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\( \int _ { 0 } ^ { \infty } E d m \quad , \quad E = \sqrt { p ^ { 2 } \cdot c ^ { 2 } + m ^ { 2 } \cdot c ^ { 2 } } \)

Ask by Franklin Hills. in Mozambique
Dec 30,2024

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The integral \( \int_{0}^{\infty} E \, dm \) where \( E = \sqrt{p^2 c^2 + m^2 c^2} \) diverges to infinity.

Solution

Certainly! Let's evaluate the integral you provided: \[ \int_{0}^{\infty} E \, dm \quad \text{where} \quad E = \sqrt{p^2 c^2 + m^2 c^2} \] ### Simplifying the Expression First, factor out \( c \) from the expression for \( E \): \[ E = \sqrt{p^2 c^2 + m^2 c^2} = c \sqrt{p^2 + m^2} \] Therefore, the integral becomes: \[ \int_{0}^{\infty} E \, dm = c \int_{0}^{\infty} \sqrt{p^2 + m^2} \, dm \] ### Evaluating the Integral The integral we need to evaluate is: \[ \int_{0}^{\infty} \sqrt{m^2 + p^2} \, dm \] This is a standard integral in calculus. The antiderivative of \( \sqrt{m^2 + a^2} \) with respect to \( m \) is: \[ \frac{m}{2} \sqrt{m^2 + a^2} + \frac{a^2}{2} \ln\left(m + \sqrt{m^2 + a^2}\right) + C \] Where \( a = p \) in our case. Applying the limits from \( 0 \) to \( \infty \): \[ \left. \left( \frac{m}{2} \sqrt{m^2 + p^2} + \frac{p^2}{2} \ln\left(m + \sqrt{m^2 + p^2}\right) \right) \right|_{0}^{\infty} \] 1. **As \( m \to \infty \):** - \( \frac{m}{2} \sqrt{m^2 + p^2} \approx \frac{m}{2} \cdot m = \frac{m^2}{2} \) → \( \infty \) - \( \frac{p^2}{2} \ln\left(m + \sqrt{m^2 + p^2}\right) \approx \frac{p^2}{2} \ln(2m) \) → \( \infty \) 2. **At \( m = 0 \):** - \( \frac{0}{2} \sqrt{0 + p^2} = 0 \) - \( \frac{p^2}{2} \ln\left(0 + \sqrt{0 + p^2}\right) = \frac{p^2}{2} \ln(p) \) ### Conclusion As \( m \to \infty \), both terms in the antiderivative tend to infinity. Therefore, the definite integral diverges: \[ \int_{0}^{\infty} \sqrt{m^2 + p^2} \, dm = \infty \] Multiplying by \( c \), the original integral also diverges: \[ \int_{0}^{\infty} E \, dm = c \cdot \infty = \infty \] ### Physical Interpretation In a physical context, this divergence suggests that the integral doesn't converge to a finite value under the given conditions. This might imply that integrating energy \( E \) over all possible mass values \( m \) from \( 0 \) to \( \infty \) isn't physically meaningful without additional constraints or considerations. If you have a specific physical scenario or additional constraints in mind, please provide more details, and I can offer a more tailored explanation or solution.

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Beyond the Answer

To understand the integral better, it’s important to recognize that \( E \), often referred to as the energy, comes from the famous relation in relativistic physics, known as the energy-momentum relation. This formula shows how energy depends not just on mass \( m \), but also on momentum \( p \) and the speed of light \( c \). It captures the core concept of how objects behave at high velocities, transcending classical mechanics. When applying this integral in real-world scenarios, think about its implications in the fields of particle physics or astrophysics. For example, this relationship helps us calculate the energy of particles under various physical conditions. Such integrals assist in understanding phenomena like cosmic rays or the behavior of particles in accelerators, leading to insights into the fundamental nature of matter and the universe!

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