Find the contraction of angles seen by an observer

[guyvsdcsniper]

Homework Statement

Three metal rods are joined to form a triangle so that the interior angles of the triangle are: 30◦, 60◦, and 90◦ The triangle is at rest in spacecraft so that is entirely in the x'y'-plane with it’s hypotenuse along the y'-axis. The spacecraft is moving away from a stationary observer in the x direction at speed c/2. . Determine the angles of the
triangle as measured by the stationary observer.

Relevant Equations

sqrt(1-v^2/c^2)

I am trying to follow the work to this question but am stumped at steps 3 and 4.

I am confused as to where the cos^2(90+θ) comes from? I can see it is used to invoke sin into the equation since we have that value. Is it because we are only measuring the x-component of the movement, so we need to find the sin equivalent of cos?

 

[haruspex]

Before I try to figure out the calculation posted, wouldn’t it be simpler to drop a perpendicular from C to AB, meeting it at D, and consider the contraction of CD?
Working that way I get $\tan(\alpha)=3/2$, giving 56.3°.

 

[TSny]

It looks like the first page of the calculations finds the unprimed length $l_{AC}$ by Lorentz contracting the primed length $l'_{AC}$. This is done by using the component of the relative velocity of the frames that is parallel to the side AC according to the primed frame: $v \sin{\alpha'}$.

That seems to work!

But, on the second page, we find

This is not correct, since the triangle ABC is not a right triangle in the unprimed frame.

 

[haruspex]

This is not correct, since the triangle ABC is not a right triangle in the unprimed frame.

2 minutes later, I arrive at the same conclusions.

 

[guyvsdcsniper]

2 minutes later, I arrive at the same conclusions.

Was my assessment of why there was a cos^2(90+θ) correct? Since the last part of the solution provided is incorrect, how can I go about getting to the right answer?

 

[haruspex]

Was my assessment of why there was a cos^2(90+θ) correct?

As @TSny wrote in post #3, that comes from finding the component of v parallel to AC.

how can I go about getting to the right answer?

As I wrote in post #2, by finding the contraction of AD instead.

 

[guyvsdcsniper]

As @TSny wrote in post #3, that comes from finding the component of v parallel to AC.

As I wrote in post #2, by finding the contraction of AD instead.

Do you think you can provide a drawing of your explanation on post #2? I am having hard time visualizing it and understanding how the angles would be effected.

 

[haruspex]

Do you think you can provide a drawing of your explanation on post #2? I am having hard time visualizing it and understanding how the angles would be effected.

Draw the right angled triangle ABC' and drop a perpendicular from C' to meet AB at D. Under the contraction, we can leave A, B where they are and contract C'D to CD.
Express $\tan(\alpha)$ in terms of AD, CD. Similarly $\tan(\alpha')$.

 

[guyvsdcsniper]

Draw the right angled triangle ABC' and drop a perpendicular from C' to meet AB at D. Under the contraction, we can leave A, B where they are and contract C'D to CD.
Express $\tan(\alpha)$ in terms of AD, CD. Similarly $\tan(\alpha')$.

Ok this is my interpretation of your description. So doing this allow me to measure just the x-direction of the triangle moving?

 

[PeroK]

Ok this is my interpretation of your description. So doing this allow me to measure just the x-direction of the triangle moving?
View attachment 296242

Yes. The distances $AD$ and $BD$ are the same in both frames and $CD$ is contracted in the frame of the "stationary" observer.

It might be worth calculating $\gamma$ as a square root before you go plugging in numbers to several decimal places.

 

[guyvsdcsniper]

Yes. The distances $AD$ and $BD$ are the same in both frames and $CD$ is contracted in the frame of the "stationary" observer.

It might be worth calculating $\gamma$ as a square root before you go plugging in numbers to several decimal places.

im a little confused on what comes next. so I have to apply the lorentz transform for length contraction. How does tan come into the lorentz transform. With how it was solved on the work I provided, they took a sin of the 30 and 60 degree angles. Here I just have tan

 

[PeroK]

im a little confused on what comes next. so I have to apply the lorentz transform for length contraction. How does tan come into the lorentz transform. With how it was solved on the work I provided, they took a sin of the 30 and 60 degree angles. Here I just have tan

You have a triangle with known side lengths. You just calculate the angles.

 

[guyvsdcsniper]

You have a triangle with known side lengths. You just calculate the angles.

So I calculated the unprimed length of CD and I got 56.3 degrees.
Im confused on what length tan30 is associated with?

 

[PeroK]

So I calculated the unprimed length of CD and I got 56.3 degrees.
Im confused on what length tan30 is associated with?

I don't know what that means.

You know $|AD|$, $|BD|$ and $|CD'|$. You know that $|CD| = |CD'|/\gamma$. So, you calculate $\alpha$ and $\beta$ from that using basic trig.

 

[PeroK]

Or, get $\tan \alpha$ in terms of $\tan \alpha'$ etc.

 

[guyvsdcsniper]

Or, get $\tan \alpha$ in terms of $\tan \alpha'$ etc.

IThis is how I am interpreting the situation. Is this still not correct?

 

[TSny]

Why do you have a factor of $\tan ^2 60^o$ inside the Lorentz contraction factor?

 

[guyvsdcsniper]

View attachment 296249
Why do you have a factor of $\tan ^2 60^o$ inside the Lorentz contraction factor?

I guess I was basing this off of what the original work had done. But thinking about it harder I can see why id doesn't make sense.

Im just not seeing what everyone else is seeing to conclude the problem.

Calculating Lcd=Lcd' \gamma I get .866. That should be the only length I need to calculate since the other two lengths are perpendicular.

Now that I have Lcd=lcd'(.866) what do I do. How do I calculate alpha and beta from this?

 

[TSny]

How would you express $\tan \alpha$ in terms of $L_{CD}$ and $L_{AD}$?

 

[guyvsdcsniper]

Tan(α) = L_AD/L_CD .

If I know L_CD=L_CD'*√ 3/2, then I just need to find L_AD?

 

[TSny]

Tan(α) = L_AD/L_CD .

This is not quite right.

If I know L_CD=L_CD'*√ 3/2, then I just need to find L_AD?

You don't need to find $L_{AD}$. The idea is that you can express $\tan \alpha$ in terms of $L_{CD}$ and $L_{AD}$. Likewise you can express $\tan \alpha'$ in terms of $L_{C'D}$ and $L_{AD}$. Since you know the relation between $L_{CD}$ and $L_{C'D}$, you should be able to see the relation between $\tan \alpha$ and $\tan \alpha'$ .

 

[PeroK]

Tan(α) = L_AD/L_CD .

If I know L_CD=L_CD'*√ 3/2, then I just need to find L_AD?

First, the tangent is opposite over adjacent.

And $\tan \alpha' = \frac{|C'D|}{|AD|}$.

That gives you a relationship between $\tan \alpha$ and $\tan \alpha'$. With the same idea for $\beta$ and $\beta'$.

PS Just saw the above post.

 

[guyvsdcsniper]

First, the tangent is opposite over adjacent.

And $\tan \alpha' = \frac{|C'D|}{|AD|}$.

That gives you a relationship between $\tan \alpha$ and $\tan \alpha'$. With the same idea for $\beta$ and $\beta'$.

PS Just saw the above post.

So then $\tan \alpha = \frac{|CD|}{|AD|}$?

 

[PeroK]

and then I can say
${|C'D|}={|CD|}$ ?

That's the one length that changes! That's what's length contracted!

 

[guyvsdcsniper]

That's the one length that changes! That's what's length contracted!

Sorry I edited my post cause I realized that was wrong.

$\tan \alpha' = \frac{|C'D|}{|AD|}$. I can see that, i got this wrong on my earlier post because I was looking at the wrong angle.

I don't see how I can relate this to $\tan \alpha$

 

[guyvsdcsniper]

Well I guess If AD is the same in both frames, then I can say AD = C'D/ $\tan \alpha'$ and also AD = C'D/ $\tan \alpha$

So then C'D/ $\tan \alpha'$ = C'D/ $\tan \alpha$ ?

 

[PeroK]

Sorry I edited my post cause I realized that was wrong.

$\tan \alpha' = \frac{|C'D|}{|AD|}$. I can see that, i got this wrong on my earlier post because I was looking at the wrong angle.

I don't see how I can relate this to $\tan \alpha$

$\tan \alpha' = \frac{|C'D|}{|AD|}$, $\tan \alpha = \frac{|CD|}{|AD|}$, $|C'D| = \gamma|CD|$

 

[PeroK]

Well I guess If AD is the same in both frames, then I can say AD = C'D/ $\tan \alpha'$ and also AD = C'D/ $\tan \alpha$

So then C'D/ $\tan \alpha'$ = C'D/ $\tan \alpha$ ?

?

 

[guyvsdcsniper]

?

Sorry, I guess I am just really struggling with this problem and the logic behind solving it.

 

[PeroK]

Sorry, I guess I am just really struggling with this problem and the logic behind solving it.

Okay, but we've done all the work except solve it. There's not much left to do except a final step which gives the answer.

You're honestly saying that you cannot finish the job from here:

$\tan \alpha' = \frac{|C'D|}{|AD|}$, $\tan \alpha = \frac{|CD|}{|AD|}$, $|C'D| = \gamma|CD|$

Note that $C' \rightarrow C$ is the only point on the triangle that changes from one frame to the the other. The line $ADB$ is unaffected by length contraction, as it is perpendicular to the direction of motion.