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rwooduk said:oh i can just square each term, time for a break!
mods please feel free to delete, sorry
Mentallic said:Be careful!
[tex]\left(\sqrt{a}+\sqrt{b}\right)^2\neq a+b[/tex]
rwooduk said:hmm, yes just realized this, no restrictions, just a basic algebra problem.
but if i take PC to the other side and square both sides i'll have:
(E-PC)^2 on the left hand side? which still leaves me questioning how to get P out?
if i multiply out (E-PC)^2 and simply i get
E^2 - 2EPC = 2P^2 C^2 + M^2 C^2
I'm not sure it should be this difficult
SammyS said:That's a quadratic equation in P.
It might work out a bit more neatly if you divide through by C2 first.
What does ##\displaystyle E_e = \left(p_e^2 c^2 + m_e^2 c^4 \right)^{1/2} ## have to do with the OP ?rwooduk said:yes thought that may be an option, still would complicate the question further. I am going to have to re-evaluate how to do the problem.
thanks for everyones input!
for those interested or if you are bored and have time here is the full solution that i am working through:
http://inside.mines.edu/~lwiencke/PH300/F13/homework/hw4_solutions.pdf
the bit I am struggling with is on page 4, where it says "2pn=pe=..." in the equations above it, it solves the 4th equation down for pe.
thanks again!
SammyS said:What does ##\displaystyle E_e = \left(p_e^2 c^2 + m_e^2 c^4 \right)^{1/2} ## have to do with the OP ?
vela said:The solution you're trying to work through is doing things the hard way. I imagine that whoever wrote it went around in circles with the algebra for a while and simply wrote down the end result. One of the tricks to doing relativity is figuring out how to avoid this unnecessary algebra in the first place. Try this approach instead:
Let ##p_e## be the magnitude of the electron's momentum. Conservation of energy gives you
$$E_\mu = E_e + E_\nu + E_\bar{\nu}.$$ Rewrite that as
$$E_\mu - E_e = E_\nu + E_\bar{\nu}.$$ Using conservation of momentum, you should be able to show the righthand side is equal to ##p_e c##. Now square both sides and use the fact that ##(m_ec^2)^2 = E_e^2 - (p_e c)^2##. You'll end up with an equation you can easily solve for ##E_e## in terms of the masses.
If you want to an even simpler method, learn how to use four-vectors.
vela said:The solution you're trying to work through is doing things the hard way. I imagine that whoever wrote it went around in circles with the algebra for a while and simply wrote down the end result. One of the tricks to doing relativity is figuring out how to avoid this unnecessary algebra in the first place. Try this approach instead:
Let ##p_e## be the magnitude of the electron's momentum. Conservation of energy gives you
$$E_\mu = E_e + E_\nu + E_\bar{\nu}.$$ Rewrite that as
$$E_\mu - E_e = E_\nu + E_\bar{\nu}.$$ Using conservation of momentum, you should be able to show the righthand side is equal to ##p_e c##. Now square both sides and use the fact that ##(m_ec^2)^2 = E_e^2 - (p_e c)^2##. You'll end up with an equation you can easily solve for ##E_e## in terms of the masses.
If you want to an even simpler method, learn how to use four-vectors.
A root in algebra is a solution to an equation, where the value of the variable makes the equation true. This means that when the value of the variable is substituted into the equation, the equation will be balanced.
To separate a term in the sum of two roots, you can use the distributive property. This means that you can multiply the term by each of the roots and then add the two resulting terms together. For example, if you have the sum of two roots, √a + √b, you can separate out the term a by multiplying it by each root: a√a + a√b. Then, you can combine the two terms to get the original sum of roots.
Yes, you can still separate a term in the sum of two roots if the term is a negative number. The same steps apply as in question 2, but you will need to be careful with the signs when combining the terms. For example, if you have the sum of two roots, -√a + √b, you can separate out the term -a by multiplying it by each root: -a√a + -a√b. Then, you can combine the two terms to get the original sum of roots: -a(√a + √b).
Yes, you can separate terms in the sum of two roots if they have different radicands. The same steps apply as in question 2, but you will need to be careful when combining the terms. For example, if you have the sum of two roots, √a + √b, you can separate out the term √a by multiplying it by the first root: √a√a + √a√b. Then, you can combine the two terms to get the original sum of roots: √a^2 + √ab. However, if the two radicands are not like terms, the terms cannot be separated.
Being able to separate terms in the sum of two roots is important because it allows us to simplify and solve equations more easily. It also helps us to better understand the relationship between different terms in an equation. Additionally, separating terms in the sum of two roots is a fundamental algebraic skill that is necessary for solving more complex equations and problems.