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The conjugate Hamiltonian can be defined from the Lagrangian as,

$$ p_i ~=~ \frac{\partial L}{\partial \dot{q}^i}$$

Typically the momenta components are given in spherical polars $(r, \theta, \phi)$.

How would one find the momenta in Cartesian coordinates?

I suspect that it will just transform as the spatial coordinates do, but am unable to show this. Example attempt at the x-component:

$$ p_x = \frac{\partial L}{\partial \dot{x}} = \frac{\partial L}{\partial \dot{r}} \frac{\partial \dot{r}}{\partial \dot{x}} = p_r \frac{\partial \dot{r}}{\partial \dot{x}}$$

But cannot see where to go from here.

Edits in response to comments

Some context: I am writing some code that takes a dataset of momentum vectors in Cartesian coordinates (not produced by me) and another set of vectors in spherical coordinates (also not produced by me), and calculates the angle between any two vectors.

To do this I need to first convert the momentum vectors in spherical coordinates, into Cartesian coordinates. Therefore I want a general transformation from momenta expressed in spherical coordinates, to Cartesian coordinates.

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    $\begingroup$ typically the momenta components are given in spherical polars citation needed. $\endgroup$ – Kyle Kanos Jun 29 '17 at 11:05
  • $\begingroup$ Two examples of Hamiltonian's in Cartesian. en.wikipedia.org/wiki/Hamiltonian_mechanics $\endgroup$ – user154420 Jun 29 '17 at 11:29
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How does one find the canonical momenta in cartesian coordinates? Easy: they're given by $$ p_i=\frac{\partial L}{\partial \dot q_i}, \tag 1 $$ where $q_i=x_i$ is a cartesian coordinate. That's all you can say in the general case, without specifying a Lagrangian.


In the initial case where $L=\frac12 m \sum_i \dot x_i^2-V(x)$, then the canonical momenta are computed in every analytical mechanics textbook, and the partial derivative is particularly simple to work out: $$ p_i=\frac{\partial L}{\partial \dot x_i}=\frac{\partial }{\partial \dot x_i}\frac12 m \sum_j \dot x_j^2 = m\dot x_i, $$ i.e. the canonical momentum coincides with the kinematic momentum $m\dot x_i$. However, this is not the only possible case (for example, it is common to have $L=\frac m2 \sum_i(\dot x_i-A_i(x))^2$, in which case the canonical momentum $p_i = m(\dot x_i-A_i(x))$ differs from the kinematic momentum by a gauge-dependent vector potential $A(x)$), but since you do not give more information, there's nothing more that one can say beyond the definition $(1)$.

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  • $\begingroup$ So without specifying a Lagrangian, there isn't a general transformation to find e.g $p_x$ from $p_r, p_{\theta}, p_{\phi}$? $\endgroup$ – user1887919 Jun 29 '17 at 12:22
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    $\begingroup$ There is one, but why would you want it? In most circumstances $L$ is given in terms of the cartesian velocities, and it's just simpler to calculate the momenta directly. Unless you have a more specific situation (the details of which you've yet to disclose) there's simply no need to be doing that transformation. $\endgroup$ – Emilio Pisanty Jun 29 '17 at 12:27
  • $\begingroup$ I am approaching it more from the applied side, of wanting to find the angle between two momentum vectors. If one of these vectors is in Cartesian coordinates and one in spherical, I am wanting to convert both into the same coordinate system so as to find the angle between them $\endgroup$ – user1887919 Jun 29 '17 at 12:52
  • $\begingroup$ Approaching it "from the applied side" does not mean that you can supply insufficient information and have answers magically pop out. If your kinetic lagrangian is just $L=\frac12 m \dot{\vec x}^2$, and if both momenta are at the same point, then you can use the standard vector transformations. If either of those fail, you need to provide a complete description of what you're doing. $\endgroup$ – Emilio Pisanty Jun 29 '17 at 12:59
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The momenta are, as you noted, defined as partial derivatives. By definition, partial differentiation holds certain quantities fixed. You have to bear in mind which ones, e.g. $p_r$ holds $\dot{\theta},\,\dot{\phi}$ fixed while $p_x$ holds $\dot{y},\,\dot{z}$ fixed. This difference complicates any effort to write Cartesian-coordinate momenta in terms of their polar counterparts with the chain rule. That's not to say this task is impossible, but it's best to just return to the Lagrangian (if you don't already know it, infer it up to a function of $\mathbf{x}$ from the known momenta). For example, consider the Lagrangian$$L=\frac{m}{2}(\dot{r}^2+r^2\dot{\theta}^2+r^2\sin^2\theta\dot{\phi}^2)-V(\mathbf{x}),$$which has momenta $$p_r=m\dot{r},\,p_\theta=mr^2\dot{\theta},\,p_\phi=mr^2\sin^2\theta\dot{\phi}.$$If you already have the Lagrangian, you can rewrite in Cartesian coordinates, giving the special case Emilio Pisanty discussed. But even if all you knew were the momenta themselves, the requirements $p_r=\frac{\partial L}{\partial\dot{r}}$ etc. prove the Lagrangian must take the above form for some $V$ dependent only on undifferentiated coordinates. From this, you can carry on as usual. Edit: I should probably mention what you should do if you want to use the chain rule instead. If $x_i,\,y_j$ are different coordinate systems then$$M_{ij}:=\frac{\partial x_i}{\partial y_j}\implies dx_i=\sum_jM_{ij}dy_j,\,\dot{x}_i=\sum_jM_{ij}\dot{y}_j,\,\frac{\partial\dot{x}_i}{\partial\dot{y}_j}=M_{ij}+\sum_k\frac{\partial M_{ik}}{\partial\dot{y}_j}\dot{y}_k.$$

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