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(Some Guy)   With a pic that makes you wonder what are his evil intentions, and no financial support from any institution or government, a scientist redoes Albert Einstein's Special Theory of Relativity   ( twocircles.net) divider line
    More: Spiffy, Einstein, Indian scientists, time dilations, special relativity, Indians, Khilji, quantum field theory, Doppler  
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6841 clicks; posted to Geek » on 26 Jul 2012 at 11:58 AM (5 years ago)   |   Favorite    |   share:  Share on Twitter share via Email Share on Facebook   more»

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2012-07-26 12:26:43 PM  
2 votes:
The mangled auto-translation is kind of hard to wade through, but basically he seems to be making the same schoolboy mistake that many people make when first confronted with the idea that [for non-accelerated frames] you can't really say which is at rest and which is moving. In other words, he has written a paper on relativity that begins by rejecting relativity.

Galloping Galoshes: On the right side of the equation, yeah, as v approaches c, c2-v2 approaches 0, so both the numerator and denominator go to 0 as well. I always understood that 0/0 was undefined. I also thought that as v approached c, m became infinite. Or was I misinformed?

Yes to the first, no to the second.

When you have an equation that has a singularity at some point, as you do here, you have to analyze what happens as it approaches the neighborhood of that singularity, i.e. find the limit. To give a very simple example, consider the formula (x/x^2). At x = 0, you get 0/0. Undefined? No. On closer examination you can see that if you divide through by x, the formula is identical to (1/x); and it's clear what that does as x approaches 0: it tends to infinity. And with a proper formal analysis you can show that it does so smoothly. (And in case you're wondering, this loose word sketch can be turned into formal, rigorous math). So this particular equation is perfectly well behaved as far as mathematicians are concerned.

Now take a look his equation: if you square (c^2-v^2)^1/2/(c^2-v^2) it is essentially the same as (x/x^2), and the same analysis works. Although it naively looks like 0/0 when v =c, it is perfectly well behaved as v approaches c.
2012-07-26 11:47:26 AM  
2 votes:
No, he doesn't. In fact it's quite possible that he's just a rambling idiot with absolutely no training in mathematics or physics. Take a look at his paper. Don't worry, there's no math!

Here's the gist of his argument:
Relativistic mass = mc/(c2-v2)1/2 = mc(c2-v2)1/2/(c2-v2)

So far this is true, but now he claims that as you approach the speed of light, both the numerator and denominator go to zero, so instead of infinite mass you get 0/0 which is zero. He then naturally concludes that all matter turns into light/photon field (i.e. zero mass) when it reaches the speed of light. No seriously, that's what he's saying.
2012-07-27 12:42:43 AM  
1 vote:

starsrift: ...Or, wait. Since apparently everyone with a professional interest in the subject is reading it, maybe we should see it denounced by someone other than a guy who "really [has] no idea what he's trying to do".

Fine. The author, to use the most charitable description possible, is tragically confused about when to use plus signs and when to use that other kind. In addition, he hangs on to the classic crackpot's conceit: that the discrepancy between his ill-considered work and reality is due not to, say, a misplaced minus sign, but rather to some vast truth which only he is privileged to see.

Not to mention, lightspeed being a conversion point from matter to light photon particles is actually a hell of a lot more believable than infinite mass at rest.

Infinite rest mass was never an issue (unless I've skimmed past something even more insane than what I remember reading), but conversion from conventional matter to light as a consequence of speed is further out than you might think. Fermions and bosons do not work that way. You might be surprised at just how thoroughly this stuff has been contemplated, calculated, and tested. If you want to revolutionize humanity's understanding of the universe, you're going to need more than a quarter ounce and some pathetically sloppy math.

/Has plenty of idea what he's doing with special relativity.
//And has had occasion to measure Doppler shifts in light. Whaddya know, they come out just as Einstein calculated.
2012-07-26 12:24:44 PM  
1 vote:

DeltaPunch: get 0/0 which is zero

Speaking of someone with no training in mathematics...
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