# Relativistic factor between coordinate acceleration and proper acceleration

I did a recent question about relativistic kinematics here: Generalizing a relativistic kinematics formula for spatial-acceleration dependence.

I have a confusion. In the textbooks I've seen, they put the relationship between proper acceleration and coordinate acceleration as

$$\alpha = \gamma^3 \frac{dv}{dt}$$

However, when I try to do the derivation myself, I get a factor of $\gamma^4$ instead. I'm not sure where the error is.

My derivation is like this:

$$\frac{dx}{d \tau} = \frac{dx}{dt} \frac{dt}{d \tau} = \gamma \frac{dx}{dt}$$

where we used $\frac{dt}{d \tau} = \gamma$

Now,

$$\frac{d^2 x}{d \tau^2} = \frac{ d}{d \tau}( \gamma \frac{dx}{dt}) = \frac{d \gamma}{d \tau} \frac{dx}{dt} + \gamma^2 \frac{ d^2 x}{dt^2}$$

$$\frac{d \gamma}{d \tau} = \gamma \frac{d \gamma}{dt} = \gamma (\frac{ \gamma^3 }{c^2} \frac{dx}{dt} \frac{d^2 x}{dt^2})$$

$$\frac{d^2 x}{d \tau^2} = \gamma^4 \frac{v^2}{c^2} \frac{d^2 x}{dt^2} + \gamma^2 \frac{d^2 x}{dt^2} = \gamma^2 \frac{d^2 x}{dt^2} (\gamma^2 \frac{v^2}{c^2} + 1) = \gamma^4 \frac{d^2 x}{dt^2}$$

where on the last step, $v= \frac{dx}{dt}$ .

I don't understand where is the mistake.

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Where's the $x'$ term? –  Larry Harson Jun 2 '13 at 18:11
not sure i follow, you mean the velocity? that is $\frac{dx}{dt}$ –  diffeomorphism Jun 2 '13 at 18:52
You have $t$ and $\tau$ to define time in the lab and proper frames, an $x$ for the space coordinates of one frame, so where's your space variable for the other frame? –  Larry Harson Jun 2 '13 at 18:58
Larry, there is only an inertial frame (x,t), the proper frame is implicit, non-inertial, and assumed to coincide with (x,t) at x=0, t=0 –  diffeomorphism Jun 2 '13 at 22:19

Proper acceleration is not the same as the four-vector acceleration. Proper acceleration is defined as a Lorentz-invariant acceleration, that is, an acceleration that all inertial observers agree upon. Let's define a few concepts:

The proper velocity $\vec{u}$ is the derivative of the position with respect to proper time: $$\vec{u} = \frac{\text{d}\vec{x}}{\text{d}\tau}=\gamma\vec{v}.$$ The proper velocity is the spacial component of the velocity four-vector $\boldsymbol{U}$ $$\boldsymbol{U} = (c\gamma,\vec{u}),$$ and the acceleration four-vector $\boldsymbol{A}$ is $$\boldsymbol{A} = \frac{\text{d}\boldsymbol{U}}{\text{d}\tau} = \left(c\frac{\text{d}\gamma}{\text{d}\tau},\frac{\text{d}\vec{u}}{\text{d}\tau} \right).$$ The relativistic scalar product $\boldsymbol{A}\bullet\boldsymbol{A} = A_0^2 - (\vec{A})^2$ is Lorentz-invariant. We get $$\boldsymbol{A}\bullet\boldsymbol{A} = c^2\left(\frac{\text{d}\gamma}{\text{d}\tau}\right)^2 - \left(\frac{\text{d}\vec{u}}{\text{d}\tau}\right)^2.$$ Now, $$\frac{\text{d}\gamma}{\text{d}\tau} = \gamma^4\frac{\vec{v}\cdot\vec{a}}{c^2},$$ where $\vec{a} = \text{d}\vec{v}/\text{d}t$ is the classical acceleration vector. On the other hand, $$\frac{\text{d}\vec{u}}{\text{d}\tau} = \frac{\text{d}\gamma}{\text{d}\tau}\vec{v} + \gamma\frac{\text{d}\vec{v}}{\text{d}\tau} = \gamma^4\left(\frac{\vec{v}\cdot\vec{a}}{c^2}\right)\vec{v} + \gamma^2\vec{a}.$$ Let's call $v_\parallel$ the component $\vec{v}$ parallel to $\vec{a}$, and $v_\perp$ the perpendicular component. It follows that $\vec{v}\cdot\vec{a}=v_\parallel a$, so that \begin{align} \boldsymbol{A}\bullet\boldsymbol{A} &= \gamma^8\frac{v_\parallel^2 a^2}{c^2} - \gamma^8\left(\frac{v_\parallel^2 a^2}{c^2}\right)\left(\frac{v^2}{c^2}\right) - 2\gamma^6\frac{v_\parallel^2 a^2}{c^2} - \gamma^4a^2\\ &= \gamma^6\frac{v_\parallel^2 a^2}{c^2} - 2\gamma^6\frac{v_\parallel^2 a^2}{c^2} - \gamma^4a^2\\ &= -\gamma^6\frac{v_\parallel^2 a^2}{c^2} - \gamma^6a^2\left(1 - \frac{v_\parallel^2}{c^2}- \frac{v_\perp^2}{c^2}\right)\\ &= -\frac{\gamma^6}{\gamma_\perp^2} a^2, \end{align} where $$\gamma_\perp = \frac{1}{\sqrt{1 - v_\perp^2/c^2}}.$$ So $\boldsymbol{A}\bullet\boldsymbol{A}$ is always negative. If we now define the proper acceleration vector $\vec{\alpha}$ $$\vec{\alpha} = \frac{\gamma^3}{\gamma_\perp} \vec{a} = \frac{\gamma^3}{\gamma_\perp}\frac{\text{d}\vec{v}}{\text{d}t},$$ then $\alpha^2 = -\boldsymbol{A}\bullet\boldsymbol{A}$ is indeed Lorentz-invariant. This is why the proper acceleration is defined like this.

If $v_\perp=0$, then the formula reduces to $$\vec{\alpha} =\gamma^3 \frac{\text{d}\vec{v}}{\text{d}t} = \frac{\text{d}\vec{u}}{\text{d}t}.$$ So in this case, the proper acceleration is the derivative of the proper velocity with respect to coordinate time $t$, not the proper time. The derivative of the proper velocity with respect to proper time $\tau$ is in fact the spatial part of the four-vector acceleration $$\vec{A} = \frac{\text{d}\vec{u}}{\text{d}\tau} = \frac{\text{d}^2\vec{x}}{\text{d}\tau^2}.$$ That's where the extra $\gamma$ comes from.

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This doesn't answer the question. He wants to know where he's made a mistake in his working –  Larry Harson Jun 2 '13 at 20:13
His mistake is that he confuses proper acceleration with four-vector acceleration. I explained that these are two different quantities. –  Pulsar Jun 2 '13 at 20:16
shouldn't the module of the non-inertial forces in the accelerated frame be defined as a Lorentz-invariant scalar? if the frame feels one-G, that one-G should be agreed by all observers, instead you are saying that the proper acceleration is a derivative of coordinate time, not proper time, so a different coordinate system will measure a different G-force on that frame. How is that possible? –  diffeomorphism Jun 3 '13 at 0:05
So does this mean the acceleration felt by an accelerating observer is $du/dt$, and not $du/d\tau$? Isn't the local accelerometer measures $du/d\tau$? –  Calmarius Jun 15 '13 at 8:53
@diffeomorphism I've written an explanation in this post: physics.stackexchange.com/a/68331/24142 –  Pulsar Jun 17 '13 at 10:50

From Wiki:

In the standard inertial coordinates of special relativity, for unidirectional motion, proper acceleration is the rate of change of proper velocity with respect to coordinate time.

You're instead finding the rate of change of proper velocity with respect to proper time and, so, are picking up an extra Lorentz factor. See Four-acceleration.

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This is how I would do the derivation using either a 3-vector or 4-vector approach:

## 3-vector approach

Let frames $S$ and $S'$ move at a constant velocity, and $p$ be an accelerated point. The strategy is to find how the acceleration of $p$ transforms between $S$ and $S'$, and then to make the velocity of $S'$ equal to the instantaneous velocity of $p$ which then gives us the relationship between the proper acceleration and lab acceleration of $p$. Without loss of generality and to simplify things, $p$ accelerates along $x$, and $S'$ moves along the $x-\text{axis}$ at constant velocity $v$ as usual.

In $S'$, the coordinates of $p$ are $(x',y',z',t')$ which can be expressed in terms of the $S$ coordinates via the Lorentz transformations. Using these we can also find $\frac{dt'}{dt}$:

$$x' =\gamma(x-vt),\quad t' = \gamma(t - vx/c^2),\quad \frac{dt'}{dt}= \gamma(1-\frac{vu_x}{c^2})$$

$p$ is accelerated along the x'-axis:

\begin{align*} \frac{d^2x'}{dt'^2} &=\frac{d}{dt'}\frac{dt}{dt'}\frac{d}{dt}\gamma(x - vt) =\frac{dt}{dt'}\frac{d}{dt}\frac{u_x - v}{1-\frac{vu_x}{c^2}} =\frac{1}{\gamma(1-\frac{vu_x)}{c^2}} \cdot \frac{1}{\gamma^2(1-\frac{vu_x}{c^2})^2} \frac{d^2x}{dt^2}\\\ &=\frac{1}{\gamma^3(1-\frac{vu_x}{c^2})^3}\frac{d^2x}{dt^2} \end{align*}

We now make $S'$ the co-moving and therefore proper frame of $p$ with $u_x$ the velocity of $p$ in $S$ which we now set equal to $v$, finally giving $$\frac{d^2x'}{dt'^2} = \gamma^3\frac{d^2x}{dt^2}$$

## 4-vector approach

The four acceleration is given by

$$\left( {\gamma_u}^4\frac{\mathbf{a\cdot u}}{c}, {\gamma_u}^2\mathbf{a}+ {\gamma_u}^4\frac{(\mathbf{a\cdot u)}}{c^2}\mathbf{u} \right)$$

In the proper frame where $\mathbf{u=0}$, this immediately simplifies to $(0,\mathbf{a_p})$. We just Lorentz tranform the time part to establish the relationship between the lab and proper accelerations \begin{align*} \gamma^4\frac{\mathbf{a_lv}}{c^2}&= \gamma\left(0 + \frac{\mathbf{va_p}}{c^2}\right)\\ \mathbf{a_p}&=\gamma^3\mathbf{a_l} \end{align*}

## method in question

It's not clear to me what your strategy is, but it could be made to work based upon the 3-vector approach above:

$$\frac{dx}{d \tau} = \frac{dx}{dt} \frac{dt}{d \tau} = \gamma \frac{dx}{dt}$$

where we used $\frac{dt}{d \tau} = \gamma$

Correct

Now,

$$\frac{d^2 x}{d \tau^2} = \frac{ d}{d \tau}( \gamma \frac{dx}{dt}) = \frac{d \gamma}{d \tau} \frac{dx}{dt} + \gamma^2 \frac{ d^2 x}{dt^2}$$

This won't work because $\gamma$ is a function of the relative velocity $v$ between the two frames where the acceleration is being measured, and we keep this velocity constant:

$$\frac{d^2 x}{d \tau^2} = \frac{ d}{d \tau}( \gamma \frac{dx}{dt}) = \gamma^2 \frac{ d^2 x}{dt^2}$$

The term on the RHS isn't an acceleration for any frame because it's mixing the space coordinate from the proper frame with the time coordinate from the lab frame. You need to transform $x$ to $x'$ using the Lorentz transformation $x=\gamma(x' + vt)$ so that finally, keeping $\gamma$ and $v$ constant:

$$\frac{d^2 x}{d \tau^2} = \gamma^2 \frac {d^2x}{dt^2} = \gamma^2\frac {d^2}{dt^2}\gamma (x' + vt)= \gamma^3\frac{d^2x'}{dt^2}$$

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there are no two frames between which acceleration is being measured and compared, there is a single inertial frame. The gamma is between the inertial frame and the rest frame, which is non-inertial (because it is being accelerated), hence the gamma is not constant –  diffeomorphism Jun 3 '13 at 5:50
There's an instantaneous inertial frame co-moving with the proper frame. –  John McVirgo Jun 3 '13 at 10:57
correct, but the non-inertial forces are not felt in such tangential co-moving frames, only on the actual rest frame. The only use i see for those frames is that, in them, proper vectors and coordinate vectors match. Are you relying on that to find the proper acceleration? –  diffeomorphism Jun 3 '13 at 13:06
@diffeomorphism non-inertial forces are felt in an inertial frame including a comoving frame that's instantaneously at rest with the accelerated particle. I've updated my answer on this. –  John McVirgo Jun 21 '13 at 22:55