# Homework Help: Electron moves from .93c to .99c in some amount of time, how far does it travel?

1. Jan 23, 2012

### IntegrateMe

An electron moves from 0.93c to 0.99c in 0.819 x 10-11 s.

Does this mean that the particle travels a distance of:

x = (0.99c-0.93c)(0.819 x 10-11 s) m?

2. Jan 23, 2012

### kunguz

What do you mean by c? is it Coulomb? If yes, please see the definition of Coulomb from wikipedia:

"One coulomb is the magnitude (absolute value) of electrical charge in 6.24150965(16)×1018 protons or electrons."

http://en.wikipedia.org/wiki/Coulomb

3. Jan 23, 2012

### iRaid

He definitely means the speed of light (c)

4. Jan 23, 2012

### Redbelly98

Staff Emeritus
That looks wrong. What if the electron's initial and final speed were both 0.93c? Then your formula would give

x = (0.93c - 0.93c)·(0.819 x 10-11 s)
= 0·(0.819 x 10-11 s)
= 0, which is clearly wrong.

If the acceleration is constant, you can use the usual kinematic equations to find the distance traveled.

EDIT: it strikes me as very odd to use relativistic speeds in a constant-acceleration problem. Exactly what topic has your class been studying lately? What formula(s) have you been given to work with -- either in the textbook or in the class lectures?

No, it's the speed of light as iRaid said. This is a problem about motion, and has nothing to do with electric charge.

Last edited: Jan 23, 2012
5. Jan 23, 2012

### Pengwuino

No it is not right at all. The distance something travels is $\Delta x = v\Delta t$. However, that's only for a constant velocity and he absolutely does not have a constant velocity setup.

@OP: As redbelly said, this question is a bit odd. Have you looked at special relativity yet? Because you're only looking for the distance traveled, there wouldn't be any relativistic effects to consider, but it's very strange to pose the question using relativistic speeds.

6. Jan 23, 2012

### Redbelly98

Staff Emeritus
Moderator's note:

I have deleted a post that provided the formula to be used (if this motion is constant acceleration.)

Please do not provide formulas that students should be able to easily find themselves. Looking up useful information in textbooks and class notes is something students should be willing to do for themselves.

Thank you.

7. Jan 23, 2012

### IntegrateMe

Sorry for the confusion. Let me explain:

"Electrons initially at rest are subjected to a continuous force of 2x10^-12 N for 2 miles."

Determine how much time is required to increase the electrons' speed from 0.93c to 0.99c. (That is, the quantity |v|/c increases from 0.93 to 0.99).

F = dp/dt
dt = [m(vf-vi)]/F

vf-vi = 0.06c; electron's m = 9.1 x 10^-31

After plugging in I got a time of 0.819 x 10^-11 s.

Approximately how far does the electron go in this time? Why is this approximate?

And here is where my problem is...

Thank you for the responses!

8. Jan 24, 2012

### Dickfore

Actually, the wording of the problem is ambiguous. The distance traveled depends on the way acceleration changes.

EDIT:

I will give a formula for the distance traveled if the proper-acceleration vs. proper-time is given as $a_0(\tau)$.

$$\frac{d v}{d t} = \frac{ \frac{ d v' \, \left( 1+ \frac{v' \, u}{c^2} \right) - (v' + u) \, \frac{u \, d v'}{c^2} }{\left( 1+ \frac{v' \, u}{c^2} \right)^2} }{ \frac{d t' + \frac{u \, d x'}{c^2}}{\sqrt{1 - \frac{u^2}{c^2}}} } = \frac{d v'}{d t'} \, \frac{ \left( 1 - \frac{u^2}{c^2} \right)^{3/2} }{ \left( 1 + \frac{u \, v'}{c^2} \right)^3 }$$
If $v' = 0$, and $u = v$ (instantaneous proper frame), then $d v'/d t' = a_0$. Also, since we want to use proper time instead of LAB time, we have:

$$\frac{d t}{d \tau} = \left( 1 - \frac{v^2}{c^2} \right)^{-1/2}$$

Combining these two equations, we have:
$$\frac{d v}{d \tau} = \frac{d v}{d t} \, \frac{d t}{d \tau} = a_0 \, \left( 1 - \frac{v^2}{c^2} \right)^{3/2} \, \left( 1 - \frac{v^2}{c^2} \right)^{-1/2} = a_0 \, \left( 1 - \frac{v^2}{c^2} \right)$$

The variables in this ODE may be separated:
$$\frac{d v}{1 - v^2 / c^2} = a_0(\tau) \, d\tau$$
and then integrated:
$$\int_{v_i}^{v}{ \frac{d \tilde{v}}{1 - \tilde{v}^2 / c^2} } = \int_{0}^{\tau}{ a_0(\tilde{\tau}) \, d\tilde{\tau} }$$
For shorthand, we denote $A(\tau) \equiv \int_{0}^{\tau}{ a_0(\tilde{\tau}) \, d\tilde{\tau} }$. The integral over velocity is performed by introducing the hyperbolic trigonometric substitution (the parameter $\eta$ is called rapidity):
$$\frac{\tilde{v}}{c} = \tanh(\eta) \Rightarrow d\tilde{v} = c \, \mathrm{sech}^2(\eta) \, d\eta, \ \frac{1}{1 - \tilde{v}^2 / c^2} = \cosh^2(\eta)$$
and we have:
$$\eta(\tau) - \eta_i = A(\tau)$$
Thus, we have the following implicity dependence of velocity on proper time:
$$v(\tau) = c \, \tanh^{-1} \left(\eta_i + A(\tau) \right), \eta_i \equiv \tanh \left(\frac{v_i}{c} \right)$$
Once we know $\eta(\tau)$, we can find the dependence of LAB time t on proper time τ:
$$\frac{d t}{d \tau} = \left( 1 - \frac{v^2}{c^2} \right)^{-1/2} = \cosh \left( \eta(\tau) \right)$$
which may be integrated:
$$t(\tau) = \int_{0}^{\tau} {\cosh \left( \eta (\tilde{\tau}) \right) \, d\tilde{\tau} }$$
Finally, the displacement is:
$$dx = v \, dt = c \, \tanh \left( \eta(\tau) \right) \, \cosh \left( \eta(\tau) \right) \, d\tau = c \, \sinh \left( \eta(\tau) \right) \, d\tau$$
$$\Delta x = c \, \int_{0}^{\tau} { \sinh \left( \eta(\tilde{\tau} ) \right) \, d\tilde{\tau} }$$

Last edited: Jan 24, 2012
9. Jan 24, 2012

### IntegrateMe

The force of 2x10^-12 N is continuous, doesn't that describe constant acceleration?

10. Jan 24, 2012

### Dickfore

It describes constant proper acceleration (see my edit to the above post). You need to use those formulas. What other quantity you need besides the force?

11. Jan 24, 2012

### IntegrateMe

I'm taking an introductory physics course. I don't understand half of the stuff you did there. The problem should take me very little time to solve.

12. Jan 24, 2012

### Dickfore

Then, your textbook sucks because at those speeds relativistic effects are truly visible.

For example, it says that the electron uniformly accelerates, increasing its velocity by $0.06 \, c$ in $8.19 \, \mathrm{ps}$. Does it mean that in another sixth of that time, $1.365 \, \mathrm{ps}$, it will increase its velocity by another $0.01 \, c$, so that its velocity becomes $1.00 \, c$?

13. Jan 24, 2012

### IntegrateMe

I have no idea. I just used the equation:

x = x0 + v0(t) + 0.5a(t)^2

And calculated the acceleration using F = ma since I know the force acting on the electron and the mass of the electron.

14. Jan 24, 2012

### Dickfore

ok, cool.

15. Jan 24, 2012

### vela

Staff Emeritus
What topics are you currently studying in your physics course? Based on the question, it seems like you're learning about special relativity.

As Dickfore noted, the equations you used simply don't apply when the electron is moving so close to the speed of light.

16. Jan 24, 2012

### Redbelly98

Staff Emeritus
The constant force suggests that relating work done with the kinetic energy may be a useful way to go here. But kinetic energy is not simply (1/2)mv2 at these speeds, you do need to use relativity to get the kinetic energy.