Why magic and physics cannot coexist peacefully? After reading many physics and math stuff, I think that the problem maybe come from the principle of least action. So, I edited the definition of action, in order to make magic and physics coexist peacefully.


Definition of action in my world: $$ S=\int_{t_1}^{t_2} L(\mathbf{q,\dot{q},t})\;dt + \frac{\int_{t_1}^{t_2} M(\mathbf{r-q,\dot{r}-\dot{q},t})\;dt}{1+|\int_{t_1}^{t_2} L(\mathbf{q,\dot{q},t})\;dt|} $$ The first term is the definition of action in real world. The second term is added by me. $$ \int_{t_1}^{t_2} M(\mathbf{r-q,\dot{r}-\dot{q},t})\;dt $$ It is the total magical power needed to change a body's trajectory.

r is the new position. (after magic influence)

r dot is the new velocity. (after magic influence) $$ 1+|\int_{t_1}^{t_2} L(\mathbf{q,\dot{q},t})\;dt| $$ is used to suppress the influence of magic in daily lives.

"1+" is used to prevent value which is smaller than one and bigger than zero to appear in the denominator.

Absolute value is used to prevent negative value.

Finally, the magical power need to have a unit $ J^2 s$ as well. lol

Is there any contradiction (physics or math) caused by this edited definition of action? (I know that many equations will change, such as F=ma, but it is ok if there is no contradiction.)

Response to Comments:

I didn't plan to define "magical power" using some well-known physics concept, such as energy. I want to make "magical power" into something new, so its unit $ J^2 s$ is the only explanation. :p

Magic in my world also affects quantum, but the calculation of quantum field theory is difficult, I need to spend more time on it.

  • $\begingroup$ I think a great deal more context is needed here? Is magic simply a form of energy? Does it only affect newtonian physics? $\endgroup$
    – knowads
    Jun 24 '16 at 19:50

For one thing, action has units, so you can't add $1$ to it as you do in the denominator. But that's easily worked around by replacing the $1$ with a constant $h_M$.

From there, let me simplify your expression by writing the modified action as $$S = S^{(L)} + \frac{S^{(M)}}{h_M + \lvert S^{(L)}\rvert}$$ According to the stationary action principle, you need the variation $\delta S = 0$ for physical paths. Variation basically follows the same rules as derivatives, so $$\delta S = \delta S^{(L)} + \frac{\delta S^{(M)}}{h_M + \lvert S^{(L)}\rvert} \pm \frac{S^{(M)}}{(h_M + \lvert S^{(L)}\rvert)^2}\delta S^{(L)}$$ The $\pm$ sign, resulting from your use of the absolute value, is already a clue that this is going to get fairly complicated.

I won't bother to explain the sign choice in detail, but it does bring up the related issue of a sort of "gauge dependence" of your action: the fact that adding a constant to the action changes the equations of motion. In standard Lagrangian mechanics, you can add any value representable as $f(t_2) - f(t_1)$, for some differentiable function $f$, to the action with no effect on the underlying physics. This property is essential in deriving the Euler-Lagrange equations.

In your case, however, significant changes could result from adding that kind of term. That sign choice, for instance, depends on whether $S^{(L)}$ is positive or negative. So let's do the following:

  1. Take some path $\mathcal{P}$ such that $S^{(L)}[\mathcal{P}] < 0$.
  2. Define $s = \lvert S^{(L)}[\mathcal{P}]\rvert$.
  3. Define $$f(t) = \frac{2st}{t_2^{[\mathcal{P}]} - t_1^{[\mathcal{P}]}}$$ and add $f(t_2) - f(t_1) = 2s$ to the action, which changes its value from negative to positive.

Now you've flipped the sign of the action, and also flipped the sign in the variation formula, which means the quantity that needs to be set to zero is different.

What this means is that the invariance to total derivatives which is present in standard Lagrangian mechanics does not exist in your framework. By itself, that isn't a dealbreaker; the magic action still presumably has stationary paths. But good luck finding them without being able to use variational calculus on the general action. You won't have the Euler-Lagrange equations to work with, and unless you get really lucky, there won't be any general way to find the paths that set the variation to zero. This makes it a very frustrating (at best) and possibly practically useless framework.

And of course, this doesn't rule out the possibility of contradictions. It just makes them harder to identify, along with everything else.

  • $\begingroup$ That means a better modification of action is $S=S^{(L)}+S^{(M)}+f(t)$ ? $\endgroup$
    – fairytale
    Jun 24 '16 at 23:34
  • $\begingroup$ There would be no point in doing that, because then you're just back to the regular action principle. Or, it's like splitting the action into two parts that you label $S^{(L)}$ and $S^{(M)}$. (You can't have $f(t)$ because $t$ is not a parameter to the action.) $\endgroup$
    – David Z
    Jun 24 '16 at 23:56
  • $\begingroup$ Then what is the better choice? $S=S^{(L)}+S^{(M)}+2s$ ? Sorry I get confused again. lol $\endgroup$
    – fairytale
    Jun 25 '16 at 9:58
  • 1
    $\begingroup$ The point is you can't really change the way the stationary action principle works without making it exceedingly difficult (perhaps impossible) to work with. You can, however, add terms to the action to change the physics. For instance, $S = S^{(0)} + S^{(M)}$, where $S^{(0)}$ is the regular action of Newtonian mechanics and $S^{(M)}$ is your magical addition. This is equivalent to making $L^{(0)} + M$ your Lagrangian. $\endgroup$
    – David Z
    Jun 25 '16 at 10:43
  • $\begingroup$ Can the least action principle principle apply to systems that have non-conservative force? $\endgroup$
    – fairytale
    Jun 26 '16 at 22:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .