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Energy-Momentum Formula Calculator

\[ E^2 = (p c)^2 + (m c^2)^2 \]

1. What is the Energy-Momentum Formula Calculator?

Definition: This calculator computes the total energy (\(E\)) of a particle using the relativistic energy-momentum relation \( E^2 = (p c)^2 + (m c^2)^2 \), where \(p\) is the momentum, \(m\) is the rest mass, and \(c\) is the speed of light. Solving for \(E\), it uses \( E = \sqrt{(p c)^2 + (m c^2)^2} \).

Purpose: It is used in special relativity to determine the total energy of a particle, applicable in particle physics, astrophysics, and relativistic mechanics.

2. How Does the Calculator Work?

The calculator uses the relativistic energy-momentum relation:

Formula: \[ E = \sqrt{(p c)^2 + (m c^2)^2} \] where:

  • \(E\): Energy (J, kJ, eV)
  • \(p\): Momentum (kg·m/s, g·m/s, lb·ft/s)
  • \(m\): Rest mass (kg, g, lb)
  • \(c\): Speed of light (m/s, km/s, mi/s)

Unit Conversions:

  • Momentum:
    • 1 kg·m/s = 1 kg·m/s
    • 1 g·m/s = 0.001 kg·m/s
    • 1 lb·ft/s = 1.382549548 kg·m/s
  • Rest Mass:
    • 1 kg = 1 kg
    • 1 g = 0.001 kg
    • 1 lb = 0.45359237 kg
  • Speed of Light:
    • 1 m/s = 1 m/s
    • 1 km/s = 1000 m/s
    • 1 mi/s = 1609.344 m/s
  • Energy (Output):
    • 1 J = 1 J
    • 1 kJ = 1000 J
    • 1 eV = \( 1.602176634 \times 10^{-19} \) J
The energy is calculated in joules (J) and can be converted to the selected output unit (J, kJ, eV). Results greater than 10,000 or less than 0.001 are displayed in scientific notation; otherwise, they are shown with 4 decimal places.

Steps:

  • Enter the momentum (\(p\)), rest mass (\(m\)), and speed of light (\(c\)) with their units (default: \(p = 0 \, \text{kg·m/s}\), \(m = 9.1093837015 \times 10^{-31} \, \text{kg}\), \(c = 299792458 \, \text{m/s}\)).
  • Convert inputs to SI units (kg·m/s, kg, m/s).
  • Validate that momentum and rest mass are non-negative, and speed of light is greater than 0.
  • Calculate the energy in joules using the formula.
  • Convert the energy to the selected output unit.
  • Display the result, using scientific notation if the value is greater than 10,000 or less than 0.001, otherwise rounded to 4 decimal places.

3. Importance of Energy-Momentum Calculation

Calculating energy using the energy-momentum relation is crucial for:

  • Particle Physics: Determining the total energy of particles moving at relativistic speeds, such as in accelerators or cosmic rays.
  • Astrophysics: Analyzing high-energy particles and objects in extreme conditions, like near black holes or in stellar processes.
  • Education: Teaching the principles of special relativity and the unification of energy and momentum in modern physics.

4. Using the Calculator

Examples:

  • Example 1: Calculate the energy for \(p = 0 \, \text{kg·m/s}\), \(m = 9.1093837015 \times 10^{-31} \, \text{kg}\), \(c = 299792458 \, \text{m/s}\), output in eV:
    • Enter \(p = 0 \, \text{kg·m/s}\), \(m = 9.1093837015 \times 10^{-31} \, \text{kg}\), \(c = 299792458 \, \text{m/s}\).
    • Momentum term: \((p c)^2 = (0 \times 299792458)^2 = 0\).
    • Rest energy term: \((m c^2)^2 = (9.1093837015 \times 10^{-31} \times (299792458)^2)^2 \approx (8.199 \times 10^{-14})^2 \approx 6.723 \times 10^{-27}\).
    • Energy in J: \(E = \sqrt{0 + 6.723 \times 10^{-27}} \approx 8.199 \times 10^{-14} \, \text{J}\).
    • Convert to output unit (eV): \(8.199 \times 10^{-14} \times \frac{1}{1.602176634 \times 10^{-19}} \approx 511644.7 \, \text{eV}\).
    • Result: \( \text{Energy} = 5.1164 \times 10^5 \, \text{eV} \).
  • Example 2: Calculate the energy for \(p = 13.8255 \, \text{lb·ft/s}\), \(m = 2.20462 \, \text{lb}\), \(c = 186282.4 \, \text{mi/s}\), output in J:
    • Enter \(p = 13.8255 \, \text{lb·ft/s}\), \(m = 2.20462 \, \text{lb}\), \(c = 186282.4 \, \text{mi/s}\).
    • Convert: \(p = 13.8255 \times 1.382549548 = 19.104 \, \text{kg·m/s}\), \(m = 2.20462 \times 0.45359237 = 1 \, \text{kg}\), \(c = 186282.4 \times 1609.344 = 2.998 \times 10^8 \, \text{m/s}\).
    • Momentum term: \((p c)^2 = (19.104 \times 2.998 \times 10^8)^2 \approx (5.728 \times 10^9)^2 \approx 3.281 \times 10^{19}\).
    • Rest energy term: \((m c^2)^2 = (1 \times (2.998 \times 10^8)^2)^2 \approx (8.988 \times 10^{16})^2 \approx 8.079 \times 10^{33}\).
    • Energy in J: \(E = \sqrt{3.281 \times 10^{19} + 8.079 \times 10^{33}} \approx \sqrt{8.079 \times 10^{33}} \approx 8.988 \times 10^{16} \, \text{J}\).
    • Output unit: J (no conversion needed).
    • Result: \( \text{Energy} = 8.9880 \times 10^{16} \, \text{J} \).

5. Frequently Asked Questions (FAQ)

Q: What is the energy-momentum relation?
A: The energy-momentum relation \( E^2 = (p c)^2 + (m c^2)^2 \) is a fundamental equation in special relativity, relating a particle’s total energy (\(E\)) to its momentum (\(p\)) and rest mass (\(m\)). It generalizes the energy of a particle, including both its rest energy (\(m c^2\)) and kinetic energy (via momentum).

Q: Why must momentum and rest mass be non-negative?
A: Momentum and rest mass are scalar magnitudes in this context, so they must be non-negative. Negative momentum would imply direction, which this calculator does not account for, and negative rest mass is physically meaningless.

Q: What happens if the momentum is zero?
A: If the momentum (\(p\)) is zero, the particle is at rest, and the formula simplifies to \( E = m c^2 \), which is the rest energy of the particle, famously derived by Einstein.

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