# Lorentz Transformation - Clock

• CGM
In summary, the equation relating the time period of a moving clock to that of a stationary clock is t'=yt-((v/c)^2)t.

## Homework Statement

Use the Lorentz Transformation equations to derive the formula relating the time period of a moving clock to that of a stationary clock

X'=y(X-vt)
Y'=Y
Z'=Z
t'=y(t-vx/c^2)

## The Attempt at a Solution

t'=1/sqrt(1-(v/c)^2) . (t-vx/c^2)

CGM said:

## Homework Statement

Use the Lorentz Transformation equations to derive the formula relating the time period of a moving clock to that of a stationary clock

X'=y(X-vt)
Y'=Y
Z'=Z
t'=y(t-vx/c^2)

## The Attempt at a Solution

t'=1/sqrt(1-(v/c)^2) . (t-vx/c^2)

You're almost there. If the clock is at rest in the primed coordinate system, that means that its location as a function of time can be chosen to be $X' = 0$. So using the first equation from your "Relevant equations", you can solve for $x$ in terms of $t$. Plug that into your equation.

stevendaryl said:
You're almost there. If the clock is at rest in the primed coordinate system, that means that its location as a function of time can be chosen to be $X' = 0$. So using the first equation from your "Relevant equations", you can solve for $x$ in terms of $t$. Plug that into your equation.
So, t'=yt?

CGM said:
So, t'=yt?

How did you get that? Take your equation, $t' = \gamma (t - \frac{vx}{c^2})$ and replace $x$ by $v t$.

stevendaryl said:
How did you get that? Take your equation, $t' = \gamma (t - \frac{vx}{c^2})$ and replace $x$ by $v t$.
Never mind I did something silly.
So 0=y(x-vt) => x=vt

t'=y(t-(v/t)^2 . t)

CGM said:
Never mind I did something silly.
So 0=y(x-vt) => x=vt

t'=y(t-(v/t)^2 . t)

I'm afraid you did something silly again. If you start with $t' = \gamma (t - \frac{vx}{c^2})$ and substitute $x=vt$, you get: $t' = \gamma (t - \frac{v^2 t}{c^2})$. Now factor out a factor of $t$ from the right-hand side, and remember what $\gamma$ is.

stevendaryl said:
I'm afraid you did something silly again. If you start with $t' = \gamma (t - \frac{vx}{c^2})$ and substitute $x=vt$, you get: $t' = \gamma (t - \frac{v^2 t}{c^2})$. Now factor out a factor of $t$ from the right-hand side, and remember what $\gamma$ is.
Yes that's what I said, maybe my notation is confusing.

t'=yt-((v/c)^2)t
t'=t(y-(v/c)^2)

Is this the answer or is there more simplification to be done?

CGM said:
Yes that's what I said, maybe my notation is confusing.

t'=yt-((v/c)^2)t
t'=t(y-(v/c)^2)

Is this the answer or is there more simplification to be done?

No, you have the expression

$t' = \gamma (t - (v/c)^2 t)$

Factor out $t$ to get:

$t' = \gamma t (1 - (v/c)^2)$

Now, if you remember what the definition of $\gamma$ is, you can write $1-(v/c)^2 = 1/\gamma^2$

stevendaryl said:
No, you have the expression

$t' = \gamma (t - (v/c)^2 t)$

Factor out $t$ to get:

$t' = \gamma t (1 - (v/c)^2)$

Now, if you remember what the definition of $\gamma$ is, you can write $1-(v/c)^2 = 1/\gamma^2$
Ah yes, I see my mistake. Thank you.

t'=t/y