🎓 Doc Freemo 🇳🇱 @freemo@qoto.org

Sabine Hossenfelder: Backreaction: Dark energy might not exist after all

http://backreaction.blogspot.com/2019/11/dark-energy-might-not-exist-after-all.html?m=1

There are some interesting comments too ..

#cosmolgy

@freemo we all look like amateurs hehehehe

@design_RG Here is mention in the history books of one of my ancestors from 1442, a direct male ancestor.

@design_RG You can also see one form of our Coat of Arms there. In this case due to marriage it is a combination of the 4 houses coat of arms. The Coat of Arms that is specific to my lineage is attached rendered with colors.

Being an Armiger myself means my male history is actually really easy to follow.

@freemo
#introductions

#programming
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To help some of the newcomers make connections: name 5-7 things that interest as tags so they are searchable. Then boost this post or repeat its instructions so others know to do the same. Add #introductions to the post.

#Photography
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#ClassicalLiberal

It’s about time.

@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