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

[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.