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.