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Galilean transformations - What am I messing up here?

  • Thread starter Achmed
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  • #1
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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 travelling 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 [itex] \Delta t= \dfrac{2l}{c} [/itex]. I tried to do 2 too, but I keep getting a somewhat incorrect answer. Here's what I do:

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

Answers and Replies

  • #2
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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, [itex]t'=\frac{4l}{\sqrt{4c^2-v^2}}[/itex], and when [itex]v<<c[/itex] (newtonian approach) then [itex]t'=t=\frac{2l}{c}[/itex].
 
  • #3
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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, [itex]t'=\frac{4l}{\sqrt{4c^2-v^2}}[/itex], and when [itex]v<<c[/itex] (newtonian approach) then [itex]t'=t=\frac{2l}{c}[/itex].

If you draw a triangle with ABC where AB is vt', then wouldn't AC + BC be equal to simply c't'?
 
  • #4
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I'm not sure if I followed you in your definition [itex]c'=c-v[/itex] (remember Einstein's discovery about speed of light as the only constant element in his Theory of Relativity). Am I misunderstanding you?
 
  • #5
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I'm not sure if I followed you in your definition [itex]c'=c-v[/itex] (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.
 
  • #6
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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' ([itex]ct'/2[/itex]—which is what you write but with c' instead of c), and Pythagoras gives the answer for t' seen above.
 
  • #7
10
0
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' ([itex]ct'/2[/itex]—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?
 
  • #8
31
3
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.
 
  • #9
vela
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Draw an isosceles triangle [itex]ABC[/itex] with [itex] |AB| = vt' [/itex], and the height is l. I figured that [itex] t' = \dfrac{2|AB|}{c'} [/itex], where [itex] c' = c-v [/itex].
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?
 

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