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Julmust
Sweden4867 Posts
From this equation ![[image loading]](http://i50.tinypic.com/vesr9u.gif) I need to get what gamma equals. My own attempts so far has failed since gamma can't be 1 or lower for the 2nd equation to work. This is what I've done so far, and it's not correct just thought I'd share what I've tried: ![[image loading]](http://i48.tinypic.com/22hm2o.gif) I know that's not right since that would mean that ![[image loading]](http://i47.tinypic.com/vwwvux.gif) So could someone PLEASE help me get gamma out of that equation. I know its really fucking simple but I feel like a dumbass. (if you wanna test it out m = 9,1094*10^-31, c = 3*10^8 and E = 3,98*10^-17)   | ||
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crate
United States2474 Posts
On May 21 2010 07:38 Julmust wrote: This is what I've done so far, and it's not correct just thought I'd share what I've tried: I know that's not right since that would mean that Your second equation here does not follow from your first. You messed up the algebra. You have g = (E + mc^2)/(mc^2) So g = E/(mc^2) + 1 | ||
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Assault_1
Canada1950 Posts
how do u get (E+mc^2)/(mc^2) = E o_o (E+mc^2)/(mc^2) = (E/(mc^2)) + 1 | ||
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Impervious
Canada4214 Posts
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tYsopz
Norway215 Posts
E = ymc^2-mc^2 => E = mc^2(y-1) => y = E/mc^2+1 y being gamma, idk how to make the symbol in windows. dont do what the above poster did, you can use that formula as well to get the answer, but that's not how the "official" formula is, the relativistic formula for kinetic energy is: Ek = (y-1)m^2) If you don't get to the official formula, chances are your answer will be deemed incorrect. | ||
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rushz0rz
Canada5300 Posts
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crate
United States2474 Posts
Relativistic energy is E^2 = (pc)^2 + (mc^2)^2; this does not simplify to E = (gamma)mc^2 - mc^2 unless I'm missing something. | ||
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Julmust
Sweden4867 Posts
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Jonoman92
United States9106 Posts
E=mc^2(y-1) Divide both sides: E/(mc^2)=y-1 Add one to both sides: (E/(mc^2))+1=y | ||
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tYsopz
Norway215 Posts
Etot = ymc^2 Formula for relativistic kinetic energy is: Ek = (y-1)mc^2 Formula for mass energy / ("hvileenergi" in norwegian) E0 = mc^2 E0 + Ek = Etot if that wasn't clear. | ||
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tYsopz
Norway215 Posts
E^2 = (pc)^2 + (mc^2)^2 => E^2 = (γmc^2)^2 + (mc^2)^2 => E = (γmc^2) + (mc^2) => E=(γ-1)mc^2 | ||
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crate
United States2474 Posts
On May 21 2010 08:07 tYsopz wrote: E^2 = (γmc^2)^2 + (mc^2)^2 => E = (γmc^2) + (mc^2) This is wrong. a^2 = b^2 + c^2 a != b + c edit: your last equation also doesn't follow, it should be (γ+1)(mc^2) instead of (γ-1)(mc^2) as you've written. If the original equation is referring to relativistic kinetic energy as E it's correct I think, but I dislike calling that E instead of E_k or T since it's misleading. | ||
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tYsopz
Norway215 Posts
Edit: you know how it is when you know what it is supposed to look like :p You sure about that formula, never seen it? Edit: also, my formula for relativistic momentum is wrong. | ||
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crate
United States2474 Posts
On May 21 2010 08:09 tYsopz wrote: that is awkward, yeah i was a bit too quick >< Edit: you know how it is when you know what it is supposed to look like :p You sure about that formula, never seen it? Relativistic total energy is E = (pc)^2 + (mc^2)^2, yes. p is relativistic momentum: p = γmv http://en.wikipedia.org/wiki/Mass_in_special_relativity#The_relativistic_energy-momentum_equation | ||
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tYsopz
Norway215 Posts
unless the fact it is 4 hours past my bedtime is putting a stopper to my ability to think. | ||
Chill
Calgary25991 Posts
Edit: To be clearer, the request doesn't even make sense. How are we going to just "get rid" of a variable? | ||
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HeavOnEarth
United States7087 Posts
@ Chill , he meant "out of the equation", instead of "get out" So like, "help me get gamma, out of that equation." Also -insert obligatory pout about homework threads in a very biased manner- | ||
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Julmust
Sweden4867 Posts
On May 21 2010 08:36 Chill wrote: i think your problem is basic math, not physics. my problem was that my brain wasn't functioning correctly, I solved the problem (which, as it shows, had nothing to do with this equation  ) and now it's back online. | ||
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tYsopz
Norway215 Posts
=> E^2 = (γmvc)^2 + m^2c^4 => E^2 = γ^2m^2v^2c^2 + m^2c^4 => E^2 = (1/(1-(v^2/c^2)))m^2v^2c^2 + m^2c^4 => E^2 = (1/((c^2-v^2)/c^2)))*m^2v^2c^2 + m^2c^4 => E^2 = (c^2/(c^2-v^2)*m^2v^2c^2 + m^2c^4 => E^2 = m^2c^4(v^2/(c^2-v^2)+1) => E^2 = m^2c^4(v^2+c^2-v^2)/(c^2-v^2) => E^2 = m^2c^4(c^2/(c^2-v^2)) => E^2 = m^2c^4(1/(c^2-v^2)/c^2) => E^2 = m^2c^4(1/(1-v^2/c^2)) => E^2 = m^2c^4γ^2 => E = γmc^2 Uggh, should have used my scanner. | ||
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Darby.mcg
United States16 Posts
@tYsopz: I'm going to assume that y^2 = 1/(1-v^2/c^2) | ||
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FiBsTeR
United States415 Posts
On May 21 2010 09:29 Darby.mcg wrote: I'm going to college next year, will these sorts of things become useful in the near or far future if I'm going into physics? It depends on the structure of the course, but I think if you can do the math, you can do the physics. But the converse might not be true... yeah I'm a math major.  | ||
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tYsopz
Norway215 Posts
(I'm of course referring to the transformation of E^2=(pc)^2 + (mc^2)^2 => E = γmc^2) and yeah y^2 = 1/(1-v^2/c^2), seeing as y = 1/(sqrt(1-v^2/c^2)) (lorentz factor) | ||
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Darby.mcg
United States16 Posts
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