Lorentz Transformation - Clock

[CGM]

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

Homework Equations

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)

 

[stevendaryl]

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

Homework Equations

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.

 

[CGM]

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?

 

[stevendaryl]

So, t'=yt?

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

 

[CGM]

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)

 

[stevendaryl]

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.

 

[CGM]

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?

 

[stevendaryl]

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

 

[CGM]

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