Boosting to an instantaneous rest frame

[Geofleur]

The four-velocity and four-acceleration of a particle may be written as

$(V^\mu) = \gamma(\textbf{v},c)$

and

$(A^\mu) = \gamma \left(\frac{d\gamma}{dt}\textbf{v}+\gamma\frac{d\textbf{v}}{dt},c\frac{d\gamma}{dt} \right)$.

where $\gamma$ is the Lorentz factor, $\textbf{v}$ is the ordinary three-velocity (which I will assume is entirely along the x-direction), and $c$ is the speed of light. In a frame of reference in which the particle is instantaneously at rest, $\gamma = 1$, $\frac{d\gamma}{dt} = 0$, and $v = 0$, giving

$(V^\mu) = (0,c)$

and

$(A^\mu) = (\frac{d\textbf{v}}{dt},0)$.

It ought to be possible to derive these same expressions by starting with a particle moving with velocity $\textbf{v}$ and then Lorentz boosting into its instantaneous rest frame. However, when I try to do this I get the wrong result for the four-acceleration. The matrix for the Lorentz transformation giving primed quantities in terms of unprimed quantities is

$[L^{\mu}_{\nu}] = \begin{bmatrix} \gamma & 0 & 0 & -\gamma v / c \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\gamma v / c & 0 & 0 & \gamma \\ \end{bmatrix}$

The four-acceleration should transform as $A'^{\mu}=L^{\mu}_{\nu}A^\nu$. But then applying this formula to $A'^1$ I get

$A'^1 = L^1_1 A^1 + L^1_2 A^2 + L^1_3 A^3 + L^1_4 A^4 = \gamma^2 \frac{d\gamma}{dt}v + \gamma^3 \frac{dv}{dt} - \gamma^2 \frac{d\gamma}{dt}v = \gamma^3 \frac{dv}{dt}$, which is different from the correct result by a factor of $\gamma^3$! Where have I gone wrong?

 

[Cryo]

The problem with boosting into the instanteneous reference frame is that the coordinate basis of that frame depends on proper time, so you have to take that into account. See the last section in https://en.wikipedia.org/wiki/Four-acceleration

Consider this. Let $S$ be the lab-fram end $\bar{S}$ the instantaneous reference frame.

The four-acceleration in $S$ is:

$a^\mu=\frac{du^\mu}{d\tau}$

Where $u^\mu$ is the four-velocity, but:

$u^\mu=\frac{\partial x^\mu}{\partial \bar{x}^\nu}\bar{u}^\nu$

So:

$a^\mu=\frac{d}{d\tau}\left(\frac{\partial x^\mu}{\partial \bar{x}^\nu}\bar{u}^\nu\right)=\frac{d}{d\tau}\left(\frac{\partial x^\mu}{\partial \bar{x}^\nu}\right)\bar{u}^\nu+\left(\frac{\partial x^\mu}{\partial \bar{x}^\nu}\right)\frac{d\bar{u}^\nu}{d\tau}$

But you made $\gamma$ on proper time, so $\frac{d}{d\tau}\left(\frac{\partial x^\mu}{\partial \bar{x}^\nu}\right)\neq 0$. To handle this properly you either need to extend the proper-time derivative on four-acceleration to be covariant (see link), or you need to make instantaneous reference frame better defined. Call it instantaneous reference at time $\tau=\tau_0$ and use $\gamma_0=\gamma_{\tau=\tau_0}$ (i.e. constant) to boost into it. In this case, of course, you will only get the 'right' answer at time $\tau=\tau_0$

 

[PAllen]

In a sense your only mistake is misinterpreting the result of your transform. What you have derived is that

ϒ³dv/dt = dv'/dt’

which is correct (for your simplification assumptions). @Cryo explains why you can’t transform the formulas in the way you expect. But at one moment, your transform is fine, and you have derived an important known result, if you interpret it correctly.

 

[Geofleur]

Thanks for your responses! I'm going to need to chew on this for a day or so; once it seems like I understand what's going on, I'll post another attempt at deriving $A^\mu = (d\textbf{v}/dt,0)$ as well as an attempt at interpreting my previous result correctly.

 

[PeroK]

Thanks for your responses! I'm going to need to chew on this for a day or so; once it seems like I understand what's going on, I'll post another attempt at deriving $A^\mu = (d\textbf{v}/dt,0)$ as well as an attempt at interpreting my previous result correctly.

Conceptually the instantaneous rest frame is an IRF that coincides with the accelerating particle's rest frame only for an instant. It's fine to say that, for example, $\gamma = 1$ in this frame. But you can't say that $\frac{d\gamma}{dt} =0$ in this frame. This is because the particle is accelerating in any IRF.

This relates also to the difference between the coordinate time $t$ in the IRF and the proper time of the particle.

 

[Cryo]

Somehow none of the helpers, including me, mentioned Rindler coordinates.

As already stated by me and @PeroK, the normal notion of intertial reference frame can only be applied to instantaneous reference frame if you define it for one proper time only. Alternatively you can introduce coordinates for the whole space-time where the world-line of your accelerated object corresponds to curve of a single changing coordinate. For uniformly accelerated observer such coordinates are Rindler Coordinates https://en.wikipedia.org/wiki/Rindler_coordinates.

In such coordinate the notion of instanteneous reference frame will become much easier, but the price to pay will be that your covariant derivative will no longer be just the partial derivative, you will have to include a non-trivial connection.

 

[PAllen]

Conceptually the instantaneous rest frame is an IRF that coincides with the accelerating particle's rest frame only for an instant. It's fine to say that, for example, $\gamma = 1$ in this frame. But you can't say that $\frac{d\gamma}{dt} =0$ in this frame. This is because the particle is accelerating in any IRF.

This relates also to the difference between the coordinate time $t$ in the IRF and the proper time of the particle.

Actually, you can say dϒ/dt is zero at this moment because this derivative is ϒ³ v⋅(dv/dt), which is clearly zero when v=0.

 

[Cryo]

Actually, you can say dϒ/dt is zero at this moment because this derivative is ϒ³ v⋅(dv/dt), which is clearly zero when v=0.

Will, go for second derivative ... or third ... best out of five :-)

 

[PAllen]

Just want to reiterate, there is nothing wrong with OP reasoning and math except the end of the last line, along with an erroneous expectation - that an LT will change variables (including what you are differentiating by). My post #3 explains what is really derived by the given reasoning.

To exemplify this for a simpler case, suppose a world line is given in some frame by (t,f(t)). Boosting to another frame will give you the coordinates for this world line in the new frame - but in terms of expressions of the original:

(t',x') = ϒ(t - v f(t), f(t) - vt) [I use c=1]

You don't get x' as a function of t' just by doing an LT.

I think the only issue in OP is a subtler form of this error of thought.

 

[PeroK]

Actually, you can say dϒ/dt is zero at this moment because this derivative is ϒ³ v⋅(dv/dt), which is clearly zero when v=0.

Yes, I overlooked that $\gamma$ must be at a minimum when $v =0$.

 

[vanhees71]

The mistake is your definition of acceleration. As you have defined it, it's no four-vector. To get a four-vector you have to take derivatives with respect to proper time, not coordinate time. That's why four-velocity is
$$c u^{\mu}=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau},$$
where
$$\mathrm{d} s^2=c^2 \mathrm{d} \tau^2=\eta_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}.$$
By definition $u^{\mu}$ is the time-like unit vector of the instantaneous reference frame of the particle along its world line, which is most conveniently parametriced (as long as you deal with only one particle!) by proper time.

Now you have $u_{\mu} u^{\mu}=1=\text{const}$ and thus
$$a^{\mu}=c \frac{\mathrm{d} u^{\mu}}{\mathrm{d} \tau},$$
the proper acceleration is a four vector with
$$u_{\mu} a^{\mu}=0,$$
because, with the dot indicating derivatives wrt. proper time,
$$u_{\mu} u^{\mu}=1 \; \Rightarrow \dot{u}_{\mu} u^{\mu}=a_{\mu} u^{\mu}/c=0.$$
Thus $a^{\mu}$ is a space-like vector. In the instantaneous rest frame (I denote vector components with respect to the instantaneous rest frame with a star on the symbol).
$$a^{\star \mu}= \begin{pmatrix}0,\vec{a}^{*} \end{pmatrix}.$$
That's because in the instantaneous rest frame $u^{*\mu}=(1,0,0,0)^{\text{T}}$.

 

[Geofleur]

@vanhees71 I'm not sure I understand - it seems to me that I did calculate the four-acceleration using the proper time, just like so:

$A^\mu = \frac{dV^{\mu}}{d\tau}=\gamma\frac{dV^{\mu}}{dt}$,

so that

$A^\mu = \gamma \left( \frac{d\gamma}{dt}\textbf{v} + \gamma\frac{d\textbf{v}}{dt},c\frac{d\gamma}{dt}\right).$

 

[Geofleur]

I think maybe I finally understand the meaning of the result I got. I started with the four-vectors

$V^\mu = \gamma(v,c)$,

and

$A^\mu = \left(\gamma \frac{d\gamma}{dt}v + \gamma^2 \frac{dv}{dt}, c\gamma \frac{d\gamma}{dt} \right),$

where I assume that both the three-acceleration and the three-velocity are entirely in the x-direction. After boosting to a frame moving along the x-direction with speed $v$ I get

$V'^\mu= (0,c)$,

and

$A'^\mu = \left(\gamma^3\frac{dv}{dt},0 \right)$. I can check these results by noting that another way of writing the un-primed four-acceleration is

$A^\mu = \gamma^4 \frac{dv}{dt}(1,\frac{v}{c})$,

because $\frac{d\gamma}{dt}=\frac{\gamma^3 v}{c^2}\frac{dv}{dt}$. Then I get

$A^\mu A_\mu = A'^\mu A'_\mu = \gamma^6 \left( \frac{dv}{dt} \right)^2,$

as expected.

The transformed four-acceleration has zero ct-component and the transformed four-velocity has zero x-component. Because the basis vectors $e_1$ and $e_4$ have components in both the $e'_1$ and $e'_4$ directions, neither the four-acceleration nor the four-velocity can have the same zero components in both frames (unless they are identically the zero vector). The $\gamma^3$ shows up in the primed four-acceleration to keep the magnitude of the four-acceleration the same to compensate for the ct-component being zeroed out.

What I've just described is taking a given four-vector and viewing it from two different frames. Back at the beginning of this thread, I started with a particle in its instantaneous rest frame, where $V'^\mu = (0,c)$, and $A'^\mu = (dv/dt,0)$. In contrast to the above calculation, the first term in the four-acceleration has no factor $\gamma^3,$ and that's because I would have to boost in a very particular way to get this result, not just from a frame going at the same speed and in the same direction as the particle as viewed from that frame. Using the inverse Lorentz transform, I would need to boost from a frame in which

$A^\mu = \gamma \frac{dv}{dt}\left(1,\frac{v}{c}\right)$.

I'm not sure, but it seems there is no such frame, because this four-acceleration seems to imply that $v=0$, that the only transformation that works is the identity transformation! I get this result by setting the generic expression for the four-acceleration of a particle in any frame (see the second equation above) equal to the expression above. It leads to $\gamma = 1$.

 

[vanhees71]

I think, it's very confusing to calculate things with non-covariant objects like $\gamma$ and $\vec{\beta}$. It's better to stay as long as possible with covariant objects like four-vector components and only at the very final step express them in terms of the non-covariant quantities, which usually is not needed though, except to illustrate something as seen from a specific observer in terms of these non-covariant objects.

 

[Geofleur]

I'll try to keep that in mind in future!

 

[PAllen]

I still think there is a misunderstanding of @Geofleur that has been unrecognized by the OP, and that most posts here are tangential to the core misunderstanding. That is that:

1) For a scalar function of some coordinates, to express it in terms of new coordinates, you just need to apply the coordinate transform itself. Thus, for inertial frames in SR, given f(t,x,y,z) substitute e.g. :
t = ϒ(t'+vx'), x= ϒ(x' + vt'), y=y', z=z’ and get some g(t',x',y',t'), after simplifying.

2) To express a vector function of some coorinates as a vector function of some other coordinates, you need to do as in (1) plus apply the vector transform matrix. The OP is ignoring that the coordinate transform is necessary.

I will make this explicit in reference to the OP. Hints have not seemed to work.

The four-velocity and four-acceleration of a particle may be written as

$(V^\mu) = \gamma(\textbf{v},c)$

and

$(A^\mu) = \gamma \left(\frac{d\gamma}{dt}\textbf{v}+\gamma\frac{d\textbf{v}}{dt},c\frac{d\gamma}{dt} \right)$.

where $\gamma$ is the Lorentz factor, $\textbf{v}$ is the ordinary three-velocity (which I will assume is entirely along the x-direction), and $c$ is the speed of light. In a frame of reference in which the particle is instantaneously at rest, $\gamma = 1$, $\frac{d\gamma}{dt} = 0$, and $v = 0$, giving

$(V^\mu) = (0,c)$

and

$(A^\mu) = (\frac{d\textbf{v'}}{dt'},0)$.

This is all correct, and similar statements can be found in many SR books. However, to clarify that the specialization of the general 4-vector formula to a frame in which the instantaneous velocity is zero, is a frame you are going to boost into, it helps to distinguish the coordinates, as I have inserted in the quoted text above (i.e. use primed coordinates for the instantaneous frame).

It ought to be possible to derive these same expressions by starting with a particle moving with velocity $\textbf{v}$ and then Lorentz boosting into its instantaneous rest frame.

It is possible, but you have to apply the vector transform plus the coordinate transform.

However, when I try to do this I get the wrong result for the four-acceleration.

This is because you never apply the coordinate transform.

The matrix for the Lorentz transformation giving primed quantities in terms of unprimed quantities is

$[L^{\mu}_{\nu}] = \begin{bmatrix} \gamma & 0 & 0 & -\gamma v / c \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\gamma v / c & 0 & 0 & \gamma \\ \end{bmatrix}$

The four-acceleration should transform as $A'^{\mu}=L^{\mu}_{\nu}A^\nu$. But then applying this formula to $A'^1$ I get

$A'^1 = L^1_1 A^1 + L^1_2 A^2 + L^1_3 A^3 + L^1_4 A^4 = \gamma^2 \frac{d\gamma}{dt}v + \gamma^3 \frac{dv}{dt} - \gamma^2 \frac{d\gamma}{dt}v = \gamma^3 \frac{dv}{dt}$

This is all correct so far. Now you have to do the coordinate transform as well to compare to the expected result. We write:

dv/dt = (d/dt) (dx/dt)

and use chain rule plus inverse derivative rule to note, for any function f:

df/dt = (df/dt') / (dt/dt')

Applying this to dx/dt with using the inverse Lorentz transform, and using u for the boost velocity (keeping v to mean dx/dt and v' to mean dx'/dt') you get just the expected velocity addition rule:

dx/dt = (v' + u)/(1 + uv')

Applying again gives a somewhat messy result that simplifies (only after differentiating so as not to lose that there is acceleration even though though we are stipulating than v'=0) given v'=0 and u=v to:

dv/dt = (1/ϒ³)dv'/dt'

Substituting this in the result of the vector transform gives simply dv'/dt', as expected.

Note that the time component of proper acceleration avoids these complications only because the result of the vector transform alone is identically zero.

I hope this helps clear up the core misunderstanding of the OP.

 

[Geofleur]

Yes, thank you for that very detailed explanation. It's almost painfully obvious now, but I just couldn't see it before for some reason.