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...making Trash the appropriate place for the thread.

By your style, you must be Mahesh Khati. Why do you need two different accounts? Anyways: [math]\frac{dE}{dt}=\vec{F}\vec{u}=F_xu_x+F_yu_u+F_zu_z[/math] so: [math]F'_x=\frac{F_x}{1-Vu_x/c^2}-\frac{V}{c^2}\frac{F_xu_x+F_yu_u+F_zu_z}{1-Vu_x/c^2}=F_x-\frac{V}{c^2}\frac{F_yu_u+F_zu_z}{1-Vu_x/c^2}[/math] If you replace V with -V you get: [math]F'_x=F_x+\frac{V}{c^2}\frac{F_yu_u+F_zu_z}{1+Vu_x/c^2}[/math] (for a different derivation, see below) You aren't "moving with constant velocity". The motion is accelerated, remember? So, you do not want to finish the calculations, after all the hints I gave you? [math]F'_x=\frac{1}{1+Vu_x/c^2}(F_x+m_0V\frac{d}{dt}\frac{1}{\sqrt{1-u^2/c^2}})=\frac{1}{1+Vu_x/c^2}(F_x+\frac{V}{c^2}(F_xu_x+F_yu_y+F_zu_z))=F_x+\frac{V}{c^2}\frac{F_yu_y+F_zu_z}{1+Vu_x/c^2}[/math] You can find the correct derivation in any book. I gave "hari" a reference. This is what he's talking about. Heck, I just got tired of waiting for him to realize his mistakes and I finished the proof for him.

sigh

Ok, gone full out woo.

Here is the data , you can try doing it by yourself. Hint, it will not change much, your theory is DOA.

Still nothing. Let me help you a little more: I made a mistake earlier (I took the partial derivative instead of the total one): [math]\frac{dt'}{dt}=\gamma(V)(1+V/c^2 \frac{dx}{dt})=\gamma(1+Vu_x/c^2)[/math] [math]F'_x=\frac{m_0}{1+Vu_x/c^2}( \frac{d}{dt}\frac{u_x}{\sqrt{1-u^2/c^2}}+V\frac{d}{dt}\frac{1}{\sqrt{1-u^2/c^2}})=\frac{1}{1+Vu_x/c^2}(F_x+m_0V\frac{d}{dt}\frac{1}{\sqrt{1-u^2/c^2}})[/math]

So, nothing? Let me help you some more: [math]u'_x=\frac{u_x+V}{1+u_xV/c^2}[/math] [math]t'=\gamma(V)(t+Vx/c^2)[/math] You will also need: [math]\frac{dt'}{dt}=\gamma(V)[/math] and [math]\frac{1}{\sqrt{1-u'^2/c^2}}=\frac{1+u_xV/c^2}{\sqrt{1-u^2/c^2}\sqrt{1-V^2/c^2}}[/math] [math] \frac{u'_x}{\sqrt{1-(u'/c)^2}}=\frac{u_x+V}{\sqrt{1-u^2/c^2}\sqrt{1-V^2/c^2}}[/math] All you have left to do is to calculate: [math]\frac{d}{dt'}\frac{u_x+V}{\sqrt{1-u^2/c^2}\sqrt{1-V^2/c^2}}=\gamma \frac{dt}{dt'}\frac{d}{dt}\frac{u_x+V}{\sqrt{1-u^2/c^2}}=\frac{d}{dt}\frac{u_x+V}{\sqrt{1-u^2/c^2}}=\frac{d}{dt}\frac{u_x}{\sqrt{1-u^2/c^2}}+V\frac{d}{dt}\frac{1}{\sqrt{1-u^2/c^2}}[/math] The first term (multiplied with [math]m_0[/math]) gives [math]F_x[/math], the second term gives out the symmetric terms dependent on [math]F_y,F_z[/math] Do you think you could do that without introducing new mistakes?

So, you do not know how to calculate derivatives. I gave you some hints, didn't seem to register with you. You must be related to Mahesh Khati. Or you are going to the same school, where they teach relativity denying based on bad calculus. I agree. It must be difficult for you to do any physics when you are unable to calculate a basic derivative. It is even more disturbing when you cannot do that after being shown how to do the calculations correctly. Let's try a different way to show what you are doing wrong. Let's admit that for a minute. Nope, this doesn't follow, you are simply making up nonsense.The correct derivation is: [math]F'_x=\frac{d}{dt'} \frac{m_0u'_x}{\sqrt{1-(u'/c)^2}}[/math]

Hilarious , I wonder why he doesn't give out his name, I know a few circuses that need good clowns.

You will have to figure where you went wrong by yourself, I narrowed it down for you. Have fun! Do you go to the same school as Mahesh Khati? What happens when you calculate: [math]\frac{d}{du_y} \frac{u_y}{\sqrt{1-(u/c)^2}}=\frac{1-u^2/c^2+u_yu/c^2}{\sqrt{1-u^2/c^2}^3}[/math] If you do it correctly, you get the above. If you do it wrong, as you keep doing it, you get what you are getting.

I bet that your example will be thoroughly ignored by the "scientists via posting on the internet". Not going to happen, he needs all his time to post on the internet, no time left for studying. Correction: learned people resist the so-called "change" coming from fringers. Instead, they debunk the fringers' ideas.

Spot on, two thumbs up. You see, learning science is tough, it is so much easier to post nonsense. With the advent of the internet, so many out of the box "scientists" have sprouted.

Heheh, citing crank, self-published "books". Nice going, mephestopheles. Bzzt, the dimensions are all wrong That much is obvious. Yep, full blown crackpot.

Well, you seem unable to get anything right <shrug>.

I'll have Ranch with this.

You are wrong.

This is a great post.

Err, it took me exactly 30 seconds to tear down Le Repeux theory. It took me a few minutes to tear down Mahesh Khati's theory (because it contained math). So, why don't you post your "theory"?

Too bad that I can only give you only one thumb up. All internet "scientists" should read the above and take heed (but it is not going to happen).

Yes, we also noticed that you try to hijack every thread with your same idea. You have a thread about "mass", stick to that thread, stop trying to hijack others. Yes, you are resiting the debunking of your claims.

Try again , with this: [math]E^2-(pc)^2=(mc^2)^2[/math] It is standard practice to set c=1.

Yes, the crackpot trademark is never to accept that his ideas have been refuted and to continue plastering them all over the internet, despite the ever increasing embarrassment caused during their refutation.

What he and his sidekicks are saying that that paradigm shifts also come from people who post on the internet. We are the bad guys for tearing down their "paradigm shifting" theories.

1. Yet , you keep pushing them. Why? 2. Why do you think that is the case? Must have something to do with point 1. 3. Your "ideas" have been shot full of holes multiple times. 4. In mainstream science we don't. If, on the other hand, you keep pushing your fringe ideas, the only solution is to start building the prototypes, running the experiments, yourself. No scientist would waste his time trying to build prototypes from flawed theories, like yours.