# Homework Help: Meaning of Negative Time

1. Aug 28, 2011

### AirForceOne

So my initial thought was that there's no such thing as negative time. Obviously, however, that's not the correct answer, because let's face it; the question can't be that easy.

The question could have started with just the velocity equation and asked us to find the meaning behind the negative time when we set v=0. However, the question did not start with just the velocity equation, leading me to believe that the answer is related to the position equation in some way.

So I'm thinking...position...negative time...well if I plug in t= -sqrt(2) into the position equation, I get -150 meters. I'm lost. I give up. Help?

Thanks.

PS- This question is from a college level dynamics course...the concepts in this chapter involve differential equations...if that helps at all.

Last edited: Aug 29, 2011
2. Aug 28, 2011

### stallionx

Just As well as you can set time a negative number, you can get a negative position, velocity and acceleration.

All above mentioned variables are Vectors and can have positive directions and negative ones.

Putting time negative, tells you what happened t secs ago.

But generally equalities are for after motion is pending.

3. Aug 28, 2011

### AirForceOne

Just because I get a negative position, velocity and acceleration, that doesn't mean negative time is what happened t seconds ago...?

4. Aug 28, 2011

### stallionx

If the functions of time are written for a beginning where t = 0 ; and then what happens, then negative time would be meaningless.

If Everything starts at t=0 and equations are written wrt that, what happened t seconds before the initial conditions would not mean much.

Please try to use the positive roots for time in order not to be confused.

Of course, that is my personal opinion :)

5. Aug 29, 2011

### AirForceOne

So what would the answer to the question be? That time was sqrt(2) secs before the the initial condition of x=20?

6. Aug 29, 2011

### stallionx

For time = t>=0

You can get Negative results for Position vector, velocity Vector and Acceleration vector.

If you take magnitudes of them, you will end up absoluting the found-out value ( location, speed, magnitude of acceleration. )

7. Aug 29, 2011

### Pengwuino

Setting t = 0 is arbitrary and nothing weird happens when you let your problem go into negative time.

Suppose you have a problem where an object is falling from an airplane. Let's say you allow the object to fall for 10 seconds so that it has an initial velocity and has been falling and all that. So at this point you can setup the differential equation or kinematic equations about an object falling down in a constant acceleration. You typically say "Ok so we start at t = 0 and initial velocity = 100m/s" or something. Those are your initial conditions.

What happens when we let t go negative? What happened as the object fell before we started the problem is encoded in the initial velocity = 100m/s (assuming the gravitational acceleration is constant). What happens when you let t go negative tells you what was happening to the object before you decided to declare t = 0 at whatever height you decided to start the problem at. It literally is winding the clock back. As time goes negative, you'll see the velocity slow down and the height (if you're doing with those kinematic equations) will become greater and greater until the object reaches probably around -10s where the velocity will become 0 and the height will come back to where the object fell from.

8. Aug 29, 2011

### JeffKoch

It's straightforward enough, dx/dt=-180-90t^2; at t=0, v=-180 and if you think of this as a moving object, it's moving in the negative x direction with speed 180 at t=0. For larger positive values of t, the speed increases, still in the negative x direction. For negative values of t, speed also increases, still in the negative x direction. So the object is moving in the same direction with variable speed - it first slows to 180, then accelerates again. There's nothing profound about negative time here, it just refers to times prior to what you (arbitrarily) define as t=0.

9. Aug 30, 2011

### 121910marj

it's as simple as" TIME is SCALAR, no direction required.

10. Aug 30, 2011

### WannabeNewton

That depends on if you are talking about time within the context of Newtonian Mechanics or General Relativity.

11. Aug 30, 2011

### Pengwuino

No it is not that simple. Something like electric charge is a scalar but the negative/positive absolutely has meaning and implications.

Even in GR, I wouldn't call time a vector. It's a component of a 4-vector, but like energy, not a vector on its own.

12. Aug 30, 2011

### WannabeNewton

I didn't intend to call it a vector, just a coordinate function. But yes, I see your point.

13. Jun 3, 2012

### treffibug

In truth, I suspect that the last term should have been -x^2 or +x^3 as the velocity in this example starts at -180 and decreases (ie gets speedier: larger in the negative direction), so at no stage does the velocity reach zero anywhere for t>0.

Assuming that the problem was meant to have at least two mathematical stationary points: one for t>0 (ie when the object was meant to slow down to a stop) and t<0 (some time before the object got out of bed that morning) the crux of the revised problem is this: The map (mathematics) is not the territory (real world).

From t=0 the real world body follows that mathematical expression, but before t=0 the body was following some other real world constraint that may have been x=0t (in the case of a stationary body) or x=f(where the person or machine carrying it was at t<0). For t<0 you are looking at a mathematical expression the body was paying no attention to. For example, if you were to walk off a cliff at t=0, there are rules your body would obey and mathematical paths it would follow, but before t=0, you would have to look at some other means than this paltry mathematical expression to predict your position (perhaps a smartphone app that tracked the position of your pocket for t<0).

The real world object only follows this paltry maths/map whilst it is between t=0 and some final t where t(final)>0 when it hits the ground or a wall. Even if it doesn't do this (for example a planet moving in "largely frictionless space") it will slowly wander off the mathematical map path due to friction from cosmic dust & rays, gravitational effects of other heavenly bodies and other real-world annoyances. The complexities of the real world will always get in the way of simplistic and flawed mathematical models.

Last edited: Jun 3, 2012
14. Jun 3, 2012

### Ray Vickson

In problems of the type you cite, negative time just means times before you "start measuring". In the example you gave, IF the same law of motion applied before you decided to start measuring the system (that is, at negative times), your solution giving t = -sqrt(2) would be perfectly OK. Negative time is no more mysterious than negative distance: if I measure distance along a line, starting from some origin, and say that positive distance is to the East, then negative distance is to the West. If, for purposes of describing some system, I start observing at 12:00 noon on Wednesday, negative times are just times before 12:00 Wednesday.

Much of classical Physics has fundamental laws that are invariant under time reversal, but this does not apply to thermodynamics, for example. Over the years numerous solutions to the problem of "the arrow of time" have been proposed, which tries to come to grips with the issue that while the fundamental mechanical laws are time-reversible, in the real world time flows one way only (from "past" to "future"). So, some of these issues are quite deep at a fundamental level, but are not deep in the context of your cited problem.

RGV