# Time-reverse symmetry of the principle of relativity

1. Jul 9, 2008

### Chrisc

Please look at the attached diagram and let me know
if there is a reason for the asymmetric dynamics due
to the relative position of rest or, if I have incorrectly
interpreted the mechanics.
Kev, I haven't forgotten.

#### Attached Files:

• ###### BodiesInCollision.jpg
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2. Jul 10, 2008

### Chrisc

Not a single response tells me either it is too confusing in the
graphic form I posted, or I've pointed out something no one can
rationalize with relativity.
In case it's the former, I've included below a less detailed
text version of the question.

The time-reverse symmetry of the dynamics in a simple two body collision seems
to hold only when the observer is initially at rest with the lessor of the two massive bodies.

If this is as straight forward as I think it is, it has significant implications for
the principle of relativity.
Have I misinterpreted the mechanics, or is this stumping everyone?

3. Jul 10, 2008

Staff Emeritus
I think it's more likely that people don't want to spend the time hunting down your mistake. Relativistic mechanics is time symmetric.

4. Jul 11, 2008

### DrGreg

I can't work out what you think the problem is. As far as I can see, everything in your diagrams is time-symmetric and you haven't explained why you think there is asymmetry.

However, your diagrams are correct only in Newtonian mechanics, not in relativistic mechanics.

In relativity, momentum is

$$p = \frac {mv} {\sqrt{1 - v^2/c^2}}$$​

and velocities are transformed from one frame to another using

$$w = \frac {u - v} {1 - uv/c^2}$$​

Rest mass $m$ is not conserved in collisions but energy

$$E = \frac {mc^2} {\sqrt{1 - v^2/c^2}}$$​

is conserved in elastic collisions.

5. Jul 11, 2008

### Chrisc

When the observer is initially at rest with B1(#5), the time-reverse symmetry of the kinematics shown in #8 violate the laws of dynamics.
B2 will not bring B1 to rest. The kinematics all appear time-reverse symmetric as kinematics are the quantitative expressions of the mechanics with T set to negative.
When the observer is initially at rest with B1, the correct time-reverse dynamics of #5 would present the same as #2.
The time-reverse symmetry of this event is only upheld when the observer is initially at rest with B2,
a situation that implies there is a problem with the symmetry of relativistic mechanics through time-reversal.

6. Jul 11, 2008

### MeJennifer

What do you mean by bringing something to rest. Rest is a relational not an absolute concept in relativity. Something is only at rest in relation to something else.

7. Jul 11, 2008

### Chrisc

I mean with respect to the observer.
When the observer is initially at rest with respect to one of the two bodies (B1 and B2)
they observe differing but valid mechanics after the collision.
The problem arises when the same event is considered through time-reversal.
In the first case where the observer is initially at rest with respect to B2, the
time-reverse mechanics are valid.
When the observer is initially at rest with respect to B1, the time-reverse
mechanics (when held to the symmetry of the forward-time kinematics)
require dynamics that violate or contradict the laws (Newton's).

8. Jul 11, 2008

### MeJennifer

Not true.

9. Jul 11, 2008

### Staff: Mentor

f = dp/dt is obviously time symmetric. If you have some scenario which is not time symmetric then it cannot follow the laws of mechanics.

10. Jul 12, 2008

### Chrisc

11. Jul 12, 2008

### Chrisc

A body B1 with mass M and velocity v wrt to the observer collides with a body B2 with mass 1/2M at rest with respect to the observer.
The mass B1 comes to rest wrt the observer and the body B2 moves away with velocity 2v wrt the observer.
This is the (invalid) time-reverse description of the (valid) time forward event. It does not happen in nature as it conserves momentum via increased kinetic energy.
So it appears as a statement of kinematics, f = dp/dt is time-reverse symmetric, but it is not necessarily always time-reverse symmetric as a statment of dynamics.

I am not pointing out a scenario that is not time-reverse symmetric and upholds the laws, I am pointing out a scenario that appears time-reverse symmetric in its kinematics but as such its dynamics must violate the laws.
If you take that to mean that it is just not time reverse symmetric in the first place, then you are recognizing what I am saying as it is the same event that is time-reverse symmetric when the observer is at rest wrt B2.

12. Jul 12, 2008

### George Jones

Staff Emeritus
Chrisc, it seems to that you're asking

"Why is there a thermodynamic arrow of time?"

13. Jul 12, 2008

### JesseM

This would be true in Newtonian mechanics but it doesn't work in SR.
Total momentum before collision: $$\frac{M*v}{\sqrt{1 - v^2/c^2}}$$
Total momentum after: $$\frac{(M/2)*(2v)}{\sqrt{1 - (2v)^2/c^2}}$$
These are not equal, so this can't be correct (momentum should be conserved in collisions in SR just like it is in Newtonian mechanics).

14. Jul 12, 2008

### George Jones

Staff Emeritus
I don't think that this affects Chrisc's argument. Chrisc's argument is that inelastic collsions occur only one way in time.

Roughly, heat (internal energy) is more disordered then translational kinetic energy of an entire object, hence, by thermodynamics, inelastic collisions only happen one way in time.

15. Jul 12, 2008

### JesseM

Ah, he didn't specifically refer to the collision as inelastic, but now I see that he mentions the kinetic energy changes (in Newtonian terms as well as relativistic ones). And yes, inelastic collisions involve a change in entropy (kinetic energy of the center of mass being transformed into heat, which is random kinetic energy of many molecules in different directions) which is why they are extremely unlikely to happen in reverse, although in terms of the fundamental non-thermodynamic laws of physics there is nothing physically impossible about the reversed scenario.

16. Jul 12, 2008

### Staff: Mentor

I'm sorry, but what you are saying here doesn't make any sense. Newton's laws are the laws of dynamics. Newton's laws are time symmetric. Therefore dynamics are time symmetric. Kinematics are just dynamics w/o the forces, so if the dynamics are time symmetric then the kinematics are also time symmetric. This is obvious and clear from the laws themselves, you don't need to worry about specific cases because the laws are symmetric in general.

In your example, the explanation is simple, in the forward case momentum is conserved through a decrease in KE (KE->thermal energy), in the reverse case momentum is conserved through an increase in KE (thermal energy->KE). The fact that the reverse case doesn't happen in nature is due to the non-time symmetry of thermodynamics, not any asymmetry in dynamics.

A minor point is that your analysis is non-relativistic.

17. Jul 12, 2008

### yuiop

Your diagrams are obviously non relativistic. There seems to be an error in the calculations in your diagrams when the calculations are done using the Newtonian equations.

The equation for a head on elastic collision is given here: http://hyperphysics.phy-astr.gsu.edu/Hbase/elacol2.html#c1

Using the notation given in that link, you have initial conditions:

Ball B1: $$m_1=2m, v_1=2v$$
Ball B2: $$m_2=1m, v_2=0v$$

The final velocity of mass m1 is:

$$v_1' = v1 \frac{m1-m2}{m1+m2} = 2v\frac{2m-1m}{2m+1m} =2/3v$$

The final velocity of mass m2 is:

$$v_2' = v1 \frac{2m_1}{m_1+m_2} = v\frac{4m}{2m+1m} =8/3v$$

The initial total momentum of the system is $$(m_1 v_1 )+(m_2 v_2) = 4mv$$

The final total momentum of the system is $$(m_1 v_1')+(m_2 v_2') = 4/3mv+8/3mv = 12/3mv = 4mv$$

Total momentum before and after the collision is conserved.

The total initial kinetic energy of the system is:

$$1/2m_1 (v_1)^2 + 1/2m_2 (v_2)^2 = 4mv^2$$

The total final kinetic energy of the system is:

$$1/2m_1 (v_1')^2 + 1/2m_2 (v_2')^2 = m(2/3v)^2 + 1/2m(8/3)^2 = 4mv^2$$

Total KE before and after the collision is conserved.

The final conditions in your diagram of:

Ball B1: $$m_1=2m, v_1'=v$$

Ball B2: $$m_2=1m, v_2'=2v$$

do not satisfy the conservation of momentum AND conservation of KE laws.

Another condition that is normally satisfied in a head on elastic collision, is that the velocity of approach equals the velocity of separation. In the example I gave the velocity of approach is 2v and the velocity of separation is also 2v (8/3v - 2/3v). The velocity of approach in your diagram is 2v while the velocity of separation is 1v.

I am pretty sure that when you do the forward calculations correctly, they will time reverse correctly.

Last edited: Jul 12, 2008
18. Jul 12, 2008

### George Jones

Staff Emeritus
Have you read posts #12 through #16? The equations in the link don't apply, since they are for elastic collisions. Chrisc analyzes inelastic collisions that are physically realistic, and that cannot be excluded from consideration.

Consider a more extreme example.

Two equal mass objects collide and stick together. Before the collision, the objects move with equal speeds in opposite directions with respect to a particular frame. By conservation of momentum, the combined object does not move after the collision.

This is a completely plausible physical scenario, i.e., think putty.

The time reverse of the collision is not plausible at all. A blob of putty does not separate into two smaller blobs spontaneously, with each of the two smaller blobs moving in different directions.

This is in accord with the laws of thermodynamics and statistical mechanics.
I think Chrisc has done the forward calculations correctly for plausible inelastic collisions. Even though Chrisc didn't make an error by choosing to analyze inelastic collisions, as they happen all the time in the real world, I think Chrisc would agree that elastic collisions are time reversible.

19. Jul 12, 2008

### yuiop

In the diagram he attached to post#1 he shows the case for the elastic collision in frames 1 to 4 and the inelastic case in frames 5 to 8. I took the time to check his calculations in frames 1 to 4 (the elastic case) and when I found them to be in error and I did not really persue the thread further. Ignoring the fact that there is an error in his elastic case I can see now that he is making the case that elastic collisions appear to be reversible while inelastic collisions do not appear to be reversible.

Looking at the equation for an inelastic collision http://hyperphysics.phy-astr.gsu.edu/Hbase/inecol.html#c1 the final velocity of the combined mass of B1 and B2 should be 4/3v and not the 1v shown in frames 4 to 8. Despite the fact Chrisc has made an error in both the elastic and inelastic cases it should not distract us from the case he is making that inelastic collisons appear to be non-reversible. As I understand it, classical dynamics does not forbid a blob of putty separating into two smaller blobs spontaneously, with each of the two smaller blobs moving in different directions when analysed at the molecular scale. All it says is that it is statistically unlikely. A freek set of unlikely collisons at the molecular level producing that sort of reverse reaction is unlikely but not imposssible. It is basically an example of the arrow of time and increasing entropy. Another example is a glass falling off a table and breaking into a million pieces. The reverse situation of the glass reassembling itself and ending back up on top of the table is not impossible, just statistically extremely unlikely in classical dynamics.

Basically it comes down to the fact that converting coherant motion (the parallel motion of the molecules that make up the ball) to incoherant random motion (heat) is more likey than the reverse in nature, (but not impossible). An example of the reverse in nature would be the thermal heat of magma beneath the surface of the Earth being converted into coherant motion of the water and steam being ejected from a geyser.

Last edited by a moderator: Jul 13, 2008
20. Jul 12, 2008

### George Jones

Staff Emeritus
I admit that I only checked Chrisc's calculation in post #11; I didn't have the stamina to examine the whole thread in detail. Post #11 was enough to show me Chrisc's point. In the real world, inelastic collisions are not reversible. I checked to see if anyone made the connection with thermodynamics.
This just isn't going to happen in the real world. The difference in phase space volumes is more than enormous.

Last edited: Jul 13, 2008