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As I've learned classical mechanics from different sources, I've seen both

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_k} \right) - \frac{\partial L}{\partial q_k} = 0,$$

and

$$\frac{\partial L}{\partial q_k} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_k} \right) = 0.$$

Obviously, they can be obtained from one another by simply multiplying by negative one. The question I ask is, why? I have seen most textbooks use the former notation, which requires you to multiply by negative one when you extremize the action. I suspect this is simply stylistic/convenient, as it's easier to deal with positive $\ddot{q}^i$ terms. Are there any deeper reasons for this seemingly cosmetic difference?

Note: this question isn't opinion-based. I asked if there were any deeper reasons behind this. If the answer is no, then that's ok. That's all I wanted to know. It's not a subjective matter; if there is genuinely no difference (which I suspected, as you might've been able to tell from what I said), then that's fine, and my question is resolved. I just wanted to make sure it's as such.

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    $\begingroup$ No deeper reason. $\endgroup$
    – Hyperon
    Commented Mar 27 at 15:39
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    $\begingroup$ The first one looks better :) $\endgroup$
    – MaximeJaccon
    Commented Mar 27 at 15:39
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    $\begingroup$ Perhaps the better question is why we don’t mimic $F=ma$ more and write $\frac{\partial L}{\partial q_k}=\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_k}\right)$ :) $\endgroup$
    – peek-a-boo
    Commented Mar 27 at 15:41
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    $\begingroup$ Voting to reopen. The question is not at all opinion-based. There is a simple objective answer - there is no reason to prefer one form over the other since they are equivalent in all respects. $\endgroup$
    – gandalf61
    Commented Mar 27 at 17:47

2 Answers 2

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If you were to prefer one over the other, perhaps you could argue for

$$\frac{\partial L}{\partial q_k} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_k} \right) = 0.$$

since that's what directly follows from the condition that the integrand must be 0 in the principle of least action.

But in physics there's generally no need for a "deep reason" to multiply an equation by -1 to make it look "nicer" or more convenient.

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    $\begingroup$ oh this reminds me: if you were to look at higher order $t$-derivatives, you would keep going and alternating signs: $+\frac{d^2}{dt^2}\left(\frac{\partial L}{\partial\ddot{q}_k}\right)$ and so on. $\endgroup$
    – peek-a-boo
    Commented Mar 27 at 15:44
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Why not?

Those two forms are mathematically equivalent, since $a=b$ is same identity as $-a=-b$, so you can choose one of those representations which look nicer to you.

For example, say we are analyzing damped oscillation where spring restoring force is in balance to damping force, so: $ -bv - kx = 0$, but personally to me, saying that $bv+kx=0$ is nicer because equation is less polluted with negative sign symbols.

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