<p><span class="math-container">$t$</span> signifies time; see the Wikipedia article for <a href="https://en.wikipedia.org/wiki/Spacetime" rel="noreferrer">spacetime</a>, and then the subsection for <a href="https://en.wikipedia.org/wiki/Spacetime#4%E2%80%91vectors" rel="noreferrer">4-vectors</a>. </p>

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

<p>In order to tell someone else about it you need the three space coordinates <span class="math-container">$x, y, z$</span> and the time coordinate <span class="math-container">$t$</span>. Without all four, you won't be able to see it happen.</p>

<hr>

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

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

CC BY-SA 4.0
2019-10-30T11:44:37.923Z
Sir I updated my question.
आर्यभट्ट
CC BY-SA 4.0
2019-10-30T11:51:01.653Z
@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
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2019-10-30T11:57:50.553Z
@yuvrajsingh by the way, in Stack Exchange there's no need for things like "Sir".
uhoh
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2019-10-30T21:45:27.593Z
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.
Tin Wizard
CC BY-SA 4.0
2019-10-31T04:50:16.180Z
I went over this very fact explaining negative numbers to my daughter. Start at 0 and go to the right - these are the numbers you know and love. Start at 0 and umm... go to the other way. There's your negative numbers. In a few years I'm going to bust out what happens when you leave the number line to enter the complex plane... :-)
corsiKa
CC BY-SA 4.0
2019-10-31T05:16:22.067Z
@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](https://space.stackexchange.com/q/19232/12102) and also  [Why doesn't Ariane-5 lift off at T=0 seconds?](https://space.stackexchange.com/q/39392/12102) and possibly [Do Blue Origin's BE-3 engines need to run for 7 seconds to “warm up”?](https://space.stackexchange.com/q/26994/12102) for sources.
uhoh
CC BY-SA 4.0
2019-10-30T11:30:08.567Z
2019-10-30T11:54:00.030Z
2019-10-30T11:54:00.030Z
uhoh
<p>A negative time coordinate means nothing more or less that </p>

<blockquote>
  <p>"this happened before the clock read zero".</p>
</blockquote>

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

CC BY-SA 4.0
2019-10-30T20:58:47.333Z
Common Era doesn't have a year zero.
EldritchWarlord
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2019-10-31T04:03:58.853Z
Can you elaborate, because I find it difficult to understand.
आर्यभट्ट
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2019-10-31T04:15:31.790Z
@yuvrajsingh why have you accepted this answer if you don’t understand it?
Tim
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2019-10-31T05:21:56.320Z
`-1` for using a discontinuous time scale as an example for a problem that requires a continuous time scale. Since [it seems that](https://en.wikipedia.org/wiki/Common_Era#Origins) year enumeration goes from -1 to +1, there's roughly a $\pi \times 10^7$ second discontinuity in your example.
uhoh
CC BY-SA 4.0
2019-10-31T14:03:18.683Z
A better example for this might be [astronomical year numbering](https://en.wikipedia.org/wiki/Astronomical_year_numbering), which does have a year 0.
ahiijny
CC BY-SA 4.0
2019-11-07T04:03:33.587Z
Even though Common Era does not have a "year zero", it does have a "moment zero."
John Canon
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2019-10-30T18:51:45.317Z
2019-10-30T18:51:45.317Z
dmckee --- ex-moderator kitten
<p>I'd like to pick up the point you ask about in your edit.</p>

<p>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:</p>

<p><a href="https://i.sstatic.net/SPHc3.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/SPHc3.gif" alt="Cube"></a></p>

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

<p>If we chose a different origin then the position vectors <span class="math-container">$\mathbf a$</span> and <span class="math-container">$\mathbf b$</span> would change but their difference <span class="math-container">$\mathbf a - \mathbf b$</span> would not, so the interval between the points does not depend on where we place our origin.</p>

<p>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.</p>

<p>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 <span class="math-container">$t, x, y, z$</span> and identify points by their position vector <span class="math-container">$\mathbf a = (a_t, a_x, a_y, a_z)$</span>. So the interval between our two points will be:</p>

<p><span class="math-container">$$ \mathbf a - \mathbf b = (a_t - b_t, a_x - b_x, a_y - b_y, a_z - b_z) $$</span></p>

<p>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:</p>

<blockquote>
  <p>suppose I have two events A and B, how can I say A occur before B</p>
</blockquote>

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

<p>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 <span class="math-container">$x$</span>, <span class="math-container">$y$</span> and <span class="math-container">$z$</span> 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 <a href="https://en.wikipedia.org/wiki/Norm_(mathematics)" rel="nofollow noreferrer">norm</a>) is not changed by the transformation, but the individual components <span class="math-container">$(t,x,y,z)$</span> will be different in the new coordinates. This rotation happens in the time direction as well, so the time component <span class="math-container">$a_t-b_t$</span> will change. If the two points <span class="math-container">$A$</span> and <span class="math-container">$B$</span> 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.</p>

CC BY-SA 4.0
2019-11-02T13:01:23.453Z
Thanks sir, brilliantly explained.
आर्यभट्ट
CC BY-SA 4.0
2019-11-02T12:52:39.183Z
2019-11-06T08:49:06.927Z
2019-11-06T08:49:06.927Z
John Rennie
<p>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" <span class="math-container">$-t$</span> as easily as we can say <span class="math-container">$-x$</span>. Assuming that we can, what will this suggest?</p>

<p>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.... </p>

CC BY-SA 4.0
2019-10-31T04:01:13.347Z
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
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2019-11-06T09:53:13.250Z
Parsecs. Haven't you seen Star Wars?  ;-)
Phil N DeBlanc
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2019-11-06T11:07:55.133Z
No. I am less interested in movies, is there any connection with my answer.
आर्यभट्ट
<space-time><time>
How is time defined in astronomy?
आर्यभट्ट
# How is time defined in astronomy?

(Asked by: username_0 on 2019-10-30)

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....

**Tags:** space-time, time
> **Comments:**
> **username_1**: 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.  
> **username_2**: Parsecs. Haven't you seen Star Wars?  ;-)  
> **username_0**: No. I am less interested in movies, is there any connection with my answer.  

---

## 3 answers

### Answer by username_11 on 2019-11-02. Score: 7 | Accepted answer

I'd like to pick up the point you ask about in your edit.

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:

[![Cube](https://i.sstatic.net/SPHc3.gif)](https://i.sstatic.net/SPHc3.gif)

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](https://en.wikipedia.org/wiki/Norm_(mathematics))) 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.

> **Comments:**
> **username_0**: Thanks sir, brilliantly explained.  

---

### Answer by username_3 on 2019-10-30. Score: 16

$t$ signifies time; see the Wikipedia article for [spacetime](https://en.wikipedia.org/wiki/Spacetime), and then the subsection for [4-vectors](https://en.wikipedia.org/wiki/Spacetime#4%E2%80%91vectors).

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.

> **Comments:**
> **username_0**: Sir I updated my question.  
> **username_3**: @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  
> **username_3**: @yuvrajsingh by the way, in Stack Exchange there's no need for things like "Sir".  
> **username_4**: 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.  
> **username_5**: I went over this very fact explaining negative numbers to my daughter. Start at 0 and go to the right - these are the numbers you know and love. Start at 0 and umm... go to the other way. There's your negative numbers. In a few years I'm going to bust out what happens when you leave the number line to enter the complex plane... :-)  
> **username_3**: @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](https://space.stackexchange.com/q/19232/12102) and also  [Why doesn't Ariane-5 lift off at T=0 seconds?](https://space.stackexchange.com/q/39392/12102) and possibly [Do Blue Origin's BE-3 engines need to run for 7 seconds to “warm up”?](https://space.stackexchange.com/q/26994/12102) for sources.  

---

### Answer by username_6 on 2019-10-30. Score: 2

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?).

> **Comments:**
> **username_7**: Common Era doesn't have a year zero.  
> **username_0**: Can you elaborate, because I find it difficult to understand.  
> **username_8**: @yuvrajsingh why have you accepted this answer if you don’t understand it?  
> **username_3**: `-1` for using a discontinuous time scale as an example for a problem that requires a continuous time scale. Since [it seems that](https://en.wikipedia.org/wiki/Common_Era#Origins) year enumeration goes from -1 to +1, there's roughly a $\pi \times 10^7$ second discontinuity in your example.  
> **username_9**: A better example for this might be [astronomical year numbering](https://en.wikipedia.org/wiki/Astronomical_year_numbering), which does have a year 0.  
> **username_10**: Even though Common Era does not have a "year zero", it does have a "moment zero."  

---

