taeto

5 hours ago, DimaMazin said:

The point on arc has coordinates([Pi-2a]/infinity ; 1), the point on chord has coordinates (cos(a)/infinity ; sin(a)).

What do you mean by dividing by "infinity"? When extending the real numbers by new elements \(\{-\infty, +\infty\},\) and extending the usual arithmetic of real numbers to the new structure, usually you get \(0\) when you divide a real number by \(\infty\). Maybe your "infinity" is not the same as \(\infty\)?

There are so many misunderstandings and factual mistakes in the two quotes that it is hard to know where to begin. If time is viewed as an object of philosophical study, then the word may just happen to describe something else than the version of time which in physics represents one of the coordinates of an event and which can be measured. And in that case I am not qualified to comment. But it is clear that the ideas that are presented about how mathematics works are somewhat misguided. What stands out is that whereas "intuitionist mathematics", which actually does not describe any well-defined theory or language, does do different things than usual mathematics: It is notoriously unable to do more than usual mathematics already does. It differs mainly by declaring that there are facts that cannot be shown by using their more restricted logic. In particular, if you can get something out of an intuitionist argument, you can get the exact same from a usual argument, quite possibly with less work.

Anyway,  as one example from the quotes, to suggest that intuitionists "reject the existence of numbers with infinitely many digits" is just hilarious. No, numbers do not have digits; it is the decimal representation of a real number that has digits. The theory of real numbers and their arithmetic is extremely well understood and known to be consistent and complete. It is the theory of the arithmetic of whole numbers that is known to be incomplete and not known to be consistent, presumably those are the kinds of numbers the existence of which is not rejected by intuitionists. 

7 minutes ago, StringJunky said:

Ignorance, probably.

So the probable meaning is that physicists are ignorant about time. Or just "ignorance surrounds time in physics", which somehow seems a more general statement. Good to know in either case. The "time" that is spoken of may or may not be the usual fourth coordinate of spacetime, or possibly neither one of those possibilities. 

Dima, your guess is that there is some value of \(a\) for which the value of \(y\) does not change if we look at all possible value of \(t\), is that right?

My guess is that this is not true. Meaning that for any value of \(a\) that we choose, there will be at least two different values of \(t\) that produce different values of \(y\).

Do you agree that my description of our guesses is correct?

Now, since you are the one who says that there exists a specific value of \(a\) which wins the argument for your case, do you not think that you should have to present this particular value of \(a\)? Are you waiting for someone who is smarter than you to do so?

1 minute ago, StringJunky said:

It's a figure of speech with the analogy of a 'fog' being that which obscures what we want to know.

That being what?

What kind of object is "the fog surrounding time in physics" exactly? This all sounds like a pseudoscience scam. But in Quanta, really?

I will try to figure out bold math symbols.

Simply typing everything in bold: S^2 = (ct)^2 - x^2 - y^2 makes \(S^2 = (ct)^2 - x^2 - y^2 \) so doesn't do anything.

Using bm: \bm{S^2 = (ct)^2 - x^2 - y^2} produces \(\bm{S^2 = (ct)^2 - x^2 - y^2}\).

With mathbf: \mathbf{S^2 = (ct)^2 - x^2 - y^2} \(\mathbf{S^2 = (ct)^2 - x^2 - y^2} \) it works almost like it should, except for the missing italics.

There is a mathrm, so why not \mathit{\mathbf{S^2 = (ct)^2 - x^2 - y^2} }? \(\mathit{\mathbf{S^2 = (ct)^2 - x^2 - y^2} }\). Nope.

Or maybe \mathbf{\mathit{S^2 = (ct)^2 - x^2 - y^2} }: \(\mathbf{\mathit{S^2 = (ct)^2 - x^2 - y^2} }\). Interesting.

And the slow approach \boldsymbol{S}^2 = (ct)^2 - x^2 - y^2: \(\boldsymbol{S}^2 = (ct)^2 - x^2 - y^2\).

But \boldsymbol{S^2 = (ct)^2 - x^2 - y^2} \(\boldsymbol{S^2 = (ct)^2 - x^2 - y^2}\) actually nearly does it. Neat. But now the = and - are bold too.

So it must be \boldsymbol{S}^2 = (\boldsymbol{ct})^2 - \boldsymbol{x}^2 - \boldsymbol{y}^2: \(\boldsymbol{S}^2 = (\boldsymbol{ct})^2 - \boldsymbol{x}^2 - \boldsymbol{y}^2\). Tedious but alright.

Maybe \bs: \bs{S}^2 = (\bs{ct})^2 - \bs{x}^2 - \bs{y}^2? \(\bs{S}^2 = (\bs{ct})^2 - \bs{x}^2 - \bs{y}^2\). Well, then not. Done with this.

46 minutes ago, jamesfairclear said:

my proposition to make a small change to the definition of speed.

It could be handy for anyone facing speeding charges in traffic court.

2 hours ago, MigL said:

underplaying his stupidity.

To promote hydroxychloroquine as a remedy against coVid-19 does not seem stupid at all, provided you own a lot of stocks in hydroxychloroquine manufacturing companies. 

Same if you replace "hydroxychloroquine" by "magic beans" in the previous sentence.

4 minutes ago, Phi for All said:

In which, hopefully, you dream of more than Heaven and Earth, but not more than is reasonable.

I do not know about dim. For myself, I am willing to accept that the big Spaghetti Monster in the sky is capable of doing anything it wants. But at least from experience, I am mostly pretty sure what it will do in any given circumstance.

10 minutes ago, dimreepr said:

unless you have learned/experienced what I have... 

That is what you call a "tough proposition" right there.

Why did you tell me that you know that \(a\) is not equal to \(\pi/6?\) I have to understand this first.

6 minutes ago, Bmpbmp1975 said:

When you say other answers are possible we are still always in the range of about 13 billion years old right. We are talking - difference of millions of years and not billions of years?

Exactly. Unless our interpretation of our observations is very much off, we are very close to the 13.8 billion year mark, relatively speaking. It would be surprising if our measurements have fooled us so much that the actual value does not lie between 13.7 and 13.9. 

6 minutes ago, StringJunky said:

The visualization problem is the same as infinitely big. At the end of the day, it's a problem with the model in question.

Sure, it depends on the theory you want to engage. A unit line segment has volume zero in the Euclidean plane, and measure one on the real line, and it has cardinality \(2^{\aleph_0}\) as a set of points in the real plane. 

1 minute ago, StringJunky said:

Picture yourself travelling between two intersecting lines and yourself getting smaller in proportion as the lines get closer. When will you reach the end?

If I will not reach the end, then never. If I will, then, at whichever time that will be. So "vanishing", I get it now. At the end, if I reach it, I would have vanished to zero. But I am not sure that your image tells us much about places that are "infinitely small".

9 minutes ago, StringJunky said:

But you said:

What is the dimension of a point at the apex? Vanishingly small. It is not zero.

I only know Euclidean geometry and its various extensions that are locally Euclidean. The dimension of a point is zero here, and if you go to measure theory, then its size in terms of volume is zero as well. 

27 minutes ago, DimaMazin said:

a is constant . It exactly cannot be equal Pi/6. sin a is constant. You can use t₁  or t₂ instead of t . t , cos t and sin t  are variables . And y of point of definition is constant. I don't understand why you don't understand your diagram. When you take t₁ and t₂ you get L₁ and L₂ . Therefore you can do equation for definition of sin( a ) and cos(a) (using my equation for "a" definition)

I understand my own diagram in the way that it was constructed: there is an angle \(t\) that can be chosen freely, and there is an angle \(\alpha,\) or \(a\) as you call it now, which can also be chosen freely, except of course their sum \(t+\alpha\) must be less than \(\pi/2.\) There is nothing there about \(t\) or \(a\) being constant. Then the diagram contains an expression for the resulting value of the \(y\)-intercept of interest.

So if you now say that your \(a\) is a constant, and you also say that it is not \(\pi/6\) (which is a constant), then how do you know that \(a\) is not \(\pi/6?\) Is it because you already know what \(a\) is? Can you explain how you know this, and what you think is the actual value of \(a,\) as a fraction of \(\pi\) or in decimal expansion?

1 minute ago, StringJunky said:

Infinitely small = Vanishing spatial extension.

I actually think that was the point.

7 minutes ago, Airbrush said:

On a popular science program, "infinitely" small is just a figure of speech, not to be taken literally.  I should have known.

As to the idea that the universe began from a "point" that was "infinitely small"...well it seems to me it could just as well started with a diameter of a googol light years, and was shaped like crusty the clown.

On the other hand, if you look inside a black hole (don't go too close), then I suppose that mathematically, it has a point of size zero in its center which contains all its mass. Somehow the exact opposite of a photon, if you will. Whether this is absolutely a correct description or not of the reality or not, it could suggest why popular depictions of the big bang start from a very similar picture.

4 hours ago, Bmpbmp1975 said:

This post has gotten confused I thought I understood and now I think I am confused again. Lol

From my understanding the our universe is 13.7 ish billions years of age five or take 20 millions. Now différence variances are also included in wiki article age of universe still out is in the same ball park of 13.8billion. 

the last few comments from what understand seems to be insinuating this can be completely wrong or am I misunderstanding 

From a statistical point of view, no, you are not misunderstanding. The figures that came up from combined measurements were 13.8 billion years plus/minus 20 million. As swansont points out, we take this to mean that 13.8 billion is the most likely true answer, but other true answers are possible, according to a normal distribution (Bell curve) with standard deviation of some million years, probably either 10 or 20 million. Because the standard deviation is so relatively small, it is very unlikely based on our measurements that the age can be as small as 13.7 billion or as large as 13.9 billion years. Although from a statistical point of view, neither is completely impossible, only extremely unlikely. The estimate is also based on our current model of the universe.

1 hour ago, Airbrush said:

Nothing can be infinitely small.

To say that something is "infinitely" small is certainly an unfortunate mode of description. And meaningless unless it really means that its size is zero.

If we were to say that the mass of a photon is "infinitely small", we would make ourselves easy targets of ridicule. However, the mass of a photon is in reality not "infinitely small", rather it is identically zero. If we take "infinitely small", as a manner of speaking, to mean "of size zero", then we can all have a night of peaceful sleep. Except of course those who insist that nothing can be "infinitely small". 

4 minutes ago, swansont said:

You mean 100%, right?
The number you arrive at is the most likely correct value, based on the information you have. But the problem is that all you can do is make more measurements - there’s no way to peek inside the black box and see what “truth” is.

I took at face value the information from a previous post that the measurements show (13,799 +/- 20) million years. In that case 13.820B lies outside, so to 0% inside. Either way it is besides the point, since the true value will always be either 100% or 0% in any given interval.

And no, the truth does not have different values with various probabilities. It is our confidence that our measurements give us a close enough estimate of the true value which is expressed. 

Just now, swansont said:

It’s both, isn’t it? If you have a normal distribution, you are indicating a probability the true value is within the error bars because your error bar represents some number of standard deviations

That has a way of sounding right. The problem is that the true value does not get picked from a random distribution. It is still an entirely reasonable possibility that the true value is very close to 13.820 billion years. And if so, then the probability that the true value lies within the error bars is exactly 0%. 

11 hours ago, pzkpfw said:

And note those are not hard limits; it's about levels of certainty. Probabilities.

And now that we are nitpicking anyway: error bars are about confidence levels, rather than probabilities.

2 hours ago, DimaMazin said:

sin(t)=s[-2y(sin(a)-y)+or- ((sin(a)-y)²+s² - 4y²s²)^1/2] / [2(sin(a) - y)²+s² ]

Assuming you worked it out correctly, this might be another version of the same identity. I have not checked it. 

I suppose that you have no additional questions?

If we let \(t = \pi/4\) and \(a = \pi/6,\) then \(\sin t = \frac{\sqrt{2}}{2}\) and \(\sin a = \frac{1}{2}.\) Can you solve to find \(y\) for that case?