1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Absurd dilatation problem

  1. Nov 19, 2007 #1
    Hi all,

    I was giving a class on dilatation and I have proposed to the students a problem where two pieces of wire of two different materialswere to be glued to form a 2 meter wire.
    The coeficient of linear dilatation of these two substances are alpha1 = 0.0004 and alpha2 = -0.0002, i.e., a negative linear coeficient. This means that the susbtance 2 contract when a heat source is connected to it.

    The question here is to determine the length of these two pieces that sum up 2 meters.

    Of course the problem has solution and the formalism yields two values L1 and L2 such that L1 + L2 = 2, as was expected. However,since the behavior of this couple is to compensate dilatation with contraction, so that the length of the junction is always 2 meters, there must be something wrong, since other combinations of values would also fit.
    Consider that the formalism has yielded L1 = 0,8 and L2 = 1,2. It is OK. But now let´s transfer some heat to this junction so that L1 (the one that increases in dimension) is now 0,9 and the L2 (the one which contracts) is now 1,1. What is wrong? Wasn´t the formalism to show this non uniqueness?

    Thank you

    DaTario
     
    Last edited: Nov 20, 2007
  2. jcsd
  3. Nov 20, 2007 #2

    Dale

    Staff: Mentor

    In your example here if you make each piece 1 m at some temperature, then when you heat it up the substance that expands will expand twice as much as the piece that contracts. Thus the total length will change. If you want the total length to stay the same you need to have the two different materials in a specific ratio so that the contraction of material 2 exactly compensates the expansion of material 1. I assume that your formalism is designed to calculate that ratio, which is unique for a given temperature.
     
  4. Nov 20, 2007 #3
    What about

    L1 * DELTA_TEMP * alpha1 = - L2 * DELTA_TEMP * alpha2

    So,
    L1 * alpha1 = - L2 * alpha2

    Since
    L1 + L2 = 2
    ...etc,etc

    :smile:
     
  5. Nov 25, 2007 #4
    I guess I have not made myself clear.
    The formalism through the equation for \Delta L yields the proportion between the two pieces, but the equation for the sum L1 + L2 = 2 (meters) produces two well defined values for these lengths.

    Nevertheless it is easy to see that, due to the very property that under variation of temperature the sum of the lengths of these two pieces doesn't vary, there must be infinite different proportions leading to the same result. My question is why the formalism chooses this specific proportion.

    Thank you anyway,

    Best Wishes

    DaTario

    It is exacly this, but the contraint that the sum must be 2 extract the real values for these two lenghts
     
  6. Nov 25, 2007 #5

    HallsofIvy

    User Avatar
    Science Advisor

    What "formalism" are you talking about? You have L1+ L2= 2 but since the pieces change length when the temperature changes, in general that will only apply at a specific temperature.

    "The question here is to determine the length of these two pieces that sum up 2 meters."
    At any temperature? At temperature change deltaT, the length of the first piece will be L1+ 0.0004L1(deltaT)+ L2-0.0002L2(deltaT)= 2. Since you already have L1+ L2= 2, in order for that to be true you must have 0.0004L1(deltaT)-0.0002L2(deltaT)= 0 or
    0.0004L1= 0.0002L2 which simplifies to 2L1= L2. Replace L2 by 2L1 in L1+ L2= 2 and you get 3L1= 2 so L1= 2/3 meter and L2= 4/3 meter. That's exactly what both Dale Spam and Rogerio said.
     
  7. Nov 25, 2007 #6
    Ok, you have solved the example. Now you have explicitly observed what I am asking. Forgive me if I am bothering you with such silly insistent argument. But once the application of the physical equations had yielded the result L1 = 2/3 and L2 = 4/3, you must agree that if you change the temperature by \Delta T then the L1 will be \Delta L larger while L2 will be \Delta L smaller, the sum of these two still yielding 2 meters. This show that L1 plus Delta L and L2 minus Delta L also constitutes a solution to this problem. No reference to any absolute or relative value of temperature is made in this problem.

    So the pair 2/3 and 4/3 must be referring to some specific temperature, which seems not to be an important part of the problem.
    The question is: "Why the solution of this problem does not show reference to this non uniqueness observed"

    I have the impression that I am not being able to communicate this question properly.
    I must appologize for it.

    Best wishes

    DaTario
     
  8. Nov 25, 2007 #7

    HallsofIvy

    User Avatar
    Science Advisor

    Part of the problem is that you haven't quoted the problem itself! You said "The question here is to determine the length of these two pieces that sum up 2 meters."
    That can only make sense if the problem was to find the length of two pieces that will be two meters in length at any temperature. What was the exact statement of the problem?
     
  9. Nov 25, 2007 #8
    Ok.

    "Let substance A have alphaA = 0.0004 and substance B have alphaB = -0.0002. We want to join two pieces of wires, each one made of one of these two substances, such that the total length is 2 meters. Determine the length of each part (A and B) of this 2 meter junction."

    My point is that if the exact statement of this problem would have said that the wires are initially at 20 Celsius, this would have had no concrete consequence on the physics of the problem.
    The problem seems to have an intrinsic non uniqueness. Perhaps it is not my fault your not being able to understand this question.
    I thank you anyway for your attention.

    Best Regards

    DaTario
     
    Last edited: Nov 25, 2007
  10. Nov 25, 2007 #9
    The crux of the problem lies in your definition.

    You are correct that as worded (less the 20 degrees C) that there is an infinite number of solutions. Let me explain. If both materials are already at some temperature 'X' the relative lengths are immaterial so long as they total 2m. It is only under some delta temperature that the coefficients of expansion become important.

    If you rephrased the question to ask, "What proportion of lengths of substance A, and susbstance B are required to maintain a length of 2 meters invariant of the temperature?" You would then be faced with only a singular solution.
     
  11. Dec 6, 2007 #10
    Thank you a lot for your answer. But let me present an argument against your statement that

    I believe that, even the proportions are not uniquely determined, since we may have, for instance, 1.1 and 0.9 meters at the initial stage and after some Delta T, 1.3 and 0.7. Notice that the proportions are not maintained.

    Thank you again,

    Best wishes

    DaTario
     
  12. Dec 6, 2007 #11
    A related question is: If one suppose a given initial temperature, the result that is obtained through the formalism must refer to some specific experimental temperature. I'm not sure about this point, but it seems reasonable to think that, at some specific temperature, in the lab, this problem is to have only one solution.

    Best wishes

    DaTario
     
  13. Dec 7, 2007 #12

    Dale

    Staff: Mentor

    I don't know exactly what the "formalism" you keep refering to is exactly. It would help if you posted it. However, I will try to guess:

    For some materials the length of the material as a function of the temperature can be expressed as:

    L(T) = L0 + L0 alpha deltaT = L(T0) + L(T0) alpha (T-T0)

    Now, if we have two materials, A and B, we can write:

    LA(T) + LB(T) = LA(T0) + LA(T0) alphaA (T-T0) + LB0 + LB(T0) alphaB (T-T0)

    and

    d( LA(T) + LB(T) )/dT = LA(T0) alphaA + LB(T0) alphaB

    In order for the length to be constant the above derivative must be equal to 0 so:
    LA(T0) alphaA + LB(T0) alphaB = 0
    LA(T0) alphaA = - LB(T0) alphaB
    LA(T0)/LB(T0) = - alphaB/alphaA

    As you can see from this derivation the lengths given by the formalism are specifically referenced to T0. So while you are correct that 1.1 + .9 = 2 and 1.3 + .7 = 2 the point is that only one proportion will work at a given T0. The other proportions also sum to 2, but only for their respective deltaT.
     
  14. Dec 7, 2007 #13

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The flaw in your reasoning is this: the expansion (or contraction) of the material depends not only on the change of temperature and the coeeficient of expansion, but also on the initial length! Therefore, only specific initial lengths will be just right so that the expansion of one piece compensates the contraction of the other. If you sit down and solve the equations (someone posted them), it is clear. And it's also clear that there is a unique solution.
     
  15. Dec 8, 2007 #14
    Dear DaleSpan, thank you for your response. The formalism I'm referring to is exactly what you induced. Let me define formalism:
    The set of formal (mathematical and/or symbolic) operations that contitutes the way this classical and naivy theory (alpha has no restrictions and is not assumed to vary with T!) has to solve the problem.

    Now, your reference to T0 was pretty elegant, but you may have noticed that it seems to appear only at the symbolic level (as a label). One does no mathematical operation with T0. This is kind of intriguing to me.

    Related to the post of NRQED, I agree that, mathematically, this classical and rather simplified theory provides one only and well defined result.

    I am really concerned with the confrontation between this simplified theoretical scenario and the vision of the experimental situation together with the assumptions that were made.

    let me try to organize my argument:
    1) alpha (A and B) do not depend on T (tempetarure). It should, but this theory is a simplified one, OK.

    2) Suppose 1.1 m and 0.9 m is the mathematically obtained solution for wires A and B respectivelly. Now, one may begin another experiment with the lab condition Delta T Celsius hotter. At this new initial temperature, the pieces of wire A and B will be now something like 1.3 m and 0.7 m. Tell me, please, why these new values for length are not allowed to represent possible, plausible and correct answers to the same problem?

    Obs. I feel that we are dealing here with consequences of assuming invariance of alpha through changes in temperature, but I am not quite sure about it.

    Thank you both.

    Best wishes

    DaTario
     
    Last edited: Dec 8, 2007
  16. Dec 8, 2007 #15

    Dale

    Staff: Mentor

    I will respond to the rest of your post later, after I have slept. But I wanted to quickly explain T0.

    You do one extremely important operation with T0, you subtract it from T to get deltaT. Any time you have a change in something (e.g. deltaT) you must have a reference state from which it is changing (e.g. T0). So T0 is not just there as an arbitrary label, but as a label for the reference state from which deltaT is defined.
     
  17. Dec 9, 2007 #16
    I don't agree with you in this point. I would not call extremely important the mathematical participation of T0 here in this problem. Of course that there is a role played by T0, I agree, but remember that T0 is being used in a way that allow an infinite number of mathematically equivalent situations. So T0 don't leave properly its signature in this problem. One could argue that in this problem, the important variable is the variation Delta T, not the temperature itself.

    Let me emphazise that I still haven't found what I am looking for (I'm still troubled with this question)

    Best wishes

    DaTario
     
  18. Dec 9, 2007 #17
    I don't agree with you in this point. I wouldn't call extremely important the mathematical participation of T0 in this context. I agree that T0 plays a role here, but remember that it enters the mathematics in a way that allow an infinite number of equivalent situations. One could argue that the important temperature parameter here is the variation Delta T.
    What I want to express is that T0 does not leave its fingerprint in this result. This may be the cause of this non-uniqueness.

    Obs.: In order to have T0 indicating effectively the initial state, it would be necessary to have, say, an injective function (x1 neq x2 -> y1 neq y2) alpha(T0), which is not the case.

    Best wishes

    DaTario
     
  19. Dec 9, 2007 #18

    Dale

    Staff: Mentor

    We are indeed assuming that alpha does not depend on T, which is a reasonable assumption for small deltaT. In other words, alpha may be approximately constant over a 50º range, but not over a 5000º range.

    However, the key point here is to realize that, even though alpha does not depend on T, it does depend on T0!

    Let's say that we have an unknown material and wish to determine alpha for this material. How would we do that experimentally? Essentially we would just measure the length at two different temperatures. Let's say that we measure it at 10º and it is 1.1 m and at 20º it is 1.3 m. Now, starting with the above equation:

    L(T) = L(T0) + L(T0) alpha (T-T0)

    and solving for alpha we obtain

    alpha = (L(T)-L(T0))/((T-T0) L(T0))

    So, if we take 10º as the reference temperature we obtain

    alpha = (1.3-1.1)/((20-10) 1.1) = .018

    But, if we use the exact same measurements but take 20º as the reference temperature we obtain

    alpha = (1.1-1.3)/((10-20) 1.3) = .015

    I won't show the details, you can work it out for yourself, but you can show that the formalism only works at T0. If you calculate the formalism, but use it to assemble a structure at some other temperature then your structure will expand or contract with temperature changes.

    Also, it should now be no surprise that if you had two materials whose lengths were as you described above their alphas at the two reference temperatures would work out such that you would get the correct ratio at each T0.

    Your intuition is very close, we were dealing with consequences of assuming invariance of alpha through changes in the reference temperature.
     
  20. Dec 9, 2007 #19
    According to what you said, alpha = Delta L / (L0 * Delta T).

    Rigorously it presents a non-uniqueness in the determination of alpha, for instance, as several different experimental situations may yield the same alpha for a given material, with a given initial length.

    Going from 10 C to 20 C has the same effect of going from 30 C to 40 C. If you disagree at this specific point, let me know.

    We would be likely to have, also, an issue of units here (Kelvin instead of Celsius) if was it the case of having a true dependence on absolute temperature (I guess that better theories for the same phenomena may present this feature).

    Let me put again the main purpose of this thread:

    Solving the problem mathematically it does appear a well defined and unique result. But experimentally we may observe that 1.1 and 0.9 is as good a solution as does 1.3 and 0.7. So, at what point we lose track of this class of equivalence?

    Best wishes

    DaTario
     
  21. Dec 9, 2007 #20

    Dale

    Staff: Mentor

    Certainly. And how would you define deltaT without a reference temperature, T0?

    I have already set up the math and the background for you to work this problem correctly. Please justify your claim that 1.1 and 0.9 is as good a solution as 1.3 and 0.7. In which situations is it "as good a solution"? In which situations is it not?
     
  22. Dec 10, 2007 #21
    First point: There is no Delta T without a T0, I know. But for several different T0 we may have a class of experiments with the same Delta T. And if the physically meaningful parameters, as for instance, the dilatation or the alpha parameter ,depends on Delta T, so they do not depend explicitly on T0 which means that different T0 would yield the same value for these parameters. Voi la the non uniqueness! Note that you were not able to demonstrate that T0, as an absolute value for temperature, participate in the calculation of any Physically Meaningful Parameter. Remember what happens with the energy. No PMP depends on the value of the energy itself. Only energy variations produce PMP.


    Second point: I am not basing my judgement on mathematics when I say that 1.1 and 0.9 is as good a solution than 1.3 and 0.7. I am just based on the absence of a reference to any absolute value for temperature. And on the fact that, no matter to what temperature you go, the ensemble will be 2 meters long, and therefore may serve also as a good starting point.

    I am really astonished with the dificulties I (or Me and You) am experiencing.
    But once more thank you for your participation on this debate.

    I hope I could make my point clear this time.

    Best wishes

    DaTario
     
    Last edited: Dec 10, 2007
  23. Dec 11, 2007 #22

    Dale

    Staff: Mentor

    You should base your judgement on math. Don't just assume that 1.1 and 0.9 is as good a solution as 1.3 and 0.7. Set up a scenario, work out the math, and find out under what conditions 1.1 and 0.9 is as good a solution as 1.3 and 0.7.

    I thought I had explained it clearly, but since you are still confused my explanation obviously wasn't clear enough. Therefore, please play around with the math, in particular do the derivations yourself. I think that will help you understand much better than another round of me repeating my same solutions and you repeating your same objections.

    Once you have done that if you have further questions then we can discuss again.
     
  24. Dec 13, 2007 #23
    Dear DaleSpan,

    I have only entered physicsforums after having done all the calculations. I will do the calculations here. First let me repeat the question with all the details (including the "useless" information of T0).

    QUESTION
    Supose we have two pieces of wire, made of substance A and B respectively. The linear dilatation coefficient of both A and B are alphaA = 0.001 and alphaB = -0.003, the last one representing a behavior where under positive Delta T the material exhibits contraction instead of dilatation. Determine the length of each piece of wire if one is interested in having a junction of A and B such that the total length is 2 meters and under any Delta T the dilatation compensates the contraction perfectly yielding always a 2 meters junction. Assume the lab is intially at 30 Celsius.

    SOLUTION:
    first equation: L0A + L0B = 2

    second equation: Delta LA + Delta LB = 0
    which implies: L0A * alphaA * Delta T = - L0B * alphaB * Delta T

    dividing both sides by Delta T (now we get the result independent of Delta T !)
    we have:
    L0A = (-alphaB / alphaA) * L0B

    substituting in first equation: L0B = 2 / (1 - alphaB / alphaA)
    which implies directly that: L0A = 2 - L0B = (-2 * alphaB / alphaA) / (1 - alphaB / alphaA)

    substituting the numerical data we get: L0B = 2/(1 + 3) = 0.5 meter => L0A = 1.5 meter

    the end.

    Notice that:
    1) being independent of Delta T, the result end up being independent of T0, which by the way wasn't used anywhere (30 Celsius).

    2) physically its presumably true that if we increase the temperature the individual lengths will change but sum don't. We may reach for instance, from 1.5 and 0.5, the new values 1.7 and 0.3 (do you agree with this?). At this new temperature what situation do we have? Having changed the temperature to a higher value, the sum of lengths will still yield 2 meters. The temperature of the lab will be of course higher that 30 Celsius, say 65 Celsius. This value will be as useless to the system's determination of phisically meaningful parameters than was the 30 Celsius.
    Therefore this new situation (1.7 and 0.3 at 65 Celsius) must be as good a initial situation than that with 1.5 and 0.5 at 30 Celsius.

    I am trying to understand why the simple mathematics of this problem don't reveal this options.


    Thank you for your colaboration.

    Best wishes

    DaTario
     
    Last edited: Dec 13, 2007
  25. Dec 14, 2007 #24

    Dale

    Staff: Mentor

    That wasn't exactly what I had in mind. As I stated twice already I wanted you to determine under what conditions 1.7 and 0.3 was as good a solution as 1.5 and 0.5. You just asserted that it was a solution at some other temperature, but made no effort to quantify or explore that idea further. But at least now we have something concrete to discuss.

    Yes, given the values provided the temperature at which the lengths will be 1.7 m and 0.3 m is 163.333ºC.

    I put in the correct temperature, but that doesn't change your point which is correct. 1.7 m and 0.3 m at 163ºC is as good a solution or initial condition as 1.5 m and 0.5 m at 30ºC. Note that each pair of lengths is only a valid solution at one temperature!

    T0 was not used explicitly, but as I described previously alpha is a function of T0 so the results are not independent of T0.

    If my previous math-based argument did not convince you that alpha is a function of T0, then consider this physics-based argument. With a piece of material A (alpha 0.001) of length 1.5 m at 30ºC (T0) each 1ºC increase in temperature yields a 1.5 mm increase in length. At 30ºC we have 1.5 mm/1.5 m = .001 = alpha(30ºC). If we continue heating it up to 163ºC it still continues to expand by 1.5 mm for each 1ºC, however now that same 1.5 mm is a smaller fraction of 1.7 m than it was at 1.5 m. In fact, at 163ºC we have 1.5 mm/1.7 m = .00088 = alpha(163ºC).

    Going back to math-based arguments, we can derive an expression for the dependence of alpha on the reference temperature by substituting

    L(T') = L(T0') + L(T0') alpha' (T'-T0')

    into

    alpha = (L(T)-L(T0))/((T-T0) L(T0))

    to obtain

    alpha = alpha'/(1 + T0 alpha' - T0' alpha')

    Where alpha' is the original dilation parameter at the original reference temperature T0' and alpha is the dilation parameter at any other reference temperature T0, i.e. alpha is clearly a function of T0, alpha(T0). So for the current problem:

    alphaA(163ºC) = .001/(1 + 163 .001 - 30 .001) = .00088
    alphaB(163ºC) = -.003/(1 - 163 .003 + 30 .003 ) = -.005

    And we see that using the correct alpha values the formalism holds at 163ºC since

    1.7/.3 = -(-.005/.00088)

    I hope that clears everything up.

    Because the simple mathematics doesn't make it obvious that the dilation parameter depends on the reference temperature and that the formalism is only valid at the reference temperature.
     
  26. Dec 14, 2007 #25
    Let me examine your idea,

    You are saying that alpha = alpha(T).

    I found your argument very convincent in deed, but let me present this objection: according to your reasoning, integration in dT of L0 * alpha(T) would be probably the correct way to determine the length variation for a piece of wire, which is not the case. So we must conclude that alpha is independent of alpha (at least in this simplified approach).

    Aditionally, consider that the question were changed to

    QUESTION (same thing except for the lab's temperature)

    Supose we have two pieces of wire, made of substance A and B respectively. The linear dilatation coefficient of both A and B are alphaA = 0.001 and alphaB = -0.003, the last one representing a behavior where under positive Delta T the material exhibits contraction instead of dilatation. Determine the length of each piece of wire if one is interested in having a junction of A and B such that the total length is 2 meters and under any Delta T the dilatation compensates the contraction perfectly yielding always a 2 meters junction. Assume the lab is intially at 163,33 Celsius.

    You would say that now I would be dealing to a different pair of materials. (correct?)


    Best wishes

    DaTario
     
    Last edited: Dec 14, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook