p. bomb paradoxfaculty.polytechnic.org/physics/1 astronomy, cosmology, relativity... ·...

BOMB PARADOX

A sta-onary bomb is a5ached to a nub-‐shaped detonator that will blow the bomb when depressed. The length of the nub is “d” units. A firing pin needed to ignite the bomb is located at the bo5om of a well-‐shaped ac-vator. The well’s depth is “2d.” Both pieces, WHILE STATIONARY, are shown to the right. At some point, the well-‐shaped ac-vator is accelerated to a velocity “v.” As presented, will the bomb go off?

bomb(stationary)

detonator

well-shaped activator

1.)

d

2d

firing pin

You really can’t answer this unless you know more, so let’s begin by assuming the well’s velocity is .866c (note that the sketch is not to scale—this problem will be remedied on the next slide). With that informa-on, we can write the rela-ve velocity as:

β = .866 = 32

γ =1

1− vc

⎛⎝⎜

⎞⎠⎟

2⎡

⎣⎢

⎤

⎦⎥

1/2

= 1

1− .866cc

⎛⎝⎜

⎞⎠⎟

2⎡

⎣⎢

⎤

⎦⎥

1/2

= 11− .75[ ]1/2

γ = 2

and the rela-vis-c factor as:

2.)

v = .866c

bomb(stationary)

detonator

well − shaped activator

firing pin

We will start by looking at the problem from the perspec-ve of the bomb being sta-onary. That means the moving ac-vator will length contract by a factor of leaving it with a well-‐depth of “d.”

β = .866c

bomb(stationary)

detonator

length-contracted well

1γ = .5

3.)

Although the scaled sketch suggests that the nub-‐end of the detonator will hit the bo.om of the ac1vator well (detona-ng the bomb) as the bomb package comes in contact with the outside edge of the ac1vator (this would otherwise stop the incursion, or so one might think), a space-‐-me diagram is really the best way to see what is happening in this problem. That is what we are about to look at.

As a ma5er of convenience, the sketch to the right iden-fies each part of the system and how each will be termed. (No-ce that with the ac-vator’s length contrac-on, the sketch is now to scale!)

bomb-end of detonator-nub (OR front-end of bomb-package)

nub-end of detonator

outside-edge of activator well

bottom of activator well

β = .866c

bomb

detonator

activator

We also need to define an origin-‐-‐an x’ = 0 point-‐-‐in the ac-vator’s frame of reference (this will be called the prime frame of reference). We will take that origin to be located at the outside edge of the well.

β = .866c

4.)

sketch of system at ct = 0 and ct ' = 0

outside-‐edge of ac1vator well just passing nub-‐end of detonator at ct = 0 and ct’ = 0.

We addi-onally need to decide when to “start the clock.” Let’s assume that when the outside-‐edge of the ac1vator-‐well just passes the nub-‐end of the detonator, both clocks start (that is, ct = 0 and ct’ = 0 at that point).

x = 0

For our purposes, the unprimed coordinate axes (x and ct) will be associated with the bomb’s frame of reference (again, we are star-ng out assuming the bomb is not what is moving). We need to define an origin. In this case, the ideal posi-on for x = 0 is the nub-‐end of the detonator (see sketch).

x ' = 0

Second, an event can simply be a point of interest on a world-‐line, though it is more commonly associated with a point where two world-‐lines cross. For instance, if the world-‐line of the bo.om of the ac1vator (where the firing pin is) were to cross the world-‐line of the nub-‐end of the detonator, the firing pin and detonator would be in the same place at the same -me . . . which is to say the bomb would blow. That event would occur when their world-‐lines crossed.

5.)

The only other two bits of amusement we need are more like reminders than anything else.

That means that even though a point on an object may be physically sta-onary with its “x” coordinate not changing, that point will s-ll be moving in -me. In other words, a point’s world-‐line will always, at a minimum, show mo-on along the “ct” axis.

First, a world-‐line tracks the mo-on of a POINT ON AN OBJECT on a space-‐-me diagram. It allows you to iden-fy where the point is (i.e., its posi-on coordinate “x” ) at a given instant in -me (i.e., its -me coordinate “ct”).

event 1(the nub-end of detonator reaches outside edge of activator-well)

1.) event 1: The outside edge of the ac1vator-‐well reaches the nub-‐end of detonator; this happens in the unprimed frame at x = 0, ct = 0.

6.)

There are three points of interest (events) to examine for this problem. They are:

β = .866c

bomb(stationary)

detonator

activator

x = 0

The ac-vator traveling with velocity moves a distance “d” in -me “t.” That means:

2.) event 2: The outside-‐edge of the ac1vator aligns with the bomb-‐end of the detonator-‐nub (it also comes in contact with the front edge of the bomb); in bomb’s frame, this happens at x = d, ct = d/ . (Minor point: how did we get this -me?)

event 2(the outside edge of well reaches the bomb and alignswith the bomb-end of the nub)

β

7.)

x = d bomb(stationary)

v = d / t ⇒ βc = d / t ⇒ ct = d /β

"v = βc"activator

event 2

β = .866c

x = 0

event 3(the nub-end of detonatorreaches bottom of well)

3.) event 3: The bo.om of ac1vator-‐well reaches the nub-‐end of the detonator; in bomb’s frame, this happens at x = 0 at -me ct = d/ .

β

8.)

bomb(stationary)

β = .866c

x = 0

Event 3 is the point where the world-‐line of the nub-‐end of the detonator crosses the world-‐line of the bo.om of the ac1vator-‐well (i.e., where the firing pin is).

Event 2 is the point where the world-‐line of the ac1vator’s front crosses the world-‐line of the bomb-‐end of the detonator-‐nub.

9.)

Comment: No-ce that the sketch for event 3 looks just like the sketch for event 2. How so? Again, events are defined as points of interest along a world-‐line. If two separate events happen simultaneously in a given frame of reference (like the bomb’s frame) but with different coordinates, the sketch that depicts each situa-on will look the same.

(bo5om of ac-vator-‐well strikes nub-‐end of detonator)

(front-‐end of ac-vator-‐well strikes bomb-‐end of detonator-‐nub—i.e., front-‐end of bomb)

(bo5om of ac-vator-‐well)

(nub-‐end of detonator)

(front-‐end of ac-vator-‐well)

(bomb-‐end of detonator-‐nub OR front-‐end of bomb)

event 2

bomb

detonator

activator

bomb

detonator

activator

The two events happen simultaneously in the bomb’s frame of reference, so the drawings that picture these two occurrences look the same.

event 3

We are going to do things in pieces to make the final space-‐-me diagram more understandable. To start, the space-‐-me diagram below tracks the world-‐line of the nub-‐end of the detonator, assumed “sta-onary” in this frame of reference, as that point proceeds through -me.

d−d10.)

β = .866c

WORLD-LINE of nub-end of the detonator(this point is physically stationary in this frame;look at sketch--its x coordinate does not change)

as the nub-end MOVES IN TIME(remember, everything is moving in time!)

x

ct

nub moving in time

nub-endof detonator

bomb-endof detonator-nub

x = 0

nub-endof detonator

x = 0

nub-endof detonator

nub-endof detonator

The space-‐-me diagram below tracks the world-‐line of the bomb-‐end of the detonator-‐nub (aka: the front of the bomb) assumed “sta-onary” in this frame of reference, as that point proceeds through -me.

d−d11.)

WORLD-LINE of bomb-end of detonator-nub(this point is physically stationary in this frame;look at sketch--its x coordinate does not change)

MOVING IN TIME

x

ct

x = 0

nub moving in time

nub-endof detonator

bomb-endof detonator-nub

β = .866c

x = 0

bomb-end of detonator-nub



d

x = d

The space-‐-me diagram below shows the world-‐lines of the nub-‐end of the detonator and the bomb-‐end of the detonator (this is the way these lines would normally be presented).

d−d12.)

x

ct

world-line ofnub-end ofdetonator

x = 0

world-line ofbomb-end ofdetonator-nub

We would like to do a similarly thoughkul presenta-on of two world-‐lines associated with the front-‐end of the ac1vator-‐well and the bo.om of the ac1vator-‐well. Before we do, we need to review a bit.

13.)

x

x '

ct

ct '1.) Shown are the unprimed axes (x and ct) associated with the “sta-onary” reference frame and the primed axes (x’ and ct’) associated with the “moving” reference frame. 2.) The moving axes have world-‐lines that are not at right angles with one another and appear squashed as viewed from the unprimed frame. This is because those axes are moving rela-ve to the “fixed” frame. 3.) The rela-ve velocity between the two frames is where the angle between the x and x’ axes is determined by .

β = .866,

β

Δx

ΔctΔxΔct

= .866

More review:

14.)

x

ct5.) A line of simultaneity (a line that includes all points in space associated with a given -me) in the unprimed coordinate system runs parallel to the x-‐axis. A number of these are shown to the right.

ct0 = 0

ct1

ct2

ct3

ct4

ct5

ct6

ct7

ct8

ct9

More review:

15.)

x

ct

6.) A line of constant posi-on (a line that includes all points in -me associated with a given point in space) in the unprimed coordinate system run parallel to the ct-‐axis. A number of these lines are shown to the right.

x0 = 0x1 x2 x3 x4 x5 x6 x7 x8 x9 x10

S-ll more review:

16.)

x

x '

ct

ct '7.) A line of simultaneity in the primed coordinate system runs parallel to the x’-‐axis. A number of these lines are shown to the right.

S-ll more review:

17.)

x

x '

ct

ct '8.) A line of constant posi-on in the primed coordinate system runs parallel to the ct’-‐axis. A number of these lines are shown to the right.

18.)

x

x '

ctct '

Note that if the rela-ve mo-on was directed toward the lem instead of toward the right, as would be the case if we were assuming it was the bomb that was moving toward the ac-vator instead of vice versa, the diagram that included lines of simultaneity would have looked like:

So what would the world-‐line of the front of the ac1vator-‐well look like? Well, as that point would be sta-onary in its own frame of reference (the primed frame), its x’ coordinate would not change and, as a consequence, its world-‐line path would parallel the ct’ axis. And as it is supposed to be at the origin of both the primed and unprimed coordinate systems at ct = 0 and ct’ = 0 (we defined the axes that way), it looks as though its path will follow along the ct’= 0 line.

d

−d

activator well's apparent width19.)

β = .866c

x

x 'ctct '

bottom of movingactivation-well

moving IN TIME

world-line offront of

activation-well

front of movingactivation-well

moving IN TIME

What would the world-‐line of the bo.om of the ac1vator-‐well look like? First, its world-‐line will parallel the world-‐line of the ac-vator’s front-‐end, which we already know. Also, at ct = 0, that point will be at x = -‐d. Punng this all together yields the line shown.

d

−d


x

x '

ct


additionallymoving IN TIME

world-line ofbottom of

activation-well

front of movingactivation-well


ct 'at ct = 0, x = −d x = 0

Punng together the world-‐lines of the front-‐end and bo.om of the ac1vator-‐well, as viewed from the bomb’s frame of reference, we get:

d

−d


x

x '

ct

ct '



world-line ofbottom of

activation-well

world-line offront-end of

activation-well

front-end of movingactivation-well


Punng everything together: The space-‐-me diagram below tracks the world-‐lines of the ends of the detonator-‐nub, assumed “sta-onary” in this frame, and the world-‐lines of the front-‐end and bo5om of the ac-vator-‐well, assumed “moving.” On it, the nub at event 2 clearly fits into the well (that is, the ac-vator’s front-‐end strikes the bomb’s front-‐end just as the nub strikes the bo5om of the ac-va-on-‐well with the bomb

d−d22.)

β = .866c

event 1(the outside edge of wellreaches the detonator)

event 2(the bottom of the wellreaches the detonator)

event 3

β = .866c

(the outside edge of well reaches the bomb)

activation well

bottom of activation-well andand end of nub come in contact: B O O M!!!

x

x '

ct ct '

event 1

event 3event 2

world-lineof nub-end of

detonator

world-lineof bomb-end of

nub

nub

going BOOM!).

This is a more bare-‐bones version of the world lines and events in ques-on. It is more like what you would see in a textbook or on a test (that is, pictures are not superimposed on the diagram).

−d

23.)

activation well

(bottom of activation-well andand end of nub come in contact: B O O M!!!)

x

x '

ct ct '

event 1

event 3event 2

world-lineof nub-end of

detonatorworld-line

of bomb-end of nub

nub

(front-end of well andnub-end align at

ct = 0, x = 0, ct ' = 0, x ' = 0)

d

β = .866c

event 1(the outside edge of wellreaches the detonator)

event 2(the bottom of the wellreaches the detonator)

event 3

β = .866c

(the outside edge of well reaches the bomb)

from the detonator’s frame of reference, that diagram shows that event A occurs later than event B with event A happening between ct’ and ct’ and event B happening at ct’ (look at the parallel -me lines for the slanted, primed coordinate -me axis).

world-line ofnub-end ofdetonator

world-line ofbomb-end ofdetonator-nub

Parenthe-cal comment: Look at the diagram from the unprimed perspec-ve. Both event 2 and event 3 occur at the same -me in that frame (LOOK AT THE SKETCH—both events sit on the same horizontal -me line at ct ). That’s what our sketch, drawn to scale, maintains has happened. What is interes-ng is that looking at those two events

d

−d


x

x '

ct

ct '

world line ofbottom of

activation well

world line offront of

activation well

event A

event B

2

1

3

B

ct B

ct A

ct B ct

B

ct A

ct A

ct C

ct C

In short, events A and B are simultaneous in the unprimed frame of reference and NOT simultaneous in the primed frame. Interes-ng, eh?

v = .866c

well(assumed stationary) bomb

(length contracted)

So far, so good. Let’s now look at the system from the frame of reference of the ac-vator well (that is, assume the ac-vator well is sta-onary and the nub and bomb are moving to the lem). The rela-vis-c factors s-ll hold at , but now the ac-vator-‐well (being assumed to be sta-onary) has a length of 2d, as originally defined, and the nub and bomb are length contracted. The nub is now d/ = d/2 units in length. See sketch below.

β = .866 and γ = 2

25.)

γ

β = .866c

event 1(the nub-end of detonator reaches the outside edge of activator)

26.)

Although it is now the bomb and nub that are length contracted, there are s-ll three points of interest (events) to examine. They are iden-fied below:

x ' = γ x − βct( )

=2 0 −3

2⎛

⎝⎜⎞

⎠⎟c 0( )

⎛

⎝⎜

⎞

⎠⎟

⇒ x' = 0

t ' = γ ct − βx( )

=2 c 0( ) − 32

⎛

⎝⎜⎞

⎠⎟0( )

⎛

⎝⎜

⎞

⎠⎟

⇒ t' = 0

1.) event 1: The outside edge of the ac1vator-‐well reaches the nub-‐end of detonator; this happens in the unprimed frame at x = 0, ct = 0. For the sake of completeness, this happens in the primed system at:

2.) event 2: The bomb-‐end of the moving detonator-‐nub aligns with the outside-‐edge of the sta1onary ac1vator (it also comes in contact with the front edge of the bomb); in bomb’s frame, this happens at x = d, ct = d/ . Again, for completeness, in the ac-vator’s primed frame this happens at:

β = .866c

event 2(the outside edge of well reaches the bomb)

β

27.)

x ' = γ x − βct( )

=2 d −3

2⎛

⎝⎜⎞

⎠⎟d3

2⎛⎝⎜

⎞⎠⎟

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

⇒ x' = 0

t ' = γ ct − βx( )

=2 d3

2⎛⎝⎜

⎞⎠⎟

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟−

32

⎛

⎝⎜⎞

⎠⎟d( )

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

= 2d 13

2⎛⎝⎜

⎞⎠⎟

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟−

32

⎛⎝⎜

⎞⎠⎟

32

⎛⎝⎜

⎞⎠⎟

32

⎛⎝⎜

⎞⎠⎟

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

= 2d 13

2⎛⎝⎜

⎞⎠⎟

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟−

34

⎛⎝⎜

⎞⎠⎟

32

⎛⎝⎜

⎞⎠⎟

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

= 2d2 − 3

23

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟= 2d 1

2 3⎛⎝⎜

⎞⎠⎟

⇒ t' = d3

Note: In the ac-vator’s frame of reference, the outside edge of the well was originally defined (see slide 4) as being at x’ = 0. In the ac-vator’s frame, that point hasn’t changed posi-on (the ac-vator is sta-onary in its frame). In other words, the calculated x’ value makes sense.

bomb ac-vator well

event 3(the nub of detonator reachesbottom of well . . . assuming it could)

28.)

x ' = γ x − βct( )

=2 0 −3

2⎛

⎝⎜⎞

⎠⎟d3

2⎛⎝⎜

⎞⎠⎟

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

⇒ x' = −2d

t ' = γ ct − βx( )

=2 d3

2⎛⎝⎜

⎞⎠⎟

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟−

32

⎛

⎝⎜⎞

⎠⎟0( )

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

⇒ t' = 4d3

ac-vator well

3.) event 3: The nub-‐end of the detonator reaches the bo.om of ac1vator-‐well and the bomb goes off; in bomb’s frame, this happens at x = 0 at -me ct = d/ . Once again, for completeness, in the ac-vator’s frame this happens at:

β

Note: Again, this x’ coordinate makes sense. The well in the ac-vator’s frame had a depth of 2d. As it was located to the lem of x’ = 0, it isn’t surprising that its calculated value would be x’ = -‐2d (hee, hee—isn’t this fun?).

In the ac-vator-‐well’s frame of reference, assumed to be sta-onary, the primed coordinates are at right angles to one another. The front of the ac-vator-‐well is defined to be at x’ = 0, so its world-‐line proceeds through the origin along the ct’ axis (remember, the world-‐line for a point that has a fixed posi-on always runs parallel to the -me axis). The world-‐line for the front of the ac1vator-‐well is shown below.

−2dactivator well depth

29.) x '

ct '

β = .866c

event 1(the nub end of detonator reaches outside edge of activator)



x ' = 0

world-line of front of activator-well

as it proceedsthrough time

As before, we are going to do the space-‐-me diagram in pieces.

The ac-vator well’s depth in the ac-vator’s frame of reference was 2d. As the bo.om of the ac1vator well is to the lem of x’ = 0, its coordinate is always x’ = -‐ 2d. The world-‐line for the bo.om of the ac1vator-‐well is shown below.


30.) x '

ct '

β = .866c




x ' = 0

world-line of bottom of activator-well



31.) x '

ct '

β = .866c




x ' = 0

Punng the ac-vator well’s situa-on together onto one space-‐-me diagram yields:

world-line of bottom of activator-well


world-line of front of activator-well


The nub-‐end of the detonator is located at x = 0 at ct = 0 and at x’ = 0 at ct’ = 0 (that is how we defined those zeros). In other words, the nub-‐end has to proceed through the origin of both coordinate axes. (This was event 1.) Also, because that point isn’t

32.)

x

x '

ctct '

event 1

β = .866c




moving in the unprimed frame, it’s x coordinate must always be the same in that frame. Remembering that a point that is fixed in space has a world-‐line that is parallel to the -me axis-‐-‐in this case, the ct = 0 line-‐-‐ we get the world-‐line shown to the right.

We know that the world-‐line for the bomb-‐end of the nub (the front of the bomb package) will parallel the world-‐line of the nub-‐end. We also know that the front of the bomb will be d/2 units ahead of the nub-‐end due to length contrac-on. Punng this all together yields the world line shown below.

33.)

x

x '

ctct '

event 1

β = .866c




d/2

Punng the four world-‐lines together with a li5le artwork (the artwork will be removed in the next slide), we get the following space-‐-me diagram.

−2d

34.)

x

x '

ctct '

event 1

event 2

activator

well's

front

d/2

β = .866c




well's

bottom nub

activator well depth

−2d

35.)

x

x '

ctct '

event 1

event 2

activator

well's

front

d/2

β = .866c




well's

bottom

In bare-‐bones form (the way you would normally see it in a textbook, or where ever):

Here comes the fun stuff: It seems reasonable to assume that when the front of the bomb hits the front of the ac1vator-‐well (event 2), the bomb and nub will crumple together the same way the hood of a sta-onary car and the hood of a speed car will crumple together if the speeding car rams into the sta-onary car. What’s more, the sketch of our ac-vator/nub situa-on makes it look as though those two crumpled surfaces will bring the ac-vator to a stop long before event 3 can come to pass (that is, it looks as though the nub-‐end will never reach the bo.om of the well so the bomb will never be ac-vated). I have only one word in response to that. OH, CONTRARE! (well, OK, maybe two words) But how can I make such an outrageous statement?

36.)

β = .866c


event 3(the nub of detonator reaches bottom of well . . . assuming it could)


x = 0

When event 2 occurs, the bomb’s front end and the ac1vator’s front surface immediately begin to slow toward rest with respect to one another. That’s all fine and well for those two surfaces, but how does the lem-‐end of the nub located “d” units from the bomb’s front end know it’s supposed to stop? That is, what is the physical mechanism that brings the nub’s end to a halt? That mechanism is the consequence of a kind of domino effect. Molecules at the collision surface slow thereby pulling back on the molecules next to them. Those molecules slow pulling back on the molecules next to them. In other words, it is as though individual molecules are responding to a signal that propagates through the material, a signal whose speed is governed by the elas-c nature of the molecular bonding of the structure.

37.)

β = .866c

this surface stops upon contact

this surface con-nues to move for a while

−2d activator well depth38.)

nub

ct '

event 1

event 2

event 3

activator

well's

front

d/2

well's

bottom

(at this point the nub-‐end comes in contact with the bo5om of the well as the signal to stop has not yet arrived—in other words, BOOM!)

(at this point the bomb’s front end comes in contact with the well’s front end)

(nub enters well)

So let’s assume this signal is emi5ed when the collision occurs. Let’s also assume the signal moves at the speed of light (being a mechanical effect, the actual “signal” is nowhere close to that fast, but assume it is). Punng a light-‐line on our space-‐-me diagram emana-ng from event 2, we see that the light will arrive at the nub-‐end (the lem ver-cal line on the space-‐-me diagram) AFTER event 3 occurs (i.e., amer the nub-‐end strikes the bo.om of the ac1vator-‐well). In other words, the nub-‐end will set off the bomb BEFORE the light beam signal reaches it signaling that it is -me to begin slowing. As the actual signal moves much, much slower than light, this effect is even more pronounced and the nub-‐end will posi-vely set off the bomb.

Put a li5le differently, during the -me it takes the signal to reach the end of the nub, the nub will have apparently GROWN to the length of the well (plus some if the bomb doesn’t go off) and its contact with the well’s bo5om will set off the bomb. Graphically (altering the sketch some to accommodate commentary), this looks like:

39.)

1.) event 2(blue bomb stopped, nub-‐end s-ll moving)

2.) event 2++

(blue part of nub stopped, aqua part of nub s-ll moving)

3.) event 2++++

(blue part of nub stopped, aqua part of nub s-ll moving)

4.) event 3(blue part of nub stopped, aqua part of nub s-ll moving)

β = .866c

β = .866c

v = 0

v = 0

v = 0

β = .866c

BOOM

As I said before,

AIN’T THAT COOL!!!

40.)

p. bomb paradoxfaculty.polytechnic.org/physics/1 astronomy, cosmology, relativity... ·...

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