How Does Relativity Affect the Length and Orientation of a Moving Rod?

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Homework Help Overview

The discussion revolves around the effects of relativity on the length and orientation of a moving rod, specifically focusing on a scenario where an observer measures the rod's length and angle while moving at a significant fraction of the speed of light (0.995c). The original poster presents calculations for the proper length of the rod and its orientation in the rest frame, expressing a desire for a clearer mathematical approach.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the proper length and orientation angle using trigonometric relationships and Lorentz contraction, while expressing concerns about the clarity of their mathematical notation and calculations. Some participants suggest improving notation for clarity, while others discuss the adequacy of the original poster's mathematical approach.

Discussion Status

The discussion is ongoing, with participants providing feedback on notation and confirming the correctness of the original poster's calculations. There is no explicit consensus on a single method, but suggestions for clearer notation have been made.

Contextual Notes

The original poster indicates that they have found the correct values but are seeking a more straightforward method. There is mention of potential ambiguity in the mathematical approach, and participants are discussing notation conventions without resolving the underlying problem.

frankR
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An observer, moving at a speed of 0.995c relative to a rod as shown measures its length to be 2.00m and sees its length to be oriented at 30.0 degrees with respect to the direction of motion.

a) What is the proper length of the rod? (Length as measured at rest)

b) What is the orientation angle in a reference frame moving with the rod? (Again, in rest frame)


I've found the correct values, however the math seems ambiguous. I was wondering if there was another way to solve it, with cleaner mathematics.

Do = the length of the rod as measure when the rod is moving a .995c
Lo = the adjacent side of the triangle formed by the rod moving at .995

Lo = DoCos(30.0)

D = the actual length of the rod as observed from the rest frame
L = the actual length of the adjacent side of the triangle formed by the rod in the rest frame
@ = the angle formed between L and D.
h = height of the rest frame triangle

L = Lo/sqrt(1-v^2/c^2) = DoCos(30)/sqrt{(1-(.995c)^2/c^2)}

L = 17.34m

To find D you can use the following equations.

Tan(@) = h/L
Sin(@) = h/D
Cos(@)=L/D
D^2 = L^2 + h^2

I ended up getting this to solve for D.

D = sqrt{L^2 + L^2*Sin^2{Cos^-1(L/D)}^2}

My calculator solved it and got D = 17.26m or 17.42m.

The answers are 17.3m and 3.30 degree.

Is there a cleaner way to do it?

Am I missing something?
 
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I would only clean up the notation. Instead of using different letters for each quantity, I would use subscripts and primes.

L=Length of rod in frame S
Lx=Lcos(30)=component of length in direction of motion in frame S
Ly=Lsin(30)=component of length perpendicular to motion in S

and use L', Lx' and Ly' for the corresponding quantities in frame S'.
 
My math is okay then?

My notation is a lot cleaner on paper, a lot cleaner the second and third time I worked the problem out. How do you make subscripts?

Thanks for your help!
 
Originally posted by frankR
My math is okay then?

Yes, there's no other way to do it. But, using a more systematic notation will help whoever reads it (like the grader).

How do you make subscripts?

Subscripts:
If you type this: v[ sub ]0[ /sub ]
without the spaces you will get: v0.

Superscripts:
If you type this: v[ sup ]2[ /sup ]
without the spaces you will get: v2.

You can also combine the two as follows: v[ sub ]0[ /sub ][ sup ]2[ /sup ]
to get v02.
 
CoolThanks!
 
Something else: Regarding Greek letters. The ones in the "Smiley" menu (the ones you've been using) are really ugly, and what's worse they don't respond well to being superscripted or subscripted.

e[mu]
e[mu]

See?

Instead, use the forum code for this, like so:

& mu ;

without the spaces gives

μ

and for capitals, simply capitalize the name of the Greek letter, like so:

& Mu ;

to get

Μ

These will super- and subscript nicely, like so:

eμ
eμ
 

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