# Thread: equivalence principle incompatible with Schwarzschild metric

1. ## equivalence principle incompatible with Schwarzschild metric

For the past couple of weeks, I had been attempting to apply the equivalence principle to determine the angle of deflection of a photon around a gravitating body. I tried several methods, but only attained the classical amount of deflection 2 m / b, where b is the point of closest approach and m = G M / c^2. Finally I applied the angle of deflection according to the Schwarzschild metric directly to find where the difference lies for what I was getting. The result was astounding. It appears that if we use the metric, the equivalence principle doesn't apply at all, which is certainly not what I was expecting. The two seem to be incompatible. Here I will apply the metric and demonstrate how it relates to the equivalence principle.

Here is a link to the derivation for the deflection of light using the metric, beginning about halfway through the link after the Newtonian derivation. The integration about 3/4 of the way down gives the angle of deflection to four orders as the photon travels from the point of closest approach b (r0) to a greater radius r. Normally this would be integrated to infinity, but for our purposes, we only want to integrate from b to an extremely small distance past b, call it db, whereas r = b + db.

Okay, so at the point of closest approach, according to a distant observer the photon will be travelling perfectly tangent to the body with a speed of c sqrt(1 - 2 m / b). After an infinitesimal time dt, the photon will have travelled to a radius r = b + db and its position will have changed by an angle of θ depending only upon b and r which can be found by applying the integration shown in the link. So if the original coordinates of the photon at b were x1 = 0 and y1 = b, then its coordinates at r become x2 = (sin θ) r and y2 = (cos θ) r. Since the tangent speed of the photon won't change much over an infinitesimal distance, we can approximate the time of travel of the photon to be just dt = x2 / (c sqrt(1 - 2 m / b)). We will also need the distance the photon has travelled in the y direction, which is just dy = b - y2.

Running the actual numbers in a computer program to four orders for the integration to find θ, then dt and dy, given b, db, and m, I found the relations to be

dt = (sqrt(2 b db) / c) / (1 - 2 m / b)^(5/4)

dy = (3 m db / b) / (1 - 2 m / b)^(3/2)

The coordinate acceleration of the photon in the y direction, then, is found to be

a_y = 2 dy / dt^2

= 2 [(3 m db / b) / (1 - 2 m / b)^(3/2)] / [((2 b db) / c^2) / (1 - 2 m / b)^(5/2)]

= 3 m c^2 (1 - 2 m / b) / b^2

For a photon starting with zero speed in the y direction and with small m / b, a local hovering observer will also measure this coordinate acceleration of the photon, differing only by some factor of 1 - 2 m / b. So now let's compare this to the equivalence principle. Let's say that an observer in an elevator begins freefalling radially from rest at b. The coordinate acceleration will be about G M / b^2 = m c^2 / b^2 to first order. The elevator is considered to be inertial, so that if the elevator observer were to emit a photon directly tangent at the moment he began to freefall, the photon should be measured by him to continue to travel in a straight line in the tangent direction, reaching the other side of the elevator at the same height above the floor as when it was emitted (ignoring the gravity gradient). But we have found that the photon will accelerate in the direction of freefall at about 3 times that rate. The elevator observer, then, according to the metric, will observe the photon to be accelerating toward the floor of the elevator with a coordinate acceleration of about 2 m c^2 / b^2. The equivalence principle does not appear to apply when it comes to the metric.

2. Here is a post that I made in another forum.

We can perform an experiment like this right here on Earth. Let's say a photon is emitted directly along the surface of the Earth at b = 6.371*10^6 m. First let's find the angle of curvature for the photon to travel from a radius of b to b + 1 micrometer. Integrating to four orders to find the angle, I get

θ = [-atan(B / sqrt(R^2 - B^2)) + pi/2]
+ [M sqrt(R^2/B^2-1) (1/(B+R) + 1/R) - 0]
+ [1.5 M^2 ((sqrt(R^2/B^2-1)/6) (3/R^2 - 2 (5 B + 4 R) / (B (B+R)^2)) - (5 / (2 B^2)) atan(B / sqrt(R^2-B^2)))
+ 1.5 M^2 (5 / (2 B^2)) (Pi/2)]
+ [(2.5 M^3 / (15 B^3)) (45 atan(B / sqrt(R^2-B^2)) + (B sqrt(R^2/B^2-1) / (R^3 (B+R)^3)) (5 B^5 + 15 B^4 R + 70 B^3 X^2 + 247 B^2 R^3 + 306 B R^4 + 122 R^5))
- (2.5 M^3 / (15 B^3)) 45 (Pi/2)]

Using UBasic (prog"lense170" for my own notes) and plugging in b = 6.371*10^6 and r = b + 1 / 10^6 with m = G M / c^2 = 4.43589170257*10^(-3), I get an angle of θ = 5.60287838354*10^(-7) radians. So if the original coordinates of the photon were x1 = 0, y1 = 6.371*10^6, they are now

x2 = (sin θ) r = 3.56959381816 m

y2 = (cos θ) r = 6370999.99999999999999791121093462 m

So let's say that we design an immensely strong metal tube, out in free space away from any gravitation, so that we can use a light beam while inertial to be sure that it is perfectly straight. Its dimensions will be 7.1391876363 m long with a radius of 2.0887890653751*10^(-15) m. Then we will bring it to the surface of the Earth and align it perfectly level with its center at b, lying perfectly along y = b. We will then use a laser to cut it in half carefully and remove one half, so that one end of the tube will now be centered on xA = 0 and yA = b while the other end is centered at xB = 3.5659381816 m and yB = b. From end A we will fire a laser perfectly tangent down the center of the tube. From earlier, we see that the beam will bend so that it strikes right at the edge of the tube at the other end, having fallen a distance in the y direction of dy = 2.0887890653751*10^(-15) m, the radius of the tube.

Now let's say we only fire a single photon down the tube. The time it will take to traverse the tube is approximately dt = dx / c = 1.19068833218*10^(-8) sec. Let's also say that at the same time that we fire the photon at end A, we drop a massive particle from rest at the center of end B. The particle will just fall due to Earth's gravity at a rate of g = G M / b^2 = 9.82216285185 m/sec^2 (for the values used). In the time dt, then, the particle will fall a distance dh = g dt^2 / 2 = 6.96263021792*10^(-16) m. This is only about 1/2.999999999999607595510104 the distance the photon falls in the same direction over the same time. So then, if we were to release the tube at the moment we fire the photon, allowing it to freefall, the tube will fall only about a third of the distance the photon will, while if it were to be considered inertial, the tube should fall at the same rate so that the photon would continue to travel along the center of the tube, but that is not what occurs according to the metric.
However, as also mentioned there, even if the tube is immensely strong so that it resists bending due to general stresses of gravity, it may be possible that the metric affects it in some way that it is still forced to bend such that the ends droop down 2/3 the distance the light travels, about 1.39252604358*10^(-15) m, so that the photon and the center of end B of the tube will coincide at the same place upon freefalling, which would preserve the equivalence principle if that is the case.

3. I made a computer program a while back that was sensitive enough to measure the precession of Mercury and other planets. From the precession program, I found the radial acceleration of a particle to be something between a = (G M / r^2) (1 + 6 m / r) and a = (G M / r^2) (1 + 3 (v/c)^2), both of which give the accurate precession. I didn't know until now, though, what the proportion of (v/c)^2 to m / r should be. It might also be angle dependent, but we know now that for a photon travelling perfectly tangent to the gravitating body at least, its radial acceleration is rather precisely 3 (G M / r^2) / (1 - 2 m / r) according to a distant observer. The precession program is also observed from the perspective of a distant observer. In order to line up the equations to match both precession and gravitational lensing, then, it would have to be a = (G M / r^2) (1 + 2 (v/c)^2) / (1 - 2 m / r) to second order for a particle travelling in the tangent direction. Of course, it must also reduce to Newton. For freefall from rest with initial tangent speed v = 0, that gives a = (G M / r^2) / (1 - 2 m / r).

So now let's try those 3 accelerations, a_c = 3 (G M / r^2) / (1 - 2 m / r) for a photon, a_p = (G M / r^2) (1 + 2 (v/c)^2) / (1 - 2 m / r) for a particle, and a_f = (G M / r^2) / (1 - 2 m / r) for the tube freefalling from rest to see what we get in terms of the bend of the tube. If the length of the tube is d, then the photon travels the length of the tube in a time of about t_c = d / c. During that time, it accelerates radially while the tube in freefall accelerates radially also at 1/3 the rate, and the equivalence principle says the two must coincide, so the distance of the radial bend at the end of the tube from the straight length must be

d_bend = a_c t_c^2 / 2 - a_f t_c^2 / 2

= [3 (G M / r^2) / (1 - 2 m / r) - (G M / r^2) / (1 - 2 m / r)] (d / c)^2 / 2

= (G M / r^2) (d / c)^2 / (1 - 2 m / r)

Now let's find out what the bend would have to be in order for a particle travelling at v to coincide with the end. The time for the particle to traverse the tube will be about t_p = d / v, so we have

d_bend = a_p t_p^2 / 2 - a_f t_p^2 / 2

= [(G M / r^2) (1 + 2 (v/c)^2) / (1 - 2 m / r) - (G M / r^2) / (1 - 2 m / r)] (d / v)^2 / 2

= [(G M / r^2) (2 (v/c)^2) / (1 - 2 m / r)] (d / v)^2 / 2

= (G M / r^2) (d / c)^2 / (1 - 2 m / r)

which is exactly the same as what we found for the photon. So the equivalence principle definitely can work out if the tube is bent in this way, and so probably does, and we just found a formula that describes the rigidity of objects in a gravitational field.

4. Oh whoops, the acceleration for the photon that was found by taking the integration to four orders was a = 3 (G M / r^2) (1 - 2 m / r), not 3 (G M / r^2) / (1 - 2 m / r), so we may still need more orders for the precession formula or something, and the time of travel was also an approximation, but the equations line up well enough to verify that the equivalence principle can apply with a bend at the end of the tube of about (G M / r^2) (d / c)^2 to first order.

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The calculations are too chaotic, I doubt that anyone has been able to analyze them;
the purpose and meaning of the whole argument is also vague.

Perhaps you compute the slope of the secant, instead of the tangent to the path.

h(t) = gt^2/2;

secant: (h(t) - h(0)) / t = gt^2/2t = gt / 2;
tangent: dh / dt = gt, two times larger (for any: t > 0).

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Given that we can show the equivalence principle to hold in the schwarzschild metric you must be doing something wrong somewhere.
Some possibilities are: the graviational field is central but you fail to account for this over an extended falling elevator (the left and right side of the elevator would be on different geodesics, they would converge rather than stay the same distance). There is a tidal gradient that works on the photon's path and which also works on the extended elevator, maybe you fail to account for that (or think you can ignore it while you could not). Or perhaps something else entirely.

But mostly this:
Originally Posted by Hetman
The calculations are too chaotic, I doubt that anyone has been able to analyze them;
the purpose and meaning of the whole argument is also vague.

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The equivalence principle is included in the Einsein field equations, the Schwarzschild metric is a solution to them and thus the Equivalence principle holds in the Schwarzschild metric.

I suspect that the problem is that your posts make no distinction between coordinates. In the Schwarzschild metric, r is not the radial distance from the object.

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@grav

a. is not a radial distance, yet you are using it as such (already pointed out by Reality Check).
b. You are trying to use the EEP in an incorrect way (this is a more serious mistake). The equivalence is between constant acceleration and a uniform gravitational field, i.e. the field created by an infinite plane. The Schwarzschild solution is for a spherical symmetric field. This is the other reason why you are getting incorrect results. Basically, your incorrect application of SR+EEP is missing the curvature effects. Hence, you get the incorrect results (basically you are getting the Newtonian answer).
c. As pointed out by caveman and Reality Check, the EEP holds for spherically symmetric solutions but you cannot use a constant acceleration, you must take into account the fact that the field is central, so the acceleration that you want to use is radial and not constant as well. You are using constant acceleration.

As an aside, it might be a good idea for you to learn LaTex, your lengthy calculations are unreadable (not that it matters in this case).
Last edited by Grassman; 2012-Jul-24 at 01:46 PM.

9. Originally Posted by Grassman
As an aside, it might be a good idea for you to learn LaTex, your lengthy calculations are unreadable (not that it matters in this case).
Seconded. There is a handy interactive editor: http://www.codecogs.com/latex/eqneditor.php and there are some good online tutorials.

10. Thanks guys. I suppose I should learn latex, although the posts take long enough to write already. As for the topic, I already worked that out. I knew I should have put a question mark at the end of the thread title lol. Someone had pointed out in another forum that the metric causes the otherwise straight tube to bend in the gravitational field according to the distant observer, with the tube placed along the tangent direction and the bend occurring radially at each end, so in post #3 I found what that bend would have to be in order to accomodate the equivalence principle for a light pulse to travel its length while the tube falls, then again for a massive particle travelling non-relativistically. They both line up the same with the freefalling tube, so the equivalence principle indeed can work with that slight bend, although the bend is extremely small which is why I didn't account for it with very small m / r, but apparently it is just enough to make up the difference.

As for r not being the radius, I'm not sure what you mean by that. I have already worked through the mathematics for infinitesimal changes in angle using the coordinate speed of light in order to find the change in radial distance and then gained the precise metric that way. One can also gain it directly from considering the radial and tangent speeds vr and vt as measured locally and vr' and vt' according to a distant observer with

vt^2 + vr^2 = c^2 locally

(vt' / sqrt(1 - 2 m / r))^2 + (vr' / (1 - 2 m / r))^2 = c^2

((sin dθ) r / dt)^2 / (1 - 2 m / r) + (dr / dt)^2 / (1 - 2 m / r)^2 = c^2

c^2 dt^2 (1 - 2 m / r) - dr^2 / (1 - 2 m / r) - dθ^2 r^2 = 0

Here r is just the radial distance according to the distant observer and dr is the change in radial distance. What I found for the equivalence principle also uses r as radial distance as well, and it works out now. Also, the program I used to run the integration uses r as a radial distance in order to plot x = (sin θ) r and y = (cos θ) r, for instance, and works out correctly. As well as that, someone in a third forum mentioned that the acceleration of the photon should be twice the Newtonian value, not 3 times, when using isotropic coordinates. So after running the original program, I had it go back and while keeping the same time and angle, change the radial distances of b and r to b' = b / 2 - m / 2 + sqrt((b / 4) (b - 2 m)) and r' = r / 2 - m / 2 + sqrt((r / 4) (r - 2 m)) using Eddington's isotropic coordinates, then finding dy' with dy' = b' - (cos θ) r'. I then gained twice the Newtonian coordinate acceleration instead of three, so that worked out using r and b as radial distances also. I've never heard of r being anything other than a radial distance and I don't see how it could be anything other than that.

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You have made quite a few mistakes, you should pay attention to what was pointed out to you.

12. Originally Posted by caveman1917
Given that we can show the equivalence principle to hold in the schwarzschild metric you must be doing something wrong somewhere.
Some possibilities are: the graviational field is central but you fail to account for this over an extended falling elevator (the left and right side of the elevator would be on different geodesics, they would converge rather than stay the same distance). There is a tidal gradient that works on the photon's path and which also works on the extended elevator, maybe you fail to account for that (or think you can ignore it while you could not). Or perhaps something else entirely.

But mostly this:
Right, that's what I did. I ignored the tidal gradient, which becomes important across the length of the tube or the width of the elevator, even with small m / r and a small length of the tube, and makes up the difference. From the perspective of the distant observer, the ends of the tube are bent slightly downward.

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And why do you assume that the deflection must be twice
greater at any point of the trajectory of light?

It would be very surprising if it could act in such a way.

The full deflection is two times larger, and it means that measured from the center to infinity (b / r << 1).

Closer to the center (mass) should be greater than 2x,
maybe just 3x, and then - away from the center, falling below 2x, and only the average is 2 x.

According to the GR correction to the force of gravity also contains 3, something like this: 3GM / c^2

Here are the equations of the two trajectories:
http://www.theory.caltech.edu/people...ia/lclens.html
Last edited by Hetman; 2012-Jul-24 at 05:43 PM.

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Originally Posted by grav
Thanks guys. I suppose I should learn latex, although the posts take long enough to write already.
The thing is that you're already almost writing in latex format. Just put tex tags around your equations and you're almost there already.

Here are the four equations you wrote in the post i'm quoting with me simply putting tex tags around them:

Escaping special characters that immediately becomes:

It's really not hard, just write in the way you're already doing, and for special symbols check this list
http://www.artofproblemsolving.com/W.../LaTeX:Symbols

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Originally Posted by caveman1917
The thing is that you're already almost writing in latex format. Just put tex tags around your equations and you're almost there already.

Here are the four equations you wrote in the post i'm quoting with me simply putting tex tags around them:

Escaping special characters that immediately becomes:

It's really not hard, just write in the way you're already doing, and for special symbols check this list
http://www.artofproblemsolving.com/W.../LaTeX:Symbols
I would simply add that \frac{numerator}{denominator} gives

Those programs sound very familiar.

16. Cool. Let me try it with both of your suggestions and one of my own from the link.

Awesome, and not difficult at all. Thanks. (← also latex lol)

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Originally Posted by grav
Cool. Let me try it with both of your suggestions and one of my own from the link.

The above is not even dimensionally correct, let alone meaningful. Of course, it doesn't matter since you have more serious mistakes that stem from your mis-application of the EEP, as explained earlier.

18. Originally Posted by Grassman
The above is not even dimensionally correct, let alone meaningful. Of course, it doesn't matter since you have more serious mistakes that stem from your mis-application of the EEP, as explained earlier.
vt and vr are the speeds in the tangent and radial directions. Perhaps I should have written them as v_t and v_r. Anyway, the issue was resolved by post #3. EEP works out fine now to first order when considering a slight bend in the tube across the gravity gradient according to the distant observer. Currently I am working on trying to find a better way to verify that through the metric and gain more precise measurements.

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Originally Posted by grav
vt and vr are the speeds in the tangent and radial directions. Perhaps I should have written them as v_t and v_r.
You mean:

?

This is false since the space is not Euclidian (flat) and what you wrote down is the relationship in Euclidian space.

Anyway, the issue was resolved by post #3. EEP works out fine now to first order when considering a slight bend in the tube across the gravity gradient according to the distant observer.
Meaning that you were miss-applying the EEP, as I explained to you.

20. Originally Posted by Grassman
You mean:

?

This is false since the space is not Euclidian (flat) and what you wrote down is the relationship in Euclidian space.

Meaning that you were miss-applying the EEP, as I explained to you.
SR applies locally, whereas the local space is Euclidean, and the locally measured speed of light is always c.

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Originally Posted by grav
SR applies locally, whereas the local space is Euclidean, and the locally measured speed of light is always c.
True, but i'm a bit lost as to what the point still is? I think the issue got resolved, you were ignoring the centrality of the field and the radial tidal gradient as well as missing the correct coordinate use for . It seems you realized this mistake, so what exactly are we still debating?

22. Originally Posted by caveman1917
True, but i'm a bit lost as to what the point still is? I think the issue got resolved, you were ignoring the centrality of the field and the radial tidal gradient as well as missing the correct coordinate use for . It seems you realized this mistake, so what exactly are we still debating?
I'm only debating grassman (adsar?) at this point lol. As for the topic, this issue is resolved as far as what amount of bend would be required to accommodate the equivalence principle, yes, but I only found a first order approximation for what it would have to be without actually applying the metric. I want to see it more precisely from the metric, so that's what I'm working on now. It's not really ATM anymore at this point I guess, for now, unless I find results from the metric that don't agree with the first order approximation, although I suspect they will.

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Originally Posted by grav
As for r not being the radius, I'm not sure what you mean by that.
What we mean by that is that r in the Schwarzschild metric is not the radial distance from the object.
One way to think of it is as a label for nested spheres:
However, note well: in general, the Schwarzschild radial coordinate does not accurately represent radial distances, i.e. distances taken along the spacelike geodesic congruence which arise as the integral curves of . Rather, to find a suitable notion of 'spatial distance' between two of our nested spheres, we should integrate g(r)dr along some coordinate ray from the origin:

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Originally Posted by grav
SR applies locally, whereas the local space is Euclidean, and the locally measured speed of light is always c.
There is no such thing as in SR, the light speed doesn't have "components". Where do you get this stuff from?

25. Originally Posted by Reality Check
What we mean by that is that r in the Schwarzschild metric is not the radial distance from the object.
One way to think of it is as a label for nested spheres:
Oh, right. Although r is still the radial distance according to a distant observer, inferred over Euclidean space for that observer, what you are referring to is what I call the "ruler" distance, the distance that would be measured by physically placing identical rulers end to end along a path. Doing that radially, according to a distant observer, the rulers would become contracted more and more as they approach the event horizon, with an infinite number of rulers fitting there, although surprisingly measuring a finite distance if measured from there because only the very end of the last ruler is contracted to zero. Finding the radial distance by just taking r' = r / sqrt(1 - r_s / r) is what I call the "inferred" distance. A local observer, however, can only physically measure locally over very short distances, so we really only have dr' = dr / sqrt(1 - r_s / r) as measured locally. To find the ruler distance, one would have to integrate for the lengths of each ruler placed along a path, whatever that path may be. Radially from r to s, that becomes

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Originally Posted by Grassman
There is no such thing as in SR, the light speed doesn't have "components". Where do you get this stuff from?
What exactly do you mean by that?

Taking minkowski for null paths

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Originally Posted by caveman1917
What exactly do you mean by that?

Taking minkowski for null paths
Dividing the null geodesic by is meaningless, and represent absolutely nothing, though the sum of their squares adds to . You won't find the above exercise in any textbook since it has no physical meaning. As a challenge, try figuring out the value for . You can't.

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Originally Posted by Grassman
Dividing the null geodesic by is meaningless, and represent absolutely nothing, though the sum of their squares adds to .
Of course they do, it's simply a statement of the pythagorean theorem in euclidean space.

You won't find the above exercise in any textbook since it has no physical meaning.
Given that the use of that equation as an axiom is sufficient to derive SR, i'd say that having no physical meaning is open to interpretation.

As a challenge, try figuring out the value for . You can't.
I don't think anybody can figure out the value for one degree of freedom in any equation relating two degrees of freedom without more information.

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Originally Posted by caveman1917

Given that the use of that equation as an axiom is sufficient to derive SR, i'd say that having no physical meaning is open to interpretation.
Physics doesn't work this way, you don't get to manipulate unphysical entities that you made up as a result of a meaningless mathematical manipulation. In your case, dividing the geodesic by is meaningless. To further prove that, do the same operation for the general geodesic: . What did you obtain?

I don't think anybody can figure out the value for one degree of freedom in any equation relating two degrees of freedom without more information.

Precisely, which further demonstrate that the operation you performed is meaningless in the realm of physics.

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Originally Posted by grav
Oh, right. Although r is still the radial distance according to a distant observer, nferred over Euclidean space for that observer,
You are wrong on both accounts. First off, the space is not Euclidian, as explained several times already. Second off, "r is NOT a distance". Here is the correct definition of r:

-If you have a circle of circumference L, then two things happen: and where is the true distance from the center (see C. Moller, "Theory of Relativity") for example (any other textbook would also do).

-If you have a sphere of area A, then two things happen: and where is the true distance from the center

-If you have ball of volume V, then.....fill in the blanks

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