Answer
385.2k+ views
Hint:Momentum of a particle is defined as the product of the mass and velocity of a particle. Momentum is a vector quantity that means it has both magnitude and direction. Here, we will use an energy-momentum relation to find the required solution.
Complete step by step answer:
The energy-momentum relation is the relativistic equation that relates the total energy of an object to the invariant mass and the momentum. We can also say that it is the extension of the mass- energy equivalence relation of a system of bodies that has zero momentum. This equation is given below
${E^2} = {\left( {pc} \right)^2} + {\left( {m{c^2}} \right)^2}$
This equation holds good for a body or system of bodies that consists of total momentum $E$ ,
invariant mass ${m_0}$ , and the momentum $p$ . Here, $c$ is the constant that is the speed of light.
Now, when the body will be massless i.e. $m = 0$ , the above relation will become
${E^2} = {\left( {pc} \right)^2}$
$ \Rightarrow \,{E^2} = {p^2}{c^2}$
This implies that, for zero rest mass, the total energy of the body is equal to the product of mass and speed of light.
Hence, option (A) is the correct option.
Additional information:
Let, the speed of the body is less than the speed of light $c$ , the above relation becomes
${E^2} = \dfrac{1}{2}{m_0}{v^2} + {m_0}{c^2}$
This implies that the total energy of the body is equal to its classical kinetic energy plus its rest energy.
Also, if the body will be at rest, that is, it will be in the center-of-momentum frame, we will have $E = {E_0}$ and $m = {m_0}$ . Therefore, the energy-momentum relation and both forms of the mass- energy relation will become the same.
Note:The energy-momentum is similar to the mass-energy equivalence relation. Now, for the bodies having zero momentum, it simplifies that the mass-energy equation is given as $E = {m_0}{c^2}$
Therefore, we can say that the total energy is equal to the rest energy.
Complete step by step answer:
The energy-momentum relation is the relativistic equation that relates the total energy of an object to the invariant mass and the momentum. We can also say that it is the extension of the mass- energy equivalence relation of a system of bodies that has zero momentum. This equation is given below
${E^2} = {\left( {pc} \right)^2} + {\left( {m{c^2}} \right)^2}$
This equation holds good for a body or system of bodies that consists of total momentum $E$ ,
invariant mass ${m_0}$ , and the momentum $p$ . Here, $c$ is the constant that is the speed of light.
Now, when the body will be massless i.e. $m = 0$ , the above relation will become
${E^2} = {\left( {pc} \right)^2}$
$ \Rightarrow \,{E^2} = {p^2}{c^2}$
This implies that, for zero rest mass, the total energy of the body is equal to the product of mass and speed of light.
Hence, option (A) is the correct option.
Additional information:
Let, the speed of the body is less than the speed of light $c$ , the above relation becomes
${E^2} = \dfrac{1}{2}{m_0}{v^2} + {m_0}{c^2}$
This implies that the total energy of the body is equal to its classical kinetic energy plus its rest energy.
Also, if the body will be at rest, that is, it will be in the center-of-momentum frame, we will have $E = {E_0}$ and $m = {m_0}$ . Therefore, the energy-momentum relation and both forms of the mass- energy relation will become the same.
Note:The energy-momentum is similar to the mass-energy equivalence relation. Now, for the bodies having zero momentum, it simplifies that the mass-energy equation is given as $E = {m_0}{c^2}$
Therefore, we can say that the total energy is equal to the rest energy.
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