# Gamma factor in legth and mass demostration

1. Jan 8, 2007

### Littlepig

1. The problem statement, all variables and given/known data

I'm studing relativity(on extraschool) and known about gamma factor influences time, after some tries, I demostrate that if A is at Va, in relation to B: "light distance"=sqrt("light distance for B"^2-"distance traveled by A"^2)

where we say that "light distance"=c*"ta"; "light distance for B"=c*"tb"; and "distance traveled by A"=Vatb resolving and we demonstrate it...i gess you know what i mean....

now my problem is who to demonstrate that leght and mass are too influenced by gamma factor.

2. Relevant equations

wish i know then...

3. The attempt at a solution

don't have a tought...:X

2. Jan 8, 2007

### Hootenanny

Staff Emeritus
You mean you wish to prove length contraction?

3. Jan 8, 2007

### Littlepig

indeed, and mass, but by maths method, not by "concept" method, what is the begining? what is the permiss that is missing me to acept that Y can too interfer in mass and leght(by math method)

4. Jan 8, 2007

### Hootenanny

Staff Emeritus
Length contraction follows directly from time dilation. Time dilation can be 'proved' using a simple thought experiment. Length contraction follows directly from the lerentz transformations, so there is little in the way of a 'mathematical' proof, as far as I know.

5. Jan 8, 2007

### Littlepig

yes yes, till there i agree with it, but i wnat to demonstrate it mathematicaly...

ok, my point is:

for have time dilatation we have that:

2 obeservers, A and B, A is moving at speed Va and B is still.
"A" have a light clock(2 perfect mirrors, 1 photosensor, 1 laser making a 90º to mirrors).

as light travels at same speed at diferent referencials, we have that B will see the light moving not at 90º but at a diferent angle.
A will jut see light moving up and down like nothing was hapening...

now, from trignometry, $$c_1=sqrt(H^2-c_2^2)$$

as $$V=d/t$$,

we have that:
distance that light takes, measured by B is H,
distance travel by light, measured by A is $$c_1$$;
distance travel A, measured by B, is $$c_2$$;

so as $$d=V \cdot t$$;

$$t_A \cdot c= \sqrt{(t_B \cdot c)^2-(t_b \cdot V_a)^2}<=>t_B= \frac{t_A}{ \sqrt{1 - \frac{v^2}{c^2}}}$$

now, i demonstrated with maths the time dilatation....how i demonstrate leght contraction and mass dilatation???

Last edited: Jan 8, 2007
6. Jan 9, 2007

### Littlepig

still looking for some clue, i've search in wiki, but everything is already demonstrated, they only put the formulas there....the rest is explication of then...:(

7. Jan 11, 2007

### robphy

consider a light clock [whose separation of mirrors is] oriented in the direction of motion, rather than perpendicular...
..and don't forget to use the principle of relativity.

8. Jan 14, 2007

### Littlepig

ok, after some tries, i think i reach a point it's correct, however, i would like a confirmation of what i've done.

so, to demonstrate a leght contration, I used this logic:

$$t_B= \frac{t_A}{ \sqrt{1 - \frac{v_a^2}{c^2}}}, V=\frac{d}{t}<=>t=\frac{d}{v}$$
with d as the leght travel by "a".

so, as v_a is equal to both observers,

$$\frac{d_b}{v_a}=\frac{d_a/v_a}{ \sqrt{1 - \frac{v_a^2}{c^2}}}$$

what is equal to:

$$\frac{d_b}{v_a}=\frac{d_a}{v_a \cdot \sqrt{1 - \frac{v_a^2}{c^2}}}$$

what is equal to:

$$d_b=\frac{d_a}{\sqrt{1 - \frac{v_a^2}{c^2}}}$$

what is equal to:

$$d_a=d_b \cdot \sqrt{1 - \frac{v_a^2}{c^2}}}$$

so, to observer "a", who is moving, leght is everytime lower than to "b", and convertor factor is "Y"

am I right??

to demonstrate mass dilatation, I dout it's correct by my though, however, i putted it here :

$$d_a=d_b \cdot \sqrt{1 - \frac{v_a^2}{c^2}}}$$
and,
$$D= \frac{m}{V}$$
where m is mass, D is density, and V is Volume

so, considering a rectangular object,

$$m= D \cdot (H \cdot W \cdot d)$$
where H is High, w is Width, and d is Leght

considering m_a the mass of "a" for the observer "a" and m_b the mass of "a" for the observer "b"

for moving observer "a"
$$m_a= D \cdot H \cdot W \cdot d_a$$

for rest observer "b":
$$m_b= D \cdot H \cdot W \cdot \frac{d_a}{\sqrt{1 - \frac{v_a^2}{c^2}}}}$$

so, taking D off:

$$\frac{m_a}{H \cdot W \cdot d_a}=\frac{m_b}{H \cdot W \cdot d_a} \cdot \sqrt{1 - \frac{v_a^2}{c^2}}}$$

as H, W, D and d_a are equal to both observers,

$$m_a=m_b \cdot \sqrt{1 - \frac{v_a^2}{c^2}}}$$

so,

$$m_b=\frac{m_a}{\sqrt{1 - \frac{v_a^2}{c^2}}}}$$

can it be done with this logic? or we need to use another way??

thanks in advance, and hope wasn't too annoying

Last edited: Jan 14, 2007