0
$\begingroup$

I'm trying to figure out when I should use the Lorentz transformation and the inverse transformation. I know this question has been asked before but the answer doesn't really help me. To illustrate what I don't understand, I will take an example.

A skateboarder rides his skateboard towards a hole. The skateboard has a rest length of L, as does the hole. The skateboarder’s friend is at rest with respect to the ground, and tries to warn the skateboarder of his horrible fate. She says: “your skateboard is going to undergo length contraction, so it’ll be shorter than L and you’ll fall into the hole”. The skateboarder objects, saying “from my perspective, it’s the hole that contracts. My skateboard is at rest with respect to me, so it’ll retain its full rest length.

A. Write an equation $y(t)$ that describes the vertical motion of the bottom edge of the skateboard, assuming that the Newtonian equations of free-fall approximately hold in this scenario $\rightarrow y(t)=-\frac{1}{2}gt^2$

B. Use the Lorentz transform to move to the skateboarder’s frame, computing $y'_{front}(t')$ for the front of the skateboard.

And that's where I was wrong. I used the Lorentz transformation equation instead of the inverse transformation to compute $t'$. I thought that we were using the Lorentz transformation when we went from a frame of reference $F$ to a frame of reference $F'$ but I quickly realized that this reasoning did not make much sense because it just depended on the way we named the frames of reference. So to come back to my question, how is it possible to know which transformation to use (the normal or the inverse)?

Thanks in advance for your answers.

$\endgroup$

1 Answer 1

1
$\begingroup$

The easiest thing is to pick an example where it's obvious what the answer should be; that tells you which calculation to do.

For example: I pass by you, traveling in the positive direction, at $(t=0,x=0)$. Consider the event $E=(t=1,x=0)$ (in your coordinates). I assign that event a location coordinate $x'$. Do you expect $x'$ to be positive or negative? Now you know which transformation to use.

$\endgroup$

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.