# Understanding the formula in an exercise

I'm stuck for quite some time now on this issue.

I have the following question.

Spaceship with velocity of $0.5c$ passes above two points $A$ and $B$ the distance between $A$ and $B$ respect to the ground is $80meter$.

A viewer on the ground measures the difference between $t_b$ and $t_a$ $t_a$ is the time that the spaceship passes above point $A$ and $t_b$ is the time that the spaceship passes above point $B$.

Somehow in the answers of this exercise they managed to reach to following formula : $$t_b-t_a=\frac{x_b-x_a}{v}=\frac{80}{0.5c}$$ which is not clear to me.

This is what I did:

According to lorentz transformation we know that

$$x_b=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}(x'+vt_b') \rightarrow 80=\frac{1}{\sqrt{1-\frac{0.25c^2}{c^2}}}(0+0.5ct_b')\rightarrow \frac{\sqrt{0.75}*80}{0.5c}$$

But that's not the formula/answer $t_b-t_a=\frac{x_b-x_a}{v}=\frac{80}{0.5c}$ Any ideas what's wrong or how to get to the formula/answer $t_b-t_a=\frac{x_b-x_a}{v}=\frac{80}{0.5c}$

Any help will be appreciated.

• This isn't an answer but a hint: what frame or frames are the measurements being done in? When do you need to use the Lorentz transformation?
– user107153
Mar 25 '16 at 21:43
• @tfb I'm not completely sure if this is what you mean but the measurements are in MKS unit system, and I use Lorentz transformation to calculate the change (time,location) between two inertial systems with different point of view EDIT : when the speed is very high to the speed of light ($c$) Mar 25 '16 at 21:48
• @tfb OHHH, I think I understand what you mean, you mean that in that case I don't need to use Lorentz transformation but instead use the "normal" (not sure about the official name) which is $x=vt$ my question is why in this case I don't need to use Lorentz transformation? Mar 25 '16 at 21:59
• @JohnC: You are most of the way there. The thing to understand is why you don't need to use the transformation, which is why I asked the question about which frames of reference (or frame of reference, hint) matter here.
– user107153
Mar 25 '16 at 22:27

Lorentz transformations are coordinate transformations used when you have to connect two frames moving at constant relative speed. Here, you are solving for time difference in ground's frame and you have distance measured in that frame only. So there is no need for LT. You would use it if lets say you want to calculate time difference/distance as measured by spaceship.

Imagine that you are the viewer and you want to measure the time that ship takes from going to A to B. You don't need Lorentz transformations, just v=d/t.

But if you want to know how long it takes from A to B for the people in the ship, then you have to use L.T and you will see that the time is lower.