## Comments for physicspages.com

This blog is for comments and reports of errors on my physicspages.com site. I will also post notifications of new articles published on the main site using this blog, so if you want  to be kept up to date, check here regularly, or else  subscribe to it for email notifications. I realize this may look like a roundabout way of organizing things, but I don’t trust WordPress with anything substantial any more.

If you’re reporting an error or want clarification of an equation in an article, PLEASE include:

• The URL of the page – just copy this from the browser’s address bar at the top.
• The equation number containing the error.
• And of course, what you think is wrong with it.

Thanks, and I hope you find the new site as useful as the old one.

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### 79 Responses to Comments for physicspages.com

1. Jordan says:

https://physicspages.com/pdf/Carroll%20&%20Ostlie/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!

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• 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.

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2. Anonymous says:

https://physicspages.com/pdf/Shankar/Shankar%20Exercises%2012.03.05.pdf
Between equation 12 and 13, P=μv^2 is kinetic energy. Momentum should be P=μv

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• growescience says:

Fixed now. Thanks.

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3. Siddhartha says:

Hey!
https://physicspages.com/pdf/Lahiri%20QFT/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).

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• growescience says:

I had put the A on the wrong side in equation 5. Fixed now. Thanks.

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4. Sandeep says:

https://physicspages.com/pdf/Griffiths%20EM/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.

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• growescience says:

The divergence is zero, since $\frac{\partial W_{x}}{\partial x}=\frac{\partial W_{y}}{\partial y}=\frac{\partial W_{z}}{\partial z}=0$.

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5. 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?

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• 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.)

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• 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).

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• 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.

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• 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.

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• 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?

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• 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?

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• 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.

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• 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.

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• 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.

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• 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.

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6. Anonymous says:

Hello!

https://physicspages.com/pdf/Griffiths%20EM/Griffiths%20Problems%2012.47.pdf
There should be an equal sign “=” in Equation 9 for the magnetic field

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• growescience says:

Fixed now. Thanks.

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7. Robert says:

https://physicspages.com/pdf/Schroeder/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

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• growescience says:

Fixed now. Thanks.

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8. 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?

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• 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.

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9. Bogidon says:

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

http://physicspages.com/pdf/Griffiths%20EM/Griffiths%20Problems%2004.19.pdf

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• growescience says:

I should have written $E_d=\sigma_d/\epsilon$ not $E_d=\sigma_d/\epsilon_0$. Fixed now. Thanks.

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10. Vehryn says:

https://physicspages.com/pdf/Griffiths%20QM/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.

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• growescience says:

Fixed now. Thanks.

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11. pp says:

http://physicspages.com/pdf/Schroeder/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?

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• pp says:

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

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12. Tianluo_Qi (Eastern) says:

http://physicspages.com/pdf/Griffiths%20QM/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)).

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13. Mongeeses says:

http://physicspages.com/pdf/Schroeder/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.

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• growescience says:

There is a factor of g in the numerator already.

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14. Petra Axolotl says:

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.

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• growescience says:

Fixed now. Thanks.

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15. Petra Axolotl says:

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

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• growescience says:

Noted, although the calculations are much the same as those given so I won’t go through the details.

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16. Petra Axolotl says:

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.

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• growescience says:

Fixed now. Thanks.

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17. Petra Axolotl says:

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).

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• growescience says:

Fixed now. Thanks.

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18. Arindam Kumar Chatterjee says:

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

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• 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.

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19. 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.

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• 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 ${bt/\hbar}$.

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20. Nick says:

Griffiths Quantum, problem 4.18.
http://physicspages.com/pdf/Griffiths%20QM/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?

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• growescience says:

The LHS of (10) is equal to (13) because of (4), so (13) is equal to (12).

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21. Aaron Stevens says:

Hello,

I am referring to the article at:

http://www.physicspages.com/pdf/Shankar/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!

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22. Danyel Cavazos says:

Hi!

I’d like to ask something in the page
http://physicspages.com/pdf/Shankar/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!

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• growescience says:

The answer is added to the end of the post.

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• Danyel Cavazos says:

That’s what I imagined, but then that means that we should beware of using this identity when A or B is replaced by vector operators like L or S, right?

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• growescience says:

See post for reply.

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23. 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
http://www.physicspages.com/pdf/Shankar/Shankar%20Exercises%2007.03.02-03.pdf

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• growescience says:

Fixed now. Thanks.

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24. hladacpravdy says:

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 ?

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• growescience says:

See comments in original post for reply.

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25. 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?
http://physicspages.com/pdf/Griffiths%20EM/Griffiths%20Problems%2002.21-24.pdf

Thanks.

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• 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.

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• Tomas says:

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26. 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

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• 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 ${U_{K}=U_{total}-Nmgz}$.

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27. 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.

http://www.physicspages.com/pdf/Shankar/Shankar%20Exercises%2001.08.09.pdf

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• growescience says:

Fixed now. Thanks.

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28. 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.

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• 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.

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29. 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.

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• 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…

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30. 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.

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• 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.

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31. Will says:

Thank you!!!

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32. Lyle Arnett, Jr says:

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!

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• growescience says:

Excellent – thanks very much! That takes the pressure off for trying to restore the site quickly.

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33. 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!

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• 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.

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• 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

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• 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.

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34. 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.

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• growescience says:

Thanks. I don’t know how long it will take, but I’m hoping maybe by Christmas…

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