15
$\begingroup$

I am trying to figure out how undefined formulas in mathematics relates to physics. Take the following formula for terminal velocity.

$$V_\text{terminal} = \sqrt{ mg \over{c \rho A}} $$

Say we have a an air density of 0; $ \rho = 0 $ (a vacuum)

Logic tells me the particle would continue to accelerate and never reach terminal velocity, but in mathematics this formula would be undefined.

Obviously this is one of many examples of what can happen in physics problems, but what does undefined actually mean in terms of physics? I hope I am explaining myself clearly.

$\endgroup$
  • 8
    $\begingroup$ Hmm, logic tells me that if you insert $\rho=0$ there, then the original premise that lead to $v\propto\sqrt{1/\rho}$ is false in this new case. $\endgroup$ – Kyle Kanos Oct 29 '15 at 16:38
  • 4
    $\begingroup$ More to the point, there are hardly any cases of an actual $0$ in physics (i.e., the vacuum is not so vacuous at all, etc.) $\endgroup$ – Hagen von Eitzen Oct 29 '15 at 22:50
44
$\begingroup$

Yes, the particle would continue to accelerate and would never reach a terminal velocity. But that is not what this equation tells you. This equation tells you what the terminal velocity is, given the parameters of the function. When in a vacuum, there is no terminal velocity. It is not zero, it is not infinity. A terminal velocity literally does not exist and that is exactly what the equation tells you.

$\endgroup$
  • $\begingroup$ When dealing with undefined in a equation does it then tell you that what you are trying to solve can't be solved because it doesn't exist with the given parameters? $\endgroup$ – Uys of Spades Oct 29 '15 at 16:16
  • 8
    $\begingroup$ Either it does not exist or you do not have enough constraints on the problem to produce an answer or you've made some sort of math mistake. $\endgroup$ – Robert Stiffler Oct 29 '15 at 16:19
  • 2
    $\begingroup$ Makes sense. In the case of $a = {F \over m}$ if $m=0$ It would be undefined, because you can't have an acceleration is there isn't anything to accelerate. Thanks! $\endgroup$ – Uys of Spades Oct 29 '15 at 16:21
  • 2
    $\begingroup$ Right, in classical physics a mass of zero blows up a bunch of equations. In special relativity, the equations are different and massless particles can exist (they always move at the speed of light) $\endgroup$ – Robert Stiffler Oct 29 '15 at 16:24
  • 3
    $\begingroup$ A useful way of characterising it in a well-defined way would be in terms of limits: you could say that the terminal speed tends to infinity as the density drops to zero. $\endgroup$ – Holographer Oct 30 '15 at 5:34
9
$\begingroup$

As in math, you get the physical meaning via the limit: it's not "undefined", it's "going to infinity" when $\rho$ goes to 0 from above. (which is the only way in the physical world, that why its well defined there).

$\endgroup$
  • 2
    $\begingroup$ Unless my math is that rusty; isn't anything divided by 0 undefined? $\endgroup$ – Uys of Spades Oct 29 '15 at 16:12
  • 5
    $\begingroup$ "undefined" often come from the fact you could argue for several different value (e.g. +infinity or -infinity). Now the physics (or any given problem) can make one meaningful and the others meaningless. Typically for a density, viscosity, pressure, it makes sense to see what happens when these go from a positive value to a lesser and lesser value arbitrary close to zero. Then you can say that pratically this give the value in zero. $\endgroup$ – Fabrice NEYRET Oct 29 '15 at 16:16
  • 2
    $\begingroup$ I'd argue that this should start out "Unlike in math" but otherwise I agree. $\endgroup$ – David Z Oct 29 '15 at 19:52
  • 2
    $\begingroup$ @Octopus - Infinity multiplied by zero is not zero. It is an indeterminate form. $\endgroup$ – David Hammen Oct 30 '15 at 8:00
  • 2
    $\begingroup$ @FabriceNEYRET - I added a tiny bit bit to your answer. I changed "when $\rho$ goes to 0" to "when $\rho$ goes to 0 from above". $\endgroup$ – David Hammen Oct 30 '15 at 8:05
6
$\begingroup$

The issue here is that the existence of terminal velocity depends on the assumption that the particle is moving through a fluid. As real fluids have only positive non-zero densities, simply plugging in 0 breaks this assumption.

You could say that the classical limit of terminal velocity as the fluid approaches vacuum (i.e., as the density approaches 0) is infinity, but this limit is not a real number, so the particle in a vacuum has no terminal velocity. (As pointed out in the comments, if we take relativity into account, this limit becomes c, the speed of light in vacuum. Incorporating relativity into the formula would make it needlessly complicated for practical purposes.)

$\endgroup$
  • $\begingroup$ Well, the classical velocity is unbounded, obviously relativity would say $V_{term}\to c$. But surely your 2nd sentence is key to OP's problem. $\endgroup$ – Kyle Kanos Oct 29 '15 at 17:31
  • $\begingroup$ @Kyle Good point. Edited to take this into account. $\endgroup$ – Jacob Hunter Oct 29 '15 at 17:47
5
$\begingroup$

Obviously this is one of many examples of what can happen in physics problems, but what does undefined actually mean in terms of physics? I hope I am explaining myself clearly.

The mathematical formulas used to model physical observables are valid in specific frameworks, where they have fitted observations and predicted new observations successfully. An undefined or an infinity means that the mathematical model fails and a new physics model/theory has to be found. In your example the model fails for 0 density and either a new formula is needed for the vacuum, or the target ( terminal velocity) is meaningless in vacuum.

Examining whether there exists a terminal velocity in vacuum leads to the theory of special relativity. A particle in vacuum can never reach the velocity of light, and it is a completely new model of particle behavior that has been validated innumerable times.

Elementary particles are point particles, the infinity of the 1/r electric potential at r=0 classically is defused by the development of quantum mechanics, a completely new framework underlying all nature.

So undefined mathematics points to new theories/models needed.

$\endgroup$
  • 1
    $\begingroup$ +1 Mathematical formulas are models, nothing more! They do not control reality, they predict it, and they can predict it imperfectly! $\endgroup$ – Cort Ammon Oct 30 '15 at 19:48
4
$\begingroup$

Your question indeed very clear,a good example, and cuts to the heart of physics - being able to define, predict the world using mathematical models. But you have to make sure your models are correct - and complete.

Fabrice and Robert are right - velocity will go to infinity without constraints. And there is the clue as to what's going on as a physics interpretation - constraints. The equation you cite is the steady state solution - velocity of a small mass accelerating towards a larger body with gravitational acceleration g, with a projected area, $A$, and in an atmosphere with density $\rho$.

Remove the atmosphere, and as you say there is no longer a drag force, and so the initial dynamic model from which you derived this steady state formula is different. But when you let $\rho$ go to zero it's (almost) giving you the right answer. I say almost because the equation you cite is not complete without an additional equation: one that constrains the path of the attracting bodies to the point at which they collide - a hard limit. Things can't fall forever - they eventually hit the ground. That's the real and complete physics of the matter.

A caveat: "Things can't fall forever" this might not be true for a black hole. Time and space are so warped it may take 'forever' for the object to reach the singularity according to clocks outside the black hole. But these physics require additional mathematics to model the behavior.

$\endgroup$
1
$\begingroup$

It essentially means "the question is nonsensical, so don't ask."

For example, in this particular case, we're trying to find terminal velocity. Terminal velocity is the speed at which acceleration due to gravity and wind resistance cancel each other out. If there is no atmosphere to provide wind resistance, then by definition such a speed can never exist, and so there is no way to define what the terminal velocity would be in such a situation.

$\endgroup$
  • 1
    $\begingroup$ But he did ask. And you made sense of his question. $\endgroup$ – Octopus Oct 29 '15 at 22:30
  • $\begingroup$ Perhaps it could be put instead as “The answer is not a (real) number.” $\endgroup$ – Kevin Reid Oct 30 '15 at 1:46
1
$\begingroup$

The equation you gave is an approximation for a situation where $\rho$ is non-zero and gravity is constant, and represents the limit of velocity that would be approached by an object (note - this value would never be reached, just approached).

How long it takes to get there can also be calculated (again on the assumption of constant density and gravity). The equations have a "characteristic time" $\tau= \frac{v_t}{g}$ in them - in other words, the concept of "terminal velocity" doesn't even start to have any meaning until a time equal to several $\tau$ has passed.

A few things to note then:

  • gravity is not constant: and if you compute the velocity achieved by an object in free fall from "infinity", and assume all friction due to atmospheric drag to be zero, the object will reach a velocity of only about 11 km/s - that's all the potential energy converted to kinetic.

  • with drag present, an object will have to fall for a length of time equal to several $\tau$ to get even close to terminal velocity (for the acceleration to slow down).

  • at extreme velocities, relativistic effects would begin to play a role

None of these things are considered in your simple treatment of the terminal velocity equation.

When a physical equation gives us what appears to be a nonsensical answer, it usually means our understanding of (or description of) the physics described by the equation is incomplete. A famous example of this was the Ultra-Violet catastrophe - the resolution to which was an important factor in the formulation of quantum mechanics.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.