# Lorentz Transformations problem

1. Jun 5, 2014

### Meekay

Hello all,

I have an exam on Monday and am having trouble with this problem, any help would be greatly appreciated!

Q: A straight stick of length L' is at rest in the moving S' frame. The stick appears to have length L in the S frame. The S' frame is moving at a velocity √(2/3) c. Calculate the following quantities:

gamma = ______
Δx' = ______L
Δy' = ______L

And calculate the length of the stick L' as observed in the S' frame.

ΔL' = ______L

And at what angle (with respect to the x'-axis) is the stick L' observed to be in the S' frame?

angle' = _____degrees

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Relevant equations:

1/Sqrt[1 - (β])^2]
Δx' = gamma(Δx - vΔt)

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My attempt:

the gamma factor is = 1/√[1 - (√[2/3])^2] which is 1.73

then I use Δx' = gamma * x since Δt = 0

so Δx' would be gamma/2 in units of L? - (divided by 2 due to the nature of a 30 60 90 triangle?) so Δx' = .865 L?

Then Δy' is equal to √(3)/2 L because there is no contraction of length in the y direction because the motion is along the x-axis, so √(3)/2 is the only factor applied due to the nature of a 30 60 90 triangle.

as for the next two im sort of lost.

Thanks for any help.

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Last edited: Jun 5, 2014
2. Jun 5, 2014

### collinsmark

Hello Meekay,

Welcome to Physics Forums!

I'm a little confused about the delta on L'. Should that just be L'?

Ignoring minor rounding differences, that looks reasonable to me.

That also looks correct.

The Pythagorean theorem should come in useful. Follow that up with your favorite inverse trigonometric function (arctan, for example).

[Edit: By the way, if you're clever and keep all answers in terms of fractions and roots, the last answer -- the angle -- should be obvious even without a calculator.]

Last edited: Jun 5, 2014
3. Jun 5, 2014

### Meekay

Thank you for your your reassurance and help. And yes it is L' and not a delta L'. I got a bit carried away with deltas I suppose. And wow that last part was very simple, I feel pretty dumb now, I was trying to use a relativistic type equation.

So i got L' = 1.22 L . If im right, that does make sense because the stick is at rest in the S' frame and its length is observed to be contracted in the S frame.

And for the angle I got ~ 45 degrees using sin^-1(delta y'/L')

Thanks again for the help!

4. Jun 6, 2014

### collinsmark

'Looks good to me!

[Edit: btw, if you keep things in terms of fractions and roots, you'll find that the angle is not just approximately 45 deg, it's exactly 45 deg.]

5. Jun 6, 2014

### Meekay

Awesome, thanks. And okay I gotcha, I will next time. I need to brush up on my trig.