Whenever we graph time along a $x$-axis, we graph it along negative and positive $x$-axis both. So what does a negative time mean in the negative case axis? What does the negative sign indicate?
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3 Answers
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Suppose you work for NASA and are the person who is given the task of announcing the time to the launch of a particular rocket. You will call this time $t=0$, and so five seconds before the launch you will announce "t minus 5..4..3..2..1..ignition". Now this time $t=0$ could have been for example 28th February 12 noon, and everything before that particular event exists on the "negative time" axis, or all points in time before this event.
We do the same thing always in classical mechanics. We can define any point to be $t=0$ and it simply defines what will happen/happened at this point, and everything prior to this point is the past and everything after it is the future.
Consider also a case where we have a position versus time graph, which shows an object with displacement according to the equation $$x(t)=(5t^2+1)\,\mathrm{m},$$ where $\mathrm{m}$ represents metres and $t$ is measured in seconds. You are told its displacement at say $t=1\,\mathrm{s}$ is $6\,\mathrm{m}$. You can confirm this by substituting $t=1\,\mathrm{s}$ into this equation. But now you want to know what was its displacement at $t=-2\,\mathrm{s}$. You can once again substitute this value for $t$ into the same equation and you will know that the object had a displacement of $21\,\mathrm{m}$, two seconds before it reached the axis defined by $t=0\,\mathrm{s}$.
"Negative time" simply means all points in time before a specific event, that we say will happen at $t=0\,\mathrm{s}$.
We do a very similar thing in relativity with spacetime diagrams. In such cases, the spacetime origin is represented by $x = 0$ and $t = 0$ and represents the present time and location of the observer (in that reference frame). Events with $t > 0$ are in the future, and events with $t< 0$ are in the past of this observer. We can choose the location of the origin to make the solution of a problem as simple and convenient as possible when dealing with regular dynamics.
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1$\begingroup$ I never before understood that t minus 5 stuff. I always interpreted it as meaning $(t - 5)$, not $t = -5$. $\endgroup$– lvellaCommented Mar 10, 2021 at 14:35
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1$\begingroup$ @lvella you understood correctly, it is a big T for time or L for launch (T-5) or (L-5) in countdowns nesdis.noaa.gov/content/… so t=T-5 or t= L+5 , not T=-5. Launch is planned for L-0 and happens at T-0. Of course, T could be arbitrarily set to be the origin, as would be done in any plot related to the launch, and t could be negative as in history.nasa.gov/afj/ap08fj/pics/buildup.gif $\endgroup$ Commented Mar 10, 2021 at 15:05
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In the simplest possible terms, negative time means last year, last month, last week, yesterday, one hour ago, one minute ago, one second ago. It refers to the past.
answered Mar 10, 2021 at 3:58
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$\begingroup$ What news with the scan? $\endgroup$ Commented Mar 10, 2021 at 13:54
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$\begingroup$ @Larsaseeidaklaxtarsa, thanks for asking- the scan came back clean this time, now we monitor. next scan will be in 4 months. $\endgroup$ Commented Mar 10, 2021 at 17:13
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$\begingroup$ ... implicit assumption $t = 0$ is "now", which is often not the case. $\endgroup$ Commented Mar 10, 2021 at 23:15
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$\begingroup$ @comptonscattering, yes, that was implicit. $\endgroup$ Commented Mar 11, 2021 at 1:31
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The convention is that positive time is the future and negative time is the past. If you assume this convention then you can explicitly show that Newton's laws are time reversible or time symmetric.
A ball is thrown.
$y=v_{oy}t - \frac{g}{2} t^2$
$x=v_{ox}t$
A film of the thrown ball is played in reverse. $t \to -t$
$y=-v_{oy}t - \frac{g}{2} t^2$
$x=-v_{ox}t$
In both cases, $m\frac{dy^2}{dt^2}=-mg$.
If you watch a film of a ball moving along such a path then you can't tell whether the film is going forward in time or backward. Both motions are consistent with Newton's Laws and nothing looks unusual if time is reversed.
So, yes, it is ultimately a convention, but it is a convention that works. Obviously, you could completely reverse the convention and that would be consistent with Newton's Laws.
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1$\begingroup$ If you're interested in the physics of time then Hans Reichenbach's The Direction of Time and Roger Penrose's Cycles of Time are worth reading. $\endgroup$– user288901Commented Mar 10, 2021 at 5:28
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$\begingroup$ And there is a lot more to be said about this kind of time symmetry in physics. Victor Stenger wrote a whole book about it. Timeless Reality. Feynman and Wheeler solved the self energy of the electron problem by assuming that electromagnetic fields propagate from the past. etc. $\endgroup$– user288901Commented Mar 10, 2021 at 5:56
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$\begingroup$ "nothing looks unusual if time is reversed." This doesn't seem consistent with experience. If throw a tennis ball across a court, it will bounce at decreasing heights until it rolls along. It's pretty obvious that it looks unusual when played backwards. How does that fit into what you are saying here? $\endgroup$ Commented Mar 10, 2021 at 16:23
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$\begingroup$ @JimmyJames - they are being imprecise or careless in their speech. If the system is lossless then nothing looks unusual if time is reversed. When you bring in friction - like when the ball hits the court ground and loses some kinetic energy - then your point becomes valid. The people who do physics using only math and rarely involve themselves with physical world tend to forget reality has friction. The equations that 'Directions in Physics' is using are nice equations but they do not model friction. $\endgroup$– anonCommented Mar 10, 2021 at 21:25
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$\begingroup$ @EricM Perhaps. Thermodynamics also seems relevant here. I guess I'm just trying to understand the claim. Is it just Netwon's laws excluding such factors? From prior reading and study, I think there's something to this but I've never been able to square these claims with real experience. I'm not asserting they are wrong, just that the 'films of motion' that would appear normal played backwards are a very small subset of all such films. $\endgroup$ Commented Mar 10, 2021 at 21:36