Notices by 🎓 Dr. Freemo :jpf: 🇳🇱 (freemo@qoto.org), page 112

@zeccano

Yes im sure...

The frame of reference is the observer not the thing he observers.

By your deformed logic everything is a single frame of reference so long as the observer can saee it, including the other observer.

Thats not how it works and its a reflection of just how grossly uneducated you are.

I will not continue to argue with someone willfully stupid.

@CCoinTradingIdeas

@CCoinTradingIdeas

Told ya that even after the equation was derived, followed his rules, and disproved him wrong, he would still try to cry fowl and make up more excuses. Dogma only at this point.

Shouldn't be surprising he tries so hard to avoid the experimental evidence all of it disproves him and not a single one supports him. So he pretends they dont exist and just acts like a fool and thinks it will get it somewhere.

@zeccano

@zeccano
What you just said makes no sense. You said it isnt in another frame of reference, but thats bullshit, anything with vleocity different than yourself is in a different frame of reference, as is anything with a different acceleration, or even something in a different physical location.

Some of those frame of referntces are similar some are less similar, but everything has its own frame of reference.

I am done debating with you, you have truly reached insane levels of absurdity and sickness to still assert this nonsense even after i showed you the math playing at your own damn rules.

So which is it, concede, or keep lying to yourself. I satisfied all your demands as you asked and proved you wrong, on your own terms. What you do next defines if you are capable of growing as a person or not...

@CCoinTradingIdeas

@CCoinTradingIdeas

Told ya that even after the equation was derived, followed his rules, and disproved him wrong, he would still try to cry fowl and make up more excuses. Dogma only at this point.

Shouldn't be surprising he tries so hard to avoid the experimental evidence all of it disproves him and not a single one supports him. So he pretends they dont exist and just acts like a fool and thinks it will get it somewhere.

So which is it, concede and start to learn, or give up and succumb to willful ignorance?

@zeccano

@zeccano
What you just said makes no sense. You said it isnt in another frame of reference, but thats bullshit, anything with vleocity different than yourself is in a different frame of reference, as is anything with a different acceleration, or even something in a different physical location.

Some of those frame of referntces are similar some are less similar, but everything has its own frame of reference.

I am done debating with you, you have truly reached insane levels of absurdity and sickness to still assert this nonsense even after i showed you the math playing at your own damn rules.

@CCoinTradingIdeas

@zeccano

1.2156205e-27 kg

@CCoinTradingIdeas

@CCoinTradingIdeas

Told ya that even after the equation was derrived, followed his rules, and disproved him wrong, he would still try to cry fowl and make up more excuses. Dogma only at this point.

Shouldnt be surprising he tries so hard to avoid the experimental evidence all of it disproves him and not a single one supports him. So he pretends they dont exist and just acts like a fool and thinks it will get it somwhere.

I am now officially done with the conversation.

@zeccano

@CCoinTradingIdeas Nos that I did exactly what @zeccano asked and derived the relatiistic mass for him, using the equations of his choosing, and shows it is non-0... which he claimed to be impossible to do mathematically...

What do you think the chance is he will admit he is wrong and start trying to learn actual relativity (or at least tell people its real!)?

I'll take some bets...

@zeccano

NOTE: for those who cant read the below equations (they aren't rendered) you can follow this link to see: http://mathb.in/38381

Here you go a full derivation from the math equations you started with to the one I provided you several times...

axioms (given by you, agreed by physicist):

\[E = {m}_{rel} \cdot {C}^{2}\]

\[{E}^{2} = {p}^{2} \cdot {C}^{2} + {{m}_{rest}}^{2} \cdot {C}^{4}\]

De Broglie's Equation for the momentum of a photon:

\[p = \frac{h}{\lambda}\]

Formula to convert wavelength \(\lambda\) into a frequency:

\[\lambda = \frac{C}{f}\]

Simplify given axioms:

\[{({m}_{rel} \cdot {C}^{2})}^{2} = {p}^{2} \cdot {C}^{2} + {0}^{2} \cdot {C}^{4}\]

\[{{m}_{rel}}^{2} \cdot {C}^{4} = {p}^{2} \cdot {C}^{2}\]

\[{{m}_{rel}}^{2} = \frac{{p}^{2} \cdot {C}^{2}}{{C}^{4}}\]

\[{{m}_{rel}}^{2} = \frac{{p}^{2}}{{C}^{2}}\]

\[{m}_{rel} = \sqrt{\frac{{p}^{2}}{{C}^{2}}}\]

then use de broglie's equation for the momentum...

\[{m}_{rel} = \sqrt{\frac{{(\frac{h}{\lambda})}^{2}}{{C}^{2}}}\]

\[{m}_{rel} = \sqrt{\frac{\frac{{h}^{2}}{{\lambda}^{2}}}{{C}^{2}}}\]

\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{{C}^{2} \cdot {\lambda}^{2}}}\]

Now convert wavelength to frequency...

\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{{C}^{2} \cdot {(\frac{C}{f})}^{2}}}\]

\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{{C}^{2} \cdot \frac{C^2}{f^2}}}\]

\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{\frac{C^4}{f^2}}}\]

\[{m}_{rel} = \sqrt{{h}^{2} \cdot \frac{f^2}{C^4}}\]

\[{m}_{rel} = \sqrt{\frac{{h}^{2} \cdot f^2}{C^4}}\]

and bam you arrive at the equation for the relativistic mass of a photon.

\[{m}_{rel} = \frac{h \cdot f}{c^2} \]

#physics #relativity #einstein #math #maths #mathematics

@CCoinTradingIdeas

@zeccano

NOTE: for those who cant read the below equations (they aren't rendered) you can follow this link to see: http://mathb.in/38381

Here you go a full derivation from the math equations you started with to the one I provided you several times...

axioms (given by you, agreed by physicist):

\[E = {m}_{rel} \cdot {C}^{2}\]

\[{E}^{2} = {p}^{2} \cdot {C}^{2} + {{m}_{rest}}^{2} \cdot {C}^{4}\]

De Broglie's Equation for the momentum of a photon:

\[p = \frac{h}{\lambda}\]

Formula to convert wavelength \(\lambda\) into a frequency:

\[\lambda = \frac{C}{f}\]

Simplify given axioms:

\[{({m}_{rel} \cdot {C}^{2})}^{2} = {p}^{2} \cdot {C}^{2} + {0}^{2} \cdot {C}^{4}\]

\[{{m}_{rel}}^{2} \cdot {C}^{4} = {p}^{2} \cdot {C}^{2}\]

\[{{m}_{rel}}^{2} = \frac{{p}^{2} \cdot {C}^{2}}{{C}^{4}}\]

\[{{m}_{rel}}^{2} = \frac{{p}^{2}}{{C}^{2}}\]

\[{m}_{rel} = \sqrt{\frac{{p}^{2}}{{C}^{2}}}\]

then use de broglie's equation for the momentum...

\[{m}_{rel} = \sqrt{\frac{{(\frac{h}{\lambda})}^{2}}{{C}^{2}}}\]

\[{m}_{rel} = \sqrt{\frac{\frac{{h}^{2}}{{\lambda}^{2}}}{{C}^{2}}}\]

\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{{C}^{2} \cdot {\lambda}^{2}}}\]

Now convert wavelength to frequency...

\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{{C}^{2} \cdot {(\frac{C}{f})}^{2}}}\]

\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{{C}^{2} \cdot \frac{C^2}{f^2}}}\]

\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{\frac{C^4}{f^2}}}\]

\[{m}_{rel} = \sqrt{{h}^{2} \cdot \frac{f^2}{C^4}}\]

\[{m}_{rel} = \sqrt{\frac{{h}^{2} \cdot f^2}{C^4}}\]

and bam you arrive at the equation for the relativistic mass of a photon.

\[{m}_{rel} = \frac{h \cdot f}{c^2} \]

#physics #relativity #einstein #math #maths #mathematics

@CCoinTradingIdeas

@zeccano

Here you go a full derivation from the math equations you started wth to the one I provided you several times...

axioms (given by you, agreed by physicist):

\[E = {m}_{rel} \cdot {C}^{2}\]

\[{E}^{2} = {p}^{2} \cdot {C}^{2} + {{m}_{rest}}^{2} \cdot {C}^{4}\]

De Broglie's Equation for the momentum of a photon:

\[p = \frac{h}{\lambda}\]

Formula to convert wavelength \(\lambda\) into a frequency:

\[\lambda = \frac{C}{f}\]

Simplify given axioms:

\[{({m}_{rel} \cdot {C}^{2})}^{2} = {p}^{2} \cdot {C}^{2} + {0}^{2} \cdot {C}^{4}\]

\[{{m}_{rel}}^{2} \cdot {C}^{4} = {p}^{2} \cdot {C}^{2}\]

\[{{m}_{rel}}^{2} = \frac{{p}^{2} \cdot {C}^{2}}{{C}^{4}}\]

\[{{m}_{rel}}^{2} = \frac{{p}^{2}}{{C}^{2}}\]

\[{m}_{rel} = \sqrt{\frac{{p}^{2}}{{C}^{2}}}\]

then use de broglie's equation for the momentum...

\[{m}_{rel} = \sqrt{\frac{{(\frac{h}{\lambda})}^{2}}{{C}^{2}}}\]

\[{m}_{rel} = \sqrt{\frac{\frac{{h}^{2}}{{\lambda}^{2}}}{{C}^{2}}}\]

\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{{C}^{2} \cdot {\lambda}^{2}}}\]

Now convert wavelength to frequency...

\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{{C}^{2} \cdot {(\frac{C}{f})}^{2}}}\]

\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{{C}^{2} \cdot \frac{C^2}{f^2}}}\]

\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{\frac{C^4}{f^2}}}\]

\[{m}_{rel} = \sqrt{{h}^{2} \cdot \frac{f^2}{C^4}}\]

\[{m}_{rel} = \sqrt{\frac{{h}^{2} \cdot f^2}{C^4}}\]

and bam you arrive at the equation for the relativistic mass of a photon.

\[{m}_{rel} = \frac{h \cdot f}{c^2} \]

#physics #relativity #einstein #math #maths #mathematics

@CCoinTradingIdeas

@Absinthe Can never be too safe

@zeccano

I will shortly. I have it written out but diuble checking my work.

@CCoinTradingIdeas

@zeccano

No one leaves any backdoors opens, it is simply that these ideas are hard to explain to lay people, physicists talk to each other in a more technical language.

The fault you see is because you refuse to learn the language to describe these things

@CCoinTradingIdeas

@zeccano Sure thing, thats easy.

@zeccano Why have you still not answered about the equations you provided, you want the derrivation of relativistic mass or not!?

@CCoinTradingIdeas

He hears what he wants...

@zeccano

@zeccano

I would e-mail dr. licoln just so he can tell you that you are wrong (He will likely respond to a fellow scientist like myself). If he tells you relativity is real will you believe him.

Also waiting for you to respond to this:

https://qoto.org/@freemo/103223393374883539

@CCoinTradingIdeas

@hansw I need more of these in my life!

@zeccano

I presume you agree with the two axioms you already stated earlier for our derivation (you provided):

\(E = {m}_{rel} \cdot {C}^{2}\)

and

\({E}^{2} = {p}^{2} \cdot {C}^{2} + {{m}_{rest}}^{2} \cdot {C}^{4}\)

Correct, if so I will be happy to use these as our axioms for the derivation.

@CCoinTradingIdeas