0
$\begingroup$

Say I have a lamp on a car $M$ in the middle of $A$ and $B$, who are both $c$ distance away from $C$. (so it takes light 1 second to travel to both $A$ or $B$ from $M$). Now the car starts traveling to $B$ with lamp turned on, with speed of $1\ \mathrm{m}/\mathrm{s}$ (so very small comparing to the speed of light). Who among $A$ and $B$ would observe the light earlier?

My understanding of the special relativity is that we have to use https://en.wikipedia.org/wiki/Velocity-addition_formula to compute the relative speed. The the velocity of the light to $A$ is $\frac{c-1}{1-(c/c^2)} = c$ where the velocity of the light to $B$ is also $\frac{c+1}{1+c/c^2} = c$. so $A$ and $B$ should observe the light at the same time.

Is my understanding correct?

If my understanding is correct, say the car is moving with the speed of light. Then the above relative velocity to $A$ is $0/0$. How do we answer that question?

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ From which reference frame are we observing this from? $\endgroup$
    – najkim
    Commented Feb 3, 2021 at 17:23
  • 2
    $\begingroup$ Are you familiar with relativity of simultaneity? $\endgroup$
    – Sandejo
    Commented Feb 3, 2021 at 17:24

1 Answer 1

Reset to default
2
$\begingroup$

Light always travels at c in every reference frame - this is one of the core tenets of special relativity. If the lamp is turned on at halfway between A and B, they will both receive the light at the same time as seen by a static observer. There is no way once the light is turned on to travel by vehicle and deliver light faster than the leading edge of the light turned on at t = 0.

Improve this answer
$\endgroup$

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.