Is Proving Einstein Wrong a Serious Challenge or Just a Fun Debate?

mruncleramos
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This site was linked at the top of the physics forums website. www.relativitychallenge.com I guess everybody has freedom of speech. Do people just get a kick out of proving Einstein wrong?
 
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mruncleramos said:
This site was linked at the top of the physics forums website. www.relativitychallenge.com I guess everybody has freedom of speech. Do people just get a kick out of proving Einstein wrong?

Non-professional physics fans sometimes think they can become famous
by proving Einstein wrong.

It not unlike trying to break a sports record, only an athlete usually
knows if they are cutting the mustard or not. Most of these folks
don't know how much they don't know.

BTW, I'm not saying he won't be shown to be incomplete at some point,
but probably not by such folks.
 
Hm, I've already discovered mathematical and physical fallacies in his argument. Some people just can't do algebra i guess.
 
Note that physicsforums.com did not choose that ad... Google did.

Google pays sites for the right to display ads of their [Google's] choosing, based on keywords that appear on the site. In turn, advertisers pay Google in order to have their sites be advertised on other sites that contain specified keywords.

Click on the "Ads by Google" link for information. There's also a feedback form. If enough people complain about crackpot ads on this site, maybe Google will try to adjust their algorithms accordingly.
 
jtbell said:
Note that physicsforums.com did not choose that ad... Google did.

We block bad ads when we find them, but new ones keep popping up.
 

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Thread 'Is the variation of the metric ##\delta g_{\mu\nu}## a tensor?'

From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...
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