Where is the mistake in this derivation of Length Contraction?

[aliinuur]

TL;DR

I've got L=L_0*sqrt(1+v^2/c^2)-v^2/c^2 instead of L=L_0*sqrt(1-v^2/c^2). Where did i make mistake?

I tried to use square box with two perpendicularly oscillating photons to derive length contraction formula. Photons are emitted, reflected from box walls and absorbed simultaneously from the box's point of view. If the box have some velocity v, then assuming photons are still emitted and absorbed(but not reflected) simultaneously, I've got L=L_0*sqrt(1+v^2/c^2)-v^2/c^2 instead of L=L_0*sqrt(1-v^2/c^2). Where did I make mistake?

 

[Ibix]

Difficult to tell if you don't show your derivation.

Note that you can use $\LaTeX$ to lay out maths legibly. (If it wasn't working when you wrote your first post it should be now because I've used LaTeX - refresh the page if you still don't see it rendered.)

 

[aliinuur]

here is my calculations for v=c/2. Diagonal arrows - path of the vertical photon, horizontal arrows - path of the horizontal photon, both photons travel distance 2*R=2*L*sqrt(1+1/4). I find contracted length by subtracting x1(path of the emitter when horizontal photon hits the wall) from the path of horizontal particle when it hits the wall.

 

[aliinuur]

Difficult to tell if you don't show your derivation.

Note that you can use $\LaTeX$ to lay out maths legibly. (If it wasn't working when you wrote your first post it should be now because I've used LaTeX - refresh the page if you still don't see it rendered.)

i've posted my derivation as reply above

 

[PeterDonis]

i've posted my derivation as reply above

Images are not acceptable for this. Please use the PF LaTeX feature to post equations directly.

 

[Ibix]

Don't use a specific $v$ - you can always put a number in later.

Let's say it takes time $T$ for the pulses to return. The vertical pulse travels the same distance on both legs, which must therefore be $cT/2$. The box travels distance $vT/2$ horizontally and is $L_0$ across. Therefore Pythagoras tells us that $(cT/2)^2=(vT/2)^2+(L_0)^2$, or $$T^2=\frac{(2L_0)^2}{c^2-v^2}$$
Now consider the horizontally propagating pulse. It doesn't travel the same distance in the positive and negative directions. If the box has length $L$, what time does the reflection happen? And then what time does the return happen?

Quote or reply to my post to see how I used LaTeX. It's easy enough to learn.

 

[Vanadium 50]

(1) As others have said, please post your derivation in a form that is easy to understand. You asked for help; make it easy for people to help you.

(2) Label all events.

(3) Did you correctly consider the relativity of simultanety?

 

[aliinuur]

Don't use a specific $v$ - you can always put a number in later.

Let's say it takes time $T$ for the pulses to return. The vertical pulse travels the same distance on both legs, which must therefore be $cT/2$. The box travels distance $vT/2$ horizontally and is $L_0$ across. Therefore Pythagoras tells us that $(cT/2)^2=(vT/2)^2+(L_0)^2$, or $$T^2=\frac{(2L_0)^2}{c^2-v^2}$$
Now consider the horizontally propagating pulse. It doesn't travel the same distance in the positive and negative directions. If the box has length $L$, what time does the reflection happen? And then what time does the return happen?

Quote or reply to my post to see how I used LaTeX. It's easy enough to learn.

Here, v=c/2.
1. When the vertical photon (photon V) hits ceiling of the box, thus traveling distance R, the emitter travels distance $$\frac{L_0}{2}$$ along the x-axis, the horizontal photon (photon H) also travels distance R along the x-axis, so that it is P-distance ahead of where it must meet the emitter to be absorbed with photon V simultaneously.
2. Photon V is reflected and travels one more distance R to meet the emitter at $$x=L_0$$ . Photon H must be reflected after some distance D and return to the emitter at $$x=L_0$$ traveling overall distance R, thus R=P+2D.
3. It turns out that $$D=\frac{L_0}{2}$$ .
4. Contracted Length L = (x-position of photon H) subtract (x-position of the emitter) when photon H hits the wall. So photon H travels distance D and hits the wall, while the emitter travels distance D/2 (since v=c/2), so x-position of the emitter is $$x_1= \frac{L_0}{2}+\frac{D}{2}$$ while photon H position is $R+D$ .
5. Expressing R, D, P, x1 by $$L_0$$ I've got $$L=L_0(\sqrt{1+\frac{1}{4}}-\frac{1}{4})$$

 

[Ibix]

View attachment 351919
Here, v=c/2.
1. When the vertical photon (photon V) hits ceiling of the box, thus traveling distance R, the emitter travels distance $$\frac{L_0}{2}$$ along the x-axis

No - if the box travels a distance $L_0/2$ at $c/2$ then the light pulse must have travelled $L_0$, which is not far enough to have reached the top of the box if it was travelling diagonally. The pulse will not reflect at the time you say. See the calculation in my last post.

I haven't checked the rest of your work, but the error in the first line will propagate through it, I'm afraid.

 

[aliinuur]

yeah, here is my wrong. Thanks!!!!!!!!! Now it fits perfectly!