Lorentz Transformations problem

AI Thread Summary
The discussion revolves around solving a Lorentz transformation problem involving a stick at rest in one frame and its observed length in another moving frame. The gamma factor was calculated as approximately 1.73, leading to Δx' being about 0.865L and Δy' being √(3)/2L, confirming no contraction in the y-direction. The correct length of the stick L' in the S' frame was determined to be 1.22L, indicating length contraction in the S frame. The angle of the stick with respect to the x'-axis was found to be exactly 45 degrees using trigonometric functions. Overall, the participants clarified concepts and confirmed calculations related to relativistic effects on length and angles.
Meekay
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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

------------
Relevant equations:

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

-----------
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 I am sort of lost.

Thanks for any help.
 

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Hello Meekay,

Welcome to Physics Forums! :smile:

Meekay said:
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

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

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

angle' = _____degrees

------------
Relevant equations:

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

-----------
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?

Ignoring minor rounding differences, that looks reasonable to me. :approve:

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.

That also looks correct. :approve:

as for the next two I am sort of lost.

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

[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.]
 
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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 I am 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!
 
Meekay said:
So i got L' = 1.22 L . If I am 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')
'Looks good to me! :smile:

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