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Originally Posted By: Paul
in order to reverse time the arrow must reverse direction.
spinning the electron in the opposite direction…


There is a very basic assumption here, and I often find myself wondering about it. The assumption is that if you reverse time, you reverse physical processes. There have been some revolting things written along those lines regarding eating and eliminating, but let’s not go there.

If reversing time involves reversing physical processes, surely this will run foul of the second law of thermodynamics. Isn’t that the kiss of death for any proposal?


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Originally Posted By: Bill S.
If reversing time involves reversing physical processes, surely this will run foul of the second law of thermodynamics. Isn’t that the kiss of death for any proposal?

Thermodynamics is a macroscopic law you can't even formally define it at the micro level. I think I have mentioned before temperature suffers the same problem there isn't actually a clean statistic called "temperature" its a macro assembly of a group of QM statistics. So the 2nd law is a consequence of not the reason for the arrow of time.

Formally it's stated like this
Originally Posted By: https://en.wikipedia.org/wiki/Second_law_of_thermodynamics
The second law is an empirical finding that has been accepted as an axiom of thermodynamic theory. Statistical thermodynamics, classical or quantum, explains the microscopic origin of the law

What the bit in red is saying you just need statistics to understand the law and it works in QM or classical. Wiki actually does a nice walk thru in all the different ways.

Go down and it takes you directly to the thing you just stated

Originally Posted By: https://en.wikipedia.org/wiki/Second_law_of_thermodynamics
The second law has been proposed to supply an explanation of the difference between moving forward and backwards in time, such as why the cause precedes the effect (the causal arrow of time)

Can you guess why it's a proposal and not accepted as the explaination .... let me write the word .... STATISTICS

You now need to create how the universe calculates and balances statistics.

Last edited by Orac; 02/04/16 09:16 PM.

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Bill S

the below math isn't really used to calculate anything
its open and there is no decision yet.

Quote:
Function X ^ 0 = Function Y ^ 0 = Function Z ^ 0 = 1


its something that is more like philosophy , there are
ongoing discussions about its having validity or not.

but its not to be confused with actual in practice math.

and using my calculator

X^0 = 0
and if the X is replaced by a 1
1^0 = 1
then the same would be true
for 0^0=0

if X is not given a value and there are no other values
that can be used to find a value for X then X has no value.

so X^0=0
assigning a 1 to X Y and Z
X Y and Z must each have a value of 1 for the equation
to be correct.







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Please noone say anything .... just don't

Paul what does you calculator say this answer is

X^(T^0) = ?

I should know this but I am really stuck.

Hint: Paul you just utterly broke calculus you cant even solve 3y = 3, it's stone dead like a blue Norwegian parrot. Think carefully what you just did above.

Last edited by Orac; 02/05/16 05:56 AM.

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Quote:
Paul what does you calculator say this answer is

X^(T^0) = ?


X has been given no value
T has been given no value

so X^(T^0)= 0

Quote:
you cant even solve 3y = 3


its already solved -->(=3)

solve for y is y = 1

3 x 1 = 3

anything else is nonsense , I don't use letters that have
no value in math unless Im writing a formula to find a value
or to determine how to find a value.
I only use letters that have known values within a formula
mainly because my math must be written into a computer
program because I mostly only use math within computer programs.

and my computer cant deliver a correct calculation
using letters that have no value.

if I use a letter I must assign a value to that letter.

and there is no calculation that I cannot correctly calculate
so the different varieties of math formulas and the ways that they are written don't mean much to me or my computer
because I can work out how the computer needs the programming
to be in order to deliver a correct precise calculation.

if you think that calculus or any of the diverse methods
used to calculate has any advantage over a computer
program then please let me know what that advantage might be.

getting back to my earlier question to you that you
answered with another question pertaining to something else...

Quote:

function X = function(mt, ut + 1/2at^2);
function Y = function(-mt, -ut - 1/2at^2);


Definitions:
m = horizontal velocity I threw rock at
u = vertical velocity I threw rock at (unlikely I got it perfectly horizontal)
a = gravity AKA 9.8 m/s/s
t = time from launch


([+or-]velocity x time , [+or-]velocity x time [+ or -] .5 gravity x time^2)

what exactly is the comma used for ... and what would you
expect the formula to calculate?



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I am trying to understand your gibberish as best that I can


function X = function(mt, ut + 1/2at^2);
function Y = function(-mt, -ut - 1/2at^2);

using v = 10 mps
using t = 5 sec

mt, ut + 1/2at^2
10 x 5 , 10 x 5 + .5 x 9.8 x 5 x 5

50 , 50 + 122.5



replacing your comma with math opperator symbols.

comma as addition
[+50] + [+50] + 122.5 = +222.5


comma as addition
[-50] + [-50] - 122.5 = -222.5

comma as subtraction
[+50] - [+50] + 122.5 = +122.5


comma as subtraction
[-50] - [-50] - 122.5 = -122.5

comma as multiplication
[+50] x [+50] + 122.5 = +2,622.5 <----


comma as multiplication
[-50] x [-50] - 122.5 = +2,377.5 <----

comma as division
[+50] / [+50] + 122.5 = +123.5 <----


comma as division
[-50] / [-50] - 122.5 = -121.5 <----

which is why I asked you what the comma was used for.

again what does the comma represent?


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OMG I take it you didn't do calculus at school ... the point of calculus is you don't need to know the numbers.

Originally Posted By: paul
so X^(T^0)= 0

So now check your answer from your logic from above smile

So normal maths we do the thing inside the bracket first, T^0 has two possibilities according to you ... either T is zero, or T is some non zero value.

a) If T = 0 then you say T^0 = 0
b) If T = any other number T^0 = 1 is what what you said above

Ok so lets put simplify these results into the X power using the same rules

a) So T=0 hence (T^0)=0 so simplified you get X^0 = ?
b) If T is any other number T^0 =1 so simplified you get X^1 = ?

b) is easy there is only one answer X^1 = X ... so your answer is X

a) We are back to needing to know if X is zero or not
So if X = 0 then X^0 = 0
If X = any other number the X^0 = 1

So Paul mathematics breaks like this with 3 answers
b) If X & T are both non zero X^(T^0) = X
a1) If X is zero but T is non zero X^(T^0) = 0^1 = 0
a2) If X is non zero but T is zero X^(T^0) = X^0 = 1

For the rest of us we set X^0 = 1
We get only one answer X^(T^0) equals X always ... your (b) answer. We agree with your answer except for T or X equal to zero.

So whats the problem with Paul's answer ... well that is where the next bit comes in

Originally Posted By: Paul
3y = 3
its already solved -->(=3)

solve for y is y = 1

3 x 1 = 3

Well you would think that but there was a little bit of information I withheld from you

Directly above it I put this line and didn't show you

Let y = 0^0

So y actually equals 0^0 I just neglected to tell you it shouldn't matter it's only a number smile

You say 0^0 = 0

ooops I get 3 x 0 = 3 .... mathematics says NO.

So either I can't use calculus at all with powers or 0^0=0 is wrong

Your decision to make 0^0 = 0 ... creates a massive discontinuity.
All your power maths simply stops working and that is why we can't and won't do that.

Small problems like that never stop you but for mathematicians it's a drop dead and so they rethink the problem
http://www.askamathematician.com/2010/12...chers-disagree/

Mathematician: Zero raised to the zero power is one. Why? Because mathematicians said so. No really, it’s true.

Yeah your mathematics teacher never let you in on this dark little secret at school, we sometimes have to intervene in mathematics. The bad news is there is quite a few of these little nasty secrets in mathematics we don't tell you about because it confuses your poor little layman heads.

Generally you can place them all category called RENORMALIZATION and the problem is always zero, irrrationals and infinities.

As I said this is not a one off problem there are alot of them just layman have it in there heads that maths is totally self consistent ... well it isn't.

Last edited by Orac; 02/05/16 04:30 PM.

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Quote:
Your decision to make 0^0 = 0 ... creates a massive discontinuity.
All your power maths simply stops working and that is why we can't and won't do that.


it certainly would create a massive discontinuity within the
BS math and BS physics realm but not in any reality realm
where reality resides.

0^0=0 check it on your scientific calculator.

I did...

Quote:
ooops I get 3 x 0 = 3


if you don't have the 3 to begin with then it cant equal 3
and 3 x 0 means that you don't have the 3 to begin with.

you cant multiply by zero , I bet you have a really
difficult time trying to arrive at any type of correct
answer because you cant use a actual calculator unless
someone from the BS math and BS physics realm has designed
one to complement the BS math you guys use.

sorry orac your math system is complete BS and must be the
type of logic that built the BS physics that you reside in.

let me ask you how would you calculate this one...

L * O = ?


I am intentionally leaving out important information so that
you can use calculus to arrive at a answer to my question
to you.

I already know the values for L and O but lets see if
you can predict the values.

and for the fith or sixth time what was the comma used for
in your previous gibberish?

you claim that you answer questions and if you don't answer this question I will get you banned from the forum for
telling lies.






Last edited by paul; 02/05/16 04:56 PM.

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Ok short one for this
Originally Posted By: Paul
what exactly is the comma used for ... and what would you expect the formula to calculate?

Paul when you write the X,Y co-ordinates for a graph or of a 2D formula they look like these example

(3,4) or (5,7) or (1.345,5.6677)

2D co-ordinates are seperated by a comma (x, y)

Now what I did was write the calculus into the formula so here is the first which we called function

function X = function(mt, ut + 1/2at^2)

Ok so what this does is plots the 2D movement over time of the rock ... watch lets plug some values in

m = I threw the rock horizontally at 3ms
u = I didn't quite get it horizontal it was 2ms downwards
a = gravity = 9.8m/s
t = time from throw

So lets do 1 second intervals of t.

t=0sec ... 2D coord = (3*0, 1/2*9.8*0^2) = (0, 0)
t=1sec ... 2D coord = (3*1, 1/2*9.8*1^2) = (3, 4.9)
t=2sec ... 2D coord = (3*2, 1/2*9.8*2^2) = (6, 19.6)
t=3sec ... 2D coord = (3*3, 1/2*9.8*3^2) = (9, 44.1)
t=4sec ... 2D coord = (3*4, 1/2*9.8*4^2) = (12, 78.4)
t=5sec ... 2D coord = (3*5, 1/2*9.8*5^2) = (15, 122.5)

So that is the position of the rock in flight for any given value of time ... that is what the function maps

I also know given how I threw the rock it landed 15m out from where I threw it and the cliff is 122.5m tall.

That is sort of the point of calculus to be able to do that and why we can't let you break it by setting 0^0 = 0.

Your answer in some ways seems logical I cant fault you on that, all I can say is everything breaks.

Last edited by Orac; 02/05/16 05:07 PM.

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Originally Posted By: paul
L * O = ?

L can be any value you like it doesn't matter and it can be a function and it can be 0^0 or anything else.

The answer to that is always zero .. its a mathematics definition.

So I am going to say ... I just made this up it doesn't matter

L = X * Y * Z + PQR - S / T

This is why you need calculus to make sense, what you just wrote is what is called indeterminate it could be any number.

If you need me to show you that my answer above multiplied by zero equals zero just say.

Last edited by Orac; 02/05/16 05:08 PM.

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Can I suggest you look at this which explains why things become undefined or indeterminate.

https://www.khanacademy.org/math/algebra...d-indeterminate

Finally I would never ask for you to be banned. I am just showing you why mathematicians are forced into making certain decisions.

If you choose not to understand this stuff that is your choice but I will ignore you because things go crazy in maths if you don't fix them.

All this stuff is why I don't think mathematics is fundamental in any way it needs to many human decisions, if you like maths seems to be a construct of the human mind at least to me.

Last edited by Orac; 02/05/16 05:05 PM.

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Quote:
So I am going to say ... I just made this up it doesn't matter

L = X * Y * Z + PQR - S / T


I asked you to determine the values that I intentionally
left out so you could find them.

you failed to find them.

L=15 and 0=20

so it doesn't work.

it has no value at all to me.

I cant think of or even dream up any reason why I should consider calculus as being a viable math method used to calculate anything.


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Originally Posted By: paul
it has no value at all to me.

I cant think of or even dream up any reason why I should consider calculus as being a viable math method used to calculate anything.

LOL you didn't give me a problem that was solvable (insufficient information) and then complain I can't solve it smile

Yes well I can see it has no value to you because you really don't want to know .. so why don't we leave it there calculus will always be a mystery to you.

Can I ask did however Paul did you not do calculus at school? I don't know the USA school system very well sorry, most school systems I know it is mandatory.

Last edited by Orac; 02/05/16 05:27 PM.

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LOL , so the rock has a final velocity of
5 mps and it has traveled 19 meters vertically when it
impacted the ground 122.5 meters downwards from
where you threw the rock.


Quote:
m = I threw the rock horizontally at 3ms
u = I didn't quite get it horizontal it was 2ms downwards
a = gravity = 9.8m/s
t = time from throw

So lets do 1 second intervals of t.

t=0sec ... 2D coord = (3*0, 1/2*9.8*0^2) = (0, 0)
t=1sec ... 2D coord = (3*1, 1/2*9.8*1^2) = (3, 4.9)
t=2sec ... 2D coord = (3*2, 1/2*9.8*2^2) = (6, 19.6)
t=3sec ... 2D coord = (3*3, 1/2*9.8*3^2) = (9, 44.1)
t=4sec ... 2D coord = (3*4, 1/2*9.8*4^2) = (12, 78.4)
t=5sec ... 2D coord = (3*5, 1/2*9.8*5^2) = (15, 122.5)


Im beginning to understand a little more about how you
believe things the way you do.


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Originally Posted By: paul
LOL , so the rock has a final velocity of
5 mps and it has traveled 19 meters vertically when it
impacted the ground 122.5 meters downwards from
where you threw the rock.

Sigh ... Paul, where are you getting that from there is no calculation of velocity in that form smile

Show me where you are getting the numbers 5mps and 19mps from?

Last edited by Orac; 02/05/16 05:31 PM.

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Ok here is what you are working out
https://hannibalphysics.wikispaces.com/Ch+2+Free+Fall

Let me take each movement individually maybe the penny will drop then

Lets just look at the horizontal first its dead easy. I have a horizontal velocity of launch of 3m/s. So every second it goes away from the cliff 3m. If the drop takes 5 seconds it must land 15m out from the cliff and it will still be going at 3m/s.

The x value of the calculus is that number m*t and it matches for every value.

The vertical movement is a bit trickier because the rock accelerates. Without going to elaborate detail there exists a solution called Newtons second equation of motion which gives you the distance over time. The equation removes the need to worry about calculating the intermediate velocities.

http://aphysicsteacher.blogspot.com.au/2010/02/newtons-second-law-of-motion.html

S = ut + 1/2at2

That is what the y coordinate is.

I am totally lost where you are getting velocities from the function doesn't calculate them its calculating distances.

I have hidden a lot of maths including the velocities behind the background which is why you use calculus.

Added: Here the blue ball plot is what we are making the 2D coords (x,y)for each time


Do you get it now the horizontal and vertical need different equations.

Last edited by Orac; 02/05/16 06:10 PM.

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Originally Posted By: Orac
Can I ask did however Paul did you not do calculus at school? I don't know the USA school system very well sorry, most school systems I know it is mandatory.

He may not have had calculus in high school. I know I didn't. The highest math that was offered was solid geometry. All the 'higher level' math we had was geometry, algebra, trigonometry, and solid geometry. They were all electives. This was a while back and I'm not sure when they added more advanced mathematics. I know they do study things in grade school that we considered to be college level.

Bill Gill


C is not the speed of light in a vacuum.
C is the universal speed limit.
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Quote:
Paul, where are you getting that from there is no calculation of velocity in that form smile


Im taking the velocity from what you originally posted

Quote:
function X = function(mt, ut + 1/2at^2);
function Y = function(-mt, -ut - 1/2at^2);


Definitions:
m = horizontal velocity I threw rock at
u = vertical velocity I threw rock at (unlikely I got it perfectly horizontal)
a = gravity AKA 9.8 m/s/s
t = time from launch


also , you did use velocity in your post...


Quote:
m = I threw the rock horizontally at 3ms
u = I didn't quite get it horizontal it was 2ms downwards
a = gravity = 9.8m/s
t = time from throw

So lets do 1 second intervals of t.

t=0sec ... 2D coord = (3*0, 1/2*9.8*0^2) = (0, 0)
t=1sec ... 2D coord = (3*1, 1/2*9.8*1^2) = (3, 4.9)
t=2sec ... 2D coord = (3*2, 1/2*9.8*2^2) = (6, 19.6)
t=3sec ... 2D coord = (3*3, 1/2*9.8*3^2) = (9, 44.1)
t=4sec ... 2D coord = (3*4, 1/2*9.8*4^2) = (12, 78.4)
t=5sec ... 2D coord = (3*5, 1/2*9.8*5^2) = (15, 122.5)


and all your doing is multiplying the 3mps times the seconds.

and this would give you the 15 meters distance when the
rock hit the ground but only if the rock fell for exactly 5
seconds...

how far vertically would the rock fall in 5 seconds?

how long would it take for the rock to fall 122.5 meters?

ahemmm ...

another previous point that you pointed out just now.

Quote:
ooops I get 3 x 0 = 3


hmmm... then how does 3*0=0 in your calculation below
if ooops I get 3 x 0 = 3

Quote:
(3*0, 1/2*9.8*0^2) = (0, 0)


3*0=0 I agree.

is this the part where we ask the dungeon master to
roll the loaded dice?

because my answers are consistent and your answers fluctuate.

as needed.







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Originally Posted By: Bill S
Is this because information is not conserved in these actions?


Perhaps a better question would be: How much of the information in this thread is worth conserving?

Then again, one could ask: Is this sort of "contest" science, mathematics or urology?


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Far be it from me to comment on something I don't understand, but as I have mentioned before, I'm a little suspicious of some of the things from this site that I do understand (sort of).


http://www.askamathematician.com/2011/07/q-does-light-experience-time/#comments

Last edited by Bill S.; 02/05/16 10:26 PM.

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