Solving Algebra Problem on Electrodynamics of Moving Bodies

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The discussion revolves around a mathematical error encountered while solving equations related to the electrodynamics of moving bodies. A participant initially struggles with a sign discrepancy in their calculations, specifically between fractions involving terms like 1/(c-v) and 1/(c+v). After reviewing their work, they discover a typo in their original PDF, which incorrectly used 1/(c+v) instead of 1/(c-v). This correction resolves the issue, leading to the correct formulation of the equations. The participant expresses gratitude for the feedback and acknowledges the importance of careful review in mathematical work.
Daminc
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Hi following a link on another thread I began to read On The Electrodynamics Of Moving Bodies

img23.gif


should equal

img24.gif


but I end up with a minus instead of a plus between the two fractions(in the last equation).

I've gone through my calculations but I cannot see where I've gone wrong.

I could post my working out if necessary but it might be hard to read in this format
 
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I assume you got this far on the coefficient of [pard]&tau/[pard]t

(1/2) * (1/(c-v) - (1/c+v))

Collecting fractions:

(1/2) * ((c+v) - (c-v)) / ((c-v)*(c+v))
(1/2) * (2v) / (c^2 - v^2)
v / (c^2 - v^2)

Hurkyl

(edit: fixed tags)
 
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It's almost impossible to catch your own error and the sillier the error is the harder it is to find it!

In the first formula you have 1/(c-v)+ 1/(c+v) on one side. You get common denominators, of course, and have ((c+v)+(c-v))/(c^2- v^2)= 2c/(c^2- v^2) which is then multiplied by 1/2: c/(c^2- v^2)

On the other side of the same formula you have 1/(c-v). Subtracting c/(c^2- v^2) from both sides: 1/(c-v)-c/(c^2- v^2)= c/(c^2-v^2)- (c+v)/(c^2-v^2)= v/(c^2- v^2).

Did you move the fraction to the wrong side?
 
After I got c/(c^2- v^2) on left side I moved the
1/(c-v)over to isolate [pard]T/[pard]x over on the right then
removed [pard]T/[pard]t from the brackets to get the fraction v/(c^2- v^2).

I them moved the whole lot across to put it equal to zero
 
I am not sure if http://home.attbi.com/~rossgr1/Specialrel.PDF addresses you exact problem, but you may find it smooths your way to the final DE.
 
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Ok that PDF works it out a different way but, even now, things don't look right

From the bottom of the second page

We have "Now recombining the RHS and LHS we have:"

1/2 ((1/c+v)+(1/c-v))[pard]T/[pard]t = [pard]T/[pard]x +(1/c+v)([pard]T/[pard]t)

Right, fine and dandy

BUT, when I collate the terms I get:

1/2((1/c-v)-(1/c-v)([pard]T/[pard]t)=[pard]T/[pard]x

NOT

1/2((1/c+v)-(1/c-v)([pard]T/[pard]t)=[pard]T/[pard]x

So when it comes to multiple both sides with -1 I still get a sign discrepancy :frown:

I'm going to start to cry real soon...
 
I am getting the same result as you, let me look at this to see if I can sort out what happened. It has been a year since I wrote that PDF up, so I need to take some time to get back into the material.
 
Well, since I could not find an obvious error in my PDF, I went back to the source. There is where I found MY error. Compare my RHS to Einstein's in my starting equaition. You will see that I have a 1/(c+v) where he has a 1/(c-v). If you replace my RHS term with the correct one it all comes out in the end. Since I worked all this out on paper then transcriped to MS Word editor and PDF, I will claim it as a typo.

My apology, and thanks all at the same time. I have posted that PDF several times, here and other places over the past year. It looks like someone finally read it!
 
Alleluia

I thought I was going mad.

It's a good paper, I bet lots of people read it but I bet they rarely actually follow the math.

Thanks again.
 
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I have posted a corrected PDF, the link above should now connect to a PDF that does not have the error discovered in this thread.
 

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