Solving for Momentum in Energy Equation

In summary, the conversation was about solving an equation involving a delta function and using the theory of distributions. The original equation was incorrect and the correct solution was provided by one of the participants. There was also a discussion about the relationship between the theorem of residues and the theory of distributions.
  • #1
RedX
970
3
[tex]\delta(\frac{p_f^2}{2m}-E_i^o-\hbar\omega)=\frac{m}{p_f} \delta(p_f-[2m(E_i^o+\hbar\omega)]^{\frac{1}{2}})[/tex]

Shouldn't the right hand side be multiplued by 2?
 
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  • #2
marlon said:
i am sorry but according to me the fraction before the delta should disappear.

marlon



Mr.QM+QFT guru,you f***ed up big time... :tongue:


Apply the THEORY:
[tex] f(x):R\rightarrow R [/tex](1)
only with simple zero-s (the algebraic multiplicies of the roots need to be 1).
Let's denote the solutions of the equation
[tex] f(x)=0 [/tex](2)
by [itex] (x_{\Delta})_{\Delta={1,...,N}} [/itex] (3)
and let's assume that:
[tex] \frac{df(x)}{dx}|_{x=x_{\Delta}} \neq 0 [/tex] (4)

Then in the theory of distributions there can be shown that:

[tex] \delta f(x)=\sum_{\Delta =1}^{N} \frac{\delta (x-x_{\Delta})}{|\frac{df(x)}{dx}|_{x=x_{\Delta}}|} [/tex] (5)

Read,Marlon...Read... :rolleyes:

Daniel.
 
Last edited:
  • #3
is there anyone else that can solve this problem in a clear and more mature manner. It's been a while since i worked with distributions in this way and it seems quite interesting to me. Can someone tell me what i did wrong ?

thanks in advance

regards
marlon
 
  • #4
Okay:
[tex] f(x)\rightarrow f(p_{f})=\frac{p_{f}^{2}}{2m}-E_{0}^{i}-\hbar\omega [/tex] (6)

Solving the equation
[tex] f(p_{f})=0 [/tex] (7)

,yields the 2 solutions (which fortunately have the degree of multiplicity exactly 1)

[tex] p_{f}^{1,2}=\pm \sqrt{2m(E_{0}^{i}+\hbar\omega)} [/tex] (8)

Computing the derivative of the function on the solutions (8) of the equation (7),we get,after considering the modulus/absolute value:
[tex] \frac{df(p_{f})}{dp_{f}}=\frac{\sqrt{2m(E_{0}^{i}+\hbar\omega)}}{m} [/tex] (9)

Combining (8),(9) and the general formula (5) (v.prior post),we get:

[tex] \delta (\frac{p_{f}^{2}}{2m}-E_{0}^{i}-m_{i})=\frac{m}{\sqrt{2m(E_{0}^{i}+\hbar\omega)}} \{\delta[p_{f}-\sqrt{2m(E_{0}^{i}+\hbar\omega)}]+\delta[p_{f}+\sqrt{2m(E_{0}^{i}+\hbar\omega)}]\} [/tex] (10)

which is totally different than what the OP had posted...

IIRC,when learning QFT,i always said to myself:theorem of residues and the theory of distributions go hand in hand...

Daniel.
 
  • #5
Indeed, i just looked up the rule at hand. I get the same solution and i see where i went wrong in my first post. thanks for the polite correction.

marlon

i deleted my erroneous post
 
  • #6
marlon said:
thanks for the polite correction.

marlon

It's always nice to encounter the two of you in a post. All this harmony and warmth...
 
  • #7
da_willem said:
It's always nice to encounter the two of you in a post. All this harmony and warmth...

yes, we really are the best of friends... :wink:

marlon
 

1. What is the energy equation used for?

The energy equation is used to calculate the total energy of a system, taking into account both kinetic and potential energy.

2. How is momentum related to energy in the energy equation?

Momentum is a component of kinetic energy in the energy equation. It represents the mass and velocity of an object and is used to calculate the kinetic energy of the system.

3. What is the formula for solving for momentum in the energy equation?

The formula for solving for momentum in the energy equation is p = mv, where p represents momentum, m represents mass, and v represents velocity.

4. How is momentum conserved in the energy equation?

In a closed system, momentum is always conserved, meaning the total momentum of the system before an event is equal to the total momentum after the event. This is reflected in the energy equation as the change in kinetic energy is equal to the change in potential energy.

5. What are some real-life applications of the energy equation and solving for momentum?

The energy equation and solving for momentum are used in various fields such as physics, engineering, and mechanics. They are essential for understanding the motion and energy of objects in systems, such as in the design of roller coasters, cars, and rockets.

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