Express this trajectory in terms of proper time

In summary, the object moves solely on the $x$-axis and its speed is not constant. The Lorentz factor $\gamma = \sqrt{1+ \kappa ^2 t^2}$ can be used to calculate the velocity $v_x = \frac{dx}{dt} = \frac{c \kappa t}{\sqrt{1+\kappa ^{2} t^2}}$. To express $x$ and $t$ as a function of proper time $\tau$, the relationship $\frac{dt}{d\tau} = \gamma$ needs to be integrated. This yields a transcendental function, which can be written as $\sinh^{-1}(\kappa t)$ or
  • #1
JTFreitas
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Homework Statement
Consider an object whose motion is described by ##x(t) = \frac{c}{\kappa}(\sqrt{1+\kappa^2 t^2} -1)##, where ##c## is the speed of light and ##\kappa## is some constant. Express ##x## and ##t## as a function of proper time.
Relevant Equations
$$\frac{dt}{d \tau} = \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$
The object moves solely on the $x$-axis, hence I calculated its speed to be $v_x = \frac{dx}{dt} = \frac{c \kappa t}{\sqrt(1+\kappa ^{2} t^2$ Because its speed is not constant, I suppose the Lorentz factor $\gamma = \gamma (t)$, and by plugging in the velocity, I obtain $\gamma = \sqrt(1+ \kappa ^2 t^2)$

I was told to integrate the relationship between $d\tau$ and $dt$ which is just $d\tau = \frac{dt}{\gamma}$. However, integrating this to obtain $\tau$ as a function of $t$ yields a transcendental function that can't exactly be inverted (to express $t$ in terms of $\tau$), and I am stuck in what to try here, in order to obtain the trajectory $x(\tau)$.

Edit: probably obvious but $\tau$ is the proper time.
 
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You need two dollar signs to delimit Latex.
 
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Homework Statement:: Consider an object whose motion is described by $$x(t) = \frac{c}{\kappa}(\sqrt{1+\kappa^2 t^2} -1)$$, where ##c## is the speed of light and ##\kappa## is some constant. Express ##x## and ##t## as a function of proper time.
Relevant Equations:: $$\frac{dt}{d \tau} = \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$

The object moves solely on the $x$-axis, hence I calculated its speed to be $$v_x = \frac{dx}{dt} = \frac{c \kappa t}{\sqrt{1+\kappa ^2 t^2}}$$ Because its speed is not constant, I suppose the Lorentz factor $$\gamma = \gamma (t)$$, and by plugging in the velocity, I obtain $$\gamma = \sqrt{1+ \kappa ^2 t^2}$$

I was told to integrate the relationship between ##d\tau## and ##dt## which is just $$d\tau = \frac{dt}{\gamma}$$. However, integrating this to obtain ##\tau## as a function of ##t## yields a transcendental function that can't exactly be inverted (to express ##t## in terms of ##\tau##), and I am stuck in what to try here, in order to obtain the trajectory ##x(\tau)##.

Edit: probably obvious but ##\tau## is the proper time.
 
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  • #4
JTFreitas said:
Homework Statement:: Consider an object whose motion is described by ##x(t) = \frac{c}{\kappa}(\sqrt{1+\kappa^2 t^2} -1)##, where ##c## is the speed of light and ##\kappa## is some constant. Express ##x## and ##t## as a function of proper time.
Relevant Equations:: ##\frac{dt}{d \tau} = \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}##

The object moves solely on the ##x##-axis, hence I calculated its speed to be $$v_x = \frac{dx}{dt} = \frac{c \kappa t}{\sqrt{1+\kappa ^{2} t^2}}$$ Because its speed is not constant, I suppose the Lorentz factor ##\gamma = \gamma (t)##, and by plugging in the velocity, I obtain ##\gamma = \sqrt{1+ \kappa ^2 t^2}##

I was told to integrate the relationship between ##d\tau## and ##dt## which is just ##d\tau = \frac{dt}{\gamma}##. However, integrating this to obtain ##\tau## as a function of ##t## yields a transcendental function that can't exactly be inverted (to express ##t## in terms of ##\tau##), and I am stuck in what to try here, in order to obtain the trajectory ##x(\tau)##.

Edit: probably obvious but ##\tau## is the proper time.
I fixed your maths - use ## to delimit inline maths, not $, and use $$ for "paragraph" maths. Also \sqrt needs curly brackets, not parentheses.

Your working seems to be correct. What did you get when you tried to integrate? I get a fairly straightforward expression. If you got an inverse sinh plus a term involving a square root, you solved ##d\tau=\gamma dt## instead of ##dt/\gamma##.
 
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  • #5
Ibix said:
I fixed your maths - use ## to delimit inline maths, not ,anduse,anduse$ for "paragraph" maths. Also \sqrt needs curly brackets, not parentheses.

Your working seems to be correct. What did you get when you tried to integrate? I get a fairly straightforward expression. If you got an inverse sinh plus a term involving a square root, you solved dτ=γdt instead of dt/γ.
Thanks for the fix. I saw the LaTeX guide now, but my brain typed with the usual inline.

Regarding the work: I found the integral through a formula: ##\ln{\kappa t + \kappa^2 t^2}##, but I just found out that the same integral can also be written as ##\sinh^{-1}(\kappa t)##. This does make the rest of the problem much easier. I am still not used to the type of functions that are common in relativity. In great part I just need pointers to make sure I am in the right path. Thank you!
 
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1. What is proper time?

Proper time is the time measured by a clock that is at rest in the same frame as the object being observed. It is the time experienced by an object or observer moving through space-time.

2. How is proper time different from coordinate time?

Coordinate time is the time measured by an observer at a distance from the object being observed. It is affected by the relative motion of the observer and the object, whereas proper time is not affected by motion.

3. Why is it important to express a trajectory in terms of proper time?

Expressing a trajectory in terms of proper time allows us to accurately describe the motion of an object without being affected by the observer's perspective. It is a more fundamental and reliable measure of time in the context of special relativity.

4. How is proper time calculated?

Proper time is calculated using the equation t = γτ, where t is the coordinate time, γ is the Lorentz factor, and τ is the proper time. This equation takes into account the time dilation effect of special relativity.

5. Can proper time be negative?

No, proper time cannot be negative. It is a physical quantity and therefore must have a positive value. A negative proper time would imply that an object is moving backwards in time, which is not possible according to our current understanding of physics.

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