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If the rod is angled with respect to A's direction of motion,how would the rod look like to him?

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- Thread starter anantchowdhary
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If the rod is angled with respect to A's direction of motion,how would the rod look like to him?

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Draw a box, perpendicular to the direction of motion, around the rod. Then the depth of the box is contracted and the width and height of the box stays the same. From there you can calculate the effective contracted length.

If the rod is angled with respect to A's direction of motion,how would the rod look like to him?

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ok thnx a lot.Just wanted to be sure

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jtbell

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[edited to change the original [itex]\theta[/itex] to [itex]\theta_0[/itex], to make clear that this angle is the "proper" angle of the rod with respect to the x-axis, that is, the angle measured in the rod's rest frame.]

For simplicity, let the velocity be in the x-direction and the rod be oriented at an angle [itex]\theta_0[/itex] with respect to the x-axis in the rod's rest frame. The proper length of the rod has components

[tex]L_{0x} = L_0 \cos \theta_0[/itex]

[tex]L_{0y} = L_0 \sin \theta_0[/itex]

The x-component undergoes length-contraction but the y-component does not:

[tex]L_x = L_{0x} \sqrt { 1 - \frac{v^2}{c^2} }[/tex]

[tex]L_y = L_{0y}[/tex]

The length of the moving rod is

[tex]L = \sqrt{L_x^2 + L_y^2}[/tex]

Putting all these together I get

[tex]L = L_0 \sqrt{1 - \frac{v^2}{c^2} \cos^2 \theta_0}[/tex]

For [itex]\theta_0 = 0[/itex] (velocity parallel to the rod) this gives the usual length-contraction formula. For [itex]\theta_0 = 90^\circ[/itex] (velocity perpendicular to the rod) there is no contraction, as expected.

For simplicity, let the velocity be in the x-direction and the rod be oriented at an angle [itex]\theta_0[/itex] with respect to the x-axis in the rod's rest frame. The proper length of the rod has components

[tex]L_{0x} = L_0 \cos \theta_0[/itex]

[tex]L_{0y} = L_0 \sin \theta_0[/itex]

The x-component undergoes length-contraction but the y-component does not:

[tex]L_x = L_{0x} \sqrt { 1 - \frac{v^2}{c^2} }[/tex]

[tex]L_y = L_{0y}[/tex]

The length of the moving rod is

[tex]L = \sqrt{L_x^2 + L_y^2}[/tex]

Putting all these together I get

[tex]L = L_0 \sqrt{1 - \frac{v^2}{c^2} \cos^2 \theta_0}[/tex]

For [itex]\theta_0 = 0[/itex] (velocity parallel to the rod) this gives the usual length-contraction formula. For [itex]\theta_0 = 90^\circ[/itex] (velocity perpendicular to the rod) there is no contraction, as expected.

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Well...shudnt ust the radius of the rod change?This is as the rod is moving parallel to the x axis even though it is angled.Its motion is not at an angle

And i just posted one of the equations :[tex] L_0_y = \frac {L_y} { \sqrt {1 - \frac {{0}^2} {c^2}}} [/tex] in another thread and a member said it was incorrect.I said that if the relative velocity in the direction of the y axis is zero, then [tex]L_0_y=L_y[/tex]

Was i wrong?

And i just posted one of the equations :[tex] L_0_y = \frac {L_y} { \sqrt {1 - \frac {{0}^2} {c^2}}} [/tex] in another thread and a member said it was incorrect.I said that if the relative velocity in the direction of the y axis is zero, then [tex]L_0_y=L_y[/tex]

Was i wrong?

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- #6

Doc Al

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The motion of the rod is along the x-axis, thus only length components parallel to the x-axis will be "contracted". (Why do you think this only affects the radius of the rod??)Well...shudnt ust the radius of the rod change?This is as the rod is moving parallel to the x axis even though it is angled.Its motion is not at an angle

Nothing wrong with this (which is exactly as jtbell explained).And i just posted one of the equations :[tex] L_0_y = \frac {L_y} { \sqrt {1 - \frac {{0}^2} {c^2}}} [/tex] in another thread and a member said it was incorrect.I said that if the relative velocity in the direction of the y axis is zero, then [tex]L_0_y=L_y[/tex]

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thanks.No i just meant to say if the rod was a cylinder its radius as viewed would change.

So wouldnt the rod just appear to get thinner and thinner?

Also the rod isnt a vector!Its velocity is a vector which is just parallel to the x axis.

Was my question misunderstood?

So wouldnt the rod just appear to get thinner and thinner?

Also the rod isnt a vector!Its velocity is a vector which is just parallel to the x axis.

Was my question misunderstood?

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jtbell

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Now imagine that your rod is embedded inside this rectangular object, extending from one corner to the diagonally opposite corner. The ends of the rod stay at the corners of the object as the object contracts as described above.

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jtbell

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In a frame in which the rod is moving, the angle is different,

To see this, start with the angle of the rod in the "moving" frame:

[tex]\cos \theta = \frac{L_x}{L}[/tex]

Using the other equations in my previous post, you can find

[tex]\cos \theta =\frac {\cos \theta_0 \sqrt {1 - \frac {v^2}{c^2}}} {\sqrt {1 - \frac {v^2}{c^2} \cos^2 \theta_0}}[/tex]

See what happens when [itex]\theta_0 = 0[/itex] and when [itex]\theta_0 = 90^\circ[/itex]?

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Doc Al

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The only dimension that "contracts" is the component parallel to the relative velocity, which is the x-component, in this example.Why would we need to take the component of the rod?Shouldnt we be just taking the component of the relative velocity [tex]\vec v[/tex] ?

And in this case as the direction of the vector is parallel to the x axis,even though the length of the rod makes an agle with the axis ,the rod should just go thinner!

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so the angle wont change isnt it?

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Doc Al

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As jtbell explained in detail, the angle the stick makes with the x-axis will be different when measured from a moving frame. How could it not be? The angle the stick makes is given by [itex]\tan\theta = L_x/L_y[/itex]--since [itex]L_x[/itex] changes while [itex]L_y[/itex] does not, [itex]\theta[/itex] must change.so the angle wont change isnt it?

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hmmm...so this is because space is contracting....

- #16

Integral

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Well...shudnt ust the radius of the rod change?This is as the rod is moving parallel to the x axis even though it is angled.Its motion is not at an angle

And i just posted one of the equations :[tex] L_0_y = \frac {L_y} { \sqrt {1 - \frac {{0}^2} {c^2}}} [/tex] in another thread and a member said it was incorrect.I said that if the relative velocity in the direction of the y axis is zero, then [tex]L_0_y=L_y[/tex]

Was i wrong?

You were not referring to the dimensions of the rod in that other thread, but to a y component of motion. That is why I said that was wrong.

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