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108 Responses to Comments for physicspages.com

  1. bilzebor says:

    hello!
    in Tom Lancaster and Stephen J. Blundell, Quantum Field Theory for the Gifted Amateur, Problem 19.3, you may have forgotten a possible diagram for question (b) at the 4-th order:

    or maybe I missed something,
    anyway, thanks a lot for your site!

    Like

  2. Agustin says:

    Hi, thanks for your amazing blog. Do you have a post about the differences between the QFT books? It would be helpful for the self-learner. I noticed a comment you did about Klauber’s book. Isn’t it a truly introductory book? which of the books you’ve read are the “most introductory”?

    Like

    • growescience says:

      I have to confess that I didn’t get far enough into most books on QFT to be able to compare them. Of the ones I did manage to penetrate to some depth, I would say that by far the best is Sidney Coleman’s lectures. Although it’s a huge book, the space is used to good effect in that he explains most things thoroughly and in depth.
      The books by Lahiri & Pal, and also by Lancaster & Blundell are also worth having a look at if it’s your first exposure to QFT, but seriously, go for Coleman’s book in the first instance.
      I didn’t really like Klauber’s book all that much. It didn’t really explain things all that well (in my opinion – others seem to like it) and it also assumed a fair bit of classical field theory on the part of the reader.
      The books by Peskin & Schroeder, Zee and Srednicki were too advanced for me to make much of them. They might be worth a look once you have a basic understanding of QFT from other books.

      Like

  3. Pedro Dardengo Mesquita says:

    I tried myself on a problem you said you were not able to solve and after lots of headache I was able to solve it.
    Tom Lancaster and Stephen J. Blundell, Quantum Field Theory for the Gifted Amateur, (Oxford University Press, 2014), Problem 25.6

    I have typed it in LaTEX and its a very lengthy and detailed explanation. You can access it here.
    PDF file
    https://www.dropbox.com/s/r354vln2ei8kkid/Solution25.6.pdf?dl=0

    .tex file so you may salvage any equations, or just copy the entire thing, no problems with that.
    https://www.dropbox.com/s/yh8e9x945vwx15f/Solution25.6.tex?dl=0

    You said in your site that you only wanted a tip but I only read that part when I was halfway in the typing process, so I decided to just end it. Maybe you are one of those that want to see as little “spoilers” as possible of the solution, even in that case I invite you to look at the very first equations of my solution and then you can finish it yourself if that’s the case.

    This solution took me a lot of time, from the begging I knew the way to do it but doing the calculation proved to be a very lengthy and tortuous path, I made many mistakes along the way that took me days to correct, specially in the Matsubara sum (that’s why I added a complete explanation of this specific sum). Without the help of friends (that also happens to know and like this blog) I would take much longer to solve and spot the reason why so many times the result I got was different from the book. Finally it’s solved now.

    Cheers

    Like

    • Pedro Dardengo Mesquita says:

      Please notice that I’ve made the solution for the general reader that’s why I spent time talking about things you probably already know.

      Like

    • growescience says:

      Many thanks for this. I’ve added your solution to my web site (with proper accreditation).
      It does make me wonder why a problem that is that difficult was put in a book for ‘amateurs’ though (even if they are ‘gifted’).

      Like

  4. Vinícius Vargas says:

    There is a mistake at equation 33 (perhaps 31 and 32 also) on Tom Lancaster and Stephen J. Blundell, Quantum Field Theory for the Gifted Amateur, (Oxford University Press, 2014), Problem 13.1.

    You can recognize these b as eigenstates of Jz. Therefore, Jz should be diag(1, 0, -1).

    Like

    • Vinícius Vargas says:

      ps: I don`t know why you blocked (or if you) me on my other account. I can`t see my comments and I can`t comment anymore on that account. Did I say something offensive? Sorry if I did.

      Like

      • growescience says:

        No, I certainly haven’t blocked you, as I welcome reports of errors on the posts. I don’t know why you can’t see your comments. I checked in the settings and there’s nothing obvious there that could have blocked you. Give it another try…

        Like

    • growescience says:

      Fixed (I think). I’m not entirely sure of how to justify the steps in equations 10 and 31, other than that it gives the right answer…

      Like

  5. Vinícius Sant Unioni Ângelo Vargas says:

    There is a wrong \mu derivative of W^\mu function inside parenthesis in equations 16 and 18 on Tom Lancaster and Stephen J. Blundell, Quantum Field Theory for the Gifted Amateur, (Oxford University Press, 2014), Problem 10.2

    Like

  6. Vinícius Sant Unioni Ângelo Vargas says:

    There is a missing i factor in equation 10 on Tom Lancaster and Stephen J. Blundell, Quantum Field Theory for the Gifted Amateur, (Oxford University Press, 2014), Problem 10.1

    Like

  7. Kristine Haley says:

    Hello Dr. Rowe,

    I wanted to extend my gratitude for your website and in particular, electrodynamics. I cannot tell you how much your lecture notes have helped me, especially during a shut down and having to self learn the topics by myself. Your explanations are clear, concise, and the best I have come across.

    Thank you.
    Kristine

    Like

  8. D dwivedi says:

    Click to access Griffiths%20Problems%2011.04.pdf

    References: Griffiths, David J. (2007), Introduction to Electrodynamics,
    3rd Edition; Pearson Education – Chapter 11, Post 4

    Here in this problem if I integrate equation 26 I get zero result. But radiated power is not zero. I will be grateful to you if you could explain this integration please. Thank you

    Like

    • growescience says:

      The \phi integral just gives 2\pi. The \theta integral is \int_{0}^{\pi}\left(1+\cos^{2}\theta\right)\sin\theta\ d\theta=\frac{8}{3}. Note that the \theta integral must be non-zero since the integrand is strictly positive over the range of integration.

      Like

  9. Anthony Ayari says:

    Hello

    concerning Problem 8.10 from Griffiths: Introduction to quantum mechanics.

    Click to access Griffiths%20Problems%2008.10.pdf

    I would say that the problem is not well formulated for 2 reasons which lead to an incorrect answer for the transmission:

    1) The term with the B coefficient represents the incident wave although there is a minus sign in the exponential. So the transmission should be T = F^2/B^2 and not F^2/A^2. This can be checked by noting that B^2 is larger than A^2 (there is a plus sign in B in front of 2exp(2*gamma) instead of the minus sign in your equation (42) for A. I am not saying that equation 42 is incorrect).
    2) F^2/B^2 leads to the correct transmission only in some specific case otherwise it is better to use the ration of the current of the wave function, or T= (B^2-A^2)/B^2.

    Like

  10. S. Eriksson says:

    This is not an error report, I simply would like to say Thank you for your superb explanations in Introduction to Electromagnetism. As you are reading this, hundreds of students all over the world are using your hard work to learn Physics. Your website is a truly remarkable achievement!

    Like

  11. Condereal says:

    Hi!

    Respect to the following problems of L&B:

    Click to access Lancaster%20Problems%2009.04.pdf


    Click to access Lancaster%20Problems%2009.03.pdf

    I’m no expert, but I think the problem with the sings that you see when obtaining the infinitesimal generators is that in the book L&B the mix an “active” Lorentz boost with a “passive” rotation”. I think they should either user both “active” or both “passive”, but not mixing. This means that equation (9.56) in the book does not have the correct signs from the beginning.

    Hope this is the case, at least for me it worked.
    cheers!

    Like

    • growescience says:

      It’s possible. Given the confusion arising from their own errata, I’m not sure what the correct result actually is.

      Like

    • growescience says:

      Thanks for the comments. I’ve added your comment to the PDF file.

      Like

      • condereal says:

        You’re welcome! Hope this helps anyone else. I will try to write my own solution in latex. In case you find this useful I can share it.

        And thank you for physicspages, it’s a really awesome project!

        Like

  12. f997 says:

    Problem 11.4 from Griffiths: Introduction to quantum mechanics.

    Click to access Griffiths%20Problems%2011.03-04.pdf

    I think equation (22) and (23) are the exact expression, they aren’t a small arguments form.

    Like

  13. Jordan says:

    Click to access Carroll%20&%20Ostlie%2004.01.pdf

    I think there is an error between equations 35 and 36. Using 31 yields

    a_{44}^2 – \frac{u^2}{c^2} a_{11}^2 = 1,

    not

    a_{44}^2 – \frac{u^2}{c^2} a_{44}^2 = 1.

    but perhaps I’m missing something!

    Like

    • growescience says:

      Quite right. I’ve now fixed the derivation, so hopefully it’s correct now. (The previous solution had the wrong sign for a_41 as well!). Thanks for pointing this out.

      Like

  14. Anonymous says:

    Click to access Shankar%20Exercises%2012.03.05.pdf


    Between equation 12 and 13, P=μv^2 is kinetic energy. Momentum should be P=μv

    Like

  15. Siddhartha says:

    Hey!

    Click to access Lahiri%20&%20Pal%20Problems%2001.01-04.pdf


    While going from equation 12 to 13, it will be equal to A^2 instead of 1/(A^2).

    Like

  16. Sandeep says:

    Click to access Griffiths%20Problems%2005.53.pdf


    The eqn. no. 17 in the solution of problem 5.53 of Griffiths electrodynamics does not seem to be divergenceless. Could you please explain.

    Like

  17. Robert says:

    For Example 7.9 in Electrodynamics, page 319 (4th edition), it seems that you are assuming the induced electric field to be zero at the location of the wire (and increasing with distance from the wire) rather than non-zero (and decreasing with distance from the wire). Both are consistent with Faraday’s law (direction of circulation of E around Amperian loop) but the latter (opposite the change in current) seems more consistent with the self-inductance. What am I missing?

    Like

    • growescience says:

      I don’t know what you’re referring to – Problem 7.9 in Griffiths is about calculating the EMF generated by varying magnetic flux through a loop. Could you please provide the exact URL of the page and equation number you’re referring to? (Also, note that I have the 3rd edition of Griffiths’s book, not the 4th.)

      Like

      • Robert says:

        In the 3rd edition it is on page 308. It is “Example 7.9”, not “Problem 7.9”. Within the solution in the 3rd edition, there is equation (7.20) and equation (7.21). In the 4th edition, the equations are numbered (7.19) and (7.20).

        Like

      • growescience says:

        The electric field depends on the rate of change of the magnetic flux through the loop, and this in turn depends on the area of the loop. In Griffiths’s example, he takes s_0 (the distance of the near edge of the loop from the wire) to be constant and varies the distance s of the outer edge. Thus as s gets larger, the area gets larger so the magnetic flux also increases, and thus the rate of change of flux increases in proportion to the area of the loop. As he points out, though, E doesn’t increase forever since the quasistatic approximation isn’t valid for very large s.

        Like

      • Robert says:

        Yes, the electric field depends on the rate of change of the magnetic flux through the loop, and this in turn depends on the area of the loop. Consequently, the rate of change of flux increases as the area of the loop increases. However, the magnetic field isn’t constant in space (it decreases with increasing s) and so the rate of change of flux is not going to increase at the same rate as the area of the the loop. I agree that the quasistatic approximation isn’t valid for very large s but that doesn’t seem to be what is causing the problem in this case. Rather, it seems the problem is in assuming E = 0 at s = 0 instead of E = 0 at infinity.

        Like

      • growescience says:

        The induced E will always keep increasing as long as we keep increasing the area of the loop, no matter what B does inside the loop (well, providing B is always in the same direction and the change dI/dt is nonzero). The variation in B is accounted for by the integral. You don’t need to assume anything about where E is 0, since E comes out of the formula. Are you perhaps confusing E with the potential V, which does need to have a zero point defined, since it’s only differences in V that are measurable?

        Like

      • Robert says:

        I’m not sure what you mean by the induced E. Do you mean the induced circulation of E? If so, I agree that the induced circulation of E increases as the loop increases in the sense that the difference between E(s) and E(s_0) must increase as the loop size increases. I just don’t see why that necessarily means that E(s) must increase as the loop size increases. What am I missing?

        Like

      • growescience says:

        The question states that we’re looking for “the induced electric field”, which is the field induced by the changing magnetic field. By your own argument, if E (s) – E(s_0) increases and E(s_0) is constant, then E (s) would have to increase, true?
        I do agree that it does seem a bit odd that E increases as you get farther from the wire even though B decreases, but the increase is slow (logarithmic) and is valid only for relatively short distances.

        Like

      • Robert says:

        I think the question is this: if E(s) – E(s_0) increases and E(s_0) is constant, then does E(s) have to increase? I agree that E(s) has to become more positive (i.e., dE(s)/ds > 0). However, suppose E(s_0) is negative. Then E(s) – E(s_0) will increase as E(s) as becomes less and less negative, approaching zero.

        Like

      • growescience says:

        E is the magnitude of a vector so must be positive. It is only the dot product of E with the length vector around the Amperian loop that can be positive or negative and this is already accounted for in the expression given in the example.
        If you’re still having problems you could try posting to physicsforums.com.

        Like

      • Robert says:

        Thanks for your help. I agree that the integral goes to infinity as s goes to infinity and I agree that that isn’t realistic and stems from the invalid assumption that dI/dt impacts the B field everywhere instantaneously. I just think E decreases, rather than increases, with distance from the wire (and, similarly, the equation should work for small s, when the quasistatic approximation is valid). I’ll see if perhaps physicsforums.com can help clear up my confusion.

        Like

  18. Anonymous says:

    Hello!

    Click to access Griffiths%20Problems%2012.47.pdf


    There should be an equal sign “=” in Equation 9 for the magnetic field

    Like

  19. Robert says:

    Click to access Schroeder%20Problems%2005.16.pdf


    Equation 11 in the write up for problem 5.16 in Schroeder
    I believe this equation should be dV=beta*V*dT not dV=beta*V*dV

    Like

  20. Brad says:

    In https://www.physicspages.com/pdf/Griffiths%20QM/Griffiths%20Problems%2001.11-12.pdf :
    Equation 15 gives a solution to the integral that has limits -r,r. The denominator in the arctan is sqrt(r^2-x^2) which seems like it would result in a divide by zero. It’s been quite a while since I’ve had trig or calculus. Am I missing something in the math here?

    Like

    • growescience says:

      The argument of the arctan does indeed go to plus/minus infinity at the limits, but remember that the tan of pi/2 is infinity, so the arctan of plus/minus infinity is plus/minus pi/2.

      Like

  21. Bogidon says:

    In Griffith’s 4.19, I don’t follow how equation (39) follows from equation (38), since \sigma_d \ne \sigma_v.

    Click to access Griffiths%20Problems%2004.19.pdf

    Like

  22. Vehryn says:

    Click to access Griffiths%20Problems%2006.27.pdf


    More an issue with typesetting/formatting than anything else. Equations (8), (10), and (11) overflow beyond the margins and even extend beyond the page. This could be made more legible by breaking those equations across multiple lines.

    Like

  23. pp says:

    Click to access Schroeder%20Problems%2003.03-04.pdf


    Problem 3.3-3.4
    Since you are using exact functions to plot S vs UA and UB, (my guess – 1/sqrt(x) and 1/2sqrt(x)) I am wondering if it is possible to show quantitatively on the graph where both the objects have the same slope. With the functions which you have used, an equilibrium state is at very low energy content. Any suggestions which function (as a trial entropy vs energy function) shall I use to show clearly?

    Like

    • pp says:

      I think I was not clear in my question. In other words, could you mark the equilibrium location in your graphs?

      Like

  24. Tianluo_Qi (Eastern) says:

    Click to access Griffiths%20Problems%2002.02.pdf


    Problem 2.2
    All the arguments here are based on the hypothesis that the state function has a maximum or minimum. If it doesn’t, or if state function doesn’t even have secondary derivative at extremum point at all, values of the state function and its secondary derivative can have the same sign, but it’s normalizable as well. As an example of the latter kind, consider the state function of the bound state of a delta function well, namely, sqrt(a)*exp(-a*abs(x)).

    Like

  25. Mongeeses says:

    Click to access Schroeder%20Problems%2005.32.pdf


    equation 14
    You have the pressure as the mass over the area, but by definition pressure is the force over the area, I think you have left out a factor of “g” to make the numerator as a force.

    Like

  26. 2.7.8 (1-3): In Eqs. (9), (10), (11), i and j should be switched in the numerator and the denominator of the second term.

    Like

  27. 2.7.6: For completeness, one should also verify {R_CM, p} = 0 and {r, p_cm} = 0.

    Like

  28. Not that it matters to the end result. But square roots are missing in two denominators in Eq. 32 in the answer key to Shankar’s 2.7.5.

    Like

  29. Regarding Shankar 2.5.1, it is fairly obvious but any way…

    One should replace the index of summation ‘i’ by something else in Eqs. (5) and (6).

    Like

  30. Arindam Kumar Chatterjee says:

    sir, please restore the Quantum field theory pages of Peskin’s book…..urgently needed.

    Like

    • growescience says:

      I’ve restored what I had on Peskin & Schroeder, but this covers only a few bits of Chapter 2. I couldn’t understand the book well enough to do any more.

      Like

  31. Chris Kranenberg says:

    Griffiths Intro to QM Solution 3.37 b): the solution can be taken a little further by replacing the vector notations with their matrices, perform the matrix addition to obtain a matrix solution with cosine and isine expressions.

    Like

    • growescience says:

      If you mean eqn (15), yes, you could factor out the {e^{-iat/\hbar}} and then substitute from (8) and (9) to get the answer in terms of sin and cos of {bt/\hbar}.

      Like

  32. Nick says:

    Griffiths Quantum, problem 4.18.

    Click to access Griffiths%20Problems%2004.18.pdf

    I can’t follow from eq. 12 to 13. What justifies adding or subtracting 1 from m on the f state?

    Like

  33. Aaron Stevens says:

    Hello,

    I am referring to the article at:

    Click to access Shankar%20Exercises%2011.04.01%20-%2011.04.04.pdf

    I am specifically referring to problem 11.4.2 in Shankar’s Quantum Mechanics book (around equation 33 in the above link).

    The final equation 38 you arrive at is valid, but then you say,

    “Since the cosine is periodic, we can’t actually calculate a unique value for
    its average, although if we do the average over an exact number of periods,
    the average is still zero. I have a feeling that I’m missing something obvious
    here, so any suggestions are welcome.”

    Let me see what you think:
    The final expression that results in the expectation value of the cosine function is not the same thing as the average of the cosine function. The only time the expectation value is equal to the average is when the probability distribution in question is the uniform distribution. In general, this expectation value is state dependent. It is a measure of what we would “expect” this cosine function to be given the probability distribution of finding the particle between x and x+dx in space (psi* times psi). We could contrive a state that gives 0 for this expectation value, but I doubt that the state will then evolve according to the Schrodinger’s equation in such a way as to keep the expectation value set at 0.

    To have momentum conservation, we need the expectation value of the commutator [P,H] to be 0 always, since momentum conservation is a property of the physical system in question, not on the states within that system. So since we get a “final answer” that is not identically 0, we must say that momentum is not conserved.

    Another way I thought to solve the problem that is simpler but I am unsure is valid is that since the potential (and therefore Hamiltonian) is not invariant under infinitesimal translations that we cannot say momentum is conserved. Or thinking a little bit differently, just because we have found points in space where the potential energy is the same does not mean momentum is conserved. It is like if you were to push a block down and back up a hill with friction so that it begins and ends at rest. If you are considering just the block as your system and then saying since it starts and stops with the same energy that its energy must have been conserved the entire time, you would not be correct.

    I grade for a quantum class that uses this book, and I think a lot of the students refer to these solutions, so this is how I came across this. Thanks!

    Like

  34. Danyel Cavazos says:

    Hi!

    I’d like to ask something in the page

    Click to access Shankar%20Exercises%2014.03.04.pdf

    How do we go from eq. 22 to eq. 23? I.e., how do we know that when we evaluate Tr(A_i s_i B_j s_j) we can take A_i and B_j out of the trace operation?

    Thank you so much!

    Like

  35. Kevin says:

    On Shankar’s exercise 7.3.2, you were asked to verify that Hermite’s polynomials follow the recursion relation 5.3.15 (on the book), you proved they follow the recursion in equation 5.3.25

    Click to access Shankar%20Exercises%2007.03.02-03.pdf

    Like

  36. hladacpravdy says:

    Example 2.25

    Click to access Griffiths%20Problems%2002.25.pdf

    Equation 21:
    = (m * alpha / hb)^2

    From this follows:
    sigma_p = sqrt( – ^2) = m * alpha / hb

    Unit analysis:
    m … kg
    alpha … J
    hb … Js
    m * alpha / hb ……. kg/s
    SI unit for momentum is kg m/s. Where I have done mistake ?

    Like

  37. tomas says:

    I have a simple question about the Electrodynamics Spherical potential example 1.
    If the path integral goes from infinity to r, why doesnt the dl point inwards? It would appear an extra minus which will make it all wrong. But aren’t the Electric Field and the dl pointing in opossite directions?

    Click to access Griffiths%20Problems%2002.21-24.pdf

    Thanks.

    Like

    • growescience says:

      I think it’s just convention that if you choose your reference point to be at infinity, then {d\mathbf{l}} always points away from the origin. You can always fix the sign of {V} if you’re calculating it from {\int\mathbf{E}\cdot d\mathbf{l}}, since you know that you must have {\mathbf{E}=-\nabla V}. Thus if {\mathbf{E}} points outwards (away from the origin), then {V} has to decrease as you move away from the origin.

      Like

  38. Victoria Jenne says:

    Your logic is wrong in problem CHEMICAL POTENTIAL OF AN IDEAL GAS problem # 3.37 Thermodynamics. The book defines U as internal energy so you can’t add potential energy to kenetic energy without previously using a quantum mechanics proof to alter the multiplicity which would entail altering the volume veriable. Because we are obviously not changing the internal energy of the system your proof in not valid, even if you did get the correct answer.

    Best Wishes,
    Vicky

    chrome-extension://ecnphlgnajanjnkcmbpancdjoidceilk/content/web/viewer.html?source=extension_pdfhandler&file=http%3A%2F%2Fphysicspages.com%2Fpdf%2FSchroeder%2FSchroeder%2520Problems%252003.37.pdf

    Like

    • growescience says:

      If I understand what you’re saying, I don’t think there’s a problem. The derivation of the Sackur-Tetrode equation given by Schroeder in section 2.5 counts the number of states available to an ideal gas by using the quantum uncertainty principle, and assumes that the gas is contained in a volume in free space (that is, no gravitational field). This leads to Schroeder’s equation 2.40 which is in turn used to derive equation 2.49. The energy {U} in this derivation is entirely kinetic, since it’s obtained from the ‘volume’ in momentum space. If we apply a gravitational field by placing the gas a distance {z} above the Earth’s surface, we don’t change the number of states available to the gas; all we do is shift their total energy by an amount {mgz} per molecule. Thus the derivation of the Sackur-Tetrode equation would still be valid, except that the kinetic energy in the formula would now be {U_{K}=U_{total}-Nmgz}.

      Like

  39. Anonymous says:

    I believe your equation 0.7 for Shankar’s problem 1.8.9 should be the ith component of L, not the L vector as a whole.

    Click to access Shankar%20Exercises%2001.08.09.pdf

    Like

  40. Arpon says:

    I have some confusion about your solution to problem 14.05.03, Principles of Quantum Mechanics, Shankar. (Link: http://physicspages.com/pdf/Shankar/Shankar%20Exercises%2014.05.03%20-%2014.05.04.pdf )
    “Now suppose we return to the case where the first apparatus transmits only spin z of +h/2 but the second (aligned along the x axis) transmits everything (no blocked beam) into a third apparatus, which is aligned again along the z axis, but now transmits only particles with spin z of -h/2. In this
    case, the middle (x axis) apparatus has no effect since it doesn’t filter the particles at all, with the result that we’re feeding a stream of +h/2 particles into an apparatus that detects only spin -h/2. In this case, nothing will get through.”
    My opinion is: When the spin +z electrons pass through the second SG (along x axis) apparatus, the outcome is electrons in the +x and -x spin state. The state is no more +z spin. So when the +x and -x spin electrons pass through the 3rd SG apparatus (along z axis) 50% of them should get through.

    Like

    • growescience says:

      I originally thought the same thing, but I think the point is that we aren’t allowed to look at the output of the middle detector, so we don’t ever measure a particle’s x-spin. In that case, all the particles remain in the +z state, so they all get blocked in the third detector. I suspect you’re right that if we did measure the x-spin in the middle detector that would place the particle in either the +x or -x state, with z-spin undetermined, so that 50% of them would indeed get through the third detector.
      Shankar gives zero as the answer at the back of the book, so my guess is that’s what’s happening.

      Like

  41. incomprehensiblething says:

    Just to say thank you for all your work on PhysicsPages, past and present. Your site is much appreciated, but don’t overstress yourself getting it back online.

    Like

    • growescience says:

      Glad you find it useful. I’m trying to get it restored ASAP, since once it’s done, I’ll be able to get back to actually writing something about physics…

      Like

  42. Electro says:

    sir when will your answers for griffiths electrodynamics be ready. It was very very helpful and i also need it now. Your explanations were brilliant. Please we all need that fast sir.

    Like

    • growescience says:

      It will take some time for me to restore everything. In the meantime, there is an archive of the old site available which contains most of the content up to March 2017. See the physicspages.com home page for the link, as it doesn’t seem possible to insert links in comments here.

      Like

  43. Will says:

    Thank you!!!

    Like

  44. Hello. There is a web page called the Wayback Machine (http://web.archive.org/). They take snapshots of the internet at various times and store them for access. There is a nearly 100% functional version of the old Word Press site from June of this year available. I have been using it to view content. You can find it here:

    http://web.archive.org/web/20170620174446/http://www.physicspages.com/welcome-to-physicspages/

    It might be a good option to share with people until you get the new website up and running? Just an idea.

    I’m an undergrad Physics student. Three semesters to go (including this one). Your website is extremely valuable, and I thank you very much for the work you have done.

    Cheers!

    Like

  45. Danyel Cavazos says:

    Although I don’t know extremely much about web design, I have done a little before. This page is and has always been extremely useful, please let me know if I can help you in any way!

    Like

    • growescience says:

      Thanks for the offer. I think the basic design is OK – I want to keep it simple so it’s easy to maintain. The only thing that would be useful is some way of reliably displaying PDF files in the various browsers, although I suspect this might not be possible. Chrome has removed support for the Adobe plugin. On Firefox, the equals signs in the equations don’t show up. And so on and on…
      I think the best solution is for users to just download the PDFs and view them offline.

      Like

      • Danyel Cavazos says:

        What were you using to display equations before? I’m sure there are ways of presenting LaTex eqns in HTML code via extensions

        Like

      • growescience says:

        I was using the Latex “Beautiful Math” plugin from Jetpack in WordPress. I think there are Latex extensions (Mathjax?) for displaying math in HTML, but as I use the Lyx editor for writing the posts in Latex and it has a Latex–>PDF converter, it’s easiest to just create PDFs and upload them.
        Thanks anyway.

        Like

  46. Antonio Victor Nascimento says:

    i knew the website the last wek, it is very great and important. I am physics student of UFES in brazil.
    i hope that the website normalize soon.
    sorry my english..
    hugs of brazil.

    Like

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