Explaining C: How Space Changes with Speed

[zoobyshoe]

Pardon me for jumping in again, but I think I see what zoobyshoe is trying to do. (I find this thread hard to follow due to all the numbers flying around.)

First off, I believe you are using the wrong value for \Delta t', the time interval between the flares according to the rod frame. From the LT, \Delta t' = -\Delta x' v/c^2, thus \Delta t' = 0.1667E-8 seconds.

Here's the full equation again:

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

It seems to me that in boiling it down you have forgotten the gamma:

\Delta t' = -\Delta x' v/c^2,

I'm not sure we can do without that. What do you think?

Is this what you were looking for zoobyshoe?

I think you may understand what I'm going for. It seems to me there should be a neat and tidy way to get from the time interval in the t' frame to the exact length that will be marked in the t frame by the flares, which should be exactly the length of a contracted meter rod at .5c, the speed I've selected as a sample.

I am kind of baffled that I have managed to get kinda, sort of close, but not exactly there. Everything is done by LT, it should be neat and consistent, on paper anyway.

But I think you are possibly right that there could be an error in some previous math somewhere which is throwing things off. I'm not sure it is the time interval, though.

Thanks for keeping an open mind!

 

[zoobyshoe]

Doc Al,

Posts #87 and #88 are where we hashed out the time interval and other stuff, if you want to look them over.

 

[Doc Al]

Here's the full equation again:

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

It seems to me that in boiling it down you have forgotten the gamma:

\Delta t' = -\Delta x' v/c^2,

I'm not sure we can do without that. What do you think?

I think you are using the wrong equations. The equation you need is this:

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

We are given that the events (the flares) occur simultaneously in the rail frame, so \Delta t = 0. Which allows us to conclude that:

\Delta t' = - v\Delta x'/c^2

As you can see, there is no gamma.

I think you may understand what I'm going for. It seems to me there should be a neat and tidy way to get from the time interval in the t' frame to the exact length that will be marked in the t frame by the flares, which should be exactly the length of a contracted meter rod at .5c, the speed I've selected as a sample.

The way to relate things is as I described in my last post. Remember that to the rod observers the marks do not represent a measurement of the length of the meterstick. To the rod observers, those flares fired at different times. But you can certainly figure out what the rod observer would measure as the distance between those marks, and then relate that to what the rail observers would measure.

Thanks for keeping an open mind!

As long as you can restrain yourself from accusing me of sophistry, I'm happy to work with you.

PS: Regarding my earlier comment "it's real... and it's spectacular": I take it you are not a Seinfeld fan?

 

[zoobyshoe]

I think you are using the wrong equations. The equation you need is this:

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

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

Remember that to the rod observers the marks do not represent a measurement of the length of the meterstick. To the rod observers, those flares fired at different times. But you can certainly figure out what the rod observer would measure as the distance between those marks, and then relate that to what the rail observers would measure.

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.

As long as you can restrain yourself from accusing me of sophistry, I'm happy to work with you.

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

PS: Regarding my earlier comment "it's real... and it's spectacular": I take it you are not a Seinfeld fan?

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!

 

[Doc Al]

an error in post #87

Posts #87 and #88 are where we hashed out the time interval and other stuff, if you want to look them over.

I looked over post #87 and I see the error made.

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

Let's plug a value of .5 c in for rod velocity.

For gamma that gives \gamma = 1.1547005

x= length of rod = 1 meter = 1

Remember that the rod frame is the primed frame and the rail frame is the unprimed frame. There is absolutely nothing wrong with the formula, but I believe you mixed up \Delta x' and \Delta x. \Delta x' = 1 meter; \Delta x does not.

 

[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!".