How to express space-dependent acceleration?

[kent davidge]

Since the acceleration $\vec a$ is given by $\vec a = \frac{d^2 \vec x}{dt^2}$, it is a function of $t$ only. Of course, the derivative implies that $t = t(\vec x)$ so we can also in principle express $\vec a$ in terms of $\vec x$. But how can we express an acceleration dependent on both time and space? I mean an acceleration of the type $\vec a = \vec a(t,\vec x)$?

 

[BvU]

As you say, $a$ is a function of $t$ only: $\vec a(t,\vec x(t)) = \vec a(t)$. Your question seems to suppose a specific trajectory or something, so that $t=t(\vec x)$ where now $\vec x$ is the independent variable ? Meaning that for every position there is a time assigned ?

How would that work with e.g. a circular trajectory ?

 

[kent davidge]

Yes, I got that. If position is time dependent, which is implied by stating that $\vec a = \frac{d^2 \vec x}{dt^2}$, then there is really only one independent variable, namely $t$. However it seems intuitive that in nature we should encounter situations where the acceleration depends both on space and time, so that $\{\vec x, t \}$ are independent of one another. In those cases, then what should we do?

 

[BvU]

$x$ and $t$ independent of each other ? That's not a case of $t(\vec x)$, then !

If we want to calculate a trajectory, e.g. for a mass $m$ in a gravity field from a mass $M >> m$, we have a dependency $\vec a = \vec a (\vec x)$ and we also have $\vec a = {d^2\vec x\over dt^2}$. After solving for $\vec x(t)$ we still don't have a $\vec a (\vec x, t)$ outside the trajectory.

 

[kent davidge]

$x$ and $t$ independent of each other ? That's not a case of $t(\vec x)$, then !

right

If we want to calculate a trajectory, e.g. for a mass $m$ in a gravity field from a mass $M >> m$, we have a dependency $\vec a = \vec a (\vec x)$ and we also have $\vec a = {d^2\vec x\over dt^2}$. After solving for $\vec x(t)$ we still don't have a $\vec a (\vec x, t)$ outside the trajectory.

then should Newton's second law be written generally as $F(\vec x, t) = m \vec a (\vec x, t)$? and only for a specified, time-parametrized trajectory $\vec x(t)$, can it be written as $F = m \frac{d^2 x}{dt^2}$?

 

[BvU]

No. $\vec F = m\vec a$ is general: it is valid independent of $\vec x$.

 

[kent davidge]

No. $\vec F = m\vec a$ is general: it is valid independent of $\vec x$.

so how can we express this for a arbritary trajectory?

 

[BvU]

arbritary trajectory

by twice differentiating the position wrt time

But I suppose that isn't what you mean ?

 

[kent davidge]

by twice differentiating the position wrt time

But I suppose that isn't what you mean ?

hmm, I think that answers my question, yes. So what can I conclude? That we can't have an acceleration of the form $\vec a(\vec x, t)$ because $\vec x$ is always $\vec x(t)$?

 

[Ibix]

hmm, I think that answers my question, yes. So what can I conclude? That we can't have an acceleration of the form $\vec a(\vec x, t)$ because $\vec x$ is always $\vec x(t)$?

A trajectory is, by definition, a smooth path through space. Thus it can be written $\vec x(\lambda)$, where $\lambda$ is some smooth parameter along the line. Since you can only be at one place at one time and can't revisit earlier time, $t$ is a possible parameter. Thus you can write $\vec x(t)$.

 

[kent davidge]

A trajectory is, by definition, a smooth path through space. Thus it can be written $\vec x(\lambda)$, where $\lambda$ is some smooth parameter along the line. Since you can only be at one place at one time and can't revisit earlier time, $t$ is a possible parameter. Thus you can write $\vec x(t)$.

So the conclusion is that $\vec a$ (and thus, also $\vec F$) cannot be dependent on both position and time?

 

[Vanadium 50]

So the conclusion is that →a\vec a (and thus, also →F\vec F) cannot be dependent on both position and time?

No, the conclusion is a particle can't be in two places at once.

 

[Ibix]

So the conclusion is that $\vec a$ (and thus, also $\vec F$) cannot be dependent on both position and time?

You can certainly write down a time and space varying force field - Newtonian gravity around a binary star for example. But a particle can only be in one place at a time, so its acceleration at time $t$ must be expressible as a function of that single parameter.

If you fill space with matter of density $\rho(\vec x,t)$ then you could write down an acceleration field $a(\vec x,t)$, but you appeared to be talking about a particle.

 

[FactChecker]

You need to be careful. If your goal is to specify the accelerations at different points for different elements, then you can write $a(t,x)$. But if you are trying to define the acceleration of a given element as time progresses, then you must remember that $x(t)$ is already defined by the double integral of acceleration and the initial position. In that case, you are not free to specify a conflicting $x(t)$.