Which of the following equation correctly represents the momentum p of

1.	Which of the following equation correctly represents the momentum p of a photon of Energy E?1. E/c2. E2c3. Ec4. Ec2
Answer» Correct Answer - Option 1 : E/c Relation between Relativistic E and\(\vec{P}\) 1) E = K + moc2 2)\(E = \dfrac{m_oc^2}{\sqrt{1-\dfrac{r^2}{c^2}}}\) 3)\(P = \dfrac{m_o}{\sqrt{1-\dfrac{r^2}{c^2}}}\) Derivation: We begin by squaring equation (2) on both sides, i.e. \(E^2 = \dfrac{m_o^2c^4}{1-r^2/c^2}\) Next, we insert the term (V2 - V2) as shown: \(E^2 = \dfrac{m_o^2 c^2 (V^2 - V^2 +c^2)}{1-r^2/c^2} \) \(=\dfrac{m_o^2c^2V^2}{1-r^2/c^2} - \dfrac{m_o^2 c^2 V^2}{1-r^2 /c^2} + \dfrac{m_o^2 c^4}{1-r^2/c^2}\) \(E^2=\underset{= \ p}{\underbrace{\left(\dfrac{m_oV}{\sqrt{1-r^2/c^2}}\right)^2}}c^2 + \dfrac{m_oc^2 c^4 - m^2 c^2 r^2}{1-r^2/c^2}\) \(E^2 =p^2 c^2 + m_o^2 c^2 \left(\dfrac{c^2 - r^2}{1-r^2/c^2}\right)\) \(E^2 = p^2 c^2 + m_o^2 c^2 \left(\dfrac{c^2 - V^2}{\dfrac{c^2 - V^2}{C^2}}\right)\) \(E^2 = p^2 c^2 + m_o^2 c^4\)where E is the total energy of the particle \(m_o^2 c^4 = E^2 - p^2 c^2\)← will be identical in all inertial reference frames.

Which of the following equation correctly represents the momentum p of a photon of Energy E?1. E/c2. E2c3. Ec4. Ec2

Answer» Correct Answer - Option 1 : E/c

Relation between Relativistic E and\(\vec{P}\)

1) E = K + moc2

2)\(E = \dfrac{m_oc^2}{\sqrt{1-\dfrac{r^2}{c^2}}}\)

3)\(P = \dfrac{m_o}{\sqrt{1-\dfrac{r^2}{c^2}}}\)

Derivation:

We begin by squaring equation (2) on both sides, i.e.

\(E^2 = \dfrac{m_o^2c^4}{1-r^2/c^2}\)

Next, we insert the term (V2 - V2) as shown:

\(E^2 = \dfrac{m_o^2 c^2 (V^2 - V^2 +c^2)}{1-r^2/c^2} \)

\(=\dfrac{m_o^2c^2V^2}{1-r^2/c^2} - \dfrac{m_o^2 c^2 V^2}{1-r^2 /c^2} + \dfrac{m_o^2 c^4}{1-r^2/c^2}\)

\(E^2=\underset{= \ p}{\underbrace{\left(\dfrac{m_oV}{\sqrt{1-r^2/c^2}}\right)^2}}c^2 + \dfrac{m_oc^2 c^4 - m^2 c^2 r^2}{1-r^2/c^2}\)

\(E^2 =p^2 c^2 + m_o^2 c^2 \left(\dfrac{c^2 - r^2}{1-r^2/c^2}\right)\)

\(E^2 = p^2 c^2 + m_o^2 c^2 \left(\dfrac{c^2 - V^2}{\dfrac{c^2 - V^2}{C^2}}\right)\)

\(E^2 = p^2 c^2 + m_o^2 c^4\)where E is the total energy of the particle

\(m_o^2 c^4 = E^2 - p^2 c^2\)← will be identical in all inertial reference frames.