Deriving the Energy-Momentum Formula

In summary, the energy-momentum relationship, E^2 = p^2 * c^2 + (m*c^2)^2, can be derived from the expressions E = (gamma)*m*c and p = (gamma)*m*u, where (gamma) = 1 / sqrt(1 - (u^2)/(c^2)) is the Lorentz transformation factor. By squaring both equations and subtracting, the (gamma) terms can be eliminated, resulting in the desired equation. This derivation does not require the concept of relativistic mass.
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Homework Statement


Show that the energy-momentum relationship, E^2 = p^2 * c^2 + (m*c^2)^2, follows from the expressions E = (gamma)*m*c and p = (gamma)*m*u

where

(gamma) = 1 / sqrt(1 - (u^2)/(c^2)) the lorentz transformation factor.
m is the rest mass.
c is the speed of light
u is the velocity of the particle
E is the total energy
p is the momentum

The book does not teach about relativistic mass, so I think I supposed to derive this without making a distinction between m and m0.

Homework Equations


(1): E^2 = p^2 * c^2 + (m*c^2)^2
(2): E = (gamma)*m*c
(3): p = (gamma)*m*u

The Attempt at a Solution


When the chapter introduces the formula E^2 = p^2 * c^2 + (m*c)^2, it does not show how it derived this equation. Instead it says that it just says "By squaring [equations (2) and (3)] and subtracting, we can eliminate u. The result after some algebra is [equation (1)]."

My first attempt was to start by squaring both sides of equation (3).

p^2 = (gamma)^2*m^2*u^2

Then get it in terms of m^2.

m^2 = (p^2)/(u^2) - (p^2)/(c^2)

In equation (2), square both sides, then substitute m^2 to get:

E^2 = (gamma)^2 * [(p^2)/(u^2) - (p^2)/(c^2)] * c^2

After some algebra I got:

E^2 = [(gamma)^4 * m^2 * c^4] - [(gamma)^2 * p^2 * c^2]

This is nearly what I'm trying to derive, however, the (gamma) terms are still there and I don't know how to get rid of them.
 
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Related to Deriving the Energy-Momentum Formula

1. What is the energy-momentum formula?

The energy-momentum formula is a fundamental equation in physics that relates an object's energy and momentum. It states that the total energy of an object is equal to its mass multiplied by the speed of light squared (E=mc²), where c is the speed of light in a vacuum, and its momentum is equal to its mass multiplied by its velocity (p=mv).

2. How is the energy-momentum formula derived?

The energy-momentum formula is derived from Albert Einstein's theory of special relativity, which states that the laws of physics are the same for all observers in uniform motion. By using mathematical equations and principles, including the Lorentz transformation and the mass-energy equivalence equation, the energy-momentum formula can be derived.

3. Why is the energy-momentum formula important?

The energy-momentum formula is important because it helps us understand the relationship between energy and momentum, two fundamental quantities in physics. It also plays a crucial role in many areas of physics, including particle physics, quantum mechanics, and cosmology.

4. Can the energy-momentum formula be applied to all objects?

Yes, the energy-momentum formula can be applied to all objects, regardless of their size or speed. However, it is most commonly used for objects that are moving at speeds close to the speed of light, such as particles in particle accelerators.

5. Is the energy-momentum formula a universal law?

No, the energy-momentum formula is not considered a universal law like Newton's laws of motion. It is a consequence of special relativity and only applies in certain situations, such as when objects are moving at high speeds or in the absence of external forces.

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