Understanding Relation of Proper & Vector Quantities

[Arman777]

Let me define the letters before because they will be confusing:

$x$: 3-vector
$v$: 3-velocity
$a$: 3-acceleration

$X$: 4-vector
$U$: 4-velocity
$A$: 4-acceleration

$\alpha$: proper acceleration
$u$: proper velocity

One can define the proper time as, $$d\tau = \sqrt{1 - v^2}dt~~(2)$$ and four velocity,

$$\vec{U} = \frac{d\vec{X}}{d\tau} = (\frac{dt} {d\tau}, \frac{d\vec{x}} {d\tau}) = (\gamma, \gamma\vec{v})$$

and four-acceleration$$A = \frac{d\vec{U}}{d\tau} = \frac{d^2\vec{X}}{d\tau^2} = (\gamma\dot{\gamma}, \vec{a}\gamma^2+ \vec{v}\gamma\dot{\gamma})$$

My confusion starts about the definitions of the proper-velocity and proper acceleration.

It seems that proper-velocity is just the one of the components of the four velocity such that,

$$\vec{u} = \frac{d\vec{x}}{d\tau} = \gamma \vec{v}$$ and the proper acceleration $$\vec{\alpha} = \frac{d\vec{u}}{dt} = \gamma^3\vec{a}$$

Whats the relation between proper velocity/acceleration vs four-velocity/acceleration ? Why they are defined like that and where we use it ?

Also If I try to do manually I get into trouble,

$$\vec{\alpha} = \frac{d\vec{u}}{dt} = \frac{d}{dt}(\gamma\vec{v}) = \dot{\gamma} \vec{v} + \gamma \vec{a}$$

$$\vec{\alpha} = (\vec{v}\cdot\vec{a})\gamma^3 \vec{v} + \gamma \vec{a}$$

which seems incorrect.

 

[Orodruin]

and the proper acceleration

→α=d→udt=γ3→a​

Where did you get this definition from? Proper acceleration is generally defined as a scalar in terms of the magnitude of the 4-acceleration.

 

[robphy]

In the (1+1) case, the proper acceleration can be defined as $\alpha = d\theta/d\tau$, the derivative of the rapidity with respect to proper time. Generally, it is the magnitude of the 4-acceleration (as @Orodruin says). It is an invariant.

The proper velocity is (as you say) the spatial component of the 4-velocity, and is thus not invariant. In terms of rapidity, it is $v\gamma=c\sinh\theta$. Its name probably arises because it is defined as a derivative with respect to proper time... but I think the name elevates it above its relative importance. It is also called the celerity, which might be a better name for it (in my opinion).

 

[Arman777]

Where did you get this definition from? Proper acceleration is generally defined as a scalar in terms of the magnitude of the 4-acceleration.

I saw it online in a site. I am just confused a bit.

So do you mean ?

$$\vec{\alpha} = \sqrt{-(\gamma^2 \dot{\gamma}^2) + [a^2\gamma^4+ 2(\vec{v} \cdot \vec{a})\gamma^3\dot{\gamma} + v^2\gamma^2\dot{\gamma}^2}$$

From here I get,

$$\vec{\alpha} = \sqrt{-\dot{\gamma}^2 + a^2\gamma^4+2(\vec{v} \cdot \vec{a})\gamma^3\dot{\gamma} }$$

$$\vec{\alpha} = \sqrt{\dot{\gamma}^2 + a^2\gamma^4 }$$ ??

 

[pervect]

and four-acceleration$$A = \frac{d\vec{U}}{d\tau} = \frac{d^2\vec{X}}{d\tau^2} = (\gamma\dot{\gamma}, \vec{a}\gamma^2+ \vec{v}\gamma\dot{\gamma})$$

I was following you up to this point, but I am confused by the notation here. When you write $\dot{\gamma}$, what variable are you differentiation $\gamma$ with respect to?

My first thought is that you might be writing $\gamma$ as a function of v, where $v=\sqrt{\vec{v} \cdot \vec{v}}$, but I don't see how this can lead to the expression on the right hand side of the above equation.

If we write $\gamma(v(t(\tau))$, then we can say that

$$\frac{d\gamma(\tau)}{d\tau} = \frac{d\gamma(v)}{dv}\frac{dv(t)}{dt}\frac{dt(\tau)}{d\tau}$$

but I don't see how this relates to what you wrote.

At any rate, I'm not quite following where this expression came from - it doesn't look familiar. And it's crucial to understand this before progressing to the rest of your post/questions.

In an instantaneous rest frame of an object, where $\vec{v}=0$, the components of the 4-acceleration are $(0, \vec{a})$, but the relations gets very messy in any other frame.

My own general approach is to work natively in the 4-vectors, in terms of the invariant dot product of 4-vectors. I would highly recommend this approach, MTW's "Gravitation" has a good treatment of the accelerated observer using this approach, though it does use some tensor notation as well.

Writing the invariant dot product isn't complicated, but there are a few preliminares. First, we need to assume that the value of c=1, something you have already implicitly done. (I assume you're aware of this, if not it could be another source of confusion). Then you need to pick a sign convention. I use the -+++ sign convention, though the +--- sign convention may be easeir.

Using the +--- sign convention, we can write for example

$$\Delta \tau^2 = \Delta X \cdot \Delta X = \Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2$$

 

[PeterDonis]

I saw it online in a site.

What site? Please give a link. You've been here long enough that you should be familiar with the PF rules about references.

 

[Arman777]

What site? Please give a link. You've been here long enough that you should be familiar with the PF rules about references.

https://physics.stackexchange.com/questions/68108/why-proper-acceleration-is-du-dt-and-not-du-d-tau

The first answer. I actually tried to understand it but it get kind of confusing for me. I understood some things but still I cannot derive it myself which that bugs me. I grasp the proper velocity but in the calculation of the proper acceleration are we using another observer ? Since 4-acceleration is invarient it has a transformatio property I guess which is the acceleration transformation but there's not much good source I can look at this map. Its also okay ıf you can give me sources so I can read these things.

 

[Arman777]

I was following you up to this point, but I am confused by the notation here. When you write ˙γγ˙\dot{\gamma}, what variable are you differentiation γγ\gamma with respect to?

by $dt$

it doesn't look familiar.

Its also like this in the https://en.wikipedia.org/wiki/Four-acceleration

In an instantaneous rest frame of an object, where →v=0v→=0\vec{v}=0, the components of the 4-acceleration are (0,→a)(0,a→)(0, \vec{a}), but the relations gets very messy in any other frame.

So what shuld we do ?

 

[Arman777]

This can be completely wrong but let me try something.

Let us set two observers $S$ and $S'$ and another object which has a four-acceleration $\vec{A} = (\gamma \dot{\gamma}, \vec{a}\gamma^2 + \vec{v}\gamma\dot{\gamma})$ w.r.t $S$. However at some point $S'$ has a velocity $v$ so that the velocity of this object w.r.t $S'$ seems zero (instantaneously co-moving inertial reference frame) . This means that,

$\vec{A'} = (0, \vec{a}')$

By using proper acceleration we can write, $$\alpha = \sqrt{A' \cdot A'} = \sqrt{A \cdot A}$$ so

$$a' = \sqrt{\dot{\gamma}^2 + a^2\gamma^4 }$$ ?

but there's also something like, $a' = \frac{a}{\gamma^3}$ https://en.wikipedia.org/wiki/Acceleration_(special_relativity) in the proper acceleration section. Where does it come from ?

 

[robphy]

Maybe it's just a coincidence...
but @Arman777 and https://physics.stackexchange.com/q...-time-proper-velocity-and-proper-acceleration seem to be asking the same kinds of questions...
at different levels.

 

[Arman777]

Maybe it's just a coincidence...
but @Arman777 and https://physics.stackexchange.com/q...-time-proper-velocity-and-proper-acceleration seem to be asking the same kinds of questions...
at different levels.

Yes :p I am not him but I saw the question and got curious. I never saw a clear-cut definiton of these terms. In my Modern Physics Courses they never teach stuff like this. So I am trying to learn things on my own.

 

[PAllen]

I was following you up to this point, but I am confused by the notation here. When you write $\dot{\gamma}$, what variable are you differentiation $\gamma$ with respect to?

My first thought is that you might be writing $\gamma$ as a function of v, where $v=\sqrt{\vec{v} \cdot \vec{v}}$, but I don't see how this can lead to the expression on the right hand side of the above equation.

If we write $\gamma(v(t(\tau))$, then we can say that

$$\frac{d\gamma(\tau)}{d\tau} = \frac{d\gamma(v)}{dv}\frac{dv(t)}{dt}\frac{dt(\tau)}{d\tau}$$

but I don't see how this relates to what you wrote.

At any rate, I'm not quite following where this expression came from - it doesn't look familiar. And it's crucial to understand this before progressing to the rest of your post/questions.

In an instantaneous rest frame of an object, where $\vec{v}=0$, the components of the 4-acceleration are $(0, \vec{a})$, but the relations gets very messy in any other frame.

My own general approach is to work natively in the 4-vectors, in terms of the invariant dot product of 4-vectors. I would highly recommend this approach, MTW's "Gravitation" has a good treatment of the accelerated observer using this approach, though it does use some tensor notation as well.

Writing the invariant dot product isn't complicated, but there are a few preliminares. First, we need to assume that the value of c=1, something you have already implicitly done. (I assume you're aware of this, if not it could be another source of confusion). Then you need to pick a sign convention. I use the -+++ sign convention, though the +--- sign convention may be easeir.

Using the +--- sign convention, we can write for example

$$\Delta \tau^2 = \Delta X \cdot \Delta X = \Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2$$

The dot is derivative by time. It is just taking the standard form of 4 velocity, and differentiating via the chain rule - first by time, then time by tau. The result is exactly as stated by @Arman777 . I’ve seen this before, and routinely rederive it - including a second ago.

 

[PeterDonis]

I grasp the proper velocity but in the calculation of the proper acceleration are we using another observer ?

No. Unfortunately the adjective "proper" is used in multiple senses in this area of physics, which makes the terminology hard to follow.

 

[Arman777]

No. Unfortunately the adjective "proper" is used in multiple senses in this area of physics, which makes the terminology hard to follow.

I see. So my post #9 is True ?

 

[robphy]

Yes :p I am not him but I saw the question and got curious. I never saw a clear-cut definiton of these terms. In my Modern Physics Courses they never teach stuff like this. So I am trying to learn things on my own.

I didn't think you were the same as that phySE poster.
But I wondered if both of you were somehow at the same spot in some textbook or course.

My reply above was a subset of my reply at phySE,
partly to address the term "proper velocity" which I am not a fan of.

 

[robphy]

but there's also something like, a′=aγ3a′=aγ3a' = \frac{a}{\gamma^3} https://en.wikipedia.org/wiki/Acceleration_(special_relativity) in the proper acceleration section. Where does it come from ?

Sorry if my symbols don't match yours.

The proper acceleration is $\displaystyle \rho =\frac{d\theta}{d\tau} = \frac{d\theta}{dt} \frac{dt}{d\tau} = \frac{d\theta}{dt}\cosh\theta$.
The coordinate acceleration is $\displaystyle \alpha =\frac{d\beta}{dt} = \frac{d}{dt}\tanh\theta = \frac{d\theta}{dt}\frac{d}{d\theta}\tanh\theta= \frac{d\theta}{dt} \mbox{sech}^2\theta$.

Eliminating $\displaystyle \frac{d\theta}{dt}$, we have $\rho =\alpha \cosh^3\theta$, which is $\rho =\alpha \gamma^3$

Geometrically, in this context,
"proper acceleration" $\rho$ [how the Minkowski-angle $\theta$ varies with Minkowski-arc-length $\tau$]
is the Minkowski analogue of the "curvature of a plane curve"
(Euclidean: https://mathworld.wolfram.com/Curvature.html )

Although these calculations can be carried out in coordinates* or with 4-vectors,
I think the above captures the geometric essence of the concept.
(These days, I need my rapidity crutch to look over a calculation done in coordinates* or with lots of $\gamma$s
[since I have to see what the various symbols mean].)

 

[PeterDonis]

So my post #9 is True ?

Not entirely.

You are correct that the general formula for the components of the 4-acceleration $A$ in an arbitrary frame are given by $A = \left( \gamma \dot{\gamma}, \vec{a} \gamma^2 + \vec{v} \gamma \dot{\gamma} \right)$, and that in the momentarily comoving inertial frame of the object with 4-acceleration $A$, we have components $A = \left( 0 , \vec{a^\prime} \right)$, where I have put the prime on the acceleration 3-vector to make it clear that the 4-vector $A$ remains the same, but its components are being given with respect to a different frame than before.

You are also correct that the norm $|A|$ of the 4-acceleration vector is an invariant, so we can compute it in any frame and it will have the same value in all frames. It's easily computed in the momentarily comoving frame, in which it is simply $a^\prime$, the magnitude of the 3-vector $\vec{a^\prime}$. So $a^\prime$ is the magnitude of the acceleration the object actually feels; it's the physical quantity.

If we compute the norm $|A|$ in an arbitrary frame, we have to use the more complicated component formulas above, which gives $|A| = \sqrt{\left[ \vec{a} \gamma^2 + \vec{v} \gamma \dot{\gamma} \right]^2 - \left[ \gamma \dot{\gamma} \right]^2}$. Expanding this out and collecting terms gives

$$
a^\prime = |A| = \sqrt{ a^2 \gamma^4 + 2 \vec{a} \cdot \vec{v} \gamma^3 \dot{\gamma} - \dot{\gamma}^2 }
$$

which is not quite the same as the formula you gave.

 

[pervect]

We had a thread a long time ago for the magnitude of the proper acceleration in terms of the components of the 3-acceleration and 3-velocity.

I believe the correct answer turned out to be by Orodruin, in https://www.physicsforums.com/threa...cceleration-in-terms-of-three-vectors.950694/

There are other answers in this thread, including one of mine, that I now believe to be incorrect.

Letting $\alpha$ be the magnitude of the proper acceleration, which as has been noted is an invariant indepedent of frame of reference, and letting $\vec{a}$ be the 3-acceleration, and $\vec{v}$ be the three -velocity, and noting that we are also assuming that c=1, we have:

$$\alpha^2 = \gamma^4 \, \vec{a} \cdot \vec{a} + \gamma^6 \, (\vec{v} \cdot \vec{a})^2$$

Let us look at the two cases. One is when the acceleration is perpendicular to the velocity. Then $\vec{a} \cdot \vec{v} = 0$, and we have:

$$\alpha^2 = \gamma^4 \, \vec{a} \cdot \vec{a} $$

If we let |a|= $\sqrt{\vec{a} \cdot \vec{a}}$ this is just $\alpha = \gamma^2 |a|$

Now lets' look at the case when the aceleration and the velocity are parallel. Letting |a| be defined as before, we can write:

$$\alpha^2 = \gamma^4 |a|^2 + \gamma^6 |a|^2 v^2 = \gamma^4 (1 + \frac{v^2}{1-v^2}) |a|^2 = \gamma^4(\frac{1-v^2+v^2}{1-v^2}) |a|^2 = \gamma^6 |a|^2$$

Thus in this calse $\alpha = \gamma^3 |a|$

I believe these two test cases are correct. The exercise of verifying this and comparing this answer with the various others presented in this thread is left as an exercise for someone else.

I personally find the geometric formulation in terms of 4-vectors MUCH simpler to work with than mucking about with 3-vectors. But for those who don't, the above results might be of some interest.

 

[PeterDonis]

which is not quite the same as the formula you gave

Actually, it does work out the same if you evaluate $\dot{\gamma}$, which gives $\dot{\gamma} = \gamma^3 \vec{v} \cdot \vec{a}$. The end result is the same as the formula @pervect posted in post #18.

 

[PeroK]

$$
a^\prime = |A| = \sqrt{ a^2 \gamma^4 + 2 \vec{a} \cdot \vec{v} \gamma^3 \dot{\gamma} - \dot{\gamma}^2 }
$$

which is not quite the same as the formula you gave.

If you continue, using:
$$\dot \gamma = \gamma^3 (\vec a \cdot \vec v)$$
This simplifies to:
$$a' = \sqrt{\gamma^4 a^2 + \gamma^6(\vec a \cdot \vec v)^2}$$

 

[PeroK]

$$a' = \sqrt{\dot{\gamma}^2 + a^2\gamma^4 }$$ ?

Likewise, if you use
$$\dot \gamma = \gamma^3 (\vec a \cdot \vec v)$$
This simplifies to:
$$a' = \sqrt{\gamma^4 a^2 + \gamma^6(\vec a \cdot \vec v)^2}$$

 

[Arman777]

Proper acceleration defined as the magnitude of the 4-acceleration. 4-acceleration defined as,

$$\vec{A} = \frac{d\vec{U}}{d\tau} =(\gamma\dot{\gamma}, \vec{a}\gamma^2+ \vec{v}\gamma\dot{\gamma})$$

where $U$ is the 4-velocity. So the proper acceleration, $\alpha$, becomes,

$$\alpha = \sqrt{\vec{A} \cdot \vec{A}} = $$

Let us say that the 4-acceleration of $S$ is $A$. For $S'$ it is $A'$. Then the proper acceleration is an invarient quantity

Which implies

$$\alpha = \sqrt{\vec{A} \cdot \vec{A}} = \sqrt{\vec{A'} \cdot \vec{A'}}$$

For $S$

$$\vec{A} = (\gamma_u\dot{\gamma_u}, \vec{a}\gamma_u^2+ \vec{u}\gamma_u\dot{\gamma_u})$$

Thus,

$$\alpha = \sqrt{-\gamma_u^2\dot{\gamma_u}^2 + a^2\gamma_u^4 + u^2\gamma_u^2\dot{\gamma_u}^2+2(\vec{a} \cdot \vec{u}) \gamma_u^3\dot{\gamma_u}}$$

Since $$\dot{\gamma_u} = (\vec{a} \cdot \vec{u}) \gamma_u^3$$ we have

$$\alpha = \sqrt{\dot{\gamma_u}^2 + a^2\gamma_u^4} = \sqrt{(\vec{a} \cdot \vec{u})^2\gamma_u^6 + a^2\gamma_u^4}$$

When we take the case where $$\vec{a} \parallel \vec{u}$$ we have,

$$\alpha = \sqrt{a^2u^2\gamma_u^6 + a^2\gamma_u^4}$$

$$\alpha = a\gamma_u^2\sqrt{u^2\gamma_u^2 + 1}$$

$$\alpha = a\gamma_u^2\sqrt{u^2\frac{1}{1-u^2}+ 1} = a\gamma_u^2\sqrt{\frac{1}{1-u^2}}=a\gamma_u^3$$

For $S'$

$$\alpha = \sqrt{\vec{A'} \cdot \vec{A'}} = $$

$$\vec{A'} = (\gamma_{u'}\dot{\gamma_{u'}}, \vec{a'}\gamma_{u'}^2+ \vec{u'}\gamma_{u'}\dot{\gamma_{u'}})$$

But we said that, $u'=0$ in an instatenous rest frame so $\gamma_{u'} = 1$ and
$\dot{\gamma_{u'}}=0$. Thus,

$$\vec{A'} = (0, \vec{a'})$$

hence,

$$\alpha = \sqrt{a'^2} = a'$$

So finally we have,

$$\alpha = a' = a\gamma_u^3$$

 

[Arman777]

Thanks for your replies

Now lets' look at the case when the aceleration and the velocity are parallel. Letting |a| be defined as before, we can write:

This was helpful. Somehow I never thought them as separate things (perpendicular, parallel etc)

Sorry if my symbols don't match yours.

The proper acceleration is $\displaystyle \rho =\frac{d\theta}{d\tau} = \frac{d\theta}{dt} \frac{dt}{d\tau} = \frac{d\theta}{dt}\cosh\theta$.
The coordinate acceleration is $\displaystyle \alpha =\frac{d\beta}{dt} = \frac{d}{dt}\tanh\theta = \frac{d\theta}{dt}\frac{d}{d\theta}\tanh\theta= \frac{d\theta}{dt} \mbox{sech}^2\theta$.

Eliminating $\displaystyle \frac{d\theta}{dt}$, we have $\rho =\alpha \cosh^3\theta$, which is $\rho =\alpha \gamma^3$

Geometrically, in this context,
"proper acceleration" $\rho$ [how the Minkowski-angle $\theta$ varies with Minkowski-arc-length $\tau$]
is the Minkowski analogue of the "curvature of a plane curve"
(Euclidean: https://mathworld.wolfram.com/Curvature.html )

Although these calculations can be carried out in coordinates* or with 4-vectors,
I think the above captures the geometric essence of the concept.
(These days, I need my rapidity crutch to look over a calculation done in coordinates* or with lots of $\gamma$s
[since I have to see what the various symbols mean].)

I ll look into this in detail, when I see cosh, sinh I get kind of lost. however I noticed something. In my previous post I claimed

$a' = a\gamma^3$ but it seems in your post $p = \alpha \gamma^3$ so $p$ is like the measured acceleration by $S'$ ?

 

[robphy]

$a' = a\gamma^3$ but it seems in your post $p = \alpha \gamma^3$ so p is like the measured acceleration by S' ?

In that reply, I prefaced with "Sorry if my symbols don't match yours. "

So, I will transcribe my equation with the terms I used to define them
$$\rho = \alpha \gamma^3$$
is
$$\left( \begin{array}{c} \mbox{ proper } \\ \mbox{ acceleration } \end{array} \right) = \left( \begin{array}{c} \mbox{ coordinate } \\ \mbox{ acceleration } \end{array} \right) \gamma^3$$
which, in your symbols, is
$$\alpha = a \gamma^3$$

I ll look into this in detail, when I see cosh, sinh I get kind of lost.

These days, when I see a plethora of $\gamma$s, I get lost...
I like to see $\cosh\theta$
because beneath every algebraic calculation in special relativity involving $v$ and $\gamma=\frac{1}{\sqrt{1-v^2}}$
is a geometrical figure on a Spacetime diagram trying to get noticed...

Relativity is often characterized as "the geometry of spacetime"...
so, I try to see the geometry through both 4-vectors and trigonometry
because relative-rapidities (relative angles) are invariant but relative-velocities (relative-slopes) are not.
(Here's my new example: cryptic formula for momentum in a collision is really the altitude of triangle.)