# Homework Help: Length Contraction of Bridge rotated 30 degrees

1. Dec 7, 2015

### marimuda

1. The problem statement, all variables and given/known data

A Train is driving with a uniformed velocity. v=1/4 c. The train is driving under a bridge, which is rotated \theta=30 degree in relation to the railway. The bridges length is 20 m long in a reference frame of the bridge (stationary bridge).

Calculate the length of the bridge, seen from the train (reference frame, follows the train)

2. Relevant equations

Not sure

Projection length

$\Delta x’=L_0 cos\theta_0$

$\Delta y’=L_0 sin\theta_0$

$]\Delta x’=]\Delta x$ And $]\Delta y’=]\Delta y$

The Length L of the bridge measured from the train.

$L=\sqrt{(\Deltax)^2+(\Deltay)^2}=\sqrt{L_0 {\frac (-(-v^2 cos^2 \theta_0}{c^2}+1)$ Or written as $L_0 (1- \beta ^2 \cos^2 \theta_0 )^(0.5)$

3. The attempt at a solution

$L=\sqrt{L_0 {\frac (-(-v^2 cos^2 \theta_0}{c^2}+1)$

$L=\sqrt{20 m {\frac (-(-0.25c^2 cos^2(30 }{c^2}+1)$

$L=4.472 m$

*note: think the result comes out a bit low, (assumption based on the Lorentz factor gamma(v)=1.0328)
*Apologize for the mess in the thread, can't get my commands to work in Latex form

2. Dec 7, 2015

### .Scott

Your computation of the Lorentz Factor is correct: 1/sqrt(1-(1/16)) = sqrt(16/15) = 1.0328
So, it should be clear that, with that factor, you aren't going to go from 20 meters to 4.472 meters.

Show me these intermediate computations and I point out your error:
1) Delta X and Delta Y of 20 meter bridge from the bridge's reference frame.
2) Delta X and Delta Y of 20 meter bridge from the trains's reference frame.
3) Length of bridge from trains reference frame.

3. Dec 7, 2015

### Staff: Mentor

What is the length contraction factor in the direction of train travel? What is the length contraction factor in the direction perpendicular to the direction of train travel?

4. Dec 7, 2015

### marimuda

Hey, here's my calculations with all intermediate computations .
Ups, can see that there is some Faroese in the image, hopefully it doesn't make to confusing.

5. Dec 7, 2015

### Staff: Mentor

There is no length contraction in the y direction. The contraction in the y direction is 1.0

Chet

6. Dec 7, 2015

### marimuda

So, I just solve Lx=L0x/gamma(v) and Ly=L0y?

7. Dec 7, 2015

Sure.

8. Dec 7, 2015

### PeroK

I get a slightly different answer.

Now that you're studying more advanced topics, you may want to think about using algebra more than arithmetic. For example, why is the value of $c$ relevant? You are given the velocity $v = c/4$ so $c$ cancels out in the expression for $\gamma$.

All the intermediate computations are unnecessary. It makes your work difficult to check, because it all boils down to putting numbers in a calculator and checking what comes out.

Also, ask yourself this: what if the angle were 40 degrees or the speed $c/5$? Or,if the length of the bridge were 50m? Would you go right back to the start and plug in some different numbers with a whole set of different intermediate computatons? Or could you use some elementary algebra to solve for any angle $\theta$ and any velocity $v = \beta c$? And then simply plug the specific numbers in once at the end?

9. Dec 7, 2015

### marimuda

Thanks for the responds.. Is your answer 19.526 m ? , that's what I got when I did the correction chestermiller pointed out. I know that I need to do some algebraic gymnastics to set it up more efficiently, was just more thinking about if I even was in the right ball park.
We don't have any special relativity course in school, and my 'final project' is about special relativity, So it's kind of new.
Isn't that difficult to cook down one equation and then isolate θ . But I can see that c cancels out since the velocity is given in terms of c units

10. Dec 7, 2015

### PeroK

It's not difficult and in fact you might learn a couple of interesting things from it.

Hint calculating $L'^2$ makes it easier.

Yes, that's what I got.

11. Dec 11, 2015

### marimuda

Apologize for bringing this post up again, but I'm still working in the problem, even tho we came to the solution and PeroK gave some brilliant advice, It may be that I need some mathematical clarity, or just mentally burned out by this project.
The thing that I'm trying to figure out is, how could I solve the same problem in 4-vector notation?
I don't even have any sensible 'attempted solution'

12. Dec 11, 2015

### PeroK

I guess you mean by using the Lorentz Transformation? Basically, any problem you can solve by using the concepts of time dilation and length contraction, you can solve more generally by using Lorentz. What do you know about the Lorentz Transformation? Do you know the term "proper" length?

13. Dec 11, 2015

### marimuda

Yes, I know the term proper length, I've watched Leonard Susskind lectures on special relativity, and checked several books on the topic so the terms don't sound that unfamiliar.
With other words, I know how to do the Lorentz transformation 'brute force', I just don't know how I can write it up efficiently into 4-vector / tensor notation and from that extract length contraction along x-axis and no contraction along y-axis. And get a length out of it.

14. Dec 11, 2015

### PeroK

Do you want to try using Lorentz on this problem?

15. Dec 11, 2015

### marimuda

I don't get what you mean? ,, Is there any other way?

16. Dec 11, 2015

### PeroK

Yes, just use length contraction.

17. Dec 11, 2015

### marimuda

Now you totally lost me.
the length contraction is a consequence of the Lorentz transformation? where does my chain drop off? :/

18. Dec 11, 2015

### PeroK

You've lost me! Perhaps you should restate your question from the beginning.

19. Dec 11, 2015

### marimuda

hehe apologize, yes maybe it needs a bit of clarification.
Alright, I've solved the problem above, (this is the same problem), where the length of the bridge observed is 19.526 m when the speed of the train is 1/4 c and the bridge is at a angle of 30 degree with a length of 20 m.
I asked my teacher if I could just use standard length contraction straight forward to solve the problem.
His answer was: " Yes, the question can be solved straight forwardly with the use of Lorentz transformation, but the problem (train problem , which I'm trying to solve) is a extension of the 4-vector objective in the project.
So solve the problem as extension of the 4-vector section

20. Dec 11, 2015

### PeroK

That means nothing to me, I'm sorry to say.

21. Dec 11, 2015

### marimuda

So, it's not only myself that gets lost by that request?
I was thinking along the lines of , if there is a Matrix (Lorentz transformation for my particular case ) which is multiplied on the four-vector x^μ, that when seen from the reference frame of the bridge ( stationary ) the length of the bridge is 20, and when seen from the moving observer, gets transformed the proper way, to give the result 19.526m . But I don't know how to do a matrix multiplication that does it for me.

22. Dec 11, 2015

### marimuda

I think that x^'μ = Λ x^μ , if μ goes from 1 to 3 (spatial components) , and Λ is the Lorentz transformation along the x-axis,is on the right track but have no clue how to get the components into there, (cos⁡(θ) , sin⁡(θ) ) and L_0 so that I can solve for L

23. Dec 11, 2015

### Staff: Mentor

You are asking what the Lorentz Transformation looks like if the direction of the relative velocity vector of the S' frame does not line up with the x direction of the S frame, correct? See Exercise 2.7 Boost in an Arbitrary Direction in Section 2.9 of Meisner, Thorne, and Wheeler, Gravitation.

Chet

24. Dec 11, 2015

### FactChecker

If you put L0 inside the square root, it is bound to be very small even at 0 relative velocity. That has to be wrong.

25. Dec 11, 2015

### marimuda

yes exactly, and how to write it.
Ouch, that was even uglier than I could imagine. My teacher must have swallowed a rock or something, we've just gone through 2 D vector algebra in class, and now he expects this from me from my final school project. hehe