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.

 

[guyvsdcsniper]

Im just confused. I understand that we need to get C'D into the mix. I want this because it runs parallel to the direction of motion. Anything perpendicular will not get contracted, only going in the direction of motion will result in contracts so that's why we need the parallel components.

I know that $\tan \alpha' = \frac{|C'D|}{|AD|}$ and $\tan \beta' = \frac{|C'D|}{|DB|}$ . Tan allows me to relate something contracted in one frame to something that remains the same in both frames

I also know that C'D = ϒCD, which is the contracted length is equal to the uncontracted length times gamma. I know that gamma is equal to √(¾) is is √(3)/2 for this problem.

So this tells me that the contracted length is √(3)/2 CD.

So then let me say $\tan \alpha' = √(3)/2* \frac{|CD|}{|AD|}$ And I know that $\frac{|CD|}{|AD|}$ is tan60 degrees, which is equivalent to √3.

so √(3)/2 * √(3) = 1/2. therefore tan^-1(1/2) = 26.6 degrees?

Is this correct?

 

[guyvsdcsniper]

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: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.

Im sorry if its coming off that way . I genuinely want to learn how to approach this problem and I am not consciously trying to ask for the answer. I understand you all have put in the effort to help set the problem up and help me understand, but its just hard for me to really grasp it.

 

[PeroK]

So then let me say $\tan \alpha' = √(3)/2* \frac{|CD|}{|AD|}$ And I know that $\frac{|CD|}{|AD|}$ is tan60 degrees, which is equivalent to √3.

so √(3)/2 * √(3) = 1/2. therefore tan^-1(1/2) = 26.6 degrees?

Is this correct?

$\alpha' = 60$ degrees. The primed frame is the rocket frame.

 

[guyvsdcsniper]

$\alpha' = 60$ degrees. The primed frame is the rocket frame.

Your right. So then I should get this as my answer. And then I can just repeat the same for beta.

Then add those two up and subtract by 180 to get the 3rd angle?

 

[PeroK]

$\alpha$ can't be greater than $60$ degrees.

You need to be more careful in your calculations.