# Algebra and physics

1. Aug 10, 2010

1. The problem statement, all variables and given/known data
When does the rules of algebra not apply in physics?
I mean in photo electric effect E = K + $$\Phi$$ right?
and phi is a constant
so if we have twice the energy 2E = 2K + 2 $$\Phi$$
and then E2=K2 + $$\Phi$$2
and then $$\Phi$$2 = 2E1 - 2K1
$$\Phi$$2 /2 =E1-K1 = $$\Phi$$2/2 = $$\Phi$$1 so phi changed...
but phi is a constant right dependent only on the type of material so it wont change? (I have very little knowledge in photo electric effects this is mainly how and when algebra does not apply in physics)

2. Relevant equations

3. The attempt at a solution

Last edited: Aug 10, 2010
2. Aug 10, 2010

Sorry I noticed that I didn't write everything when I made the original post so my argument made no sense I edited it .

3. Aug 10, 2010

### hikaru1221

It's like if you feed one puppy and one kitty on one, well, cake (!?), your puppy and kitty are full; then if you want to throw away 2 cakes in a meaningful way (), then you must get another puppy and another kitty, so 2 puppies and 2 kitties totally for 2 cakes
If we understand one cake as E, one kitty as K and one puppy as P (let P denote phi, I'm lazy), then actually P2 = 2 puppies. That is, puppy is still puppy, it doesn't grow bigger or smaller, just the number of puppies increases. Likewise, P2 doesn't stand for the work function; P2 is the amount of energy given up by 2 electrons, while P1 (or P) is the amount of energy given up by 1 electron. It's just that we define the work function = the amount of energy given up by 1 electron, or simply P.

EDIT: In conclusion, algebra in particular and math in general concur with physics (or is it the way around, that physics concurs with them? ). I mean, in this case, math does apply.

4. Aug 10, 2010

Umm thanks for the reply. But I shamefully say that I didn't get it completely I mean let say we want the kinetic energy K2 we know where we increased the energy of the photons
we know the values of E , K , phi in the original case and we doubled E
by using another type of light
would K be doubled and phi? I mean we didn't change anything in the material it self but phi can't be doubled because it is constant so the algebriac equation wont hold right?
because unless K2=2K1 and phi2 = 2phi 1 , E2 wont be equal to 2E1

5. Aug 10, 2010

### hikaru1221

Ah, I see. So that really is a problem with math. I mean, your math
So first you have an electron with kinetic energy K, so you need a photon of energy E. The relation is: E = K + P. Then if you want another electron with 2K, then what you need is NOT a photon of energy 2E! That is, if you shoot a photon of energy 2E, you won't get an electron of 2K.
(1): In order to get an electron of 2K, what you need is a photon of (2K + P), which is NOT equal to 2E.
(2): If you shoot a photon of 2E, what you get is an electron of (2E - P), which is NOT equal to 2K.
And what the mathematical equation 2E = 2K + 2P means, I have already explained. That is the case when you shoot 2 photons of E and consequently, get back 2 electrons of K. This case is not the same as the two cases (1) and (2) above.

6. Aug 10, 2010