# Length contraction derivation

1. Jan 15, 2008

### bernhard.rothenstein

Textbooks I know derive the formula that accounts for the time dilation by measuring simultaneously the space coordinates of a moving rod.
Is there another way to derive it?

2. Jan 15, 2008

### kdv

I am guessing you meant "lenght" instead of time in your first sentence. And I am assuming that you mean "deriving from the Lorentz transformations".

That's the most direct way to measure the lenght of a moving object: measuring the positions of its extremities at the same time in the frame of the measurement. Butthis requires two observers (it could be done with a single observer but then th eobserver is not local to the two events and must take into account the finite speed of light and that makes things much more complicated).

A second way requiring only a single observer is to do the following: Notice the time it takes for the moving object to pass in front of you (i.e. measure the time elapsed between the front of the object being aligned with you and the back of the object being aligned with you. Use the fact that this time you measured must be equal to the length of the object (in your frame) divided by the speed of the object. The speed and the lengths are two unknowns. Now use this in the Lorentz transformations and you can solve for the lenth contraction formula.

3. Jan 15, 2008

### clem

$$ds^2=dt^2-dx^2$$ is an invariant.
Choosing dx=vdt in one frame and dx=0 in another gives the TD formula.

4. Jan 15, 2008

### bernhard.rothenstein

length contraction

Please tell me if that can be done in the case of length contraction?

5. Jan 15, 2008

### bernhard.rothenstein

length contraction time dilation

I think there are to complementary experiments. In the first one in I we use a stationary rod of proper length L(0) and two synchronized at its ends. We measure the speed of a clock moving between its two ends as as a quotient between a proper length L(0) and a coordinate time interval T
V=L(0)/T. (1) Reverse the situation considering that an observer using the moving clock as a wrist watch measure the speed of the rod used in the previous experiment. He measures a proper time interval T(0) but a distorted(?) length L and considers that the speed of the rod is
V=L/T(0) . (2)

The result is
L(0)/T=L/T(0) (3)
Is it correct to conclude that if we have length contraction in our hand we have time silation as well and viceversa?

6. Jan 17, 2008

### morrobay

On page two there is a derivation with x' =0 and x = v/c(ct).
web.bryant.edu/~bblais/pdf/relativity.pdf

7. Jan 17, 2008

### pam

No. $$dx'^2-dt'^2=dx^2-dt^2$$.
In the moving frame, classical intuition is used to assume that setting dt'=0 gives the correct length of a moving rod. No formula results from the equation because dt can be anything in the rest system. The Lorentz transformation is needed to discuss length.

8. Jan 17, 2008

### 1effect

Very nice proof. The above may also be a good explanation why there are no experimental tests of length contraction to date: http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html

9. Jan 17, 2008

### bernhard.rothenstein

length contraction

I think there are two complementary experiments. In the first one in I we use a stationary rod of proper length L(0) and two synchronized clocks at its ends. We measure the speed V of a clock moving between its two ends as as a quotient between a proper length L(0) and a coordinate time interval T
V=L(0)/T. (1) Reverse the situation considering from I' that an observer using the moving clock as a wrist watch measures the speed of the rod used in the previous experiment. He measures a proper time interval T(0) but a distorted(coordinate) length L and considers that the speed of the rod is
V=L/T(0) . (2)

The result is
L(0)/T=L/T(0) (3)
Consider the invariance of the space-time interval in the conditions in which in I' we measure a proper time interval T(0) i.e.
VVTT-ccTT=-ccT(0)T(0) (4)
or
VV-cc=-cc[T(0)/T]^2=-cc[L/L(0)] (5)
where we have taken into account (3)
where from the formula that accounts for the length contraction
L=L(0)sqrt(1-VV/cc) (6)
Do you discover a flow in the lines above?