Lorentz Transformations problem

[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

------------
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.

 

[collinsmark]

Hello Meekay,

Welcome to Physics Forums!

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.

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.

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).

[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.]

 

[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 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!

 

[collinsmark]

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!

[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.]

 

[Meekay]

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