I was only agreeing with the final statement that was quoted. I agree that the link's discussion of 'proper time' is sloppy. (He's only talking of inertial frames, not the general case.) Thanks for the heads up.ghwellsjr said:Doc Al, did you read the link that aintnuthin referenced? He only quoted the very end of it.
This same link came up in another thread and my comment was that it is total nonsense:
https://www.physicsforums.com/showpost.php?p=3407874&postcount=16
DaleSpam said:Interesting how you are so willing to ask others to put in a lot of effort to answer a large string of questions, but so unwilling to put in even a minimal amount of effort of your own to do a single exercise.
How about this: Pick one set of questions, then wait for the answers. Don't stack up post after post. (It's all basically the same problem.)aintnuthin said:Doc Al, I have posed a number of questions, but, so far, at least, you have responded rather selectively. Is there a particular reason for that? For example:
1. Am I asking too many questions? More than you care (or have time) to answer?
2. Do you intend to answer them, later, but just don't have sufficient time right now?
3. Do some of my questions strike you as too incomprehensible and/or too ill-composed to respond to?
4. Do some of my questions strike you as being "too stupid" to even merit a response?
I appreciate your assistance, and I am willing to be patient and wait for your answers, if they will be forthcoming. Any answers you give me will, I hope, help me to better understand what you are saying, and many eliminate other questions.
Doc Al said:But in the rocket frame that 'pole' (the stars) is moving, thus Earth observers cannot use the simple length contraction formula on it...You need to use the full LT. Then you'll get the correct result that Earth observers measure a distance of 8 light years.
Doc Al said:No, it's got to be the length of something that is stationary in one frame. That's the proper length of the object. All other frames see a shorter object.
How is an explicit mathematical derivation from the Lorentz transform an "unsupported assertion" and not "a consistent, rational explanation" for this question? Your idea of "a consistent, rational explanation" needs some revision.aintnuthin said:Given the nature of your "answers," which to me are little more than unsupported assertions, without any attempt at providing a consistent, rational explanation
Yes, based on the evidence in this thread, I believe you are lazy. Regarding ability, I think you are perfectly able to understand, I am confident that the math is within your capability; you are just unwilling to make the effort required. Do you think that those of us who understand relativity did so without doing exercises such as the one I suggested for you?aintnuthin said:The thrust of your comments have basically been in the vein of assailing my character (e.g. "lazy'), or my ability to understand.
You can always use the full LT, if you know what you're doing and what the LT represents.aintnuthin said:Is it sometimes impossible to use what you are calling the "full LT,"
Exactly. For special cases, you can use 'shortcuts' such as time dilation and length contraction.or is it just that's it's unnecessary because convenient shortcuts are available?
It's simple. Say you are in a rocket that you measure to be 100 m long. You are traveling at .866c with respect to me. When I measure the length of your rocket, I get 50 m. Any frame moving with respect to you will measure a shorter length for your rocket than you do.Your reference to the "length of an object" raises some questions in my mind, but I really don't want to get sidetracked on them right now. For now I have a question about this statement in particular: " All other frames see a shorter object."
Shorter than what? The very form of the word implies a comparison between two things, and therefore is impossible to interpret, standing alone.
Doc Al said:You can always use the full LT, if you know what you're doing and what the LT represents... For special cases, you can use 'shortcuts' such as time dilation and length contraction.
Doc Al said:It's simple. Say you are in a rocket that you measure to be 100 m long. You are traveling at .866c with respect to me. When I measure the length of your rocket, I get 50 m. Any frame moving with respect to you will measure a shorter length for your rocket than you do.
Right. It cannot be 'at odds' with it, since those shortcuts are derived from the full LT. But you must know when and where those 'shortcuts' can be applied.aintnuthin said:OK, thanks. Would it be fair to assume, then, that using a shortcut (properly) would never give you an answer which was at odds with the one you would arrive at if you had taken the longer route and used the full LT (properly)?
Exactly.So, by implication, if the diameter of the Earth was, say, 8,000 miles, and I was in a rocket traveling .866c relative to you, then I would measure that diameter to be 4000 miles? Is that correct?
It may be useful to highlight the "special cases":Doc Al said:Exactly. For special cases, you can use 'shortcuts' such as time dilation and length contraction.
I'll respond to that post below.aintnuthin said:OK, thanks. I'm still trying to put all of this together. Post 45 ended with this question: "They are two different questions, with two different answers, right?"
My thinking there was that if you were given a certain piece of information, then two, not just one, questions could arise about the relationship between the two objects in question. Do you spot some flaw or improper assumption in that post?
The statement 'measures a distance' is somewhat ambiguous. Shall we assume that he's measuring the length of something at rest with respect to him?aintnuthin said:One last way of approaching this:
GIVEN: Henry measures a distance to be 6LS in his frame. With respect to an object traversing that same distance at a speed of .6c with respect to him, TWO separate questions arise.
It's got nothing to do with Henry being stationary. It's got to do with the object he's measuring being at rest in his frame.Question 1: Using the LT, what, according to Henry, will the other party measure that same distance to be in their frame?
Answer: 4.84 seconds. Why? Because he will assume that he is stationary, and will therefore expect the "moving" party to measure that same distance to be shorter. Their distance will be contracted, relative to him, or so he calculates, using the LT.
To use the length contraction formula correctly, the 'thing' whose length is being measured must be at rest in some frame. Obviously, it can only be at rest in one frame. So it's either at rest in Henry's frame or Harriet's frame. Pick one.Question 2: Using the LT, and assuming that the other party calculates Harry's distance of 6 LS to be contracted, relative to themself, what will the other party (Harriet, here) calculate their own distance to be?
Answer: 7.5 light seconds. Compared to Harriet the "moving party" (Henry) will have the shorter distance (of 6 LS), as she calculates it.
Not just two different questions, but two different scenarios.They are two different questions, with two different answers, right?
This is very strange. You said you had a discussion of relativity on your blog. Now you appear to be saying you consider relativity to be "imputted by fiat". The symmetry of this situation is exactly what relativity says. It is, indeed, the reason for the name!aintnuthin said:It's fine to recognize symmetry, when it exists. It is wrong to impute it by fiat when it does not exist.
Doc Al said:The statement 'measures a distance' is somewhat ambiguous. Shall we assume that he's measuring the length of something at rest with respect to him?
Seems OK. I'll summarize: Henry measures the distance between the clocks (which are at rest in his frame) to be 6 LS. Anyone who knows relativity can use simple length contraction to deduce what some other frame would measure as the distance between those clocks. If they move at .6c, they will measure the distance between the clocks to be contracted to 4.8 LS.aintnuthin said:Given that distinction, I would say that Henry "measures" the distance between the two clocks in his frame to be 6LS. If my understanding is correct, he could then "transform" those very same measurements onto a frame moving with respect to him. By "transform," I mean impute his measurements, via calculation, to another frame.
In that case he is not "measuring" anything in the other frame. All measurements are made in his own frame. He nonetheless "calculates" that if he measures that distance to be 6 LS, then a person in a frame moving at .6c with respect to him would perceive it to be merely 4.84 LS.
Is there something wrong with the way I view this?
HallsofIvy said:This is very strange. You said you had a discussion of relativity on your blog. Now you appear to be saying you consider relativity to be "imputted by fiat". The symmetry of this situation is exactly what relativity says. It is, indeed, the reason for the name!
Doc Al said:Seems OK. I'll summarize: Henry measures the distance between the clocks (which are at rest in his frame) to be 6 LS. Anyone who knows relativity can use simple length contraction to deduce what some other frame would measure as the distance between those clocks. If they move at .6c, they will measure the distance between the clocks to be contracted to 4.8 LS.
No it doesn't, for the nth time. (His speed with respect to himself is of course zero.)aintnuthin said:Now all that presupposes that he is stationary, which, by hypothesis, he is not.
I'm not getting your logic. Of course he 'acknowledges' that he and Harriet are in relative motion. But that's got nothing to do with his measurement of the distance between the clocks, since the clocks are at rest with respect to him.If he acknowledged (rather than denied) that he was moving at .6c and assumed that Harriet was stationary, how would he impute his own measurement to her frame in that case?
Doc Al said:No it doesn't, for the nth time. (His speed with respect to himself is of course zero.)
I'm not getting your logic. Of course he 'acknowledges' that he and Harriet are in relative motion. But that's got nothing to do with his measurement of the distance between the clocks, since the clocks are at rest with respect to him.
Doc Al said:Of course he 'acknowledges' that he and Harriet are in relative motion. But that's got nothing to do with his measurement of the distance between the clocks, since the clocks are at rest with respect to him.
Nope. The concept of "really" moving is meaningless. There is only relative motion.aintnuthin said:Well, I guess I'm not sure what is unclear here. My logic is kinda like this.
1. If any two objects are moving with respect to each other, at least one of them must be "really" moving, even if both are always "at rest" with respect to themselves.
Nope.2. Maybe you know who's really moving, maybe you don't, but, either way, you know one of the two is.
Nope.3. If you don't "know" then let's say the odds are 50-50 that you are moving with respect to the other, and that it is "stationary" with respect to you.
Nope. An observer always measures things from his own frame.Everyday example: I am going down the freeway at 60 mph. As between me and the road, I know ONE of us is (relatively) stationary and that one of us is (relatively) moving. In that case, I would assume that I'm the one moving relative to the road (and all other fixed things on the planet that I see), not that they are all coming to, and going past, me. In such a case I would NOT employ the LT from the perspective that *I* was stationary and that something fixed (say a tree) that was "approaching" me was "really moving."
Yep. All wrong.On the contrary, I would employ my knowledge of SR (such as it is) to calculate times and lengths from "in reverse." I would assume that Earth's clocks would be faster (not slower) than mine and that it's lengths would be longer (not shorter) than mine are in my frame. That, for example, there would be some (however infintesimal) difference in the rate at which my watch keeps time, and the rate it keeps time when I come to a stop sign. As I sped up, I would assume that my watch was running slower than the one I saw the guy standing at the stop sign wearing. I would assume that his watch was now faster than mine, and mine slower than his. Would I be wrong to make such assumptions?
All you need to know is the relative speed between the two frames.aintnuthin said:I agree, it doesn't affect the measurements that I make in my frame in the least. But it could affect how I attempt to impute (transform) those measurements to her frame via calculation, couldn't it?
This is wrong, you must get rid of this idea. It completely misses the point of relativity, not just Einstein's special relativity but also Galilean relativity of classical physics.aintnuthin said:Well, I guess I'm not sure what is unclear here. My logic is kinda like this.
1. If any two objects are moving with respect to each other, at least one of them must be "really" moving, even if both are always "at rest" with respect to themselves.
Doc Al said:Nope. The concept of "really" moving is meaningless. There is only relative motion.
Nope.
Nope.
Nope. An observer always measures things from his own frame.
The only 'assumption' is the relative motion between the frames.aintnuthin said:As i said (next), I agree that an observer always measures things from his own frame. But, likewise, he always imputes (to others) from his own frame, too. How, and what, he imputes would seem to depend on his assumptions.
Yes. You are mixing up measurements in different frames. The key problem is your assumption that one of the frames is 'really' at rest. See DaleSpam's response.Doc, that's sure a lot of "nopes," without any elaboration on your part. Is what I'm suggesting impossible?
DaleSpam said:This is wrong, you must get rid of this idea. It completely misses the point of relativity, not just Einstein's special relativity but also Galilean relativity of classical physics.
Even in Newtonian physics there is no such thing as "really" moving, only moving relative to something else. This is called the "first postulate" or the "principle of relativity" and has been experimentally verified fact since Galileo's time:
http://en.wikipedia.org/wiki/Principle_of_relativity
It seems that all of your errors stem from this point. This is the fundamental symmetry I was describing above. If you want to learn relativity you definitely need to learn this point well.
Sidetracked?? Until you disabuse yourself of the notion of something 'really' being at rest, don't expect to make much progress.aintnuthin said:I don't agree with them all, but don't wish to get sidetracked on those topics.
Doc Al said:The only 'assumption' is the relative motion between the frames.
Where exactly did I say that? All one can ever say is that something is or isn't stationary with respect to something else. (Of course Harriet, and everyone else, treats themselves as being stationary in their own frame.)aintnuthin said:Doc, this assertion that there is only one assumption has confused me from the outset. After asserting that, initially, you proceded to say Harriet's measurements were determined "because she was stationary."
Your analysis is so simple. I used the Lorentz Transform to analyze the same scenario but as you can see it is very much more complicated although I do get the same answer. Could you provide insight into how you do it so much more easily?DaleSpam said:This conversation has become boring so I will work the problem.
Worldline of clock 1
x'=0
1.25 (0.6 c t + x) = 0
x=-0.6 c t
Worldline of clock 2
x'=6
1.25 (0.6 c t + x) = 6
x = 4.8 - 0.6 c t
So we immediately see that the distance between the clocks is 4.8 lightseconds.
Worldline of Harriet
x' = 0.6 c t
1.25 (0.6 c t+x)=0.75 c \left(\frac{0.6 x}{c}+t\right)
x = 0
Finding the intersection of Harriet's worldline with clock 1 we find:
0 = -0.6 c t
t = 0
Finding the intersection of Harriet's worldline with clock 2 we find:
0 = 4.8 - 0.6 c t
t = 8/c
So we see that the time was 8 seconds.
Claimant B is correct. QED.
I am glad that you liked it!ghwellsjr said:Your analysis is so simple. I used the Lorentz Transform to analyze the same scenario but as you can see it is very much more complicated although I do get the same answer. Could you provide insight into how you do it so much more easily?
One tip for making it more concise. If you are using four-vectors then you can cast the Lorentz transforms as matrices instead of as equations.ghwellsjr said:So now we use the Lorentz Transform on each of these six events in Henry's FoR to see what they are in Harriet's FoR.
First we want to calculate gamma, γ, which is equal to 1/√(1-ß²) where ß is the speed as a fraction of c. So:
γ = 1/√(1-ß²) = 1/√(1-0.6²) = 1/√(1-0.36) = 1/√(0.64) = 1/0.8 = 1.25
Then we use these two formulas for each t and x coordinate:
t' = γ(t-vx/c²)
x' = γ(x-vt)
That makes all of these into matrix multiplications.ghwellsjr said:Now we do the detailed calculations. Some of the events are used twice so we only have to do four sets of calculations.
[0,0]:
t=0 and x=0
t' = γ(t-vx) = 1.25(0-0.6*0) = 0
x' = γ(x-vt) = 1.25(0-0.6*0) = 0
[0,6]:
t=0 and x=6
t' = γ(t-vx) = 1.25(0-0.6*6) = 1.25(-3.6) = -4.5
x' = γ(x-vt) = 1.25(6-0.6*0) = 7.5
[10,0]:
t=10 and x=0
t' = γ(t-vx) = 1.25(10-0.6*0) = 12.5
x' = γ(x-vt) = 1.25(0-0.6*10) = -7.5
[10,6]:
t=10 and x=6
t' = γ(t-vx) = 1.25(10-0.6*6) = 1.25(10-3.6) = 1.25(6.4) = 8
x' = γ(x-vt) = 1.25(6-0.6*10) = 0
For the time dilation your approach is nice and easy.ghwellsjr said:And substituting the coordinates from Henry's FoR to Harriet's FoR:
[0,0] Location of clock1 at start of scenario
[-4.5,7.5] Location of Henry and clock2 at start of scenario
[12.5,-7.5] Location of clock1 at end of scenario
[8,0] Location of Henry and clock2 at end of scenario
[0,0] Location of Harriet and her clock at start of scenario
[8,0] Location of Harriet and her clock at end of scenario
From this you can see that the time from start to end of the scenario for Harriet is 8 seconds.
This is probably the most direct way to get the distance from what is given, but it is still somewhat indirect. The problem is, of course, the relativity of simultaneity. Length is the distance between two worldlines, so when I do a problem that involves length contraction I like to use worldlines instead of events. You can still do that using the four-vector notation:ghwellsjr said:There are many ways that Harriet can measure the distance between Henry's two clocks. Probably the simplest is for her to measure how long it takes between clock1 and clock2 arriving at her location and knowing the speed of the clocks (and Henry) she can calculate the distance as simply the speed multiplied by the time interval. This would be 0.6 times 8 or 4.8 light-seconds.
No worries, it is hard for someone who doesn't understand the principle of relativity to understand the rest of relativity.One Brow said:When I first posted, aintnuthin pretty much assured me you would correct my erroneous notions. Thank you all for the explanations offered. I was not aware he planned on following me here. I won't trouble you again.
