The Lance Duel Paradox: Who Will Win?

[gatusso]

Hello everyone,

If two knights, one with a 5m long lance (A) and the other one with a 3m long lances (B) horizontally hold their lances in the direction of each other and move towards each other with a speed close to speed of light, who will win? because if the speed is really high, for each of them the other one has a shorter lance and will predict he will be the winner. Actually B with a shorter lance at rest than A will claim he will win. I read the ladder paradox but i can't relate it to this problem. can someone help me.(this is not homework question tho, my friend and I were debating)

 

[Nugatory]

If by "win", you mean "knocks the other one off his horse, so that he can't do the same to me", then there is a certain closing speed (try calculating it - it's fun!) below which the guy with the 5m lance wins, above which neither win because they both get hit by the opposing lancepoint.

In any case, all observers everywhere, regardless of their speed, will agree about the result. There will never be a paradox where some observers see an unhorsed knight and other don't.

 

[gatusso]

But then how can u reconcile this with one observer's view who sees his lance to be longer than his rival's? can u explain the speed after which, they both get hit

 

[Nugatory]

But then how can u reconcile this with one observer's view who sees his lance to be longer than his rival's? can u explain the speed after which, they both get hit

You might try googling for the "bug-rivet paradox", which is closely related.

There are two important things going on here:

First, what happens at one end of a lance does not instantaneously affect what happens at the other end, so when my lancepoint hits the other guy in the chest and knocks him off his horse, his lancepoint is not immediately deflected away from my chest. As I view events, it is possible for his lancepoint, already very near to my chest, to hit my chest after my lancepoint has hit his chest (but before I've seen my lancepoint hit his chest, because I can't see that until light from that event has traveled to my eyes).

Second, there's the relativity of simultaneity. If the closing speed is high enough (I'll repeat myself - try calculating it - it's fun!) some observers will see my lance hitting him to happen first, other observers will see his lance hitting me happening first.

You can also try drawing a space-time diagram tracking the four interesting objects through space-time: The two lancepoints and the two targets. (Ghwellsjr has software for drawing these, and if he's watching this thread he's probably working on it already).

 

[ghwellsjr]

Ghwellsjr has software for drawing these, and if he's watching this thread he's probably working on it already.

Yes, I have started but can't continue right now, maybe I'll have something by tomorrow.

 

[Meir Achuz]

Sir Longlance defeats Sir Shortlance in any coordinate system.
This is seen most easily in the Earth system where they approach each other at the same speed,
so there is no preference for either.
The situation is a bit tricky in the rest system of Shortlance, especially if he tries to keep all times the same.
What happens is that he gets unhorsed before he knows it. He gets fooled by trying o do things at equal time in his rest system. This takes a bit of algebra to show.
Take the positions where Longlance's spear strikes Shortlance's armour (to use the knightly spelling).
to be x_1=0 and y_1=0. x_1 is the position of the tip of Longlance's lance,
and y_1 is the position of Shortlance's body.
x_2=-L/gamma is the position of Longlance's body,
and y_2=-S/gamma is the position of the tip of Shortlance's lance.
All of this is at the same time, which can be taken as t=0.
Then a Lorentz transformation to Shortlances's rest system gives t'_2L=-vL and t'_2S=-vS,
while t'_1L=t'_1S=0. These times are not equal, so the usual Lorentz contraction is not applicable.

 

[yuiop]

Sir Longlance defeats Sir Shortlance in any coordinate system.
This is seen most easily in the Earth system where they approach each other at the same speed,
so there is no preference for either.

What about when the velocity of Longlance relative to Shortlance is such that the the length contraction factor is 0.6 and the length of Longlance's spear is the same length as the Shortlance's spear as measured in Shortlance's rest frame? In this reference frame they would both strike each other simultaneously and that would count as a draw. In another reference frame moving in the opposite direction to Longlance relative to Shortlance's rest frame, Shortlance strikes first. As Nugatory pointed out, you have to careful how you define 'win'.

 

[ghwellsjr]

As promised, here is a set of spacetime diagrams for the speed of each knight traveling at 0.6c in the "earth" frame. I have used the unit of light-meter to refer to the length of time that light travels in one meter. The green knight is traveling to the right with the red tip of his lance extended out 5 meters in front of him (as will be apparent later in his rest frame). The black knight is traveling to the left with the blue tip of his lance extended out 3 meters in from of him.

Here is the diagram for the "earth" frame:

Note that the red tip of the longer lance contacts the black knight before the blue tip of the shorter lance contacts the green knight, as expected. However, the shock wave of the impact cannot travel through each lance to the respective knight nearly as fast as the speed of light and I have drawn in two signals to show this limit for each lance. Thus, even though the green knight "wins" in this reference frame, he still gets impaled by the blue tip of the black knight's lance.

If we look at the green knight's rest frame, we see that there is even more time between the red tip of his lance impacting the black knight but the maximum shock wave still takes the same amount of the black knight's time to reach his blue tip and again, it's too late:

Now for the rest frame of the black knight:

In this frame, the length of the green knight's lance is contracted to about 2.5 meters, shorter than the black knight's 3-meter lance, and so the black knight "wins" but since it takes a long time for the shock wave to travel down the green knight's lance back to the red tip, the black knight still gets knocked off his horse.

 

[Meir Achuz]

Who gets knocked off cannot depend on coordinate system no matter what space-time diagrams are drawn.
The four times (for the front and back of each lance) can be the same in only one Lorentz system.
Trying to set them all equal in a different Lorentz system leads to the errors above.

 

[TumblingDice]

Who gets knocked off cannot depend on coordinate system no matter what space-time diagrams are drawn.
The four times (for the front and back of each lance) can be the same in only one Lorentz system.
Trying to set them all equal in a different Lorentz system leads to the errors above.

What errors do you see? George's diagrams and commentary seem valid, without any "who gets knocked off" issues.

The space-time diagrams illustrate how the concept of simultaneity in different reference frames and the maximum speed of a wave in a rigid object can influence the perception of how the event 'transpires'. (Which tip touches first, and how each knight suffers the consequences of the collisions.)

 

[Nugatory]

Who gets knocked off cannot depend on coordinate system no matter what space-time diagrams are drawn.

We aren't getting different results by varying the coordinate system here, we're getting different results by varying the physical situation. For any relative velocity of the two knights, there is exactly one possible physical outcome and all observers, regardless of their velocity and choice of coordinates will observe that outcome - no one doubts that. But different relative velocities of the knights will produce different outcomes, just as the relative velocity between a bullet and its target affects the coordinate-independent damage done to the target by the bullet.

 

[TumblingDice]

We aren't getting different results by varying the coordinate system here, we're getting different results by varying the physical situation. For any relative velocity of the two knights, there is exactly one possible physical outcome and all observers, regardless of their velocity and choice of coordinates will observe that outcome - no one doubts that.

From your post #4, I thought that the 'sizzle' behind this thought experiment involved the calculation of simultaneity when combined with the maximum signal speed in a rigid object. So at .6 c, different reference frames will see different tip strike first, however speed of rigid object resolves any paradoxes and same knights end up on the ground.

Have I been thinking too quickly or just plain wrong?

 

[Meir Achuz]

The space-time diagrams are all wrong, because (in the Earth diagram) the red lines both stop when the red knight is dehorsed. The red lines must therefor also stop in all the other diagrams at the time given by the Lorentz transformation to those frames.

 

[Meir Achuz]

Another way to see that only Sir Shortlance is dehorsed is to agree that this happens in the Earth frame.
The other Lorentz frames correspond to observing the joust from a moving anachronistic car. Whichever direction the car is traveling, you will see Sir Shortlance dehorsed. Your driving by the scene cannot save him.
You may observe some funny time sequences, but he will be dehorsed.

 

[PAllen]

Who gets knocked off cannot depend on coordinate system no matter what space-time diagrams are drawn.
The four times (for the front and back of each lance) can be the same in only one Lorentz system.
Trying to set them all equal in a different Lorentz system leads to the errors above.

Nugatory is correct on this and his explanation was good so I chose not to add to this thread till now. Note that their is no frame dependence on what happens: either long lance wins, or both lose. The correct outcome is produce in any frame with correct analysis.

As an intuition for such situations, I have long felt SR books should have a section on the interaction of matter at high relative speeds. The key point is that even 'unobtainium' with a speed of sound and bonding influence equal to the speed of light behaves, to first order, as a collection of independent dust particles when the relative speed is high enough. The modern concept (relative to when I first learned this stuff) of congruence of world line makes this precise. Model an object (lance) as a congruence of world lines. Posit an interaction at one world line (the initial contact between a lance and and a knight). Consider the future directed light cone from this event. The other world lines in the congruence must behave as if this collision event did not happen until they cross that light cone.

Understanding this, the resolution is simple. Above a certain relative speed, the two events: Knight A getting skewered, knight B getting skewered, have spacelike separation. This, of course, means there cannot be any preferred time ordering of these events.

This gets to the core intuition which this 'paradox' forces you to reconsider: "If I get hit first, I don't get to knock off the other knight". This has a whole bunch of caution words for SR: 'first' only has invariant meaning for causally connected events. Then, you are talking about events in two different places, which implies simultaneity (invariantly: spacelike separation) is going to be key to analysis. This 'obvious' statement becomes false in principle above a certain relative speed.

 

[yuiop]

If by "win", you mean "knocks the other one off his horse, so that he can't do the same to me", then there is a certain closing speed (try calculating it - it's fun!) below which the guy with the 5m lance wins, above which neither win because they both get hit by the opposing lancepoint.

I get the critical minimum relative speed of the 2 knights to be (8/17)c for a draw (both unseated). Is that what you get?

This relative speed is determined by whether or not the 'potential event' (e2) of Longlance being hit by Shortlance is in the future light cone of the event of Shortlance being hit by Longlance (e1) and calculated by solving $v >= c(1-3/5 \sqrt{1-v^2/c^2})$ for v (if I have done it right).

I am assuming that the deflection shock wave travels along the spear of Shortlance at the speed of light, but in reality the shock wave is slower than c, so the minimum speed will be a bit higher.

 

[PAllen]

I get the critical minimum relative speed of the 2 knights to be (8/17)c for a draw (both unseated). Is that what you get?

This relative speed is determined by whether or not the 'potential event' (e2) of Longlance being hit by Shortlance is in the future light cone of the event of Shortlance being hit by Longlance (e1) and calculated by solving $v >= c(1-3/5 \sqrt{1-v^2/c^2})$ for v (if I have done it right).

I am assuming that the deflection shock wave travels along the spear of Shortlance at the speed of light, but in reality the shock wave is slower than c, so the minimum speed will be a bit higher.

I get this as well, by a different (but equivalent) calculation.

 

[PeterDonis]

The space-time diagrams are all wrong, because (in the Earth diagram) the red lines both stop when the red knight is dehorsed.

I assume you mean "the red line and the green line both stop when the green knight is dehorsed". That is not correct; the *green* line stops (actually, it changes trajectory) when the green knight is dehorsed, but the *red* line (the worldline of the tip of the green knight's lance) does *not*, because it takes time for the change in the green knight's trajectory to propagate down the green knight's lance to its tip (the red line), since the change cannot propagate faster than the speed of light. Until that happens, the green knight's lance tip (the red line) will remain on its original trajectory, and will therefore still unhorse the black knight if the relative velocities of the two knights are large enough that the two unhorsing events (blue line crosses green line, and red line crosses black line) are spacelike separated. That's what ghwellsjr's spacetime diagrams (correctly) show.

 

[PeterDonis]

in reality the shock wave is slower than c, so the minimum speed will be a bit higher.

Wouldn't slower shock wave propagation (for real materials, a *lot* slower--typical sound speeds in solids are about $10^{-5} c$) make the minimum speed for a draw *lower*, not higher?

 

[WannabeNewton]

OP, work out the following exercise from Schutz "A First Course in General Relativity" and then move on to your scenario; the main idea behind the two is the same (finite propagation speed of sound).

 

[Dale]

Another way to see that only Sir Shortlance is dehorsed is to agree that this happens in the Earth frame.

The other respondents are correct. Both knights are dehorsed at sufficient speed. Any frame is fine, including the Earth's frame.

In order to stop the opponent's lance tip a knight must hit the opponent and then the tension wave from the hit must travel through the opponent's lance to the opponent's lance tip. In any frame where the two hits are spacelike separated there is no time for the tension wave to travel and both knights are dehorsed.

 

[yuiop]

Wouldn't slower shock wave propagation (for real materials, a *lot* slower--typical sound speeds in solids are about $10^{-5} c$) make the minimum speed for a draw *lower*, not higher?

Ummmm ... yes you're right!

Using the figure you gave for the shock wave speed and using a rounded figure of 300,000,000 m/s for the speed of light, the minimum critical speed is 1200 m/s or 0.0004c which is pretty slow in relativity terms.

 

[PeterDonis]

1200 m/s

Are you sure? The condition for a draw is that the time for the shock wave to travel 3 m up the shorter lance at 3000 m/s (10^-5 c) must be greater than the time for the shorter lance's tip to travel 2 m, the difference in lance lengths (at slow enough speeds, which these will be, length contraction of the lances is negligible, and so is time dilation, so it doesn't matter to which frame we refer the times). That gives 2000 m/s as the minimum closure rate--the speed of the shorter lance's tip relative to the other knight, and therefore the speed of either knight relative to the other at the start of the joust. At these low speeds, we can just add velocities linearly, so each knight is riding at 1000 m/s towards the other, as seen in the rest frame of the jousting field (assuming both knights ride at the same speed).

or 0.0004c

I think you left out a couple of zeros: 1200 m/s = 1.2 x 10^3 m/s = 4 x 10^-6 c = 0.000004c. My answer for the closure rate is 2000 m/s = 6.67 x 10^-6 c; each knight rides at 1000 m/s = 3.33 x 10^-6 c.

 

[PeterDonis]

My answer for the closure rate is 2000 m/s = 6.67 x 10^-6 c; each knight rides at 1000 m/s = 3.33 x 10^-6 c.

This calculation does bring up another issue, though: with sound speeds and closure speeds this low, the two "skewering" events--Longlance's lance tip strikes Shortlance, and Shortlance's lance tip strikes Longlance--are no longer spacelike separated! (The time separation between the two events at the critical speed is a millisecond, but light travels the distances in question in microseconds.) So a full analysis of this type of scenario must take more into account than just the causal relationship between the two "skewering" events.

One key point that has been left out so far is that what actually unhorses a knight is not just the other knight's lance tip striking him, but the other knight's lance exerting enough force on him to knock him off his horse. In other words, the striking knight's lance has to transfer sufficient momentum to the struck knight. So to really evaluate whether Shortlance can still knock Longlance off his horse, we need to evaluate, not whether Shortlance's lance tip can reach Longlance, but whether Shortlance's lance can transfer enough momentum to Longlance.

Since the necessary momentum transfer requires some time, the condition for the minimum velocity for a draw must look at, not when Shortlance's lance tip first strikes Longlance, but when it has made contact for long enough to unhorse Longlance. For large enough relativistic velocities, where the momentum transfer events are clearly spacelike separated, I don't think this changes the analysis too much, since at these speeds Longlance's lance, viewed from Shortlance's rest frame, will be length contracted so that it appears shorter than Shortlance's lance.

However, for low-speed cases, such as the one yuiop and I were just discussing, I think looking at momentum transfer will make a big difference. (Note that "low speed" here is still a kilometer per second, i.e., still much too fast for our ordinary intuitions about jousting to be applicable.) As soon as Longlance's lance hits Shortlance, Shortlance will start being pushed backwards off his horse, and therefore Shortlance's lance will start stretching, reducing its ability to transfer momentum. Since the two lance tip striking events are timelike separated, the stretching affects the result in all reference frames, i.e., it is an invariant property of the scenario. For speeds near the critical speed, Shortlance's lance will be significantly stretched by the time its tip reaches Longlance, so almost no momentum will actually be transferred, and Longlance will not actually be unhorsed; Shortlance's lance tip will bump against him just before it gets pulled backward as the shock wave in the lance reaches its tip and the lance starts to re-contract in response to the stretching.

I haven't tried to calculate how much faster the knights would have to be traveling for there still to be significant momentum transfer for lance sound speeds much less than the speed of light: but I suspect that the actual "critical speed" will still be relativistic, i.e., that in order for Longlance to actually be unhorsed, the two "skewering" events *will* have to be spacelike separated.

 

[PAllen]

Are you sure? The condition for a draw is that the time for the shock wave to travel 3 m up the shorter lance at 3000 m/s (10^-5 c) must be greater than the time for the shorter lance's tip to travel 2 m, the difference in lance lengths (at slow enough speeds, which these will be, length contraction of the lances is negligible, and so is time dilation, so it doesn't matter to which frame we refer the times). That gives 2000 m/s as the minimum closure rate--the speed of the shorter lance's tip relative to the other knight, and therefore the speed of either knight relative to the other at the start of the joust. At these low speeds, we can just add velocities linearly, so each knight is riding at 1000 m/s towards the other, as seen in the rest frame of the jousting field (assuming both knights ride at the same speed).
I think you left out a couple of zeros: 1200 m/s = 1.2 x 10^3 m/s = 4 x 10^-6 c = 0.000004c. My answer for the closure rate is 2000 m/s = 6.67 x 10^-6 c; each knight rides at 1000 m/s = 3.33 x 10^-6 c.

Let's at least give them diamond lances, with speed of sound 12,000 m/s. Using the simplified analysis above (not Peter's points in the next post) I get the condition is merely: 3/s = 2/v, s=speed of sound in lance, v = closing speed (easiest to see in rest frame of Sir Short Lance). Then , for 12000 m/s you get closing speed of 8000 m/s = about 18,000 mph.

 

[yuiop]

Are you sure? The condition for a draw is that the time for the shock wave to travel 3 m up the shorter lance at 3000 m/s (10^-5 c) must be greater than the time for the shorter lance's tip to travel 2 m, the difference in lance lengths (at slow enough speeds, which these will be, length contraction of the lances is negligible, and so is time dilation, so it doesn't matter to which frame we refer the times). That gives 2000 m/s as the minimum closure rate--the speed of the shorter lance's tip relative to the other knight, and therefore the speed of either knight relative to the other at the start of the joust. At these low speeds, we can just add velocities linearly, so each knight is riding at 1000 m/s towards the other, as seen in the rest frame of the jousting field (assuming both knights ride at the same speed).

The original equation I posted was:

$v >= c(1-3/5 \sqrt{1-v^2/c^2})$

When the speed of the shock wave is s rather than c the equation becomes:

$v >= s(1-3/5 \sqrt{1-v^2/c^2})$

The calculation substituting s = c/10^5 and c = 3*10^8 can be seen here and gives 1200 m/s.

I think you left out a couple of zeros: 1200 m/s = 1.2 x 10^3 m/s = 4 x 10^-6 c = 0.000004c.

Yes, it appears I did miss couple of zeroes and our answers at least have similar magnitude. However your result is still over 60% greater than my result. It may be that the peculiar relativistic effects in this scenario are significant at relatively low speeds (like relativistic electromagnetic effects) and so low speed approximations might cause greater errors than anticipated.

It might be interesting to consider how to exaggerate the effect even to more, to the stage where the relative velocities are low enough that it might be a doable experiment in a real laboratory (without real knights) :P

 

[PeterDonis]

your result is still over 60% greater than my result. It may be that the peculiar relativistic effects in this scenario are significant at relatively low speeds (like relativistic electromagnetic effects) and so low speed approximations might cause greater errors than anticipated.

No, I think it's because your equation is incorrect. Let me write what I think is the correct equation using symbols; I'll use units in which $c = 1$, and define $s$ as the sound speed in the lance, $L_S$ as the rest length of the shorter lance, $L_L$ as the rest length of the longer lance, and $v$ as the velocity of either knight relative to the other.

In Shortlance's rest frame, Longlance is traveling at $v$ towards him, so when Longlance's lance tip hits Shortlance, Shortlance's lance tip is a distance $L_L \sqrt{1 - v^2} - L_S$ from Longlance. (Note that this distance can be negative if $v$ is large enough, indicating that Longlance's lance is length contracted to a smaller length than Shortlance's lance, but as we'll see, that just means that *any* $s >= 0$ is consistent with the duel being a draw.) The time it takes the shock wave to traverse Shortlance's lance is $L_S / s$, and the time it takes Shortlance's lance tip to cover the distance is $\left(L_L \sqrt{1 - v^2} - L_S \right) / v$. For Shortlance's lance tip to reach Longlance, the first time must be greater than the second, so we must have

$$
L_S / s > \left(L_L \sqrt{1 - v^2} - L_S \right) / v
$$

which can be rearranged to

$$
v > s \left(\frac{L_L}{L_S} \sqrt{1 - v^2} - 1 \right)
$$

Note that if the factor in parentheses on the RHS is negative (meaning Longlance's lance is length contracted to a shorter length than Shortlance's lance), then as noted above, any $s >= 0$ will satisfy the condition.

For $v << 1$, $\sqrt{1 - v^2} \approx 1$, and we have

$$
v > s \left(\frac{L_L}{L_S} - 1 \right)
$$

which for $L_L = 5$, $L_S = 3$ gives $v > 2s / 3$, as I posted.

 

[Nugatory]

From your post #4, I thought that the 'sizzle' behind this thought experiment involved the calculation of simultaneity when combined with the maximum signal speed in a rigid object. So at .6 c, different reference frames will see different tip strike first, however speed of rigid object resolves any paradoxes and same knights end up on the ground.

That's right.
However, with a sufficiently lower speed of approach between the two knights (this speed has nothing to do with any reference frames or the speed of any observer - don't be confused by the fact that we might choose to use a reference frame in which one or the other knight is at rest) the result will be only one knight being unhorsed. If the knights are moving slowly enough, the various impacts will be timelike-separated not spacelike-separated, and despite the relativity of simultaneity all observers will agree on their relative ordering.

 

[yuiop]

For $v << 1$, $\sqrt{1 - v^2} \approx 1$, and we have

$$
v > s \left(\frac{L_L}{L_S} - 1 \right)
$$

which for $L_L = 5$, $L_S = 3$ gives $v > 2s / 3$, as I posted.

It appears I stand corrected again (gulp). I traced the problem in my equation (which was calculated in the rest frame of the long lance) as being due to my assumption that the shock wave would travel at speed s in that reference frame, which is not correct because the short lance is moving in that reference frame.

 

[PAllen]

I think the space like separation case is beaten to death. The case of speeds on the order sound in steel or diamond boil down to what the rules are. If it is first to be skewered (let's face it - given a lance hitting you much faster than bullet - for steel or diamond speed of sound - armor is irrelevant and you are skewered, not de-horsed ) the answer is unambiguous. For timelike events the temporal ordering is invariant. If the rule is you have to remain alive, then for speeds where lances are moving thousands of mph, both die.