What is proper time, proper velocity and proper acceleration?

Question

• I am trying to derive the relativistic rocket equations found here [(4),(5),(6),(7),(8)] but I do not understand proper time, proper velocity and proper acceleration.

Define a point $P$ with spacetime coordinates $(t,x,y,z)$ in reference frame $S$ and $(t',x',y',z')$ in frame $S'$ which is moving at velocity $v$ relative to frame $S$ (parallel to the x-axis). Using the Lorentz transformation, the coordinates are related in the following way: $$t'=\gamma(t-vx/c^2)$$ $$x'=\gamma(x-vt)$$ $$y'=y$$ $$z'=z$$ where $$\gamma = \frac{1}{\sqrt{1-v^2/c^2}}.$$

• From my understanding, if $S$ is the Earth reference frame and $S'$ is the reference frame of a rocket with velocity $v$ moving in the x-axis direction relative to Earth then; a clock on Earth would measure an event E at time $t$ and a clock on the rocket would measure event E at time $t'$. Is this correct? Is $t'$ proper time?

The inverse Lorentz transformation is given by: $$t=\gamma(t'+vx'/c^2)$$ $$x=\gamma(x'+vt')$$ $$y=y'$$ $$z=z'$$ Taking the differentials of $x$ and $t$, $$dx=\gamma(dx'+vdt')=\gamma(v'+v)dt'$$ $$dt=\gamma(dt'+dx'v/c^2)=\gamma(1+v'v/c^2)dt'$$ Dividing $dx$ by $dt$, $$\frac{dx}{dt}=v=\frac{v'+v}{1+v'v/c^2}$$ Differentiating gives, $$\frac{dv}{dt}=a=\frac{dv'}{\gamma^2(1+v'v/c^2)^2dt}$$ Substituting $dt = \gamma(1+v'v/c^2)dt'$, $$a=\frac{a'}{\gamma^3(1+v'v/c^2)^3}$$

• How does constant acceleration work with special relativity and the Lorentz transformation?

• Is $v' = \frac{dx'}{dt'}$ proper velocity?

• From my understanding, $a$ is the acceleration of point $P$ measured from reference frame $S$ but what is $a'$? Is it the acceleration of point $P$ measured from reference frame $S'$? If $S'$ was the reference frame of a rocket with constant acceleration, is $a'$ the acceleration measured inside the rocket?

• Is $a'$ proper acceleration? Wikipedia states that proper acceleration is $a' = \gamma^3a$. How did they get that?

Using $a' = \gamma^3 a = \gamma^3\frac{dv}{dt}$ and integrating with respect to $t$, $$\frac{v}{\sqrt{1-v^2/c^2}} = a't$$ Rearranging for $v$ gives, $$v = \frac{dx}{dt} = \frac{a't}{1+(a't/c)^2}$$ Integrating again with respect to $t$, $$x=\frac{c^2}{a'}(\sqrt{1+(a't/c)^2}-1)$$ Both of these equations can be found here.

• I don't know how to derive these two formulas: $$t=\frac{c}{a'}\sinh{(\frac{a't'}{c})}$$ $$T=\frac{c}{a'}\sinh^{-1}{\frac{a't}{c}}$$

Answer 1

If two events are at $(t,x,y,z)$ and $(d+dt,\; x+dx,\; y+dy,\; z+dz)$ in some given inertial frame then the proper time between them is $d \tau$, given by: $$ c^2 d\tau^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2 $$ which gives $$ d\tau = \sqrt{dt^2 - (dx^2 + dy^2 + dz^2)/c^2} $$ If you are not used to writing things like $dt^2$ and $dx^2$ then don't worry; keep reading and I will explain a little more in a moment. Physically, the quantity $d\tau$ is the amount of time between the events, as registered by a clock which moves at constant velocity from one event to the other.

If such a clock moves for some longer period, then it will access events which are further apart, and then the total proper time is the integral of all the little bits of $d\tau$ along the path. $$ \tau = \int_{\mbox{path}} d\tau $$ The "path" here is a path in spacetime. It is called the worldline.

To do this integral in practice, first we divide the equation for $d\tau$ by $dt$, giving $$ \frac{d\tau}{dt} = \sqrt{1 - \frac{1}{c^2} \left( \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2 \right) } \\ = \sqrt{1 - v^2/c^2} $$ and then $$ \tau = \int d\tau = \int \frac{d\tau}{dt} dt = \int \sqrt{1-v^2/c^2} \; dt $$ If the speed is constant then this integral can be done straight away. It is $$ \tau = \sqrt{1 - v^2/c^2} \; \Delta t $$ where $\Delta t$ is the elapsed time in some inertial frame, and $v$ is the speed of the clock relative to that inertial frame. Notice that if the frame is the one in which the clock is at rest, then we get that the proper time is equal to the time as measured in that frame, which is consistent with what I said previously about the physical interpretation of proper time.

Proper time is an absolutely central concept in special relativity, so it is well worth investing the effort to think it through carefully.

The Lorentz factor $\gamma$ is defined $$ \gamma \equiv \frac{1}{\sqrt{1-v^2/c^2}} $$ and so we have $$ \Delta t = \gamma \tau . $$ Since $\gamma \ge 1$, this result shows that the time between two events, as measured in some reference frame, is in general longer than the proper time between those two events. This is called time dilation. For an example, consider particles such as muons travelling at high speed through Earth's atmosphere. The proper time between the creation and decay of such a muon is about 2 micro-seconds, and the time as observed by time-measuring devices fixed relative to Earth is about 50 micro-seconds.

The term "proper velocity" is not standard terminology so I will not try to define it. The velocity (relative to some inertial frame) is a 3-vector given by $d{\bf x}/dt$ and the 4-velocity is a 4-vector given by $$ v^\mu \equiv \frac{d x^\mu}{d\tau}. $$

The term "proper acceleration" is usually understood to mean the ordinary 3-acceleration as observed in the instantaneous rest frame of the entity in question. That is, $$ \mbox{proper acceleration } \; {\bf a}_0 = \frac{d {\bf v}}{dt} \;\;\;\mbox{in frame where} \;\; {\bf v} = 0. $$

The 4-acceleration is defined as $$ a^\mu \equiv \frac{d v^\mu}{d\tau}. $$ That is, $$ \left( \begin{array}{c} a^0 \\ a^1 \\ a^2 \\ a^3 \end{array} \right) = \frac{d}{d\tau} \left( \begin{array}{c} \gamma c \\ \gamma {\bf v} \end{array} \right) = \gamma \frac{d}{dt} \left( \begin{array}{c} \gamma c \\ \gamma {\bf v} \end{array} \right) = \left( \begin{array}{c} \gamma \frac{d\gamma}{dt} c \\ \gamma \frac{d\gamma}{dt} {\bf v} + \gamma^2 \frac{d\bf v}{dt}\end{array} \right). $$ In the instantaneous rest frame one has $\gamma = 1$ and $\dot{\gamma} = 0$ so in this frame one finds $$ a^\mu = \left( \begin{array}{c} 0 \\ \frac{d\bf v}{dt} \end{array} \right) = \left( \begin{array}{c} 0 \\ {\bf a}_0 \end{array} \right) $$ Hence the invariant magnitude of $a^\mu$ is equal to the magnitude of the proper acceleration.

Answer 2

Note: I ll use another name for the velocities because I could not understand yours.

How does constant acceleration work with special relativity and the Lorentz transformation?

Let me take two Frames, $S(t,x)$ and $S'(t',x')$, where $S'$moves with a velocity $v$ and the rocket moves with a velocity $u$ w.r.t $S$. Where the rocket moves with a velocity $u'$ w.r.t $S'$.

To obtain an expression for the acceleration transformation we need to consider an inertial reference frame $S'$ such that it always moves with the rocket. This implies that $u'=0$ and $u=v$ at all times.

In the situation we say that $S'$ is the instantaneous rest frame of the accelerating observer.

In this sense the acceleration between two reference frames becomes,

$$a' = \frac{(1-v^2/c^2)^{3/2}}{(1-uv/c^2)^3}a$$

for $u'=0$ and $u=v$,

$$a' = \frac{(1-u^2/c^2)^{3/2}}{(1-u^2/c^2)^3}a = (1-u^2/c^2)^{-3/2}a$$

or $$a' = a\gamma^3$$

Is v′=dx′/dt′ proper velocity?

In general the proper velocity defined as,

$$w = \frac{dx}{d\tau}$$ where $x$ is just the position of the particle. So we have,

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

In this sense I do not think $v'$ (in my notation $u'$) is the proper velocity.

From my understanding, a is the acceleration of point P measured from reference frame S but what is a′? Is it the acceleration of point P measured from reference frame S′? If S′ was the reference frame of a rocket with constant acceleration, is a′ the acceleration measured inside the rocket?

Yes, $a'$ is the measured acceleration by the $S'$ which is the instantenous rest frame of the accelerating rocket.

Is $a′$ proper acceleration? Wikipedia states that proper acceleration is $a′=γ^3a$. How did they get that?

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,

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

I don't know how to derive these two formulas:

So we have, $$a = a'( 1 - u^2/c^2)^{3/2}$$ and $\frac{dt}{d\tau} = \gamma$

Now, $$a = \frac{du / d\tau}{dt / d\tau} $$

From here you can write,

$$\frac{du} {d\tau} = a \frac{dt} {d\tau} = a'(1-u^2/c^2)$$

wehn you integrate to find $u$ you'll get,

$$u = ctanh(a'\tau /c)$$ Hence we can write, $$\gamma = cosh(a'\tau /c)$$

Since $$dt = \gamma d\tau $$ we have,

$$t = \int cosh(a'\tau / c)d\tau$$

$$t = \frac{c}{a'}sinh(\frac{a'\tau}{c})$$

Answer 3

I believe your derivation of $a$ is correct, at least according to this wikipedia page. However, $a$ is not the proper acceleration, since you took the derivative $dx/dt$ and not $dx/d\tau$. Here, $d\tau = dt^\prime$. A derivation of proper acceleration is further on that wikipedia page.

Conceptually, the proper time of a moving particle is defined as the time recorded by a clock that follows the trajectory of that particle in spacetime. It is the time recorded by a clock in the particle's own reference frame, even if that reference frame is non-inertial. In that case, the proper time is recorded by a family of inertial frames $S_t$, where at any $t$, $S_t$ is instantaneously co-moving with the particle. If we use the 4-position, the 4-velocity and 4-acceleration are derivatives $d/d\tau$.

Answer 4

“ What is proper time, proper velocity and proper acceleration?”

Proper time is the time elapsed along a given observer worldline, measured by a wristwatch carried by the observer. It is akin to what an odometer measures along given a path. It is an invariant quantity of the given path between two events. (This agrees with @Danny Kong’s description.)

Proper velocity (called the celerity) is the spatial component of an observer’s 4-velocity. In terms of the velocity, it is $v/\sqrt{1-(v/c)^2}=v\gamma$. In terms of rapidity, it is $v\cosh\theta=c\tanh\theta\cosh\theta=c\sinh\theta$. It is not an invariant—-it depends if the frame of reference making the measurement. Its name derives from being a derivative with respect to proper time... although it is not an invariant.)

Proper acceleration $\rho$ is the magnitude of the 4-acceleration, which measures the “curvature of a curve”, its deviation from being a geodesic (from being inertial). It is the derivative of the rapidity with respect to proper time $\rho=d\theta/ds$. In terms of coordinate acceleration, it is $\rho=a\cosh^3\theta=a\gamma^3$. It is an invariant.

(In terms of your calculation, I’ll have to look at more closely. What might an issue in your calculation is distinguishing the "magnitude of a 4-vector" and the "x-component of a 4-vector".)