# On the Relativity of Lengths and Times

1. Jul 22, 2005

### paco1955

let there be given a stationary rigid rod; and let its length be l as measured by a measuring-rod which is also stationary. we now imagine the axis of the rod lying along the axis of x of a stationary system of co-ordinates, and that a uniform motion of parallel translation with velocity v along the axis of x in the direction of increasing x is then imparted to the rod. we now inquire as to the length of the moving rod, and imagine its length to be ascertained by the following operation:

by means of stationary clocks set up in the stationary system and synchronizing, an observer ascertains at what points of the stationary system the two ends of the rod to be measured are located at a definite time. the distance between these two points, measured by the measuring-rod, is a length which may be designated "the length of the rod".

how does the observer ascertains the points of the stationary system at which the two ends of the rod to be measured are located at a definite time?

i do not know how to ascertain the two points on the axis of x in order to measure the distance between them.

need help

thanks

Last edited: Jul 22, 2005
2. Jul 22, 2005

### learningphysics

I believe your question amounts to deriving length contraction?

You can do this with Lorentz transformations... let S' be the moving frame, and S the stationary frame. $$\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$$

The left end of the rod (the rod is moving towards the right) has equation:
$$x_1' = 0$$

so

$$\gamma (x_1 -vt_1) = 0$$ (eq 1)

The right end:
$$x_2' = L$$

so
$$\gamma (x_2 - vt_2) = L$$ (eq 2)

eq. 2 - eq. 1 gives:

$$\gamma (x_2 - x_1 - v(t_2 - t_1)) = L$$

but $$t_2 = t_1$$

so

$$\gamma(x_2 - x_1) = L$$

So $$x_2 - x_1 = L/\gamma$$

So the length as measured in the stationary frame is: $$L/\gamma$$

Last edited: Jul 22, 2005
3. Jul 25, 2005

### robert Ihnot

learningphysics: I believe your question amounts to deriving length contraction?

I am not sure that is the question. He wants to know, or I am wondering about, the physical aspects of how the stationary crew can be so positioned that they know the exact spots on the ground at time t_0 that the length is being accurately measured. (It seems it would be easier to measure the time elapsed from the same stationary point, but then we need to know the velocity.)

Last edited: Jul 25, 2005
4. Jul 25, 2005

### jdavel

paco1955,

You can assume there are lots of stationary clocks along the x axis and an observer standing by each one.

As the rod goes by, one observer will see the front of the stick at his location at t=0 (or whaterver). Another observer will see the back of the rod at his location at t=0. the distance between these two observers in the stationary frame is "the length of the rod" as measured in the stationary frame. this length will of course be less than the proper length, the length of the rod in the moving frame.

5. Jul 26, 2005

### robert Ihnot

jdavel: You can assume there are lots of stationary clocks along the x axis and an observer standing by each one.

Sartori in Understanding Relativity seems to take this for granted.

He also talks about the proper time, which is two measurements of time at the same clock. Yet to measure the length this way we would need to know the velocity. This seems to require a reading of length?