Algebra, how to separate a term in the sum of two roots

[rwooduk]

My mind has gone blank and I've suddenly forgotten basic algebra, please could someone give me direction on how to make P the subject of this equation?

E = (P^2 C^2 + M^2 C^4)^1/2 + (P^2 C^2)^1/2

thanks for any help

 

[rwooduk]

oh i can just square each term, time for a break!

mods please feel free to delete, sorry

 

[Mentallic]

That image is terribly low quality, but what I could make of it is that it's

E=\sqrt{P^2C^2+M^2C^2}+\sqrt{P^2C^2}

This equation seems similar to the energy-mass equivalence, but anyway, are there any restrictions on these variables? If P,C\geq 0 for example, then \sqrt{P^2C^2}=PC and then you could move that to the other side and then square both sides to get rid of the square root.

 

[Mentallic]

oh i can just square each term, time for a break!

mods please feel free to delete, sorry

Be careful!

\left(\sqrt{a}+\sqrt{b}\right)^2\neq a+b

 

[rwooduk]

Be careful!

\left(\sqrt{a}+\sqrt{b}\right)^2\neq a+b

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

im not sure it should be this difficult

 

[SammyS]

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

That's a quadratic equation in P.

It might work out a bit more neatly if you divide through by C² first.

 

[rwooduk]

That's a quadratic equation in P.

It might work out a bit more neatly if you divide through by C² first.

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]

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!

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]

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 ?

its the equation below it that he has rearranged, its very much complicated by the sudden change in notation (why oh why?) , so that how come i was trying to figure out the rearrangement of the eqn in the op.

ive circled it below, and indicated what it goes to.

 

[vela]

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.

 

[rwooduk]

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.

Excellent, will try this thank you!

And I agree, I think the person writing the solution had similar difficulties and gave it to someone else who simply wrote down the end result, hence the change in notation lol

But you are right, it is only a 4 mark question and shouldn't be this complicated. thanks for everyones help!

 

[rwooduk]

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.

hm spoke too soon, i get the first bit but then if i square both sides i.e. (E(u) - E(e))^2 i end up with E(u)^2 + E(e)^2 - 2E(u)E(e), see photo below.

 

[vela]

That's right. Keep going. Eliminating $(p_e c)^2$ from both sides, you have
$$(m_\mu c^2)^2 + (m_e c^2)^2 - 2(m_\mu c^2)E_e = 0,$$ which you can solve for $E_e$ to get
$$E_e = \frac{(m_\mu c^2)^2 + (m_e c^2)^2}{2 m_\mu c^2}.$$ Now solve for $(p_e c)^2$ using $(m_e c^2)^2 = E_e^2 - (p_e c)^2$ and plugging in the expression for $E_e$.

 

[rwooduk]

ahh i substituted for E(e) for some reason which complicated the problem, not sure why i did that since E(e) is what I'm trying to find :-/

but anyhow THANKS for all your help!