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I Parametrization and kinematics

  1. Sep 22, 2016 #1
    Hello Forum,
    In kinematics, the important variables are the velocity v, the acceleration a, and the object's position x. These variables are usually presented as functions of time: x(t), v(t) and a(t).

    The acceleration can either be constant, or vary with time, i.e. a(t), or vary with position, a(x) or even vary with velocity, i.e. a(v).
    An object that moves occupies different spatial positions x at different instants of time t. If the acceleration a(t) varies with time, the object will have a different accelerations at every different spatial position, we can interpret the acceleration also as a(x). But an acceleration that depends on time and one that depends on space are different things. Parametrically, we can always express the acceleration in terms of any other independent variable (t, v, x). But how can we distinguish between the different cases where the acceleration has a time or space dependence that is explicit or implicit if parametrization can always convert the function from one to the other?

    For example, say a(t) = 3t. We solve for x(t) and express t(x). The replace t in the equation for a(t) and get a(x). That is different from starting with an acceleration a(x)...

    Thanks.
     
  2. jcsd
  3. Sep 22, 2016 #2

    andrewkirk

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    It's best to avoid writing anything like ##a(t)## and ##a(x)##. It often leads to hopeless confusion, especially when it comes to differentiation. We have amounts (aka variables) and we have functions, and the two should not be confused. A function is a rule that, given one amount, returns another.

    So we need to decide whether we will regard ##a## as an amount or a function and stick to that. If it's an amount, we should avoid writing anything like ##(t)## or ##(x)## after it.

    With that out of the way, here's how to address the question. Let ##a,v,x,t## be amounts with the usual meanings and let there be functions ##f## and ##g## such that ##a=f(t)## and ##x=g(t)##. Then if we want to express ##a## in terms of ##x## we write
    $$a=f(g^{-1}(x))$$
    where ##g^{-1}## is the inverse function of ##g##, ie ##g^{-1}(g(t))=t##.

    Note that a function ##h## only has an inverse function if it is one-to-one, ie ##h(u_1)=h(u_2)\Rightarrow u_1=u_2##. But often when that does not hold we can get a 'local' inverse function by restricting the domain of the function.
     
  4. Sep 24, 2016 #3

    Stephen Tashi

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    It's possible for that to be true, but, for example, an object might move in figure-eight path, so that there is a point in space where it occupies the same spatial position at different times.

    If quantity A is parameterized by parameter s and quantity B is parameterized by s, we can't necessarily parameterize quantity A by quantity B since the same value of B might occur at two different values of s. For example suppose ## A = s + 3, \ B = s^2## The general mathematical approach to finding whether A can be parameterized by B would involve "the implicit function theorem".

    Is your question motivated by a topic in physics like Hamiltonian mechanics or generalized coordinates? For example, when texts deal with differential equations like ## \frac{dp}{dt} = -\frac{\partial H(q,p,t)}{\partial q} ## they treat ##p## and ##q## as independent variables. But there are possible paths for particle where ##q## might be (mathematically) a function of ##p##.
     
  5. Oct 3, 2016 #4
    Hello,
    thanks for the replies. I guess I am envisioning two different situations and trying to understand how an object would behave kinematically in each of them. The first situation is described by $$ \frac{dx^2} {dt^2} = a(x) $$ and the second by the $$ \frac{dx^2} {dt^2} = a(t) $$.

    In the first case the acceleration is a function of time t and in the other a function of position x. If we considered force fields, which extend over wide region of space, a force field ##F ## could be ##F(x)##, i.e. it the force and the acceleration ##a## would vary from point to point but remain constant in time at any position ##x##. In the case of a force field ##F(t)##, the force and acceleration are the same everywhere in space but changes in time.

    But let consider situations not described by a force field. For instance, a car can accelerate and its acceleration vary with time. As the car moves, it occupies different positions ##x## at which the acceleration will be different. That seems to imply that the acceleration can be a function of time and a function of position at the same time. Another example is Hooke's law for an ideal spring which states the acceleration is a function of the position ##x##: the object connected to the spring experiences a different acceleration at different positions ##x##. But the object is at different ##x## positions at different instants of time## t## which makes me think that its acceleration is also a function of time...

    Thanks.
     
  6. Oct 4, 2016 #5

    Stephen Tashi

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    The terminology of physics permits an ambiguity that pure mathematics doesn't like. In physics we think of something like "acceleration" as the result of a measurement not as denoting a specific function. So the letter "a" is used ambiguously to denote acceleration as a function of different things. From the point of view of pure mathematics we shouldn't name two different functions with the letter "a". Trying to make the distinction between two different functions by using a different name for their arguments (e.g. "a(t)" vs "a(x)") isn't correct from the point of view of pure mathematics because the name of the "dummy variable" in a function is arbitrary. The function a(x) = x^2 + 2 is the same function as the function a(t) = t^2 + 2.

    The different situations you describe deal with different functions that share the name of "a( )".

    Where things get confusing (to me) is when physicists differentiate expressions involving variables that have suggestive names like "acceleration", "position", "momentum", "velocity" and have in mind that these variable are specific functions. (e.g. "momentum" might mean "momentum as a function of time" as opposed to "momentum as a function of velocity".) You have to look at the derivatives and partial derivatives that they write and those that they leave out in order to determine what is a function of what.
     
  7. Oct 4, 2016 #6

    PeroK

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    There's definitely a relationship between any two of the quantities: time, position, velocity and acceleration. In many cases, you could express any one of these as a function of any other. But note that a "function" implies that there is a single well-defined value for every value fo the argument.

    The good thing about time is that you can't get the same time twice, so everything is always a function of time. But, position can be repeated. For example, if we throw an object upwards under gravity, we get the same position twice - on the way up and on the way down - with equal but opposite velocities in each case. Here there is a relationship between position and velocity, which is a function on the way up and a slightly different function on the way down:

    ##v = u - gt##

    Gives velocity as a function of time.

    ##v = \pm \sqrt{u^2 - 2gx}##

    Is a relationship between velocity and position, but it's not (strictly speaking) a function of position. Instead:

    ##v = \sqrt{u^2 - 2gx}##

    Gives velocity as a function of position on the way up. And:

    ##v = -\sqrt{u^2 - 2gx}##

    Gives velocity as a function of position on the way down.
     
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