The Red Shift Mechanism

[Greg H.]

My sole contention here has been that you were incorrect when you said

 

 

A ball going east at 1 mile per hour and rebounded

at 1 mile per hour going west would have the same

kinetic energy but different momentum.

 

The other examples were simply more of the same.

 

You can derive the equations down to however many factors you like, but based on the provided, [math]p[/math], or [math](mv)[/math] if you prefer, cannot change without changing [math]E_k[/math] unless [math]m[/math] also changes. Either all three remain the same, or at least two of them have to change at the same time. This isn't even physics - it's math.

 

Basically what you said was [math]8 = \frac{16}{2}[/math] and then later that, [math]8 = \frac{8}{2}[/math]. (Same kinetic energy {8}, different momentum {8})

[imatfaal]

My sole contention here has been that you were incorrect when you said

 

 

 

The other examples were simply more of the same.

 

You can derive the equations down to however many factors you like, but based on the provided, [math]p[/math], or [math](mv)[/math] if you prefer, cannot change without changing [math]E_k[/math] unless [math]m[/math] also changes. Either all three remain the same, or at least two of them have to change at the same time. This isn't even physics - it's math.

 

Basically what you said was [math]8 = \frac{16}{2}[/math] and then later that, [math]8 = \frac{8}{2}[/math]. (Same kinetic energy {8}, different momentum {8})

 

 

Technically you can change a ball's momentum quite easily if you bounce a ball off a wall - momentum is a vector (whereas kinetic energy is a scalar); so you flip the sign. the linear momentum of a system with no external forces is conserved - but if your system is just the ball (which if the wall is solid and connected to earth is a reasonable approximation) then your momentum will be changed . Similarly if your comet started breaking up due to internal forces then your momentum would remain the the same - but your kinetic energy would increase

[Lazarus]

Meanwhile, back at the ranch, can someone tell me whats

wrong with the electron to stars calculation? Other than

the electron is supposed to radiate.

[Strange]

Meanwhile, back at the ranch, can someone tell me whats

wrong with the electron to stars calculation? Other than

the electron is supposed to radiate.

 

Can you explain what your program is supposed to calculate ... i.e. describe the steps it goes through -- it is a bit hard to read code formatted like that (I assume it is some dialect of Basic?); and not the best way to present a calculation anyway.

 

It looks like you are calculating the velocity of an electron (assume a classical model) and then using Doppler effect to calculate the frequency shift. And then relating the change in this velocity (and hence frequency shift) to the change in energy of energy level (or just kinetic energy?). Is that correct?

 

I don't see the connection to stars. Unless they are orbiting galaxies at the same velocity - which seems unlikely.

[Greg H.]

 

 

Technically you can change a ball's momentum quite easily if you bounce a ball off a wall - momentum is a vector (whereas kinetic energy is a scalar); so you flip the sign. the linear momentum of a system with no external forces is conserved - but if your system is just the ball (which if the wall is solid and connected to earth is a reasonable approximation) then your momentum will be changed . Similarly if your comet started breaking up due to internal forces then your momentum would remain the the same - but your kinetic energy would increase

That really doesn't make any sense to me at all. I don't want to hijack this thread anymore than I already have, but if you've got a more in depth explanation you could shoot me in a private message, I'd love to read it.

 

Edit:

Imatfaal was kind enough to provide a more in depth explanation that has corrected the error in my interpretation as to what was going on. Carry on without my prattling.

Edited July 31, 2013 by Greg H.

[Lazarus]

 

Can you explain what your program is supposed to calculate ... i.e. describe the steps it goes through -- it is a bit hard to read code formatted like that (I assume it is some dialect of Basic?); and not the best way to present a calculation anyway.

 

It looks like you are calculating the velocity of an electron (assume a classical model) and then using Doppler effect to calculate the frequency shift. And then relating the change in this velocity (and hence frequency shift) to the change in energy of energy level (or just kinetic energy?). Is that correct?

 

I don't see the connection to stars. Unless they are orbiting galaxies at the same velocity - which seems unlikely.

 

 

 

THE CONCEPT

 

The velocity of stars and galaxies can be calculated from their Red Shift wave length.

 

The wave length vs velocity can be calculated from the classical elliptical Bohr hydrogen atom.

 

 

The results are similar.

 

 

 

THE ASSUMPTIONS

 

Using the classical elliptical Bohr hydrogen atom:

 

The velocity of the electron at the apogee of shell 1 is 2,188,000 times .0030394. An arbitrary number that works.

 

The velocity of the electron is inversely proportional to the shell number.

 

The photon is released at the apogee.

 

 

 

THE CALCULATION STEPS

 

 

1 Calculate the energy of a photon caused by an electron moving from shell 3 to shell 2 using the Rydberg constant.. This is the total energy change of the electron, kinetic and latent.

 

2 Calculate the kinetic energy change between shell 2 and 3 with the atom not moving.

 

3 Give some velocity to the atom.

 

4 Add the atom's velocity to the initial velocities for the electrons at the apogee.

 

5 Calculate the new difference of kinetic energies for the 2 shells.

 

6 Adjust the initial energy by the kinetic energy change.

 

7 Calculate the wave length associated with that energy.

 

8 Use the wave length to calculate the velocity of the galaxy that sent the photon to us.

 

 

THE COMPARISON

 

Now use the calculated wave length to calculate the velocity with the "Standard" method.

 

The results are remarkably close.

[Strange]

The velocity of the electron at the apogee of shell 1 is 2,188,000 times .0030394. An arbitrary number that works.

Does this mean you have chosen a couple of arbitrary numbers in order to get the right answer?

 

8 Use the wave length to calculate the velocity of the galaxy that sent the photon to us.

Can you explain how you calculate the velocity of a galaxy from the wavelength?

 

Now use the calculated wave length to calculate the velocity with the "Standard" method.

How does the "standard method" calculate the velocity of a galaxy from a wavelength?

[Lazarus]

1 Does this mean you have chosen a couple of arbitrary numbers in order to get the right answer?

2 Can you explain how you calculate the velocity of a galaxy from the wavelength?

3 How does the "standard method" calculate the velocity of a galaxy from a wavelength?

 

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

 

1 The velocity of the electron in shell 1 at the apogee was determined

by trial and error.. A number different from it will give different results.

 

2 The post above shows the steps of the calculation.

 

3 The usual way of computing the velocity of galaxies is:

z=wobserve/wemit-1 z = observed wave length over emitted minus 1

v=(((z+1)**2-1)/((z+1)**2+1)) * c Velocity of source

[Strange]

1 The velocity of the electron in shell 1 at the apogee was determined [by trial and error.. A number different from it will give different results.

 

Right. So just numerology then. You say, "isn't it amazing that this process gives the right answer." No, not when you choose a fudge factor to give the answer you want.

 

2 The post above shows the steps of the calculation.

 

Yes, but as I said, it is very hard to follow the code.

 

3 The usual way of computing the velocity of galaxies is:

z=wobserve/wemit-1 z = observed wave length over emitted minus 1

v=(((z+1)**2-1)/((z+1)**2+1)) * c Velocity of source

So what I don't understand here, is that you are generating a single energy (and, therefore, wavelength) from your electron calculation. But the red-shift of galaxies depends on two wavelengths: the original and the shifted. How does this relate to your electron calculation?

OK. I have just had a look at your first post again. It looks like you use the 'z' calculation to work out a velocity for the electron and you find this comes to the velocity you started with. That doesn't seem in the least bit surprising as you have fudged the velocity corresponding to the energy level in step 1.

 

It is like one of those mathematical tricks: "think of a number, now multiply by ... take away the number you first thought of ... your number was ... !!"

Edited July 31, 2013 by Strange

[Lazarus]

Strange said:

"It is like one of those mathematical tricks: "think of a number, now multiply by ... take away the number you first thought of ... your number was ..."

 

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

 

Thank you for reading this thread.

 

It is hard to follow it around the loop of calculations

but there really is a distinction between the

result of the calculation and the testing of the result.

 

The calculation from the electron ends with the new

wave length. The calculated wave length is called

wave and the intial wave length is called lamda.

Lamda is the emitted wave length and wave is

the observed wave length.

 

Now to test it use the customary equations to see

how the results compare.

 

The program does the calculations for several

different atom velocities..

[swansont]

"An arbitrary number that works" disqualifies this as legitimate science.

 

How is it that we can get multiple redshift values from hydrogen?

[Lazarus]

 

Tom, you are one of the people whose opinion that I

value the most so I appreciate your comments.

 

*******

 

Swansont said:

"An arbitrary number that works" disqualifies this

as legitimate science.

 

The Hubble constant, the Gravitational contant, c

and other constants are arbitrary constants that work.

 

They would all be different arbitrary constants that work

in a different universe or if we used Klingon units.

 

*******

Swansont said:

How is it that we can get multiple redshift values from

hydrogen?

 

The choice of shells that the electron moves to and from

determines the color of the light on emission and the

velocity of the atom determines the shifted wave length.

 

********

 

The other objection that you have is that electrons in a

curved path have to radiate.

 

The "proof" that they radiate in a curved path is that the

electrons in a Synchrotron radiate. That "proof" is suspect.

 

The curvature of the path of the electrons is caused by the

magnetic field from the coils at the corners of the Synchrotron.

 

The approach of the electrons is at an angle to the coils so

there is a component of the velocity straight toward the coil

which would cause the electron to slow down, lose

kinetic energy and radiate.

 

Doesn't that bring the "proof" that electrons have to radiate

when in a curved path into question?

[swansont]

Tom, you are one of the people whose opinion that I

value the most so I appreciate your comments.

 

*******

 

Swansont said:

"An arbitrary number that works" disqualifies this [/size]

as legitimate science.[/size]

 

The Hubble constant, the Gravitational contant, c

and other constants are arbitrary constants that work.

 

They would all be different arbitrary constants that work

in a different universe or if we used Klingon units.

Their value does depend on the unit system, but that is not "arbitrary". An arbitrary constant takes on any value you want; you can choose it to be convenient. You can't choose H, G or c to be any value you want.

 

 

*******

Swansont said:

How is it that we can get multiple redshift values from

hydrogen?

 

The choice of shells that the electron moves to and from

determines the color of the light on emission and the

velocity of the atom determines the shifted wave length.

 

The emission spectrum is not continuous. Not all colors are represented.

 

********

 

The other objection that you have is that electrons in a

curved path have to radiate.

 

The "proof" that they radiate in a curved path is that the

electrons in a Synchrotron radiate. That "proof" is suspect.

 

The curvature of the path of the electrons is caused by the

magnetic field from the coils at the corners of the Synchrotron.

 

The approach of the electrons is at an angle to the coils so

there is a component of the velocity straight toward the coil

which would cause the electron to slow down, lose

kinetic energy and radiate.

 

Doesn't that bring the "proof" that electrons have to radiate

when in a curved path into question?

 

Explain how magnetic force does work to slow down a charged particle.

 

F = qv X B, and Work is a dot product: dW = F.dx How can dx and F not be perpendicular? (which makes the dot product zero)

[John Cuthber]

Well, it looks like I was right.

You can't do what he said, i.e. " the velocity of galaxies and stars can be computed by just assuming an electron orbiting a proton"

Even to get "close" you need to add all these data as well.

 

"hubble=4.675e-17' Hubble constant

rh=1.0973731e7' Rydberg constant

plank=6.626e-34' Plank's constant

c=3e8' Velocity of light in meters

shell1v=2.188e6' Velocity of electron in shell 1

v1=shell1v*velfact' Modified velocity of electron in shell 1

m=.911e-30' Mass of an electron

facte7=1e7' 10 to the 7th power

facte9=1e9' 10 to the 9th power

facte20=1e20' 10 to the 20th power"

And adding things in randomly till they give the answer you were looking for is numerology, not science.

[Lazarus]

Well, it looks like I was right.

You can't do what he said, i.e. " the velocity of galaxies and stars can be computed by just assuming an electron orbiting a proton"

Even to get "close" you need to add all these data as well.

 

"hubble=4.675e-17' Hubble constant

rh=1.0973731e7' Rydberg constant

plank=6.626e-34' Plank's constant

c=3e8' Velocity of light in meters

shell1v=2.188e6' Velocity of electron in shell 1

v1=shell1v*velfact' Modified velocity of electron in shell 1

m=.911e-30' Mass of an electron

facte7=1e7' 10 to the 7th power

facte9=1e9' 10 to the 9th power

facte20=1e20' 10 to the 20th power"

And adding things in randomly till they give the answer you were looking for is numerology, not science.

 

 

Are these things wrong? The seem to be used in lots of equations.

 

 

 

Swansont said:

Their value does depend on the unit system, but that is not "arbitrary". An arbitrary constant takes on any value you want; you can choose it to be convenient. You can't choose H, G or c to be any value you want.

 

Reply:

Ok. I shouldn't say arbitrary. That is just the value the velocity of the

electron must have at the apogee to give useful results.

 

 

************

 

Swansont Said:

Explain how magnetic force does work to slow down a charged particle.

F = qv X B, and Work is a dot product: dW = F.dx How can dx and F not

be perpendicular? (which makes the dot product zero)

 

Reply:

The electron is accelerated in the Synchrotron by magneiic forces

in a different part of the Synchrotron. At the corners the approach

of the electron to the coil is equivalent to an increasing magnetic

force which changes the kinetic energy of an electron.

 

********

 

Swansont said:

The emission spectrum is not continuous. Not all colors are represented.

 

Reply:

Yes, this is a biggie.

 

However, The distinct shells are possible with a rotating or pulsing

magnetic field from the nucleus. The orbit of the electron would

sync with the changing field. That means the second possible

orbit would take twice as long to orbit as the smallest orbit. The

trhird orbit would take 3 times as long, etc.

 

An interesting observation about the circular Bohr atom is that

in an atom with many electrons orbiting.every electron circling

back to the same spot would find the electrons is the innermost

shell in ether the same positions that they were or in reversed

positions. With elliptical orbits that would be a nice way for

electrons to take turns appoaching the nucleus.

 

Further, the circular distance between any two adjacent electrons

in any shell is the same.

 

As I said earlier, It seemed very improbable tor the nucleus

to have a rotating magnetic field but the Handbook of Nuclear

Properties of 1997 mentions nuclei with magnetic forces having a

rotational character.

 

 

It appears possible that the elliptical Bohr atom could produce results

consistant with the equations that are known to work.

[swansont]

Swansont said:

Their value does depend on the unit system, but that is not "arbitrary". An arbitrary constant takes on any value you want; you can choose it to be convenient. You can't choose H, G or c to be any value you want.

 

Reply:

Ok. I shouldn't say arbitrary. That is just the value the velocity of the

electron must have at the apogee to give useful results.

 

 

************

 

Swansont Said:

Explain how magnetic force does work to slow down a charged particle.

F = qv X B, and Work is a dot product: dW = F.dx How can dx and F not

be perpendicular? (which makes the dot product zero)

 

Reply:

The electron is accelerated in the Synchrotron by magneiic forces

in a different part of the Synchrotron. At the corners the approach

of the electron to the coil is equivalent to an increasing magnetic

force which changes the kinetic energy of an electron.

When I ask you to explain, I'm not looking for a repeat of what you just said. I'm asking how it's possible a magnetic field changes the kinetic energy of an electron, given (as I showed) that magnetic forces do no work.

 

 

 

********

 

Swansont said:

The emission spectrum is not continuous. Not all colors are represented.

 

Reply:

Yes, this is a biggie.

 

However, The distinct shells are possible with a rotating or pulsing

magnetic field from the nucleus. The orbit of the electron would

sync with the changing field. That means the second possible

orbit would take twice as long to orbit as the smallest orbit. The

trhird orbit would take 3 times as long, etc.

What "pulsing magnetic field from the nucleus" are you talking about?

 

 

An interesting observation about the circular Bohr atom is that

The Bohr model is wrong, as I stated earlier.

 

 

in an atom with many electrons orbiting.every electron circling

back to the same spot would find the electrons is the innermost

shell in ether the same positions that they were or in reversed

positions. With elliptical orbits that would be a nice way for

electrons to take turns appoaching the nucleus.

 

Further, the circular distance between any two adjacent electrons

in any shell is the same.

No such thing. The Bohr model is wrong.

 

As I said earlier, It seemed very improbable tor the nucleus

to have a rotating magnetic field but the Handbook of Nuclear

Properties of 1997 mentions nuclei with magnetic forces having a

rotational character.

Yes, nuclei have magnetic moments, which are related to their spin. They do not pulsate. The fields have the effect of splitting the energy of spin up vs spin down electrons. This effect is small. In hydrogen, for example, it means the 2 electron orientations are different by a frequency of 1420 MHz, which is a smidge under 5.9 micro-eV of energy.

 

 

It appears possible that the elliptical Bohr atom could produce results

consistant with the equations that are known to work.

Bohr orbits (which are wrong) are circular.

 

You're jumping the gun (in many ways). A semi-classical model with elliptical orbits would have to be tested first, before you can apply it and claim it to be responsible for a redshift. For starters, explain how you get an S state in hydrogen, which has zero orbital angular momentum, with this model.

[Arnaud Antoine ANDRIEU]

 

What "pulsing magnetic field from the nucleus" are you talking about?

 

 

It's something beyond your understanding.

[hypervalent_iodine]

!

Moderator Note

Arnaud Antoine ANDRIEU,

It would most wise of you not to go insulting other members. Besides it not being a tolerable form of behavior, you are skating on very thin ice here as it is and any further infractions may/will see you banned.

[Lazarus]

Swansont said:

When I ask you to explain, I'm not looking for a repeat of what you just said. I'm asking how it's possible a magnetic field changes the kinetic energy of an electron, given (as I showed) that magnetic forces do no work.

 

 

Reply:

I was totally wrong on this. Tom's insisting made me rethink it and the problem was that I was only looking at one dimension of the force vector. Please excuse.

 

I will go back into my cave and meditate on why the orbiting electron doesn't have to radiate. I should have it worked out in an eon or two.

 

 

 

Before I head for the cave I would like to mention a few more pieces of this model.

 

 

The Nuclear Binding Force does not exist. The nuclei consist of protons held together by electrons.

 

The neutron consists of a proton, 2 electrons and a positron.

 

The proton consists of electrons and positrons. The mass of the proton is evenly divisible by the difference of the mass of the proton and the neutron which implies a building block.

 

All matter and radiation consists of quite small entities that can be represented by positive and negative vectors that always travel at the speed of light.

 

 

 

This is the first time I have been able to get anyone knowledgeable to discuss any of this.

 

Thanks to all of you and especially thanks to Tom.

 

Ciao,

 

Bob

[Klaynos]

Nuclear physics works. So you are again wrong. Sorry.

[Strange]

The Nuclear Binding Force does not exist. The nuclei consist of protons held together by electrons.

 

The trouble is, you still have a net positive charge in the nucleus. So you need to explain how the protons stay together given the electrostatic repulsion.

 

 

The proton consists of electrons and positrons. The mass of the proton is evenly divisible by the difference of the mass of the proton and the neutron which implies a building block.

 

I don't think so. Using the values from the Particle Data Group:

Neutron mass: 1.008664916

Proton mass: 1.0072764668

Difference (delta): 0.0013884492

Proton/delta: 725.4687283608

 

Not exactly 'evenly divisible". But maybe you are thinking that looks close to the mass of an electron? Nope. It is 1.3224485911 times the mass of the electron.

 

 

This is the first time I have been able to get anyone knowledgeable to discuss any of this.

 

I can't imagine why.

[swansont]

The Nuclear Binding Force does not exist. The nuclei consist of protons held together by electrons.

 

The neutron consists of a proton, 2 electrons and a positron.

How do you keep electrons and positrons confined, in light of the Heisenberg Uncertainty Principle? To confine them to something the size of a proton, they must have a large uncertainty in momentum, which means they aren't all that confined. How do you reconcile that?

[Lazarus]

Klaynos said:

Nuclear physics works. So you are again wrong. Sorry.

 

Reply:

The equation named Roulette did a good job

of predicting the motion of the sun and planets

around the earth. Did that really prove that the

sun orbits the earth?

 

An equation can be made for any goofy concept

and a goofy concept can be made for any equation.

For example:

 

d=vt distance = velocity times time.

 

Postulate:

 

The Mouse God is always in the same absolute

location in space.

 

Collary:

 

When the Mouse God runs the universe moves

at velocity, v.

 

Consquence:

 

A to find the absolute velocity of a human

walking you must add the velocity of the

Mouse God to the velocity of the human

in ratty space.

 

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

 

Strange said:

I can't imagine why.

 

Reply:

 

You certainly understand my problem.

 

********

 

Strange said:

 

I don't think so. Using the values from the

Particle Data Group:Neutron mass: 1.008664916

Proton mass: 1.0072764668

Difference (delta): 0.0013884492

Proton/delta: 725.4687283608

Not exactly 'evenly divisible". But maybe you are

thinking that looks close to the mass of an electron?

Nope. It is 1.3224485911 times the mass of the electron.

 

 

Reply:

I goofed big time. It is the hydrogen atom not

the proton that should be compared with the

neutron.. See if I did it right this time.

 

The weight of the hydrogen atom is 1.007825

Atom Mass Units (938 mev) and the neutron is

1.008665 AMU (939 mev). The difference i

.000840 AMU (.789 mev) but a better number is

.00083985429 AMU.

 

If you multiply 1200 times .00083985429 you get

1.007825148, the weight of the hydrogen atom and

1201 times .00083985429 is 1.00866500229, the

weight of the neutron.

 

********

 

Strange said:

 

The trouble is, you still have a net positive

charge in the nucleus. So you need to

explain how the protons stay together given

the electrostatic repulsion.

 

 

Reply:

If you put 2 protons on opposite sides of and electron

the net charge is +1. However, the attraction between

the electron and a proton is much greater than the

repulsion of the protons.

 

Same is true if you put 3 protons around and electron.

The net charge is +2 but it will implode, not explode.

 

Even 4 protons will implode but just barely.

 

Over 20 years ago one of my judo buddies forced

a friend of his who had a PHD in nuclear physics

to talk to me. I described this and he said that he

had never heard it or thought of it.

 

 

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

 

Swansont Said:

 

You're jumping the gun (in many ways). A

semi-classical model with elliptical orbits

would have to be tested first, before you

can apply it and claim it to be responsible

for a redshift. For starters, explain how

you get an S state in hydrogen, which has

zero orbital angular momentum, with this model.

 

Reply:

Let's rotate the electron enough to zero the

rotational momentum. We could name it

spin.

 

********

 

Swansont said:

"Heisenberg Uncertainty Principle"

 

Reply:

I am somewhat in agreement with Bohr's

"Don't tell God what to do" but I agree

with Einstein's contention which is

something like, "The cause and

effect should not be thrown out. The lack

of knowledge or ability to measure is

behind the need for probability."

 

.

Tom has me snookered on justifying why

orbiting electrons don't radiate . I really

don't have a eon to figure it out. It would

be nice if someone could give me a hint.

[Strange]

See if I did it right this time.

 

The weight of the hydrogen atom is 1.007825

Atom Mass Units (938 mev) and the neutron is

1.008665 AMU (939 mev). The difference i

.000840 AMU (.789 mev) but a better number is

.00083985429 AMU.

 

If you multiply 1200 times .00083985429 you get

1.007825148, the weight of the hydrogen atom and

1201 times .00083985429 is 1.00866500229, the

weight of the neutron.

 

Close, but no banana.

 

Weight of H atom: 1.007825

Weight of Neutron: 1.008664916

Delta: 0.000839916

H / Delta: 1199.9116578323

N / Delta: 1200.9116578323

P / Delta: 1199.2585768099

 

Using your less accurate numbers gives a result even further from what you claim.

 

And what does "but a better number is .00083985429 AMU" mean?

It almost sounds like you have made it up to give the right answer...

 

None of the ratios appear to be exact or have any physical meaning.

 

If you put 2 protons on opposite sides of and electron

the net charge is +1. However, the attraction between

the electron and a proton is much greater than the

repulsion of the protons.

 

Same is true if you put 3 protons around and electron.

The net charge is +2 but it will implode, not explode.

 

Even 4 protons will implode but just barely.

Please show the math to support this. And show that it applies to atoms with large numbers of protons, such as lead and uranium.

 

 

It would be nice if someone could give me a hint.

 

Here's one: you're wrong.

[Unity+]

Postulate:

The Mouse God is always in the same absolute

location in space.

Collary:

When the Mouse God runs the universe moves

at velocity, v.

Consquence:

A to find the absolute velocity of a human

walking you must add the velocity of the

Mouse God to the velocity of the human

in ratty space.

This thread definitely belongs in the trash.

 

 

I am somewhat in agreement with Bohr's

"Don't tell God what to do" but I agree

with Einstein's contention which is

something like, "The cause and

effect should not be thrown out. The lack

of knowledge or ability to measure is

behind the need for probability."

Yes, but unexplained events do not warrant irrational thinking.