Velocity definition and parameter dependence

[fog37]

Hello Forum,

Limiting our discussion to 1D motion, it is clear that the concept of instantaneous velocity is defined as the covered displacement dx divided by the time interval elapsed dt:

$$ v = \frac {dx}{dt}$$

However, mathematically, the velocity $v$ can be made to depend on any parameter $t,x,a$. For example the velocity $v$ can be expressed a function of time $v(t)$ but also as a function position $v(x)$. If we know the position function $x(t)$, we can substitute that into $v(x) = v( x(t))$ to obtain $v(t)$. If $x(t) =3t+2$ and $v(x) = x ^2$, then $v(t)= (3t+2)^2$.

• The three functions $v(t)$ and $x(t)$ $v(t)$ cannot be independent of each other because the object's motion is one and only one and the three functions cannot provide contradictory information about the motion, correct?
• Would it be possible to have a situation where the velocity is a function of both position and time $v(x,t)$? Would you have a simple example of functions that would result in that?

Thank you!

 

[etotheipi]

the velocity $v$ can be expressed ... also as a function position $v(x)$.

Only sometimes. Consider SHM along the $x$ axis, i.e. $x = x_0 \cos{\omega t}$. Then $v = -x_0\omega \sin{\omega t}$. If, for instance, $x = 0$, the velocity could be either $x_0 \omega$ or $-x_0 \omega$, so in this case the map is one-to-many and you cannot construct a function $v(x)$.

 

[vanhees71]

By definition the only unique dependence for 1D point-particle motion is $x(t)$ (and thus $v(t)=\ddot{x}(t)$ and $a(t)=\dot{v}(t)=\ddot{x}(t)$, where the dot means derivative with respect to time.

In 1D motion it can be sometimes advantageous to look at the equations of motion from a different point of view. Often the force is given by a potential, i.e.,
$$F(x)=-V'(x),$$
where the prime denotes a derivative with respect to position. Then the equation of motion reads
$$m a=m\ddot{x} =F(x)=-V'(x).$$
Now consider an arbitrary motion $x=x(t)$. So plug this into the equation of motion and multiply the equation by $\dot{x}$, which gives
$$m \dot{x} \ddot{x}=-\dot{x} V'(x)=-\frac{\mathrm{d}}{\mathrm{d} t} V(x),$$
where in the final step the chain rule has been used. Now also the left-hand side is a total time derivative, because
$$m \dot{x} \ddot{x} = \frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{x}^2 \right).$$
Plugging all this in again, one finally gets the energy-conservation law
$$\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{x}^2+V(x) \right)=0.$$
Now a function of time has a vanishing derivative if it's a constant, i.e.,
$$\frac{m}{2} \dot{x}^2 + V(x)=E=\text{const},$$
i.e., the total energy is conserved for any motion the body may make. Now you can write
[EDIT: Corrected a typo; thanks to @etotheipi for pointing it out via PM]
$$\dot{x}^2=v^2=\frac{2}{m}[E-V(x)],$$
and thus you have
$$v=v(x)=\pm \sqrt{\frac{2}{m}[E-V(x)]},$$
and this differential equation can be solved by separation, i.e.,
$$\pm \int_{x_0}^x \frac{1}{\sqrt{\frac{2}{m}[E-V(x)]}} = t-t_0.$$
Note, however, that one has to take some care about the meaning of this integral. First of all you have to use the right sign for the square root, which you usually get from the initial conditions $x(t_0)=x_0$, $v(t_0)=v_0$, which you can also use to get the constant energy $E$, which is one integration constant, and $x_0$ occurring in the above integral is the other. In this example it makes sense to consider $v$ as a function of $x$, though it's not a unique function.

E.g., take as an example the harmonic oscillator which obeys the equation of motion
$$m\dot{v}=m\ddot{x}=-k x$$
and try to solve it using the energy-conservation law calculating this integral. It's a good exercise though the direct integration is much more straight forward in this case.

 

[fog37]

Only sometimes. Consider SHM along the $x$ axis, i.e. $x = x_0 \cos{\omega t}$. Then $v = -x_0\omega \sin{\omega t}$. If, for instance, $x = 0$, the velocity could be either $x_0 \omega$ or $-x_0 \omega$, so in this case the map is one-to-many and you cannot construct a function $v(x)$.

Thank you. I am thinking that if an object moves, it occupies a certain position in time and has a certain velocity when it occupies that position at the same instant t. So, the physical situation can be described by $x(t)$ and $v(t)$ with $v(t) = \frac {dx}{dt}$. Simply substitution can always take us from $v(t)$ to $v(x)$...

 

[fog37]

By definition the only unique dependence for 1D point-particle motion is $x(t)$ (and thus $v(t)=\ddot{x}(t)$ and $a(t)=\dot{v}(t)=\ddot{x}(t)$, where the dot means derivative with respect to time.

In 1D motion it can be sometimes advantageous to look at the equations of motion from a different point of view. Often the force is given by a potential, i.e.,
$$F(x)=-V'(x),$$
where the prime denotes a derivative with respect to position. Then the equation of motion reads
$$m a=m\ddot{x} =F(x)=-V'(x).$$
Now consider an arbitrary motion $x=x(t)$. So plug this into the equation of motion and multiply the equation by $\dot{x}$, which gives
$$m \dot{x} \ddot{x}=-\dot{x} V'(x)=-\frac{\mathrm{d}}{\mathrm{d} t} V(x),$$
where in the final step the chain rule has been used. Now also the left-hand side is a total time derivative, because
$$m \dot{x} \ddot{x} = \frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{x}^2 \right).$$
Plugging all this in again, one finally gets the energy-conservation law
$$\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{x}^2+V(x) \right)=0.$$
Now a function of time has a vanishing derivative if it's a constant, i.e.,
$$\frac{m}{2} \dot{x}^2 + V(x)=E=\text{const},$$
i.e., the total energy is conserved for any motion the body may make. Now you can write
$$\dot{x}^2=v^2=2m[E-V(x)],$$
and thus you have
$$v=v(x)=\pm \sqrt{2m[E-V(x)]},$$
and this differential equation can be solved by separation, i.e.,
$$\pm \int_{x_0}^x \frac{1}{\sqrt{2m[E-V(x)]}} = t-t_0.$$
Note, however, that one has to take some care about the meaning of this integral. First of all you have to use the right sign for the square root, which you usually get from the initial conditions $x(t_0)=x_0$, $v(t_0)=v_0$, which you can also use to get the constant energy $E$, which is one integration constant, and $x_0$ occurring in the above integral is the other. In this example it makes sense to consider $v$ as a function of $x$, though it's not a unique function.

E.g., take as an example the harmonic oscillator which obeys the equation of motion
$$m\dot{v}=m\ddot{x}=-k x$$
and try to solve it using the energy-conservation law calculating this integral. It's a good exercise though the direct integration is much more straight forward in this case.

Thank you vanhees71.

When the force, hence the acceleration $a(x)$, depend solely on position (and not on time), it practically means that the object will always experience a certain acceleration $a$ and velocity $v$ when it is located at position $x$. We usually end up obtaining the position function $x(t)$ which can plug in into $a(x)$ to get the time function for $a(t)$. So we can get both $a(x)$ and $a(t)$. I guess that would be an "implicit" dependence of the acceleration on time and an explicit dependence of position...

For the SHO example, the acceleration is a function $a(x)$ only. That is the structure of the problem: whenever (in the past, now, in the future, so no time dependence) the acceleration explicitly depends on position only. Surely, we can always convert $a(x)$ to $a(t)$ even in the SHO motion to determine the acceleration of the body with time $t$.

A force $F(x)$ can be a function of position only, or a force $F(t)$ that is a function of time only or a force $F(x,t)$ that is a function of both. In all case, the object to which the force is applied moves and has a certain position, acceleration, velocity at any particular position x and time t.

 

[vanhees71]

Yes, but the mapping doesn't need to be unique. The "state of the particle", i.e., its point in "phase space" $(x,v)$ is a unique function of time, but it is not necessarily to be expressed in terms of a unique function $v=v(x)$. As I said, a pretty simple but not trivial example, is the harmonic oscillator!

 

[etotheipi]

Simply substitution can always take us from $v(t)$ to $v(x)$...

You can't do that in the general case. If the position increases monotonically, then yes, you can construct a function $v(x)$. But if you, for instance, imagine wiggling the particle along the $x$-axis, crossing over the origin many times, each time with a different speed (and a reversed direction, too), then many different $v$ will be associated with each $x$, and you cannot define a function $v(x)$.

If we go back to the simpler example of SHM, then$$x(t) = x_0 \cos{\omega t},\quad v(t) = - x_0 \omega \sin{\omega t}$$Eliminating $t$, as you suggest, takes you to$$v(x) = \pm \omega \sqrt{{x_0}^2 - x^2}$$Notice that you cannot give a unique $v(x=0)$, for instance. This $v(x)$ is not a function.

 

[fog37]

You can't do that in the general case. If the position increases monotonically, then yes, you can construct a function $v(x)$. But if you, for instance, imagine wiggling the particle along the $x$-axis, crossing over the origin many times, each time with a different speed (and a reversed direction, too), then many different $v$ will be associated with each $x$, and you cannot define a function $v(x)$.

If we go back to the simpler example of SHM, then$$x(t) = x_0 \cos{\omega t},\quad v(t) = - x_0 \sin{\omega t}$$Eliminating $t$, as you suggest, takes you to$$v(x) = \pm \omega \sqrt{{x_0}^2 - x^2}$$Notice that you cannot give a unique $v(x=0)$, for instance.

Ahhh...Ok. I see now. Thank you!

 

[vanhees71]

Note the typo in my original posting #2, which I've now corrected there!

 

[fog37]

Hello again. I am marinating the concepts discussed in this thread. It may be just semantics and concepts, but when the acceleration $a$ in Newton's second law is exclusively a function of position, $a(x)$, i.e.
$$\frac {d^2 x} {dt}^2= a(x)$$
Conceptually, object experiencing this type of acceleration will move in such a way that and its acceleration may vary in position and time but when the object passes through the same spatial position $x$ its acceleration will be the same. From the perspective of an observer watching the object move, the object has a velocity $v$ and an acceleration $a$ that change with both time and position since the object occupies different positions at different instants of time even if the acceleration value is the same at the same locations.
Eventually, the goal is to solve the equation to find the function $x(t)$ from which we can easily find $v(t)$ through the 1st derivative. Is my reasoning correct?

Example 2: different situation, when $a(t)$, leads to $$\frac {d^2 x} {dt}^2= a(t)$$

This would mean that as the object moves, its acceleration varies with time but the acceleration behavior is not tied to where the object is found at a particular moment in time. For example, car erratically moving with varying acceleration. Or a rocket taking off and increasing its speed as it climbs in altitude is an object. Both are examples of time-dependent acceleration $a(t)$. The SHO, on the other hand, imposes that the acceleration is $a(x)$ a function of position.

Thanks for the patience. If we looked at the velocity and acceleration of an object from the perspective of the moving object itself, say a car, (imagine reading the output from a speed and from an acceleration sensors attached to the moving object) both velocity and acceleration seem to be both made to be both functions of position and time by simply substituting $x(t)$ into $a(x)$. The car driver would be able to communicate either at every location or moment in time both the car velocity and acceleration.

 

[vanhees71]

The correct notation is
$$a=\ddot{x}=\frac{\mathrm{d}^2 x}{\mathrm{d} t^2}.$$
The acceleration-vector component is the 2nd time derivative of the position-vector component. Here the components are read as functions of time, i.e., the description of the trajectory the particle moves as a function of time.

What you usually have in Newtonian physics, however, is that you need to solve a differential equation, i.e.,
$$m \vec{a}=m \ddot{\vec{x}}=F(\vec{x},\dot{\vec{x}},t).$$
Only after solving this equation of motion you know the trajectory $\vec{x}=\vec{x}(t)$.

 

[Redbelly98]

Summary:: Understanding velocity v as a function of time t and position x

Hello Forum,

Limiting our discussion to 1D motion, it is clear that the concept of instantaneous velocity is defined as the covered displacement dx divided by the time interval elapsed dt:

$$ v = \frac {dx}{dt}$$

However, mathematically, the velocity $v$ can be made to depend on any parameter $t,x,a$. For example the velocity $v$ can be expressed a function of time $v(t)$ but also as a function position $v(x)$. If we know the position function $x(t)$, we can substitute that into $v(x) = v( x(t))$ to obtain $v(t)$. If $x(t) =3t+2$ and $v(x) = x ^2$, then $v(t)= (3t+2)^2$.
.
.
.

Note that for this example $x(t) =3t+2$, v must equal 3 at all times. So you can't set v equal to just any arbritary function of x.

 

[vanhees71]

This doesn't make sense dimensionally. Perhaps you mean
$$x(t)=v t + x_0, \quad v=3 \text{m}/\text{s}, \quad x_0=2 \;\text{m}?$$

 

[fog37]

Note that for this example $x(t) =3t+2$, v must equal 3 at all times. So you can't set v equal to just any arbritary function of x.

Yes, sorry, bad example.

I guess the end goal is always to solve the equation of motion (Newton's 2nd law) to obtain the kinematic behavior of the body which is given by the functions $x(t), v(t), a(t)$. These functions are the position, velocity an acceleration parametrized by the parameter time $t$. The three functions are related by time derivatives.

We can surely re-parametrize both velocity and acceleration $v,a$ using the position $x$ as parameter to get $v(x), a(x)$ by simple substitution.

It appears that the acceleration can then be either expressed as a function of time or as function of position: $a(t)$ or $a(x)$. However, when we use the acceleration in the equation of motion for the SHO, the acceleration $a$ is truly $a(x)$, i.e. a function of position ONLY. It is like one case is the acceleration we would obtain if we asked the moving body and the other (only position of position) is the acceleration if we asked its cause which is the force... That has been the source of my confusion...

 

[fog37]

Note that for this example $x(t) =3t+2$, v must equal 3 at all times. So you can't set v equal to just any arbritary function of x.

Sorry, that example does not make sense...

 

[fog37]

Hello! I am back on the same topic. I truly appreciate the contributors' patience. Sorry if I am beating a dead horse but something will click soon. I know I am almost there :)
There is one single acceleration $a$ experience by the body and that acceleration's origin is a force.

STEP 1: Every 1D motion problem starts with Newton's 2nd law, which is a 2nd order ODE.

$$\frac{d^2 x}{dt^2} = a$$
The right end side is the acceleration function $a$ which can be a function (does not have to be a one-one function) like:

1. $a =$ constant
2. $a(t)$
3. $a(x)$
4. $a(v)$
5. $a(x,t)$
6. $a(x,t,v)$
7. $a=(x,t,v, etc.)$ i.e. the acceleration can also depend on higher order derivatives (jerk, etc.)

STEP2: Once we have the acceleration and its dependence, we solve the ODE above to obtain the functions $x(t)$, and $v(t)$ and $a(t)$ by differentiation. The three equations $x(t), v(t), a(t)$ do not need to be one-one functions, correct?

My dilemma has been, for example, that when $a(x)$, the acceleration is solely position dependent and this specific dependence "drives" the equation to general forms of the solution functions $x(t)$, $v(t)$, $a(t)$ whose particular form depends on the initial conditions ICs.

For the acceleration $a(x)$, would we call its position dependence its direct and explicit dependence?
When we analyze $a(t)$, would we call its time dependence implicit?

 

[vanhees71]

Again: The function $\vec{r}(t)$ describes the trajectory of a particle described by this position vector in an inertial frame of reference. Mathematically it is a function, i.e., for any $t$ there's a unique vector $\vec{r}(t)$.

The equation of motion according to Newton's laws is a differential equation,
$$\ddot{\vec{r}}=\vec{F}(\vec{r},\vec{v},t),$$
where $\vec{F}$ is given as a function of position, velocity $\vec{v}=\dot{\vec{r}}$, and maybe explicitly on $t$. It is understood that you look for a solution $\vec{r}(t)$ for the trajectory of the particle for given initial conditions $\vec{r}(t_0)=\vec{r}_0$ and $\vec{v}(t_0)=\vec{v}_0$.

The only example in the literature I'm aware of, where the function depends on higher derivatives of $\vec{r}$ is the Lorentz-Abraham-Dirac equation for the motion of a charged particle including its reaction with its own electromagnetic field ("radiation reaction"). This equation is, however, flawed by severe problems and must be seen as an exception in the usual equations of physics. So don't worry about such exotics right now.

 

[fog37]

Again: The function $\vec{r}(t)$ describes the trajectory of a particle described by this position vector in an inertial frame of reference. Mathematically it is a function, i.e., for any $t$ there's a unique vector $\vec{r}(t)$.

The equation of motion according to Newton's laws is a differential equation,
$$\ddot{\vec{r}}=\vec{F}(\vec{r},\vec{v},t),$$
where $\vec{F}$ is given as a function of position, velocity $\vec{v}=\dot{\vec{r}}$, and maybe explicitly on $t$. It is understood that you look for a solution $\vec{r}(t)$ for the trajectory of the particle for given initial conditions $\vec{r}(t_0)=\vec{r}_0$ and $\vec{v}(t_0)=\vec{v}_0$.

The only example in the literature I'm aware of, where the function depends on higher derivatives of $\vec{r}$ is the Lorentz-Abraham-Dirac equation for the motion of a charged particle including its reaction with its own electromagnetic field ("radiation reaction"). This equation is, however, flawed by severe problems and must be seen as an exception in the usual equations of physics. So don't worry about such exotics right now.

I see how the function $r(t)$ is unique since the particle cannot be in two places at the same time but the function $r(t)$ does not have to be one-one since the particle can return to the same position later.

 

[fog37]

So when we state that the acceleration is a only a function of this or of that variable, for example $a(x)$, we are describing how the acceleration of the object generally behaves in a way that is independent of the initial and particular condition of motion. This information is what capture the actual nature of the acceleration. In the case of SHM, the acceleration at each position $x$ remains the same every time the object passes through each position.

Nonetheless, an fixed observer looking at the moving object would correctly state that the object's position, velocity, and acceleration are changing with time as the object moves back and forth in SHM if we could read from afar an accelerometer mounted on the object. The fact that we can express $a((x(t))= a(t)$ is a true fact but it does not highlight the true nature of the acceleration...Would we call the time dependence of the acceleration? Is it a "trivial" dependence?

 

[jbriggs444]

So when we state that the acceleration is a only a function of this or of that variable, for example $a(x)$, we are describing how the acceleration of the object generally behaves in a way that is independent of the initial and particular condition of motion. This information is what capture the actual nature of the acceleration. In the case of SHM, the acceleration at each position $x$ remains the same every time the object passes through each position.

Nonetheless, an fixed observer looking at the moving object would correctly state that the object's position, velocity, and acceleration are changing with time as the object moves back and forth in SHM if we could read from afar an accelerometer mounted on the object. The fact that we can express $a((x(t))= a(t)$ is a true fact but it does not highlight the true nature of the acceleration...Would we call the time dependence of the acceleration? Is it a "trivial" dependence?

Are you trying to ask which is correct, whether acceleration in simple harmonic motion is really a function of time or really a function of position?

The answer is yes. In this case it really is a function of both. There is no underlying "true nature", whatever that term is supposed to mean.

 

[fog37]

Thanks jbriggs444. I get the "true nature" term is quite vague and meaningless.

So when the acceleration is stated to be a function of a certain variable, for example, SHM has $a(x)$, air drag has $a(v)= - C t$, etc., why there is no mention that acceleration is also a function of time, even if acceleration is also a function of time if the object has acceleration and is moving?

What would be an example of a situation where the object has acceleration that depends on both position and time, i.e. $a(x,t)$?

 

[jbriggs444]

why there is no mention that acceleration is also a function of time

It is unhelpful to mention everything that could be mentioned every time that thing could be mentioned.

It is useful to derive the fact of simple harmonic motion from a situation where mass is constant and force is directly proportional to displacement from an equilibrium position. Mentioning the relationship between acceleration and the growth rings in the Sequoias in California would be unhelpful, even though such a relationship exists.

 

[fog37]

Fair enough, thanks!