# How is time defined in astronomy?

I was amazed when my teacher told me that in space, time is define as a coordinate. It means we can define something with a four coordinate system with "negative time" $$-t$$ as easily as we can say $$-x$$. Assuming that we can, what will this suggest?

Edit:) yesterday I saw so many answer on it, create one more doubt in me, that how do I define a origin in space, because there nothing to which I compare, suppose I have two events A and B, how can I say A occur before B, if I say this, from which reference I am comparing it....

• I don't think your teacher meant "space" as related to astronomy. "Spacetime" is a physical term, and it applies to events happening everywhere, including around and inside you. So this looks like a question for physics, not astronomy, site.
– IMil
Oct 31 '19 at 4:01
• Parsecs. Haven't you seen Star Wars? ;-) Nov 6 '19 at 9:53
• No. I am less interested in movies, is there any connection with my answer. Nov 6 '19 at 11:07

$$t$$ signifies time; see the Wikipedia article for spacetime, and then the subsection for 4-vectors.

The basics are pretty natural to understand. Suppose something happens, an event, like an apple falling off of a tree.

In order to tell someone else about it you need the three space coordinates $$x, y, z$$ and the time coordinate $$t$$. Without all four, you won't be able to see it happen.

update: The question was clarified to focus on the meaning of "negative time" or $$-t$$. That's no big deal. There's no problem with $$-x$$ for example. We don't think of that as negative distance or negative space. It's just a coordinate relative to some origin $$(0, 0, 0)$$, and we can put that anywhere, arbitrarily.

So you should really thing of the time coordinate as $$t-t_0$$ where $$t_0$$ is some arbitrary origin in time. It might be midnight yesterday, or midnight next Tuesday. That would make $$t_{Now}$$ positive in the first case and negative in the second case, but either way it's still Now.

• Sir I updated my question. Oct 30 '19 at 11:44
• @yuvrajsingh Oh! I see that the minus sign was a key part of your question, not just a typo! Okay I've edited further - I hope it's still clear - and I'll revise my question
– uhoh
Oct 30 '19 at 11:51
• @yuvrajsingh by the way, in Stack Exchange there's no need for things like "Sir".
– uhoh
Oct 30 '19 at 11:57
• A good example of negative time is in situations like spacecraft launches, where they'll say things like "T minus 5 minutes" meaning the current time is 5 minutes before "T", generally defined as the liftoff time. Oct 30 '19 at 21:45
• @TinMan that is an excellent example! It's such a familliar and good one that I think you can write it up as a separate supplemental answer. You can cite answers to What “actually” happens at T-minus-0 and also Why doesn't Ariane-5 lift off at T=0 seconds? and possibly Do Blue Origin's BE-3 engines need to run for 7 seconds to “warm up”? for sources.
– uhoh
Oct 31 '19 at 5:16

Suppose we have a homogeneous cube and we want to construct some axes so we can locate points in the cube. The obvious choice is to put the origin at the centre and use axes that are like this:

Suppose we have two points $$A$$ and $$B$$ then we can locate them by their position vectors $$\mathbf a$$ and $$\mathbf b$$ where $$\mathbf a$$ would be something like $$\mathbf a = (a_x, a_y, a_z)$$ and likewise for $$\mathbf b$$. Then the interval between the events would just be $$\mathbf a - \mathbf b$$.

If we chose a different origin then the position vectors $$\mathbf a$$ and $$\mathbf b$$ would change but their difference $$\mathbf a - \mathbf b$$ would not, so the interval between the points does not depend on where we place our origin.

Now suppose we have an infinite universe instead of a finite cube. An infinite universe doesn't have a centre so there isn't an obvious choice of where to put our origin. But as explained above it doesn't matter where we put the origin because the interval between the points is independent of the choice of origin. So we can put the origin anywhere we find convenient.

So far my axes have been spatial, but in relativity time is just another axis and the universe is a four dimensional manifold. That is, we choose four axes $$t, x, y, z$$ and identify points by their position vector $$\mathbf a = (a_t, a_x, a_y, a_z)$$. So the interval between our two points will be:

$$\mathbf a - \mathbf b = (a_t - b_t, a_x - b_x, a_y - b_y, a_z - b_z)$$

And as before this interval vector does not depend on our choice of where to put the origin. We can move the origin in space and in time and the interval stays the same. So when you ask:

suppose I have two events A and B, how can I say A occur before B

you just look at the time component of the interval, $$a_t - b_t$$. If this is negative then $$A$$ happened before $$B$$ and if it's positive then $$A$$ happened after $$B$$. And this result doesn't depend on what point in space or time we put the origin.

I'll mention one last point just for interest. In the above I have assumed we stay in the same inertial frame, and in that case what I've said is true that the time component of our interval is always the same. And likewise the $$x$$, $$y$$ and $$z$$ components of the interval don't change when we move our origin. However if we transform to a different inertial frame, i.e. do a Lorentz transformation, this causes our interval vector to rotate in the new axes. It's still the same vector, so it's length (technically its norm) is not changed by the transformation, but the individual components $$(t,x,y,z)$$ will be different in the new coordinates. This rotation happens in the time direction as well, so the time component $$a_t-b_t$$ will change. If the two points $$A$$ and $$B$$ are spacelike separated then the time component can even change sign. That is for spacelike separated points a Lorentz transformation change the time order of the events.

• Thanks sir, brilliantly explained. Nov 2 '19 at 13:01

A negative time coordinate means nothing more or less that

"this happened before the clock read zero".

This is exactly the same as years "BC" (which, stipped of cultural baggage means "before the time we've assigned as calandar zero", right?).

• Common Era doesn't have a year zero. Oct 30 '19 at 20:58
• Can you elaborate, because I find it difficult to understand. Oct 31 '19 at 4:03
• -1 for using a discontinuous time scale as an example for a problem that requires a continuous time scale. Since it seems that year enumeration goes from -1 to +1, there's roughly a $\pi \times 10^7$ second discontinuity in your example.
– uhoh
Oct 31 '19 at 5:21
• A better example for this might be astronomical year numbering, which does have a year 0. Oct 31 '19 at 14:03
• Even though Common Era does not have a "year zero", it does have a "moment zero." Nov 7 '19 at 4:03