Question

Take the ratio of relativistic rest energy, \[{\bf{E = \gamma m}}{{\bf{c}}^{\bf{2}}}\], to relativistic momentum, \[{\bf{p = \gamma mu}}\], and show that in the limit that mass approaches zero, you find\[\frac{{\bf{E}}}{{\bf{p}}}{\bf{ = c}}\].

Short answer

The value of energy to the ratio momentum is \[\mathop {\lim }\limits_{m \to 0} \frac{E}{p} \approx c\].

Step-by-step solution

Definition of Concept

Energy: The ability to perform work in physics. It can take many different forms, including potential, kinetic, thermal, electrical, chemical, nuclear, and others.

Find the value of energy to the ratio momentum

EM waves, according to Maxwell and others who studied them, would carry momentum. The photoelectric effect, in which photons knock electrons out of a substance, suggests photon momentum.

Considering the given information,

\[\begin{array}{l}E \approx \gamma m{c^2}\\p \approx \gamma mu\end{array}\]

The energy-to-momentum ratio has now been determined.

\[\begin{array}{l}\frac{E}{p} \approx \frac{{\gamma m{c^2}}}{{\gamma mu}}\\\frac{E}{p} \approx \frac{{{c^2}}}{u}\end{array}\]

As a particle's mass approaches zero, its velocity v approaches u, so the energy-to-momentum ratio in this limit is:

\[\begin{array}{l}\mathop {\lim }\limits_{m \to 0} \frac{E}{p} \approx \frac{{{c^2}}}{c}\\\mathop {\lim }\limits_{m \to 0} \frac{E}{p} \approx c\end{array}\]

As a result, the required equation is established.

Therefore,the value of energy to the ratio momentum is\[\mathop {\lim }\limits_{m \to 0} \frac{E}{p} \approx c\].