12
   

The Red Shift without Expansion

 
 
layman
 
  -1  
Reply Tue 7 Feb, 2017 12:47 pm
@McGentrix,
Are you familiar with the lorentz transformations, Gent?
McGentrix
 
  1  
Reply Tue 7 Feb, 2017 12:49 pm
@layman,
yes
layman
 
  -1  
Reply Tue 7 Feb, 2017 12:52 pm
@McGentrix,
OK. This is standard theoretical stuff. You want to get into the mechanics of how things are measured, but that's irrelevant here.

Do you agree that, theoretically, there is a relative speed (offhand I'd guess about .8c) where the yardsticks of the moving object to shrink to the 50% of the size of those on the RF which is not moving?
McGentrix
 
  0  
Reply Tue 7 Feb, 2017 01:17 pm
@layman,
Sure, close enough.
layman
 
  -1  
Reply Tue 7 Feb, 2017 01:25 pm
@McGentrix,
OK. The point (or one of them) is that if you took a yardstick from B (18 inches) and put it on A, then you would measure a football field to be 200 yards.

Right?
McGentrix
 
  1  
Reply Tue 7 Feb, 2017 02:27 pm
@layman,
layman wrote:

OK. The point (or one of them) is that if you took a yardstick from B (18 inches) and put it on A, then you would measure a football field to be 200 yards.

Right?



But we've just established that we know that length contraction is a thing and we would adjust for that. So, no.
layman
 
  -1  
Reply Tue 7 Feb, 2017 02:50 pm
@McGentrix,
McGentrix wrote:

layman wrote:

OK. The point (or one of them) is that if you took a yardstick from B (18 inches) and put it on A, then you would measure a football field to be 200 yards.
Right?


But we've just established that we know that length contraction is a thing and we would adjust for that. So, no.


Well, George we're just talking theoretically here. If we actually went to B and brought a yardstick back, it would also be 3' long again once we it got back. But we could just make a "replica" of what we "know" about it's length.

So, just stay with me a minute. You're making my point for me, I think.

Here was one of the questions I first posed in the illustration. We use one 18" stick. If we "call" it a yard, then we will measue 200 yards, BUT

We would just then say it wasn't "really" a yard. The 200 yards measurement was a mis-measurement caused by a distorted yardstick. We wouldn't say it was "really" 200 yards just because we had a yardstick that told us it was.

That's what you're saying too, isn't it?
McGentrix
 
  1  
Reply Tue 7 Feb, 2017 03:05 pm
@layman,
No, what I am saying is that there is a bunch of math involved and to do the calculation you are looking for requires that the math gets done and done correctly. That's why they have formulas that take length contraction and time dilation into account.

Math, this is why it exists.

This is one of those times where you explain what your point is instead of continually asking leading questions.
layman
 
  -1  
Reply Tue 7 Feb, 2017 03:07 pm
@McGentrix,
McGentrix wrote:

No, what I am saying is that there is a bunch of math involved and to do the calculation you are looking for requires that the math gets done and done correctly. That's why they have formulas that take length contraction and time dilation into account.

Math, this is why it exists.

This is one of those times where you explain what your point is instead of continually asking leading questions.


Ok, then explain your point to me--the one you just made, can you? Because I don't understand it at all. If by "bunch of math" you mean the LT calculation, we've accounted for that. It was the application of that formula that told us the yardsticks on B would only be 50% of what ours are, remember?
layman
 
  -1  
Reply Tue 7 Feb, 2017 03:18 pm
@layman,
It doesn't add anything at all to the conversation, but, if you somehow think the "calculations" are crucial to answering the question (they aren't) I will say this:

The calculations show that if the moving object is traveling at 259627884.49098 m/s, then the yardsticks on B will shrink to 18 inches. That appears to be about .866c. I had "guessed" that it would be about .8c--and you said "close enough," remember?
0 Replies
 
layman
 
  -1  
Reply Tue 7 Feb, 2017 03:47 pm
@McGentrix,
McGentrix wrote:

No, what I am saying is that there is a bunch of math involved and to do the calculation you are looking for requires that the math gets done and done correctly. That's why they have formulas that take length contraction and time dilation into account.

Math, this is why it exists.

This is one of those times where you explain what your point is instead of continually asking leading questions.


So, having done the calculations, can you answer this question now?

Quote:
Here was one of the questions I first posed in the illustration. We use one 18" stick. If we "call" it a yard, then we will measue 200 yards, BUT

We would just then say it wasn't "really" a yard. The 200 yards measurement was a mis-measurement caused by a distorted yardstick. We wouldn't say it was "really" 200 yards just because we had a yardstick that told us it was.

That's what you're saying too, isn't it?

McGentrix
 
  0  
Reply Tue 7 Feb, 2017 05:16 pm
@layman,
Imagine that you are in space on a space station and the USS Enterprise is drifting away from the space station at .25c and the chief engineer tells you that they want to use a tractor beam to catch it but they need to know the distance away that it is. They give you a 1 meter stick and place you at an air lock...

How are you going to determine the distance away the Enterprise is?
layman
 
  0  
Reply Tue 7 Feb, 2017 05:22 pm
@McGentrix,
McGentrix wrote:

Imagine that you are in space on a space station and the USS Enterprise is drifting away from the space station at .25c and the chief engineer tells you that they want to use a tractor beam to catch it but they need to know the distance away that it is. They give you a 1 meter stick and place you at an air lock...

How are you going to determine the distance away the Enterprise is?


Heh, Gent.

Can we stay on topic? Tell you what. Answer my question, first--I asked first. Then I'll see if I can answer yours--even though yours is not relevant to the theoretical topic I'm trying to discuss with you. How's that?
McGentrix
 
  2  
Reply Tue 7 Feb, 2017 05:26 pm
@layman,
My question is your question phrased as an actual question. Answer it and you answer your own question at the same time.
layman
 
  0  
Reply Tue 7 Feb, 2017 05:34 pm
@McGentrix,
McGentrix wrote:

My question is your question phrased as an actual question. Answer it and you answer your own question at the same time.


No, there's no possible way that me answering your question would even begin to answer mine. I can see that you're not following me at all.

The speed of light is often stated in terms of meters per second. You know what a meter is? It's approximately one yard. You know what 1,000 meters is? It's the equivalent of 1,000 meter sticks laid end to end. The units of measurement are not important here.

The question of just exactly HOW you would determine that something was moving away from you at approx. 260 million meters per second is also completely irrelevant to the question.

We're talking theory here, not lab experimentation. Theory, not practice. Or at least that's what I'm talking about.

Can you answer the question?

You can try to do it by answering your own question, if you think that's a relevant answer, but I don't see how it could be.
Krumple
 
  1  
Reply Tue 7 Feb, 2017 05:46 pm
@layman,
layman wrote:

McGentrix wrote:

My question is your question phrased as an actual question. Answer it and you answer your own question at the same time.


No, there's no possible way that me answering your question would even begin to answer mine. I can see that you're not following me at all.

The speed of light is often stated in terms of meters per second. You know what a meter is? It's approximately one yard. You know what 1,000 meters is? It's the equivalent of 1,000 meter sticks laid end to end. The units of measurement are not important here.

The question of just exactly HOW you would determine that something was moving away from you at approx. 260 million meters per second is also completely irrelevant to the question.

We're talking theory here, not lab experimentation. Theory, not practice. Or at least that's what I'm talking about.

Can you answer the question?

You can try to do it by answering your own question, if you think that's a relevant answer, but I don't see how it could be.


If you are talking about the length of the ship decreasing, its only a visual from any reference point NOT on the ship. If you are on the ship with a .25c velocity, everything will seem normal.

Measuring the distance to a stationary object behind you is complicated if you are traveling a way at .25c velocity. It will act as if space is stretched further behind the ship from the ship's reference point however if you were to measure from another reference frame youll get a different result.
layman
 
  0  
Reply Tue 7 Feb, 2017 05:49 pm
@Krumple,
Quote:
If you are talking about the length of the ship decreasing, its only a visual from any reference point NOT on the ship.


What do you mean by "only a visual?" Are you saying that the LT formula doesn't tell us anything? That it's just irrelevant math that doesn't tell us anything except what an eye would see, or what?

Edit: Oops. I though that was Gent I was responding to, at first. I see you're back, Krumps. I asked you a question too. Did you see it?
Krumple
 
  1  
Reply Tue 7 Feb, 2017 06:42 pm
@layman,
layman wrote:

Quote:
If you are talking about the length of the ship decreasing, its only a visual from any reference point NOT on the ship.


What do you mean by "only a visual?" Are you saying that the LT formula doesn't tell us anything? That it's just irrelevant math that doesn't tell us anything except what an eye would see, or what?

Edit: Oops. I though that was Gent I was responding to, at first. I see you're back, Krumps. I asked you a question too. Did you see it?


Quickly restate the question for me so I'm clear which one.

If you were to measure the ship's length from a reference point that is not the ship you would measure it shorter than its stationary length. However inside the ship everything would have its normal length.

This is another aspect where two reference points give two different data results but both are correct from their respective reference points.

The reason the ship shrinks as it nears C is due to time not allowing two points of space to overlap. This also causes its mass (the ship) to increase.
Krumple
 
  1  
Reply Tue 7 Feb, 2017 06:48 pm
@layman,
https://youtu.be/lPoGVP-wZv8
layman
 
  0  
Reply Tue 7 Feb, 2017 07:22 pm
@Krumple,
Krumple wrote:



Quickly restate the question for me so I'm clear which one.

The question pertains to objective reality, and was posted shortly after your last post prior to today

If you were to measure the ship's length from a reference point that is not the ship you would measure it shorter than its stationary length. However inside the ship everything would have its normal length.

OK, I agree with this, insofar as SR goes, anyway

This is another aspect where two reference points give two different data results but both are correct from their respective reference points.

That's an assumption, but not a necessary one. But, either way, the qualifying phrase "from their respective reference points" does not even pretend to address the question of which frame, if either, IS correct with respect to objective reality (and not just claims about subjective perspective). You may better understand what I'm getting at once you answer the question I already asked, and related posts made after that.

The reason the ship shrinks as it nears C is due to time not allowing two points of space to overlap. This also causes its mass (the ship) to increase.

I really don't know what you're getting at here, but "the reason" why it shrinks is an entirely different issue than I'm raising, anyway.
0 Replies
 
 

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