How does relativity affect the observation of moving objects?

[Pushoam]

Homework Statement

A.P. French q.no. 4.3

Homework Equations

t = L\c, (1)
L₀ = ϒ L

The Attempt at a Solution

The observer's frame is that frame w.r.t. which the meter stick is moving with speed 0.8c.
The observer sees the mid-point at (x,y) at a time t_o(measured from the clock situated at the observer's position)
while mid-point passes the point (x,y) at time t [measured from the clock kept at (x,y)] .
And it is assumed that all clocks are synchronized.

Now ,

t_o = (√(x²+y²) )/c + t, (3)
x = vt, v = 0.8 c

At t_o =0,
t = - (√(x²+y²) )/c, note that t is negative here.
c²t² = v²t² + y²
t² = 1/(1-0.64)c,
t =- 1/(0.6c)

So, At t_o =0, x= -4/3 m, y =0m
now, length of the meter-stick is L = 1(√(1-0.64)) m = 0.6 m
So, the left end will be at - (0.3+4/3)m and the right end will be -(4/3-0.3) at y=0.

Is this correct?

B)

The observer sees the mid-point at t_o= 1/c s.

C)To the observer the end points appear at 0.3 m left and right to the origin.What do I learn from solving this question?
Measurement of position of ends of stick at time t
And measurement of the time at which an event happens
will be different for observers located at different positions in the same reference frame.

Is my learning correct?

 

[TSny]

I believe part (a) is asking for the locations of the ends of the stick in the observer's frame at t = 0. It is not asking for where the ends appear (to the observer) to be located at t = 0. The answer to (a) does not depend on the particular location of the observer in this frame. (At least that's how I interpret the question).

I agree with your answer to (b).

Part (c) is asking for where the observer sees the ends of the stick to be located at the instant she sees the midpoint pass the origin.

 

[Pushoam]

Thank you for the reply.

I believe part (a) is asking for the locations of the ends of the stick in the observer's frame at t = 0. It is not asking for where the ends appear (to the observer) to be located at t = 0. The answer to (a) does not depend on the particular location of the observer in this frame. (At least that's how I interpret the question).

Then, in the observer's frame,
the length of the metrestick is 0.6m.
At t=0, the mid- point passes through the origin, so the ends will be 0.3 m left and right to the origin along the x-axis.

Part (c) is asking for where the observer sees the ends of the stick to be located at the instant she sees the midpoint pass the origin.

This is what I answered.

To the observer the end points appear at 0.3 m left and right to the origin.

Isn't this correct?

 

[TSny]

OK for part (a). For part (c) you must take into account the time it takes for light to travel from points of the stick to the observer.

 

[Pushoam]

For part (c)
At time t=0s, the mid- point of the stick passes through the origin w.r.t. observer's frame.
At time t = (1/c )s the observer sees the mid- point of the stick passing through the origin.
Certainly, the observer will see the left end at 0.3 m left to the origin at time t = [ { √[(0.3)²+1]}/c +0 ]s
Let's say that at time t =( 1/c )s, the observer sees the the left end at x_lm left to the origin.

w.r.t. observer's frame

The left end will be at x_lm left to the origin at time t_l .
Here, t_l = -| t_l |

At t_l,the mid point will be at a distance 0.8 c * | t_l | left to the origin.
As the length of the stick is 0.6m ,the left end will be [ 0.8 c * | t_l | +0.3 m] left to the origin.

So, the observer sees the left end at [ 0.8 c * | t_l | +0.3 m] left to the origin at time t = 1/c s.

Now, the equation is

(1/c)s = [ { √[([ 0.8 c * | t_l | +0.3 m] )²+1]}/c + (- | t_l | )s

Solving this equation gives,
| t_l | = 1.946 × 10^-10 s

Negative root can't be taken as by definition | t_l | has to be positive.

Similarly , for right end ,
Here, t_r = -| t_r |
the equation is
(1/c)s = [ { √[([ 0.8 c * | t_r | - 0.3 m] )²+1]}/c + (- | t_r | )s

Solving this equation gives,

| t_r | = 1.203 × 10^-10 s
Negative root can't be taken as by definition | t_r | has to be positive.

So, the observer sees the left end at [ 0.8 c * 1.946 × 10^-10 s +0.3 m] = 0.347 m left to the origin at time t = 1/c s.
So, the observer sees the right end at [0.3 m - 0.8 c * 1.203 × 10^-10 s ] = 0.271 m right to the origin at time t = 1/c s.Is this correct?

I didn't get what the question tries to teach us i.e significance of the question from physics point of view.
Will you please tell me something about it?

 

[TSny]

Your work looks correct. I got the same answers.

I think the purpose of the question is to show that the size and shape of a relativistically moving object as seen by the eye of an observer (or a camera) can be very different from the size and shape of the object as defined by the location of the points of the object at a particular time in the observer's frame of reference.

In your exercise, the length of the meter stick in the observer's frame is 0.60 m. But the particular observer sees the length to be 0.618 m at the instant the observer sees the midpoint pass the origin. At other instants, the observer would see a different length. Observers at different locations would see different lengths for the stick at the instant they see the midpoint pass the origin.

In relativity, we are almost always interested in the length as defined by the distance between the ends corresponding to one instant of time. All observers in the same inertial frame will agree on this length even though they see the stick to be of different lengths. Physicists are not usually concerned with the visual appearance of the object. It is interesting that it wasn't until around 1960 that any significant attention was drawn to the visual appearance of objects in relativity (whereas, of course, the concept of length contraction was in Einstein's 1905 paper and even earlier in the work of Lorentz, Fitzgerald, etc.).

If this topic interests you, then you can search the web for articles. Here's a couple
http://www.phy.pmf.unizg.hr/~npoljak/files/clanci/weisskopf.pdf
https://stuff.mit.edu/afs/athena/course/8/8.20/www/m44.pdf