# Is this real maths or somebody winding me up?

That point in Cartesian coordinates system would be at 0 on the x-axis, 0 on the y-axis and 1 on the z-axis.
Your use of the term vector doesn't make sense in the context you presented.

There are many ways to represent a vector. You could say that you have a vector a, defined as a = [0, 0, 1], which would be a vector of length 1 in the z direction. This particular vector is also known as a unit vector in the z direction; $$\hat{k}$$

If I wanted to displace an object a = [0, 0, 1], one space to the right , a=[1,0,1] , how would I show this?

Would we put Δa = [0, 0, 1]= a=[1,0,1]?

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If I wanted to displace an object a = [0, 0, 1], one space to the right , a=[1,0,1] , how would I show this?
The symbol 'a' (bolded) indicates a vector. I think you are talking about a point since you said object.

Would we put Δa = [0, 0, 1]= a=[1,0,1]?
[0, 0, 1] + [1, 0, 0]

The symbol 'a' (bolded) indicates a vector. I think you are talking about a point since you said object.

[0, 0, 1] + [1, 0, 0]
Which is 1z1x?

or z1,x1

So in practise :

[0,0,0]+[0,5,6] = x0, y5, z6 position?

[0,5,6]-[0,5,6] = x0,y0,z0?

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It's existence is a word we use to explain and quantify change.

Two Pings wasted.

You are thinking of the word AGE

Which is what I take from It's existence is a word we use to explain and quantify change

Merely saying Time's existence is a word we use to explain and quantify change

(Time's existence, the combination of the two, is NOT a word) is not a explanation of time's existence)

Poor English yes explanation no X

It is not something magical.

Irrelevant

Last Ping

I don't think you care anyway so for the third and last time

My position is Time does not exist

Put forward a discussion points which tells or / explains or / convinces me it does

PS since this off topic from your original post if if if you wish to continue you might think about going to the thread "Does time exist" after your next reply here

And spell check didn't scream about "Concessness"? You need a new spell check.

Could be but it's built in

It also could be a virus

Also the mobile phone has a cracked screen and you should see the weird outputs I sometimes get if I select the suggested word

Two Pings wasted.

You are thinking of the word AGE

Which is what I take from It's existence is a word we use to explain and quantify change

Merely saying Time's existence is a word we use to explain and quantify change

(Time's existence, the combination of the two, is NOT a word) is not a explanation of time's existence)

Poor English yes explanation no X

Irrelevant

Last Ping

I don't think you care anyway so for the third and last time

My position is Time does not exist

Put forward a discussion points which tells or / explains or / convinces me it does

PS since this off topic from your original post if if if you wish to continue you might think about going to the thread "Does time exist" after your next reply here
You said put forward a discussion point, so If I provide a discussion point , you would have to discuss and not ping without discussing.

Time is a quantifiable measurement directly proportional to a change in entropy. I will 'see' you over in the other thread.

amber said:
Time is a quantifiable measurement directly proportional to a change in entropy.
I think you need to substantiate this claim with a lot more than you've managed so far.

If it's true, then the difference between two positions of the moon as it orbits the earth, correspond to a change in entropy (in the earth-moon 'system'). Can you prove this?
I also think, if you really want to understand what vectors are and how to use them in a proper analytical setting, you should buy or borrow some decent textbooks. Wikipedia does provide some reasonably good stuff although it's usually a bit terse (i.e. not especially expansive, devoid of "assignments", etc). Youtube is another reasonable source of video lectures, but again there aren't any homework questions or labs.

Textbooks with questions in them (and possibly answers to selected questions) are better because you can find out if you've learnt how to answer questions like the ones you're asking here.
But by all means, keep thinking you're going to learn things here or at other forums--this could well be true. In my experience, you sometimes get ideas about stuff, but it really does pay to hit the books. For instance, Einstein does a fairly good job of explaining the mathematics of his general theory in his original paper, but it's pretty abstract and if you don't understand vector algebra/calculus I'd say you'll find it somewhat opaque.

But learning about physics and vector spaces (which correspond to physical systems) is entirely up to you. If you enrol at a university, say, I imagine you'll be told that; you won't get detention if you don't do your homework, you'll probably just fail and then won't be re-enrolling (because they won't accept you). Tsk tsk.

I think you need to substantiate this claim with a lot more than you've managed so far.

If it's true, then the difference between two positions of the moon as it orbits the earth, correspond to a change in entropy (in the earth-moon 'system'). Can you prove this?

Does ΔS=ΔE/t not prove it?

Thermodynamic entropy is defined as the total energy of some system of "particles" divided by the temperature of that system. It is not defined as energy divided by time.

1) Can you see why the definition means entropy is a dimensionless number?

1a) Can you see why a change in entropy, ΔS, can be defined as a change in total energy, or a change in temperature (or both)?

2) Can you see why "information" entropy is proportional to what isn't known about the states of individual particles in the total?

1a) Can you see why a change in entropy, ΔS, can be defined as a change in total energy, or a change in temperature (or both)?

Yes, because that is what I said ΔS=ΔE/t . A change in entropy is equal to the total energy changing in time/over time. It changes at the rate of c, light in , light out?

ΔT=ΔkE? A change in temperature is equal to a change in kinetic energy?

Kmax (S)=
/k......? The kinetic energy max of entropy is energy divided by the entropy volume, then divided by k space.?

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amber said:
Yes, because that is what I said ΔS=ΔE/t .
Well it's wrong. ΔS=ΔE/T, when T is fixed (not changed). Also ΔS=E/ΔT, when E is fixed. Also, ΔS=ΔE/ΔT, when neither E nor T are fixed. It depends on how the energy and/or temperature are changed (or not).
In thermodynamics, of course, you also have volume and pressure, and both of these are variable (or not). It's not a simple thing, the average "system of particles" and time may or may not be important--there are ways to derive "time-independent" relations.

Well it's wrong. ΔS=ΔE/T, when T is fixed (not changed). Also ΔS=E/ΔT, when E is fixed. Also, ΔS=ΔE/ΔT, when neither E nor T are fixed. It depends on how the energy and/or temperature are changed (or not).
In thermodynamics, of course, you also have volume and pressure, and both of these are variable (or not). It's not a simple thing, the average "system of particles" and time may or may not be important--there are ways to derive "time-independent" relations.
Thank you , I understand what you are saying and I am sure you are correct, but I am not saying what you think I was saying, it is a bit abstract.

ΔS = change of entropy

ΔE = change of energy

/t = does not mean divide by time, means over a period of time.

A change of entropy equals a change of energy over a period of time

Just thought I would clarify what I meant.

added- should that not be ΔE/T over a period of time?

Surely thermodynamic transfer takes an amount of time and even has a speed?

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amber said:
/t = does not mean divide by time, means over a period of time.
Not in any physics textbook I've ever read.

Usually when you use 't', you mean a variable that has a value which is a real number. A period of time is often notated Δt (but not necessarily). When you see notation like $$Δt = t_1 - t_0$$, it's obvious that Δt is a difference. Since everyone knows time doesn't go backwards, $$Δt = t_0 - t_1$$ "looks like" a mistake, if (and only if) the indices mean the o "happens" before the 1, over the "time interval" Δt.

M'kay? And:
Surely thermodynamic transfer takes an amount of time and even has a speed?
That "changes" of physical variables take time is a given, often this is ignored without losing anything (i.e. losing information about the system being analysed) . . .

Not in any physics textbook I've ever read.

Usually when you use 't', you mean a variable that has a value which is a real number. A period of time is often notated Δt (but not necessarily). When you see notation like $$Δt = t_1 - t_0$$, it's obvious that Δt is a difference. Since everyone knows time doesn't go backwards, $$Δt = t_0 - t_1$$ "looks like" a mistake, if (and only if) the indices mean the o "happens" before the 1, over the "time interval" Δt.

M'kay?
It was abstract and probably not in any physics books, it was to represent that a change in entropy takes/uses an amount of time.

I think I understand your time formula/equation. I have several different ways I represent a change in time.
Here is one example

Δt=←0

The arrow of time pointing backwards, now moves forward leaving a past, past memories, past geometrical positions and ageing on the way. The space behind us unchanging.

That "changes" of physical variables take time is a given, often this is ignored without losing anything (i.e. losing information about the system being analysed) . . .

Thanks for clarifying that.

-there are ways to derive "time-independent" relations.

Hopefully we can derive sometime, that time is dependent to matter and independent of space.

I keep putting quotes around words such as "changes", "happens" etc, because they depend on our concept of time. Physics is one discipline (apart from mathematics) where this concept can change in subtle ways. In ordinary languages like English, there aren't many ways to communicate without invoking time (verbs have past, present and future tenses, for one).

Thus, physics and its mathematical 'structure' is almost a way to escape this time-dependence. Calculus defines a time differential, dt, which is supposed to be an infinitesimal change in time, and defines other differentials, dx, dV etc. An infinitesimal can be thought of as a "non-standard" number, one which is nonzero but impossible to ever calculate.

I keep putting quotes around words such as "changes", "happens" etc, because they depend on our concept of time. Physics is one discipline (apart from mathematics) where this concept can change in subtle ways. In ordinary languages like English, there aren't many ways to communicate without invoking time (verbs have past, present and future tenses, for one).

Thus, physics and its mathematical 'structure' is almost a way to escape this time-dependence. Calculus defines a time differential, dt, which is supposed to be an infinitesimal change in time, and defines other differentials, dx, dV etc. An infinitesimal can be thought of as a "non-standard" number, one which is nonzero but impossible to ever calculate.
Thanks
Well! It just so happens I have explained before the smallest length (L) possible. You may or may not accept this to be true

dt=0+0=1x

0 being a 0 point , I am then adding another 0 point adjoined to the first 0 point to make a really small L.

added- I know it is hard to imagine that

(0+0)*(0+0)*(0+0)=x,y,z

0r

(0+0)³

By stating 0+0 , we are stating there is two individual 0 points.

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amber said:
dt=0+0=1x
No, 0 is well-defined. A differential is too, being a non-zero value which is always smaller than any other value, but restricted to a domain (the time domain, the domain of volumes, etc). You're attempting to re-define something which has an accepted definition which hasn't changed in centuries. Moreover, it remains as useful now as it did then, a good reason to not re-define it.

One way to think about a differential such as dx, is as a value which is close enough that it's as close as you "need" it to be, to some point (maybe to the point where a line is tangent to a curve, or where a plane is tangent to a curved surface, and so on). If you want to preserve the structure of a circle, you can't give it a radius of 0. You can however "shrink" the radius until it's very close to 0 (almost 0, however you need to define that), and still preserve the relation between the radius and circumference.

No, 0 is well-defined. A differential is too, being a non-zero value which is always smaller than any other value, but restricted to a domain (the time domain, the domain of volumes, etc). You're attempting to re-define something which has an accepted definition which hasn't changed in centuries. Moreover, it remains as useful now as it did then, a good reason to not re-define it.

One way to think about a differential such as dx, is as a value which is close enough that it's as close as you "need" it to be, to some point (maybe to the point where a line is tangent to a curve, or where a plane is tangent to a curved surface, and so on). If you want to preserve the structure of a circle, you can't give it a radius of 0. You can however "shrink" the radius until it's very close to 0 (almost 0, however you need to define that), and still preserve the relation between the radius and circumference.
I was not trying to redefine anything, I should of not put the dt, I confused you sorry. It was just meant to show the smallest possible conceivable measurement.

Thank you for your help and helping to teach me new things and clarifying my understandings and misunderstandings. I need to get some sleep now, good night.