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
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)
So, t'=yt?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 [itex]X' = 0[/itex]. So using the first equation from your "Relevant equations", you can solve for [itex]x[/itex] in terms of [itex]t[/itex]. Plug that into your equation.
CGM said:So, t'=yt?
Never mind I did something silly.stevendaryl said:How did you get that? Take your equation, [itex]t' = \gamma (t - \frac{vx}{c^2})[/itex] and replace [itex]x[/itex] by [itex]v t[/itex].
CGM said:Never mind I did something silly.
So 0=y(x-vt) => x=vt
t'=y(t-(v/t)^2 . t)
Yes that's what I said, maybe my notation is confusing.stevendaryl said:I'm afraid you did something silly again. If you start with [itex]t' = \gamma (t - \frac{vx}{c^2})[/itex] and substitute [itex]x=vt[/itex], you get: [itex]t' = \gamma (t - \frac{v^2 t}{c^2})[/itex]. Now factor out a factor of [itex]t[/itex] from the right-hand side, and remember what [itex]\gamma[/itex] is.
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?
Ah yes, I see my mistake. Thank you.stevendaryl said:No, you have the expression
[itex]t' = \gamma (t - (v/c)^2 t)[/itex]
Factor out [itex]t[/itex] to get:
[itex]t' = \gamma t (1 - (v/c)^2)[/itex]
Now, if you remember what the definition of [itex]\gamma[/itex] is, you can write [itex]1-(v/c)^2 = 1/\gamma^2[/itex]
A Lorentz transformation is a mathematical equation that describes how the measurements of space and time change between two different reference frames that are moving relative to each other. It is a key concept in Einstein's theory of Special Relativity and is used to explain the effects of time dilation and length contraction.
A Lorentz transformation affects clocks by causing them to appear to run at different rates for observers in different reference frames. This is known as time dilation, and is a result of the fact that the speed of light is constant for all observers, regardless of their relative motion.
Time dilation is the phenomenon where time appears to pass slower for an observer who is moving at high speeds relative to another observer. This is because the faster an observer moves, the more their time becomes warped or dilated. This effect has been confirmed through various experiments, such as the famous Hafele-Keating experiment.
The Lorentz transformation is related to the concept of spacetime because it shows how time and space are interconnected and can be affected by an observer's relative motion. It is a fundamental concept in Special Relativity, which states that space and time are not absolute but are part of a unified concept known as spacetime.
Lorentz transformations have many practical applications, such as in GPS technology, where they are used to account for the effects of time dilation on the precise timing needed for accurate location tracking. They are also crucial in particle physics, where high-speed particles moving at close to the speed of light require the use of Lorentz transformations to accurately calculate their behavior and interactions.