Explaining C: How Space Changes with Speed

[Doc Al]

Halliday and Resnik say this equation is for transforming from the primed into the unprimed frame. We are definitely going the other way.

Here are the complete set of Lorentz Transformations:

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

$$\Delta x' = \gamma(\Delta x - v\Delta t)$$

$$\Delta t = \gamma(\Delta t' + v\Delta x'/c^2)$$

$$\Delta x = \gamma(\Delta x' + v\Delta t')$$

Sure, the first two allow you to go from unprimed measurements to primed measurements. And vice versa, for the last two, which are called the inverse LTs. But remember, these are just equations like any other--you can use them any way you want, as long as you plug in the right values.

The main objective is to find the length that someone will measure the marks left by the rod in the rail frame to be. There should be two burn marks left by the flares which are .866025 apart, which prove the rod has "really" contracted while at speed .5c.

Right. So we need to find $\Delta x$. What are we given? We know that its a meterstick, so $\Delta x'$ = 1m. We also know that the flares go off simultaneously in the unprimed frame, so $\Delta t$ = 0. Now which LT relates those three values? Try the second one:

$$\Delta x' = \gamma(\Delta x - v\Delta t)$$

Plugging in what we know, it becomes:

$$\Delta x' = \gamma\Delta x$$

or:

$$\Delta x = \Delta x'/\gamma$$

Which tells you that the measured length is the expected 0.866m.

I can promise you that, yes, but it won't help you: I have an extensive vocabulary of alternatives. :-)

Lucky for you I'm a mentor here. In "real life" I'd rip you to shreds without mercy! (Just kidding. )

I love Seinfeld, but that line doesn't ring a bell at all. I just figured you were witnessing your Faith in The Church Of Relativity. Praise Einstein!

The original line is: "They're real... and they're spectacular!".