Explaining C: How Space Changes with Speed

[zoobyshoe]

Anyway, whoever's reading this:

my next question is how far apart, exactly, will the marks left by the flares be according to Tom's method.

First I asked myself, how long would it take something to travel one meter at 150,000 km/s?

150,000 km/s = 1.5 ⁰⁸ m/s

1/1.5⁰⁸ = 6.6667^-09

The speed of the rod at .5c is 1 meter every 6.6667^-09 seconds.

We've determined the time interval between the flashes in the t' frame will be 1.9245^-09 seconds.

How far wll any point on the rod travel in 1.9245^-09 seconds at 1 meter each 6.6667^-09 seconds?

1.9245^-09 times 6.6667^-09 = 1.283006415^-17 meters.


If event 1 (the rear flare) is first, as Tom says, then it goes off, the rod moves 1.283006415^-09 meters, then the front flare goes off.

This will leave marks 1.00000000000000001283006415 meters apart. Or perhaps even more if we consider that the track is contracted to the guy on the rod.

If, perhaps, we've been wrong about event 1 (the rear flare) being first, then the flares will leave marks that are .99999999999999998716993584 of a meter apart, or perhaps more if we consider the track is contracted to the rod guy.

In all cases these marks are too far apart to stand in support of Tom's argument. By the Lorentz Transformation for length contraction, a meter-rod traveling at .5c should leave marks that are .8660254 of a meter apart (if you want to claim that length contraction is real).

 

[quantumdude]

Whoops, I hadn't noticed that you'd responded.

This is because I failed to keep my units consistent, right? If I choose meters for the rod I can't use kilometers for the velocities. (DOH!) I could have stuck with the kilometers had I used .001 for x, I guess.

Right. Notice that you were off by a power of 10. When you're working in SI units, that's a strong indicator that you mixed up "kilo-units", "milli-units", etc..., with the "regular" units.

For some reason, I can't grasp any equation till I can plug values into it and work them out. Till I can do that, it's a meaningless abstraction to me.

Nothing wrong with that.

So the events in frame t' take place 1.9245 ^-09 seconds apart? That is: .0000000019245 of a second apart?

Yes.

I'll respond to your other post later.

edit: fixed a bracket

 

[Doc Al]

length contraction verified!

my next question is how far apart, exactly, will the marks left by the flares be according to Tom's method.

By "Tom's method" I presume you mean having the flares go off simultaneously according to the track frame (the unprimed frame) and then measuring where the marks appear on the track. This would give you a meaningful measurement of the length of the rod according to the track observers.

First I asked myself, how long would it take something to travel one meter at 150,000 km/s?

Uh oh. Let's put this aside for the moment.

We've determined the time interval between the flashes in the t' frame will be 1.9245^-09 seconds.

OK, I'll take your word for that. But the time interval in the track (unprimed) frame is zero.

How far wll any point on the rod travel in 1.9245^-09 seconds at 1 meter each 6.6667^-09 seconds?

Careful. In the rod (primed) frame, the rod doesn't move at all! (And in the track (unprimed frame) the flares are simultaneous so the rod doesn't have time to move.)

If event 1 (the rear flare) is first, as Tom says, then it goes off, the rod moves 1.283006415^-09 meters, then the front flare goes off.

You are mixing up frames. The rear flare goes off first according to the rod frame, not according to the track frame.

This will leave marks 1.00000000000000001283006415 meters apart. Or perhaps even more if we consider that the track is contracted to the guy on the rod.

Nope.

If, perhaps, we've been wrong about event 1 (the rear flare) being first, then the flares will leave marks that are .99999999999999998716993584 of a meter apart, or perhaps more if we consider the track is contracted to the rod guy.

The right way to find the \Delta x measured by the track frame is to use the LT for distance: \Delta x' = \gamma (\Delta x - v\Delta t). Since \Delta t = 0, \Delta x = \Delta x'/\gamma--thus the usual length contraction is seen, as expected.

In all cases these marks are too far apart to stand in support of Tom's argument. By the Lorentz Transformation for length contraction, a meter-rod traveling at .5c should leave marks that are .8660254 of a meter apart (if you want to claim that length contraction is real).

Oh, it's real... and it's spectacular!

 

[zoobyshoe]

By "Tom's method" I presume you mean having the flares go off simultaneously according to the track frame (the unprimed frame) and then measuring where the marks appear on the track. This would give you a meaningful measurement of the length of the rod according to the track observers.

Correct.

But the time interval in the track (unprimed) frame is zero.

I'm perfectly aware of that.

In the rod (primed) frame, the rod doesn't move at all! (And in the track (unprimed frame) the flares are simultaneous so the rod doesn't have time to move.)

Sophistry.

You are mixing up frames. The rear flare goes off first according to the rod frame, not according to the track frame.

Yes, I know. I haven't mixed the frames.

Nope.

Yup.

The right way to find the \Delta x measured by the track frame is to use the LT for distance:

We are exploring Tom's gedanken and methods here. So far they don't seem to work.

Oh, it's real... and it's spectacular!

A bald assertion.

 

[quantumdude]

I agree to a certain extent that zoobyshoe has not mixed frames, but only unwittingly and by the wrong reasoning. He takes a time interval from S' and multiplies it by (what he thinks is) the speed of the rod. Of course, we all know that the speed of the rod is zero in S', but by reciprocity S' does see the track moving backwards at the same speed that S sees the rod moving forwards. So, multiplying that time by that speed is physically meaningful. It is the length of track that S' sees go by.

The real problems in zoobyshoe's analysis are these.

First:

How far wll any point on the rod travel in 1.9245^-09 seconds at 1 meter each 6.6667^-09 seconds?

1.9245^-09 times 6.6667^-09 = 1.283006415-17 meters.

In a word: No.

You are being careless with units again, and this time it is more severe than the last, because you aren't just off by a numerical factor. This time, your answer doesn't even have the same dimensions as the left hand side. The number 1.9245*10^-9 is a number of seconds. And the number 6.6667*10^-9 is also a number of seconds. You do not get a distance by multiplying two times together.

Second:

If event 1 (the rear flare) is first, as Tom says, then it goes off, the rod moves 1.283006415-09 meters, then the front flare goes off.

You're mistaken. I said that Event 1 occurs later in S'. Of course, if it occurs first, then this analysis will fail.

edit: fixed superscript bracket

 

[quantumdude]

Doc AI: By "Tom's method" I presume you mean having the flares go off simultaneously according to the track frame (the unprimed frame) and then measuring where the marks appear on the track. This would give you a meaningful measurement of the length of the rod according to the track observers.

zoobyshoe: Correct.

In that case, you are mistaken: the method is not mine, but Einstein's.

All I have done is stipulate the setup. The method of analysis is SR. If you say the method doesn't work, then you are saying that SR doesn't work. Needless to say, that statement would require a great deal of explanation and evidence before being moved to Theory Development.

zoobyshoe: This will leave marks 1.00000000000000001283006415 meters apart. Or perhaps even more if we consider that the track is contracted to the guy on the rod.

Doc AI: Nope.

zoobyshoe: Yup.

Nope.

And furthermore, there is no "perhaps" about the issue. The Lorentz transformation is not vague or fuzzy.

We are exploring Tom's gedanken and methods here. So far they don't seem to work.

They do if you don't screw up the math.

A bald assertion.

Again: Nope.

Do you think you could possibly tone down the know-it-all attitude until you've demonstrated some ability to properly conduct this analysis? At this stage of your education, when you get a result that contradicts SR the first thing that should enter your mind is the question, "Where did I go wrong?"

If you want to ask questions, go right ahead. I'm here. But if you want to tell this Forum "how it is", then you can teach yourself relativity as far as I'm concerned. I, for one, certainly don't need to hear how SR works from someone who would flunk a relativity course in real life.

PS: The points that you label "sophistry" and "bald assertion" were exactly right, on the part of Doc AI.

edit: fixed bracketing errors

 

[zoobyshoe]

I agree to a certain extent that zoobyshoe has not mixed frames, but only unwittingly and by the wrong reasoning. He takes a time interval from S' and multiplies it by (what he thinks is) the speed of the rod. Of course, we all know that the speed of the rod is zero in S', but by reciprocity S' does see the track moving backwards at the same speed that S sees the rod moving forwards. So, multiplying that time by that speed is physically meaningful. It is the length of track that S' sees go by.

My lack of rigour in referring to the speed of the rod as opposed to the speed of the track is noted. But, as you point out, this is not what is causing the problem at all:

The real problems in zoobyshoe's analysis are these...You are being careless with units again, and this time it is more severe than the last, because you aren't just off by a numerical factor. This time, your answer doesn't even have the same dimensions as the left hand side. The number 1.9245*10^-9 is a number of seconds. And the number 6.6667*10^-9 is also a number of seconds. You do not get a distance by multiplying two times together.

I meter per 6.6667^-09 seconds is a rate of speed. As far as I can tell, it is a valid expression of rate of speed. Somehow, I am misunderstanding how to relate it too 1.9245 ^-09 to get the distance traveled by the rod in that time.

You're mistaken. I said that Event 1 occurs later in S'. Of course, if it occurs first, then this analysis will fail.

You are absolutely correct, that is what you said. I have no conception of how I reversed it in my mind. My bad, though.

 

[zoobyshoe]

Do you think you could possibly tone down the know-it-all attitude until you've demonstrated some ability to properly conduct this analysis?

What comes off as a know-it-all attitude is actually something else: a profound desperation not to be further confused.

At this stage of your education, when you get a result that contradicts SR the first thing that should enter your mind is the question, "Where did I go wrong?"

This is, in fact, the first thing I suspected and always do suspect. However, I'm generally not in a position to figure out where I went wrong. I can only present the contradiction to you as it looks to me, and let you spot my mistake. When you're right, when you've hit the nail on the head about what I'm doing wrong or what I'm misconstruing, your corrections are suddenly quite clear and meaningful. When someone's confidently wrong about what I'm doing wrong, it's just painful and confusing. I develope an attitude because I have been run ragged in other threads by people pushing me to follow red herrings about what is confusing me.

At the same time I am aware you have been gratuitously stressed in other threads, arguing with people who arrive announcing the death of relativity and such like. I'm sure if it weren't for those people I wouldn't look like such a pain in the neck.

 

[zoobyshoe]

OK. I have discovered that the correct way to relate them is division: 1.9245^-09/6.6667^-09 = .288673557 meters.

This means that the front flare will go off, the rail beneath the rod will move .288673557 meters, then the rear flare will go off. 1 meter-.288673557 meters = .711326443 meters.

The marks will be .711326443 meters apart, but I figure what seems to be .711326443 meters to the rod guy is actually a contracted version of the proper length in the rail frame.

It seems the .288673557 and .711326443 have to be uncontracted somehow to arrive at the distance the marks will be from each other when measured in the rail frame.

If I take .866025 which is the length were supposed to end up measuring in the rail frame, and contract it for a speed of .5c, I get .749999301, which is kinda, sort of promising, being kinda, sort of close to .711326443.

So, If I can contract a length by multiplying by .866025 is seems I should be able to uncontract a length by dividing by .866025

.711326443/.866025 = .82136941. Hmmmm. Kinda, sort of close to .866025, but off enough to suggest there's something else to be done.

 

[Doc Al]

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.

So let's look at the events from the view of the rod frame:
The first flare goes off and makes its mark. Then the rails move a distance D = VT = (0.5c)(0.1667E-8s) = 0.25m. Then the second flare makes its mark. So, according to the rod frame, the distance between the marks is only 0.75m.

Of course, the rail frame thinks that distances measured by the rod frame will be too short by a factor of \gamma = 1.1547. So the rail frame will measure the distance between the marks to be (1.1547)x(0.75m) = 0.866m.

Of course, for the rail observers, the distance between the marks is the length of the meterstick seen contracted by \gamma, or (1m)/(1.1547) = 0.866m.

So, as far as I can see, everything makes sense. Is this what you were looking for zoobyshoe?

 

[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!".