Explaining C: How Space Changes with Speed

[zoobyshoe]

He could have identical wires connected to the flares, and send an electrical signal from a common source to ignite the flares simultaneously.

Yes, but if he's interested in visually confirming that the flares are simultaneous in his rest frame, he has to be positioned equidistant from the flares when they go off. The light must have a path from the front flare to him that is equal in distance from the rear flare to him. Only by knowing that these distances are equal when he causes the emission (and we can stipulate he has the ability to hit the button at the right time to make it so) will he know that the flares were simultaneous because the light from both will reach him simultaneously, no?

 

[quantumdude]

Yes, but if he's interested in visually confirming that the flares are simultaneous in his rest frame, he has to be positioned equidistant from the flares when they go off. The light must have a path from the front flare to him that is equal in distance from the rear flare to him.

OK, fine let him be equidistant from the flares.

Only by knowing that these distances are equal when he causes the emission (and we can stipulate he has the ability to hit the button at the right time to make it so) will he know that the flares were simultaneous because the light from both will reach him simultaneously, no?

No. He doesn't have to actually see the light pulses at the same time. In fact, he needn't see them at all. He can know that the pulses ignited simultaneously with photodetectors that are connected to clocks. The photodetectors need not even be equidistant from the simultaneous flash. All that needs to be known is the position of each photodetector, the position of the flash, and the time at which the button was hit.

 

[zoobyshoe]

OK, fine let him be equidistant from the flares.



No. He doesn't have to actually see the light pulses at the same time. In fact, he needn't see them at all. He can know that the pulses ignited simultaneously with photodetectors that are connected to clocks. The photodetectors need not even be equidistant from the simultaneous flash. All that needs to be known is the position of each photodetector, the position of the flash, and the time at which the button was hit.

Ok, I'm confused about in which reference frame you want the flares to be simultaneous. It sounds like you want them to be simultaneous in the rods reference frame.

edit: You mean the detector/clocks are in the track reference frame with him?

 

[quantumdude]

Ok, I'm confused about in which reference frame you want the flares to be simultaneous. It sounds like you want them to be simultaneous in the rods reference frame.

No, the flares should ignite simultaneously in the frame of the track (the same frame as the guy who is watching the rod speed by). That's the only way his measurement could be called a "length" measurement.

edit: You mean the detector/clocks are in the track reference frame with him?

Yes.

 

[zoobyshoe]

No, the flares should ignite simultaneously in the frame of the track (the same frame as the guy who is watching the rod speed by). That's the only way his measurement could be called a "length" measurement.

Good.

The clock/detectors are more realistic, but for simplicity I'd rather stick with him being equidistant from the flares when they go off.

It is no problem to arrange it so the midpoint of the rod is directly in front of him when the flares ignite. This way the light has a path of equal length from each flare to him. If both beams of light reach him at the same instant, he knows the emission was simultaneous. Which it is.

The guy on the rod could be anywhere on the rod, but if we put him exactly at the midpoint of the rod something very interesting results: he too, receives the beams of light simultaneously.

That seems ridiculous because he is in motion and should, intuitively reasoning, encounter the forward beam before the rear beam. However, that would mean you are adding and subtracting velocities, which you can't do with light. The speed of light is the same to all observers in all inertial frames. He can't encounter the forward beam at c plus his own velocity. He must regard himself as being at rest with the light approaching him at c, and the end of the rod approaching him at the velocity of the rod over the track.

The beam from the rear cannot approach him at less than c just because he is moving away from it. He must regard himself as at rest with the light approaching him at c and the back end of the rod moving away from him at the velocity of the rod over the track.

He is at the exact centerpoint of the rod, both beams of light have the same distance to travel to reach him, both will do it at the same velocity, c, both will reach him simultaneously.

Looking down, he will see a contracted rail and will expect the flare marks to measure greater than a meter apart when he goes back to measure them later in the rail rest frame.

 

[quantumdude]

The guy on the rod could be anywhere on the rod, but if we put him exactly at the midpoint of the rod something very interesting results: he too, receives the beams of light simultaneously.

No, he doesn't, because the flares don't ignite simultaneously in his frame.

 

[zoobyshoe]

No, he doesn't, because the flares don't ignite simultaneously in his frame.

Let the time t when the midpoint of the rod is directly in front of the track observer be the point of synchrony between any clock that observer may have and any clock the rod observer may have such that t=o, the time of emission, is the point of synchrony of clocks for both observers.

 

[quantumdude]

Let the time t when the midpoint of the rod is directly in front of the track observer be the point of synchrony between any clock that observer may have and any clock the rod observer may have such that t=o, the time of emission, is the point of synchrony of clocks for both observers.

It doesn't matter when you zero both clocks. The flares ignite simultaneously in exactly one frame, and no other. If the flares were at the same location, then their ignition would be simultaneous in all frames. But that isn't what is happening here--these flares are spatially separated.

Let Event 1 be the rear flare igniting and Event 2 be the front flare igniting.

For observer S (the guy watching the rod zip by):
Δx=x₂-x₁=L (the length of the rod according to him)
Δt=t₂-t₁=0 (the flares ignite simultaneously according to him)

For observer S' (the guy standing on the rod):
Δt'=t₂'-t₁'=γ(Δt-vΔx/c²)
Δt'=γ(0-vL/c²)
Δt'=-γvL/c²

See? Δt' 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.

 

[zoobyshoe]

Tom,

I'm having question mark troubles again. I'm going to type in an example of what your equations look like, including question marks:

?t'=t₂ ' -1₁'=?(?t-v?x/c²)

Does this show up on your screen with 4 question marks in the equation? If so, that's what it looks like to me. If not my browser lacks some ability to properly translate whatever it is you are typing in. Latex seems to work fine on my browser.

 

[quantumdude]

?t'=t₂ ' -1₁'=?(?t-v?x/c²

Does this show up on your screen with 4 question marks in the equation?

Yes, I see 4 question marks. I haven't learned how to use LaTeX yet ( ), but the equation should look like this:

Delta(t')=t₂'-t₁'=(gamma)(Delta(t)-vDelta(x)/c²)

It's the Lorentz transformation for a time interval. It's the "interval" version of equation 1d on this page:

http://www.physics.nyu.edu/courses/V85.0020/node45.html

 

[zoobyshoe]

It's the Lorentz transformation for a time interval. It's the "interval" version of equation 1d on this page:

http://www.physics.nyu.edu/courses/V85.0020/node45.html

Tom,

That site is very interesting. I have never seen so many different manifestations of the Lorentz Transformation.

Now the one you used, you say, is the one for a time interval. The time interval you are applying it to is the interval between event one and event two, the times of emission of the rear and forward beams of light.

My first thought is to wonder if this is the right interval to be applying it to. It seems to me that the simultaneity shouldn't be obviously relative to anyone till the time of detection. I think the interval we're supposed to be applying the transformation to, is the interval between detection of the light from the two separate sources by the rail observer: the interval between his detection of the light from the rear flare and his detection of the light from the forward flare. What do you think?

 

[quantumdude]

My first thought is to wonder if this is the right interval to be applying it to. It seems to me that the simultaneity shouldn't be obviously relative to anyone till the time of detection.

No, we should be using the emissions as events, because it is the simultaneous emissions that mark the length of the rod. Whether or not the detctions are simultaneous is irrelevant to the thought experiment. Indeed, as has been noted, the pulses need not be seen by anyone at all.

I think the interval we're supposed to be applying the transformation to, is the interval between detection of the light from the two separate sources by the rail observer: the interval between his detection of the light from the rear flare and his detection of the light from the forward flare. What do you think?

As I said, the time interval between the flares being detected is not relevant. However, it just so happens that in this case (because you've got both observers equidistant from the flares) that the chronological order of emission will be the same as the chronological order of detection.

So the guy in the frame of the track sees the pulses at the same time, and the guy on the rod doesn't.

 

[zoobyshoe]

No, we should be using the emissions as events, because it is the simultaneous emissions that mark the length of the rod. Whether or not the detctions are simultaneous is irrelevant to the thought experiment. Indeed, as has been noted, the pulses need not be seen by anyone at all.

Yes, the simultaneous emissions mark the length of the rod, but the whole point of the experiment is to compare what the rail observer sees with the marks, and ask the question: will the marks be separated by a length equal to the length the rod looked to be when it was in motion?

The observer can't see the emissions at the time of emission. He is ignorant about any apparent length interval between them until their light reaches him.

As I said, the time interval between the flares being detected is not relevant.

I can't agree with this. The detection is all important. The detection tells the observer when the emission occured. Detecting the light simultaneously when he is equidistant from the sources, assures him that the emissions were simultaneous in his frame.[/color]

In Halliday and Resnick the thought experiment goes like this:

Put flares on the ends of a rod of proper length L₀, and connect them to a switch so that an observer can ignite them. Let the rod move by at a velocity v on a track, so that the ingited flares can leave marks on the track. Now let the observer ignite the flares simultaneously, in his frame[/color] (The reason for simultaneous ignition is that it is the only way you could correctly say that the distance between the marks is equal to the length of the rod).

When you say nobody even has to be there to see them, that is only because you offer a mechanical substitute for the observer which is the same thing: an observer. There has to be an observer, human or mechanical, to detect the light simultaneously. Otherwise the conditions of the experiment are not fullfilled. The whole thing is about the difference, if any, between what happens and what an observer sees.

However, it just so happens that in this case (because you've got both observers equidistant from the flares) that the chronological order of emission will be the same as the chronological order of detection.

I can't claim to understand the formula you're using very well but it seems, from what you say, to be one you can use to transform the value of a measured time interval in one frame to what that interval will measure in another frame. If that's the case, then yes it should apply equally well to the detection interval.

However, I believe that transforming the emission interval is barking up the wrong tree. The time of emission is a time of ignorance for both observers: the rail guy and the guy on the rod. I believe the thought experiment, as presented, allow us only one objective piece of information which can be transformed from the rail frame to the rod guy: the interval of detection for the rail guy.

 

[quantumdude]

Yes, the simultaneous emissions mark the length of the rod, but the whole point of the experiment is to compare what the rail observer sees with the marks, and ask the question: will the marks be separated by a length equal to the length the rod looked to be when it was in motion?

The observer can't see the emissions at the time of emission. He is ignorant about any apparent length interval between them until their light reaches him.

Whether or not he can see the flashes is irrelevant. All he needs to know is that they are simultaneous (and he does not need to see them to know that) and how far apart they are. If he knows that they were simultaneous, then he can go up to the marks at his leisure and measure the distance between them.

Tom: As I said, the time interval between the flares being detected is not relevant.

zoobyshoe: I can't agree with this. The detection is all important. The detection tells the observer when the emission occured. Detecting the light simultaneously when he is equidistant from the sources, assures him that the emissions were simultaneous in his frame.[/color]

I mean that the time interval between the two lights being detected is not relevant to the calculation with the Lorentz transformation. Yes, we need to know the time and location of the detection so that we can go back and determine when and where the emission occured. But you were inquiring about what we should be applying the Lorentz transformation to, and the time interval for detection of the pulses is not it. We need to apply it to the time interval for emission to get the length of the rod in each frame.

When you say nobody even has to be there to see them, that is only because you offer a mechanical substitute for the observer which is the same thing: an observer. There has to be an observer, human or mechanical, to detect the light simultaneously. Otherwise the conditions of the experiment are not fullfilled. The whole thing is about the difference, if any, between what happens and what an observer sees.

No, the light does not have to be detected at all. The reason is that the light is not the only indicator of the ignition. For instance, you can have each flare trip a switch that stops a stopwatch at the moment of ignition.

Yes[/color], we have to take experimental data of some sort on the time and place of ignition in each frame.

No[/color], the light from the flares need not be observed by anyone or anything. The flares and the light from them are purely incidental. The exact same thought experiment could be done with lightless markers, and it wouldn't change a thing.

I can't claim to understand the formula you're using very well but it seems, from what you say, to be one you can use to transform the value of a measured time interval in one frame to what that interval will measure in another frame. If that's the case, then yes it should apply equally well to the detection interval.

The Lorentz transformation does apply to any interval, including the detection interval. But I am not bothering with Lorentz transforming the detection interval, because that interval is not related to the length of the rod in any frame. The observer could be anywhere in his frame, and the time interval between detections could be anything according to him. It makes no never mind whatsoever to the simultaneity (or lack thereof) of the events, except of course that whatever time interval he does measure between detected light pulses will be consistent with the simultaneity/nonsimultaneity of the events in his frame.

However, I believe that transforming the emission interval is barking up the wrong tree. The time of emission is a time of ignorance for both observers: the rail guy and the guy on the rod. I believe the thought experiment, as presented, allow us only one objective piece of information which can be transformed from the rail frame to the rod guy: the interval of detection for the rail guy.

No. You can use the information about the detection to calculate back to the information about the emission, and indeed you must. That's the only way we can say anything about the length of the rod according to any observer. It's not the detections that are used to measure the length of the rod, it's the emissions.

 

[zoobyshoe]

Tom,
I think I found the equations you are using in one of my books.

Are these them? :

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

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

 

[quantumdude]

Tom,
I think I found the equations you are using in one of my books.

Are these them? :

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

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

Those are them. I'll learn to TeX this weekend.

 

[zoobyshoe]

Here's gamma:

\gamma = \frac{1}{\sqrt{1- \beta^2}}

\beta = \frac {v}{c}

 

[zoobyshoe]

Or:

\gamma = \frac{1}{\sqrt{1 -(\frac{v}{c})^2}}



Tex tips:

gamma = \gamma

fraction = \frac followed by {stuff in numerator}{denominator} (that is: you indicate the difference with these: {}{} numerator first, then denominator

Square root = \sqrt followed by: {everything you want under the square root sign}

beta =\beta

Delta (different than \delta) = \Delta

squared = number to be squared followed by ^2

[/ tex] to end the whole line, and to begin it.<br /> <br /> Add more to your store of abilities as you go by clicking on the quote button of any post with some Tex you want to be able to use, and observe how they did it, or look through the Tex thread till you find what you need.<br /> <br /> Always good to preview your post before submitting because it&#039;s easy to miss a \ or an ending } when you&#039;re typing along. The &quot;preview post&quot; feature in any thread can be used to practise to your heart&#039;s content, I also realized. (Just abandon it when you&#039;re done, don&#039;t hit &quot;submit reply&quot;.)

 

[zoobyshoe]

In Halliday and Resnick the thought experiment goes like this:

Interestingly enough, I found these:

\Delta t=\gamma(\Delta t&#039;+v\Delta x&#039;/c^2)

\Delta t&#039;=\gamma(\Delta t-v\Delta x/c^2)

in Halliday and Resnik. I didn't even realize I had this book. I pick up physics texts all the time at the swap meet for peanuts just in case one will have a explanation of something that's easier to grasp than some other source.

Any way, what edition do you have? I can't find the flare gedanken in mine, at least not in the relativity part. They spread some relativity into other parts, though.

 

[quantumdude]

Any way, what edition do you have?

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

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".

 

[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&#039; = \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&#039; = 1.1547005(-150,000/9^1^0)

so: \Delta t&#039; = 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\&#039;\; = \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\&#039;\; = \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\&#039;\; = \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\&#039;\; = \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&#039; = \gamma (\Delta x - v\Delta t). Since \Delta t = 0, \Delta x = \Delta x&#039;/\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&#039;, the time interval between the flares according to the rod frame. From the LT, \Delta t&#039; = -\Delta x&#039; v/c^2, thus \Delta t&#039; = 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?