shape of a relativistic wheel

[Tach]

Why are you asking me to prove your case?

I am not asking you since I know you will never put down the math. I am asking przyk.

 

[Tach]

Consider the worldlines of the rim ends of opposite spokes, horizontal at t'=0:

$$x_1' = -R.cos(\omega t') \\
y_1'-R = R.sin(\omega t')$$

$$x2' = R.cos(\omega t') \\
y_2'-R = -R.sin(\omega t')$$

This still doesn't seem right. In the frame of the axle you should have:

$$x'=R cos (\omega t' + \phi)$$
$$y'=R sin (\omega t' + \phi)$$

therefore:

$$\frac{y'}{x'}=tan (\omega t' + \phi)$$


The transformation between the axle frame and the ground frame is:

$$x'=\gamma (x-Vt)$$
$$y'=y-R$$
$$t'=\gamma(t-Vx/c^2)$$

so:

$$y-R=\gamma (x-Vt) tan(\omega \gamma (t-Vx/c^2)+\phi)$$

This is the general equation (I get the cotangent because I reverse the parametrization for x and y). For the particular case $$t=0$$

$$y-R=\gamma x tan(\phi-\omega \gamma Vx/c^2)$$

If you use the parametrization:

$$y'=R cos (\omega t' + \phi)$$
$$x'=R sin (\omega t' + \phi)$$

you should get the other form:

$$y-R=\gamma x ctan(\phi-\omega \gamma Vx/c^2)$$

 

[przyk]

But what both you and przyk fail to acknowledge is that the notion of what is "upper half" is frame dependent.

This is completely wrong. I did acknowledge that (I'm not sure I really agree with the terminology, but let's go with it). In fact it should have been clear I was well aware of this way back in my [POST=2848590]first post[/POST] in this thread, when I asserted that most of the mass would be in the top half in the ground frame, and from my subsequent elaboration at the end of [POST=2848712]post #59[/POST].

So either you're lying, or [POST=2848890]continuing[/POST] to prove you don't even bother to read what people say. (Have you noticed I routinely link to past posts, by the way? I can do this because the record in this thread is actually on my side.)

It is you who has failed to explain why the notion of "upper half" being "frame dependent" should be an argument against it. The kinetic energy is also frame dependent. Do you think that means it's "invalid" to ask what the kinetic energy is in a given frame?

You made no attempt, przyk made a sketch that I have shown to contain an invalid parametrization of the length of the halves as a function of the arc length.

First of all, stones and glass houses: you have never shown any calculations. Your single lazy [POST=2848455]attempt[/POST] at a derivation contained not one but two conceptual errors.

Second, you have not "shown" that my parameterisation was invalid, you merely said it was in a pathetic attempt to redefine the problem. Since RJBeery has now clarified that everyone except you understood him correctly, your attempted evasion is revealed for the strawman it always was.

But you know what, let's forget that for a second, because your strawman backfires spectacularly. Let's say we do define the top half "invariantly", meaning that the collection of wheel elements constituting the "top half" at some instant in the ground frame is the same as the collection of wheel elements in the top half at some "corresponding" instant in the axle frame. Then the approach you tried in [POST=2848455]post #51[/POST] does become valid (with the correction $$p_{\mathrm{upper}} = - p_{\mathrm{lower}}$$), because it applies to an invariantly defined collection of wheel elements. So setting $$E \equiv E_{\mathrm{upper}} = E_{\mathrm{lower}}$$ and $$p \equiv p_{\mathrm{upper}} = - p_{\mathrm{lower}}$$, you find

$$
\begin{align}
E'_{\mathrm{upper}} &= \gamma (E + pv) \,, \\
E'_{\mathrm{lower}} &= \gamma (E - pv) \,.
\end{align}
$$​

which are clearly different unless $$p = 0$$ (i.e. a non-rotating wheel).

Whoops.

 

[RJBeery]

You know, if you don't even understand the subject of the debate after so many posts...

That's an odd comment since I'm the one that posed the question under discussion.

First you claimed that $$P_{upper}=P_{lower}$$ so ANY change in the measuring frame would leave this equality unchanged. When you realized that was wrong you apparently switched the definition of the midline to be exactly the line which divides the energy. Forget that. Deal with the midline which is created by the infinitely long fence of height R behind which the wheel is rotating. Would the gravitational field become unbalanced "above" and "below" the top of the fence?

 

[przyk]

I said "gravitational effects" as in "differences" between the "upper" and "lower" half. Do you think that there are any differences for an observer located in the center of the wheel?

No.

If there aren't any, why would such differences appear for a moving observer?

You tell me, since I never said such differences would appear.

This is what your approach claims, so I am asking you to prove it.

No, it is you who is claiming that the fact the upper half has more energy would imply "gravitational effects". I do not take your word for it that "the effects increase with the relative speed between the observer and the axle". So, derivation or reference please.

By the way, consider one of your own questions. Surely you would agree with me that the energy of a dense mass is frame dependent and increases with velocity (so it increases if I switch into a frame in which the mass is moving, for instance), right? Does the mass turn into a black hole if it is moving fast enough? You've already indicated "no". If you think this, why should you expect that the imbalance in energy in the top and bottom halves should lead to these supposedly paradoxical "gravitational effects" of yours?

 

[RJBeery]

Whoa Przyk, to be fair to Tach, I don't think anyone actually thinks the paradoxical gravitational effects would occur based upon who is making the observation. My question is simply, how does the framework avoid them from occurring? Did you see my question's reformulation 2 posts ago?

(That being said, if I had my way Tach would be completely barred from this discussion. Arrogant condescension, cherry picking arguments, nit-picking irrelevancies, rewriting history and being plain wrong is a bad combination)

 

[Tach]

Second, you have not "shown" that my parameterisation was invalid, you merely said it was in a pathetic attempt to redefine the problem.

Sure I showed you that the parametrisation is invalid, you blindly assume that each half maps in a L length interval (one into [0,L] and the other one into [L,2L]). I pointed out repeatedly to you that you have no proof of that. Once you start throwing invectives around, you know that you lost the argument.

But you know what, let's forget that for a second, because your strawman backfires spectacularly. Let's say we do define the top half "invariantly", meaning that the collection of wheel elements constituting the "top half" at some instant in the ground frame is the same as the collection of wheel elements in the top half at some "corresponding" instant in the axle frame.

But this is precisely what you got backwards and you continue to get backwards. What I have been telling you is that the separation between the two "halves" is dynamic. Its equation is the equation of the spoke corresponding to $$\phi_i=0$$. You can see that its orientation in the moving frame changes as a function of the relative speed $$V$$ between the axle frame and the ground frame. As such, there is no such thing as an "invariant" separation between halves. I understood your contention to my energy-momentum based explanation and I abandoned the approach. So, the only valid approach left is to integrate $$\gamma=\frac{1}{\sqrt {1-(v/c)^2}} $$ over the ever varying "halves" of the ellipse. In another post I showed the functions $$v_x=v_x(V,\phi,t)$$ and $$v_y=v_y(V,\phi,t)$$. If you do the calculations, the integrand has the very nice expression $$\gamma=\frac{1+(V/c)^2 cos (\omega t+\phi)}{1-(V/c)^2}$$. Now, all that is left is finding out the integration limits. And, no, they aren't [0,L] and [L,2L] despite what you might have come up with earlier. Your attempt at parametrization with respect to arc length is not the correct one, the natural parametrization is with respect to the angle $$\phi$$.

Then the approach you tried in [POST=2848455]post #51[/POST] does become valid (with the correction $$p_{\mathrm{upper}} = - p_{\mathrm{lower}}$$), because it applies to an invariantly defined collection of wheel elements. So setting $$E \equiv E_{\mathrm{upper}} = E_{\mathrm{lower}}$$ and $$p \equiv p_{\mathrm{upper}} = - p_{\mathrm{lower}}$$, you find

$$
\begin{align}
E'_{\mathrm{upper}} &= \gamma (E + pv) \,, \\
E'_{\mathrm{lower}} &= \gamma (E - pv) \,.
\end{align}
$$​

Err, no, $$(E,p)'_{upper}$$ isn't the transform of $$(E,p)_{upper}$$. See post 60.

which are clearly different unless $$p = 0$$ (i.e. a non-rotating wheel).

Whoops.

Whoops indeed, you produced a strawman.

 

[DRZion]

Yeah, nasty thread.
Just consider the concept - a tank, proper length L driving along the road at speed v.

The tank's track runs forward and back along the length of the tank, around infinitesimal driving wheels at each end.

The bottom half of the track is in contact with the road.

In the tank frame, the top and bottom halves of the track are both moving at speed v.

The track's total proper length is $$2\gamma L$$.

In the road frame, the bottom half of the track is at rest, and has proper length $$L/\gamma$$.


Since track density is proportional to proper length, it is clear that in the road frame less than half the track's mass is in the bottom half.

Yes, this track around a tank is the same concept! But it is a lot better because the effects of terrell penrose rotation won't affect it so neatly, or that is my first impression.

...because there is no PROOF to that. Just some armwaving and no math. Neither of you calculated the mass of the "upper" half. We can't even agree what "upper" half is. Put up a calculation and we'll talk.

przyk did no calculation, he showed a generic formula that proves nothing.

That is false. For the third time, you need to use the transformed (aberrated) line, i.e. the inclined line that is the transformation of the horizontal line passing through the axle.

The proof is there, but it is not a mathematical proof! Me, przyk, RJBeery, and pete all see it while you for the longest time are arguing against it. Are you really not convinced by simple word proofs based on common sense?

The aberrated mid-line is a red herring; the main focus is the way the wheel appears to an outside observer, and what gravitational effects it will have on an outside observer.

The aberration of the distribution of the rim is inherent in the thread from the very first post, so you 'pointing it out' is kind of redundant.

As late as post 48, you were still questioning whether the transformed midline would be S-shaped or U-shaped.

So, while we are talking about aberration, and were from the very beginning, it is not 100% that the wheel will look distorted when observed from a side view (where the wheel is in the plane of the monitor):

The wheel will not appear like an eclipse. There is a funny combination of light delay effects which h means the wheel will remain looking like a circle. Its named after Penrose and someone else. Length contraction happens but visually a second opposite effect conspires to cancel it in the case of circles and spheres, not anything else. Its to do with how the boost is like a complex rotation in time. Ill Google when I am not using a phone to post.

In addition to this, the movies that I have shown also support that a relativistic wheel will not look distorted from a side view, because of some other relativistic spookiness. Granted, I put a lot of faith in these videos. From the videos it seems that a wheel spinning at relativistic speeds will look differently distorted depending on what angle it is viewed from.

This does not defeat the purpose of the argument, which is to show that the center of mass of the wheel/tread will shift without any apparent force.

However, since the hypothetical involved here, goes far beyond any pratical real situation, how is it possible to know with any certainty what impact frame dragging might play? The hypothetical involves conditions that cannot exist, except within a hypothetical, science fiction or one's imagination. At least within the context of current knowledge and technologies.

That is not true, it can exist, and it can even exist on earth.

While a wheel spinning at relativistic speeds cannot exist, because of the relativistic forces, it can exist in another case.
Take the tank-tread. The tank tread is accelerated to 30 m/s. Then, the entire tank is accelerated to c - (1 m/s). The thread on the tank will not snap because that would mean different rules in the frame of the tank. But, to us it will still appear as if most of the mass was either in the lower or upper halves.

On earth, it can exist in a particle accelerator. Hello? What are atoms? They are nuclei which are orbited by electrons in circular paths.. right?

Whoa Przyk, to be fair to Tach, I don't think anyone actually thinks the paradoxical gravitational effects would occur based upon who is making the observation. My question is simply, how does the framework avoid them from occurring? Did you see my question's reformulation 2 posts ago?

(That being said, if I had my way Tach would be completely barred from this discussion. Arrogant condescension, cherry picking arguments, nit-picking irrelevancies, rewriting history and being plain wrong is a bad combination)

While I agree with you here, he is the only one who still adheres to the main issue in the thread:

Yes, looks right now.
Except that a problem surfaces: if you put up a "fence" of height R , more than half the spokes will be visible above the "fence". This doesn't seem right. This is where this thread started from. So, something is terribly wrong with this "picture".

If this is true, than we can produce scenarios in which reality has two courses which it can take, and no way to determine which course will be taken.

 

[Tach]

Yes, this track around a tank is the same concept! But it is a lot better because the effects of terrell penrose rotation won't affect it so neatly, or that is my first impression.

This is of course false, the Terrell effect is just as present, you can't "wish it away".

The proof is there, but it is not a mathematical proof!

In other words, there is no proof.

Me, przyk, RJBeery, and pete all see it while you for the longest time are arguing against it. Are you really not convinced by simple word proofs based on common sense?

Not if they aren't backed up by solid math. There isn't any backing up this claim.

The aberrated mid-line is a red herring; the main focus is the way the wheel appears to an outside observer, and what gravitational effects it will have on an outside observer.

So, what gravitational effects will it have on a moving observer?

So, while we are talking about aberration, and were from the very beginning, it is not 100% that the wheel will look distorted when observed from a side view (where the wheel is in the plane of the monitor):

Actually , the wheel looks distorted, I even gave you the precise equation, it is a family of ellipses moving at speed $$V$$. What is still in debate, it the form of distortion of the spokes, my equations clearly differe from Pete's. I think that I pointed out the source of error in Pete's equations.

In addition to this, the movies that I have shown also support that a relativistic wheel will not look distorted from a side view, because of some other relativistic spookiness.

Actually the movies show the distortion, all of them.

Granted, I put a lot of faith in these videos. From the videos it seems that a wheel spinning at relativistic speeds will look differently distorted depending on what angle it is viewed from.

You are contradicting what you just said.

This does not defeat the purpose of the argument, which is to show that the center of mass of the wheel/tread will shift without any apparent force.

It will? What proof do you have for this claim?

Hello? What are atoms? They are nuclei which are orbited by electrons in circular paths.. right?

Maybe 60 years ago this view was valid, not today.

 

[OnlyMe]

That is not true, it can exist, and it can even exist on earth.

While a wheel spinning at relativistic speeds cannot exist, because of the relativistic forces, it can exist in another case.
Take the tank-tread. The tank tread is accelerated to 30 m/s. Then, the entire tank is accelerated to c - (1 m/s). The thread on the tank will not snap because that would mean different rules in the frame of the tank. But, to us it will still appear as if most of the mass was either in the lower or upper halves.

On earth, it can exist in a particle accelerator. Hello? What are atoms? They are nuclei which are orbited by electrons in circular paths.. right?

The tank is not a practical example either, as excellerating the entire tank to c - (1 m/s) is itself problematic..., beyond our current technology.

Particle accellerators should be a good alternative, relativistic velocities are attainable, in that case. But I don' see how it fits the current hypothetical.

Particle accelerators accellerate particles and ions, bare nuclei.

Has there been any test involving whole atoms at relativistic velocities? Something like the LHC would likely strip the electrons away. Would seem to have had to happen in a linear accelerator. Even then I would expect the atom to become unstable, torn apart by the EM fields of the accelerator.

 

[James R]

Over 24 hours after I ask Tach to post the derivation he said he'd already done yesterday, and what do I see? Nothing from Tach.

We all know what that means. It means Tach can't do it and never did it in the first place.

 

[przyk]

Sure I showed you that the parametrisation is invalid

No, you called it "invalid", asserting that it was "not invariant", which is not an argument against it. This is pathetic revisionism. Now that you're cornered, suddently you're trying to tell everyone what the problem really was, even after RJ, the person who brought it up, corrected you about that.

you blindly assume that each half maps in a L length interval (one into [0,L] and the other one into [L,2L]).

No, it follows from the way everyone, except you, is defining the top and bottom halves of the wheel. The silhouette of the wheel in the ground frame is a length contracted ellipse with mirror symmetry about the horizontal axis, so each half gets half the perimeter. It's really not complicated.

But this is precisely what you got backwards and you continue to get backwards.

When it's you who doesn't understand the problem everyone else is discussing, you are hardly in a position to tell anyone else what they're doing is backwards.

What I have been telling you is that the separation between the two "halves" is dynamic. Its equation is the equation of the spoke corresponding to $$\phi_i=0$$.

Are you saying you're just defining the "top" half as the collection of wheel elements that happened to be at the top of the wheel initially in the axle frame? Does the "top half" of the wheel periodically touch the ground? If so, can you explain how in the world you got that from [POST=2847579]RJ's post that started all this[/POST]?

 

[James R]

I note that this is the sum total of Tach's argument on the matter so far:

Except that a problem surfaces: if you put up a "fence" of height R , more than half the spokes will be visible above the "fence". This doesn't seem right. This is where this thread started from. So, something is terribly wrong with this "picture".

It "doesn't seem right" to Tach's intuition. Therefore "something is terribly wrong".

Tach, being god-like, apparently isn't required to back up claims with evidence or calculations. Sure, occasionally he will claim to have done a calculation. In such cases he almost invariably leaves this is an "exercise" for others to "repeat". In fact, it usually turns out that he can't do and hasn't done the calculation himself, as is apparently the case here.

 

[Pete]

I've missed a fair bit of conversation, so some of this might no longer be relevant...

Here, remember?

Thanks for the link. I'd forgotten about that thread. Classic Tach.
You should read it again, and the link you used in your first post...

Then, you have to deal with the very disturbing issue that , if you covered half the wheel with a fence of height R, you would be seeing 6 spokes (not 4) above the fence. I pointed out this to you twice.

...especially where it describes this image (visual appearance of the rolling wheel in the ground frame):

(movie)
...and this image (diagram of the rolling wheel at an instant in a) the axle frame, and b) the ground frame):

You didn't seem to think it absurd then.

But what both you and przyk fail to acknowledge is that the notion of what is "upper half" is frame dependent.

No, Tach, we're quite clear on what is frame dependent.

The shape of the line dividing the upper and lower half-masses of the wheel is frame dependent. That's never been in question - it's inherent in the opening post.

The set of wheel elements above the fence at any instant is also frame dependent... and that's what we've been talking about ever since the OP.

You mean that you still maintain that you can get gravitational effects by moving by the wheel? And your answer is just "GR spookiness"?

No, I've never said that you can get gravitational effects by moving the wheel.
I've explicitly said that I have no idea how to to the required mathematics.
"GR spookiness" = "I don't know how to calculate the gravitational effects, and I suspect that you don't either"

Well , for one, you all claim that the "upper half" of the wheel has more energy (kinetic, total) than the "lower half".

Correct, bearing in mind that we're talking about the geometric upper half in the moving frame.

So, since total energy does gravitate, it follows that any moving frame could detect this disparity while the axle frame will obviously not detect anything.

Do you have any calculations to support that assertion?
RJBeery did suggest it as a possibility, but not with any certainty.
I personally suspect that this does not follow at all... certainly not in any paradoxical fashion.
That particular question (what frame dependent gravitational effects follow from the frame dependent mass/energy distribution) is what this thread should be about, once we can all agree on what the actual mass/energy distribution is.

It seems to me that you are arguing against an asymmetrical distribution based only on a hand-waving argument about gravitational effects.

NO. None of you posted a formal calculation supporting this claim.

It seems to me Przyk's integral is explicitly about that particular claim. Perhaps you didn't read it properly, so here it is again:
Are we all agreed that in the moving frame, the portion of the wheel above the level of the axle has more kinetic energy than the portion of the wheel below the level of the axle?
Note that this claim isn't about the rim above the transformed midline... it's about the portion above the axle's y-coordinate in the moving frame.
Which is what Przyk's integral is about.

I will post my calculation AFTER you post yours supporting the opposite claim.

Why?
You made an unsupported claim that in the moving frame, the KE of the portion of the wheel above the transformed midline is the same as the rest of the wheel.
I doubt that claim, but since it's peripheral to the thread I don't care enough to check it thoroughly.
Do you have a calculation to support your claim?
If you do, then show it.
If not, fine. I'll just continue to suspect you're wrong.

Yes, you are right. I get that the spokes are on the same side of the diameter, though my math is different since I have the parametrization of x and you the reverse of yours:

$$x'=R sin (\omega t' +\phi)$$
$$y'=R cos (\omega t' +\phi)$$

leading to the exact equation of the spokes given early in the thread:

$$y-R=\gamma (x-Vt) ctan[\omega \gamma (t-Vx/c^2)+\phi_i ]$$

t=0 results into:

$$y-R=\gamma x ctan (\phi_i-\omega \gamma Vx/c^2 )$$
$$x'=\gamma x$$

Therefore:

$$y-R=x' ctan (\phi_i-\omega V x'/c^2)$$

So, for:

$$\phi_1=0$$
$$y_1-R=x'_1 ctan (-\omega V x'_1/c^2)<0$$

$$\phi_2=\pi$$
$$y_2-R=x'_2 ctan (\pi-\omega V x'_1/c^2)<0$$

Your parametrization is for spokes that are vertical at $$t'=0, \phi=0, \phi=\pi$$
For those spokes, x'=0 at t=0, making your expression for y'-R undefined.

This still doesn't seem right. In the frame of the axle you should have:

$$x'=R cos (\omega t' + \phi)$$
$$y'=R sin (\omega t' + \phi)$$

I was using $$\phi=0$$ to be parallel to the x-axis at t'=0 to make the equations a little neater.
I should probably have kept with your parametrization for consistency, but it quickly simplifies into the expressions I gave anyway.
Also, we should probably decide whether the ground is at y=0 or y=-R, and stick to that convention.

It would be neater with the ground at y=-R, but we've mostly had the ground at y=0 so far so I'm easy either way.

 

[Pete]

If this is true, than we can produce scenarios in which reality has two courses which it can take, and no way to determine which course will be taken.

There are three questions:

1) How does the y-axis mass-energy distribution of the rolling wheel depend on our reference frame?
2) What frame-dependent gravitational effects are implied by the answer to 1)?
3) Does the answer to 2) imply any contradictory outcomes?

Question 1) is being thrashed out.
Tach says the answer should be no, but the calculations so far suggest (to me at least) that the answer is yes, that in the ground frame there is more mass-energy above the fence than below.

The second and third questions haven't been addressed with any kind of formalism.

I think that the reason Tach is disagreeing with przyk, RJBeery, and myself about the first question is that he thinks that would imply that the answer to the third question is yes.


But I'm sure we can all agree that if the answer to the third question turns out to be 'yes', then we're doing something wrong.

 

[Pete]

Let's say we do define the top half "invariantly", meaning that the collection of wheel elements constituting the "top half" at some instant in the ground frame is the same as the collection of wheel elements in the top half at some "corresponding" instant in the axle frame. Then the approach you tried in [POST=2848455]post #51[/POST] does become valid (with the correction $$p_{\mathrm{upper}} = - p_{\mathrm{lower}}$$), because it applies to an invariantly defined collection of wheel elements. So setting $$E \equiv E_{\mathrm{upper}} = E_{\mathrm{lower}}$$ and $$p \equiv p_{\mathrm{upper}} = - p_{\mathrm{lower}}$$, you find

$$
\begin{align}
E'_{\mathrm{upper}} &= \gamma (E + pv) \,, \\
E'_{\mathrm{lower}} &= \gamma (E - pv) \,.
\end{align}
$$​

which are clearly different unless $$p = 0$$ (i.e. a non-rotating wheel).

Err, no, $$(E,p)'_{upper}$$ isn't the transform of $$(E,p)_{upper}$$.

What do you mean? Isn't that true by definition, since przyk is explicitly talking about the transformed upper elements?
And you are directly contradicting what you said in [post=2848455]post 51[/post]:

In any other frame S', moving with speed V wrt S', the energy-momentum transforms according to SR:

$$E'=\gamma(E+pV)$$

Therefore:

$$E'_{upper}=\gamma(E_{upper}+p_{upper}V)$$

See post 60.

What part of [post=2848722]post 60[/post] is relevant?

 

[Masterov]

Can you show a fixed wheel which rotates in the plane of the monitor screen?

 

[Tach]

Your parametrization is for spokes that are vertical at $$t'=0, \phi=0, \phi=\pi$$
For those spokes, x'=0 at t=0, making your expression for y'-R undefined.

No it doesn't. Remember, the limit $$\frac{x}{sin x}$$ when $$x->0$$ is well defined.

I was using $$\phi=0$$ to be parallel to the x-axis at t'=0 to make the equations a little neater.

Still doesn't explain why you got the results that you got. I explained that if change the parametrization this exchanges the $$cotan$$ for a $$tan$$. Certainly not the $$sin$$ that you have.

I should probably have kept with your parametrization for consistency, but it quickly simplifies into the expressions I gave anyway.
Also, we should probably decide whether the ground is at y=0 or y=-R, and stick to that convention.

Ground is y=0, the axle is at y=R. How about you did your derivation over , without skipping details, the way I did it? Then, we can figure out why you got a different result. You should be getting eith $$tan$$ or $$cotan$$ but not the $$sin$$ you are showing.

No, Tach, we're quite clear on what is frame dependent.

This is the first time that either of you admits it. I explained it to przyk since post 60 (bottom of the page), he never admitted it.

The set of wheel elements above the fence at any instant is also frame dependent... and that's what we've been talking about ever since the OP.

This is not good enough. In the axle frame the fence blocks any view of the lower 4 spokes. Through what miracle does the light evade the fence in the ground frame? Saying "it is frame dependent" doesn't solve the issue.

 

[Tach]

I note that this is the sum total of Tach's argument on the matter so far:



It "doesn't seem right" to Tach's intuition. Therefore "something is terribly wrong".

Yes, because the light coming from the spokes is blocked in the axle frame, so it cannot be received in the ground frame unless it somehow manages to "evade" the frame. So, in order to do something useful (instead of your standard trolling), why don't you work in explaining this paradox?

Tach, being god-like, apparently isn't required to back up claims with evidence or calculations.

Actually, you are outright lying, adding this to your stalking. Very unbecoming. Instead of wasting your life stalking and wasting bandwidth with such posts, why don't you try to pick a part of the problem and solve it? There are enough loose ends that need answers.

 

[Tach]

Over 24 hours after I ask Tach to post the derivation he said he'd already done yesterday, and what do I see? Nothing from Tach.

From post 127:

"So, the only valid approach left is to integrate $$\gamma=\frac{1}{\sqrt {1-(v/c)^2}} $$ over the ever varying "halves" of the ellipse. In another post I showed the functions $$v_x=v_x(V,\phi,t)$$ and $$v_y=v_y(V,\phi,t)$$. If you do the calculations, the integrand has the very nice expression $$\gamma=\frac{1+(V/c)^2 cos (\omega t+\phi)}{1-(V/c)^2}$$. "

Now, all that is left is finding out the integration limits. Do you think you can do that , James? All by yourself? Instead of wasting your life stalking, try to roll up your sleeves and do some science for a change?