# Homework Help: World line of a particle

1. Oct 11, 2017

### Silviu

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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!

2. Oct 11, 2017

### Orodruin

Staff Emeritus
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.

3. Oct 11, 2017

### Silviu

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?

4. Oct 11, 2017

### Orodruin

Staff Emeritus
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.

5. Oct 11, 2017

### Silviu

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.

6. Oct 11, 2017

### Orodruin

Staff Emeritus
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?

7. Oct 11, 2017

### Silviu

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.

8. Oct 11, 2017

### Orodruin

Staff Emeritus
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.

9. Oct 11, 2017

### Silviu

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

10. Oct 11, 2017

### Orodruin

Staff Emeritus
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?

Last edited: Oct 11, 2017
11. Oct 11, 2017

### Silviu

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

12. Oct 11, 2017

### Orodruin

Staff Emeritus
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.

13. Oct 11, 2017

### Silviu

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

14. Oct 11, 2017

### Orodruin

Staff Emeritus
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 $\theta$ .

15. Oct 11, 2017

### Orodruin

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

16. Oct 11, 2017

### Silviu

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

17. Oct 11, 2017

### Orodruin

Staff Emeritus
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.

18. Oct 11, 2017

### Silviu

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

19. Oct 11, 2017

### Orodruin

Staff Emeritus
Indeed. And number 5?

20. Oct 11, 2017

### Silviu

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