What does d represent in the context of \(\frac{dX}{dY}\) in Relativity?

[Stephanus]

Dear PF Forum,
I'm studying Relativity,
And there's some term in math that I don't know. And I am even not an English native.
in $\frac{dX}{dY}$ What does "d" mean?
I know that dX is the tiny slice of X. What do we say it in English?

 

[Dr. Courtney]

I know that dX is the tiny slice of X. What do we say it in English?

That's pretty good. I like the word infinitesimal, but tiny is good.

https://en.wikipedia.org/wiki/Infinitesimal

 

[Stephanus]

That's pretty good. I like the word infinitesimal, but tiny is good.

I know I should have asked this in "Langauge forum"?
So, "infinitesimal" is the antonym of "infinite"?

 

[Dr. Courtney]

https://en.wikipedia.org/wiki/Infinitesimal

I know I should have asked this in "Langauge forum"?
So, "infinitesimal" is the antonym of "infinite"?

The words "opposite" and "antonym" are too imprecise in english to be of much use in mathematics.

What is the opposite of x? Is it -x? Or is it 1/x?

What is the opposite of INF? Is it -INF? Or is it 1/INF? Something else?

 

[Stephanus]

https://en.wikipedia.org/wiki/Infinitesimal
The words "opposite" and "antonym" are too imprecise in english to be of much use in mathematics.

What is the opposite of x? Is it -x? Or is it 1/x?

What is the opposite of INF? Is it -INF? Or is it 1/INF? Something else?

THAT'S RIGHT. Another good point for me. Thanks.
So dX is if you slice X again and again and again,... to infinity.
[Add: Just for fun,
If you challenge me, "How do you slice X? In half or 1/3 or 1/4 or...?
Supposed we divide the biggest part of X with n = 1/3
1. There is X -> A: 1/3 and B: 2/3
2. Divide the biggest part; B: 2/3 -> C: 2/9 and D: 4/9
3. Divide the biggest part; D: 4/9 -> E: 2/27 and F: 4/27
4. Divide the biggest part: A: 1/3 -> G: 1/9 and H: 2/9
5. Divide the biggest part; A or C: 2/9 -> etc...
No matter what method you choose for n, dX still infinitesimal]

 

[HallsofIvy]

Your mistake is in thinking that "d" has a separate meaning! "dy/dx" is defined as a single entity, the "derivative of function y with respect to variable x. That is defined early in an introductory Calculus course. A little later in such a course, we define the quantities "dy" and "dx", separately, in such a way that "dy" divided by "dx" is equal to the single quantity "dy/dx" originally defined. But we do NOT define "d" as a separate quantity.

(You can calculate an approximation to "dy/dx" by using "small changes in y divided by small changes in x" giving the idea that "d" means "small changes in" but that is just a "mnemonic".)

 

[paisiello2]

Isn't d/dy considered as an operator on x?

 

[Stephanus]

Your mistake is in thinking that "d" has a separate meaning! "dy/dx" is defined as a single entity, the "derivative of function y with respect to variable x. That is defined early in an introductory Calculus course. A little later in such a course, we define the quantities "dy" and "dx", separately, in such a way that "dy" divided by "dx" is equal to the single quantity "dy/dx" originally defined. But we do NOT define "d" as a separate quantity.

(You can calculate an approximation to "dy/dx" by using "small changes in y divided by small changes in x" giving the idea that "d" means "small changes in" but that is just a "mnemonic".)

Yes, HallsofIvy.
If you divide dX by dY in Y = X³ + n, you'll get the gradient, or if you touch a ruler to that point. Y = 3X²
If you divide dX by dY in C² = x²+y², you'll get the tangent.
But as a non English speaker, I want to know what is the "word" for d.
"Derivative" is a good one.
I just can't keep saying for dX/dY like this:
"You should take the very very small part of X and divide it by the very very small part of Y and ..."
This one is better
"Take the derivative of X/Y and..."
Dr. Courtney choosen of word is good "Infinitesimal", and yours too. "Derivative".
Perhaps I would use both of them. Derivative seems good.Now, I can go back to Special Relativity (SR) and ask this question
"In $U^\mu = \frac{dX^\mu}{d\tau}$ why do we have to use the derivative of X instead of X only". Seems good sentence? And it just hits me with this SR thing. Am I being fooled by SR. In the past 3 months I was just doing math, and math, and math. Not a single thing about physics??

 

[Mark44]

Yes, HallsofIvy.
If you divide dX by dY in Y = X³ + n, you'll get the gradient, or if you touch a ruler to that point. Y = 3X²

No, you don't divide dX by dY. If Y is a function of X, as above, you can find the derivative of Y with respect to X, or dY/dX (usually read as "dee Y dee X"). This new function gives you the slope of the tangent line on the original function at an arbitrary point.

If you divide dX by dY in C² = x²+y², you'll get the tangent.

Here you can use implicit differentiation to find either dX/dY or dY/dX. The latter, dY/dX, gives the slope of the tangent line at an arbitrary point on the circle that the equation represents. It's called implicit differentiation because the equation doesn't give either X or Y as a function of the other variable.

Since mathematicians normally use lower case letters, I'm going to switch to the more-used form, dy/dx etc.

But as a non English speaker, I want to know what is the "word" for d.
"Derivative" is a good one.

No, it isn't. If you are given that, say, y = x², for example, the differential of y, denoted dy, is often defined as $dy = \frac{dy}{dx} \cdot dx = 2x dx$. Here dy is the differential of y and dx is the differential of x.

I just can't keep saying for dX/dY like this:
"You should take the very very small part of X and divide it by the very very small part of Y and ..."
This one is better
"Take the derivative of X/Y and..."
Dr. Courtney choosen of word is good "Infinitesimal", and yours too. "Derivative".
Perhaps I would use both of them. Derivative seems good.Now, I can go back to Special Relativity (SR) and ask this question
"In $U^\mu = \frac{dX^\mu}{d\tau}$ why do we have to use the derivative of X instead of X only". Seems good sentence?

I'm not familiar with this equation, but it seems to be defining $U^{\mu}$. On the right side is the derivative of $x^{\mu}$ (not the derivative of x) with respect to $\tau$.

And it just hits me with this SR thing. Am I being fooled by SR. In the past 3 months I was just doing math, and math, and math. Not a single thing about physics??

 

[pbuk]

For dy/dx we say "dee wye bye dee eks" or sometimes "dee bye dee eks ov wye". We can also say "the derivative of y with respect to x", or if it is obvious from the context what we are differentiating by just "the derivative of y". I think Dr Courtney misunderstood what you were asking because we would not normally use "infinitesimal" in this way.

 

[Mark44]

For dy/dx we say "dee wye bye dee eks" or sometimes "dee bye dee eks ov wye".

This might be a cultural thing. In my experience of the past 50 years, in the US we say for dy/dx, "dee wye dee eks" without the "by" or "of."

 

[Dr. Courtney]

I have found it useful to focus on d/dx as an operator that differentiates what it acts on with respect to x. dy/dx = (d/dx) y

 

[pbuk]

This might be a cultural thing. In my experience of the past 50 years, in the US we say for dy/dx, "dee wye dee eks" without the "by" or "of."

My experience of calculus in the UK only goes back 35 years, and "dee wye dee eks" is certainly something that would always have been well understood, and possibly the prefererence now.

 

[aikismos]

My experience of calculus in the UK only goes back 35 years, and "dee wye dee eks" is certainly something that would always have been well understood, and possibly the prefererence now.

I worked as a math teacher in high school up to two years ago and "dee wye dee eks" was the norm in the department.

 

[Dr. Courtney]

Being aware of how students can get the wrong impression from certain notations, I try and mix in a fair amount of:

d/dt [x(t)]

and

d/dx [f(x)]

and other forms where the letter specifying the function and the input variable are frequently different from y and x.

Students need to understand calculus with a wide variety of functions and input variables. Sticking to religiously to y as the output and x as the input works against this understanding.

 

[Stephanus]

For dy/dx we say "dee wye bye dee eks" or sometimes "dee bye dee eks ov wye". We can also say "the derivative of y with respect to x", or if it is obvious from the context what we are differentiating by just "the derivative of y". I think Dr Courtney misunderstood what you were asking because we would not normally use "infinitesimal" in this way.

Yes, in voice . How do you type it in written? Because I was trying to understand SR, and there's something like that. $U^{\mu} = \frac{dX^{\mu'}}{d\tau}$. And I wanted to express my question in 'English'.
Something like: "Why we should divide the dee eks(distance) by the dee tau (time)?"
But I read many useful math concepts from this thread beside the question that I asked here.

 

[Dr. Courtney]

For dy/dx we say "dee wye bye dee eks" or sometimes "dee bye dee eks ov wye". We can also say "the derivative of y with respect to x", or if it is obvious from the context what we are differentiating by just "the derivative of y". I think Dr Courtney misunderstood what you were asking because we would not normally use "infinitesimal" in this way.

See: http://www.sjsu.edu/faculty/watkins/infincalc.htm

My answer was intentional. The word "infinitesimal" probably should be used more often in the teaching and learning of Calculus. Newton and Leibniz did intend the derivative to be understood as an [(infinitesimal of the input variable) divided by an (infinitesimal of the output variable) ].

Thus, I believe understanding "d" as an infinitesimal is correct and what I intended.

 

[Stephanus]

No, you don't divide dX by dY. If Y is a function of X, as above, you can find the derivative of Y with respect to X, or dY/dX (usually read as "dee Y dee X"). This new function gives you the slope of the tangent line on the original function at an arbitrary point.

Ahh, I put it upside down. After 30 years from high school, I should have joined "Are you smarter than fifth grader"
Yes, I remember. The gradient of vertical line is infinity right. And horizontal line is zero.
Should have written $\frac{dY}{dX}$

Here you can use implicit differentiation to find either dX/dY or dY/dX. The latter, dY/dX, gives the slope of the tangent line at an arbitrary point on the circle that the equation represents. It's called implicit differentiation because the equation doesn't give either X or Y as a function of the other variable.

Okay, okay.

Since mathematicians normally use lower case letters, I'm going to switch to the more-used form, dy/dx etc.

Ok.

I'm not familiar with this equation, but it seems to be defining $U^{\mu}$. On the right side is the derivative of $x^{\mu}$ (not the derivative of x) with respect to $\tau$.

Yes! In math language.
But what is in 'English'
My problem is...
From $\frac{dx^{\mu'}}{d\tau}$ It's the (a)tiny/(b)slice/(c)derivative distance divided by (a)tiny/(b)slice/(c)derivative time so it's velocity at that particular time. It's the English that I want to know before I go back to SR forum again and ask my question in a better sentence.
I'm not an English speaker, I wasn't definitely went to English school before. So I don't know the word for this dX.
I'm aware if you have a curve, say y = x². Then you touch a ruler to that curve in some arbitrary point, say in x=5, so the gradient of your ruler is $\frac{dY}{dX}$, right. = 6, correction: 10
Thanks for your answer.
Gradient = $\frac{dx}{dy}$

[Add: I know the number, I know the result, I don't know the 'English']

 

[Mark44]

Yes! In math language.
But what is in 'English'
My problem is...
From ##\frac{dx^{\mu}}{d\tau}## It's the (a)tiny/(b)slice/(c)derivative distance divided by (a)tiny/(b)slice/(c)derivative time so it's velocity at that particular time.

No, it's the rate of change of $x^{\mu}$ with respect to $\tau$ or the derivative of $x^{\mu}$ with respect to $\tau$. I don't see how this would be a velocity at all, even if $\tau$ happens to represent time.

It's the English that I want to know before I go back to SR forum again and ask my question in a better sentence.

I can't say it any better than I did just above without knowing what $x^{\mu}$ represents.

I'm not an English speaker, I wasn't definitely went to English school before. So I don't know the word for this dX.

As I said already, dx is the differential of x.

I'm aware if you have a curve, say y = x². Then you touch a ruler to that curve in some arbitrary point, say in x=5, so the gradient of your ruler is $\frac{dY}{dX}$, right. = 6, correction: 10

Right, the slope of the tangent line to y = x² at x = 5 is 10. In other words, $\frac{dy}{dx} |_{x = 5} = 10$.

Thanks for your answer.
Gradient = $\frac{dx}{dy}$

No, this would normally be dy/dx, not dx/dy.

[Add: I know the number, I know the result, I don't know the 'English']

 

[Stephanus]

No, it's the rate of change of $x^{\mu}$ with respect to $\tau$ or the derivative of $x^{\mu}$ with respect to $\tau$. I don't see how this would be a velocity at all, even if $\tau$ happens to represent time.

This equation is from this: http://www.lecture-notes.co.uk/susskind/special-relativity/lecture-6/relativistic-kinematics/

[..]If we suppose that the velocity of a particle is changing over time - that is, it can accelerate and decelarate - then the particle's rest frame is not necessarily an inertial reference frame. However, at any given time (either proper time or coordinate time), we can consider a frame that, at that instant, is moving at the same velocity as the particle and has the particle at its origin.

When we talk about the particle frame, we shall really mean this ‘momentarily co-moving reference frame’, and use primed coordinates to refer to it. In this frame, we have

Position: $X^{\mu'} \rightarrow (\tau,0,0,0)$
Velocity: $U^{\mu'} = \frac{dX^{\mu'}}{d\tau} \rightarrow (1,0,0,0)$
Momentum: $P^{\mu'} = mU^{\mu'} \rightarrow (m,0,0,0)$

And I wonder, why Velocity is $U^{\mu'} = \frac{dX^{\mu'}}{d\tau} \rightarrow (1,0,0,0)$?It's not the equation that I ask, it's the English.
Does velocity change? Is it acceleration? I wanted to ask the question in a better English, that's why I ask about this "d".

Then I saw one paragraph above:

[..]If we suppose that the velocity of a particle is changing over time - that is, it can accelerate and decelarate - then the particle's rest frame is not necessarily an inertial reference frame. However, at any given time (either proper time or coordinate time), we can consider a frame that, at that instant, is moving at the same velocity as the particle and has the particle at its origin.

So I have already canceled my question in SR Forum, but then I discussed math instead.
It's not the math that I ask, it's the English.
Okay, one more example:
the $\frac{sin(dx)}{dx} = cos(x)$ It's just the 'English' that I don't know.
The ... sin(dx)/dx is cos(x)?
You say, 'derivative', I like it. Other says diferential, infinitesimal. I, the math ignorance , say "slice".
Btw, thanks for your answer. Your answer has given me other knowledge about math, too.

 

[Mark44]

This equation is from this: http://www.lecture-notes.co.uk/susskind/special-relativity/lecture-6/relativistic-kinematics/And I wonder, why Velocity is $U^{\mu'} = \frac{dX^{\mu'}}{d\tau} \rightarrow (1,0,0,0)$?It's not the equation that I ask, it's the English.

$X^{\mu}$ is the position. As a 4-vector it is ($\tau$, 0, 0, 0) - as given in what you quoted from the lecture notes.
The velocity, is given as $\frac{dX^{\mu}}{d\tau}$. This is just the rate of change of the position with respect to $\tau$, so the 4-vector would be (1, 0, 0, 0). The first coordinate of the position vector is $\tau$, so differentiating with respect to $\tau$ gives 1 in that position. All other coordinates of the position vector are 0, so their derivatives are zero as well.

Does velocity change? Is it acceleration?

Yes and yes.

I wanted to ask the question in a better English, that's why I ask about this "d".

$d\tau$ (for example) is the differential of $\tau$, an infinitesimally small quantity.

Then I saw one paragraph above:So I have already canceled my question in SR Forum, but then I discussed math instead.
It's not the math that I ask, it's the English.
Okay, one more example:
the $\frac{sin(dx)}{dx} = cos(x)$ It's just the 'English' that I don't know.
The ... sin(dx)/dx is cos(x)?

I don't think you understand the math, either. What you have above should be $\frac{d(sin(x))}{dx} = cos(x)$, or $\frac{d}{dx} \{sin(x)\} = cos(x)$, NOT $\frac{sin(dx)}{dx} = cos(x)$. In words, the derivative with respect to x of sin(x) is cos(x). This relationship shows up in the first quarter or semester of calculus.

You say, 'derivative', I like it. Other says diferential, infinitesimal. I, the math ignorance , say "slice".
Btw, thanks for your answer. Your answer has given me other knowledge about math, too.

 

[Stephanus]

I don't think you understand the math, either. What you have above should be $\frac{d(sin(x))}{dx} = cos(x)$, or $\frac{d}{dx} \{sin(x)\} = cos(x)$, NOT $\frac{sin(dx)}{dx} = cos(x)$. In words, the derivative with respect to x of sin(x) is cos(x). This relationship shows up in the first quarter or semester of calculus.

Yes, it's not sin (dx) but d sin(x)

 

[pbuk]

See: http://www.sjsu.edu/faculty/watkins/infincalc.htm

My answer was intentional. The word "infinitesimal" probably should be used more often in the teaching and learning of Calculus. Newton and Leibniz did intend the derivative to be understood as an [(infinitesimal of the input variable) divided by an (infinitesimal of the output variable) ].

Thus, I believe understanding "d" as an infinitesimal is correct and what I intended.

I don't agree. I believe that it is important to understand that $\frac{d}{dx} f(x)$ is shorthand for $\lim_{h \to 0} \frac {f(x+h)-f(x)}{h}$; talking about "an infinitesimal of f(x)" before you have confidently grasped that concept is, I believe, confusing.

 

[Dr. Courtney]

I don't agree. I believe that it is important to understand that $\frac{d}{dx} f(x)$ is shorthand for $\lim_{h \to 0} \frac {f(x+h)-f(x)}{h}$; talking about "an infinitesimal of f(x)" before you have confidently grasped that concept is, I believe, confusing.

There is a lot of calculus left in a good course after the limit definition of derivative has been mastered. Some teachers might choose to wait to introduce the word infinitesimal to the discussion, but I believe well-trained students will have heard it a lot by the end.

 

[pbuk]

Are you talking about a rigorous introduction of hyperreal analysis or the naive use of infinitesimals by Leibniz and Newton's "fluxions"? If it is the latter then I can't agree with you: the proof of the chain rule is not as simple as writing $$ \frac{dy}{dx} = \frac{dy}{dx} \cdot 1 = \frac{dy}{dx} \cdot \frac{dz}{dz} = \frac{dy}{dz} \cdot \frac{dz}{dx} $$ until you have proved some properties of the hyperreals.

 

[pbuk]

It would be less confusing if Euler's notation $D_x y$ had stuck.

 

[Stephanus]

I don't think you understand the math, either. What you have above should be $\frac{d(sin(x))}{dx} = cos(x)$, or $\frac{d}{dx} \{sin(x)\} = cos(x)$, NOT $\frac{sin(dx)}{dx} = cos(x)$. In words, the derivative with respect to x of sin(x) is cos(x). This relationship shows up in the first quarter or semester of calculus.

Yes, what I wrote was wrong sin (dx).
Thinking it over while driving 90kmh on highway in my way home. It's like
(sin(2.0000001)-sin(2)/0.000001
It is not
sin(0.000001)/0.00001, it's different.
I think you know what I mean by 0.00001, it's d = lim -> 0.
(Could have written 0.00000000000000000001, but the concept is the same).

 

[Stephanus]

I don't agree. I believe that it is important to understand that $\frac{d}{dx} f(x)$ is shorthand for $\lim_{h \to 0} \frac {f(x+h)-f(x)}{h}$; talking about "an infinitesimal of f(x)" before you have confidently grasped that concept is, I believe, confusing.

Come on MrAnchovy. Like Dr. Courtney said, it's the word "infinitesimal" that I'm looking for. It's not the math. But well, I choose "derivative". But the concept is exactly what you write.
Limit d ≈ 0. $\frac{sin(x+d)-sin(x)}{d}$

 

[Mark44]

Limit d ≈ 0. $\frac{sin(x+d)-sin(x)}{d}$

This makes no sense. d by itself has no meaning here.

 

[Stephanus]

This makes no sense. d by itself has no meaning here.

Yes, sorry. Not d, but h.
$lim_{h \rightarrow 0} \frac{sin(x+h)-sin(x)}{(x+h)-(x)}$.
the (x+h)-(x). It is in the form of $\frac{f(a)-f(b)}{a-b}$, right.
Should have change the letter d.

and this one is what I think that doesn't makes sense, see below.
$lim_{h \rightarrow 0} \frac{sin(h)}{h}$.

[Add: could have written $lim_{d \rightarrow 0} \frac{f(x+d)-f(x)}{d}$
But "d" here doesn't have any meaning right. It's just a variable.
But in this case, $V^{\mu} = \frac{dx}{d\tau}$, now in this context, d has a meaning.

]

 

[Stephanus]

Just want to add one thing.
I just realized that writing sin(x+h)-sin(x) = sin(h) is somewhat forgiveable.
But, writing cos(x+h)-cos(x) = cos(h) is really unforgiven. Should have remembered it "religiously" like the above post said.

 

[Mark44]

Yes, sorry. Not d, but h.
$lim_{h \rightarrow 0} \frac{sin(x+h)-sin(x)}{(x+h)-(x)}$.
the (x+h)-(x). It is in the form of $\frac{f(a)-f(b)}{a-b}$, right.
Should have change the letter d.

and this one is what I think that doesn't makes sense, see below.
$lim_{h \rightarrow 0} \frac{sin(h)}{h}$.

$lim_{h \rightarrow 0} \frac{sin(h)}{h} = 1$, a fact that is proven in most calculus textbooks. It's not immediatedly obvious, but you can check that it is reasonable by evaluating $\frac{sin(h)} h$ for smaller and smaller values of h (in radians).

[Add: could have written $lim_{d \rightarrow 0} \frac{f(x+d)-f(x)}{d}$
But "d" here doesn't have any meaning right. It's just a variable.
But in this case, $V^{\mu} = \frac{dx}{d\tau}$, now in this context, d has a meaning.

In your first example above, d does have a meaning -- as you say it's a variable.
In your second example, d by itself has no meaning, but $\frac{dx}{d\tau}$ means the derivative of x with respect to $\tau$. By definition $\frac{dx}{d\tau} = \lim_{h \to 0}\frac{x(\tau + h) - x(\tau)}{h}$

Just want to add one thing.
I just realized that writing sin(x+h)-sin(x) = sin(h) is somewhat forgiveable.

Why? This is not true.

But, writing cos(x+h)-cos(x) = cos(h) is really unforgiven. Should have remembered it "religiously" like the above post said.

I don't see the difference here. Neither one is true.

 

[Stephanus]

$lim_{h \rightarrow 0} \frac{sin(h)}{h} = 1$, a fact that is proven in most calculus textbooks[..]

Yes, this is a FACT!. I don't argue about it either.

I just realized that writing sin(x+h)-sin(x) = sin(h) is somewhat forgiveable.

Why? This is not true.

But, writing cos(x+h)-cos(x) = cos(h) is really unforgiven. Should have remembered it "religiously" like the above post said.

I don't see the difference here. Neither one is true.

Come on Mark114, give me some slack
$lim_{h \rightarrow 0} sin(h)$ is almost zero, too. But of course $lim_{h \rightarrow 0} sin(h) ≠ (sin(x+h)-sin(x))$
And of course $cos(h)$ is a light year away from $cos(x+h)-cos(x)$
While $sin(h)$ is, I could say, in the neighborhood of $sin(x+h)-sin(x)$, that's why I made a mistake with the statement above. sin(dx) while what I meant is d sin(x)

 

[Stephanus]

In your second example, d by itself has no meaning, but $\frac{dx}{d\tau}$ means the derivative of x with respect to $\tau$. By definition $\frac{dx}{d\tau} = \lim_{h \to 0}\frac{x(\tau + h) - x(\tau)}{h}$

Oh.
Thanks, that helps me much in understanding SR.

Btw, about this $\frac{dx}{d\tau}$. As you know in SR (Special Relativity) they often use X/T as their function not T/X (or Y/X) as in real 'math'.
That's why I made an error by saying gradient is X/Y, while what I meant is Y/X.
Of course in 'math', the gradient for vertical length is undefined, but in SR a vertical line as you know, is a rest frame.
While in SR, the horizontal line is undefined, velocity is instant, but in 'math' a horizontal line is Y = n, the gradient is zero.

 

[Mark44]

Yes, this is a FACT!. I don't argue about it either.
Come on Mark114, give me some slack
$lim_{h \rightarrow 0} sin(h)$ is almost zero, too.

No, not almost zero. $\lim_{h \to 0} \sin(h)$ IS zero.

But of course $lim_{h \rightarrow 0} sin(h) ≠ (sin(x+h)-sin(x))$
And of course $cos(h)$ is a light year away from $cos(x+h)-cos(x)$
While $sin(h)$ is, I could say, in the neighborhood of $sin(x+h)-sin(x)$, that's why I made a mistake with the statement above. sin(dx) while what I meant is d sin(x)

 

[Stephanus]

No, not almost zero. $\lim_{h \to 0} \sin(h)$ IS zero.

I'm sorry Mark44.
if $lim_{h \to 0} \sin(h)$ is zero, how can

$lim_{h \rightarrow 0} \frac{sin(h)}{h} = 1$

I think $lim_{h \to 0} \sin(h) = h$. and also tan(h) = h

 

[Mark44]

I'm sorry Mark44.
if $lim_{h \to 0} \sin(h)$ is zero, how can

$lim_{h \rightarrow 0} \frac{sin(h)}{h} = 1$

As I said in a previous post, this limit is proved in many calculus textbooks.

I think $lim_{h \to 0} \sin(h) = h$. and also tan(h) = h

No, both of these are wrong. After you take a limit involving h, h will not appear in the result.
What is true is that for small h, $\sin(h) \approx h$ and $\tan(h) \approx h$, but both limits you showed above are zero.
$lim_{h \to 0} \sin(h) = 0$
and $lim_{h \to 0} \tan(h) = 0$ as well.

Before attempting to study advanced physics topics, you should get a calculus textbook and study it, or study it in an online course..

 

[micromass]

I think $lim_{h \to 0} \sin(h) = h$. and also tan(h) = h

This makes no sense since on the left side, $h$ is a dummy variable, while on the right it appears to be a real number.

 

[micromass]

Of course in 'math', the gradient for vertical length is undefined, but in SR a vertical line as you know, is a rest frame.
While in SR, the horizontal line is undefined, velocity is instant, but in 'math' a horizontal line is Y = n, the gradient is zero.

You probably don't want to use the word gradient here...

 

[jtbell]

You probably don't want to use the word gradient here...

To be more explicit:

In English-language math terminology, "gradient" is a concept from multivariable (vector) calculus, usually in three dimensions.

When we refer to the inclination of a line on a two-dimensional graph, e.g. a spacetime diagram for x and t in SR, we use the word "slope."

 

[Stephanus]

To be more explicit:

In English-language math terminology, "gradient" is a concept from multivariable (vector) calculus, usually in three dimensions.

When we refer to the inclination of a line on a two-dimensional graph, e.g. a spacetime diagram for x and t in SR, we use the word "slope."

Yeah, I"m an Indonesian and in high school time, we used the word "gradien", no "t" there
But that was 30 years ago. I don't know what the word now.
And in space time diagram for an accelerated world line, this "slope" is the velocity at that particular time?
Can I ask here.
In $Y = 4X + n$, the "slope" is "4"? Please confirm, so I can understand the explanations in SR Forum.
Thanks.

 

[jtbell]

And in space time diagram for an accelerated world line, this "slope" is the velocity at that particular time?

This is sneaky...

In classical (non-relativistic) physics we always draw these diagrams with t on the horizontal axis and x on the vertical axis. The slope is dx/dt which is indeed the velocity, v.

However, in relativistic physics, we customarily draw spacetime diagrams the other way around: x on the horizontal axis and t on the vertical axis. The slope is dt/dx which is the reciprocal of the velocity, 1/v. I don't remember why people started doing it this way. Someone in the relativity forum probably knows.