Derive p^2/2m from relativistic equations

  • #1
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1
Given the relativistic equation for energy E2 = (pc)2 + (mc2)2
I want to find the non-relativistic approximation for kinetic energy in non-relativistic terms,
Knr = p2/2m

I start off with subtracting the rest energy
E0=mc2
from the above equation.

So K = E - E0

and assume that c is very large.
I've messed around for hours on the algebra and I need help.
I want to show that K ≈ Knr

I am doing this using a linear approximation. I've written the energy as E = E0 √1+x

And using the function f(x)=√1+x about x = 0

I've derived the linearization as L(x) = 1 + x/2

But I am struggling with relating to the equations above to show that K ≈ Knr
 
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Answers and Replies

  • #2
TSny
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I am doing this using a linear approximation. I've written the energy as E = E0 √1+x

And using the function f(x)=√1+x about x = 0

I've derived the linearization as L(x) = 1 + x/2
You have the right approach. Can you tell us how x is related to p when you write E = E0 √(1+x) ?
 
  • #3
13
1
That's where I get stuck, relating the two equations

1. E = E0√(1+x) and
2. E = √( (pc)2 + (mc2)2)

I've arranged the relativistic equations as such
3. K = E - E0
3. K = √( (pc)2 + (mc2)2) - mc2

I don't know what x represents in equation 1 and so I don't know where I go next.

I tried a lot of algebraic manipulation of equation 2. including substituting mv for p
In an effort to find a structural relationship, but I'm lost. I would appreciate help relating x
to p.
This question comes from a numerical methods section of a calculus course. I completely understand the numerical methods, but I'm struggling with the physics and how to apply a linear approximation to relate the two equations. I don't understand which part of equation 3. I am substituting with Eq. 1 and if any variables should be eliminated or rearranged.
 
  • #4
TSny
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1. E = E0√(1+x) and
2. E = √( (pc)2 + (mc2)2)
Can you rewrite equation 2 using E0 instead of mc2? That might help you see how to identify x in equation 1.
 
  • #5
13
1
Yes,

mc2 = √(E2 - (pc)2)

sub into eq 1?

E = √(E2 - (pc)2)⋅√(1 + x)

I still dont see it. Please help
 
  • #6
TSny
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Your equation 2 can be written as ##E = \sqrt{(pc)^2 + E_0^2}##. Is there any way to "pull" the ##E_0## outside the square root so that it looks more like equation 1?
 
  • #7
13
1
Yes!

E = E 0 √((pc)2/E02 +1)

So x represents (pc)2/E02

This problem has taken up my entire night and it's only worth 0.2% of my final mark haha. The problem was so interesting that I could not put it down. I will tackle the rest of the problem in the morning, might bug you again for a hint if I get stuck. Thanks so much for your help, you're awesome ! :approve:
 
  • #8
TSny
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E = E 0 √((pc)2/E02 +1)

So x represents (pc)2/E02
Yes, good.
 

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