Problem with differential form of Maxwell's third equation?

[vanesch]

literature) - http://www.angelfire.com/sc3/elmag/files/EM05FL.pdf Tomorrow I am going to take it to some professors and discuss it with them.

Well, the first equation is already wrong, so I stopped reading. You cannot permute the integral and the d/dt if the surface is moving.

This is easily verified: take the vector field a to be equal to (3,0,0) when (x,y,z) is within the cube (0<x<1, 0<y<1, 0<z<1) and consider a unit square parallel to the yz plane, which is in the plane x = 0.5 and which touches the Y and Z axis at t = 0, and is moving with a velocity = 5 in the z-direction.
Clearly, d a / dt = 0 because it is a static field, independent of t, so the right hand is 0. However, the left hand isn't: the flux over this moving surface is 3 at t=0, and decreases linearly to 0 at t = 0.2, so d phi/dt = -15.

 

[vanesch]

What about this website?

http://www.wolfram-stanek.de/maxwell_equations.htm

Sad.

There is a way to interpret this confusing story, and the hint of where this comes from is given here:

2. IF we are using this replacement ( ∂ / ∂ t ) by (d / dt) in case of moved bodies
never forget the necessary relations for field transformations
(i.e. Lorentz' transformation) hidden in the left sides of the equations (1a) - (4a)

The ' fields are the fields in the locally co-moving frame. THEN of course all this is correct: the E' field is the E-field AS SEEN BY A LOCALLY CO-MOVING OBSERVER, and not by the observer in the coordinate frame where we are "working from" (in which, for instance, the velocity is specified). So all this is correct IN A NON-INERTIAL FRAME.
This is what I hinted at in the beginning of this thread:
a constant B field for observer A is seen as an E-field plus a B field for a moving observer (the E field being v x B with v the velocity of B wrt A, and B the same, if the transformation is non-relativistic).
BUT IT IS EXTREMELY CONFUSING to use an E' field, which is defined for DIFFERENT OBSERVERS in different places !
Of course, for this co-moving observer, the seen E-field is the only one acting upon the test charge there, because relative to this observer, this test-charge is at rest (so the vxB term for the test charge is 0, given that the v OF THE CHARGE WRT THE LOCAL CO-MOVING OBSERVER is 0).

That is why normally, when talking about the E-field, we assume it to be a field *as seen by one and the same observer everywhere*.

 

[leright]

Well, the first equation is already wrong, so I stopped reading. You cannot permute the integral and the d/dt if the surface is moving.

This is easily verified: take the vector field a to be equal to (3,0,0) when (x,y,z) is within the cube (0<x<1, 0<y<1, 0<z<1) and consider a unit square parallel to the yz plane, which is in the plane x = 0.5 and which touches the Y and Z axis at t = 0, and is moving with a velocity = 5 in the z-direction.
Clearly, d a / dt = 0 because it is a static field, independent of t, so the right hand is 0. However, the left hand isn't: the flux over this moving surface is 3 at t=0, and decreases linearly to 0 at t = 0.2, so d phi/dt = -15.

no no no...there are direct quotes from texts like Jackson in that article that claim otherwise. Remember, total derivatives are different than partial derivatives and total derivatives CAN be commuted into the integral when there is a time varying loop. This is known as the convective derivative. http://mathworld.wolfram.com/ConvectiveDerivative.html

 

[leright]

Well, the first equation is already wrong, so I stopped reading. You cannot permute the integral and the d/dt if the surface is moving.

This is easily verified: take the vector field a to be equal to (3,0,0) when (x,y,z) is within the cube (0<x<1, 0<y<1, 0<z<1) and consider a unit square parallel to the yz plane, which is in the plane x = 0.5 and which touches the Y and Z axis at t = 0, and is moving with a velocity = 5 in the z-direction.
Clearly, d a / dt = 0 because it is a static field, independent of t, so the right hand is 0. However, the left hand isn't: the flux over this moving surface is 3 at t=0, and decreases linearly to 0 at t = 0.2, so d phi/dt = -15.

you're missing the difference between partial time derivatives and total time derivatives.

 

[leright]

Sad.

There is a way to interpret this confusing story, and the hint of where this comes from is given here:The ' fields are the fields in the locally co-moving frame. THEN of course all this is correct: the E' field is the E-field AS SEEN BY A LOCALLY CO-MOVING OBSERVER, and not by the observer in the coordinate frame where we are "working from" (in which, for instance, the velocity is specified). So all this is correct IN A NON-INERTIAL FRAME.
This is what I hinted at in the beginning of this thread:
a constant B field for observer A is seen as an E-field plus a B field for a moving observer (the E field being v x B with v the velocity of B wrt A, and B the same, if the transformation is non-relativistic).
BUT IT IS EXTREMELY CONFUSING to use an E' field, which is defined for DIFFERENT OBSERVERS in different places !
Of course, for this co-moving observer, the seen E-field is the only one acting upon the test charge there, because relative to this observer, this test-charge is at rest (so the vxB term for the test charge is 0, given that the v OF THE CHARGE WRT THE LOCAL CO-MOVING OBSERVER is 0).

That is why normally, when talking about the E-field, we assume it to be a field *as seen by one and the same observer everywhere*.

I am pretty confident that the expressions on that website work even when the inertial reference frame is not the same velocity as the moving charge. If this is the case then the original -partial of B wrt t will work just fine and there's no need to this mumbo jumbo.

Also, I keep thinking back to the Lorentz force equation, that most all textbook state, complementing the above 'modified' maxwell equations. IF it is really an E-field resulting from this motional EMF then the v x B term in should not be there in the Lorentz force equation, and the total force on a charge be reduced to qE. This is what happens when you call every damn field an E-field.

Also, the Vemf is DEFINED as the integral (E.dl) (work that is done on a charge, per unit charge, and the force per unit charge they HAPPEN to use is the E-field, so ANY EMF is caused by some kind of E-field, BY DEFINITION. So, Lorentz force equation CONTRADICTS the definition of EMF. I am so angry at the fellow that created this nonsense (not these websites but the original creators, failing to carefully formulate meaningfulterminology) right now it isn't even funny (this is including maxwell and hertz)

THis confusion alone has really shaken my confidence in the framework of physics, and I am afraid most every other facet of physics is the same way. It's a shame really. People (even the greatest physicists of all time) get too lost in the math, and forget to use their intuition, and think beyond the damn greek letters and equations. It is just a damned mess. Communicating the information in a sensible way is just as important as being able to calculate things to high degrees of accuracy.

I wouldn't be so dissappointed if it were only a few testbooks with this nonsensical garble, but when the most well respected textbooks of the field display the same nonsense, it makes me hurl.

 

[leright]

http://en.wikipedia.org/wiki/Convective_derivative

look at the identity here, regarding to convective derivative

 

[leright]

http://www.andrijar.com/phipps/index.html

ok, I take this back...this website clarifies almost all of my confusion, thank god. Here, in Hertzian electrodynamics, F = qE . Hertz calls everything an electric field.

So there are two camps you can lie in when you talk about electrodynamics. You can talk in terms of Hertzian electrodynamics, or Maxwellian electrodynamics. You cannot mix the two ideas. The Lorentz force is different in Herzian electrodynamics, and is actually called Hertzian force.

Personally, I like the Herzian theory much better, since it describes much more ideas in the same amount of equations, but it also expands the meaning of the electric field, claiming that all forces on charges are due to E-fields, and the B-fields present only produce E-fields that exert the forces. I think this is the best way to do things since EMF is defined as E integrated over a path. Wonderful. I wish textbooks would make this distinction more clear though.

I have learned a lot from this discussion. Vanesch, I appreciate you help, and I think we both gained from the struggles in this thread.

 

[vanesch]

http://www.andrijar.com/phipps/index.html
ok, I take this back...this website clarifies almost all of my confusion, thank god. Here, in Hertzian electrodynamics, F = qE . Hertz calls everything an electric field.

Yes, you can do that, it is what I said: you can define the "electric field" E' the electric field in the locally co-moving frame, which means you have a different coordinate frame in each point of space. In that co-moving frame, of course, the v of your test charge is 0, and the Lorentz force E + vxB reduces to E.

But the problem with that approach is that it only works for ONE velocity in a point. What do you do when you have two velocities in a point (like in a plasma, where you may have two species of particles with same or different charge, and which have different hydrodynamical velocities) ? Which frame are you now going to take to calculate E' ? You now have TWO E' in a single point, because you have two velocities.
In other words, E' is not a vector field (mapping from space into a vector space).

It is much, much easier to say that in the observer frame (whatever that is, but defined everywhere), there are two fields, E and B, and that the force on a moving charge with velocity v in that frame equals q (E + vxB).
E and B are then well-defined vector fields for this observer, independent of the motion of any test charge.

 

[jackiefrost]

Thanks to everyone participating in this discussion. This is a facinating subject!

As a sidelight, if any of you have access to the Feynman Lectures (vol 2, 17-1 and 17-2), he delves into this very topic though on a fairly elementary level.

 

[vanesch]

Also, I keep thinking back to the Lorentz force equation, that most all textbook state, complementing the above 'modified' maxwell equations. IF it is really an E-field resulting from this motional EMF then the v x B term in should not be there in the Lorentz force equation, and the total force on a charge be reduced to qE. This is what happens when you call every damn field an E-field.

Yes, of course, you'll have to choose. Or you define the E-field to be the E-field in the frame which is co-moving with the charge (so that the charge is stationary in this frame) and then of course the total force on the charge is always q.E ;
OR you define the E and B field wrt to an observer, and define the motion of a charge wrt that same observer (velocity v), and then the force on that charge is q(E + vxB).

But the first approach misses something important which the second one includes. The first approach needs to define a new E field for every moving charge around, even if they are in the same space point (cfr plasma physics for instance).
The second (Maxwell) approach is much more elegant, for the following reason. Maxwell says that, for a given observer (independent of any state of motion of a particle), to each point in space you can attach two vectors, called E and B. And if a particle is moving through that point with a velocity v, then it will experience a force q(E(x) + v x B(x)). It is the same E(x) field and the same B(x) field, no matter what the state of motion of the particle. We only need position and velocity. It doesn't matter, for instance, what its acceleration is. The force is a function of position, and velocity. You give me position and velocity and I tell you the force.
Newtonian gravity, for instance, is only dependent on position: m.g(x). It doesn't depend on velocity. But the electromagnetic force DOES depend on position and velocity.

The Hertzian approach needs to assign a field to each different state of motion, without the elegance of seeing that the contributions can be written in the form E + v x B.

EMF is simply the integral of the total force on charge carriers on the loop (which, hence, co-move with the loop). This force is then given by E + v x B of course. If the loop is stationary, it reduces to E, otherwise it includes the v x B contribution.

Also, the Vemf is DEFINED as the integral (E.dl) (work that is done on a charge, per unit charge, and the force per unit charge they HAPPEN to use is the E-field, so ANY EMF is caused by some kind of E-field, BY DEFINITION. So, Lorentz force equation CONTRADICTS the definition of EMF.

Yes, and no, because you have to consider each piece of loop in its own reference frame, in which it is at rest, so in which the velocity is 0 (although that velocity is not 0 of course wrt the original observer). So even if you add a v x B term, it doesn't contribute because v is by definition always 0. In this Hertzian approach, the velocity of a charged particle is always 0 wrt to the frame in which the field is defined, and hence the lorentz force always reduces to q E.



I am so angry at the fellow that created this nonsense (not these websites but the original creators, failing to carefully formulate meaningfulterminology) right now it isn't even funny (this is including maxwell and hertz)

THis confusion alone has really shaken my confidence in the framework of physics, and I am afraid most every other facet of physics is the same way. It's a shame really. People (even the greatest physicists of all time) get too lost in the math, and forget to use their intuition, and think beyond the damn greek letters and equations. It is just a damned mess. Communicating the information in a sensible way is just as important as being able to calculate things to high degrees of accuracy.

I wouldn't be so dissappointed if it were only a few testbooks with this nonsensical garble, but when the most well respected textbooks of the field display the same nonsense, it makes me hurl.

You don't find this kind of confusion in standard graduate texts. Most of these confusions are "pedagology" to make things "easier".

 

[leright]

Yes, you can do that, it is what I said: you can define the "electric field" E' the electric field in the locally co-moving frame, which means you have a different coordinate frame in each point of space. In that co-moving frame, of course, the v of your test charge is 0, and the Lorentz force E + vxB reduces to E.

But the problem with that approach is that it only works for ONE velocity in a point. What do you do when you have two velocities in a point (like in a plasma, where you may have two species of particles with same or different charge, and which have different hydrodynamical velocities) ? Which frame are you now going to take to calculate E' ? You now have TWO E' in a single point, because you have two velocities.
In other words, E' is not a vector field (mapping from space into a vector space).

It is much, much easier to say that in the observer frame (whatever that is, but defined everywhere), there are two fields, E and B, and that the force on a moving charge with velocity v in that frame equals q (E + vxB).
E and B are then well-defined vector fields for this observer, independent of the motion of any test charge.

No no no, F = qE also works when a charge is moving relative to a inertial reference frame when you are using the hertzian definition of the E-field, which claims that the magnetic force should really be considered an electric force due to an electric field, and the magnetic field simply induces the E-field that exerts the force. This, to me anyway, makes more sense simply because EMF is defined as the integral of E over a closed path the work that an E-field does on a charge). This definition of EMF claims the E-field is a mathematical description of anything that causes a force on a charged object, and this is how the Hertzian description of the E-field treats the subject.

Also, I would think F = qE still holds in Hertzian electrodynamics when there are multiple charges moving at different relative velocities wrt an inertial reference frame due to the superposition principle. The "E-field" (as it is defined by Hertz) due to charges q1, q2, q3, etc, moving in a B-field with different velocities would simply be the sums of the forces due to the individual charges, and each force would be divided by the respective charge that caused it.

If you want to use the idea that EMF is the integral of E-field over a closed loop (whcih everyone does) then the Hertz school of thought fits better, I would think. If you want to want to stick with Maxwell then you might want to define EMF at least as integral(E + V x B).dl, since EMF is not only caused by an E-field.

All of this really comes down to clear definitions of key terminology. There is no doubt we could correctly solve the problems without clear definitions, but physics isn't just about solving problems.

And the part you mentioned about it be impossible to define the E-field with the V x B term when there are distributions of charge (ring, for example) because all of the differential charges are moving at differnet speeds...well, in this case you would simply integral V x B over the contour path. The E-field would vary with time since the velocity at the different points on the ring varies with time, and also the B-field might vary with time, but this is problem. However, this really doesn't matter (outside of philosophical discussion) since the problems become rather trivial when solved with good ol' faraday's law (minus the rate of change of magnetic flux.)

 

[leright]

Thanks to everyone participating in this discussion. This is a facinating subject!

As a sidelight, if any of you have access to the Feynman Lectures (vol 2, 17-1 and 17-2), he delves into this very topic though on a fairly elementary level.

He delves into the problems with Maxwell's equations, or he delves into the subject of electrodynamics?

 

[leright]

But yes, having an expression that allows the E-field and B-field to be the same regardless of the motion of the reference frame or the velocity of the charge moving through the field may seem to be more elegant. However, how do you know this description is physically accurate.

But then another thought occurred to me. What if the charge is what we might consider stationary, but you are moving, and therefore the charge is moving. How do we think through this? Therefore a force acts on this charge (regardless of whether it's due to an E-field or not)? Maybe this velocity should be interpreted as the velocity relative to the B-field, instead? But what about if the B-field is time-varying? Then it is moving relative to the charge, regardless of the movement of the inertial reference frame. Maybe this idea encompasses the fact that curlE is minus the partial of B wrt t. The changing B-field (even if the inertial reference frame is not changing) causes an E-field that is perpendicular to the motion of the B-field, which is changing (which is what curlE implies...the curl of E points in the opposite direction of the changing B-field, so the E-field "swirls" around the changing B-field).

So, maybe, this should be interpreted as meaning all of these velocities are wrt to the B-field. Faraday did his experiments, I assume, only in a stationary reference frame, and therefore the reference frame of the B-field and the observer is the same speed.

So, maybe the reference frame of the observer means nothing in these problems, and V x B is the velocity of the charge movement wrt the B-field.

So, maybe the lesson here is (which I think is known to most people), is that induced E-fields are produced in such a way as to work against changing B-fields (so the magnetic flux produced by the E-field resists the B-field's changing flux), and induced B-fields are produced in such a way as to work against changing E-fields (so the electric flux produced by the B-field resists the E-field's changing flux). After this is all said and done, it is the resulting E-field that exerts all of the force on test charges.

I think I am getting somewhere, and I bet I can eventually explain special relativity in an intuitive way based on nothing more than the fundamental experimental observations in classical electrodynamics.

 

[vanesch]

Also, I would think F = qE still holds in Hertzian electrodynamics when there are multiple charges moving at different relative velocities wrt an inertial reference frame due to the superposition principle. The "E-field" (as it is defined by Hertz) due to charges q1, q2, q3, etc, moving in a B-field with different velocities would simply be the sums of the forces due to the individual charges, and each force would be divided by the respective charge that caused it.

I think we're talking next to each other here. In order for something to be a field, it needs to take on a value at a certain point in space. Now imagine that there are 3 particles, each with charge 1, moving through point (0,0,0). Particle A has velocity (1,0,0), Particle B is at rest, and Particle C has velocity (0,0,1). The Maxwell B-field is (0,1,0) and the Maxwell E-field is (2,0,0).

What's the value of the Hertzian "E-field" in point (0,0,0) so that I can calculate the forces on particles A,B and C ?
For particle A, it is (2,0,0) + (0,0,1) = (2,0,1)
For particle B, it is (2,0,0)
For particle C, it is (2,0,0) + (-1,0,0) = (1,0,0)

So for each different particle, we'd need a DIFFERENT value of the hertzian E-field in one and the same point (0,0,0).

If you want to use the idea that EMF is the integral of E-field over a closed loop (whcih everyone does) then the Hertz school of thought fits better, I would think.

Of course, because the starting point is wrong of course, and not everybody does that. With this definition of EMF, there would for instance, not be an EMF in a circuit with a battery. The EMF is the total force exerced on charges around the loop. These forces can be of electromagnetic, or of other origin (such as electrochemical).

If you want to want to stick with Maxwell then you might want to define EMF at least as integral(E + V x B).dl, since EMF is not only caused by an E-field.

Yes, and that's the correct way of course.

And the part you mentioned about it be impossible to define the E-field with the V x B term when there are distributions of charge (ring, for example) because all of the differential charges are moving at differnet speeds...well, in this case you would simply integral V x B over the contour path.

Look at my example above. I'm not talking about different V at different points, I' m talking about different V at the same point.

The E-field would vary with time since the velocity at the different points on the ring varies with time, and also the B-field might vary with time, but this is problem. However, this really doesn't matter (outside of philosophical discussion) since the problems become rather trivial when solved with good ol' faraday's law (minus the rate of change of magnetic flux.)

Yes, of course, for electric machines, Faraday's law is entirely correct.

 

[vanesch]

But yes, having an expression that allows the E-field and B-field to be the same regardless of the motion of the reference frame or the velocity of the charge moving through the field may seem to be more elegant. However, how do you know this description is physically accurate.

Well, because it is experimentally confirmed, and hence postulated as a law of nature! As all other parts of physics...

But then another thought occurred to me. What if the charge is what we might consider stationary, but you are moving, and therefore the charge is moving. How do we think through this?

Imagine that in the frame (A) where the charge is stationary, the E field is 0, and the B field is constant. The lorentz force is 0 in this case.
Now, go to a frame B which is moving with velocity V wrt A. In this frame (B), there is an electric and a magnetic field: E = V x B and B remains what it was. Now, the total Lorentz force in this frame, on the charge, which is moving with velocity -V in this frame, is given by:
F = q (E + (-V) x B) = q ( V x B + (-V) x B) = 0 too.

 

[lalbatros]

understanding electromagnetic water flow meters

Hello,

I posted the following a few weeks ago:

https://www.physicsforums.com/showthread.php?t=118872"

Actually, asking this is equivalent to ask:

how can a motional field be measured with a voltmeter,
and in the flowmeter, is the motional field really measured,
and is a voltmeter measuring and electric field,
and if yes does it mean that a motional field ...​

Any insight ?

Michel

 

[lalbatros]

is it a coïncidence ...

Let's have a look at the well known identity:

$$\frac{d}{dt} \int_{St} B.dS = \int_{St} \frac{\partial B}{\partial t}.dS + \int_{\partial St}v \times B .ds$$

The vXB term that you see there, this is just the motional field.
Is it a coïncidence that it is precisely this term that appears in the Lorentz force ?
And finally, why is called 'motional field' ?

 

[leright]

Imagine that in the frame (A) where the charge is stationary, the E field is 0, and the B field is constant. The lorentz force is 0 in this case.
Now, go to a frame B which is moving with velocity V wrt A. In this frame (B), there is an electric and a magnetic field: E = V x B and B remains what it was. Now, the total Lorentz force in this frame, on the charge, which is moving with velocity -V in this frame, is given by:
F = q (E + (-V) x B) = q ( V x B + (-V) x B) = 0 too.

It must be late, because I don't see how you got to the conclusion that E = V x B in the non-stationary reference frame. I see how you got the magnetic force though...

 

[leright]

Let's have a look at the well known identity:

$$\frac{d}{dt} \int_{St} B.dS = \int_{St} \frac{\partial B}{\partial t}.dS + \int_{\partial St}v \times B .ds$$

The vXB term that you see there, this is just the motional field.
Is it a coïncidence that it is precisely this term that appears in the Lorentz force ?
And finally, why is called 'motional field' ?

That's exactly what I've been bashing my head against the wall for days about. When Maxwell originally formulated the original equations he made the assumption that motional EMF was not occurring and the conditions were free space and when he commuted the derivative into the integral he only brought in a partial derivative. This inherently leaves a part of the EMF out of the picture if you define EMF as a force integrated around a closed path (motional EMF), but it is compatible with the Lotentz force equation commonly reffered to in textbooks. Also, as Vanesch pointed out, if an E-field is considered to be the cause of the force due to a moving charge in a b-field then 3 different charges moving at three different velocities pass a common point in space, they will all have different forces, or in other words they will all have differing field desciptions. This makes it difficult to deal with large groups of charge with different charges moving at varying velocities (like in a plasma), because each charge will have its own unique field description. A field should be a description of the means by which various particles are acted upon in space, regardless of the size, orientation, or velocity of said particles. Maxwell's E-fields allow such a consistent description. In Maxwell's formulation, the E-field is the same for all charges. Also, the B-field is the same for all moving charges. However, in Hertz's formulation, the E-field for each particle differs and depends on the respective velocities of each particle.

What I am curious about though, is if there is a charge moving in a uniform B-field with a force acting upon it, and this charge passes by another stationary charge, will the additional stationary charge feel the force set up by the B-field and the other moving charge, in addition to the force the stationary charge feels by the moving charge itself (due to the coulomb force). In addition, in what direction does this occur.

OK, I am getting exhausted and I need to get up at 6am tomorrow for work. peace.

 

[vanesch]

It must be late, because I don't see how you got to the conclusion that E = V x B in the non-stationary reference frame. I see how you got the magnetic force though...

The transformation rules of the (E,B) field from one coordinate frame to another are given by:

http://farside.ph.utexas.edu/teaching/jk1/lectures/node24.html

In the non-relativistic limit, this comes down to:

If (E,B) is the field in a coordinate frame 1, and (E',B') is the field in a coordinate frame 2 which has velocity V wrt the coordinate frame 1, then this reduces to:

E' = E + V x B
B' = B

This must be the case of course, if the Lorentz force is to be consistent:

If you want to have the same force to act in the two reference frames, you have to have, for a particle which has velocity v in frame 1 (and hence, non-relativistically, v - V in frame 2):

F = q (E + v x B) = q (E' + (v - V) x B')

for all possible velocities v, then you quickly find the above-cited transformation rules.This is what I meant with "the E and B fields do not transform as vectors under boosts" in my post #3 in this thread...cheers,
Patrick.

 

[creat]

Well, the first equation is already wrong, so I stopped reading. You cannot permute the integral and the d/dt if the surface is moving.

This is easily verified: take the vector field a to be equal to (3,0,0) when (x,y,z) is within the cube (0<x<1, 0<y<1, 0<z<1) and consider a unit square parallel to the yz plane, which is in the plane x = 0.5 and which touches the Y and Z axis at t = 0, and is moving with a velocity = 5 in the z-direction.
Clearly, d a / dt = 0 because it is a static field, independent of t, so the right hand is 0. However, the left hand isn't: the flux over this moving surface is 3 at t=0, and decreases linearly to 0 at t = 0.2, so d phi/dt = -15.

I am the author of the site. In the meantime I have updated the site:
http://www.angelfire.com/sc3/elmag
and the page:
http://www.angelfire.com/sc3/elmag/files/EM05FL.pdf

You are wrong with this example. The field on the unit square is not static, for
a = (3,0,0), for v*t <= z < 1
a = (0,0,0), for 1 <= z <= 1+v*t.
So d a / dt = -15 delta(v*t), (delta is Dirac's delta function)
After integration we have that left side is equal to the right.
Nobody's perfect, except the Heavenly Father