# Classical Physics with Time going Backwards

1. Feb 10, 2008

### greeniguana00

First, what does it mean for time to be going backwards? Basically, any time there is a reference to time, we just need to negate it - so time will have a negative effect of what it usually would.

Velocity (1D): There is an object that moved from A to B in time t at a constant speed. If we define the direction from A to B as positive, then using the definition of velocity, its velocity would be (B-A)/t. If time were to go backwards, it would still make a trip between A and B, but this time, time would be going in reverse. So, its velocity would be (B-A)/(-t), which is the negative of its velocity when time goes forwards. By multiplying this by (-1)/(-1), we get (A-B)/t.

To look at what happens more accurately, I think it is necessary to assign a time value to the object at points A and B. So now instead of just the displacement (x) of the object, we are looking at the object at points (x1, t1) and (x2, t2). All the points of the existence of the object are on a 2d plane, with one coordinate being time and the other displacement. This object's entire existence is a static 2d graph. Not only this, but we know the relationship that determines which points on this graph describe the object. We know that the velocity (v) is constant, and the definition of velocity is dx/dt. v=dx/dt, so the velocity of the object is (x2-x1)/(t2-t1). In the previous paragraph, we were saying we were negating (t) to make time go backwards. Here it is very clear what we are doing: we are negating the (t) value for every point of existence of the object. Instead of the points (x1, t1) and (x2, t2) describing the object, the points (x1, -t1) (x2, -t2) describe the object. Velocity is still dx/dt, but this time dx/dt would be (x2-x1)/((-t2)-(-t1)) = (x2-x1)/(t1-t2), which is the negative of what we would get if time weren't negated.

So, an object moving at a constant velocity v would be moving at -v if time were to go backwards, and this is still determined by the equation v=dx/dt.

Velocity (2D): The velocity in two dimensions is the vector sum of the velocities of the object in the x and y direction. The velocities of the object in the x direction and y direction are still determined by dx/dt and dy/dt respectively, and both of those negate when time goes backwards, as previously determined. The vector sum of the velocities when time goes forward would be <dx/dt, dy/dt>. When time goes backwards, it would be <-dx/dt, -dy/dt> which equals -<dx/dt, dy/dt>, so the velocity in two dimensions would also negate if time were to go backwards. You can see this would expand to n dimensions as well.

Acceleration: The best definition of acceleration is (dx/dt)/dt. Unlike velocity, this requires three points to calculate its value if it is constant. So, we know dx/dt represents velocity, so (dx/dt)/dt would represent dv/dt. If we have three sequential points on the graph with coordinates (x1, t1), (x2, t2) and (x3, t3), then average velocity between the first point and second point would be (x2-x1)/(t2-t1), and between the second and third would be (x3-x2)/(t3-t2). Because dv/dt is constant, we know that if v changes some amount from one point to another, it will change that same amount travelling that same distance again. So, while the velocity at points 1 and 2 is not (x2-x1)/(t2-t1), the velocity at their midpoint is the same amount higher than at point 1 as it is lower than at point 2, and that is equal to the average velocity between points 1 and 2. Velocity is not constant, but because we know acceleration is, we know that the midpoint between two points has the same velocity as the average velocity between those two points. So, now we know the velocity at the midpoint of points 1 and 2 is (x2-x1)/(t2-t1) and the velocity at the midpoint of points 2 and 3 is (x3-x2)/(t3-t2). We also know that time at the midpoint between (1) and (2) is (t1+t2)/2 and the time at the midpoint between (2) and (3) is (t2+t3)/2. Therefore, the acceleration of the object from the first midpoint to the second (which is also equal to the acceleration throughout all of its existence) is equal to ((x3-x2)/(t3-t2) - (x2-x1)/(t2-t1)) / ((t2+t3)/2) - (t1+t2)/2). This is for time going forwards, of course.

But what about if time is going backwards? Well, let's do what we did before: negate t for every point. This gets us ((x3-x2)/((-t3)-(-t2)) - (x2-x1)/((-t2)-(-t1))) / ((((-t2)+(-t3))/2) - ((-t1)+(-t2))/2) which equals -((x3-x2)/(t3-t2) - (x2-x1)/(t2-t1)) / -((t2+t3)/2) - (t1+t2)/2) which equals ((x3-x2)/(t3-t2) - (x2-x1)/(t2-t1)) / ((t2+t3)/2) - (t1+t2)/2). This means the acceleration when time goes forwards is exactly the same as when time goes backwards! Neat eh?!

Because it is exactly the same, the vector sum of accelerations (for multiple dimensions) will also remain the same whether time goes forwards of backwards.

Force: F=ma says that the net force applied on an object is equal to its mass times its acceleration. Because acceleration remains the same whether time goes forwards or backwards, all forces remain the same magnitude and direction whether time goes forwards or backwards.

Energy: In terms of kinetic energy, the quantity (1/2)(m)(v^2) is the same with time going forwards or backwards. As we determined earlier, the velocity of an object when time is going backwards is the negative of the velocity when time is going forwards. This means the amount of kinetic energy would be (1/2)(m)((-v)^2) = (1/2)(m)(v^2).

In terms of gravitation potential energy, the quantity (m)(g)(h), where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object, will also remain the same whether time is going forwards or backwards. This is because (as we determined earlier) acceleration remains constant whether time is going forwards or backwards.

Momentum: Momentum is mass*velocity, and since velocity negates when time goes backwards, momentum negates when time goes backwards.

2. Feb 10, 2008

### ZapperZ

Staff Emeritus
... and the point of all this is.... ?

Zz.

3. Feb 10, 2008

### greeniguana00

Well, any scenario where time goes forwards also follows the laws of physics with the same scenario running backwards.

4. Feb 10, 2008

### ZapperZ

Staff Emeritus
You have never heard of broken time reversal symmetry events?

Zz.

Last edited: Feb 10, 2008
5. Feb 10, 2008

### greeniguana00

Of course that goes beyond classical physics. As far as we are concerned, the exact path taken by a boulder as it falls off of a cliff and smashes is reversible.

6. Feb 10, 2008

### ZapperZ

Staff Emeritus
Actually, it doesn't. Many aspect of classical magnetism breaks such symmetry. If you do a search on "broken time reversal symmetry", you'll find several.

Zz.

7. Feb 10, 2008

### greeniguana00

I thought magnetism just negated.

8. Feb 10, 2008

### ZapperZ

Staff Emeritus
I have no clue what you just said.

Have you done the search that I suggested?

Zz.

9. Feb 10, 2008

### greeniguana00

http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000074000004000313000001&idtype=cvips&gifs=yes [Broken]

From abstract: "the equations of motion of charged particles are invariant under time reversal"

If only I could read it without paying \$. Was there something else I was supposed to find when googling?

Last edited by a moderator: May 3, 2017
10. Feb 10, 2008

### ZapperZ

Staff Emeritus
Note that I didn't say ALL of classical E&M has broken time reversal symmetry. I said that in some system, even classically, you can have broken time reversal symmetry.

http://info.ifpan.edu.pl/~masli/Publications/Li_PRL_78_1556_1997.pdf

http://www3.interscience.wiley.com/cgi-bin/abstract/110577399/ABSTRACT?CRETRY=1&SRETRY=0

http://ieeexplore.ieee.org/Xplore/l...0/04202649.pdf?tp=&isnumber=&arnumber=4202649

etc...

Zz.

Last edited by a moderator: May 3, 2017
11. Feb 10, 2008

### monish

Oh come on now.