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Consider a transformation from one set of generalized coordinate and momentum (q, p) to another set (Q, P) denoted by,
Q = pqs; P = qr
where s and r are constants. The transformation is canonical if
Q = pqs; P = qr
where s and r are constants. The transformation is canonical if
- a)s = 0 and r = 1
- b)s = 2 and r = −1
- c)s = 0 and r = −1
- d)s = 2 and r = 1
Correct answer is option 'B'. Can you explain this answer?
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Consider a transformation from one set of generalized coordinate and m...
Concept:
Canonical transformations allow us to change the phase-space coordinate system that we use to express a problem, preserving the form of Hamilton's equations.
Canonical transformations allow us to change the phase-space coordinate system that we use to express a problem, preserving the form of Hamilton's equations.
Calculation:
The correct answer is option (B).
The correct answer is option (B).
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Consider a transformation from one set of generalized coordinate and m...
Canonical Transformation Explanation:
Canonical transformations are transformations that preserve the form of Hamilton's equations. In other words, the new set of generalized coordinates and momenta must satisfy the following condition:
$$\frac{\partial Q_i}{\partial q_j} \frac{\partial P_i}{\partial p_j} - \frac{\partial Q_i}{\partial p_j} \frac{\partial P_i}{\partial q_j} = \delta_{ij}$$
where i, j = 1, 2, ..., n. Here, n is the number of degrees of freedom.
Given transformation:
Q = pqs; P = qr
Checking for Canonical Transformation:
To check if the transformation is canonical, we need to calculate the partial derivatives and substitute them into the condition above.
$$\frac{\partial Q}{\partial q} = ps; \frac{\partial Q}{\partial p} = qs$$
$$\frac{\partial P}{\partial q} = r; \frac{\partial P}{\partial p} = 0$$
Substitute these into the canonical transformation condition:
$$ps \times 0 - qs \times r = -qr = -\delta_{ij}$$
This implies that s = 2 and r = -1 for the transformation to be canonical.
Therefore, the correct answer is option B: s = 2 and r = -1.
Canonical transformations are transformations that preserve the form of Hamilton's equations. In other words, the new set of generalized coordinates and momenta must satisfy the following condition:
$$\frac{\partial Q_i}{\partial q_j} \frac{\partial P_i}{\partial p_j} - \frac{\partial Q_i}{\partial p_j} \frac{\partial P_i}{\partial q_j} = \delta_{ij}$$
where i, j = 1, 2, ..., n. Here, n is the number of degrees of freedom.
Given transformation:
Q = pqs; P = qr
Checking for Canonical Transformation:
To check if the transformation is canonical, we need to calculate the partial derivatives and substitute them into the condition above.
$$\frac{\partial Q}{\partial q} = ps; \frac{\partial Q}{\partial p} = qs$$
$$\frac{\partial P}{\partial q} = r; \frac{\partial P}{\partial p} = 0$$
Substitute these into the canonical transformation condition:
$$ps \times 0 - qs \times r = -qr = -\delta_{ij}$$
This implies that s = 2 and r = -1 for the transformation to be canonical.
Therefore, the correct answer is option B: s = 2 and r = -1.
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Consider a transformation from one set of generalized coordinate and momentum (q, p) to another set (Q, P) denoted by,Q = pqs; P = qrwhere s and r are constants. The transformation is canonical ifa)s = 0 and r = 1b)s = 2 and r = −1c)s = 0 and r = −1d)s = 2 and r = 1Correct answer is option 'B'. Can you explain this answer?
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