Modern Physics: Length Contraction


by smithg86
Tags: contraction, length, modern, physics
smithg86
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#1
Sep11-07, 03:51 PM
P: 60
1. The problem statement, all variables and given/known data

A rod of length L_0 moves with a speed v along the horizontal direction. The rod makes an angle of [tex]\vartheta[/tex]_0 with respect to the x'-axis.

(a) Show that the length of the rod as measured by a stationary observer is given by

L = L_0 [1-(v/c)[tex]^{2}[/tex] cos [tex]^{2} ( [/tex] [tex]\vartheta[/tex]_0 ) ] [tex]^{1/2}[/tex]

(b) Show that the angle that the rod makes with the x-axis is given by the expression
tan [tex]\vartheta[/tex] = [tex]\gamma[/tex] tan [tex]\vartheta[/tex]_0.
These results show that the rod is both contracted and rotated. (Take the lower end of the rod to be at the origin of the primed coordinate system.)

2. Relevant equations

[tex]\gamma[/tex] = [1- (v/c)^2]^(-1/2)

(Length contraction formula) L_0 = L / [tex]\gamma[/tex]

3. The attempt at a solution

The horizontal component of the rod in the x'-axis is:
x_0 = L_0 cos ( [tex]\vartheta[/tex]_0 )

Applying the length contraction formula, I was able to show (which differs from what I was supposed to show):

L = L_0 cos [tex]\vartheta[/tex]_0 / [1- (v/c)^2]^(1/2)

I do not understand why this is not the correct answer. I did not attempt the second part of the question.
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Doc Al
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#2
Sep11-07, 04:31 PM
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You found the horizontal component of the stick's length in the stationary frame, but you need the entire length. (What happens to the vertical component?)

Note: You made an error in computing the contraction of the horizontal component; see my comment in post #5.
smithg86
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#3
Sep11-07, 04:52 PM
P: 60
The vertical component does not contract. Using the Pythagorean theorem,

L^2 = L_0 ^2 sin^2 [tex]\vartheta[/tex]_0 + [L_0 ^2 cos [tex]\vartheta[/tex]_0 ] / [1 - (v/c)^2 ]

or

L = L_0 { sin^2 [tex]\vartheta[/tex]_0 + [ cos [tex]\vartheta[/tex]_0 ] / [1 - (v/c)^2 ] }^(1/2)

I've tried to get this into the desired form but I can't. Am I correct up to this point, or did I make a mistake somewhere?

smithg86
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#4
Sep11-07, 06:21 PM
P: 60

Modern Physics: Length Contraction


After playing around with the last expression for L, I was able to get something that looked a little more like what I am supposed to show:

L = L_0 * [tex]\gamma[/tex] [ 1-(v/c)^2 +(v/c)^2 cos^2([tex]\vartheta[/tex]_0) ]^(1/2)

...am I heading in the right direction? I can't help but notice that there is no gamma in the correct answer.
Doc Al
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#5
Sep11-07, 06:24 PM
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Quote Quote by smithg86 View Post
The vertical component does not contract. Using the Pythagorean theorem,

L^2 = L_0 ^2 sin^2 [tex]\vartheta[/tex]_0 + [L_0 ^2 cos [tex]\vartheta[/tex]_0 ] / [1 - (v/c)^2 ]
Yes, the vertical component does not contract. But you are making an error with the horizontal component (which I didn't spot before). The contraction of that component should be [itex]L_0 \cos \vartheta /\gamma[/itex], but you have it as [itex]L_0 \cos \vartheta \gamma[/itex].

Fix that and keep going. Hint: Use a trig identity to eliminate the sines.
smithg86
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#6
Sep11-07, 06:57 PM
P: 60
Quote Quote by Doc Al View Post
Yes, the vertical component does not contract. But you are making an error with the horizontal component (which I didn't spot before). The contraction of that component should be [itex]L_0 \cos \vartheta /\gamma[/itex], but you have it as [itex]L_0 \cos \vartheta \gamma[/itex].

Fix that and keep going. Hint: Use a trig identity to eliminate the sines.
Doc Al,

Thanks for your help, I was able to get the correct answer after fixing that error. However, I am more confused than before. Maybe I was misinterpreting the problem statement, but this is what I was thinking:

Draw 2 right triangles. The hypotenuse represents the rod, the two legs represent the horizontal and vertical components of the rod. The rod is moving to the right at velocity v. The bottom left vertex angle is [tex]\vartheta[/tex]_0, the bottom right vertex angle is a right angle. Assume both horizontal and vertical components are in the positive direction, from the origin.

Choose one of the triangles to have a smaller horizontal component. The rod represented by this triangle is in reference frame S'; the axes in S' are x' and y'. The other is in reference frame S; the axes in S are x and y. Call the vertical component of S' y_0, the horizontal x_0, the angle [tex]\vartheta[/tex]_0 and the hypotenuse L_0. Similarly, call the components in S: x,y,L and [tex]\vartheta[/tex]. S' is the frame of a stationary observer looking at the rod.

We can say:
x_0 = L_0 cos [tex]\vartheta[/tex]_0
y_0 = L_0 sin [tex]\vartheta[/tex]_0
x^2 + y^2 = L^2

Since x is greater than x_0 and gamma is greater than 1, I said:

(gamma) * x_0 = x

Which led me to the wrong answer. But I don't understand WHY x_0 = (gamma)*x is correct.
Can you explain your reasoning or correct my faulty logic?
Doc Al
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#7
Sep11-07, 07:12 PM
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Quote Quote by smithg86 View Post
We can say:
x_0 = L_0 cos [tex]\vartheta[/tex]_0
y_0 = L_0 sin [tex]\vartheta[/tex]_0
x^2 + y^2 = L^2
So far, so good. If I understand your notation, x is the horizontal component of the length as measured in the stationary frame (the frame that sees the stick move at speed v).

Since x is greater than x_0 and gamma is greater than 1, I said:

(gamma) * x_0 = x
You have it exactly backwards. Yes, gamma > 1; but according to the stationary frame, the horizontal component of the moving stick is contracted, thus x = x_0/(gamma).

(I just noticed that in your first post you had the "length contraction" formula backwards. Sorry for not spotting that!)

Let me know if this is unclear.

FYI: I have a little formula "cheat sheet" that you might find helpful: Basic Equations of Special Relativity
smithg86
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#8
Sep14-07, 10:36 AM
P: 60
thanks!
JayKo
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#9
Oct10-07, 05:15 PM
P: 128
Quote Quote by smithg86 View Post
thanks!


Ok i manage to crack the 1st part, the contracted length is basically square root of {contracted vertical length squared} + {vertical component squared without contraction}

but how about the 2nd part to prove that the angle is contracted and rotated? any1? thanks
JayKo
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#10
Oct10-07, 06:02 PM
P: 128
Ooops. i manage to get it done, thanks.
aj4
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#11
Sep14-10, 07:39 PM
P: 2
I understand how you got to the equation in post #5 but I dont understand the trig identity used in eliminating the sines from the equation
aj4
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#12
Sep14-10, 07:46 PM
P: 2
I was thinking sin^2 + cos^2 = 1 but I don't think that is correct


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