describe the path the particle takes by 4 functions

t($\tau$), x($\tau$), y($\tau$), z($\tau$)

and the 4-acceleration is given by

$$\left(\frac{d^2 t}{d \tau^2}, \frac{d^2 x}{d \tau^2}, \frac{d^2 y}{d \tau^2}, \frac{d^2 z}{d \tau^2} \right)$$

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.

Last edited: Feb 21, 2008
4. Feb 22, 2008

gordonblur

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 $$x(\tau)$$ (a 4-vec relative to some inertial frame $$X$$) that the proper acceleration

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

where $$\Lambda$$ is the Lorentz transform that maps from frame $$X$$ to the particle's instantaneous co-moving frame?

5. Feb 22, 2008

Staff: Mentor

Yes, that is correct.

6. Feb 22, 2008

gordonblur

Great, so here is the bit I don't understand.

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

$$y=\Lambda \, x$$

where $$\Lambda$$ is the Lorentz transform that maps from $$X$$ to $$Y$$ and the proper acceleration is

$$a=(0,a_1,a_2,a_3)=\Lambda_X \ddot{x}=\Lambda_Y \ddot{y}$$

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

However, $$\ddot{y}=\Lambda \, \ddot{x} \Rightarrow a=\Lambda_Y \Lambda \, \ddot{x}$$, but in general $$\Lambda_X \neq \Lambda_Y \Lambda$$.

Where have I gone wrong?

7. Feb 23, 2008

gordonblur

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 $$g$$ has world line:

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

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

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

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

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

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

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

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

So $$a_x\cdot a_x=a_y\cdot a_y=-g^2$$ but $$a_x\neq a_y$$.

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?

Last edited: Feb 24, 2008
8. Feb 23, 2008

Staff: Mentor

Be careful here. If I am reading correctly $$\ddot{y}$$ denotes differentiation wrt coordinate time in Y and $$\ddot{x}$$ denotes differentiation wrt coordinate time in X, so I am not 100% sure that $$\ddot{y}=\Lambda \, \ddot{x}$$ is true. You may be correct, but I have never tried to transform derivatives wrt coordinate time before so I am unsure.

9. Feb 23, 2008

gordonblur

$$\ddot{y}=\frac{d^2 y(\tau)}{d \tau^2}$$

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. Feb 23, 2008

gordonblur

The example I gave demonstrates that the problem lies in this line from post #6.

In general $$\Lambda_X \ddot{x} \neq \Lambda_Y \ddot{y}$$.

11. Feb 23, 2008

Staff: Mentor

That should transform correctly then. Let me look into it in detail.