B Are the following two formulas equivalent?

[south]

TL;DR Summary

The first one is well known. The second one is not and motivates the question.

$$T = m \hspace{0.03 cm}C^2 \left( \dfrac{1}{\sqrt{1-\dfrac{v^2}{C^2}}} -1 \right) $$

$$ T = \dfrac{m \ v^2}{ \left( 1+\sqrt{1-\dfrac{v^2}{C^2}} \right) \sqrt{1-\dfrac{v^2}{C^2}}} $$

 

[Orodruin]

Common denominator:
$$
\frac{mC^2}{\sqrt{1-v^2/C^2}}\left( 1 - \sqrt{1-v^2/C^2}\right)
$$
Multiply denominator and numerator by the conjugate $1 + \sqrt{1-v^2/c^2}$:
$$
\frac{mC^2}{\sqrt{1-v^2/C^2}}\frac{ 1 - (1-v^2/C^2)}{(1+\sqrt{1-v^2/C^2})}
=
\frac{mC^2}{\sqrt{1-v^2/C^2}}\frac{ v^2/C^2}{(1+\sqrt{1-v^2/C^2})}
$$
Yes, they are the same and it is relatively easy to show algebraically.

 

[south]

Yes, they are the same and it is relatively easy to show algebraically.

Thanks Orodium for help me.

The second formula makes it easy to demonstrate that for $$v \rightarrow 0 $$ the Newtonian formula appears, because the denominator directly tends to 2. This has a didactic advantage in secondary school, because it does not require series development.

Cordial greetings.


Cordial greetings.

 

[robphy]

Possibly useful reading:
https://physics.stackexchange.com/q...e-that-kinetic-energy-equals-k-fracp2-gamma1m

While true, the issue I have with these attempts is to motivate WHY these forms arise as such...
what is the interpretation, beyond being a convenient way to find the non-relativistic case (using substitution, rather than a limit)?

 

[Orodruin]

While true, the issue I have with these attempts is to motivate WHY these forms arise as such...
what is the interpretation, beyond being a convenient way to find the non-relativistic case (using substitution, rather than a limit)?

Why does there have to be a deeper interpretation than being a didactic trick for people who are not yet familiar with Taylor expansion?

 

[south]

Possibly useful reading:
https://physics.stackexchange.com/q...e-that-kinetic-energy-equals-k-fracp2-gamma1m

While true, the issue I have with these attempts is to motivate WHY these forms arise as such...
what is the interpretation, beyond being a convenient way to find the non-relativistic case (using substitution, rather than a limit)?

Tell me what you think, does the following sentence seem sensible or stupid to you?If a and B are equivalent, they admit the same interpretation and mean the same thing. The only difference could be in how a and B mathematically inspire whoever decides to develop something based on A and/or B.

Regarding the formula displayed in the link, if I interpreted the nomenclature correctly, it has an error. It is corrected by multiplying the second term by

$$ \sqrt{ 1 - \dfrac{v^2}{C^2} } $$

and with that you obtain the formula for my question.

 

[south]

Why does there have to be a deeper interpretation than being a didactic trick for people who are not yet familiar with Taylor expansion?

Taking the formula in my question as true, I continued to operate (including a couple of changes of variables) and arrived at an equation that has an extremum. I found the extremum and the result is the following.
$$e^\delta = \dfrac{2+\delta}{1+\delta} $$
with
$$\delta = \sqrt{1-\dfrac{v^2}{C^2} } $$
Numerically I obtained an approximation to the solution.
δ=0.508554724060375(5)
which corresponds to
$$ \dfrac{v}{C}=0.7487202289405... $$
The extremum occurs for
$$v \simeq \dfrac{3}{4} \hspace{0.1 cm}C$$
Mathematically that extreme exists, it corresponds to almost 3/4 of C, but I have no idea what the physical situation is like at that extreme.

If I had access to the resources necessary to accelerate particles up to v= 0.74872(...) C , I would do it with great curiosity.

 

[Ibix]

Take$$y=x$$Change variable to $u=x+1$ to get$$y=u-1$$Change variable again to $v=\sqrt{u}$ to get$$y=v^2-1$$Differentiate to get$$\frac{dy}{dv}=2v$$So there's an extremum at $v=0$, which is $u=0$ and $x=-1$.

Have I proved that there's an extremum in $y=x$? Or is it an artefact of my manipulation and some thoughtless differentiation? Now go and look at a graph of $K=(\gamma-1)mc^2$. Are there any stationary points except $v=0$?

 

[robphy]

While true, the issue I have with these attempts is to motivate WHY these forms arise as such...
what is the interpretation, beyond being a convenient way to find the non-relativistic case (using substitution, rather than a limit)?

Why does there have to be a deeper interpretation than being a didactic trick for people who are not yet familiar with Taylor expansion?

Of course, there doesn't have to be...
but I think we would prefer to have a deeper interpretation [using physical principles and methods developing intuition, not a collection of tricks]
rather than...
here's a mathematically equivalent form [motivated algebraically because we know the limit we expect].

Does this algebraic form teach us anything else?
Maybe.
It would be nice if the result of the "didactic trick" ("didactrick")
naturally fell out of a physics calculation (like the work-energy theorem)...
or had a Minkowski-spacetime-geometric interpretation.

But if such a "didactic trick" is all that one has, then it would be better than not having one.

When developing relativistic kinetic energy, Kleppner (An Introduction to Mechanics) notes
"This expression for kinetic energy bears little resemblance to its classical counterpart."
The works I cited above attempt to develop a resemblance algebraically...
but I don't think those attempts tell us anything more.


(I am writing an article that develops the resemblance using Minkowski-spacetime-geometry,
with analogues in Euclidean geometry and Galilean-spacetime-geometry.)

 

[robphy]

If a and B are equivalent, they admit the same interpretation and mean the same thing. The only difference could be in how a and B mathematically inspire whoever decides to develop something based on A and/or B.

It depends on what "equivalent" means, and what "same interpretation" means.
Certainly an alternative form can suggest a different interpretation. That's fine.. and possibly very good.
So, I am asking for the different interpretation....
what new physical or mathematical interpretations and intuitions arise?
(In the works I cited above... I don't see anything new
other than they are conjured up algebraically (but sadly not physically)
to give the non-relativistic limit easily.)

 

[SammyS]

He muses as follows.
. . .

@south .
In your above Quote of @robphy 's post#10, only the first sentence is @robphy 's. The rest of that inset is your writing. It is as follows and belongs in the main body of your post#11

"
I understand what you say and after thinking about it I agree. If the two equivalent forms are different, it is reasonable to look for details in each one that you does not suspect in the other.
My Physics 3 teacher, Ladislao Nicolás Bodnár, detected something in the second formula, that is, in this one:-
$\displaystyle \quad\quad T = \dfrac{m \hspace{0.1 cm} v^2}{\left( 1+ \sqrt{1- \dfrac{v^2}{C^2}} \right) \sqrt{1- \dfrac{v^ 2}{C^2}}}$
He says that the denominator has a similar shape to what appears in an equation corresponding to two coupled systems, such as two pendulums, two mass-spring oscillators, etc. I use Bodnár's words because I did not need to study coupled systems to pass exams and I cannot corroborate or deny what my teacher said.
"

He muses as follows.
. . .

 

[south]

Thanks SammyS for letting me know about the error. I'll try to fix that.
Edit: I have not been able to correct it. Robphy's writings were always mixed up with mine. I finally deleted the post.

 

[south]

Of course, there doesn't have to be...
but I think we would prefer to have a deeper interpretation [using physical principles and methods developing intuition, not a collection of tricks]
rather than...
here's a mathematically equivalent form [motivated algebraically because we know the limit we expect].

Does this algebraic form teach us anything else?
Maybe.
It would be nice if the result of the "didactic trick" ("didactrick")
naturally fell out of a physics calculation (like the work-energy theorem)...
or had a Minkowski-spacetime-geometric interpretation.

But if such a "didactic trick" is all that one has, then it would be better than not having one.

When developing relativistic kinetic energy, Kleppner (An Introduction to Mechanics) notes
"This expression for kinetic energy bears little resemblance to its classical counterpart."
The works I cited above attempt to develop a resemblance algebraically...
but I don't think those attempts tell us anything more.


(I am writing an article that develops the resemblance using Minkowski-spacetime-geometry,
with analogues in Euclidean geometry and Galilean-spacetime-geometry.)

My Physics 3 teacher, Ladislao Nicolás Bodnár, found some similarity between the denominator of the second formula
$$ T = \dfrac{m \hspace{0.1 cm} v^2}{\left( 1+ \sqrt{1- \dfrac{v^2}{C^2}} \right) \sqrt{1- \dfrac{v^2}{C^2}}} $$
and an equation referring to two coupled systems, of the mass-spring type, pendulums, L-C, for example.

In the cases mentioned, a mass-spring system is coupled to the other, a pendulum is coupled to the other and an L-C oscillator is coupled to the other.

What does a solitary mobile object traveling in a vacuum couple with? To couple with the mobile object there is nothing, except the vacuum. If we take the case seriously, there is no other option than to conceive the vacuum as a physical entity endowed with properties that allow coupling.

Some people claim that $$ \varepsilon_o \ and \ \mu_o $$ are simply factors that serve to adapt units, without the slightest physical function. Bodnár never accepted this claim and always conceived of these two constants as real physical properties of the vacuum, inductance per unit length and capacitance per unit length.

Bodnár also assumed that this pair of constants implies directivity of electrodynamic phenomena, since in the omnidirectional case they would be inductance and capacitance per unit volume.

 

[Herman Trivilino]

The second formula makes it easy to demonstrate that for v→0 the Newtonian formula appears, because the denominator directly tends to 2. This has a didactic advantage in secondary school, because it does not require series development.

Check out this post https://www.physicsforums.com/posts/5635657/bookmark

 

[Orodruin]

Check out this post https://www.physicsforums.com/posts/5635657/bookmark

https://www.physicsforums.com/threa...-a-relativistic-particle.895897/#post-5635657

 

[south]

Check out this post https://www.physicsforums.com/posts/5635657/bookmark

Hi Mister T. I don't know what bookmark is here inthe forum.

 

[SammyS]

Hi Mister T. I don't know what bookmark is here in the forum.

Simply follow the link that @Orodruin gave. It looks like Oro gave that so you didn't have a problem with that 'bookmark' .

 

[Herman Trivilino]

Hi Mister T. I don't know what bookmark is here inthe forum.

Yeah, sorry about that. Follow the link in Post #15.