Explaining C: How Space Changes with Speed

[zoobyshoe]

I have the 2nd, 4th, and 5th editions.

We're clearly fated to be out of phase, Tom. I have the third edition, 1988.

Mind you, the thought experiment in H+R may not have actually involved flares. I really don't remember, because it's been awhile. As I said, the flares are totally incidental. But the basic idea is taught there: Only simultaneous measurements of the position of the ends of the rod constitute a "length measurement".

Yeah, hehehehehe. I think the flare experiment is more of a Tom Mattson original than you think. In my copy all they do is warn you to mark the positions of the head and tail of your goldfish simultaneously (in your reference frame) rather than arbitrarily. They don't propose any mechanism to do this.

What they actually say about the "reality" of length contraction is essentially what Eddington says:

"The questions, `Does the rod really shrink?' and `Do the atoms in the rod really get pushed closer together?' are not proper questions within the framework of relativity. The length of the rod is what you measure it to be and motion affects measurements."

p.962 3rd edition 1988.

They don't have any gedakens where the moving thing actually leaves physical marks on the stationary one. It is clear that you fully believe your flare and rod thing is an obvious extrapolation of what they're saying, but they, like Eddington, are carefully avoiding saying exactly what you think they're saying.

If you proposed your flare thing to Halliday and Resnik, I believe they would say what they said: "It's not a proper question in the framework of relativity."

 

[quantumdude]

Yeah, hehehehehe. I think the flare experiment is more of a Tom Mattson original than you think. In my copy all they do is warn you to mark the positions of the head and tail of your goldfish simultaneously (in your reference frame) rather than arbitrarily. They don't propose any mechanism to do this.

Well, the flares certainly provide a means to do it.

They don't have any gedakens where the moving thing actually leaves physical marks on the stationary one. It is clear that you fully believe your flare and rod thing is an obvious extrapolation of what they're saying, but they, like Eddington, are carefully avoiding saying exactly what you think they're saying.

It is an obvious extrapolation of what they say. It makes no comment whatsoever on what "really" happens to the rod. It simply provides a means of simultaneously recording the positions of the two ends of it.

If you proposed your flare thing to Halliday and Resnik, I believe they would say what they said: "It's not a proper question in the framework of relativity."

They would not say that at all, because it's a perfectly proper question in the framework of relativity.

"Proper questions" in the framework of relativity are questions that relate to intervals in spacetime. That much should be clear from even a glance at the Lorentz transformation. The measurement scheme I have used does only one thing: It records the places and times of events.

If that thought experiment is not a proper question in the framework of relativity, then there are no proper questions in the framework of relativity.

What makes you think otherwise?

 

[pmb_phy]

I'm new to this thread so please bear with me. I took a look at the first posts in this thread but there are too many for me to read (due to my back and the pain of sitting and reading). But I agree with the original post to some extent, and to the extent that I understood what he was trying to say.

He was suggesting that the reason that the speed of light is constant has something to do with space changing. That is true. The reason why c is invariant is due to a combination of Lorentz contraction and time dilation. A close and thoughful look at the transformation rules for velocity shows that.

Also there is nothing electromagnetic about Lorentz contraction. Its distances that contract and that's why the lengths of rods contract. All one need to do in order to see that is true is to consider the distance between two point particles which are separated in space. The distance between them is contracted and there is no physical connection between them.

Pete

 

[zoobyshoe]

If that thought experiment is not a proper question in the framework of relativity, then there are no proper questions in the framework of relativity.

What makes you think otherwise?

Ah! Don't ask me! Ask them. I am merely quoting their respose:"The questions`Does the rod really shrink?' and `Do the atoms in the rod really get pushed closer together?' are not proper questions in the framework of relativity"

Is your thought experiment proper in my estimation? Is it "proper" to try and pin down the most literal possible proof of length contraction? Absolutely.

My point is that "In Halliday and Resnik the thought experiment goes like this" ought to have been something more like: "Extrapolating from what Halliday and Resnik say we can use flares and a rod", and so on, and so forth. The way you phrased it gives the unequivocal impression that very set up is to be found in their book. (They do have a thing with a train and clocks, and they have a goldfish, but they don't have a rod and flares, or anything where the moving body leaves physical marks on the stationary one.)

 

[zoobyshoe]

He was suggesting that the reason that the speed of light is constant has something to do with space changing. That is true.

Hi Pete,

Yes, but he also had some point about the angle at which you encounter the light changing until at c you would theoretically be encountering it at 90º. Could you make any sense out of that?

-Zooby

 

[pmb_phy]

Hi Pete,

Yes, but he also had some point about the angle at which you encounter the light changing until at c you would theoretically be encountering it at 90º. Could you make any sense out of that?

-Zooby

No. I was totally lost on what he was talking about. Too bad we can't have a black board here huh?

Pete

 

[zoobyshoe]

No. I was totally lost on what he was talking about. Too bad we can't have a black board here huh?

I'm not sure it wouldn't just give people an even greater capacity to confuse me. :-)

 

[quantumdude]

Ah! Don't ask me! Ask them. I am merely quoting their respose:"The questions`Does the rod really shrink?' and `Do the atoms in the rod really get pushed closer together?' are not proper questions in the framework of relativity"

But I don't need to ask them. "Proper questions" in SR pertain to spacetime intervals, and that's precisely what this thought experiment is about. I could go down the hall and ask Resnick (he's at my school) to verify that, but he'd probably think I'm a simpleton.

Is your thought experiment proper in my estimation? Is it "proper" to try and pin down the most literal possible proof of length contraction? Absolutely.

It would help both this discussion and your personal understanding of relativity a great deal if you would have a good look at the Lorentz transformation. Then, you could see for yourself that questions of spacetime intervals are proper questions in SR, in the strictest sense.

My point is that "In Halliday and Resnik the thought experiment goes like this" ought to have been something more like: "Extrapolating from what Halliday and Resnik say we can use flares and a rod", and so on, and so forth. The way you phrased it gives the unequivocal impression that very set up is to be found in their book.

I double checked, and the scenario in fact does not appear exactly as I described it (more on this below). But so what? This isn't about flares or identical wires, it's about the Lorentz transformation, simultaneity, and length measurements. In other words, it's about everything that isn't being discussed, because we are so lost in these irrelevant tangents about the actual mechanisms of the thought experiments and who saw which light pulse in what order. I'm sorry I ever mentioned "flares", because it has become a focal point of the discussion.

(They do have a thing with a train and clocks, and they have a goldfish, but they don't have a rod and flares, or anything where the moving body leaves physical marks on the stationary one.)

They do in the 4th and 5th editions.

Upon double checking, I found that the length contraction thought experiments are done with stopwatches and trains, as you say. But flipping back a page to the section on the relativity of simultaneity, we find in my copies of the text the "flare" thought experiment that I tried unsuccessfully to recall in all its details. The scenario has 2 rocket ships passing each other. They are close enough so that 2 flares ignite simultaneously in one frame, but not in another, and they leave permanent marks on the ships. The thought experiment stops with the discussion of simultaneity, but since length contraction can be derived from it, I see no reason not to say what I said before: This is an obvious extrapolation of what Halliday and Resnick does say.

 

[pmb_phy]

Hi Pete,

Yes, but he also had some point about the angle at which you encounter the light changing until at c you would theoretically be encountering it at 90º. Could you make any sense out of that?

-Zooby

Perhaps the angles he was speaking of were from his vizualization of the moving body. We think in terms of math and coordinate systems. Perhaps he's thinking of what he sees as a body passes him by, like a train passing a person staning on one side of the track. Our heads are atr first looking to one direction, the direction the train is comming. Then that line of sight changes angularly. Perhaps that's the angle he was speaking of. And it is true that angles do change upon Lorentz transformation.

Pete

 

[zoobyshoe]

This is an obvious extrapolation of what Halliday and Resnick does say.

Only in the sense that it's an obvious extrapolation of any good text on relativity.

Anyway, what is the signifigance of a negative value for the time interval in the t' frame? What's negative time?

 

[Chronos]

Anyway, what is the signifigance of a negative value for the time interval in the t' frame? What's negative time?

Short answer is, the past. Negative t is a problem when you try to do the math. The solutions end up being the square root of a negative number. Those results do not have predictive value [using the term 'predictive' in the future tense].

 

[quantumdude]

Only in the sense that it's an obvious extrapolation of any good text on relativity.

Well, I'm sorry that swapping two rocket ships for a rod and rail has confused you so much, but the fact of the matter is that the thought experiment described in Halliday and Resnick can be used to teach length contraction with just a line or two of math. But if you don't understand the Lorentz transformation, then it would seem as though they were worlds apart.

Anyway, what is the signifigance of a negative value for the time interval in the t' frame? What's negative time?

If you set your stopwatch to t=0 when a flare ignites, then t<0 corresponds to those moments before the flare ignited. I'm sure that at some point you've heard the countdown to a rocket launch: "T-minus-10...T-minus-9...etc."

It's the same thing.

However, I wasn't talking about negative time. I was talking about a negative time interval. If t₂-t₁ is less than zero, it means that t₁ is greater than t₂. That means that t event 1 occurs later than event 2.

 

[zoobyshoe]

Short answer is, the past. Negative t is a problem when you try to do the math. The solutions end up being the square root of a negative number. Those results do not have predictive value [using the term 'predictive' in the future tense].

Thanks Chronos,

The specific situation is this, from Tom's breakdown of the flare/rod gedanken:

For observer S' (the guy standing on the rod):
?t'=t₂'-t₁'=?(?t-v?x/c²)
?t'=?(0-vL/c²)
?t'=-?vL/c²

See? ?t' is negative (not zero). This means that, according to the guy on the rod, event 1 (the rear flare igniting) occurs later than event 2 (the front flare igniting). If the two flares ignite at different times, and the observer is at their midpoint, then there is no way that he is going to receive both pulses simultaneously.

The interval in the t frame, an interval of 0, becomes a negative interval in the t' frame, which is not 0. (It has some very small value which, given a meter rod, varies according to what specific speed it has.)

How do we "locate" this negative, "past" interval of time with respect to t? Does one end of the negative interval fall on t=0, and the other at t= -.0000000001 (for instance)?

 

[quantumdude]

How do we "locate" this negative, "past" interval of time with respect to t?

The same way the folks at NASA locate "T-minus-10": with a clock.

Does one end of the negative interval fall on t=0, and the other at t= -.0000000001 (for instance)?

Yes.

 

[zoobyshoe]

The same way the folks at NASA locate "T-minus-10": with a clock.

I think you were composing your post to me at the same time I was composing mine to chronos. Your answer wasn't up yet when I submitted. I wasn't ignoring it.

So, you are saying if we get a value of, for example, t' = -.00000001 one end of the interval would be located at that position and the other at 0'?

 

[quantumdude]

So, you are saying if we get a value of, for example, t' = -.00000001 one end of the interval would be located at that position and the other at 0'?

If the event to which you ascribe t' = -.00000001 occurs .00000001 time units earlier than the event to which you ascribe 0', then the time coordintate of the first event is as you say.

 

[zoobyshoe]

OK. You want me to be more familiar with the LT. Let's practise:

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

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

$$\Delta t = 0, c^2 = 9^1^0$$

Therefore: $$\Delta t' = 1.1547005(-150,000/9^1^0)$$

so: $$\Delta t' = 1.1547005(1.6667^-^0^6) = 1.9245^-^0^6$$

I am reasonably sure this is right, but it would be best if you checked it thoroughly to see if I have misconstrued anything.

 

[quantumdude]

Almost. c²=9*10¹⁶ m²/s², so your final answer should be 1.9245*10^-9 s. But I didn't intend for you to get bogged down in numerical calculations. What I think you need to do is learn what the Lorentz transformation means.

First, look at what it accepts as inputs: spacetime coordinates. So, experiments designed to test SR (aka the "proper questions in the framework of SR") are those experiments that record spacetime coordinates. This can be done with clocks, meter sticks, flares ( ), electrical switches, what have you. If it accurately captures the spacetime coordinates of events, then it is a pertinent issue in SR.

Second, think about what your numerical result means. As has been discussed, a time interval in frame S' given by $\Delta t\'\; = \gamma(\Delta t - v\Delta x/c^2)$ (look ma, I'm TeX-ing!) is taken to be the difference between times $t_2-t_1$. So, if $\Delta t\'\; = \gamma(\Delta t - v\Delta x/c^2)$ comes out negative, then it means that $t_1$ is greater than $t_2$. This means that Event 1 happens later than Event 2 in the frame S'.

With me so far?

 

[zoobyshoe]

Almost. c²=9*10¹⁶ m²/s², so your final answer should be 1.9245*10^-9 s.

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.

But I didn't intend for you to get bogged down in numerical calculations. What I think you need to do is learn what the Lorentz transformation means.

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.

First, look at what it accepts as inputs: spacetime coordinates.

Yep, I get this aspect.

Second, think about what your numerical result means. As has been discussed, a time interval in frame S' given by $\Delta t\'\; = \gamma(\Delta t - v\Delta x/c^2)$ (look ma, I'm TeX-ing!) is taken to be the difference between times $t_2-t_1$. So, if $\Delta t\'\; = \gamma(\Delta t - v\Delta x/c^2)$ comes out negative, then it means that $t_1$ is greater than $t_2$. This means that Event 1 happens later than Event 2 in the frame S'.

Glad you explained that. I was wondering how you'd determined which was first.

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

 

[zoobyshoe]

Ground control to Major Tom. We have lost your signal. Do you read us?

 

[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.