Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Use this scenario:

An electron gains speed in the Stanford Linear Accelerator (SLA) across 3000 meters, reaching a final velocity of 0.95c due to a constant force pushing on the electron. Given the mass of the electron and the constant force pushing on the electron, what is the function for velocity in terms of time?

How do we derive this?

share|cite|improve this question
Wikipedia puts the energy at 50 GeV which is going to be rather a lot closer to c than a measly 95%. Electrons are so light that any decent accelerator can put right up against c. – dmckee Dec 10 '11 at 22:52
I'm so stuck on this. I realize what you're saying, but lets just assume the numbers I mentioned. There must be some constant force pushing the electron, which we can find based on knowing the final speed of the electron and the distance it travels. If we know this force, we can find the speed of the electron at any given moment and thus the length contraction that the electron observes at any given moment and then calculate the rate of change of the contraction which is proportional to the rate of change of the velocity of the electron. – trusktr Dec 10 '11 at 23:56
You'll probably find the problem easier if you think in terms of energy rather than speed and force. Where you'd have assumed constant force before assume constant $dE/dx$ (this is consistent because of the work--energy theorem). Note that particle physicists use units where $c =1$, so that the mass of the electron is 511 keV. – dmckee Dec 11 '11 at 2:15
trusktr, perhaps you could edit some of your reasoning from your last comment into the question? Ordinarily I would close this kind of question, where you present your homework problem and ask "How do I do this?" but your comment shows that you have put some thought into it. – David Z Dec 11 '11 at 2:45
I figured it out. I'll come back and post the answer ASAP. – trusktr Dec 11 '11 at 3:14

There are two different interpretations of "constant force" possible

  • The momentum gained by the particle per unit coordinate time is constant
  • The acceleration in the particle's rest frame is constant.

In terms of trajectories, the first says that ${d\over dt} {dx\over d\tau} = f$ where f is constant, t is time, x is position and $\tau$ is proper time. The time in this situation is the total electron's momentum divided by the rate of momentum increase, f. Then

$$p(t) = {dx\over d\tau} = {mv\over \sqrt{1-v^2}} = f t $$

solving for v,

$$ v = {at\over \sqrt{1+(at)^2}}$$ where $a={f\over m}$, which can be integrated to give the constant momentum-rate trajecotry x(t) explicitly.

$$ x(t) = {1\over a} \sqrt{1+(at)^2}$$

so that

$$ x^2 - t^2 = {1\over a^2}$$

This is the hyperbola in spacetime, so it has constant curvature, and so it is also the solution to the second interpretation: $||{d^2x\over d\tau^2}|| =f$, where f is constant. The two interpretations, although it is not immediately obvious, are the same.

To understand the coincidence needs a little bit of hyperbolic geometry (it works for circles in Euclidean space just as well). Given a trajectory with rapidity (relativistic analog of angle) $\gamma(t)$, the rate of change of the momentum with respect to time is the time rate of change of $\sinh(\gamma)$, which is $\cosh(\gamma){d\over dt}{\gamma}$, where the derivative is with respect to time, but this is also $\dot{\gamma}$, where the dot is derivative with respect to proper time.

This allows you to solve the stated problem very simply. The momentum of the electron at the end of the motion, when $v=.95$, is just over 3 times the mass of the electron (times c). This divided by the force is the time required to accelerate the electron to that speed.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.