Centipede inertial frame problem

[Adam564]

Homework Statement

A high-speed centipede in S’ is 10.0 cm long measured at rest in S’. A butcher in S holds two cleavers (A and B) 9.00 cm apart (measured in S). The centipede runs at such high speed v across a chopping block that the butcher measures the length of the centipede to be 8.00 cm.

(a) How fast is the centipede moving across the chopping block?

(b) According to the centipede, how far apart are the cleavers? Let both cleavers strike the chopping block when t=0 according to the butcher.

(c) What are the spacetime coordinates of these events in S and S’?

(d) Based on your calculations, is the centipede still intact after the cleavers strike the chopping block? Why or why not?

(e) Compare the spacetime intervals measured in the two frames.

Please help me solving these (a) to (e).

Homework Equations

v=Δx/Δt
Δt=ϒΔt'
Δt'=x'/c
ϒ=1/√1-β^2
v/c=β

The Attempt at a Solution

(what I have so far)
a)
S' at rest:
10.0 cm (length of centipede)

S while centipede is moving:
8.00 cm (length of centipede)
distance the centipede needs to cover (x)=0.01 m (0.09-0.08) (cleaver spacing minus length of centipede)
v=Δx/Δt=0.01 m/ϒΔt'
v=0.01/ϒ(x'/c)
ϒβ=0.01/x'=0.01/0.1=0.1
(ϒβ)^2=((1/√1-β^2)^2)(β^2)=0.1
β=0.3015
v=0.3015c

 

[kuruman]

According to forum rules, you need to show the relevant equations and tell us what you think should be your approach to the problem and what questions you have. We do not give answers away. Please read our guidelines
https://www.physicsforums.com/threads/guidelines-for-students-and-helpers.686781/

 

[haruspex]

what I have so far)
a)
S' at rest:
10.0 cm (length of centipede)

S while centipede is moving:
8.00 cm (length of centipede)
distance the centipede needs to cover (x)=0.01 m (0.09-0.08) (cleaver spacing minus length of centipede)
v=Δx/Δt=0.01 m/ϒΔt'
v=0.01/ϒ(x'/c)
ϒβ=0.01/x'=0.01/0.1=0.1
(ϒβ)^2=((1/√1-β^2)^2)(β^2)=0.1
β=0.3015
v=0.3015c

The trouble with adding this as an edit, instead of a new post, is that many readers might see post #2 and read no further, expecting your attempt to appear as another post.
Unfortunately I am ignorant of SR so cannot pass comment on your attempt, but hope that adding this post might alert someone more qualified to respond.

 

[kuruman]

The trouble with adding this as an edit, instead of a new post, is that many readers might see post #2 and read no further, expecting your attempt to appear as another post.

Thanks for pointing this out. I was unaware of the edit.
@Adam564: Your approach is incorrect. Consider this. If the proper length of the centipede is 10 cm but according to the butcher it is contracted to 8 cm, what is γ?

 

[Adam564]

Thanks for pointing this out. I was unaware of the edit.
@Adam564: Your approach is incorrect. Consider this. If the proper length of the centipede is 10 cm but according to the butcher it is contracted to 8 cm, what is γ?

@kuruman As far as I know, γ=1/√(1-(v/c)^2)
Are you suggesting that γ can be determined some other way? What I am getting from your comment is it could be 2 cm, since that is the change in length of the centipede between the reference frames.

 

[Adam564]

The trouble with adding this as an edit, instead of a new post, is that many readers might see post #2 and read no further, expecting your attempt to appear as another post.
Unfortunately I am ignorant of SR so cannot pass comment on your attempt, but hope that adding this post might alert someone more qualified to respond.

Ok, thanks. I suppose once I am done all the parts I'll make a new post altogether.

 

[kuruman]

@kuruman As far as I know, γ=1/√(1-(v/c)^2)
Are you suggesting that γ can be determined some other way? What I am getting from your comment is it could be 2 cm, since that is the change in length of the centipede between the reference frames.

Not what you should be looking at. Please read section "Basis in relativity" here and study the equations therein.
https://en.wikipedia.org/wiki/Length_contraction
You are given ##L_0## and ##L##, no?
Before you get too involved with parts c, d, e there is an issue that needs to be resolved that I did not mention earlier. The key question is "In the butcher's frame, where is the worm's tail (or head) at the time the cleavers are dropped (t = 0)?" If that is not given by the problem, you have to make some assumptions before you can answer the last three parts. Obviously the butcher wants to chop the worm, but he can drop the cleavers at any time from the moment the worm's head is past the first cleaver until the worm's tail is past the second cleaver.

 

[Adam564]

Updated solution to part a):
L=L₀√(1-((v/c)^2))
L=0.08m
L₀=0.1m
v=0.6c

 

[kuruman]

Updated solution to part a):
L=L₀√(1-((v/c)^2))
L=0.08m
L₀=0.1m
v=0.6c

That is correct.

 

[Adam564]

b)
L=L0√(1-((v/c)^2))
L₀=0.09m
v=0.6c
L=0.072 m

c)
S: cleavers are 9 cm apart
speed of centipede is 0.6c
centipede measures 8 cm

S':
centipede is 10 cm long
cleavers are 0.072 m apart

d) the centipede will be cut

 

[kuruman]

c)
S: cleavers are 9 cm apart
speed of centipede is 0.6c
centipede measures 8 cm

In part (c) you are asked to provide the space time coordinates. That means use the Lorentz transformation equations. What coordinates are appropriate to express? However before you do that, as I pointed out earlier in #7, "In the butcher's frame, where is the centipede's tail (or head) at the time the cleavers are dropped (t = 0)?" If the centipede is one light year away either to the left or to the right of the cleavers, it will not be cut. You have to make an assumption here.

 

[PAllen]

To me, the only interesting interpretation of the problem is if the question asked is whether it is possible for the centipede to be cut into 3 pieces, given the rest of the problem statement. The problem as stated clearly seems to miss the issue raised by @kuruman.

 

[kuruman]

To me, the only interesting interpretation of the problem is if the question asked is whether it is possible for the centipede to be cut into 3 pieces, given the rest of the problem statement. The problem as stated clearly seems to miss the issue raised by @kuruman.

I believe it is possible but not if the butcher drops the cleavers simultaneously in his frame. The problem states that the cleavers hit the block without specifying whether this happens simultaneously or not.

 

[PAllen]

I believe it is possible but not if the butcher drops the cleavers simultaneously in his frame. The problem states that the cleavers hit the block without specifying whether this happens simultaneously or not.

The problem states both cleavers hit the block at t=0 according to the butcher. Another way to complete the speciation, more direct (but less elegant?) than the three pieces question, would be to simply state that the midpoint of the centipede is at the midpoint between the cleavers at t=0 per the butcher.

 

[kuruman]

The problem states both cleavers hit the block at t=0 according to the butcher.

That is true, I forgot seeing it when I first read the problem.

... would be to simply state that the midpoint of the centipede is at the midpoint between the cleavers at t=0 per the butcher.

That's a good choice and in fact the one I used when I solved the problem.

 

[PAllen]

Just to be clear, given what is specified in the problem, it is impossible for the centipede to be cut into 3 pieces. Understanding that, in both frames of reference, is the core of the problem. Extra credit for clearly understanding what to change in problem statement to make it possible.

 

[kuruman]

Just to be clear, given what is specified in the problem, it is impossible for the centipede to be cut into 3 pieces. Understanding that, in both frames of reference, is the core of the problem. Extra credit for clearly understanding what to change in problem statement to make it possible.

This is assuming that the cleavers drop when the midpoints of cleaver-to-cleaver distance and centipede length are coincident in the butcher's frame, correct?

 

[PAllen]

This is assuming that the cleavers drop when the midpoints of cleaver-to-cleaver distance and centipede length are coincident in the butcher's frame, correct?

No, as long as cleavers hit block at same time per butcher, as specified in problem, there is no way at all for the centipede to be cut in 3 pieces, no matter what the timing of its inertial motion is. Of course its speed relative to butcher is determined by the problem statement, which is important.

 

[Adam564]

i think the solution for c) is like this
(x₁',x₂')=(0,0.1125)
(x₁,x₂)=(0,0.09)
(t₁',t₂')=(0,0.0675)
(t₁,t₂)=(0,0)
x₁'=γ(x₁-vt₁)
x₂'=γ(x₂+vt₂)
t₁'=γ(t₁-(v/c²)x₁)
t₂'=γ(t₂-(v/c²)x₂)

 

[kuruman]

i think the solution for c) is like this
(x₁',x₂')=(0,0.1125)
(x₁,x₂)=(0,0.09)
(t₁',t₂')=(0,0.0675)
(t₁,t₂)=(0,0)
x₁'=γ(x₁-vt₁)
x₂'=γ(x₂+vt₂)
t₁'=γ(t₁-(v/c²)x₁)
t₂'=γ(t₂-(v/c²)x₂)

These numbers don't meant much if you don't define the events that they are supposed to be the space time coordinates of. Also, it will be much easier for someone to check your work in (c) if you used symbols for the given numbers. For example, t'₂ cannot be 0.0675. Besides being way too high, if put in t₂ = 0 in t₂'=γ(t₂-(v/c²)x₂), you should get a negative number.

x₂'=γ(x₂+vt₂) is incorrect. There should be a minus sign on the right hand side instead of a plus.

The Lorentz transformations are written in this standard form assuming that the origins of frames S and S' are coincident at time t = t' = 0. The centipede is an extended object of proper length 10 cm. Where, with respect to the centipede, have you chosen the origin of S' to be? Its head? Its tail? Its midpoint? Without specifying this you will not be able to ascertain whether the centipede gets cut or not.

When I look at (x₁,x₂)=(0,0.09), I assume that these are the space coordinates at t = 0 of the cleavers in the butcher's frame. OK. Then (x₁',x₂')=(0,0.1125) are what? They cannot be the cleavers' coordinates in the centipede's frame because their separation is greater than the proper length of 10 cm. Maybe this is the doing of the plus sign that should have been minus.

 

[Adam564]

I just got the answers, everything is correct, except for some of c) and whole d).
t₂' is in fact -2.25*10^-10 s.
this is because t₂'=[t₂-((vx₂)/c²)]/√1-((v/c)²)

therefore centipede is not intact because the centipede passes before the cleavers strike the table
which is in fact what I wrote so I must've done a calculation error, otherwise I am grateful for your help

 

[kuruman]

I just got the answers, everything is correct, except for some of c) and whole d).
t₂' is in fact -2.25*10^-10 s.
this is because t₂'=[t₂-((vx₂)/c²)]/√1-((v/c)²)

therefore centipede is not intact because the centipede passes before the cleavers strike the table
which is in fact what I wrote so I must've done a calculation error, otherwise I am grateful for your help

If you are satisfied with your answer, so be it. However, it's still not clear to me where the centipede is relative to the butcher at t = 0 when the cleavers come down.

 

[PAllen]

I just got the answers, everything is correct, except for some of c) and whole d).
t₂' is in fact -2.25*10^-10 s.
this is because t₂'=[t₂-((vx₂)/c²)]/√1-((v/c)²)

therefore centipede is not intact because the centipede passes before the cleavers strike the table
which is in fact what I wrote so I must've done a calculation error, otherwise I am grateful for your help

Whether the centipede is intact or not depends entirely on where it is at t=0 in the butcher frame, which is not given at all in the problem as you’ve stated it. There is a range of positions where the centipede is cut in two pieces by one cleaver, another range where it is cut in two by the other cleaver, and three ranges where it is intact. In particular, if the midpoint of the centipede is midway between the cleavers at t=0 in the butcher frame, the centipede will be intact. Further, as I have stated, there is no way for both cleavers to cut the centipede, so it cannot be cut into 3 pieces.

 

[kuruman]

Whether the centipede is intact or not depends entirely on where it is at t=0 in the butcher frame, which is not given at all in the problem as you’ve stated it. There is a range of positions where the centipede is cut in two pieces by one cleaver, another range where it is cut in two by the other cleaver, and three ranges where it is intact. In particular, if the midpoint of the centipede is midway between the cleavers at t=0 in the butcher frame, the centipede will be intact. Further, as I have stated, there is no way for both cleavers to cut the centipede, so it cannot be cut into 3 pieces.

Amen to that.