# Magnitude of proper acceleration in terms of three-vectors

• A
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## Main Question or Discussion Point

I've been wondering if there is an expression for the magnitude of the proper acceleration of an observer in terms of three-velocities and three-accelerations.

I didn't find any, so I took a stab at calculating it. However, it was not at all convenient to do the calculations without 4-vectors, so I used them in my caculations. Even so perhaps the results could be mildly useful to someone who asks a question about acceleration in special realtivity (of which we get large numbers ) without a background in 4-vectors (of which there are also large numbers. The point is that one can see that given the 3-acceleration and 3-velocity of an observer, one can compute the invariant magnitude of the 4-acceleration which is just the magnitude of the acceleration that is experienced by an instantaneously co-moving observer.

Onto the calculations.

We start with defining the components of the 3-velocity as vx, vy, vz, and also the 3-acceleration ax=dx/dt, ay=dy/dt, az=dz/dt. I used geometric units, with c=1, though I'll attempt to put the necessary factors of c back in in the final result.

We can write the components of the 4-velocity as:

$$u^t = \frac{1}{\sqrt{1-vx^2-vy^2-vz^2}} \quad u^x=\frac{vx}{\sqrt{1-vx^2-vy^2-vz^2}}$$$$u^y=\frac{vy}{\sqrt{1-vx^2-vy^2-vz^2}} \quad u^z=\frac{vz}{\sqrt{1-vx^2-vy^2-vz^2}}$$

Then we can compute the 4-accerlation $\alpha$ as:

$$\alpha^t =\frac{d}{d\tau} u^t = \frac{du^t}{dt}\frac{dt}{d\tau} = u^t \,at \quad \alpha^x = \frac{d}{d\tau} u^x = \frac{du^x}{dt} \frac{dt}{d\tau} = u^t\,ax$$
$$\alpha^y = u^t \, ay \quad \alpha^z = u^t \, az$$

Then the square magnitude of the proper acceleration, $A^2$ is giving by

$$A^2 = (\alpha^x)^2+(\alpha^y)^2+(\alpha^z)^2-(\alpha^t)^2$$

Much computer algebra later we get the following expression for $A^2$:

$$\frac{(1-\beta_y^2-\beta_z^2)\,ax^2 + (1-\beta_x^2-\beta_z^2)\,ay^2+(1-\beta_x^2-\beta_y^2)\,az^2 + 2\beta_x\beta_y\,ax\,ay +2\beta_x\beta_z\,ax\,az+2\beta_y\beta_z\,ay\,az}{ (1-\beta_x^2-\beta_y^2-\beta_z^2)^3}$$

where we have introduced $\beta_x =\frac{vx}{c} \quad \beta_y = \frac{vy}{c} \quad \beta_z = \frac{vz}{c}$.

Hopefully this will serve as a motivation for why it's much easier to use 4-vectors to calculate proper accelerations, as the 4-vector expession $(\alpha^x)^2+(\alpha^y)^2+(\alpha^z)^2-(\alpha^t)^2$ is much simpler than the 3-vector expression.

As far as whether or not I made any errors - it's hard to say, but it seems to pass a few quick tests, namely the 1d case where only $\beta_x$ is nonzero, and the 2d case of the sliding block.

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Orodruin
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I am pretty sure the computation you are looking for is available in Rindler's "Introduction to Special Relativity". I also have a vague memory of deriving it in my lecture notes. The correct expression is
$$\alpha^2 = \gamma^4 \vec a^2 + \gamma^6 (\vec u \cdot \vec a)^2,$$
where $\alpha$ is the proper acceleration, $\vec u$ the 3-velocity, and $\vec a$ the 3-acceleration.

Dale
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Much computer algebra later
This would be my approach too!

Orodruin
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This would be my approach too!
Well, it becomes significantly simpler if you actually use the 3-vector formalism instead of writing everything out in components.

Dale
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Well, it becomes significantly simpler if you actually use the 3-vector formalism instead of writing everything out in components.
I am not sure what you mean by that.

Orodruin
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I am not sure what you mean by that.
Instead of writing out all components explicitly, use $\vec u$ and $\vec a$.

Here is a rapidity-based approach.

$\vec{\beta} = \vec{v}/c$
$\gamma = (1 - \beta^2)^{-1/2}$
$\vec{\omega} = \gamma \vec{\beta}$
$\phi = \tanh^{-1}{\beta}$
Little circle means a proper-time-times-$c$ derivative: $\mathring{\phi} = (d \phi / d \tau)/c$

Four-acceleration (in my preferred units) is:

$\vec A = \mathring {\vec V} = (\mathring{\gamma}, \mathring{\vec \omega})$.

Its squared magnitude under (-+++) signature is:

$\vec A^2 = - \mathring{\gamma}^2 + \mathring{\vec \omega}^2 = - \left( \dfrac{d}{c \, d \tau} \, \cosh{\phi} \right)^2 + \left( \dfrac{d}{c \, d\tau} \left( \hat{\omega} \sinh{\phi} \right) \right)^2$.

Then:

$\vec A^2 = - (\sinh{\phi} \, \mathring{\phi})^2 + (\hat{\omega} \cosh{\phi} \, \mathring{\phi} + \mathring{\hat{\omega}} \sinh{\phi})^2$,

and since $\hat{\omega} \cdot \mathring{\hat{\omega}} = 0$, $\hat{\omega}^2 = 1$, and $\cosh^{2}{\phi} - \sinh^{2}{\phi} = 1$:

$\vec A^2 = \mathring{\phi}^2 + \big( \mathring{\hat{\omega}} \sinh{\phi} \big)^2$.

If I didn't make any mistakes, that should be the squared proper acceleration. It would take some algebra to verify it. On first glance it checks out though—the second term vanishes in the instantaneous rest frame, as it should.

If I didn't make any mistakes, that should be the squared proper acceleration. It would take some algebra to verify it.
Verified it by rewriting @Orodruin's expression like this:

$\gamma^4 \left( \dfrac{d}{c \, dt} (\beta \hat{\beta} ) \right)^2 + \gamma^6 \beta^2 \left( \dfrac{d \beta}{c \, dt} \right)^2$,

and subbing in hyperbolic functions.

I've been wondering if there is an expression for the magnitude of the proper acceleration of an observer in terms of three-velocities and three-accelerations.
That article has several expressions relating proper acceleration and 3-acceleration.
https://en.wikipedia.org/wiki/Acceleration_(special_relativity)#Proper_acceleration

At the end of that section, a formula equivalent to that of Orodruin appears, which is derived by noticing that the magnitude of proper acceleration corresponds to the magnitude of 4-acceleration.

Staff Emeritus
I am pretty sure the computation you are looking for is available in Rindler's "Introduction to Special Relativity". I also have a vague memory of deriving it in my lecture notes. The correct expression is
$$\alpha^2 = \gamma^4 \vec a^2 + \frac{\gamma^6}{c^2} (\vec u \cdot \vec a)^2,$$
where $\alpha$ is the proper acceleration, $\vec u$ the 3-velocity, and $\vec a$ the 3-acceleration.
I'm getting the above result now, which I format as $\gamma^4 \,( \vec{a} \cdot \vec{a}) + \frac{\gamma^6}{c^2} \,( \vec{v} \cdot \vec{a})^2$, denoting the 3-velocity with $\vec{v}$ and the 3-acceleration with $\vec{a}$ - a slightly different choice of symbols. I don't think it's the same as what I was getting earlier :(. As of yet, I don't quite see if it's the same expression wiki gets. I don't have Rindler's textbook (I have seen it in the past). If I get to the library again I'll look it up though. I'd like to thank everyone for a good & helpful thread.

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I'm getting the above result now, which I format as $\gamma^4 \,( \vec{a} \cdot \vec{a}) + \gamma^6 \,( \vec{v} \cdot \vec{a})$, denoting the 3-velocity with $\vec{v}$ and the 3-acceleration with $\vec{a}$ - a slightly different choice of symbols. I don't think it's the same as what I was getting earlier :(. As of yet, I don't quite see if it's the same expression wiki gets. I don't have Rindler's textbook (I have seen it in the past). If I get to the library again I'll look it up though. I'd like to thank everyone for a good & helpful thread.
Try this:

$\gamma^4 a^2 + \gamma^6 (\vec{v} \cdot \vec{a})^2 = \gamma^4 a^2 + \gamma^6 (va \cos{\theta})^2 = \gamma^6 a^2 \left( \gamma^{-2} + (v \cos{\theta})^2 \right)$,

where $\theta$ is the angle between $\vec{v}$ and $\vec a$. In the parentheses, you have $1 - v^2 + (v \cos{\theta})^2 = 1 - (v \sin{\theta})^2$, but $v \sin{\theta}$ is just the component of $\vec{v}$ that's perpendicular to $\vec{a}$, so:

$\gamma^6 a^2 \left( 1 - v_{\perp \vec a}^2 \right)$.

Now define $\gamma_{\perp} \equiv (1 - v_{\perp \vec a}^2)^{-1/2}$, and your (unsquared) proper acceleration is:

$\dfrac{\gamma^3 \vec a}{\gamma_{\perp}}$.

That matches what you see in Wikipedia, since the denominator is $\gamma$ for $\vec v \perp \vec a$ and unity for $\vec v \parallel \vec a$.