Galilean transformations - What am I messing up here?

[Achmed]

This is a follow-up to a question I asked earlier. We have the following exercise:

We have two parallel mirrors, which are located at y=0 and y=l in the (x,y) plane. A photon is traveling between the mirrors, up and down along the y-axis. Consider an observer O at rest w.r.t. the mirrors.

1. What's the time (Δt) measure by O for the photon to make a full period.

Consider an observer O' which moves along the x-axis with a speed v, which is constant. Assume l' = l.

2. Using Galilean transformations from Newt. mech. , what's the time measured by O' for the photon to make a full period (draw a picture to illustrate the logic). Compare this with the time measured by O.

So for 1, the answer is $\Delta t= \dfrac{2l}{c}$. I tried to do 2 too, but I keep getting a somewhat incorrect answer. Here's what I do:

Draw an isosceles triangle $ABC$ with $|AB| = vt'$, and the height is l. I figured that $t' = \dfrac{2|AB|}{c'}$, where $c' = c-v$. So we try to solve the following equation: $(c't')^2 = 4l^2 + (vt')^2$, but if I solve for t', I get an almost correct answer, which of course is still incorrect. You get $t' = \dfrac{2l}{\sqrt{c^2-2cv}}$, and of course we want $t= \dfrac{2l}{c}$ because in Newtonian mechanics t=t'. What have I done wrong?

 

[cwasdqwe]

If I'm not wrong, you've missed a factor 2 in t' (you gave the solution for just AC (or BC), and it should be both to consider the path that the photon would do for the O' observer. So it is, $t'=\frac{4l}{\sqrt{4c^2-v^2}}$, and when $v<<c$ (Newtonian approach) then $t'=t=\frac{2l}{c}$.

 

[Achmed]

If I'm not wrong, you've missed a factor 2 in t' (you gave the solution for just AC (or BC), and it should be both to consider the path that the photon would do for the O' observer. So it is, $t'=\frac{4l}{\sqrt{4c^2-v^2}}$, and when $v<<c$ (Newtonian approach) then $t'=t=\frac{2l}{c}$.

If you draw a triangle with ABC where AB is vt', then wouldn't AC + BC be equal to simply c't'?

 

[cwasdqwe]

I'm not sure if I followed you in your definition $c'=c-v$ (remember Einstein's discovery about speed of light as the only constant element in his Theory of Relativity). Am I misunderstanding you?

 

[Achmed]

I'm not sure if I followed you in your definition $c'=c-v$ (remember Einstein's discovery about speed of light as the only constant element in his Theory of Relativity). Am I misunderstanding you?

This question is a sort of intro question for a special relativity course, in the next question you have to assume c' = c. But in this question you have to assume that everything is Newtonian, i.e. you want to obtain that t=t'. But, as I say in the OP, I can't seem to figure out what I've done wrong.

 

[cwasdqwe]

Hope I'm getting it well (the axis and the mirror positions): I consider two parallel mirrors (y=0 and y=l) —parallel to the xz plane, although we don't need to consider z axis here. The observer O' travels perpendicular to these in x-axis (AB in my triangle) —or equivalent, the mirrors move in x-axis with constant speed v. In this scenario, O' looks how the photon doing a path which is not a straight line in y (case 1), but two diagonal lines (AC+CB). These three paths close the triangle of height l. Now, we know the two cathetus (AB/2 and the height, l) and the hypothenuse is half the distance that the photon travels for O' ($ct'/2$—which is what you write but with c' instead of c), and Pythagoras gives the answer for t' seen above.

 

[Achmed]

Hope I'm getting it well (the axis and the mirror positions): I consider two parallel mirrors (y=0 and y=l) —parallel to the xz plane, although we don't need to consider z axis here. The observer O' travels perpendicular to these in x-axis (AB in my triangle) —or equivalent, the mirrors move in x-axis with constant speed v. In this scenario, O' looks how the photon doing a path which is not a straight line in y (case 1), but two diagonal lines (AC+CB). These three paths close the triangle of height l. Now, we know the two cathetus (AB/2 and the height, l) and the hypothenuse is half the distance that the photon travels for O' ($ct'/2$—which is what you write but with c' instead of c), and Pythagoras gives the answer for t' seen above.

Shouldn't it be c't'/2 = (c-v)t/2?

 

[cwasdqwe]

Keep in mind that the photon is moving in the y-axis, and the second observer is moving in the x-axis. The speed of the photon is always c, so the path would be AC+CB=t'c. If the photon moved in x axis, then we would face intuition against relativity, and the photon moves always at c, but this luxury costs something to the Universe: to deform space-time in order to make it happen.

 

[vela]

Draw an isosceles triangle $ABC$ with $|AB| = vt'$, and the height is l. I figured that $t' = \dfrac{2|AB|}{c'}$, where $c' = c-v$.

Where exactly are A, B, and C? Is AB the base of the triangle and C the apex (so that AC = BC)? If so, I don't see why you think that t' = 2 AB/c'. Shouldn't it be t' = 2 AC/c'?

If B is the apex so that AB = BC and AC is the base of the triangle, then shouldn't you have AC = vt'?

Finally, c' is not c-v. Think about the case where v=c. You'd have c'=0, but the photon would be moving along a 45-degree line, right?