Time-reverse symmetry of the principle of relativity

[atyy]

Does everyone also agree with the following:
The force required to accelerate a 1kg mass to 2m/s from rest, is equal to the force required to bring a 1kg mass to rest from an initial velocity of 2m/s?

That isn't true. Newton's 2nd law (F=ma) governs acceleration, not velocity. In principle, any force (no matter how small) can bring a 1kg mass to any speed (no matter how large). A small force will produce a small acceleration, and require a longer time to bring the mass up to speed than a large force.

F=ma cannot be used without specifying how the force depends on the properties (like charge or spring constant) of the colliding masses. But in many situations, we don't have a good equation that describes the force. So instead of using Newton's 2nd and 3rd laws directly, we use conservation of energy and conservation of momentum, because they are principles derived from Newton's laws.

An important thing to note is that the conservation of momentum is equivalent to Newton's third law.

F1=m2a2 (Force applied by m1 on m2, Newton's 2nd law)
F2=m1a1 (Force applied by m2 on m1, Newton's 2nd law)
F1=-F2 (Newton's third law)

Combining the above three equations, and remembering that acceleration is the rate of change of velocity (a=dv/dt):

m2a2=-m1a1
m2a2+m1a1=0
d(m2v2+m1v1)/dt=0 ===> m2v2+m1v1 is constant, ie. momentum is conserved.

This is why the collisions satisfy Newton's 3rd law automatically, even though we don't have a good description of what is happening at the impact - there's no such thing as two rigid masses colliding and rebounding instantly, despite the impression some textbooks give.

 

[JesseM]

Everyone agrees the 2kg/m/s momentum of the smaller incident mass imparts a 8/3kg/m/s momentum on the larger mass.
Everyone agrees that Newton's third law requires the collision be an exchange of equal and opposite force.
Does everyone also agree with the following:
The force required to accelerate a 1kg mass to 2m/s from rest, is equal to the force required to bring a 1kg mass to rest from an initial velocity of 2m/s?

No, as atyy says, you're ignoring the fact that Newton's second law, F=m*a, means that if two masses experience the same force, the smaller mass will experience a greater acceleration. Maybe this will be a little clearer if you rearrange the equation to be a = F/m. Obviously if you keep F constant but decrease m, F/m will increase.

 

[JustinLevy]

Just to point it out if it hasn't been already: special relativity does not require time reversal symmetry. It only requires Poincare symmetry.

There has already been a discussion on this on this forum. (if not more than one)

 

[Chrisc]

The measured forces depend on the frame of reference you choose to measure them in.

Unless I'm mistaken, no one has suggested the observer changes frames simply by observing the collision.

You must have a direct line to a higher power.

Not that I'm aware of.

How can you understand physics without the maths ?

I don't claim to. I struggle with the math constantly because I don't accept the math as defining the physical principles, I accept it as proof of them.

I suggest your 'principles' are just wrong but you won't accept it.

I have been wrong countless times in the past and will undoubtedly be wrong countless times in the future.
But I can't believe you honestly think physics should be learned by "accepting" popular opinion instead of reason.

 

[Chrisc]

atyy and JesseM, I don't understand why you both disagree with this.
It looks like both of you have made the same point I did in the statement you disagree with.
Either you have both misread my last post or I am missing something far more fundamental than I thought.
If it's the latter please explain the difference in the force required to accelerate 1kg from rest to a constant velocity of 2m/s in the same time as the force required to bring 1kg to rest from a constant velocity of 2m/s.

 

[JustinLevy]

If it's the latter please explain the difference in the force required to accelerate 1kg from rest to a constant velocity of 2m/s in the same time as the force required to bring 1kg to rest from a constant velocity of 2m/s.

Your question is a bit too vague to answer directly, but let me try to guess what you are having trouble with:

In the inertial rest frame of a 1kg object, I push on it with a force of 1 Newton.
From the viewpoint of an inertial frame in which the object is moving at 2m/s, is the force still 1 Newton?

The answer is NO.

If that surprises you, then that is were your mistake lies.
I don't remember the force transformations off hand. There's probably a better source than this, but here is what google gave me after some short searching: http://www.geocities.com/physics_world/sr/force_trans.htm

;-------------
And again, as I mentioned previously, special relativity does NOT require timer reversal symmetry. So I don't really know what the over-arching idea is that you are searching for here, as the title of your thread is implying something incorrect to begin with.

 

[atyy]

Either you have both misread my last post or I am missing something far more fundamental than I thought. If it's the latter please explain the difference in the force required to accelerate 1kg from rest to a constant velocity of 2m/s in the same time as the force required to bring 1kg to rest from a constant velocity of 2m/s.

Yes, I misunderstood your post - I did not realize you meant in the same time.

If so, how does a 2kg mass at rest impart more force (8/3kg/m/s^2) on the 1kg incident mass than the 1kg incident mass imparts on the 2kg mass (6/3kg/m/s^2) and uphold Newton's third law?

Are you referring to the following calculation by kev? My comments below assume this is what you mean, let me know if you meant something else.

To find out the final velocities when the incident mass is less than the rest mass you use the exact same equations and just change the variables as required, so for the case where the incident mass (B1) is 1 Kg moving at 2 m/s and the mass at rest (B2) is 2 Kgs:

Ball B1: m_1 = 1 Kg, v_1 = 2 m/s
Ball B2: m_2 = 2 Kg, v_2 = 0 m/s

The final velocity of B1 is:

v_1 ' = v_1 \frac{m_1-m_2}{m_1+m_2} = 2 m/s \frac{1 Kg - 2 Kg}{1 Kg + 2 Kg} = -2/3 m/s

The final velocity of B2 is:

v_2 ' = v_1 \frac{2*m_1}{m_1+m_2} = 2 m/s \frac{2 Kg}{(1 Kg + 2 Kg)} = +4/3 m/s

The negative sign for the final velocity of the smaller mass B1 simply means the smaller mass rebounds in the opposite direction to its initial velocity.

I am not sure exactly what you are doing, but I think you must be using 2kg(4/3-0)=8/3, and 1kg(-2/3+2)=-6/3. I think it is just a sign error and you should use 1kg(-2/3-2))=-8/3 ?

 

[yuiop]

I am not sure exactly what you are doing, but I think you must be using 2kg(4/3-0)=8/3, and 1kg(-2/3+2)=-6/3. I think it is just a sign error and you should use 1kg(-2/3-2))=-8/3 ?

Minor correction: 1kg(-2/3+2)=-6/3 should be 1kg(-2/3+2) = 1kg(-2/3+6/3) = 4/3

The force required to accelerate a 1kg mass to 2m/s from rest, is equal to the force required to bring a 1kg mass to rest from an initial velocity of 2m/s?
The force required to accelerate a 2kg mass to 1m/s from rest, is equal to the force required to accelerate a 1kg mass to 2m/s from rest?
If so, how does a 2kg mass at rest impart more force (8/3kg/m/s^2) on the 1kg incident mass than the 1kg incident mass imparts on the 2kg mass (6/3kg/m/s^2) and uphold Newton's third law?

Chrisc has gone from units of momentum to units of force. To do that he would have to take the derivative with respect to time. One way to do that is to use the idea of impulses and assume the forces are equal and opposite for a very brief and equal time during the collison while the masses are in physical contact. Analysing coliisions using the impulse concept can be very useful in complex situations such as where the collision includes both angular and linear momentum but I have not the impulse calculations for this particular example yet.

Just giving a hint where to look for a solution to this problem ;)

 

[Chrisc]

To do that he would have to take the derivative with respect to time. One way to do that is to use the idea of impulses and assume the forces are equal and opposite for a very brief and equal time during the collison while the masses are in physical contact.

Thanks kev, I think you've solved my problem. It is now also clear that I should have realized this when JesseM mentioned earlier that an instantaneous collision results in infinite force.
If I want to consider the mechanics of the collision as perfectly elastic and involving perfectly rigid bodies, I am essentially saying no loss of energy and no time of collision.
The combination of these two (unreal) conditions does not accentuate the principles of the laws as I thought, but makes them less applicable.
The instantaneous exchange of energy and momentum between masses makes any calculation of energy and momentum impossible. The only way to continue under these mechanics (as atyy and others have mentioned) is to work backward from the conservation laws and let the mechanics be determined by validating the equations.
There is no point then in concerning myself with the conflict that arises in the principles under these conditions because the restrictions I've put on the mechanics does not allow the principles to translate across inertial frames or through time-reversal. Essentially my mistake was to removed time from the collision so as to make any direction in time irrelevant and then question the conflict that arises when time is put back in without changing the mechanics to "real" or inelastic collisions.
If I understand the mistake I've made (it's still not fully distilled in my mind yet) it appears there is a correlation between the principles of the laws involved that is more direct than I thought. I'll leave that for a future post.

Thanks JesseM, kev, DrGreg, atyy and everyone else who had the patience and invested the time to help me understand this.

 

[atyy]

Thanks JesseM, kev, DrGreg, atyy and everyone else who had the patience and invested the time to help me understand this.

Chrisc, I have a confession to make which I thought should wait until you were satisfied that the collisions obey Newton's 3rd law: strictly speaking, it is not possible to derive conservation of energy from Newton's 2nd and 3rd laws.

1. Newton's 2nd and 3rd laws together imply conservation of momentum.

2. Newton's 2nd and 3rd laws alone, without further specification of F are not time reversible. We need to add the further assumption that F does not change sign when time and the velocity of the particles are reversed - for example, F generally should not depend on the velocity of the particles (it can depend on other things like position, mass, charge etc). Once we have added this assumption, we can also derive conservation of energy, which as you have seen, is necessary for time reversibility.

All the fundamental forces are believed to be time-reversible, so everything should be time-reversible if we consider the motion of all particles, including those that take away heat. But it is often useful to neglect their motion, and pretend that there are forces that are not time reversible, such as friction.

http://math.ucr.edu/home/baez/noether.html

 

[Chrisc]

Wow atyy, that's a very lucid discussion of the problem I have with the mechanics discussed above.
Caution: what follows is simply the ramblings of someone who almost understands a very interesting new idea.
The symmetry (time or Galilei boosts) exists (are fully conserved quantities) for anyone inertial observer only in the state of the system at t=0.
The instant one tries to separate or make quantifiable distinctions between momentum and energy there is no relative time derivative. (They are transient "identities" between frames.)
In other words, one has no choice but to resort to composite systems of statistically ideal quantities or as JesseM suggested, look at the center of momentum and mass not the individual bodies through time.
I don't know if this makes sense to anyone, but it makes my problem very clear in terms of the units (physical dimensions) I keep messing up.
It essentially means the time component of the equations distinguishes energy and momentum, but does not qualify the distinction between them. It makes the total conservation of the example I gave only possible when the potential energy of the mass at rest is considered as existing through time even though it has no motion. Sorry, I'm rambling, but there is something very intriguing here that will take me some time to fully grasp.
Thanks again atyy.