A puzzle that seems interesting to me

[jajrussel]

I have thought of something that has become somewhat of a puzzle game for me. How I got to this thought is not that important, but involved some frustration with the fact that the square root of 2 is what it is. It was somewhat disconcerting that the answer is not 1, when the thought I was following needed an answer of 1. Then I saw a triangle that represented the Pythagorean Theorem. Then something that Michel 123456 wrote in a thread some time ago came to mind; “The diagonal is the distance”.

I looked the thread up to see what he was talking about, and it does seem to me that his diagram applies to the thought I was having; well somewhat. Not quite... Maybe...

The thing is that if I allowed that the diagonal is the distance using a triangle it seemed to have solved my problem, but not really after I thought it through. Still, the concept seemed interesting enough and in truth at this moment I can not actually remember what the original thought was accept as I have stated. I needed an answer of 1.

It no longer seems to be a game, but an obsession.

I would try to use the Pythagorean Theorem to apply a view and see what I could come up with.

Trying to play by the rules of the Theorem I needed values squared, and the very first application I tried turned out seeming to not need the squaring of values. This would require an addendum to the rules of the Theorem that says if the value of any line is zero then squaring is not needed.

I say that the diagonal is equal to c

Note - The triangle is only a diagram so any line can have any value except that in this case I said the diagonal has a value of c (the speed of light). The other line values therefore follow the rules of the Theorem.

So if the diagonal is equal to c, and if the horizontal is equal to c, then the vertical ends up with a value of zero, so squaring doesn't seem to be necessary if one of the lines ends up with a value of zero.

The diagonal is c the speed of light.

The horizontal is velocity which is said to be c.

The vertical is mass which can only be zero.

This agrees with science and as a diagram seemed to be cool. Except for the fact that if I don’t allow that the triangle is only a diagram the end result would in this case become a straight line, which in itself my say something interesting.

Now to plug in different numbers.

The diagonal is c.

The horizontal is, half c

This becomes c squared for the diagonal. = 89875517873681764 m/s

And ((c/2) squared) for the horizontal. = 22468879468420441 m/s

The result for the vertical becomes 67406638255365099. What? Mass value?

The root value of the vertical comes out to be 259627884.2023044. Again what?

It seems that I may be trying to add apples and oranges.

Am I trying to add apples and oranges?

If the answer is yes is there a way to convert the vertical value to m/s using an equation?

I do not see anything wrong with using an equation to represent a line so long as the equations value is what it is supposed to be following the Theorem.

I am not nearly qualified to do this, but was thinking along the lines of inertia. An object's mass value is related to its velocity which means that its mass should be able to be stated in terms of m/s. I could be saying this wrong so I will try again.

Gravitational force can be given in terms of m/s, and this reasoning seems to imply that an objects velocity should give it inertial force that might be able to be expressed in m/s. My reasoning is not always true, so I could be wrong, and if I am, this becomes a question. Is the reasoning wrong? And if it is, does it affect what I am trying to do here? Of course it would also be nice to know why?

Actually, I am starting to have second thoughts about the apples and oranges thing.

I am thinking that the root value of each line has to be legit, which leads me to think that in this instance the root value has to represent mass. The diagram has to express a thought, and the thought is that (0 mass equals c, so a positive mass results in a velocity of less than c). Therefore the root value of the line has to be in line with the thought, it has to be legit. This means that when the root value is squared the resulting value which may be expressed as an equation has to be legit in that the physics of the equation has to be legitimate, and the same equation has to apply in every instance. This means that if there is a need for the equation to change for different mass values, then the allowance for the change has to be written into the equation. You can not use one equation for one mass value, and then a different equation for a different mass value.

Anyway, this is a puzzle game that seems to have become an obsession for me, and I am not really happy with the fact that I am not qualified to play my own game.

The game does not have to be the way I have stated. The diagonal could be time, the horizontal distance, with the vertical being velocity. It may be necessary for clarity to invert the scales of any of the lines, but remember the triangle is just a diagram. The goal is to use values or equations for the diagonal, horizontal or the vertical lines that when finished follows the Theorem.

I also tried to make a time dilation statement by saying the vertical line is time, the diagonal is the speed of light, and the horizontal line is a spaceship traveling at whatever velocity. This seemed pretty straight forward until I started thinking about it. Actually, this may have been the original reason I started on a thought that seems to be turning out to be insanity. I wanted to show somehow that if the speed of light was invariant then time should also be invariant. The Pythagorean Theorem didn't come into play in this thought.

All I wanted was a triangle shaped diagram. It sort of, kind of worked. The faster I went the bigger the triangle became. For me time never changed. Everything stretched out. The faster I went the bigger the diagonal and the vertical lines became in proportion to my speed. Since c is invariant regardless of how fast I was moving I could not find a time dilation. Time for me was invariant. The clock kept ticking at the same pace. An observer however, would see the triangle getting bigger and bigger and use this as an excuse to play around with time. Which seems irrational to me, because my observation results with me very near to c, and time just keeps right on ticking one second at a time.

Maybe if I can go c then I would see what everyone is talking about. I realize that physics says that so long as I have mass I am not going to reach c, so I seem to have a conundrum. A riddle that I can not solve no matter what spin I put on the word play. I also realize that if time is ticking steadily away for me one second at a time without regard to how fast I am going if I managed to reach c and time stops, well that is a pretty wicked time curve for me, because I have gone from a steady ticking clock to a dead stop at an instant. This would seem to mean to me that I went from going really very fast to a dead stop in and instant.

To me it would be much more rational for the observer to assume that he is delusional, that his calculations are flawed, and that maybe somehow I could show it to him. Then I saw the diagram representing the Pythagorean Theorem, then remembered Michel 123456 saying the diagonal is the distance, then rationalized that time had to be the diagonal, which plays into his diagram, and that velocity and distance are subsets of time represented by the horizontal, and vertical lines. Then I realized that I personally am not that clever.

I can not do it. On realizing that I can not do it I started thinking of other ways I might use the Theorem, and the original thought got lost in the quagmire. I am beginning to have a better understanding of physics, but I really am not that clever. It has become an obsessive game for me. It seems to me that the Pythagorean Theorem should be the perfect vessel for presenting the thought, but I can not seem to get past just figuring out the rules.

It may be something simple. It may end up being as complex as using more triangles to represent each line of the triangle, and more triangles representing each line of those triangles. With each following the rules of the Theorem.

It may be impossible to do, but I suspect it is just simply too difficult for me to reason through and that someone in the forum if they become interested enough can do it, and likely without all the extra triangles.

I apologize for this being somewhat wordy. I started out just trying to present a puzzle. Then I started remembering, and the more I wrote the more I remembered. I could try to make the whole presentation more concise, but I am afraid that if I do I will become frustrated, and too caught up in showing my own ignorance, then click select all, and hit the delete button.

Edited May 16, 2013 by jajrussel

[michel123456]

 

I have thought of something that has become somewhat of a puzzle game for me. How I got to this thought is not that important, but involved some frustration with the fact that the square root of 2 is what it is. It was somewhat disconcerting that the answer is not 1, when the thought I was following needed an answer of 1. Then I saw a triangle that represented the Pythagorean Theorem. Then something that Michel 123456 wrote in a thread some time ago came to mind; “The diagonal is the distance”.

I looked the thread up to see what he was talking about, and it does seem to me that his diagram applies to the thought I was having; well somewhat. Not quite... Maybe...

The thing is that if I allowed that the diagonal is the distance using a triangle it seemed to have solved my problem, but not really after I thought it through. Still, the concept seemed interesting enough and in truth at this moment I can not actually remember what the original thought was accept as I have stated. I needed an answer of 1.

It no longer seems to be a game, but an obsession.

I would try to use the Pythagorean Theorem to apply a view and see what I could come up with.

Trying to play by the rules of the Theorem I needed values squared, and the very first application I tried turned out seeming to not need the squaring of values. This would require an addendum to the rules of the Theorem that says if the value of any line is zero then squaring is not needed.

I say that the diagonal is equal to c

Note - The triangle is only a diagram so any line can have any value except that in this case I said the diagonal has a value of c (the speed of light). The other line values therefore follow the rules of the Theorem.

So if the diagonal is equal to c, and if the horizontal is equal to c, then the vertical ends up with a value of zero, so squaring doesn't seem to be necessary if one of the lines ends up with a value of zero.

The diagonal is c the speed of light.

The horizontal is velocity which is said to be c.

The vertical is mass which can only be zero.

This agrees with science and as a diagram seemed to be cool. Except for the fact that if I don’t allow that the triangle is only a diagram the end result would in this case become a straight line, which in itself my say something interesting.

Now to plug in different numbers.

The diagonal is c.

The horizontal is, half c

This becomes c squared for the diagonal. = 89875517873681764 m/s

And ((c/2) squared) for the horizontal. = 22468879468420441 m/s

The result for the vertical becomes 67406638255365099. What? Mass value?

The root value of the vertical comes out to be 259627884.2023044. Again what?

(...)

Read again what you wrote:

"The diagonal is c."

If the diagonal is c then why do you calculate the diagonal again stating : This becomes c squared for the diagonal. = 89875517873681764 m/s

 

this is wrong.

 

you should have stated otherwise, do you know trigonometry?

What is the angle needed in order to make correspond c for the diagonal and c/2 on the X axis?

 

 

------------

(edit)

You' re not trying to add apples and oranges, you are multiplying apples by oranges, I think physicists do that all the time.

[jajrussel]

Note - The triangle is only a diagram so any line can have any value except that in this case I said the diagonal has a value of c (the speed of light). The other line values therefore follow the rules of the Theorem.

 

I believe this is where I may have confused you; see the red text. I am sorry. In actuality all three lines follow the rules of the Theorem. My intent was to say that the root value of the diagonal is c. Making this statement would leave the other two line values open for determination.

 

If either the horizontal or the vertical is also equal to c then the line not given a stated value will be = 0

The diagonal is always the greater value, so simple math is all that is required to calculate the unknown line value.

 

The Pythagorean Theorem allows one to calculate the diagonal if the horizontal and vertical is known, so if the diagonal is stated, and the horizontal is stated, then Pythagorean method allows one to discover the unknown line value.

 

You asked; If the diagonal is c then why do you calculate the diagonal again stating : This becomes c squared for the diagonal. = 89875517873681764 m/s

 

Again I may have been confusing, for this I am sorry. When I stated the diagonal is c I should have been more clear by noting that c is the diagonals root value when using the Pythagorean method.

 

You stated; You' re not trying to add apples and oranges, you are multiplying apples by oranges, I think physicists do that all the time.

I believe that you have made this incorrect assumption because I have not made it clear to you that I am using the Pythagorean Method.

There is an order of operation. There is a time to multiply, and there is a time to add, or subtract. When I was speaking of apples and oranges it was time to add, or subtract.

 

Perhaps you should make it clear to me why knowing the value of an angle is necessary when using the Pythagorean Method for calculating a diagonal.

[michel123456]

If the diagonal is c squared then units are m^2/sec^2

 

You are correct that the pythagorean method uses addition. That's an issue, yes. Usually you cannot add apples and oranges. But many diagrams use different units on X and Y axes and work well. Of course in this case what the diagonal represents is not always so evident.

 

I understand your idea about the horizontal axis being the same as the diagonal, in which case the vertical value is zero.

I understand your idea about representing mass zero (a photon).

That gives you units m^2/sec^2 on the X axis (the units of c^2)

And units of Mass on the Y axis.

 

Having Mass multiplied by c^2 then the surface of the diagram is Energy (e=mc^2)

I remember having seen that somewhere.

That makes sense.

 

(edit)

But I doubt the diagonal has a value of c^2. I don't know what would represent the length of the diagonal.

 

The other problem is that c^2 is a constant. Only mass is a variable.

 

The next problem is that in this diagram, for m=0, Energy=0 which is not what happen for a photon (a photon has energy)

[Delta1212]

If one of the sides has a length of 0, it's not a triangle. It's a line.

[michel123456]

A segment.

[Delta1212]

Well, yes.

[jajrussel]

It is a segment with a mass value of zero...

 

I squared c because is what the theorem required. The point being to get to a place where addition and subtraction can be used to determine a value. After the line value is determined we then need to return to the root value of each line.

 

c is a constant; I am not sure if c^2 is a constant except in the sense of its relationship to c. Its value will always be the square of c.

 

c is also invariant so I am accepting that regardless what the value of the horizontal or the vertical line is the type of triangle I have is always the same; A right triangle... So it can be said it is a segment of a right triangle, with a mass value of 0.

 

The problem that I arrive at here is that the diagonal can be no greater than c, and the horizontal line which is velocity can be no greater than c, but the vertical can seemingly be infinite in value. To me this seems like a problem because I am not sure of how I can label the mass unit except to say that it is equal to 1/c.

 

As I think about this it does not seem to be any different than accepting a relationship of c to an infinite distance. So in terms of mass there can be a view of infinite mass, but how does the unit value of 1/c relate to mass?

 

Again, my ignorance here is limiting.