Time-reverse symmetry of the principle of relativity

[yuiop]

kev, you are talking about composite matter again. We are talking about the principles of the laws as they pertain
to the "theoretical" transfer of energy between point-masses in collision.
What you have demonstrated can be thought of as a "sling shot". The smaller mass is in contact with the larger
for a period of time while it acquires a greater velocity. It can remain in contact because of its composite nature.
Its constituent parts, flex, decelerate, change direction and accelerate, all the while it is still a ball in macro terms.
Change the super balls to point-masses and the principles of the laws dictate the smaller will not acquire a greater velocity than the larger had before collision.

It might be as well to note that there are probably no real point masses in nature, which have mass and zero volume. Physicists think black holes might be an example of genuine point masses (I disagree) and in the case of colliding black holes I would imagine the collision is very inelastic.

 

[JesseM]

Again, you are right. I should have said the mass of the smaller times the meters per second per second of the larger.

Meters per second per second are units of acceleration. Are you just saying that F=m*a, i.e. Newton's second law of motion? Is so why didn't you say so? And again, in no way can you use F=m*a to show that the velocity of the smaller mass can't be larger than the velocity of the larger one. For example, if both received the same constant force F during the time t of the collision, then F=m*a tells us that the smaller mass will experience a larger constant acceleration during this time interval, and since for constant acceleration (change in velocity) = (acceleration)*(time interval), the smaller mass will experience a larger change in velocity.

Newton's third law does dictate the limit of the forces by virtue of the necessity that they be equal.

The only limit is that one be the same size as the other, Newton's third law alone doesn't set any limit on the magnitude of the force that both experience. In fact for an idealized collision between point masses, we must assume that each experiences an infinite force because the collision only lasts for an instant!

As the motion between the masses is relative, the only property that determines the energy, momentum of each is their mass. The smaller mass then sets the limit.

Their energy and momentum are determined by both mass and velocity.

As the smaller increases velocity, the larger decreases velocity.

No, as the smaller gains velocity to the left, the larger gains velocity to the right. Since the smaller started out moving to the right and the larger started out moving to the left, if they're colliding for a finite time, each one's speed is decreasing until it hits zero, then increasing as it continues to acquire velocity in the direction opposite to the one it was moving before the collision.

You are talking about a different collision here. I stated the observer was at rest with the larger or smaller in each case. You are considering them both initially moving.

Sorry, I was just thinking about the collision from the perspective of a different reference frame. From the perspective of the center-of-mass rest frame, their momenta must be equal and opposite at every moment, including during the collision, so if the collision happens over an extended time because their shapes are compressing and uncompressing like perfectly elastic rubber balls, in the center-of-mass frame both their momenta are symmetrically going to zero as they compress, they're both at rest at the point of maximum compression, and then increasing symmetrically as they decompress...since momentum is mass*velocity, the velocity of the smaller mass must be dropping more rapidly during the compression, and increasing more rapidly during the decompression (i.e. the smaller mass experiences a larger change in velocity in both phases). We can then switch to the frame where the smaller mass was initially at rest, and in this frame the center of mass of the system will be moving to the left at some constant velocity v which is smaller than the velocity V of the larger mass to the left. So in this frame, during the compression the smaller mass' velocity must be increasing to v to the left, while the larger mass' velocity is decreasing to v (since at maximum compression they were both at rest in the center-of-mass frame, and in this frame the center of mass is moving at v, they both must be moving at v at the point of maximum compression in this frame). Then during the decompression the smaller mass' velocity must be increasing symmetrically, so if it went from 0 to v during the compression it must go from v to 2v during the decompression, while if the larger mass' velocity decreased by (V - v) during the compression (since V - (V - v) = v), it must decrease by a further (V - v) during the decompression, so since it was moving at v at the point of maximum compression it will now be moving at v - (V - v) = 2v - V.

So, we conclude from this that in the frame where the smaller mass is initially at rest, after the collision the smaller mass is moving at 2v to the left, whereas the larger mass is moving at 2v - V to the left. Provided that 2v - V is always positive, this is just another way of showing that the smaller mass will always have a larger speed after the collision (if 2v - V could be negative, it might turn out that its magnitude to the right was actually larger than v to the left). And in fact it is possible to show 2v - V will always be positive--remember that V was the velocity of the larger mass, and v was the velocity of the center of mass, in the frame where the small mass is initially at rest, and center of mass velocity = (total momentum)/(total mass). In this case the small mass had zero momentum so the total momentum is just the momentum of the larger mass, MV, and the total mass is (m + M). So if v = VM/(m + M), then 2v - V = [2VM/(m + M)] - V = [2VM/(m + M)] - [V*(m + M)/(m + M)] = (2VM - Vm - VM)/(m + M) = (VM - Vm)/(M + m). Since M is larger than m, VM - Vm must be positive, so that means the velocity to the left of the larger mass after the collision, 2v - V, must always be positive in this frame, and "ALWAYS SMALLER" than the velocity to the left of the smaller mass after the collision, 2v.

I used your term "constituent" part, then your term "point-like" constituents. I have continued to use the more recognized term "point-mass" as you had seemed to have recognized what we are talking about with this term.
Now you are right back to composite matter, electromagnetic fields and springs.
For clarification, a "point-mass" is not a "point-particle", it is an idealized, infinitely small object.

I know what a point-mass is, but I didn't catch that you wanted to talk exclusively about point masses in your last two posts to me, I thought you just wanted to avoid talking about situations where kinetic energy was lost to heat (i.e. you wanted to focus on elastic collisions), and we can imagine perfectly elastic non-point masses which is what I was doing. Of course your comments about the limits on the magnitude of forces makes even less sense in the context of point masses--in order for energy and momentum to be conserved throughout all finite time-intervals, in Newtonian physics the collisions of point masses must always be infinitely brief (if they stayed in contact for any finite time, then for momentum to be conserved they'd have to be at rest in their center-of-mass frame, but that would mean their kinetic energy would drop to zero during this time-interval), which naturally means the force needed to achieve a finite change in velocity during that instant must be treated as infinitely large (a dirac delta function).

Nope, your argument is handwavey and not based on anything in Newtonian mechanics.

Consider the masses as Point-masses and you will see it is true and based on Newtonian mechanics.

How? What part of Newtonian mechanics? Do you understand that in Newtonian mechanics, for point masses the collision must be infinitely brief, so the force and acceleration at that instant must be infinite? This is why it would be much better to talk about elastic collisions of non-point masses which behave like idealized springs during the collision, compressing when they meet and then decompressing until they begin to move apart.

I did not say it was. I said the energy of mass, and the forces they exert. When the small mass is at rest it has no kinetic energy and no momentum, yet it exerts a force on the larger. That would be the "force" exerted by the energy of mass. Call it resistance to motion or inertia if you prefer.

This is all totally confused, neither inertia nor energy exerts a force in Newtonian mechanics, every specific force has a name, whether it's the spring force or the normal force or whatever. In an elastic collision of a deformable non-point-mass like a perfect rubber ball, the force is basically a type of spring force, caused by the compression of the ball during the collision. The more rigid the spring is (i.e. the greater the spring constant), the more brief the time will be between when they first touch and when they depart (because it the force rises more quickly as they compress and thus they accelerate in opposite directions more quickly and begin moving apart more quickly), and the infinite force which happens at a single instant in the collision of two point masses (or two perfectly rigid non-point masses) is simply an idealized limiting case of this.

Now, for a collision which takes place over some finite time the size of the two masses does enter into considerations of how they long they remain in contact, but that's just because F=m*a, so the same force doesn't accelerate a larger mass as much. Suppose we have a collection of perfectly elastic rubber balls that all have the same radius, and also all have the same internal "spring constant", so that if the radius of each ball is 5 cm, if you smoosh two balls together so their centers are only 8 cm apart (meaning each one's radius in that direction is squashed to 4 cm), the force that each ball is exerting on the other will be the same regardless of which two balls you chose. It would nevertheless be true that if you collided two lower-mass balls, they would become less compressed before starting to move apart and would remain in contact for less total time, the reason just being that by F = m*a they will be accelerating faster at a given amount of compression than larger balls would be, so it will take less time until their velocities have become matched, which must be the point of maximum compression since afterwards their velocities begin to increase in opposite directions.

This statement is true when the correction you mentioned is applied. M*m/s/s.

OK, so the forces are equal and opposite by the third law, and by the second law the magnitude of each object's acceleration is given by F = m*a, where F is the force on it and m is its own mass. You still haven't given any coherent argument as to why you think this implies there is a limit on the size of the force, or why it implies that the velocity of the smaller mass will be smaller than the velocity of the larger one after the collision (when in fact any Newtonian analysis will show the opposite must be true in an elastic collision).

You seem far more comfortable with the math than the principles.
If you are willing to consider the theory and principles behind the mechanics of point-masses in collision, you will understand my point.

You seem unwilling to consider the possibility that your grasp of the principles is poor, and that your overall point doesn't make sense. I'm sure if you try to make your argument in the classical physics forum, all the experts there would agree you don't know what you're misunderstanding the principles of Newtonian mechanics.

 

[Chrisc]

You seem unwilling to consider the possibility that your grasp of the principles is poor, and that your overall point doesn't make sense. I'm sure if you try to make your argument in the classical physics forum, all the experts there would agree you don't know what you're misunderstanding the principles of Newtonian mechanics.

OK let's assume I'm wrong and you're right.
From post #95 you said:

"whenever we have a collision where the larger mass comes to rest after the collision, it must be true that the velocity v1 of the larger mass before the collision is "ALWAYS LESS" than the velocity v2 of the smaller mass after the collision."

From the first post on, I have been talking about one or the other masses being initially at rest with respect to the observer.
I have been talking about perfectly elastic collisions of perfectly rigid mass points.
In my example, the larger (2-kg) mass was moving from left to right toward the smaller (1-kg) mass that was at rest with respect to the observer.
Will this collision result in the larger coming to rest?
If not, can you give me an example where the larger mass comes to rest after the collision when the smaller is initially at rest before the collision?

 

[atyy]

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.

Let's work within Newtonian physics, since your question makes sense there.

The collision conserves momentum.

However it does not conserve kinetic energy, and so it is not elastic.

The formula for kinetic energy is mass(velocity)²/2

The initial kinetic energy is m(2v)²/2=2mv².

The final kinetic energy is 2m(v)²/2=mv².

This means that energy is lost as heat, which consists of vibrations in random directions or photons flying random directions. So it will not be time reversible, unless you also time reverse all those random movements (which you did not include in your diagram).

It's interesting why momentum can be conserved even though energy is not. The formulas for momentum and energy both contain velocity, so how can you change velocity without changing momentum? The answer is that momentum is a number and a direction (vector), but energy is just a (positive) number. The random vibrations take away the energy (ordinary addition and subtraction), and also the momentum - but the vibrations are random, so the momentum in one direction is taken away as much as in the opposite direction, and there is no loss of momentum (vector addition and subtraction) .

To see what happens if the collision is elastic, I suggest you solve for the final velocities of both balls by assuming both momentum (mv) and kinetic energy conservation (mv²/2).

 

[DrGreg]

OK let's assume I'm wrong and you're right.
From post #95 you said:

"whenever we have a collision where the larger mass comes to rest after the collision, it must be true that the velocity v1 of the larger mass before the collision is "ALWAYS LESS" than the velocity v2 of the smaller mass after the collision."

From the first post on, I have been talking about one or the other masses being initially at rest with respect to the observer.
I have been talking about perfectly elastic collisions of perfectly rigid mass points.
In my example, the larger (2-kg) mass was moving from left to right toward the smaller (1-kg) mass that was at rest with respect to the observer.
Will this collision result in the larger coming to rest?
If not, can you give me an example where the larger mass comes to rest after the collision when the smaller is initially at rest before the collision?

The maths isn't that difficult (in this specific case) and it's the only rigorous way to solve this.

Consider mass m with velocities v₁ and v₂ before and after collision, and mass M with velocities V₁ and V₂ before and after collision. Assume there are no other forces acting, and we'll use Newtonian formulas rather than relativistic.

Conservation of momentum always holds, so

$$mv_1 + MV_1 = mv_2 + MV_2$$ ...(1)​

If you have finally decided you want to consider elastic collisions only then conservation of kinetic energy holds too, so

$$\frac{1}{2} m v_1^2 + \frac{1}{2} M V_1^2 = \frac{1}{2} m v_2^2 + \frac{1}{2} M V_2^2$$ ...(2)​

These are the only two equations you need. Once you have specified "elastic" then it doesn't matter whether these are "rigid point masses" or not.

Now you have asked if it is possible to have v₁ = V₂ = 0. The equations now simplify to

$$MV_1 = mv_2$$ ...(3)
$$\frac{1}{2} M V_1^2 = \frac{1}{2} m v_2^2$$ ...(4)​

Divide (4) by ½(3) to get V₁ = v₂, and substitute back into (3) to get M = m. That wasn't hard, was it?

So the situation you describe can happen only for equal masses (assuming elastic collisions). (Example: Newton's cradle.)

 

[JesseM]

OK let's assume I'm wrong and you're right.
From post #95 you said:

"whenever we have a collision where the larger mass comes to rest after the collision, it must be true that the velocity v1 of the larger mass before the collision is "ALWAYS LESS" than the velocity v2 of the smaller mass after the collision."

From the first post on, I have been talking about one or the other masses being initially at rest with respect to the observer.
I have been talking about perfectly elastic collisions of perfectly rigid mass points.
In my example, the larger (2-kg) mass was moving from left to right toward the smaller (1-kg) mass that was at rest with respect to the observer.
Will this collision result in the larger coming to rest?
If not, can you give me an example where the larger mass comes to rest after the collision when the smaller is initially at rest before the collision?

As DrGreg said, if mass #1 is initially at rest and mass #2 is coming towards it, and they collide elastically, the only way that mass #2 can come to rest is if their masses are equal. You can also see this by looking at the formula I derived in my previous post--if the mass initially at rest has mass m, and the mass moving towards it to the left has mass M and velocity V, then the center of mass of the system will always be moving to the left at v = VM/(m + M) since this is the total momentum divided by the total mass. With the variables defined this way, after the collision the mass m initially at rest will always be moving at 2v to the left, and the mass M will always be moving at (2v - V) to the left. So, the only way to have 2v - V = 0 is if [2VM/(m + M)] - V = 0, which means [2VM/(m + M)] - [V(m + M)/(m + M)] = (2VM - Vm - VM)/(m + M) = V(M - m)/(m + M) = 0. Assuming V is not zero, the only way V(M - m)/(m + M) can be 0 is if (M - m) = 0, i.e. M = m.

 

[Chrisc]

Thank you atyy, DrGreg and JesseM.
That is very logical mathematical reasoning.
In plain English you're saying that the incident mass will only come to rest if the mass it collides with is of equal mass.
In the case of the incident mass being greater than the mass at rest, both will continue to move in the same direction after collision. In the example JesseM gave their velocities will be 2/3m/s and 8/3m/s respectively, when the incident mass is 2-kg with velocity 2m/s and the other is 1-kg at rest. This also conserves momentum and kinetic energy.
How do you conserver the momentum and kinetic energy when the incident mass is less than the mass at rest?
It would appear from the reasoning you've given, if the incident mass is smaller than the mass at rest, the incident mass would never come to rest either.

 

[atyy]

It would appear from the reasoning you've given, if the incident mass is smaller than the mass at rest, the incident mass would never come to rest either.

Yes. And since in this case, you are conserving energy and momentum, we will not be able to attribute lack of reversibility to random vibrations of particles that you did not explicitly include in your diagram and calculation. So if this new situation fails to be time reversible, we will have no way out, and we shall have to conclude that Newton's laws are not time reversible. (Let me hedge my bets, and say there is also a possibility I didn't think this reply through carefully - since I shall be quite upset if I find out Newton's laws are not time reversible)

Why don't you try it and see?

--------------------
BTW, there is a subtlety about Newton's laws for point particles. If you just ask "There are 2 point particles with masses and velocities m1,m2,v1,v2, what happens if we apply F=ma?" - then there is actually NO answer.

F=ma requires the explicit specification of what F is.
For example, if there is gravity, then F=Gm1m2/r²
Or if the particles have charges q1,q1, then F=Kq1q1/r².
The reason you may have been confused is you tried to solve the problem without explicit specification of F.

The reason we solve the problem using energy and momentum conservation is that it is equivalent to solving the problem with only "partial", non-explicit specification of F. I can try to explain more if you're interested, but first you should calculate the case where energy and momentum are conserved, and see if it is time reversible.
--------------------

 

[DrGreg]

Thank you atyy, DrGreg and JesseM.
That is very logical mathematical reasoning.
In plain English you're saying that the incident mass will only come to rest if the mass it collides with is of equal mass.
In the case of the incident mass being greater than the mass at rest, both will continue to move in the same direction after collision. In the example JesseM gave their velocities will be 2/3m/s and 8/3m/s respectively, when the incident mass is 2-kg with velocity 2m/s and the other is 1-kg at rest. This also conserves momentum and kinetic energy.
How do you conserver the momentum and kinetic energy when the incident mass is less than the mass at rest?
It would appear from the reasoning you've given, if the incident mass is smaller than the mass at rest, the incident mass would never come to rest either.

I won't go through the proof (exercise for the reader!) but, unless I've made a silly mistake, the solution to equations (1) and (2) in my post #110 is:

$$v_2 = \frac{m-M}{m+M} v_1 + \frac{2M}{m+M} V_1$$ ...(5)
$$V_2 = \frac{2m}{M+m} v_1 + \frac{M-m}{M+m} V_1$$ ...(6)​

If $V_1 = 0$ and $m < M$ (small mass m hitting larger stationary mass M),

$$v_2 = \frac{m-M}{m+M} v_1 < 0$$ ...(7)
$$V_2 = \frac{2m}{M+m} v_1 < v_1$$ ...(8)​

The smaller incident mass m rebounds in the opposite direction.


On the other hand, if $v_1 = 0$ and $m < M$ (large mass M hitting stationary small mass m),

$$v_2 = \frac{2M}{m+M} V_1 > V_1$$ ...(9)
$$V_2 = \frac{M-m}{M+m} V_1 > 0$$ ...(10)​

The smaller stationary mass m is shot with a velocity greater than the incident mass M originally had, the larger mass slows down but does not stop.

 

[yuiop]

Thank you atyy, DrGreg and JesseM.
That is very logical mathematical reasoning.
In plain English you're saying that the incident mass will only come to rest if the mass it collides with is of equal mass.
In the case of the incident mass being greater than the mass at rest, both will continue to move in the same direction after collision. In the example JesseM gave their velocities will be 2/3m/s and 8/3m/s respectively, when the incident mass is 2-kg with velocity 2m/s and the other is 1-kg at rest. This also conserves momentum and kinetic energy.
How do you conserver the momentum and kinetic energy when the incident mass is less than the mass at rest?
It would appear from the reasoning you've given, if the incident mass is smaller than the mass at rest, the incident mass would never come to rest either.

If you go back to https://www.physicsforums.com/showpost.php?p=1799644&postcount=17"you will see that I showed way back then that the final velocities would be "be 2/3m/s and 8/3m/s respectively" as shown in this quote:

...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:

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

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.

The formulas automatically conserve energy and momentum because they are derived assuming the conservation of energy and momentum as shown in the hyperphysics link.

... So if this new situation fails to be time reversible, we will have no way out, and we shall have to conclude that Newton's laws are not time reversible. (Let me hedge my bets, and say there is also a possibility I didn't think this reply through carefully - since I shall be quite upset if I find out Newton's laws are not time reversible)

Why don't you try it and see?

-------------------

To run the movie backwards and find the time reverse of the above situation simply make the final velocities the initial velocities and reverse the signs. I am using the full equations here because the equations used above are the simplified version where it assumed the target ball B2 is stationary. See http://en.wikipedia.org/wiki/Elastic_collision

Ball B1: $$m_1 = 1 Kg, v_1 = +2/3 m/s$$
Ball B2: $$m_2 = 2 Kg, v_2 = -4/3 m/s$$

The final (time reversed) velocity of B1 is:

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

The final (time reversed) velocity of B2 is:

$$v_2 ' = v_1 \frac{v_2(m_2-m_1)+2*m_1*v_1}{m_1+m_2} = \frac{2/3 m/s (2 Kg-1 Kg) +2*1 Kg*-2/3 m/s}{1 Kg + 2 Kg} = 0 m/s$$

which is the exact time reversed symmetry.

 

[yuiop]

In my last post (#117) I showed how to do the time reversed calculation using the full hairy momentum calculation but there is a simpler, quicker way to do it if you are comfortable with switching reference frames (which you should be if you are interested in relativity )

The calculaton for the incident mass having less mass than the stationary mass was:

...
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:

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

Taking the sign reversed final velocities as the new initial velocities for the time reversed situation the initial conditions are now:

Ball B1: $$m_1 = 1 Kg, v_1 = +2/3 m/s$$
Ball B2: $$m_2 = 2 Kg, v_2 = -4/3 m/s$$

The equations are simpler when one of the masses is at rest so we switch to new reference frame (F2) where the observer is moving at -4/3 m/s relative to the original reference frame (F1) and we do that by adding 4/3 m/s to all the velocities to obtain the initial conditions in F2 which are :

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

We note that we already know the answer to this calculation because these are the same initial conditions described in the my quote above so the final velocities in frame F2 are:

$$v_1 ' = -2/3 m/s$$

$$v_2 ' = +4/3 m/s$$

Now we switch back to the original reference frame to obtain the final velocities by subtacting 4/3 m/s from all the velocities to obtain the final answer in F1:

$$v_1 ' = -2 m/s$$

$$v_2 ' = 0 m/s$$

which is the exact time reversed symmetry.

 

[JesseM]

Thank you atyy, DrGreg and JesseM.
That is very logical mathematical reasoning.
In plain English you're saying that the incident mass will only come to rest if the mass it collides with is of equal mass.
In the case of the incident mass being greater than the mass at rest, both will continue to move in the same direction after collision. In the example JesseM gave their velocities will be 2/3m/s and 8/3m/s respectively, when the incident mass is 2-kg with velocity 2m/s and the other is 1-kg at rest. This also conserves momentum and kinetic energy.
How do you conserver the momentum and kinetic energy when the incident mass is less than the mass at rest?

The formulas I gave for the velocities after the collision works regardless of whether M is larger than m or vice versa. My formulas said that after the collision you'd have a velocity of 2v to the left for the mass m that was initially at rest, and (2v - V) to the left for the mass M that was initially approaching the other one at V to the left, with v as the velocity of the center of mass of the system, i.e. VM/(M + m) to the left (total momentum/total mass). If you work things out, you see that (2v - V) works out to V(M - m)/(m + M). Obviously if M is smaller than m, then (M - m) will be negative, so this means the velocity of the mass M in the left direction must be negative, i.e. after the collision it will always be moving to the right.

 

[atyy]

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:

In my last post (#117) I showed how to do the time reversed calculation using the full hairy momentum calculation but there is a simpler, quicker way to do it if you are comfortable with switching reference frames (which you should be if you are interested in relativity )

Beautiful! I suppose we should wait for Chrisc's calculation, since even Feynman and Schwinger got the same wrong result when they calculated the Lamb shift.

 

[Chrisc]

Thanks everyone.
The math works fine in the sense that it conserves both momentum and kinetic energy.
I checked the relativity of the math and mechanics by setting the observer co-moving with each mass.
But I don't imagine anyone lost sleep waiting for my answer.
As kev said, it conserves because the math is designed to conserve. If the mechanics are (theoretically) considered valid because they meet the criteria of conservation, then there is no point in questioning the math.
But my problem is that the mechanics do not seem (theoretically) to uphold the principles.

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?
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?

 

[Mentz114]

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?

The measured forces depend on the frame of reference you choose to measure them in. It's been explained to you several times that there's no contradiction here.

But my problem is that the mechanics do not seem (theoretically) to uphold the principles.

You must have a direct line to a higher power. How can you understand physics without the maths ? I suggest your 'principles' are just wrong but you won't accept it.

 

[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.