]> Flickering Bulb Spacetime Paradox

# The Flickering Bulb Paradox

## Proposed by G.P. Sanstry

Two long conducting rails are open at one end but connected at the other end through a lamp and battery in series. One of the rails has a square vertical offset 2m long. Between the rails a H-shaped slider moves frictionlessly. The vertical bars of the sliders are conducting and the cross piece is insulating. If either of bars of the slider is touching both of the rails simultaneously then the circuit is completed and the bulb lights.

If you take the slider and move it along the rails then the bulb will stay constantly lit because there is a constant connection until bar A hits the gap and immediately when that happens bar B makes connection completing the circuit, so there is no break.

Spacetime physics however has a phenomena known as Lorentz contraction such that an object shrinks along the axis of motion. This so called "stretch factor" is represented by γ where:

$γ = 1 1-vrel2$

Where vrel is the relative velocity between the observer and the observed as a fraction of the speed of light.

This concept of contraction of length introduces an apparent paradox. Say for instance that the slider is moving at 259627884 m/s, that's approximately 32c. At that speed γ=2, so the length of the slider would be reduced by a factor of two, making it 1m.

If the slider is contracted then there will be a length of time during which the circuit is completely broken and the light goes off.

The theory of relativity rests heavily on the concept that there is no absolute frame of observation, so it is equally valid to see the experiment as the slider holding still and the rails moving in the opposite direction at vrel. From this frame though the rails will be Lorentz contracted and so it will appear as:

The gap is now only 1m, so the slider maintains contact the entire time and the bulb will never go off. Relativity does not allow for different events to occur for different observers. The order of the events can be different, but the bulb either goes off or it doesn't and that fact is the same for all frames of reference. So, there must be some way to resolve this paradox, or the contraction of length does not exist.

The resolution comes with the recognition that the electromotive force, ε, like everything else, cannot move faster than the speed of light. This means that when a rail makes (or breaks) contact there is a time lapse of at least 4 light-meters before that change in force will reach the bulb. (It is 4m to the side of the gap closest to the bulb, 6m to the far side.) To better examine this, it is convenient to define some events. These events will take place in all frames (though not necessarily at the same time or even in the same order.) They are shown here in the frame with the rails at rest, though again this is arbitrary and could have been shown just as easily in the frame with the slider at rest.

Event #1: Bar A enters the gap (breaking contact with the rails):

Event #2: Bar B exits the gap (reestablishing contact with the rails):

You will notice that the bulb is not out in either event even though during the interval between then there was no circuit. This seems incorrect at first, but again the changes in ε are not simultaneous; they require time for the change is force to propagate through the wire. These changes in ε are represented with *s. To better analyze this we will define two more events.

Event #3: The drop in ε from event #1 reaches the bulb:

Event #4: The restoration of ε from event #2 reaches the bulb:

If event #4 occurs at or before the time event three occurs then ε is never diminished and the bulb will not flicker.

## How long is the bulb off in the frame where the rail is at rest?

The bulb will be off for the period of time between events 3 and 4 ($Δt3,4$). Using the following notations:

Obviously there are relationships between events 1 and 3, 2 and 4 since 1 causes 3 and 2 causes 4. The time between 1 and 3 is just:

$Δt1,3 = t3-t1 = -x vpulse$

Since those events are separated by the time that it takes the pulse to cover the distance from x to 0. The distance is negative because the pulse is traveling toward 0. (The value of vpulse is negative as well.) Similarly:

$Δt2,4 = t4-t2 = -x+J vpulse$

The time between 1 and 2 is a little more sophisticated. The length if the gap is J. At event #1 the position of bar B is $x+S′$, that means the distance from bar B and the other side of the gap is $J-S′$. The time that it will take the slider to move that distance is:

${\Delta t}_{1,2}={t}_{2}-{t}_{1}=\frac{J-{S}^{\prime }}{{v}_{rel}}$

Since $t1$ is the first event, it is convenient to assign:

$t1 = 0$

So:

$\begin{array}{rcl}{\Delta t}_{3,4}{|}_{{t}_{1}=0}& =& {t}_{4}-{t}_{3}\\ & =& \left({\Delta t}_{2,4}+{t}_{2}\right)-\left({\Delta t}_{1,3}+{t}_{1}\right)\\ & =& \left({\Delta t}_{2,4}+\left({\Delta t}_{1,2}+{t}_{1}\right)\right)-\left({\Delta t}_{1,3}+{t}_{1}\right)\\ & =& \frac{-x-J}{{v}_{pulse}}+\frac{J-{S}^{\prime }}{{v}_{rel}}+0-\frac{-x}{{v}_{pulse}}-0\\ & =& \frac{-J}{{v}_{pulse}}+\frac{J-{S}^{\prime }}{{v}_{rel}}\\ & =& \frac{J{v}_{pulse}-J{v}_{rel}-{S}^{\prime }{v}_{pulse}}{{v}_{pulse}{v}_{rel}}\end{array}$

Substituting in the values that were given:

• $J=S=2m$
• $vrel= 32$
• $vpulse =-1$
• $γ= 1 1-vrel2 = 1 1-322 =2$
• $S′ = Sγ = S2 = 1m$
$Δt3,4= 23+2 3 ≊3.1547m$

In spacetime it is not uncommon to express lengths of time in units commonly associated with distance. The two converge with the speed of light which is constant in all frames of reference at $c≡299792458ms$.

## How long is the bulb off in the frame where the slider is at rest?

This scenario is very similar to the one where the rails are at rest: the time that the bulb will be off is still the time between events 3 and 4 ($Δt′3,4$). Using the following notations:

One important difference in this scenario is that relative velocity between the change in ε and the bulb (vpulse) is effectively reduced. The pulse is essentially "chasing" the bulb and so:

$v′pulse= vpulse- vrel$

Based on this change, the previous equations are altered slightly:

${{\Delta t}^{\prime }}_{1,3}={{t}^{\prime }}_{3}-{{t}^{\prime }}_{1}=\frac{-{x}^{\prime }}{{{v}^{\prime }}_{pulse}}$ ${{\Delta t}^{\prime }}_{2,4}={{t}^{\prime }}_{4}-{{t}^{\prime }}_{2}=\frac{-\left({x}^{\prime }+{J}^{\prime }\right)}{{{v}^{\prime }}_{pulse}}$

The other major difference in this frame of reference is that event #2 happens before event #1. This means that the time difference between 1 and 2 is the time it takes bar A to traverse the distance S-J. (As opposed to the time for bar B to pass J-S in the frame with the rails at rest.)

${{\Delta t}^{\prime }}_{2,1}={{t}^{\prime }}_{1}-{{t}^{\prime }}_{2}=\frac{-\left(S-{J}^{\prime }\right)}{{v}_{rel}}$

Since event #2 is now the first, it is convenient to combine the equations with $t′2=0$

$\begin{array}{rcl}{{\Delta t}^{\prime }}_{3,4}{|}_{{{t}^{\prime }}_{2}=0}& =& {{t}^{\prime }}_{4}-{{t}^{\prime }}_{3}\\ & =& \left({{\Delta t}^{\prime }}_{2,4}+{{t}^{\prime }}_{2}\right)-\left({{\Delta t}^{\prime }}_{1,3}+{{t}^{\prime }}_{1}\right)\\ & =& \left({{\Delta t}^{\prime }}_{2,4}+{{t}^{\prime }}_{2}\right)-\left({{\Delta t}^{\prime }}_{1,3}+\left({{\Delta t}^{\prime }}_{2,1}+{{t}^{\prime }}_{2}\right)\right)\\ & =& \frac{-\left({x}^{\prime }+{J}^{\prime }\right)}{{{v}^{\prime }}_{pulse}}+0-\frac{-{x}^{\prime }}{{{v}^{\prime }}_{pulse}}-\frac{-\left(S-{J}^{\prime }\right)}{{v}_{rel}}-0\\ & =& \frac{{-J}^{\prime }}{{{v}^{\prime }}_{pulse}}+\frac{S-{J}^{\prime }}{{v}_{rel}}\\ & =& \frac{S{{v}^{\prime }}_{pulse}-{J}^{\prime }{{v}^{\prime }}_{pulse}-{J}^{\prime }{v}_{rel}}{{{v}^{\prime }}_{pulse}{v}_{rel}}\end{array}$

Substituting in the values that were given:

• J=S=2m
• $vrel= -32$
• $vpulse =-1$
• $γ= 1 1-vrel2 = 1 1--322 =2$
• $J′ = Jγ = J2 = 1m$
• $v′pulse = vpulse - vrel = -1- -32 = 32-1$
$Δt′3,4 = 43-4 23-3 ≊6.3094m$

## In which frame is the proper time found?

The proper time is the time separation in the frame where the space separation is 0m. Event #3 and #4 occur in the same place in the frame where the rails are at rest. The rails frame is therefore the frame with the proper time.

## What is the ratio of the times between these frames?

According to my figures:

$\frac{{{\Delta t}^{\prime }}_{3,4}}{{\Delta t}_{3,4}}=\frac{\frac{4\sqrt{3}-4}{2\sqrt{3}-3}}{\frac{2\sqrt{3}+2}{\sqrt{3}}}=\frac{2\left(3-\sqrt{3}\right)}{3-\sqrt{3}}=2$

## What should this ratio be according to the time dilation effect?

The relationship between two frames of reference is represented by γ:

$Δt′ = γΔt Δt′ Δt = γ$

This is agrees with the computed value of 2 for $Δt′3,4Δt3,4$.

# Further Questions

The transmission of electrical impulses through different mediums is generally less than the speed of light. For the next questions, assume that the pulses travel at a speed of .75c in the frame where the rail is at rest.

## At what speed do they travel in the slider frame?

This is found using the law of combination of velocities. The velocity of C relative to B:

$vC,B = vC,A-vB,A 1-vB,AvC,A$
• $vpulse,rail=- 34$
• $vslider,rail= 32$
$vpulse,slider = vpulse,rail-vslider,rail 1-vslider,railvpulse,rail = -34-32 1+ 34 32 = - 6+43 8+33 ≊ -.97969$

## How long is the lamp off as observed in the rail frame?

$\begin{array}{rcl}{\Delta t}_{3,4}{|}_{{t}_{1}=0}& =& \frac{J{v}_{pulse,rail}-J{v}_{rel}-{S}^{\prime }{v}_{pulse,rail}}{{v}_{pulse,rail}{v}_{rel}}\\ & =& \frac{2\left(-\frac{3}{4}\right)-2\left(\frac{\sqrt{3}}{2}\right)-1\left(-\frac{3}{4}\right)}{\left(-\frac{3}{4}\right)\left(\frac{\sqrt{3}}{2}\right)}\\ & =& \frac{6+8\sqrt{3}}{3\sqrt{3}}\\ & \approxeq & 3.8124m\end{array}$

## How long is the lamp off as observed in the slider frame?

$v′pulse,slider = vpulse,slider-vrel = - 6+43 8+33 --32 = 32 - 6+43 8+33 = - 3 16+63 ≊ 0.1137m$ $Δt′3,4 |t′2=0 = Sv′pulse,slider - J′v′pulse,slider - J′vrel v′pulse,slider vrel = 2 -316+63 - 1 -316+63 - 1 -32 -32 -316+63 = 12+163 33 ≊ 7.6427$

## What is the ratio of these off times?

$Δt′3,4 Δt3,4 = 12+163 33 6+83 33 =2 =γ$

## How long does the slider need to be to eliminate any flicker?

When there is no flicker, the resumption of ε reaches the bulb before the drop, i.e. $Δt3,4≤0$.

$Δt3,4 ≤ 0 Jvpulse - Jvrel - S′vpulse vpulse vrel ≤ 0 Jvpulse - Jvrel - S′vpulse ≤ 0 S′vpulse ≥ Jvpulse - Jvrel Svpulse γ ≥ J vpulse-vrel S ≥ Jγ vpulse-vrel vpulse S ≥ J vpulse-vrel vpulse 1-vrel2$