Question on the Newtonian displacement of light

[brpetrucci]

Hi everyone! Hope your day is going well. I’m an ex-physics student who recently wanted to go back to studying the subject (as a hobby, mostly). So I picked up Zee’s GR book since GR is the thing I’m the most interested in. I expected to hit a wall on some basic things since I’m rusty, and did so already in exercise I.1.3.

The question asks to consider light as corpuscles and calculate the deflection of light by the sun. I was confused even where to start, so I peeked at the solutions and saw that Zee mentions applying equation (19) with r_min < 0. I’m confused on why that’s the case. He later says that we can consider u_min = 1/r_max as being 0, which I think I understand (light’s the only thing fast enough to escape the gravitational pull into infinity?), but I was confused by that statement of his. Can anyone explain why r_min would be < 0? Mathematically, of course, since he himself mentions physically it doesn’t make sense.

I’m not sure what the rules are about posting PDFs of books here, so let me know if I can. Otherwise, the problem is in page 33 and equation (19) in page 30. Thanks in advance!

 

[Ibix]

Welcome to PF.

We have $\LaTeX$ specifically so that you can typeset equations. There's a howto guide linked below the reply box if you don't know it.

 

[brpetrucci]

Welcome to PF.

We have $\LaTeX$ specifically so that you can typeset equations. There's a howto guide linked below the reply box if you don't know it.

Thank you. It seems like I can't edit that post, but I'll make sure to remember that for the future. I guess the equations I mentioned would be $r_{min} < 0$, and $u_{min} = 1/r_{max}$.

 

[berkeman]

I’m not sure what the rules are about posting PDFs of books here, so let me know if I can. Otherwise, the problem is in page 33 and equation (19) in page 30.

We follow the "Fair Use" copyright laws, which allow you to post a small portion of a page for non-commercial purposes, as long as you include attribution to where you got it. So it should be fine to post a picture (clear, well-lit, upright, etc.) of the problem and the solution. Use the "Attach files" link below the Edit window to upload PDF and JPEG files.

 

[brpetrucci]

We follow the "Fair Use" copyright laws, which allow you to post a small portion of a page for non-commercial purposes, as long as you include attribution to where you got it. So it should be fine to post a picture (clear, well-lit, upright, etc.) of the problem and the solution. Use the "Attach files" link below the Edit window to upload PDF and JPEG files.

Thanks! Will do. The equation in question is here, in page 30 of Zee's Einstein Gravity in a Nutshell. In his solution in page 793, he starts "We could still use (19) except that the root $r_{min}$ is now negative, which is not physical since $u = 1/r > 0$. A moment’s thought indicates that the lower limit for the $u$ integral in (19) should be set to 0." (LaTeX mine). This should make this question self-contained, given one is familiar with the derivation of the Newtonian orbit. Thank you both! Hope someone can help now that I've made my post less badly formatted and it's in the correct place, lol.

To reiterate, the question is why $r_{min} < 0$. Furthermore, if we assume that that root should be negative, why is the solution to that to make $u_{min} = 0$ and keep $u_{max}>0$, if $u_{max} = 1/r_{min}$?

 

[PeroK]

I'm struggling to relate the extract from Zee in post #5, which appears to be about closed (elliptical) orbits, and your opening post about deflection of light by the Sun.

 

[PeroK]

PS It's hard without seeing the relevant equations 17-19.

 

[brpetrucci]

I'm struggling to relate the extract from Zee in post #5, which appears to be about closed (elliptical) orbits, and your opening post about deflection of light by the Sun.

The extract is to give background to what Zee references in the solution to the question as equation 19. The question about the deflection of light by the sun is question 3 of this chapter--I didn't post a specific picture because that's effectively the extent of the question, "Calculate the deflection of light by the sun, applying what you learned in the text to the case $\epsilon > 0$."

PS It's hard without seeing the relevant equations 17-19.

I was trying to minimize the amount of pictures of the book, but you make a fair point. Here is the previous part of the derivation of the orbit. Equation 19 is in the picture on post #5. I can clarify other points if something is still confusing. Thank you for your attention!

 

[Ibix]

So $u=1/r$ and therefore $u_{min}=1/r_{max}$. What's $r_{max}$ for an open orbit?

 

[brpetrucci]

So $u=1/r$ and therefore $u_{min}=1/r_{max}$. What's $r_{max}$ for an open orbit?

Infinity! Yeah, that part makes sense. I guess I don’t know how to explain why we expect the orbit of the light to be open from Newtonian terms only. But even then, I guess the thing that tripped me up the most here was the $r_{min} < 0$ part. I guess it’s not essential to the final solution, but I still wanted to try and understand what Zee mentioned there

 

[Ibix]

I guess I don’t know how to explain why we expect the orbit of the light to be open from Newtonian terms only.

Because it's manifestly faster than escape velocity.

But even then, I guess the thing that tripped me up the most here was the $r_{min} < 0$ part. I guess it’s not essential to the final solution, but I still wanted to try and understand what Zee mentioned there

Right, sorry. Got lost on what your question was going backwards and forwards between posts on my phone. Look at the extract you posted in #5 where he re-writes $2\epsilon-l^2u^2+2\kappa u$ in terms of its roots, which is how he defines $u_{min}$. Note that $\epsilon$ is the "energy at infinity" - the amount of energy (or actually energy per unit mass, to be precise) that the particle has when it reaches infinity. What can you say about $\epsilon$ for a closed orbit, an orbit that just reaches infinity, and an open orbit? What implication does this have for $u_{min}$ in those cases?

 

[brpetrucci]

Because it's manifestly faster than escape velocity.

Oh yeah of course—the escape velocity is by definition the velocity the particle needs to reach infinity at 0 energy, so being faster than the escape velocity implies an open orbit. Fair enough!

Right, sorry. Got lost on what your question was going backwards and forwards between posts on my phone. Look at the extract you posted in #5 where he re-writes $2\epsilon-l^2u^2+2\kappa u$ in terms of its roots, which is how he defines $u_{min}$. Note that $\epsilon$ is the "energy at infinity" - the amount of energy (or actually energy per unit mass, to be precise) that the particle has when it reaches infinity. What can you say about $\epsilon$ for a closed orbit, an orbit that just reaches infinity, and an open orbit? What implication does this have for $u_{min}$ in those cases?

Yeah not your fault no worries, I confused everyone in this post haha ok that all makes sense. For an orbit that just reaches infinity that’s 0, that’s the escape velocity case. A closed orbit means we never reach infinity, so $\epsilon$ negative, right? That sounds weird to say since there’s no negative energy but I guess the point is that potential energy is higher in absolute value than kinetic energy. Then for an open orbit it should be positive, I see now why Zee himself says that.

We can find the roots then as

$u = \frac{-\kappa \pm \sqrt{\kappa^2+2l^2\epsilon}}{l^2}$

So if $\epsilon = 0$, we have just one negative root and one zero, which makes sense since $r$ goes to infinity. If it is negative I guess it depends on the specific value, but we know it just has either one or two roots depending on whether the orbit is circular or elliptical, ok. If it is positive, it has two roots.

OH I see—the sqrt is always greater in absolute value than $\kappa$, so one of the roots will be negative. Though that shows $u_{min} <0$, which means $r_{max}<0$ right? But I guess max and min have weird meanings anyway here, I see why he wrote it like that. The “min” there probably intends to talk about absolute value anyway, so the $u$ that’s negative is probably the max one. Anyway, in any case it makes sense. Thanks a lot! I feel bad for not seeing this for myself, but I guess I have to accept that I’m rusty haha I appreciate the help a lot!

 

[brpetrucci]

Sorry to bother you again Ibix, just continuing to get confused on this solution :/ not sure if it's me being rusty or something else. But here, Zee explains how to go from equation (19) with $u_{min}=0$ to the final angle. I understand what he says up until the equation he gets for $\sin^2{\xi_{min}}$, great. But then he describes the solution when $\kappa = 0$ and then approximates the final result by setting $\xi$ to the same value of the sine squared, and $\kappa^2=0$. I'm assuming the premises behind those are 1. $\kappa^2 << 2\epsilon l^2$, and 2. $\frac{\pi}{2} - \xi_{min} \sim \frac{1}{2} - \sin^2{\xi_{min}}$, which means we're assuming the deviation is really low. Is that the case? I'm just not sure why we can make those approximations, and it seems kind of loose. I guess it might be the only way to do it with Newtonian physics.