What wins?
A) An unstoppable force
B) A unmovable object
Unstoppable force pierces through the unmovable object and doesn't stop, meanwhile while the object was pierced it didn't move so it remains unmovable.
What wins?
A) An unstoppable force
B) A unmovable object
Unstoppable force pierces through the unmovable object and doesn't stop, meanwhile while the object was pierced it didn't move so it remains unmovable.
no... because then the part that was pierced would have been move :p
----
The enemy is in front of us, the enemy is behind us, the enemy is to the right and left of us. They cant get away this time! - General Douglas Mcarthur
What wins?
A) An unstoppable force
B) A unmovable object
Unstoppable force pierces through the unmovable object and doesn't stop, meanwhile while the object was pierced it didn't move so it remains unmovable.
no... because then the part that was pierced would have been move :p
Atoms are mostly empty space. It is not hard to imagine if such a scenario existed, the atoms could pass through the empty space.
What wins?
A) An unstoppable force
B) A unmovable object
The common answer to this problem is that both unstoppable forces and immovable objects are imaginary, and so it is impossible to answer the question. That's not a problem in and of itself. I can answer lots of questions about imaginary objects as long as they follow well defined imaginary rules. Immovable objects and unstoppable forces fail in this regard because they are contradictory terms. If our imaginary world contains an an object whose motion cannot be altered by any force, it cannot, by definition, contain a force that can move any object.
To say anything about immovable objects and unstoppable forces, then, I'll have to amend their definitions to avoid this contradiction. The simplest way is to put these two objects in a category unto themselves:
An immovable object is an object whose motion cannot be changed by any force, except possibly by an unstoppable force (see below).
An unstoppable force is a force strong enough to change the motion of any object, except possibly an immovable object (see above).
With these amended definitions, I can work with the both concepts simultaneously and rigorously. The idea is that the "immovability" of the object and "unstoppability" are described by yet-to-be-defined energy scales on the order of some huge energy Λ, which is far greater than the energy scales of any of the other forces and potentials in our imaginary world. So huge, in fact, that in the end I will take the limit Λ→∞Λ→∞ , in effect stating that the immovable object and unstoppable force are way above the scale of anything else in our imaginary world. Moreover, while both the "immovability" and "unstoppability" will be nearly infinite, their ratio will be finite and well-defined.
Dimensional analysis
Consider an immovable object of mass M. Mass is only an impediment to being accelerated quickly; even a tiny force can move an object of immense mass, albeit slowly. So, I don't interpret immovable to mean having immense mass. Instead, to be immovable by any finite force the object must be at the bottom of some infinite potential well. This means that it would take an infinite amount of energy to perturb the object and permanently alter its motion - something only the unstoppable force will be capable of doing. This potential well must have a characteristic width L, which is the width of the region to which the object is confined. The potential well must also have a depth V, a quantity with dimensions of energy. While the spirit of an immovable object implies that the width of the potential should be small, it is really the depth of the well that characterizes the object's immovability. The depth of the well V should accordingly be infinite, on the scale of Λ.
In non-relativistic classical mechanics, there is only one way of doing so. I must say that V ~ Λ. This is because V is the only quantity with dimensions of energy that can be formed out of M, L, and V.
This changes when allowing relativistic effects. To do so, I add the constant speed of light c to our bag of parameters. Now there is an extra independent quantity with dimensions of energy, Mc2Mc2 . This is the rest energy due to the objects mass. From V and Mc2Mc2 the two of these I can form the dimensionless ratio
vM=VMc2vM=VMc2
(Read this notation as "the potential V measured in units of the rest mass energy Mc2Mc2 "). I am now no longer completely restricted to setting V ~ Λ. In principle, I could let Vf(vM)∼ΛVf(vM)∼Λ for any real function f(vM)f(vM) . However, it is hard to imagine quantities of the form Vf(vM)Vf(vM) with more natural meaning than simply V, and it seems artificial to set some "unnatural" Vf(vM)Vf(vM) to be on the order of Λ. If you wanted, you could repeat this analysis, perhaps setting V2/M∼ΛV2/M∼Λ , although I don't know how this could be interpreted. If V was a momentum scale, this could be interpreted as kinetic energy, but V is the scale of the confining potential.
This is somewhat of a moot point, though, because I will only be using non-relativistic classical mechanics and quantum mechanics to analyze the interaction of the object and force, so the relativistic energy scale Mc2Mc2 is irrelevant. This is a lie; the entire idea of the problem at hand (see the definitions), that Λ is by far the largest energy scale in our toy world, requires that Λ≫Mc2Λ≫Mc2 . This makes the scenario highly relativistic. For simplicity I'm ignoring this. An interesting extension of this problem might consider relativistic effects, or might consider M ~ Λ so that the non-relativistic treatment is a good approximation.
What about including quantum mechanical effects? Then I can include ℏℏ among our constants, and now I can form the energy
E=ℏ2ML2E=ℏ2ML2
Up to some proportionality constant on the order of unity, this is the zero-point energy of the object confined to its well. In quantum mechanics, objects never have precisely zero momentum (see: Heisenberg's uncertainy principle) and thus can never have exactly zero energy. So, instead of resting exactly at the bottom of its potential well, the quantum object shuffles around with some nonzero kinetic energy. It is a natural energy scale in the problem, and just as before, I can form the dimensionless ratio
vE=VE=VML2ℏ2,vE=VE=VML2ℏ2,
which is just the potential V measured in units of E. Again, this opens up the possibility of letting Vf(vE)∼ΛVf(vE)∼Λ for any real function f(vE)f(vE) rather than just V ~ Λ. And again, there seems to be little reason for doing so. Naively, it seems to me that these quantities Vf(vE)Vf(vE) are artificial, and setting any of them to be on the scale of Λ (as I implicitly did in my original answer) is unnatural.
Next I'll consider an unstoppable force. The force must have some magnitude F, naturally, which is the quantity directly responsible for the unstoppability of the force. However, it must also have associated width and time scales. The first is the width of the region over which the force acts. There are really only three possibilities for the size LFLF of this width: LF≪LLF≪L , LF∼LLF∼L , or LF≫LLF≫L . Qualitatively, the first leads to a force with infinitesimal range and which is thus "weak," in a certain sense. The last leads to a force with effectively infinite range, and which "strong" in a similar sense. In my opinion, this is a different type of strength than that our unstoppable force should possess. For example, gravity and QCD (before color confinement) are both infinite range forces, but are vastly different in strength. So, I'll take the middle option and for simplicity let LF=LLF=L .
The second is the timescale over which the force turns on. I have a before-and-after problem, where I want to see what the effect of the unstoppable force is on the immovable object. Before some time t0t0 , the force will be turned off, but it will then then turn on, and it will take some time T (possibly very short) for it to turn on (it could turn on and then off again in this timescale, but I'm not examining this possibility). In my original answer, this time scale was T = 0. For simplicity, I will stick with that choice.
Classically, there is again only one intrinsic energy scale, which is FL, so I must take
FL∼ΛFL∼Λ
Since classically it also turned out that V ~ Λ, the problem will end up being defined by the ratio
α=FL2Vα=FL2V
Although I'm skipping a relativistic treatment, I'll still go through the dimensional analysis in this case. The quantities L and T can be combined to form a dimensionless ratio r = cT/L. This is essentially a measure of how far information can travel across the range L of the force during the duration T of the force. Presumably this could be important in a relativistic treatment, especially if r is much less than one. However, repeating myself yet again, it makes little sense to let FL f(r) ~ Λ for some non-trivial function f(r).
Turning on quantum mechanics, I get a new inherent energy scale
Eon=ℏT=ℏω,Eon=ℏT=ℏω,
where ω = 1/T is a frequency corresponding to the inverse of the time interval T. At this point, I shouldn't even mention that I'm discounting the possibility of setting FLf(V/Eon)∼ΛFLf(V/Eon)∼Λ rather than simply FL ~ Λ. However, the energy EonEon is important for other reasons. I've chosen T to be the turn-on time of the force, and the sudden perturbation approximation I'll be making is only valid when the change in energy ΔE is much less than EonEon . An unfortunate and unexpected problem arises. As stated, I've taken T = 0. This means that ΔE≪EonΔE≪Eon is automatically satisfied, but it also means that EonEon is infinite and thus on the same scale as Λ. In a certain sense, I've made the force "unstoppable" in two different ways. However, I'm fairly sure that that this "strength" is similar to the strength of an infinite range force. It would probably be better to approach this problem with a finite turn-on time T, but for simplicity I will not.
Picking a model
Now its time to pick an appropriate model for our confining potential well. The simplest possibility is
−VLδ(x)−VLδ(x)
Remember that V ~ Λ, and I want to take the limit Λ→∞Λ→∞ . It won't make sense to do this in the middle of the problem; instead I'll take the limit at the end (when I'll calculate a transition rate after the particle is hit with an unstoppable force) and hope it yields a finite quantity.
What about the unstoppable force? Recall that I want to describe a collision (i.e. a scattering event) where a force of infinite magnitude turns on at time t0t0 and acts over a region of width M. Well, a force of infinite magnitude in the positive x direction will suffice:
FLδ(x)θ(t−t0)x^FLδ(x)θ(t−t0)x^
This is the gradient of the scattering potential
−FLθ(x)θ(t−t0)−FLθ(x)θ(t−t0)
So, here is my plan of attack: I'll find the probability of a transition from the ground state after a perturbation representing the force. Then, I'll take the limit Λ→∞Λ→∞ and try to find an answer that depends only on α.
Classical
Originally, I only did this within the framework of quantum mechanics, but in principle, I could do this within the classical framework. A quick peak suggested that it might be very difficult, however. The Lagrangian is
I have no idea how to approach this differential equation, or if it is even well formed. Perhaps treating the well as a delta function works in quantum mechanics, but not classical mechanics.
Quantum
In the framework of quantum mechanics, the generator of dynamics is the Hamiltonian
Rather than try to directly solve the equations of motion for this time dependent Hamiltonian, I'll use the sudden perturbation approximation. At times t<t0t<t0 , the Hamiltonian is
Hi=−ℏ22M∇2−VLδ(x)Hi=−ℏ22M∇2−VLδ(x)
and at t=t0t=t0 is instantaneously perturbed to
Hf=−ℏ22M∇2−VLδ(x)−FLθ(x)Hf=−ℏ22M∇2−VLδ(x)−FLθ(x)
Is it safe to treat this as a perturbation when the applied force is very strong? Yes - the sudden perturbation approximation is valid as long as the change in energy is much less than ℏ/Tℏ/T , where T is the turn-on time of the force. Since I've taken T to be 0, this condition is automatically satisfied. This is a bit fast and loose, and admittedly a weak point in my analysis.
First, I'll need to calculate the initial state of the object for times t<t0t<t0 , which is the ground state of the Hamiltonian HiHi . The ground state wavefunction ψ(x,t)ψ(x,t) satisfies Hiψ=E0.Hiψ=E0. Away from x = 0 in either direction this is just the free Hamiltonian. Since we're looking for a bound ground state, the wavefunction should die off at either infinity. Moreover, there's a basic principle in QM that given a symmetric wavefunction and asymmetric wavefunction, the symmetric wavefunction is lower in energy (doesn't cross x axis -> less curvature -> less energy). So, the wavefunction must have the form Ae−λ|x|Ae−λ|x| for x≠0x≠0 (and by continuity must take the value A at x = 0). Now I can integrate around x = 0 to relate these arbitrary coefficients, getting
Substituting in the ansatz forms for ψ, this equality yields
λ=MVLℏ2=vEL≫1Lλ=MVLℏ2=vEL≫1L
The wavefunction is exponentially suppressed outside of the well with an infinitesimal decay length. This is just what I expect for an immovable object - its wavefunction overlaps negligibly with regions outside its confining well. Meanwhile, A is just a constant to make the norm of the wavefunction equal to one. Sadly I'll probably need it. You can check that A=λ√A=λ .
Now I need to know the possible final states for t>t0t>t0 . This means finding the eigenstates of the final Hamiltonian HfHf . Actually I won't need all of them - just the bound states, of which I believe there is only one. This can be solved much the same way as the previous one. For x < 0, the Hamiltonian is free and again must have the form Be−λ1|x|Be−λ1|x| . For x > 0, the Hamiltonian is again free, just with some constant potential term -FL. So, the wavefunction must have the form Be−λ2|x|Be−λ2|x| . The proportionality term is B on both sides because the wavefunction must be continuous, but the two decay lengths λ1λ1 and λ2λ2 are not necessarily equal, since the two sides have different constant potential terms (0 and -FL, respectively). The parameters λ1λ1 and λ2λ2 are related to the energy of the state:
and solving these two equations leaves, fascinatingly,
λ1,2=λ±F2V=λ±αLλ1,2=λ±F2V=λ±αL
Already you can see that the dimensionless ratio α = FL/2V will come into play. Both M and M are on the order of the infinite energy scale Λ, but the problem is starting to involve only their finite ratio. Moreover, things aren't looking good for the unstoppable force. It has only managed to modify the ground state wavefunction by splitting λ ~ Λ in two with a difference Δλ = 2α/L ~ 1/L.
Another thing can can be seen right away - α must be less than λL, or this wavefunction will be unbound. However, this is trivial since α ~ 1 and λL=VL=vE≫1λL=VL=vE≫1 . In my original, more confused answer, I didn't see that α must be much less than λL and devoted a lot of time trying to imbue meaning to the case that α ~ λL. This was because I short-shrifted a dimensional analysis of what characterizes the immovable object and unstoppable force (an analysis presented above in great detail).
Anyways, I'll sadly need the normalization constant B (it's great when you don't). Working it out, it is
Finally, onto the solution. We'll use the sudden perturbation approximation - since the Hamiltonian changes nearly instantly, the wavefunction doesn't have time to evolve. So, the amplitude that it stays in this bound state is simply the overlap of the two states:
What wins?
A) An unstoppable force
B) A unmovable object
The common answer to this problem is that both unstoppable forces and immovable objects are imaginary, and so it is impossible to answer the question. That's not a problem in and of itself. I can answer lots of questions about imaginary objects as long as they follow well defined imaginary rules. Immovable objects and unstoppable forces fail in this regard because they are contradictory terms. If our imaginary world contains an an object whose motion cannot be altered by any force, it cannot, by definition, contain a force that can move any object.
To say anything about immovable objects and unstoppable forces, then, I'll have to amend their definitions to avoid this contradiction. The simplest way is to put these two objects in a category unto themselves:
An immovable object is an object whose motion cannot be changed by any force, except possibly by an unstoppable force (see below).
An unstoppable force is a force strong enough to change the motion of any object, except possibly an immovable object (see above).
With these amended definitions, I can work with the both concepts simultaneously and rigorously. The idea is that the "immovability" of the object and "unstoppability" are described by yet-to-be-defined energy scales on the order of some huge energy Λ, which is far greater than the energy scales of any of the other forces and potentials in our imaginary world. So huge, in fact, that in the end I will take the limit Λ→∞Λ→∞ , in effect stating that the immovable object and unstoppable force are way above the scale of anything else in our imaginary world. Moreover, while both the "immovability" and "unstoppability" will be nearly infinite, their ratio will be finite and well-defined.
Dimensional analysis
Consider an immovable object of mass M. Mass is only an impediment to being accelerated quickly; even a tiny force can move an object of immense mass, albeit slowly. So, I don't interpret immovable to mean having immense mass. Instead, to be immovable by any finite force the object must be at the bottom of some infinite potential well. This means that it would take an infinite amount of energy to perturb the object and permanently alter its motion - something only the unstoppable force will be capable of doing. This potential well must have a characteristic width L, which is the width of the region to which the object is confined. The potential well must also have a depth V, a quantity with dimensions of energy. While the spirit of an immovable object implies that the width of the potential should be small, it is really the depth of the well that characterizes the object's immovability. The depth of the well V should accordingly be infinite, on the scale of Λ.
In non-relativistic classical mechanics, there is only one way of doing so. I must say that V ~ Λ. This is because V is the only quantity with dimensions of energy that can be formed out of M, L, and V.
This changes when allowing relativistic effects. To do so, I add the constant speed of light c to our bag of parameters. Now there is an extra independent quantity with dimensions of energy, Mc2Mc2 . This is the rest energy due to the objects mass. From V and Mc2Mc2 the two of these I can form the dimensionless ratio
vM=VMc2vM=VMc2
(Read this notation as "the potential V measured in units of the rest mass energy Mc2Mc2 "). I am now no longer completely restricted to setting V ~ Λ. In principle, I could let Vf(vM)∼ΛVf(vM)∼Λ for any real function f(vM)f(vM) . However, it is hard to imagine quantities of the form Vf(vM)Vf(vM) with more natural meaning than simply V, and it seems artificial to set some "unnatural" Vf(vM)Vf(vM) to be on the order of Λ. If you wanted, you could repeat this analysis, perhaps setting V2/M∼ΛV2/M∼Λ , although I don't know how this could be interpreted. If V was a momentum scale, this could be interpreted as kinetic energy, but V is the scale of the confining potential.
This is somewhat of a moot point, though, because I will only be using non-relativistic classical mechanics and quantum mechanics to analyze the interaction of the object and force, so the relativistic energy scale Mc2Mc2 is irrelevant. This is a lie; the entire idea of the problem at hand (see the definitions), that Λ is by far the largest energy scale in our toy world, requires that Λ≫Mc2Λ≫Mc2 . This makes the scenario highly relativistic. For simplicity I'm ignoring this. An interesting extension of this problem might consider relativistic effects, or might consider M ~ Λ so that the non-relativistic treatment is a good approximation.
What about including quantum mechanical effects? Then I can include ℏℏ among our constants, and now I can form the energy
E=ℏ2ML2E=ℏ2ML2
Up to some proportionality constant on the order of unity, this is the zero-point energy of the object confined to its well. In quantum mechanics, objects never have precisely zero momentum (see: Heisenberg's uncertainy principle) and thus can never have exactly zero energy. So, instead of resting exactly at the bottom of its potential well, the quantum object shuffles around with some nonzero kinetic energy. It is a natural energy scale in the problem, and just as before, I can form the dimensionless ratio
vE=VE=VML2ℏ2,vE=VE=VML2ℏ2,
which is just the potential V measured in units of E. Again, this opens up the possibility of letting Vf(vE)∼ΛVf(vE)∼Λ for any real function f(vE)f(vE) rather than just V ~ Λ. And again, there seems to be little reason for doing so. Naively, it seems to me that these quantities Vf(vE)Vf(vE) are artificial, and setting any of them to be on the scale of Λ (as I implicitly did in my original answer) is unnatural.
Next I'll consider an unstoppable force. The force must have some magnitude F, naturally, which is the quantity directly responsible for the unstoppability of the force. However, it must also have associated width and time scales. The first is the width of the region over which the force acts. There are really only three possibilities for the size LFLF of this width: LF≪LLF≪L , LF∼LLF∼L , or LF≫LLF≫L . Qualitatively, the first leads to a force with infinitesimal range and which is thus "weak," in a certain sense. The last leads to a force with effectively infinite range, and which "strong" in a similar sense. In my opinion, this is a different type of strength than that our unstoppable force should possess. For example, gravity and QCD (before color confinement) are both infinite range forces, but are vastly different in strength. So, I'll take the middle option and for simplicity let LF=LLF=L .
The second is the timescale over which the force turns on. I have a before-and-after problem, where I want to see what the effect of the unstoppable force is on the immovable object. Before some time t0t0 , the force will be turned off, but it will then then turn on, and it will take some time T (possibly very short) for it to turn on (it could turn on and then off again in this timescale, but I'm not examining this possibility). In my original answer, this time scale was T = 0. For simplicity, I will stick with that choice.
Classically, there is again only one intrinsic energy scale, which is FL, so I must take
FL∼ΛFL∼Λ
Since classically it also turned out that V ~ Λ, the problem will end up being defined by the ratio
α=FL2Vα=FL2V
Although I'm skipping a relativistic treatment, I'll still go through the dimensional analysis in this case. The quantities L and T can be combined to form a dimensionless ratio r = cT/L. This is essentially a measure of how far information can travel across the range L of the force during the duration T of the force. Presumably this could be important in a relativistic treatment, especially if r is much less than one. However, repeating myself yet again, it makes little sense to let FL f(r) ~ Λ for some non-trivial function f(r).
Turning on quantum mechanics, I get a new inherent energy scale
Eon=ℏT=ℏω,Eon=ℏT=ℏω,
where ω = 1/T is a frequency corresponding to the inverse of the time interval T. At this point, I shouldn't even mention that I'm discounting the possibility of setting FLf(V/Eon)∼ΛFLf(V/Eon)∼Λ rather than simply FL ~ Λ. However, the energy EonEon is important for other reasons. I've chosen T to be the turn-on time of the force, and the sudden perturbation approximation I'll be making is only valid when the change in energy ΔE is much less than EonEon . An unfortunate and unexpected problem arises. As stated, I've taken T = 0. This means that ΔE≪EonΔE≪Eon is automatically satisfied, but it also means that EonEon is infinite and thus on the same scale as Λ. In a certain sense, I've made the force "unstoppable" in two different ways. However, I'm fairly sure that that this "strength" is similar to the strength of an infinite range force. It would probably be better to approach this problem with a finite turn-on time T, but for simplicity I will not.
Picking a model
Now its time to pick an appropriate model for our confining potential well. The simplest possibility is
−VLδ(x)−VLδ(x)
Remember that V ~ Λ, and I want to take the limit Λ→∞Λ→∞ . It won't make sense to do this in the middle of the problem; instead I'll take the limit at the end (when I'll calculate a transition rate after the particle is hit with an unstoppable force) and hope it yields a finite quantity.
What about the unstoppable force? Recall that I want to describe a collision (i.e. a scattering event) where a force of infinite magnitude turns on at time t0t0 and acts over a region of width M. Well, a force of infinite magnitude in the positive x direction will suffice:
FLδ(x)θ(t−t0)x^FLδ(x)θ(t−t0)x^
This is the gradient of the scattering potential
−FLθ(x)θ(t−t0)−FLθ(x)θ(t−t0)
So, here is my plan of attack: I'll find the probability of a transition from the ground state after a perturbation representing the force. Then, I'll take the limit Λ→∞Λ→∞ and try to find an answer that depends only on α.
Classical
Originally, I only did this within the framework of quantum mechanics, but in principle, I could do this within the classical framework. A quick peak suggested that it might be very difficult, however. The Lagrangian is
I have no idea how to approach this differential equation, or if it is even well formed. Perhaps treating the well as a delta function works in quantum mechanics, but not classical mechanics.
Quantum
In the framework of quantum mechanics, the generator of dynamics is the Hamiltonian
Rather than try to directly solve the equations of motion for this time dependent Hamiltonian, I'll use the sudden perturbation approximation. At times t<t0t<t0 , the Hamiltonian is
Hi=−ℏ22M∇2−VLδ(x)Hi=−ℏ22M∇2−VLδ(x)
and at t=t0t=t0 is instantaneously perturbed to
Hf=−ℏ22M∇2−VLδ(x)−FLθ(x)Hf=−ℏ22M∇2−VLδ(x)−FLθ(x)
Is it safe to treat this as a perturbation when the applied force is very strong? Yes - the sudden perturbation approximation is valid as long as the change in energy is much less than ℏ/Tℏ/T , where T is the turn-on time of the force. Since I've taken T to be 0, this condition is automatically satisfied. This is a bit fast and loose, and admittedly a weak point in my analysis.
First, I'll need to calculate the initial state of the object for times t<t0t<t0 , which is the ground state of the Hamiltonian HiHi . The ground state wavefunction ψ(x,t)ψ(x,t) satisfies Hiψ=E0.Hiψ=E0. Away from x = 0 in either direction this is just the free Hamiltonian. Since we're looking for a bound ground state, the wavefunction should die off at either infinity. Moreover, there's a basic principle in QM that given a symmetric wavefunction and asymmetric wavefunction, the symmetric wavefunction is lower in energy (doesn't cross x axis -> less curvature -> less energy). So, the wavefunction must have the form Ae−λ|x|Ae−λ|x| for x≠0x≠0 (and by continuity must take the value A at x = 0). Now I can integrate around x = 0 to relate these arbitrary coefficients, getting
Substituting in the ansatz forms for ψ, this equality yields
λ=MVLℏ2=vEL≫1Lλ=MVLℏ2=vEL≫1L
The wavefunction is exponentially suppressed outside of the well with an infinitesimal decay length. This is just what I expect for an immovable object - its wavefunction overlaps negligibly with regions outside its confining well. Meanwhile, A is just a constant to make the norm of the wavefunction equal to one. Sadly I'll probably need it. You can check that A=λ√A=λ .
Now I need to know the possible final states for t>t0t>t0 . This means finding the eigenstates of the final Hamiltonian HfHf . Actually I won't need all of them - just the bound states, of which I believe there is only one. This can be solved much the same way as the previous one. For x < 0, the Hamiltonian is free and again must have the form Be−λ1|x|Be−λ1|x| . For x > 0, the Hamiltonian is again free, just with some constant potential term -FL. So, the wavefunction must have the form Be−λ2|x|Be−λ2|x| . The proportionality term is B on both sides because the wavefunction must be continuous, but the two decay lengths λ1λ1 and λ2λ2 are not necessarily equal, since the two sides have different constant potential terms (0 and -FL, respectively). The parameters λ1λ1 and λ2λ2 are related to the energy of the state:
and solving these two equations leaves, fascinatingly,
λ1,2=λ±F2V=λ±αLλ1,2=λ±F2V=λ±αL
Already you can see that the dimensionless ratio α = FL/2V will come into play. Both M and M are on the order of the infinite energy scale Λ, but the problem is starting to involve only their finite ratio. Moreover, things aren't looking good for the unstoppable force. It has only managed to modify the ground state wavefunction by splitting λ ~ Λ in two with a difference Δλ = 2α/L ~ 1/L.
Another thing can can be seen right away - α must be less than λL, or this wavefunction will be unbound. However, this is trivial since α ~ 1 and λL=VL=vE≫1λL=VL=vE≫1 . In my original, more confused answer, I didn't see that α must be much less than λL and devoted a lot of time trying to imbue meaning to the case that α ~ λL. This was because I short-shrifted a dimensional analysis of what characterizes the immovable object and unstoppable force (an analysis presented above in great detail).
Anyways, I'll sadly need the normalization constant B (it's great when you don't). Working it out, it is
Finally, onto the solution. We'll use the sudden perturbation approximation - since the Hamiltonian changes nearly instantly, the wavefunction doesn't have time to evolve. So, the amplitude that it stays in this bound state is simply the overlap of the two states:
What wins?
A) An unstoppable force
B) A unmovable object
The common answer to this problem is that both unstoppable forces and immovable objects are imaginary, and so it is impossible to answer the question. That's not a problem in and of itself. I can answer lots of questions about imaginary objects as long as they follow well defined imaginary rules. Immovable objects and unstoppable forces fail in this regard because they are contradictory terms. If our imaginary world contains an an object whose motion cannot be altered by any force, it cannot, by definition, contain a force that can move any object.
To say anything about immovable objects and unstoppable forces, then, I'll have to amend their definitions to avoid this contradiction. The simplest way is to put these two objects in a category unto themselves:
An immovable object is an object whose motion cannot be changed by any force, except possibly by an unstoppable force (see below).
An unstoppable force is a force strong enough to change the motion of any object, except possibly an immovable object (see above).
With these amended definitions, I can work with the both concepts simultaneously and rigorously. The idea is that the "immovability" of the object and "unstoppability" are described by yet-to-be-defined energy scales on the order of some huge energy Λ, which is far greater than the energy scales of any of the other forces and potentials in our imaginary world. So huge, in fact, that in the end I will take the limit Λ→∞Λ→∞ , in effect stating that the immovable object and unstoppable force are way above the scale of anything else in our imaginary world. Moreover, while both the "immovability" and "unstoppability" will be nearly infinite, their ratio will be finite and well-defined.
Dimensional analysis
Consider an immovable object of mass M. Mass is only an impediment to being accelerated quickly; even a tiny force can move an object of immense mass, albeit slowly. So, I don't interpret immovable to mean having immense mass. Instead, to be immovable by any finite force the object must be at the bottom of some infinite potential well. This means that it would take an infinite amount of energy to perturb the object and permanently alter its motion - something only the unstoppable force will be capable of doing. This potential well must have a characteristic width L, which is the width of the region to which the object is confined. The potential well must also have a depth V, a quantity with dimensions of energy. While the spirit of an immovable object implies that the width of the potential should be small, it is really the depth of the well that characterizes the object's immovability. The depth of the well V should accordingly be infinite, on the scale of Λ.
In non-relativistic classical mechanics, there is only one way of doing so. I must say that V ~ Λ. This is because V is the only quantity with dimensions of energy that can be formed out of M, L, and V.
This changes when allowing relativistic effects. To do so, I add the constant speed of light c to our bag of parameters. Now there is an extra independent quantity with dimensions of energy, Mc2Mc2 . This is the rest energy due to the objects mass. From V and Mc2Mc2 the two of these I can form the dimensionless ratio
vM=VMc2vM=VMc2
(Read this notation as "the potential V measured in units of the rest mass energy Mc2Mc2 "). I am now no longer completely restricted to setting V ~ Λ. In principle, I could let Vf(vM)∼ΛVf(vM)∼Λ for any real function f(vM)f(vM) . However, it is hard to imagine quantities of the form Vf(vM)Vf(vM) with more natural meaning than simply V, and it seems artificial to set some "unnatural" Vf(vM)Vf(vM) to be on the order of Λ. If you wanted, you could repeat this analysis, perhaps setting V2/M∼ΛV2/M∼Λ , although I don't know how this could be interpreted. If V was a momentum scale, this could be interpreted as kinetic energy, but V is the scale of the confining potential.
This is somewhat of a moot point, though, because I will only be using non-relativistic classical mechanics and quantum mechanics to analyze the interaction of the object and force, so the relativistic energy scale Mc2Mc2 is irrelevant. This is a lie; the entire idea of the problem at hand (see the definitions), that Λ is by far the largest energy scale in our toy world, requires that Λ≫Mc2Λ≫Mc2 . This makes the scenario highly relativistic. For simplicity I'm ignoring this. An interesting extension of this problem might consider relativistic effects, or might consider M ~ Λ so that the non-relativistic treatment is a good approximation.
What about including quantum mechanical effects? Then I can include ℏℏ among our constants, and now I can form the energy
E=ℏ2ML2E=ℏ2ML2
Up to some proportionality constant on the order of unity, this is the zero-point energy of the object confined to its well. In quantum mechanics, objects never have precisely zero momentum (see: Heisenberg's uncertainy principle) and thus can never have exactly zero energy. So, instead of resting exactly at the bottom of its potential well, the quantum object shuffles around with some nonzero kinetic energy. It is a natural energy scale in the problem, and just as before, I can form the dimensionless ratio
vE=VE=VML2ℏ2,vE=VE=VML2ℏ2,
which is just the potential V measured in units of E. Again, this opens up the possibility of letting Vf(vE)∼ΛVf(vE)∼Λ for any real function f(vE)f(vE) rather than just V ~ Λ. And again, there seems to be little reason for doing so. Naively, it seems to me that these quantities Vf(vE)Vf(vE) are artificial, and setting any of them to be on the scale of Λ (as I implicitly did in my original answer) is unnatural.
Next I'll consider an unstoppable force. The force must have some magnitude F, naturally, which is the quantity directly responsible for the unstoppability of the force. However, it must also have associated width and time scales. The first is the width of the region over which the force acts. There are really only three possibilities for the size LFLF of this width: LF≪LLF≪L , LF∼LLF∼L , or LF≫LLF≫L . Qualitatively, the first leads to a force with infinitesimal range and which is thus "weak," in a certain sense. The last leads to a force with effectively infinite range, and which "strong" in a similar sense. In my opinion, this is a different type of strength than that our unstoppable force should possess. For example, gravity and QCD (before color confinement) are both infinite range forces, but are vastly different in strength. So, I'll take the middle option and for simplicity let LF=LLF=L .
The second is the timescale over which the force turns on. I have a before-and-after problem, where I want to see what the effect of the unstoppable force is on the immovable object. Before some time t0t0 , the force will be turned off, but it will then then turn on, and it will take some time T (possibly very short) for it to turn on (it could turn on and then off again in this timescale, but I'm not examining this possibility). In my original answer, this time scale was T = 0. For simplicity, I will stick with that choice.
Classically, there is again only one intrinsic energy scale, which is FL, so I must take
FL∼ΛFL∼Λ
Since classically it also turned out that V ~ Λ, the problem will end up being defined by the ratio
α=FL2Vα=FL2V
Although I'm skipping a relativistic treatment, I'll still go through the dimensional analysis in this case. The quantities L and T can be combined to form a dimensionless ratio r = cT/L. This is essentially a measure of how far information can travel across the range L of the force during the duration T of the force. Presumably this could be important in a relativistic treatment, especially if r is much less than one. However, repeating myself yet again, it makes little sense to let FL f(r) ~ Λ for some non-trivial function f(r).
Turning on quantum mechanics, I get a new inherent energy scale
Eon=ℏT=ℏω,Eon=ℏT=ℏω,
where ω = 1/T is a frequency corresponding to the inverse of the time interval T. At this point, I shouldn't even mention that I'm discounting the possibility of setting FLf(V/Eon)∼ΛFLf(V/Eon)∼Λ rather than simply FL ~ Λ. However, the energy EonEon is important for other reasons. I've chosen T to be the turn-on time of the force, and the sudden perturbation approximation I'll be making is only valid when the change in energy ΔE is much less than EonEon . An unfortunate and unexpected problem arises. As stated, I've taken T = 0. This means that ΔE≪EonΔE≪Eon is automatically satisfied, but it also means that EonEon is infinite and thus on the same scale as Λ. In a certain sense, I've made the force "unstoppable" in two different ways. However, I'm fairly sure that that this "strength" is similar to the strength of an infinite range force. It would probably be better to approach this problem with a finite turn-on time T, but for simplicity I will not.
Picking a model
Now its time to pick an appropriate model for our confining potential well. The simplest possibility is
−VLδ(x)−VLδ(x)
Remember that V ~ Λ, and I want to take the limit Λ→∞Λ→∞ . It won't make sense to do this in the middle of the problem; instead I'll take the limit at the end (when I'll calculate a transition rate after the particle is hit with an unstoppable force) and hope it yields a finite quantity.
What about the unstoppable force? Recall that I want to describe a collision (i.e. a scattering event) where a force of infinite magnitude turns on at time t0t0 and acts over a region of width M. Well, a force of infinite magnitude in the positive x direction will suffice:
FLδ(x)θ(t−t0)x^FLδ(x)θ(t−t0)x^
This is the gradient of the scattering potential
−FLθ(x)θ(t−t0)−FLθ(x)θ(t−t0)
So, here is my plan of attack: I'll find the probability of a transition from the ground state after a perturbation representing the force. Then, I'll take the limit Λ→∞Λ→∞ and try to find an answer that depends only on α.
Classical
Originally, I only did this within the framework of quantum mechanics, but in principle, I could do this within the classical framework. A quick peak suggested that it might be very difficult, however. The Lagrangian is
I have no idea how to approach this differential equation, or if it is even well formed. Perhaps treating the well as a delta function works in quantum mechanics, but not classical mechanics.
Quantum
In the framework of quantum mechanics, the generator of dynamics is the Hamiltonian
Rather than try to directly solve the equations of motion for this time dependent Hamiltonian, I'll use the sudden perturbation approximation. At times t<t0t<t0 , the Hamiltonian is
Hi=−ℏ22M∇2−VLδ(x)Hi=−ℏ22M∇2−VLδ(x)
and at t=t0t=t0 is instantaneously perturbed to
Hf=−ℏ22M∇2−VLδ(x)−FLθ(x)Hf=−ℏ22M∇2−VLδ(x)−FLθ(x)
Is it safe to treat this as a perturbation when the applied force is very strong? Yes - the sudden perturbation approximation is valid as long as the change in energy is much less than ℏ/Tℏ/T , where T is the turn-on time of the force. Since I've taken T to be 0, this condition is automatically satisfied. This is a bit fast and loose, and admittedly a weak point in my analysis.
First, I'll need to calculate the initial state of the object for times t<t0t<t0 , which is the ground state of the Hamiltonian HiHi . The ground state wavefunction ψ(x,t)ψ(x,t) satisfies Hiψ=E0.Hiψ=E0. Away from x = 0 in either direction this is just the free Hamiltonian. Since we're looking for a bound ground state, the wavefunction should die off at either infinity. Moreover, there's a basic principle in QM that given a symmetric wavefunction and asymmetric wavefunction, the symmetric wavefunction is lower in energy (doesn't cross x axis -> less curvature -> less energy). So, the wavefunction must have the form Ae−λ|x|Ae−λ|x| for x≠0x≠0 (and by continuity must take the value A at x = 0). Now I can integrate around x = 0 to relate these arbitrary coefficients, getting
Substituting in the ansatz forms for ψ, this equality yields
λ=MVLℏ2=vEL≫1Lλ=MVLℏ2=vEL≫1L
The wavefunction is exponentially suppressed outside of the well with an infinitesimal decay length. This is just what I expect for an immovable object - its wavefunction overlaps negligibly with regions outside its confining well. Meanwhile, A is just a constant to make the norm of the wavefunction equal to one. Sadly I'll probably need it. You can check that A=λ√A=λ .
Now I need to know the possible final states for t>t0t>t0 . This means finding the eigenstates of the final Hamiltonian HfHf . Actually I won't need all of them - just the bound states, of which I believe there is only one. This can be solved much the same way as the previous one. For x < 0, the Hamiltonian is free and again must have the form Be−λ1|x|Be−λ1|x| . For x > 0, the Hamiltonian is again free, just with some constant potential term -FL. So, the wavefunction must have the form Be−λ2|x|Be−λ2|x| . The proportionality term is B on both sides because the wavefunction must be continuous, but the two decay lengths λ1λ1 and λ2λ2 are not necessarily equal, since the two sides have different constant potential terms (0 and -FL, respectively). The parameters λ1λ1 and λ2λ2 are related to the energy of the state:
and solving these two equations leaves, fascinatingly,
λ1,2=λ±F2V=λ±αLλ1,2=λ±F2V=λ±αL
Already you can see that the dimensionless ratio α = FL/2V will come into play. Both M and M are on the order of the infinite energy scale Λ, but the problem is starting to involve only their finite ratio. Moreover, things aren't looking good for the unstoppable force. It has only managed to modify the ground state wavefunction by splitting λ ~ Λ in two with a difference Δλ = 2α/L ~ 1/L.
Another thing can can be seen right away - α must be less than λL, or this wavefunction will be unbound. However, this is trivial since α ~ 1 and λL=VL=vE≫1λL=VL=vE≫1 . In my original, more confused answer, I didn't see that α must be much less than λL and devoted a lot of time trying to imbue meaning to the case that α ~ λL. This was because I short-shrifted a dimensional analysis of what characterizes the immovable object and unstoppable force (an analysis presented above in great detail).
Anyways, I'll sadly need the normalization constant B (it's great when you don't). Working it out, it is
Finally, onto the solution. We'll use the sudden perturbation approximation - since the Hamiltonian changes nearly instantly, the wavefunction doesn't have time to evolve. So, the amplitude that it stays in this bound state is simply the overlap of the two states:
What wins?
A) An unstoppable force
B) A unmovable object
Unstoppable force pierces through the unmovable object and doesn't stop, meanwhile while the object was pierced it didn't move so it remains unmovable.
no... because then the part that was pierced would have been move :p
Atoms are mostly empty space. It is not hard to imagine if such a scenario existed, the atoms could pass through the empty space.
atoms never really touch willingly. you think you are touching your keyboard right now but you actually aren't. What is actually happening is the atoms in the keyboard are repelling the atoms in your fingers.Want to know what happens when two atoms touch? Nuclear fusion. Atom bombs. All the sun's energy. Why does this happen? Because atoms repel each other and you have to overcome great force to get them to collide. At a certain distance, atoms can "link", and vibrate in place( if its a solid. For a hypothetical, true immovable object to exist, it would have to be a solid at 0 degrees Kelvin ( absolutely no vibration/heat which is never oralmost never found in space anywhere. So while atoms are 99% space, it is the ENERGY within the atom that repels other atoms, not the atom itself.
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The enemy is in front of us, the enemy is behind us, the enemy is to the right and left of us. They cant get away this time! - General Douglas Mcarthur
What wins?
A) An unstoppable force
B) A unmovable object
Unstoppable force pierces through the unmovable object and doesn't stop, meanwhile while the object was pierced it didn't move so it remains unmovable.
no... because then the part that was pierced would have been move :p
Atoms are mostly empty space. It is not hard to imagine if such a scenario existed, the atoms could pass through the empty space.
atoms never really touch willingly. you think you are touching your keyboard right now but you actually aren't. What is actually happening is the atoms in the keyboard are repelling the atoms in your fingers.Want to know what happens when two atoms touch? Nuclear fusion. Atom bombs. All the sun's energy. Why does this happen? Because atoms repel each other and you have to overcome great force to get them to collide. At a certain distance, atoms can "link", and vibrate in place( if its a solid. For a hypothetical, true immovable object to exist, it would have to be a solid at 0 degrees Kelvin ( absolutely no vibration/heat which is never oralmost never found in space anywhere. So while atoms are 99% space, it is the ENERGY within the atom that repels other atoms, not the atom itself.
Cool story bro. Trying to use real physics to explain abstract ideas such as unstoppable objects.
What wins?
A) An unstoppable force
B) A unmovable object
Unstoppable force pierces through the unmovable object and doesn't stop, meanwhile while the object was pierced it didn't move so it remains unmovable.
no... because then the part that was pierced would have been move :p
Atoms are mostly empty space. It is not hard to imagine if such a scenario existed, the atoms could pass through the empty space.
atoms never really touch willingly. you think you are touching your keyboard right now but you actually aren't. What is actually happening is the atoms in the keyboard are repelling the atoms in your fingers.Want to know what happens when two atoms touch? Nuclear fusion. Atom bombs. All the sun's energy. Why does this happen? Because atoms repel each other and you have to overcome great force to get them to collide. At a certain distance, atoms can "link", and vibrate in place( if its a solid. For a hypothetical, true immovable object to exist, it would have to be a solid at 0 degrees Kelvin ( absolutely no vibration/heat which is never oralmost never found in space anywhere. So while atoms are 99% space, it is the ENERGY within the atom that repels other atoms, not the atom itself.
Cool story bro. Trying to use real physics to explain abstract ideas such as unstoppable objects.
LOL "using real physics to describe abstract ideas" cmon man, you did the exact same thing xD... albeit in a much less formal and educated way.... All I was trying to do was put my two cents in on the topic and I proved one of your points wrong and you got all defensive. Ofc im going to use real physics to describe any idea, WHAT ELSE AM I SUPPOSED TO USE? FAKE PHYSICS? Abstract ideas? For real dude, are you one of those people who get all defensive ANY time you get proven wrong? I hate those people. It is ok to be incorrect sometimes you know. There was absolutely no need to criticize my post unless you have something insightful to say, which you didn't. So please rethink your approach to talking to people for the love of everyone you will ever talk to.
----
The enemy is in front of us, the enemy is behind us, the enemy is to the right and left of us. They cant get away this time! - General Douglas Mcarthur
What wins?
A) An unstoppable force
B) A unmovable object
The common answer to this problem is that both unstoppable forces and immovable objects are imaginary, and so it is impossible to answer the question. That's not a problem in and of itself. I can answer lots of questions about imaginary objects as long as they follow well defined imaginary rules. Immovable objects and unstoppable forces fail in this regard because they are contradictory terms. If our imaginary world contains an an object whose motion cannot be altered by any force, it cannot, by definition, contain a force that can move any object.
To say anything about immovable objects and unstoppable forces, then, I'll have to amend their definitions to avoid this contradiction. The simplest way is to put these two objects in a category unto themselves:
An immovable object is an object whose motion cannot be changed by any force, except possibly by an unstoppable force (see below).
An unstoppable force is a force strong enough to change the motion of any object, except possibly an immovable object (see above).
With these amended definitions, I can work with the both concepts simultaneously and rigorously. The idea is that the "immovability" of the object and "unstoppability" are described by yet-to-be-defined energy scales on the order of some huge energy Λ, which is far greater than the energy scales of any of the other forces and potentials in our imaginary world. So huge, in fact, that in the end I will take the limit Λ→∞Λ→∞ , in effect stating that the immovable object and unstoppable force are way above the scale of anything else in our imaginary world. Moreover, while both the "immovability" and "unstoppability" will be nearly infinite, their ratio will be finite and well-defined.
Dimensional analysis
Consider an immovable object of mass M. Mass is only an impediment to being accelerated quickly; even a tiny force can move an object of immense mass, albeit slowly. So, I don't interpret immovable to mean having immense mass. Instead, to be immovable by any finite force the object must be at the bottom of some infinite potential well. This means that it would take an infinite amount of energy to perturb the object and permanently alter its motion - something only the unstoppable force will be capable of doing. This potential well must have a characteristic width L, which is the width of the region to which the object is confined. The potential well must also have a depth V, a quantity with dimensions of energy. While the spirit of an immovable object implies that the width of the potential should be small, it is really the depth of the well that characterizes the object's immovability. The depth of the well V should accordingly be infinite, on the scale of Λ.
In non-relativistic classical mechanics, there is only one way of doing so. I must say that V ~ Λ. This is because V is the only quantity with dimensions of energy that can be formed out of M, L, and V.
This changes when allowing relativistic effects. To do so, I add the constant speed of light c to our bag of parameters. Now there is an extra independent quantity with dimensions of energy, Mc2Mc2 . This is the rest energy due to the objects mass. From V and Mc2Mc2 the two of these I can form the dimensionless ratio
vM=VMc2vM=VMc2
(Read this notation as "the potential V measured in units of the rest mass energy Mc2Mc2 "). I am now no longer completely restricted to setting V ~ Λ. In principle, I could let Vf(vM)∼ΛVf(vM)∼Λ for any real function f(vM)f(vM) . However, it is hard to imagine quantities of the form Vf(vM)Vf(vM) with more natural meaning than simply V, and it seems artificial to set some "unnatural" Vf(vM)Vf(vM) to be on the order of Λ. If you wanted, you could repeat this analysis, perhaps setting V2/M∼ΛV2/M∼Λ , although I don't know how this could be interpreted. If V was a momentum scale, this could be interpreted as kinetic energy, but V is the scale of the confining potential.
This is somewhat of a moot point, though, because I will only be using non-relativistic classical mechanics and quantum mechanics to analyze the interaction of the object and force, so the relativistic energy scale Mc2Mc2 is irrelevant. This is a lie; the entire idea of the problem at hand (see the definitions), that Λ is by far the largest energy scale in our toy world, requires that Λ≫Mc2Λ≫Mc2 . This makes the scenario highly relativistic. For simplicity I'm ignoring this. An interesting extension of this problem might consider relativistic effects, or might consider M ~ Λ so that the non-relativistic treatment is a good approximation.
What about including quantum mechanical effects? Then I can include ℏℏ among our constants, and now I can form the energy
E=ℏ2ML2E=ℏ2ML2
Up to some proportionality constant on the order of unity, this is the zero-point energy of the object confined to its well. In quantum mechanics, objects never have precisely zero momentum (see: Heisenberg's uncertainy principle) and thus can never have exactly zero energy. So, instead of resting exactly at the bottom of its potential well, the quantum object shuffles around with some nonzero kinetic energy. It is a natural energy scale in the problem, and just as before, I can form the dimensionless ratio
vE=VE=VML2ℏ2,vE=VE=VML2ℏ2,
which is just the potential V measured in units of E. Again, this opens up the possibility of letting Vf(vE)∼ΛVf(vE)∼Λ for any real function f(vE)f(vE) rather than just V ~ Λ. And again, there seems to be little reason for doing so. Naively, it seems to me that these quantities Vf(vE)Vf(vE) are artificial, and setting any of them to be on the scale of Λ (as I implicitly did in my original answer) is unnatural.
Next I'll consider an unstoppable force. The force must have some magnitude F, naturally, which is the quantity directly responsible for the unstoppability of the force. However, it must also have associated width and time scales. The first is the width of the region over which the force acts. There are really only three possibilities for the size LFLF of this width: LF≪LLF≪L , LF∼LLF∼L , or LF≫LLF≫L . Qualitatively, the first leads to a force with infinitesimal range and which is thus "weak," in a certain sense. The last leads to a force with effectively infinite range, and which "strong" in a similar sense. In my opinion, this is a different type of strength than that our unstoppable force should possess. For example, gravity and QCD (before color confinement) are both infinite range forces, but are vastly different in strength. So, I'll take the middle option and for simplicity let LF=LLF=L .
The second is the timescale over which the force turns on. I have a before-and-after problem, where I want to see what the effect of the unstoppable force is on the immovable object. Before some time t0t0 , the force will be turned off, but it will then then turn on, and it will take some time T (possibly very short) for it to turn on (it could turn on and then off again in this timescale, but I'm not examining this possibility). In my original answer, this time scale was T = 0. For simplicity, I will stick with that choice.
Classically, there is again only one intrinsic energy scale, which is FL, so I must take
FL∼ΛFL∼Λ
Since classically it also turned out that V ~ Λ, the problem will end up being defined by the ratio
α=FL2Vα=FL2V
Although I'm skipping a relativistic treatment, I'll still go through the dimensional analysis in this case. The quantities L and T can be combined to form a dimensionless ratio r = cT/L. This is essentially a measure of how far information can travel across the range L of the force during the duration T of the force. Presumably this could be important in a relativistic treatment, especially if r is much less than one. However, repeating myself yet again, it makes little sense to let FL f(r) ~ Λ for some non-trivial function f(r).
Turning on quantum mechanics, I get a new inherent energy scale
Eon=ℏT=ℏω,Eon=ℏT=ℏω,
where ω = 1/T is a frequency corresponding to the inverse of the time interval T. At this point, I shouldn't even mention that I'm discounting the possibility of setting FLf(V/Eon)∼ΛFLf(V/Eon)∼Λ rather than simply FL ~ Λ. However, the energy EonEon is important for other reasons. I've chosen T to be the turn-on time of the force, and the sudden perturbation approximation I'll be making is only valid when the change in energy ΔE is much less than EonEon . An unfortunate and unexpected problem arises. As stated, I've taken T = 0. This means that ΔE≪EonΔE≪Eon is automatically satisfied, but it also means that EonEon is infinite and thus on the same scale as Λ. In a certain sense, I've made the force "unstoppable" in two different ways. However, I'm fairly sure that that this "strength" is similar to the strength of an infinite range force. It would probably be better to approach this problem with a finite turn-on time T, but for simplicity I will not.
Picking a model
Now its time to pick an appropriate model for our confining potential well. The simplest possibility is
−VLδ(x)−VLδ(x)
Remember that V ~ Λ, and I want to take the limit Λ→∞Λ→∞ . It won't make sense to do this in the middle of the problem; instead I'll take the limit at the end (when I'll calculate a transition rate after the particle is hit with an unstoppable force) and hope it yields a finite quantity.
What about the unstoppable force? Recall that I want to describe a collision (i.e. a scattering event) where a force of infinite magnitude turns on at time t0t0 and acts over a region of width M. Well, a force of infinite magnitude in the positive x direction will suffice:
FLδ(x)θ(t−t0)x^FLδ(x)θ(t−t0)x^
This is the gradient of the scattering potential
−FLθ(x)θ(t−t0)−FLθ(x)θ(t−t0)
So, here is my plan of attack: I'll find the probability of a transition from the ground state after a perturbation representing the force. Then, I'll take the limit Λ→∞Λ→∞ and try to find an answer that depends only on α.
Classical
Originally, I only did this within the framework of quantum mechanics, but in principle, I could do this within the classical framework. A quick peak suggested that it might be very difficult, however. The Lagrangian is
I have no idea how to approach this differential equation, or if it is even well formed. Perhaps treating the well as a delta function works in quantum mechanics, but not classical mechanics.
Quantum
In the framework of quantum mechanics, the generator of dynamics is the Hamiltonian
Rather than try to directly solve the equations of motion for this time dependent Hamiltonian, I'll use the sudden perturbation approximation. At times t<t0t<t0 , the Hamiltonian is
Hi=−ℏ22M∇2−VLδ(x)Hi=−ℏ22M∇2−VLδ(x)
and at t=t0t=t0 is instantaneously perturbed to
Hf=−ℏ22M∇2−VLδ(x)−FLθ(x)Hf=−ℏ22M∇2−VLδ(x)−FLθ(x)
Is it safe to treat this as a perturbation when the applied force is very strong? Yes - the sudden perturbation approximation is valid as long as the change in energy is much less than ℏ/Tℏ/T , where T is the turn-on time of the force. Since I've taken T to be 0, this condition is automatically satisfied. This is a bit fast and loose, and admittedly a weak point in my analysis.
First, I'll need to calculate the initial state of the object for times t<t0t<t0 , which is the ground state of the Hamiltonian HiHi . The ground state wavefunction ψ(x,t)ψ(x,t) satisfies Hiψ=E0.Hiψ=E0. Away from x = 0 in either direction this is just the free Hamiltonian. Since we're looking for a bound ground state, the wavefunction should die off at either infinity. Moreover, there's a basic principle in QM that given a symmetric wavefunction and asymmetric wavefunction, the symmetric wavefunction is lower in energy (doesn't cross x axis -> less curvature -> less energy). So, the wavefunction must have the form Ae−λ|x|Ae−λ|x| for x≠0x≠0 (and by continuity must take the value A at x = 0). Now I can integrate around x = 0 to relate these arbitrary coefficients, getting
Substituting in the ansatz forms for ψ, this equality yields
λ=MVLℏ2=vEL≫1Lλ=MVLℏ2=vEL≫1L
The wavefunction is exponentially suppressed outside of the well with an infinitesimal decay length. This is just what I expect for an immovable object - its wavefunction overlaps negligibly with regions outside its confining well. Meanwhile, A is just a constant to make the norm of the wavefunction equal to one. Sadly I'll probably need it. You can check that A=λ√A=λ .
Now I need to know the possible final states for t>t0t>t0 . This means finding the eigenstates of the final Hamiltonian HfHf . Actually I won't need all of them - just the bound states, of which I believe there is only one. This can be solved much the same way as the previous one. For x < 0, the Hamiltonian is free and again must have the form Be−λ1|x|Be−λ1|x| . For x > 0, the Hamiltonian is again free, just with some constant potential term -FL. So, the wavefunction must have the form Be−λ2|x|Be−λ2|x| . The proportionality term is B on both sides because the wavefunction must be continuous, but the two decay lengths λ1λ1 and λ2λ2 are not necessarily equal, since the two sides have different constant potential terms (0 and -FL, respectively). The parameters λ1λ1 and λ2λ2 are related to the energy of the state:
and solving these two equations leaves, fascinatingly,
λ1,2=λ±F2V=λ±αLλ1,2=λ±F2V=λ±αL
Already you can see that the dimensionless ratio α = FL/2V will come into play. Both M and M are on the order of the infinite energy scale Λ, but the problem is starting to involve only their finite ratio. Moreover, things aren't looking good for the unstoppable force. It has only managed to modify the ground state wavefunction by splitting λ ~ Λ in two with a difference Δλ = 2α/L ~ 1/L.
Another thing can can be seen right away - α must be less than λL, or this wavefunction will be unbound. However, this is trivial since α ~ 1 and λL=VL=vE≫1λL=VL=vE≫1 . In my original, more confused answer, I didn't see that α must be much less than λL and devoted a lot of time trying to imbue meaning to the case that α ~ λL. This was because I short-shrifted a dimensional analysis of what characterizes the immovable object and unstoppable force (an analysis presented above in great detail).
Anyways, I'll sadly need the normalization constant B (it's great when you don't). Working it out, it is
Finally, onto the solution. We'll use the sudden perturbation approximation - since the Hamiltonian changes nearly instantly, the wavefunction doesn't have time to evolve. So, the amplitude that it stays in this bound state is simply the overlap of the two states:
Taking the limit Λ→∞Λ→∞ , vE=V/EvE=V/E approaches infinity, but the ratio α = FL/2V remains constant. Thus
P→Λ→∞1P→Λ→∞1
The immovable object wins!
Wow... either you copied that from somewhere or you are a genius :p
Whenever I get asked this question I usually just counter ask a question such as:
Isn't and unstoppable force an immovable object?
Isn't an immovable abject an unstoppable force?
I thought about it for a little bit and it made sense to me :/
----
The enemy is in front of us, the enemy is behind us, the enemy is to the right and left of us. They cant get away this time! - General Douglas Mcarthur
What wins?
A) An unstoppable force
B) A unmovable object
>>Snipped<<
The immovable object wins!
Just kidding though
I'm not familiar with Quantum Mechanics so you lost me there.
How long did it take you to type this? Unless it's copy-pasted?
Also, I've noticed that people here have assumed the unstoppable force to be kinetic or potential energy. What about gravitational? Like a black hole? Nothing known can escape a black hole's event horizon where the gravitational force is strongest, and thus can tentatively be classified as an unstoppable force.
Black holes are only detected by the radiation given off as it devours a....lets say, star. While the star matter revolves around the black hole like water in a giant sinkhole (think maelstrom), it gives off radiation which is detected by instruments on Earth and in orbit. But once it passes the event horizon, there's no escape for anything. Not even light. It'll be drawn in and reduced to trace Hawking radiation.
The super blackhole at the center of the galaxy, is slowly consuming it.
Admittedly, it's speculated, repeat: speculated that faster than light (FTL) objects might, might be able to escape a black hole's event horizon. However, even if, in our imaginary world, it is possible to go FTL (unlikely since Khal referred to Relativistic Physics), no one knows if they'd be able to escape.
So...gravity of a blackhole = unstoppable force? Discuss.
Moving on to the immovable object, to be truly immovable, it would require an infinite amount of inertia at rest, and moving it would require an infinite amount of force (not the same as an unstoppable force as an unstoppable force by definition cannot be stopped, while an infinite force can be stopped with infinite inertia).
Inertia is the tendency of an object to resist a change in state of motion or rest.
Now to acquire an infinite force, since Force(F)=ma, you'd require either an infinite mass, or an infinite acceleration.
Taking infinite mass, which can be acquired while travelling at the speed of light as per Einstein's e=mc2. Assuming it is possible to travel at the speed of light (and not above lightspeed as that would possibly cancel out my above black hole theory), we have our infinite force.
Infinite inertia at rest would also require infinite mass.
Thus Inertia at rest=∞, and F=∞×c=∞
Dividing the two, we get ∞/∞=1 meaning that the two forces cancel each other out.
However, in practice this is not so, since it is impossible for an object at rest to have infinite mass (ignoring that you can't travel at c-speed)
Thus, there is no such thing as an immovable object. If it existed in our imaginary world, then since it's inertia at rest and thus it's mass would be infinite, it would probably be possible to go c-speed i.e lightspeed, meaning we automatically have our infinite force which cancels out the immovability of the object in question.
More on topic, I'm now genuinely interested to try and work through that proof when I have a spare weekend (assuming I ever get out of this tabpocolypse)
so you lost me there.
Unless it's copy-pasted? 
Guys there really put the effort..
