# Time-reverse symmetry of the principle of relativity

## Main Question or Discussion Point

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.

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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?

Staff Emeritus
2019 Award
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.
I think it's more likely that people don't want to spend the time hunting down your mistake. Relativistic mechanics is time symmetric.

DrGreg
Gold Member
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.
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.

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.
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.

B2 will not bring B1 to rest.
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.

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.
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).

IWhen 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).
Not true.

Dale
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.

Not true.

f = dp/dt is obviously time symmetric.
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.

If you have some scenario which is not time symmetric then it cannot follow the laws of mechanics.
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.

George Jones
Staff Emeritus
Gold Member
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.
Chrisc, it seems to that you're asking

"Why is there a thermodynamic arrow of time?"

JesseM
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 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).

George Jones
Staff Emeritus
Gold Member
This would be true in Newtonian mechanics but it doesn't work in SR.
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.

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

Dale
Mentor
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'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.

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...
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.

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George Jones
Staff Emeritus
Gold Member
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
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.
kev said:
I am pretty sure that when you do the forward calculations correctly, they will time reverse correctly.
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.

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.
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.

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 collision are time 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.

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George Jones
Staff Emeritus
Gold Member
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.
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.
kev said:
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 nto ot impossible, just statistically extremely unlikely in classical dynamics.
This just isn't going to happen in the real world. The difference in phase space volumes is more than enormous.

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Hi George and JesseM,
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.
OK, I will restate it as, the reverse process is more than enormously statistically improbable.

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George Jones
Staff Emeritus
Gold Member
So, Chrisc, why are inelastic collisions not time reversible? Because of the second law of thermodynamics.

Why is there a second law of thermodynamics? I don't know if there is agreement on this, but some physicists, including Roger Penrose and Sean Carroll, think that the second law has a cosmological origin. In the blog entry

Sean Carroll concludes
Sean Carroll said:
But if you want to describe why the Second Law actually works in the real world in which we actually live, cosmology inevitably comes into play.

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George Jones
Staff Emeritus
Gold Member
I have split this thead. The rest appears as the new thread Entropy and Cosmology

in the Cosmology forum.

Any comments about the entropy and the second law of thermodynamics with respect to cosmology should be placed in the new thread.

Dale
Mentor
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.
I just realized that the reverse case can in fact happen in nature. All that is necessary is that in the forward case the energy is stored, e.g. in a locking spring, instead of dissipated into thermal energy. Then in the forward case momentum is conserved through a decrease in KE (KE->elastic energy), in the reverse case momentum is conserved through an increase in KE (elastic energy->KE).

I had hoped to get back to this sooner but I'm in the middle of a "money pit" renovation that is
taking all my free time, so I can't address every post right now, but George Jones has made it easier
for me to address the point I'm trying to make.
These are inelastic collisions designed to show the principle of time-reversal is not only or always
a simple matter of mathematical symmetry that conserves momentum (total energy of the system).
There is a difference between the time-reverse symmetry of the laws and the time-reverse symmetry of mechanics.
The laws must, in principle, uphold under time-reversal or they would be expressions of or indications of
faulty axiomatic foundations.
Mechanics on the other hand are not so easily reversed. The kinematics of an event are the measurable dimensions
of the system, which are easily reversed as they are simply "quantitative" expression of dimension.
To reverse the direction of time you simply flip the sign to negative and everything runs (equates) backwards.
The dynamics are the problem, as they define the forces (classically) that give rise to the kinematics.
This means a translation of momentum between differing masses must consider the "empirical" evidence
of the second law of thermodynamics. Just because Newton's laws are "quantitatively" symmetric through
time-reversal, (i.e.: equal and opposite) does not mean we will ever see a fly stop a freight train.

The problem as I see it is the "qualitative" expression of the laws under time-reversal.
My point, or question is not why is there a second law of thermodynamics, it is that the second law
conditions the mechanics according to our frame of reference.
The mechanics of time-reversal measured by an observer are different according to their frame of reference.
This is a trivial observation in most cases, but in the example I've given it makes the difference between
the law of conservation of momentum displaying increased entropy or decreased entropy.
In other words, from one of only two frames, both involved in the event, the second law is upheld
in one and contradicted in the other.
This seems to indicate a "preferred" frame with respect to the laws of mechanics.
More importantly it says something fundamentally significant about the principle of relativity
and time. The (ideal) instantaneous exchange of momentum in the collision of two differing
masses, presents a "qualitative" change in dimension depending on the frame of the observer.