Angled Length Contraction (Relativity)

[Blanchdog]

Homework Statement

Find the length of a meter stick if in the laboratory frame it is angled at 60 degrees relative to the direction of motion, and it is traveling at .8 c.

I have calculated gamma to be 1.667

Homework Equations

Lorentz transformations

The Attempt at a Solution

a' = l' cos(x), b = l' sin(x)
a = cos(60), b = sin(60)
a' = a/gamma, b' = b
l' cos(x) = cos(60)/gamma, l' sin(x) = sin(60)
combine by eliminating l'
cos(60)/(cos(x) gamma) = sin(60)/sin(x)
tan(x) = tan(60) gamma
x = arctan(tan(60) gamma)
plug x back into solve for l'

l' = cos(60)/(cos(x) gamma)

This gives me a result of about 91 cm, but the book says it should be 82 cm. Where did I go wrong?

 

[PeroK]

Homework Statement

Find the length of a meter stick if in the laboratory frame it is angled at 60 degrees relative to the direction of motion, and it is traveling at .8 c.

I have calculated gamma to be 1.667

Homework Equations

Lorentz transformations

The Attempt at a Solution

a' = l' cos(x), b = l' sin(x)
a = cos(60), b = sin(60)
a' = a/gamma, b' = b
l' cos(x) = cos(60)/gamma, l' sin(x) = sin(60)
combine by eliminating l'
cos(60)/(cos(x) gamma) = sin(60)/sin(x)
tan(x) = tan(60) gamma
x = arctan(tan(60) gamma)
plug x back into solve for l'

l' = cos(60)/(cos(x) gamma)

This gives me a result of about 91 cm, but the book says it should be 82 cm. Where did I go wrong?

I can't follow what you are trying to do at all. When you say ##a = \cos(60)##, what is ##a##?

 

[Blanchdog]

I can't follow what you are trying to do at all. When you say ##a = \cos(60)##, what is ##a##?

I'm treating the meter stick as the hypotenuse of a right triangle. a is aligned with the x-axis (the direction of motion) and b is aligned with the y-axis (perpendicular to the direction of motion).

 

[PeroK]

I'm treating the meter stick as the hypotenuse of a right triangle. a is aligned with the x-axis (the direction of motion) and b is aligned with the y-axis (perpendicular to the direction of motion).

In what frame is this? If you are in the rest frame of the stick, then the angle is not ##60°##. And if you are in the lab frame, then the lengths are not simply ##\cos## and ##\sin## as the stick is not of unit length in this frame.

 

[Blanchdog]

`

In what frame is this? If you are in the rest frame of the stick, then the angle is not ##60°##. And if you are in the lab frame, then the lengths are not simply ##\cos## and ##\sin## as the stick is not of unit length in this frame.

The angle is 60 degrees in the lab frame, meaning that the angle in the frame of the meter stick will be larger than 60 degrees. I denote this by x in the problem. And I can see why the I cannot use a unit length there, but I'm not sure how I can solve it since I know either l and don't know the angle or I know the angle and don't know l, depending on the reference frame. In this case, I need to figure out l (l') in the lab reference frame. I feel like there should be some sort of Lorentz angle transformation, but I've only see that for velocities, not distances.

 

[PeroK]

`The angle is 60 degrees in the lab frame, meaning that the angle in the frame of the meter stick will be larger than 60 degrees. I denote this by x in the problem. And I can see why the I cannot use a unit length there, but I'm not sure how I can solve it since I know either l and don't know the angle or I know the angle and don't know l, depending on the reference frame. In this case, I need to figure out l (l') in the lab reference frame. I feel like there should be some sort of Lorentz angle transformation, but I've only see that for velocities, not distances.

What do you know in the lab frame? Write down what you can. Hint: take the length in the lab frame to be ##L## with ##L^2 = a^2 + b^2##, where ##a, b## are the components of length in the direction of motion and perpendicular to the direction of motion respectively.

Hint: forget the angle in the rest frame of the rod - you don't need it.

Suggestion: use ##v, \gamma, \theta## and solve the problem generally. At this level, you should be using algebra - for many reasons - in preference to plugging in numbers at the outset.

 

[Blanchdog]

What do you know in the lab frame? Write down what you can.

Going to change notation here from above, everything denoted with a subscript 0 like this (X₀) will be the value for the meter stick's frame, everything without a subscript will be in the lab reference frame.

Lab Reference Frame:
L² = a² + b²
v = .8 c
γ = 1.667
a = a₀ / γ
b = b₀
a = L cos (θ)
b = L sin (θ)

Meter stick reference frame:
L₀ = 1
L₀² = a₀² + b₀²

Substituting into the final expression above:

L₀² = (aγ)² + b²
L₀² = (L cos(θ) γ)² + (L sin(θ))²
L₀² = L² ((cos(θ) γ)² + (sin(θ))²)
L = √(L₀²/(cos(θ)² γ² + sin(θ)²))

Plugging in numbers:

L = √(1/(cos(60)² * 1.667² + sin(60)²))
L = .832 m, which is the correct answer! Thanks for the help.

 

[PeroK]

As an aside, you can see the relationship between the angles is:

##\tan(\theta) = \frac{b}{a} = \frac{b_0}{a_0/\gamma} = \gamma \frac{b_0}{a_0} = \gamma \tan(\theta_0)##