Possible Problem with Feynman Lecture derivation of gravitation field

[Annemarie]

Chapter 13 part 4, Gravitational field of large objects, (page 13-8 in the commemorative edition)

Just a foreword, I'm not trying to prove Feynman wrong and I'm expecting to be shown where my own judgement has let me down.

I don't understand why Feynman has used p (roe, density) for the length if the mass per unit area is constant. I can see how length wise r^2=p^2+a^2, it's just pythagerous, but how can a length unit be equivalent to p, density?

I don't understand.

Whats more interesting is that when I use my own method of finding the field, I find exactly the same answer. Odd or what?

 

Oh and if anyones wondering why I'm still studying the Feynman lectures even though exams were ages ago, well I've decided that before my third year starts I want to have studied as much as possible so I can concerntrate on my dissatation, but I still have no idea what to do it on...

[Tom Mattson]

I don't understand why Feynman has used p (roe' date=' density) for the length if the mass per unit area is constant. I can see how length wise r^2=p^2+a^2, it's just pythagerous, but how can a length unit be equivalent to p, density?

[/quote']

 

Are you sure it's a density in this situation? Perhaps he is using it as a distance.

 

Oh and if anyones wondering why I'm still studying the Feynman lectures even though exams were ages ago, well I've decided that before my third year starts I want to have studied as much as possible so I can concerntrate on my dissatation, but I still have no idea what to do it on...

 

Good luck!

[Annemarie]

Theres really no point in any replying unless they can see a copy of the lecture notes. He's definatly not just using it as legnth because he actually uses the length p to cancel with a real density p.

[rajama]

I think roe is just a length in the diagram on the left - he talks about the mass between p and p+dp being dm. The mass per unit area here is - well, I guess that's mu? It's in the integral result Cx = ... 2.pi.G.mu

 

I don't know why he used roe (maybe the co-ordinate system?) but that's probably why your version worked out too - just different symbols..?

[Annemarie]

I'm going to get a scanner, I think I have pretty much considered all the possibilities, so you need to see it.

I do have an additional problem. As I mentioned before I'm reading the Feynman lectures to expand my own knowledge and I've just foudn something that doesn't make so much sense again.

Chapter 15 part 1, Feynman is talking about relativity, and he says that the speed of light waves is unaffected by the velocity of the emitter, which I'm totally cool with, then he says "This is analogous to the case of sound,. the speed of sound waves being likewise independant of the speed of the source".

Erm, that doesn't make sense to me, I thought the Doopler shift was the key thing there to do with sound waves being emitted from a moving source, where it's wavelength and frequency change according to the speed of the source, and therefore the speed of the emitted waves is changed....?

[swansont]

Chapter 15 part 1' date=' Feynman is talking about relativity, and he says that the speed of light waves is unaffected by the velocity of the emitter, which I'm totally cool with, then he says "This is analogous to the case of sound,. the speed of sound waves being likewise independant of the speed of the source".

Erm, that doesn't make sense to me, I thought the Doopler shift was the key thing there to do with sound waves being emitted from a moving source, where it's wavelength and frequency change according to the speed of the source, and therefore the speed of the emitted waves is changed....?[/quote']

 

 

The speed of sound is dictated by the medium. The motion of the source gives you a Doppler shift in frequency and wavelength but the speed is unchanged.

[MetaFrizzics]

I'm going to get a scanner' date=' I think I have pretty much considered all the possibilities, so you need to see it.

I do have an additional problem. As I mentioned before I'm reading the Feynman lectures to expand my own knowledge and I've just foudn something that doesn't make so much sense again.

Chapter 15 part 1, Feynman is talking about relativity, and he says that the speed of light waves is unaffected by the velocity of the emitter, which I'm totally cool with, then he says "This is analogous to the case of sound,. the speed of sound waves being likewise independant of the speed of the source".

Erm, that doesn't make sense to me, I thought the Doopler shift was the key thing there to do with sound waves being emitted from a moving source, where it's wavelength and frequency change according to the speed of the source, and therefore the speed of the emitted waves is changed....?[/quote']

 

The simple answer here is that once the 'sound' (impact) is released upon the air molecules (say by a speaker cone striking the air), the speed of the propagation of the motion (think of a bucket-brigade) goes on independantly. The speaker and its velocity are not relevant to how fast the sound actually travels to someplace where it did not exist before.

 

The Doppler Effect in the case of Sound is something totally different: In this case you can think of a long series of repeating strikes. But if the speaker cabinet is moving (say mounted on an ambulance), each time the speaker strikes, it is hitting the air from a new location! So if approaching, the beating of the air arrives sooner and sooner for each 'pop'.

 

If you have a repeating beat that is fast enough, it is no longer percieved as separate beats, but as a musical note (say faster than 20 cycles/beats per second). The musical note from the approaching ambulance sounds like a higher pitched note, since it is getting compressed like groceries bunching up at the end of the conveyer-belt.

 

If the approaching ambulance speed is constant, the note goes up a specific and fixed amount in pitch. It is perceived as a higher frequency sound than the one actually being generated in the frame of reference of the moving ambulance. The sound doesn't actually travel through the air faster than before. The beginning and end of the note travel at a fixed speed through the air.

 

You might think from this that the speed of sound is constant for everyone and that only the pitch or frequency is relative. This is only partly true. The speed of the sound is fixed relative to the air it is travelling through (that is the medium at rest). But the ambulance/speaker is moving through the air.

 

This means that from the speaker's rest-frame (lets say a long train, with microphones at each end and the speaker in the middle) the sound takes longer to go forward than backward, because the air is moving backward. In this case, the beginning and ending of the sound arrive at a different time, depending upon which way the air moves. The pitch is again affected, but for a different reason now. The pitch at the front of the train seems lower because the microphone is receding from the air (and getting the pops at a slower rate).

 

The speed of the sound is affected by the relative movement of the air to the microphone/ear. The pitch is affected by the relative motion of the source regardless of the air motion.

 

If the microphone is on the ground (with air stationary) sound speed is constant, but pitch changes. The approaching sound is higher in pitch, and after the vehicle passes, it lowers in pitch. (pure Doppler effect).

 

mic on ground ------ pitch higher, then lower when passed, speed constant

mic on trainfront --- pitch lower, start and stop delayed. (speed lower)

mic on trainback --- pitch higher, start and stop arrive early. (speed up )

 

I hope this helps. Doppler effects strictly affect frequency (pitch), while

motion of the medium affects the speed of propagation (and pitch too).

 

This is why the Michelson-Morley experiment is important: They were hoping to measure the speed of the aether by finding both frequency AND speed effects for light, but they only found frequency effects! No motion, and no medium!

[MetaFrizzics]

I don't understand why Feynman has used p (roe, density) for the length if the mass per unit area is constant. I can see how length wise r^2=p^2+a^2, it's just pythagerous, but how can a length unit be equivalent to p, density?

Actually the answer here is simpler than it appears:

 

First let's talk about 'normalization', a habit that physicists do without thinking that confuses almost everyone. Here's an example.

 

Newton's gravity Force = Gm1m2/d^2

 

One way I might simplify this is to just leave out m2, and call it potential energy or something instead of force, because I want to generalize or talk about the contributions from everything else except the 'testmass', isolating those components.

 

For a series of experiments or discussions, I might also completely simplify the equation, by just setting both masses at '1' (one unit of mass in some choice of sizes). Now I can also set G (the Gravitational constant) to '1' also, by choosing the right unit of distance to work with. After all this clowning, I have now 'normalized' the equation for the task at hand, and gotten to the essential (simplified) law of gravity:

 

F = 1/d^2 ........( G x m1 x m2 = 1 x 1 x 1 = ...1 ! )

 

viola! It is now clearly the simple Inverse Square law. (Notice that masses don't usually change, nor do the units we have chosen, so the only variable is distance, in an appropriate choice of units.)

 

Now I can talk about a series of experiments, as long as everyone has figured out from the simple form I am using that I have automatically assumed the masses are all '1', and the units appropriately match.

 

Feynman has done something quite similar in your example. He has noticed that the mass is proportional to the density, and that it is constant, and he has assumed that the actual 'mass' for the discussion is just '1' unit in some appropriate scale, and now he can just directly substitute in 'rho', the density. This is possible, because normally (with the full blown version) he would have had to multiply the number of units X the density to get the real mass in an arbitrary object. With the mass = 1, there is no calculation required. Its just now 'rho', or whatever the variable is called that is holding the density value (in the proper units).

 

Feynman just skipped a step without thinking out loud and left thousands of beginners frustrated for the next 20 years.