GR9677 #49



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syreen 20130911 11:48:17  Is it just me, or are your primes & unprimes swapped? Primed should be the moving frame, shouldn't it?
See GR8677 #22 for the same transformations, but with primes and unprimes switched.
Or does it not matter for this E&M transformations?
syreen 20130911 11:50:43 
Oops, just saw that the two are different.
Still, though, shouldn't we be solving for primed?

Jovensky 20130923 22:18:49 
syreen, we are looking from the rest frame of second observer, so the first observer is moving relative to the second observer, thus we are solving for the rest frame variables (the unprimed one).

  mpdude8 20120420 15:09:34  Hmm... I just said that as you increase your speed toward c, the electric field that you measure should increase.
As you increase v towards c, the square root term gets very small. Thus, it should be on the denominator. C is the only choice that has this feature  and A should be eliminated right off the bat because it has no dependence on v in the first place.   wittensdog 20091103 17:47:18  I'm pretty confident that ETS is getting at the issue of length contraction here. I would say that the field transformation formulas are a nice thing to know, if you can remember them, but in my opinion understanding how length contraction and other relativistic effects come into play has more to do with the real physics going on. If you remember the formula sigma / 2*epsilon, and you realize that the charge density should go up by a factor of gamma from length contraction, you have your answer in about two seconds.
The fact that the plane is infinite means that regardless of your velocity with respect to it, you still see it as infinite, so we know we can treat the problem as an infinite plane with a different charge density.
WoolfianOperator 20091105 18:27:55 
i agree. the other options have either no effect (A) on the field or the field getting smaller because of the terms (B,D,E). remember gamma is always greater than or equal to 1

flyboy621 20101024 16:51:33 
good solution!

blairwitch 20110312 09:32:24 
could you expound on how you get (derive) sigma/2* epsilon

asdfuogh 20111008 18:13:42 
Great point, looking at it as if the charge density became greater due to length contraction.

  tin2019 20071016 07:54:20  I find it simpler to deal with a potential 4vector which has to be invariant under Lorentz transformations. The field in question has a potential because, for using and the fact that potential must be independent of both x and y we get or, because . Now applying Lorentz transformations to we getrnrnrnrnand since we can take the vector potential to be , and thus we havernrnrnrnNow , for there is magnetic field now. component of the vector potential isrnrnrnrnHowever this is time independent, and so our equation for electric field reduces tornrn. rnrnTaking the gradient of relation we see that the electric field in primed system must have only a component, as is independent of and . This yieldsrnrnrnrnI realize that this is not the simplest solution, but it is easier for me to remember that the 4vector is Lorentz invariant. It could be that I'm using a cannon to kill a mosquito.
dumbguy 20071016 19:10:00 
yeah but you killed that mosquito

kronotsky 20181022 02:53:30 
I think this is a lot easier than remembering how the faraday tensor transforms (thanks for the tip!), but lorentz contraction is the way to go.

  hungrychemist 20071005 03:19:52  E = sigma/2*epsilon
Consider sigma (charge density) = Q/Area = Q / X * Y but respect to second observer length in X must be shorten.
find the shorten length by X' = sqrt[1(v/c)^2] * X
Then the new sigma is Q/(sqrt[1(v/c)^2] * X*Y)
plug new sigma above to E equation, to get C
greenfruit 20081101 10:06:03 
awesome.

mike 20091106 13:07:05 
This is the best way to do it, and is a nice fact to remember

 

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