Some help in understanding energy conservation

[user1139]

Homework Statement

Consider the decay of particle $A$ into particles $B$ and $T$ where $T$ is the tachyon particle. Show that a value $p$ exists to satisfy the conservation of energy equation. Note that $c=1$

Relevant Equations

Relevant equation: $m_A=\sqrt{p^2+m^2_B}-\sqrt{p^2-m^2_T}$

While I am working through proving the homework statement, I encountered a problem. The problem is as follows:

From the energy equation above, one can see that the minimum value of $p$ is $m_T$. However, how does one explain why when $p=m_T$, $\sqrt{m^2_B+m^2_T}>m_A$?

 

[PeroK]

Homework Statement:: Consider the decay of particle $A$ into particles $B$ and $T$ where $T$ is the tachyon particle. Show that a value $p$ exists to satisfy the conservation of energy equation. Note that $c=1$
Relevant Equations:: Relevant equation: $m_A=\sqrt{p^2+m^2_B}-\sqrt{p^2-m^2_T}$

While I am working through proving the homework statement, I encountered a problem. The problem is as follows:

From the energy equation above, one can see that the minimum value of $p$ is $m_T$. However, how does one explain why when $p=m_T$, $\sqrt{m^2_B+m^2_T}>m_A$?

Do you mean why is there a solution for $p$ if and only if $\sqrt{m_b^2 + m_T^2} \ge m_A$?

 

[user1139]

Do you mean why is there a solution for $p$ if and only if $\sqrt{m_b^2 + m_T^2} \ge m_A$?

I meant why if we set $p=m_T$, then the RHS = $\sqrt{m^2_B+m^2_T}$ is greater than LHS = $m_A$.

 

[PeroK]

I meant why if we set $p=m_T$, then the RHS = $\sqrt{m^2_B+m^2_T}$ is greater than LHS = $m_A$.

That's fairly simple mathematics, surely? $p= m_T$ is, therefore, not a solution.

 

[user1139]

That's fairly simple mathematics, surely? $p= m_T$ is, therefore, not a solution.

How do you show it though?

 

[PeroK]

How do you show it though?

You seem to be confused by the mathematical logic here. There are three possibilities:

1) $m_A > \sqrt{m_B^2 + m_T^2}$

In which case, there is no solution for $p$.

2) $m_A = \sqrt{m_B^2 + m_T^2}$

In which case $p = m_T$ is a solution.

3) $m_A < \sqrt{m_B^2 + m_T^2}$

In which case there is a solution $p > m_T$.

A simple way to prove all this is to consider the function $$f(p ) = \sqrt{p^2 + m_B^2} - \sqrt{p^2 - m_T^2}$$ and show that it is decreasing for $p \ge m_T$. Hint: $f(p) \rightarrow 0$ as $p \rightarrow \infty$.

 

[user1139]

Oh so it is because we want a solution for $p$ that's why we insist that when $p=m_T$, $m_A<\sqrt{m^2_B+m^2_T}$?

 

[PeroK]

Oh so it is because we want a solution for $p$ that's why we insist that when $p=m_T$, $m_A<\sqrt{m^2_B+m^2_T}$?

Um, no! If $p=m_T$, then $m_A = \sqrt{m^2_B+m^2_T}$.

 

[user1139]

Oh but I am trying to show that there is a value for $p$ that satisfies $m_A=\sqrt{p^2+m^2_B}-\sqrt{p^2-m^2_T}\,$.

 

[PeroK]

Oh but I am trying to show that there is a value for $p$ that satisfies $m_A=\sqrt{p^2+m^2_B}-\sqrt{p^2-m^2_T}\,$.

I understand that. But, as I pointed out above, you require the assumption that $m_A \le \sqrt{m_B^2 + m_T^2}$.

 

[user1139]

So I will first have to declare the assumption that $\sqrt{m^2_B+m^2_T}\geq m_A$ before showing that under the assumption, there is a $p$ for which $m_A=\sqrt{p^2+m^2_B}-\sqrt{p^2-m^2_T}$?

 

[PeroK]

So I will first have to declare the assumption that $\sqrt{m^2_B+m^2_T}\geq m_A$ before showing that under the assumption, there is a $p$ for which $m_A=\sqrt{p^2+m^2_B}-\sqrt{p^2-m^2_T}$?

That assumption and the proof go hand in hand. You need to start thinking about how you prove it. I gave you a hint above:

A simple way to prove all this is to consider the function $$f(p ) = \sqrt{p^2 + m_B^2} - \sqrt{p^2 - m_T^2}$$ and show that it is decreasing for $p \ge m_T$. Hint: $f(p) \rightarrow 0$ as $p \rightarrow \infty$.

 

[user1139]

Yes, I was able to prove from your hint. However, I am wondering if I need to declare that assumption before showing the proof.

 

[PeroK]

Yes, I was able to prove from your hint. However, I am wondering if I need to declare that assumption before showing the proof.

The assumption is revealed by the proof, in the sense that $f(p)$ cannot be greater than $\sqrt{m^2_B+m^2_T}$.

 

[user1139]

But, if I don't first declare what values $m_A$ can take, how do I show then that there is a $p$ such that LHS=RHS for $m_A=\sqrt{p^2+m^2_B}-\sqrt{p^2-m^2_T}$? This is because the proof only shows what the RHS can be.

 

[PeroK]

But, if I don't first declare what values $m_A$ can take, how do I show then that there is a $p$ such that LHS=RHS for $m_A=\sqrt{p^2+m^2_B}-\sqrt{p^2-m^2_T}$? This is because the proof only shows what the RHS can be.

You can do it either way. You can notice up front that a certain condition on $m_A$ must be satisfied. Or, you can identify the condition during the working of the proof.

To take an example. If you are looking for real solutions to $ax^2 + bx + c = 0$, then you can either assume up front that $b^2 - 4ac \ge 0$; or, you can identify this condition during your working.