Question:
Two non-relativistic particles with masses \(m_1\) and \(m_2\) move with momenta \(\mathbf{p}_1\) and \(\mathbf{p}_2\), respectively, in an inertial frame S. In another inertial frame S', moving with a constant speed with respect to S, the same particles are observed to have momenta \(\mathbf{p'}_1\) and \(\mathbf{p'}_2\), respectively.
Galilean invariance implies that
Two non-relativistic particles with masses \(m_1\) and \(m_2\) move with momenta \(\mathbf{p}_1\) and \(\mathbf{p}_2\), respectively, in an inertial frame S. In another inertial frame S', moving with a constant speed with respect to S, the same particles are observed to have momenta \(\mathbf{p'}_1\) and \(\mathbf{p'}_2\), respectively.
Galilean invariance implies that
Galilean invariance implies that
Show Hint
The quantity \(m_2\mathbf{p}_1 - m_1\mathbf{p}_2\) is proportional to the relative momentum of the two-particle system. Specifically, it is \((m_1+m_2)\) times the momentum of particle 1 in the center-of-mass frame. Physical quantities related to relative motion (like relative position, relative velocity, and relative momentum) are often invariant under Galilean transformations.
Updated On: Sep 8, 2025
- \(m_2\mathbf{p'}_1 - m_1\mathbf{p'}_2 = m_2\mathbf{p}_1 - m_1\mathbf{p}_2\)
- \(m_2\mathbf{p'}_1 + m_1\mathbf{p'}_2 = m_2\mathbf{p}_1 + m_1\mathbf{p}_2\)
- \(m_1\mathbf{p'}_1 - m_2\mathbf{p'}_2 = m_1\mathbf{p}_1 - m_2\mathbf{p}_2\)
- \(m_1\mathbf{p'}_1 + m_2\mathbf{p'}_2 = m_1\mathbf{p}_1 + m_2\mathbf{p}_2\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
This problem deals with Galilean transformations, which describe how physical quantities like velocity and momentum change when observed from different inertial reference frames moving at a constant velocity relative to each other. A quantity is said to be a Galilean invariant if its value is the same in all inertial frames. We need to find which of the given combinations of momenta is invariant.
Step 2: Key Formula or Approach:
Let frame S' move with a constant velocity \(\mathbf{V}\) with respect to frame S. According to the Galilean transformation for velocity, if a particle has velocity \(\mathbf{v}\) in frame S, its velocity \(\mathbf{v'}\) in frame S' is given by:
\[ \mathbf{v'} = \mathbf{v} - \mathbf{V} \] For a non-relativistic particle of mass \(m\), its momentum is \(\mathbf{p} = m\mathbf{v}\). The momentum in frame S', \(\mathbf{p'}\), is:
\[ \mathbf{p'} = m\mathbf{v'} = m(\mathbf{v} - \mathbf{V}) = m\mathbf{v} - m\mathbf{V} = \mathbf{p} - m\mathbf{V} \] Step 3: Detailed Explanation:
We have two particles with masses \(m_1\) and \(m_2\). Their momenta in frames S and S' are related as follows:
For particle 1: \(\mathbf{p'}_1 = \mathbf{p}_1 - m_1\mathbf{V}\)
For particle 2: \(\mathbf{p'}_2 = \mathbf{p}_2 - m_2\mathbf{V}\)
Now, let's test the expression given in option (A): \(m_2\mathbf{p'}_1 - m_1\mathbf{p'}_2\).
Substitute the transformed momenta into this expression:
\[ m_2\mathbf{p'}_1 - m_1\mathbf{p'}_2 = m_2(\mathbf{p}_1 - m_1\mathbf{V}) - m_1(\mathbf{p}_2 - m_2\mathbf{V}) \] Distribute the masses \(m_2\) and \(m_1\):
\[ = m_2\mathbf{p}_1 - m_1m_2\mathbf{V} - m_1\mathbf{p}_2 + m_1m_2\mathbf{V} \] The terms involving the relative velocity \(\mathbf{V}\) cancel each other out:
\[ = m_2\mathbf{p}_1 - m_1\mathbf{p}_2 \] Thus, we have shown that:
\[ m_2\mathbf{p'}_1 - m_1\mathbf{p'}_2 = m_2\mathbf{p}_1 - m_1\mathbf{p}_2 \] The quantity \(m_2\mathbf{p}_1 - m_1\mathbf{p}_2\) is a Galilean invariant.
Step 4: Final Answer:
The relation implied by Galilean invariance is \(m_2\mathbf{p'}_1 - m_1\mathbf{p'}_2 = m_2\mathbf{p}_1 - m_1\mathbf{p}_2\). This matches option (A).
This problem deals with Galilean transformations, which describe how physical quantities like velocity and momentum change when observed from different inertial reference frames moving at a constant velocity relative to each other. A quantity is said to be a Galilean invariant if its value is the same in all inertial frames. We need to find which of the given combinations of momenta is invariant.
Step 2: Key Formula or Approach:
Let frame S' move with a constant velocity \(\mathbf{V}\) with respect to frame S. According to the Galilean transformation for velocity, if a particle has velocity \(\mathbf{v}\) in frame S, its velocity \(\mathbf{v'}\) in frame S' is given by:
\[ \mathbf{v'} = \mathbf{v} - \mathbf{V} \] For a non-relativistic particle of mass \(m\), its momentum is \(\mathbf{p} = m\mathbf{v}\). The momentum in frame S', \(\mathbf{p'}\), is:
\[ \mathbf{p'} = m\mathbf{v'} = m(\mathbf{v} - \mathbf{V}) = m\mathbf{v} - m\mathbf{V} = \mathbf{p} - m\mathbf{V} \] Step 3: Detailed Explanation:
We have two particles with masses \(m_1\) and \(m_2\). Their momenta in frames S and S' are related as follows:
For particle 1: \(\mathbf{p'}_1 = \mathbf{p}_1 - m_1\mathbf{V}\)
For particle 2: \(\mathbf{p'}_2 = \mathbf{p}_2 - m_2\mathbf{V}\)
Now, let's test the expression given in option (A): \(m_2\mathbf{p'}_1 - m_1\mathbf{p'}_2\).
Substitute the transformed momenta into this expression:
\[ m_2\mathbf{p'}_1 - m_1\mathbf{p'}_2 = m_2(\mathbf{p}_1 - m_1\mathbf{V}) - m_1(\mathbf{p}_2 - m_2\mathbf{V}) \] Distribute the masses \(m_2\) and \(m_1\):
\[ = m_2\mathbf{p}_1 - m_1m_2\mathbf{V} - m_1\mathbf{p}_2 + m_1m_2\mathbf{V} \] The terms involving the relative velocity \(\mathbf{V}\) cancel each other out:
\[ = m_2\mathbf{p}_1 - m_1\mathbf{p}_2 \] Thus, we have shown that:
\[ m_2\mathbf{p'}_1 - m_1\mathbf{p'}_2 = m_2\mathbf{p}_1 - m_1\mathbf{p}_2 \] The quantity \(m_2\mathbf{p}_1 - m_1\mathbf{p}_2\) is a Galilean invariant.
Step 4: Final Answer:
The relation implied by Galilean invariance is \(m_2\mathbf{p'}_1 - m_1\mathbf{p'}_2 = m_2\mathbf{p}_1 - m_1\mathbf{p}_2\). This matches option (A).
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