# Taylor expanding a physics formula

1. May 7, 2015

### SU403RUNFAST

< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

So the original problem was that a stationary hydrogen atom changed states from excited to lower state and emitted a photon, i solved for the energy of the photon hf taking into account the kinetic energy of the recoiling atom since the photon released and the atom recoiling have equal and opposite momenta.
I came up with the formula, sorry i tried latexing everything but it wouldnt work for me:
DeltaE=-pc+(p^2)/2m, both momentum p's are the atoms. The problem says I need to taylor expand this formula for small deltaE and keep the first two terms. I dont know how to taylor expand something random like this, a physics formula.

Last edited by a moderator: May 7, 2015
2. May 7, 2015

### Staff: Mentor

I think you have to find DeltaE in terms of E and m first, without the (unknown) momentum in it. Then assume E/(mc^2) is small.

3. May 7, 2015

### SU403RUNFAST

But there is no E. This is part b of a problem, it says use the formula found in part a and taylor expand it.[my first formula that i solved for DeltaE=-pc+(p^2)/2m] I can replace p with mv, but there will be no E only deltaE. I dont know how to taylor expand something really. DeltaE=-mvc+.5mv^2 is after replacing p though. This question totally threw me off.

4. May 7, 2015

### SU403RUNFAST

Oh and it says that the first two terms after expanding will provide the correction for the formula deltaE=hf=hc/lambda.

5. May 7, 2015

### fzero

Conservation of momentum implies that $p$ is proportional to $f$. Once you account for $\hbar$s and $\pi$s, energy conservation gives a quadratic equation for $f$ in terms of $\Delta E$. I think the problem wants you to determine $f$ by finding the roots of the equation. The Taylor expansion is for the square root in the quadratic formula.

6. May 7, 2015

### SU403RUNFAST

It looks like you saying my equation is wrong, so I have been solving for a new one, but am going in circles. I know that hf=deltaE-(p^2)/2m, this is my energy equation, and using momentum to plug in for deltaE or p in terms of other variables is not giving me a quadratic in any of my attempts

7. May 7, 2015

### fzero

The equation in your OP has an incorrect minus sign in the $pc$ term, but the version above looks correct. I believe that $p=hf/c$, so you need to use that to write the equation in terms of $f$, $\Delta E$ and $m$. You can then solve this for $f$ and use the Taylor expansion to compare to the approximation $f = \Delta E/h$.

8. May 7, 2015

### SU403RUNFAST

Ive been getting this same equation but two terms cancel out.... hf=deltaE-[(hf/c)^2]/(2m), when you simplify hf's cancel out if you plug in deltaE=hf. If i leave the deltaE alone i get hf=deltaE-{([h^2][f^2])2m} all over c^2. It doesnt ever make a quadratic

9. May 7, 2015

### fzero

You don't plug in $\Delta E =hf$, you are trying to solve for $f$ since your analysis corrects that formula. The equation

$$\frac{h^2}{2m} f^2 + h f - \Delta E=0$$

is the quadratic equation to be solved.

10. May 7, 2015

### SU403RUNFAST

I know that but where did the c^2 go?

11. May 7, 2015

### fzero

I was being sloppy and copied the 2nd formula you wrote. The denominator should be $2mc^2$.

12. May 7, 2015

### SU403RUNFAST

okay thanks im working on it

13. May 7, 2015

### SU403RUNFAST

Well i solved the equation for two values, i set hf=x to simplify and make it easier, then using quadratic formula x=-mc^2 plus or minus the root of ([m^2][c^4]+[2mc^2][deltaE]), i have no idea on how to taylor expand the equation though....

14. May 7, 2015

### vela

Staff Emeritus
So you have
$$hf = -mc^2 \pm \sqrt{(mc^2)^2 + 2 mc^2 \Delta E} = mc^2 \left(-1 \pm \sqrt{1+\frac{2 \Delta E}{mc^2}}\right).$$ As fzero said earlier, "The Taylor expansion is for the square root in the quadratic formula."

15. May 8, 2015

### SU403RUNFAST

So I got 1 + DeltaE/mc^2 - (DeltaE^2)/(m^2c^2), so to keep only the first two terms i ignore the last one i believe. If this is correct how is this a correction to DeltaE=hf=hc/lambda?

16. May 8, 2015

### vela

Staff Emeritus
Your expansion isn't quite correct – probably just algebra mistakes. You want to keep all three terms. If you keep only the first two terms, you recover $hf = \Delta E$. The third term of the expansion is what leads to the first correction to the energy of the photon.

17. May 8, 2015

### SU403RUNFAST

okay i see my mistake and fixed it, 1 + DeltaE/mc^2 - (DeltaE^2)/(2m^2c^4) how did you see the correction? As in how do you see that hf=deltaE from the taylor expansion

18. May 8, 2015

### vela

Staff Emeritus
You have to plug the expansion into your expression for hf.

19. May 8, 2015

### SU403RUNFAST

thank you i have solved for the correction