0
$\begingroup$

I'm studying some lecture notes on special relativity and at some point one considers an inertial system in which a particle has $4$-velocity and $4$-acceleration given by

$$U = (c\frac{dt}{ds}, \frac{dx}{ds}, 0, 0), \quad A = (c\frac{d^2t}{ds^2}, \frac{d^2x}{ds^2}, 0, 0) \, .$$

We also know the acceleration is constant with magnitude $\kappa$.
I understand these $4$-vectors have to satisfy the equations $$c^2(dt/ds)^2 - (dx/ds)^2 = c^2 , \quad c^2(d^2t/ds^2)^2 - (d^2x/ds^2)^2 = - \kappa^2,$$ though I don't understand why there is a minus sign in front of $\kappa$ (does one implicitly assume the acceleration is opposite to the x-direction?).
However, I have no clue how to deduce the following pair of equations which seem to come out of the blue, namely $$c\frac{d^2t}{ds^2} = \kappa \sqrt{(dt/ds)^2 - 1} , \quad \frac{d^2x}{ds^2} = \kappa \frac{dt}{ds}.$$ Any hint?

Improve this question
$\endgroup$
4
  • $\begingroup$ What is your source? Can you check your equations for typos? $\endgroup$
    – robphy
    Commented Dec 4, 2020 at 0:02
  • $\begingroup$ @robphy I checked my lecture notes and there is no typos in my post. So, If there is a typo then it's in the lecture notes themselves. $\endgroup$
    – Ansonī Bōdo
    Commented Dec 4, 2020 at 16:30
  • $\begingroup$ (I see that an edit was done... so the units are more consistent now.) $\endgroup$
    – robphy
    Commented Dec 4, 2020 at 17:52
  • $\begingroup$ I meant no typos in the last two equations. I fixed a forgotten square in one equation. $\endgroup$
    – Ansonī Bōdo
    Commented Dec 4, 2020 at 20:30

1 Answer 1

Reset to default
1
$\begingroup$

The 4-velocity is timelike. The 4-acceleration is spacelike. So their square-norms have opposite signs. (These 4-vectors are actually orthogonal to each other. Since the square-norm of the 4-velocity is a constant, taking its derivative with respect to proper time reveals this orthogonality condition.)

Improve this answer
$\endgroup$
4
  • $\begingroup$ I understand that g(U,U) = c^2 > 0 and g(A, U) = 0, but now how do you deduce g(A,A) < 0, ie. that the 4-acceleration is spacelike? $\endgroup$
    – Ansonī Bōdo
    Commented Dec 4, 2020 at 17:21
  • $\begingroup$ What happens when you symbolically compute $d/ds( g(U,U) )$? Can you write this in terms of U and A? Then use what you know about $g(U,U)$. (Hint: An analogue would be the relationship between the acceleration vector and the velocity vector for uniform circular motion.) $\endgroup$
    – robphy
    Commented Dec 4, 2020 at 17:40
  • $\begingroup$ d/ds(g(U, U)) = 2 g(U, A) = 0, but this equation does not help determining the sign of g(A, A) $\endgroup$
    – Ansonī Bōdo
    Commented Dec 7, 2020 at 18:11
  • $\begingroup$ @AnsonīBōdo Can you see that the set of 4-vectors that are orthogonal to the unit timelike vector $\hat U$? Consider the event on the unit hyperbola at the tip of U. Can you see that the tangent hyperplane is "purely spatial" according to U? Can you see that $\hat U + \hat A$ is lightlike? $\endgroup$
    – robphy
    Commented Dec 7, 2020 at 18:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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