Force-Energy Equation

As a semi-response to Rpenner, who asked exactly what use, or even better for terms of words, what valuable use would \(\frac{1}{2}Mc^2=\frac{1}{2}Mc^2\) have, i decided it had importance due to the following relationships and generally for creating a new force-energy equation of equivalance.

Plugging \(\frac{1}{2}Mc^2=\frac{1}{2}Mc^2\) into \(E=Fvt\), we get;

\(\frac{1}{2}(Fvt)^2=\frac{1}{2}E^2\) [1]

This only tampers with the equation's numerical contents.

From another equation i derived, that solved for the time, i wanted to solve now for the energy content;

\(t=(\frac{vt}{E})mv\)

\(Mv=\frac{\frac{t}{E}}{vt}\)

\(\frac{t}{vt}E=Mv\)

\(\frac{\frac{Mv}{vt}}{t}=E\)

Finally

\(E=(\frac{Mv}{t})vt\)

Now plugging this into [1] gives

\(\frac{1}{2}(Fvt)^2=\frac{1}{2}Mc^2(\frac{Mv}{t})\)

Which says that force is due to the energy of the system, but more interestingly i think is that;

\(\frac{1}{2}Mc^2=\frac{\frac{\frac{1}{2}(Fvt)^2}{ \frac{1}{2}Mv}}{\frac{1}{2}t}\)

 

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Oh dear.

 

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What? I've plugged real values into the equations and derived the correct results. Tell me what is your problem with it?

 

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Wait, i might have made a mistake with the last one. Just checking it.

 

Why do you bother typing in so many complicated equations?

 

I don't mean to. If i could have made them simpler i would have. It just kinda turned out that way.

 

For example, you don't need to type fractions like 3 or 4 story tower, unless you are writing continued fraction or something you can get away with only one horizontal strike.

 

I can?

 

Oh dear, oh dear.

 

Once upon the time, Thomas Reid and Dugald Stewart promoted a useful idea called Scottish Realism, where metaphysical quibbles were set aside and an emphasis was placed on a joint development of science and engineering in the pursuit of what worked. American development during the 19th century owed much to this idea, even though it eventually collided with Presbyterian beliefs in a simple and literal view of biblical inerrancy resulting in that evil known as scientific creationism.

It has been suggested that wartime-hardships destroyed the traditional Scottish garden plot, acclimatized the Scottish people to tinned vegetables and set back Scottish cuisine back 100 years.

Here we see Reiku's report from the North, and we are informed that this is just an example of top work in the field. Since it is demonstrably inferior to preserved cuneiform examples, according to Reiku even top men in Scotland lag thousands of years behind the state of the art.

Why would this person want to return Scottish mathematics to scratches on stone? I believe the only conclusion for a reasonable is that Reiku is, indeed, no true Scotsman! Though he walks among them, he is intent on their destruction as a people.

Following the Mike Myers principle of "If it ain't Scottish, it's crap!" I must petition for this thread, in entirety, to be moved to the cesspool, if it has not already been placed there.

 

This is wrong since \(\Delta E=\int Fdr\), so that from the dimensional POV, this is wrong.
But hey, I got a better one:
If you take this equation:

and divide it by this equation:

(which is the same as:\( Frt = E \))
you get \(\frac{\frac{1}{2}F^2rt}{Frt} = \frac{\frac{1}{2}E^2}{E}\), from there you get after simplification: \(F=E\).
Now, you know from electrodynamics that \(F = qE\) so that you get \(E = qE\) or \(q = 1\).
You certainly know from thermodynamics that \(q=mc\Delta t\) and using the fact (just proven) the \(q = 1\), you can see that \(mc\Delta t = 1\) from this you get \( \Delta = \frac{1}{mct}\).

 

I'm not scottish.

On a side note, i know the equations are fine because the numerically work. So why not explain to me why you petitioned for it to be cesspooled? I mean, i am confused because i know the algebra works, and if it didn't, the thread would have been flocked with comments like, ''look at this mistake, and that one, and that one.''

 

:roflmao:

 

Sorry, it was supposed to be F^2vt. I am going to fix this right now. My mistake.

 

And I must object. I think that this post should become a sticky one, so than when Reiku will receive his nobel prize, we will know where it all started.

 

(By the way, Fvt=E is not my own equation -- it's a well used one in physics.)

 

Yes, it's something I helped teach a class of 1st years about a fortnight ago. Except that you applied it to a relativistic system. And assumed, without realising, that the force and velocity are both constant (or at least their scalar product is). And you applied it to a photon which cannot change it's velocity. So not just quantitative mistakes but conceptual ones too.

And Prometheus is right to laugh at you. You do realise that \(\Delta t\) is not \(\Delta \times t\) but the change in time, ie \(t_{\textrm{final}} - t_{\textrm{initial}}\). That is a well known bit of notation too.... to 14 year olds.

You keep saying "It's well known" but where are you actually getting your information about what the mainstream does or doesn't know? You don't read textbooks, you don't seem to even read pop science books other than by your beloved quack Wolf.

Yes, E=Fvt is well known but generally those who know it know how to use it. Not in relativistic systems, not when the force and velocities do not give a constant dot product and not on a photon.

No, it is not supposed to be that. You went from E = Fvt to \(\frac{1}{2}E^{2} = \frac{1}{2}F^{2}vt\). This is precisely the same mistake you did with E=pc when you said \(E^{2} = cp^{2}\). You forgot the brackets, ie if E=Fvt then \(E^{2} = (Fvt)^{2} = F^{2}v^{2}t^{2}\). The fact you, yet again, show you struggle to even multiple out brackets makes everyone see you for the fraud you are. Do you expect us to believe you can do vector calculus or integration or quantum mechanics or GR when you can't even expand out brackets?! Come on, even you must see how foolish trying to peddle such lies must be?

No, they don't work numerically. Nor do they work by dimensional analysis, something I explained to you only yesterday (and not for the first time). You must be unable to do even 'Put the numbers in and check' because you failed to do so here and you failed to do so many times for \((a+ib)(a-ib)\). I gave numerous examples of why \((a+ib)(a-ib) \not= a^{2}-b^{2}+2abi\) but you didn't listen. (a=b=1 gives 2 in the left hand side and 2i on the right so the equation is not true in general). Let's try it with E=Fvt

Let F = 10, v = 1 and t = 2. This gives E = 20. \(\frac{1}{2}E^{2} = 200\). \(\frac{1}{2}F^{2}vt = \frac{1}{2}10^{2} \times 1 \times 2 = 100\). Not equal. Therefore \(\frac{1}{2}E^{2} \not= \frac{1}{2}F^{2}vt\). Therefore you are wrong.

Do you understand or shall I explain a bit more why 200 and 100 are not equal numbers? :shrug:

 

First, prom wasn't laughing at me. And the equation is right, as i will soon take you through it.

 

(I want it known, like a right tit, i have placed the square over the wrong symbol, and it should have taken the form \((Fvr)^2\), but i have been very tired today.)

First we have the given equations, \(\frac{1}{2}Mc^2=\frac{1}{2}Mc^2\) and \(Fvt=E\). Now, since \(Mc^2\) is equal to the energy \(E\), then if we plugged in the expression \(\frac{1}{2}Mc^2\) on the left hand side of \(Fvt=E\), you must have an answer that is twice the magnitude, but then creeps in that fraction as well. \(\frac{1}{2}(Fvr)^2\). I mean, its quite obvious the rest is correct as well. Apart from me not noticing i had placed the sqaure in the wrong place, but cei la vie.

 

Fair point, you didn't make the mistake about the delta.

Unfortunately, for you, you did make the mistake with the E=Fvt equation.

How can it be right when I've given a counter example? How do you not understand that if A = BC then \(A^{2}=B^{2}C^{2}\). If \(10 = 2 \times 5\) then \(10^{2} = 2^{2} \times 5^{2}\).

Just put in numbers such that v*t is not 1 and you'll find that your two equations don't match. The equation \(E^{2} = F^{2}vt\) isn't even right on it's own because the units don't match. The left hand side has units of energy squared, which in SI units are \((kg m^{2}s^{-2})^{2} = kg^{2}m^{4}s^{-4\). The right hand side has units \((kg m s^{-2})^{2}(ms^{-1})(s) = kg^{2} m^{3}s^{-4}\). You're missing a factor with units of length, which the v*t I told you you missed out has!

So you're proven wrong via a numerical example and by dimensional analysis. There is nothing which you can say which will suddenly make that error right. The only way you can somehow show the result is to make a second error, trying to make two wrongs make a right.

I thought you were scrapping the bottom of the barrel ignorance-wise when you couldn't multiply out (A+B)(C+D) properly but now you don't seem to be able to multiply out (AB)(AB) !!

Seriously, think about if. If \(E = Fvt\) then \(E^{2} \equiv E \times E = (Fvt) \times (Fvt) = F^{2}v^{2}t^{2}\). How is it that you're unable to grasp that? I didn't think it was possible for you to be worse at maths and physics than I thought you were this time yesterday but congratulations, you've managed to limbo under my expectations. I don't know where else you can go from here except up but something tells me you'll find a way to slide under my expectations at some point in the future again....

/edit

It would seem while I was typing that you posted. At least you've seen your error. The fact you didn't realise it when you read my post, where I explicitly stated the mistake, didn't do you any favours. You did precisely the same mistake in another thread claiming that three physicists you know said that \(cp^{2} = (E \frac{v}{c})^{2}\) was correct. Hence why I (and others) don't give you the benefit of the doubt in this thread. Once is an accident, twice means you don't know you're incorrect.

/edit again

And \(E=Mc^{2}\) is the rest mass equation. If v is non-zero then you have to be using either \(E = \sqrt{(Mc^{2})^{2} + |p|^{2}c^{2}}\) or \(E = m_{0}\gamma c^{2}\). You're using neither but talking about the energy of a moving object while simultaneously ignoring relativistic kinematic effects. This is a common mistake for you.