# Derive p^2/2m from relativistic equations

• Mancuso
In summary: Now, since you are assuming that c is very large, you can also assume that (pc)2/E02 is very small. And if something small is added to 1, the result is approximately 1. So, using the linear approximation you derived, you can say that √(1 + x) ≈ 1 + x/2. Can you see how this relates to the equation K ≈ Knr?In summary, by using a linear approximation and assuming c is very large, we can show that the relativistic equation for energy can be approximated as K ≈ Knr, where K represents the total energy and Knr represents the non-relativistic approximation for kinetic energy. This
Mancuso
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

Last edited:
Mancuso said:
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) ?

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.

Mancuso said:
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.

Yes,

mc2 = √(E2 - (pc)2)

sub into eq 1?

E = √(E2 - (pc)2)⋅√(1 + x)

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?

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 !

Mancuso said:
E = E 0 √((pc)2/E02 +1)

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

## 1. How is p^2/2m derived from relativistic equations?

The expression p^2/2m can be derived from the relativistic energy-momentum relation, E^2 = p^2c^2 + m^2c^4, by solving for p^2 and then dividing by 2m.

## 2. What is the significance of p^2/2m in relativistic equations?

The expression p^2/2m represents the kinetic energy of a particle in relativistic mechanics. It is derived from the energy-momentum relation and is a fundamental quantity in understanding the behavior of particles at high speeds.

## 3. How does the expression p^2/2m relate to the mass-energy equivalence equation?

The expression p^2/2m is derived from the energy-momentum relation, which is closely related to the famous mass-energy equivalence equation, E=mc^2. Both equations are fundamental in understanding the behavior of particles in the context of relativity.

## 4. Can p^2/2m be used for all types of particles in relativistic mechanics?

Yes, p^2/2m is a universal expression that can be used for all types of particles in relativistic mechanics, including photons, electrons, and protons.

## 5. Are there any limitations to using p^2/2m in relativistic equations?

While p^2/2m is a fundamental quantity in relativistic mechanics, it may not accurately describe the behavior of particles at extremely high energies or in extreme gravitational fields. In these cases, more advanced equations and theories such as quantum mechanics and general relativity must be used.

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