Rod travelling at near c

1. Feb 27, 2007

anantchowdhary

Suppose person A and rod R are moving with a relative velocity nearing 'c'
If the rod is angled with respect to A's direction of motion,how would the rod look like to him?

2. Feb 27, 2007

MeJennifer

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.

Last edited: Feb 27, 2007
3. Feb 27, 2007

anantchowdhary

ok thnx a lot.Just wanted to be sure

4. Feb 27, 2007

Staff: Mentor

[edited to change the original $\theta$ to $\theta_0$, 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 $\theta_0$ with respect to the x-axis in the rod's rest frame. The proper length of the rod has components

$$L_{0x} = L_0 \cos \theta_0[/itex] [tex]L_{0y} = L_0 \sin \theta_0[/itex] The x-component undergoes length-contraction but the y-component does not: [tex]L_x = L_{0x} \sqrt { 1 - \frac{v^2}{c^2} }$$

$$L_y = L_{0y}$$

The length of the moving rod is

$$L = \sqrt{L_x^2 + L_y^2}$$

Putting all these together I get

$$L = L_0 \sqrt{1 - \frac{v^2}{c^2} \cos^2 \theta_0}$$

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

Last edited: Feb 28, 2007
5. Feb 27, 2007

anantchowdhary

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 :$$L_0_y = \frac {L_y} { \sqrt {1 - \frac {{0}^2} {c^2}}}$$ 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 $$L_0_y=L_y$$

Was i wrong?

Last edited: Feb 27, 2007
6. Feb 27, 2007

Staff: Mentor

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??)

Nothing wrong with this (which is exactly as jtbell explained).

7. Feb 27, 2007

anantchowdhary

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?

Last edited: Feb 27, 2007
8. Feb 27, 2007

Staff: Mentor

Consider a rectangular object whose sides are parallel to the x- and y- axes. The object moves along the x-direction. The width along the x-direction contracts, but the width along the y-direction does not, right?

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.

9. Feb 28, 2007

anantchowdhary

thats wat i had thought.But someone suggested that the angle would change?I dont think that would happen.Is it so?

10. Feb 28, 2007

Staff: Mentor

I've edited my post above to make clear that the angle in my equations is the angle that the rod makes with the x-axis, in the rod's rest frame. That is, it's the "original" angle.

In a frame in which the rod is moving, the angle is different, except if the angle is either 0 or 90 degrees to begin with. That is, if the rod is parallel to the relative velocity, in its own rest frame, it is parallel in the other frame. Likewise if the rod is perpendicular to the relative velocity.

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

$$\cos \theta = \frac{L_x}{L}$$

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

$$\cos \theta =\frac {\cos \theta_0 \sqrt {1 - \frac {v^2}{c^2}}} {\sqrt {1 - \frac {v^2}{c^2} \cos^2 \theta_0}}$$

See what happens when $\theta_0 = 0$ and when $\theta_0 = 90^\circ$?

11. Feb 28, 2007

anantchowdhary

Why would we need to take the component of the rod?Shouldnt we be just taking the component of the relative velocity $$\vec v$$ ?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!

12. Feb 28, 2007

Staff: Mentor

The only dimension that "contracts" is the component parallel to the relative velocity, which is the x-component, in this example.
All dimensions of the rod parallel to the x-axis will be contracted. In particular, the x-component of the rod's length will be contracted. (Sure, it will get thinner too, so what?)

13. Feb 28, 2007

anantchowdhary

so the angle wont change isnt it?

14. Feb 28, 2007

Staff: Mentor

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 $\tan\theta = L_x/L_y$--since $L_x$ changes while $L_y$ does not, $\theta$ must change.

15. Feb 28, 2007

anantchowdhary

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

16. Feb 28, 2007

Integral

Staff Emeritus

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.

17. Feb 28, 2007

anantchowdhary

umm.even if i was referring to the component of velocity,i meant to say as there is no velocity in the y direction,ther would be no length contraction in the y direction.Isnt that so?I was just seeing that as a reason to explain why there is no length contraction in the case given in the y direction