# Measurement of Angles and Lengths With Length Contraction

• Ascendant0
Ascendant0
Homework Statement
A meter stick is moving with speed 0.8c relative to frame S. What is the stick's length if the stick is at ## 60^{\circ} ## to its velocity v, as measured in frame S?
Relevant Equations
## l = l_\circ / \gamma ##
I'm having a hard time since the angle is being measured in frame ##S##.

What I have put together so far is ## l_y = l_{y0} = 1m(sin(\theta_\circ)) ##
And ## l_x = l_{x0}/\gamma ##

Where ## l_{y0} ## and ## l_{x0} ## are the respective x and y values of the meter stick in the proper lengths (in ##S'## reference frame moving at the same speed as the meter stick)

I know the y values won't change because there's no movement in that direction. What I can't figure out is how to figure out what the angle would be in the ##S'## system, so I can find the proper angle to calculate the length based off that.

I tried using tan to relate the equations, but it hasn't gotten me anywhere yet. Help would be greatly appreciated, as I've been racking my brain on this one for a while, and just can't seem to figure it out.

Length contraction occurs in the direction of motion, so imagine the meter stick as a triangle with a component parallel to the motion and a component perpendicular to the motion.

Notice that the side parallel to the motion changes length; the side perpendicular does not, so use:

##tan(\theta) = perpendicular.to.motion.side/parallel.to.motion.side##

Using the Pythagorean theorem, you can compute the length contraction of the meter stick.

PeroK and Ascendant0
jedishrfu said:
Length contraction occurs in the direction of motion, so imagine the meter stick as a triangle with a component parallel to the motion and a component perpendicular to the motion.

Notice that the side parallel to the motion changes length; the side perpendicular does not, so use:

##tan(\theta) = perpendicular.to.motion.side/parallel.to.motion.side##

Using the Pythagorean theorem, you can compute the length contraction of the meter stick.
I appreciate the help, and I have tried to use tan for the eq, but I'm still not getting there. I tried swapping around values for tan, but I can't seem to solve for any of it

I have:
## l_y=l_{y0} = lsin(60) ##

## l_x = l_{x0}\gamma ##

## tan(60^{\circ}) = l_y/l_x = lsin(60)/l_x ## and from that:

## l_x=1/2l ##

But, I still don't see any way to combine them, or plug them into a Pythagorean Theorem to solve for any of the sides (or the length ##l##) that I don't have yet. I mean conceptually, I know very well I have enough information here to solve for all the sides, but for some reason, I just can't piece it together right to get there. Very frustrating.

I think you aren't organising your information well, which is important to do when you are doing frame changes because it's very easy to lose track of what you do and don't know in which frame. Try to work in one frame, then transform, then fill in the blanks in the other.

Say the quantity you are looking for is ##l##. Working solely in ##S##, what are the ##x## and ##y## extents of the rod?

Transform these into ##S'##, the rest frame of the rod, so you know the ##x'## and ##y'## extents.

What additional information do you have in ##S'## that you can use with these values?

I'd strongly recommend using notation consistent with the frame labelling. So for example the ##x## extent of the rod might be ##l_x## in ##S## and it becomes either ##l'_x## or ##l_{x'}## in ##S'## (the latter is more consistent, I think, but possibly easier to "lose" the prime so the former might be better).

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PeroK and Ascendant0
Ibix said:
I think you aren't organising your information well, which is important to do when you are doing frame changes because it's very easy to lose track of what you do and don't know in which frame. Try to work in one frame, then transform, then fill in the blanks in the other.

Say the quantity you are looking for is ##l##. Working solely in ##S##, what are the ##x## and ##y## extents of the rod?

Transform these into ##S'##, the rest frame of the rod, so you know the ##x'## and ##y'## extents.

What additional information do you have in ##S'## that you can use with these values?

I'd strongly recommend using notation consistent with the frame labeling. So for example the ##x## extent of the rod might be ##l_x## in ##S## and it becomes either ##l'_x## or ##l_{x'}## in ##S'## (the latter is more consistent, I think, but possibly easier to "lose" the prime so the former might be better).
I appreciate the feedback, but I have all that as well. I have written out both triangles, all sides, all the trig for all sides, all pythagorean theorems for both underneath each individual triangle, all possible alternative eqs for each side for both, literally every single thing I could come up with. It's just I have to find 4 sides, albeit 2 of them are equal, but still unknown. So, essentially two triangles, three unknowns, and I can't see how to set them up so that I can cancel out other variables to get to one single variable for any given equation.

I have no doubt that tan is required, but I don't see how to set it up in a way to get down to just one variable. I'm thinking it's some property of tan that I'm not aware of. I'm coming back to this stuff after an 8yr layoff, so while I brushed up as best as I could before the semester, I'm still really rusty at this point.

I don't think tan is required anywhere.

If you have ##l'_x## and ##l'_y##, how do they relate to the length of the rod in ##S'##?

Ascendant0 said:
I have:
## l_y=l_{y0} = lsin(60) ##

## l_x = l_{x0}\gamma ##

## tan(60^{\circ}) = l_y/l_x = lsin(60)/l_x ## and from that:

## l_x=1/2l ##
You took the scenic route getting there. You could have reached the same conclusion in one step: ##l_x = l\cos 60^\circ = l/2##.

Ascendant0 said:
I appreciate the feedback, but I have all that as well. I have written out both triangles, all sides, all the trig for all sides, all pythagorean theorems for both underneath each individual triangle, all possible alternative eqs for each side for both, literally every single thing I could come up with. It's just I have to find 4 sides, albeit 2 of them are equal, but still unknown. So, essentially two triangles, three unknowns, and I can't see how to set them up so that I can cancel out other variables to get to one single variable for any given equation.

Could you write out the equations you have for the two triangles so we can see what you have? Also, any equations relating the lengths of the sides between the two frames as well.

PeroK
Note that you don't have to find the angle in the stick's rest frame, but that should be a simple calculation using tangents.

There is a standard trig identity for tan and sec, which allows you to get cos and sin from tan.

vela said:
You took the scenic route getting there. You could have reached the same conclusion in one step: ##l_x = l\cos 60^\circ = l/2##.

Could you write out the equations you have for the two triangles so we can see what you have? Also, any equations relating the lengths of the sides between the two frames as well.

Here is 1 of 4 different times I've written it all out now. I'm just not seeing where to make the link. I use pythagorean theorem, and when I substitute, I end up canceling everything out and just getting ## l_x = l_x ##, which is obviously worthless.

**CORRECTION TO ABOVE IMAGE** - In ##S'## list, it should be ## l'^2 = lx'^2 + ly'^2 ##

I have substituted with tangents, that gets me nowhere as well.

I've put hours and hours into this and it's driving me nuts.

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##l'^2=l_{x'}+l_{y'}## is wrong.

That has one thing you know and two you don't. Can you correct that and eliminate the things you don't know?

Then can you eliminate the two variables you don't know in the equation you get?

Then is there anything left to eliminate?

Post each answer you get if you get stuck.

Ascendant0
I'm on my phone, so it's easier to use ##L## for the length of the stick in its rest frame and ##l## in the frame where it's moving.

First, we have:$$L^2 = L_x^2 + L_y^2$$Then, we know$$L_x = \gamma l_x, \ L_y = l_y$$Finally, we know$$l_x = l\cos \theta, \ l_y = l\sin \theta$$That gives you ##L^2## in terms of ##l^2##, ##\gamma## and ##\theta##.

Ascendant0
I FINALLY got it when I got home this afternoon. I did end up using the tangents after I finally saw how to cancel things out and get it down to one variable

In case anyone else has this problem, the direction I went was this:

## tan(60) = l_y/l_x ## and ## tan(\theta') = l_y'/l_x' ##

## l_y = l_y' ## and ## l_x\gamma = l_x' ##

Using that information, I set the two tangents as a ratio over each other, the sides all cancel each other out, and I'm left with ## \theta ##. Just basic trig from there out.

Thank you so much for everyone's help. I did correct the typo you cited earlier Ibix (I was in a rush at the time), and I do plan to try to go in the direction both you and PeroK were citing above, just so I can get down a different method as well for future problems. Hopefully if anyone else runs into this issue down the road, this will get them through it

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