What is the world line of a uniformly accelerated particle along the x-axis?

[Silviu]

Homework Statement

A particle is moving along the x-axis. It is uniformly accelerated, in the sense that the acceleration measured in its instantaneous rest frame is always g, a constant. Find x and t as a function of the proper time $\tau$ assuming that the particle passes through $x_0$ at time t = 0 with zero velocity.

Homework Equations

The Attempt at a Solution

I am not sure I understand the meaning of "in the sense that the acceleration measured in its instantaneous rest frame is always g, a constant". The way I thought of doing it was $dt = \gamma d \tau = \frac{d\tau}{\sqrt{1-(gt)^2}}$ which implies $\sqrt{1-(gt)^2}dt=d\tau$, then I integrate and get $t(\tau)$ and a similar reasoning for $x(\tau)$. However I am not sure this is correct, as this implies that g is constant in the frame where we calculate x and t and I am not sure this is equivalent to what the problem states. Can someone tell me how to approach this correctly? Thank you!

 

[Orodruin]

The way I thought of doing it was $dt = \gamma d \tau = \frac{d\tau}{\sqrt{1-(gt)^2}}$

This would imply that the velocity was given by $gt$ which is not the case. What it means is that, in the instantaneous rest frame $S'$,
$$
\frac{d^2 x'}{dt'^2} = g.
$$
From there it is up to you to relate the quantities in the instantaneous rest frame to the original frame.

 

[Silviu]

This would imply that the velocity was given by $gt$ which is not the case. What it means is that, in the instantaneous rest frame $S'$,
$$
\frac{d^2 x'}{dt'^2} = g.
$$
From there it is up to you to relate the quantities in the instantaneous rest frame to the original frame.

The way I was thinking to do it was to consider the 4 acceleration, which in MCRF would be $(0,g,0,0)$, while in an inertial frame, call it S (where we want to compute x and t), it would be, based on Lorentz transformations, $g \gamma (v,1,0,0)$, where v is the instantaneous velocity. Thus, in S we would have for the 1 component $g \gamma = \frac{dv}{dt}=\frac{dv}{\gamma d\tau}$, which implies $g d\tau = \frac{dv}{\gamma^2}$ and by integration I can get $v(\tau)$ and similar reasoning for x and t. Is this correct?

 

[Orodruin]

Well, if you are familiar with 4-vectors it all becomes much simpler. Try to answer the following questions:

• What is the relationship between 4-acceleration and 4-velocity?
• What is always true about the 4-velocity?
• Based on the answer to the previous question, how can you generally parametrise the 4-velocity?
• What happens when you insert that parametrisation into the relationship between the 4-acceleration and 4-velocity?
• What happens when you then demand that $A^2 = -g^2$?

The answer to those questions should give you the 4-velocity as a function of the proper time, which you can integrate to get the space-time coordinates as a function of the proper time. No reference to the instantaneous rest frame will ever be necessary.

 

[Silviu]

Well, if you are familiar with 4-vectors it all becomes much simpler. Try to answer the following questions:

• What is the relationship between 4-acceleration and 4-velocity?
• What is always true about the 4-velocity?
• Based on the answer to the previous question, how can you generally parametrise the 4-velocity?
• What happens when you insert that parametrisation into the relationship between the 4-acceleration and 4-velocity?
• What happens when you then demand that $A^2 = -g^2$?

The answer to those questions should give you the 4-velocity as a function of the proper time, which you can integrate to get the space-time coordinates as a function of the proper time. No reference to the instantaneous rest frame will ever be necessary.

I tried something related to this, but got stuck on the way. So we have $a^\alpha = \frac{d u^\alpha}{d\tau}$. In the MCRF, $u^\alpha = (1,0,0,0)$ and in all frames $u^\alpha u_\alpha = -1$ and in another frame, $u^\alpha = \gamma(1,v_x, v_y, v_z)$. But I am not sure how to connect all these.

 

[Orodruin]

So first off, so you are using a different sign convention than I am for the inner product, so the last bullet should read $A^2 = g^2$.

So we have $a^\alpha = \frac{d u^\alpha}{d\tau}$.

Correct. This ticks off the first bullet.

In the MCRF, $u^\alpha = (1,0,0,0)$ and in all frames $u^\alpha u_\alpha = -1$ and in another frame, $u^\alpha = \gamma(1,v_x, v_y, v_z)$. But I am not sure how to connect all these.

So the $u^2 = -1$ ticks off the second bullet point. Now, your particle is only moving along the x-axis so you can neglect the y and z components completely. It is true that you can parametrise the 4-velocity as $u = \gamma(1,v)$, where $v$ is the parameter, but there is a better way for this problem. Consider the components of $u$. In terms of those components, which form does $u^2 = -1$ take?

 

[Silviu]

So first off, so you are using a different sign convention than I am for the inner product, so the last bullet should read $A^2 = g^2$.Correct. This ticks off the first bullet.So the $u^2 = -1$ ticks off the second bullet point. Now, your particle is only moving along the x-axis so you can neglect the y and z components completely. It is true that you can parametrise the 4-velocity as $u = \gamma(1,v)$, where $v$ is the parameter, but there is a better way for this problem. Consider the components of $u$. In terms of those components, which form does $u^2 = -1$ take?

But $u^2 = - 1$ means that $\gamma^2 v^2 - \gamma^2 = - 1$ which is obvious. Parametrizing in terms of proper time $\tau$, I guess would be better? We have $u^\alpha = \frac{d x^\alpha}{d \tau}$, but not sure exactly how to proceed.

 

[Orodruin]

No, I am saying to write down the constraint $u^2 = -1$ in terms of the components $u^0$ and $u^1$. Once you have done that we can discuss how to parametrise it better.

Edit: It is true that $u^2 = -1$ can be parametrised with $v$ in the way you mention and differentiating it would give you a differential equation for $v$ in terms of $\tau$. However, there is a much more elegant parametrisation that will essentially hand you the answer on a silver platter.

 

[Silviu]

No, I am saying to write down the constraint $u^2 = -1$ in terms of the components $u^0$ and $u^1$. Once you have done that we can discuss how to parametrise it better.

Edit: It is true that $u^2 = -1$ can be parametrised with $v$ in the way you mention and differentiating it would give you a differential equation for $v$ in terms of $\tau$. However, there is a much more elegant parametrisation that will essentially hand you the answer on a silver platter.

Do you mean $u_1^2-u_0^2 = -1$?

 

[Orodruin]

Yes. So how can you parametrise that equation in a straightforward manner? Does it remind you of something?

Edit: If it doesn't remind you of something. Would it remind you of something if you changed the minus signs for plusses?

 

[Silviu]

Yes. So how can you parametrise that equation in a straightforward manner? Does it remind you of something?

Oh, so I can write $u^\alpha=(cosh(\tau),sinh(\tau))$ and so $a^\alpha = (sinh(\tau),cosh(\tau))$

 

[Orodruin]

Almost. The parameter is not necessarily the proper time $\tau$, but generally a function of the proper time (since $u$ is - also it would not match dimensionally to put a dimensional argument of the hyperbolic functions). I suggest you call the parameter $\theta$ for nomenclature reasons.

 

[Silviu]

Almost. The parameter is not necessarily the proper time, but generally a function of the proper time (since $u$ is - also it would not match dimensionally to put a dimensional argument of the hyperbolic functions). I suggest you call the parameter $\theta$ for nomenclature reasons.

But still how can I use the initial condition? In the MCRF $u^alpha$ is still $(1,0,0,0)$

 

[Orodruin]

But still how can I use the initial condition? In the MCRF $u^alpha$ is still $(1,0,0,0)$

Patience, we will get there. The initial conditions you have on the position and the velocity will turn into integration constants for the coordinates and an initial condition for the parameter $\theta$ .

 

[Orodruin]

Also note that we will make no reference whatsoever to the instantaneous rest frame. Everything here is being done in the original inertial frame.

 

[Silviu]

Also note that we will make no reference whatsoever to the instantaneous rest frame. Everything here is being done in the original inertial frame.

So zero velocity initially means that $\theta(0)=0$. Then we would have $x - x_0 = \int_0^{\theta_1} sinh(\theta) = cosh(\theta_1)-1$

 

[Orodruin]

So zero velocity initially means that $\theta(0)=0$.

Yes.

Then we would have $x - x_0 = \int_0^{\theta_1} sinh(\theta) = cosh(\theta_1)-1$

Not so fast. The 4-velocity is the derivative of $x^\mu$ with respect to proper time, not with respect to the parameter we used. In order to integrate $dx^\mu/d\tau = u$, you must first find $u$ as a function of proper time $\tau$, which means finding an expression for $\theta$ in terms of $\tau$. I suggest going back to the bullet list of post #4. You have ticked off the third bullet by letting $u^\mu = (\cosh(\theta),\sinh(\theta))$. Bullets four and five remain.

 

[Silviu]

Yes.Not so fast. The 4-velocity is the derivative of $x^\mu$ with respect to proper time, not with respect to the parameter we used. In order to integrate $dx^\mu/d\tau = u$, you must first find $u$ as a function of proper time $\tau$, which means finding an expression for $\theta$ in terms of $\tau$. I suggest going back to the bullet list of post #4. You have ticked off the third bullet by letting $u^\mu = (\cosh(\theta),\sinh(\theta))$. Bullets four and five remain.

Ok, so for number 4 we have $a^\alpha = \frac{v^\alpha}{d\tau}=(sinh(\theta)\frac{d\theta}{d\tau},cosh(\theta)\frac{d\theta}{d\tau})$

 

[Orodruin]

Ok, so for number 4 we have $a^\alpha = \frac{v^\alpha}{d\tau}=(sinh(\theta)\frac{d\theta}{d\tau},cosh(\theta)\frac{d\theta}{d\tau})$

Indeed. And number 5?

 

[Silviu]

Indeed. And number 5?

$\frac{d\tehta}{d\tau}=g$, so we get $\theta = g\tau$ and then I can integrate $u^\alpha$ to get x

 

[Orodruin]

$\frac{d\tehta}{d\tau}=g$, so we get $\theta = g\tau$ and then I can integrate $u^\alpha$ to get x

Right, both $x$ and $t$ as functions of $\tau$.

 

[Silviu]

Right, both $x$ and $t$ as functions of $\tau$:

Thank you so so much!

 

[Orodruin]

Just some final thoughts regarding the parametrisation of the 4-velocity.

The parameter $\theta$ is called rapidity and is very often a much more convenient variable to work with than velocity. It relates to velocity according to $v = \tanh(\theta)$ (in units where c = 1) and, unlike velocity, it is additive for motions in the same direction. If object A has rapidity $\theta_1$ relative to B and B has rapidity $\theta_2$ relative to C, then A has rapidity $\theta_1 + \theta_2$ relative to C. In many situations, working with hyperbolic functions rather than $\gamma$ factors will greatly simplify your computations.

 

[Silviu]

Just some final thoughts regarding the parametrisation of the 4-velocity.

The parameter $\theta$ is called rapidity and is very often a much more convenient variable to work with than velocity. It relates to velocity according to $v = \tanh(\theta)$ (in units where c = 1) and, unlike velocity, it is additive for motions in the same direction. If object A has rapidity $\theta_1$ relative to B and B has rapidity $\theta_2$ relative to C, then A has rapidity $\theta_1 + \theta_2$ relative to C. In many situations, working with hyperbolic functions rather than $\gamma$ factors will greatly simplify your computations.

Does this works, even in multiple dimensions? In the problems we had just x, so we could use the trigonometric identity, but does it work in general?

 

[Orodruin]

Also note the similarity of this problem to a kinematic problem in two-dimensions where we are given that an object moves at constant speed $v$ with an acceleration $a$ perpendicular to its velocity. You would parametrise the velocity according to
$$
\vec v = \cos(\theta) \vec e_1 + \sin(\theta)\vec e_2,
$$
differentiate it to obtain
$$
\vec a = \dot\theta [-\sin(\theta)\vec e_1 + \cos(\theta)\vec e_2],
$$
which squares to
$$
a^2 = \dot\theta^2 \quad \Longrightarrow \quad \dot\theta = a.
$$
Integrating the velocity then gives a circular motion in the plane in the same manner as the world-line you hopefully obtained is a hyperbola in Minkowski space.

 

[Orodruin]

Does this works, even in multiple dimensions? In the problems we had just x, so we could use the trigonometric identity, but does it work in general?

No, in general not only the magnitude but also the direction of the acceleration matters and here you were essentially given the direction. You therefore need to specify the direction of the acceleration as well as its magnitude. The same goes for kinematic problem in #25. In three or more dimensions you will need to know the direction of the acceleration as well. This boils down to the condition $u^2 = -1$ needing three parameters for a general parametrisation in 4D Minkowski space.