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Imagine that you want to shoot something that is placed diametrically opposed to you. Should you aim straight down the diameter? You can analyze the problem from either the perspective of the inertial frame of the ground or from the POV of the non-inertial rotating platform. The answer will be the same: 1. From the ground frame POV, if you aim straight down the diameter , you will miss. The reason is that while the bullet travels down the straight line of the diameter, the target moves, so you'll miss. It is an aberration effect. 2. From the POV of the platform things are a little more complicated. The bullet is subjected to the fictitious Coriolis force, so its trajectory will be "bent" and will not follow the diameter drawn on the platform. So, you'll miss. Not a bad question....

Using buzzwords is not going to cover your lack of understanding.

Looks like you do not understand what "midpoint" and or "simultaneous" means. There are two (simultaneous) events : [math](x_1,t)[/math] [math](x_2,t)[/math] The observer located at midpoint has the coordinate [math]\frac{x_1+x_2}{2}[/math]. This observer is located on the x-axis, contrary to your claim. He or she will observe the two events as being simultaneous, contrary to your claim.

This is false. Any observer situated at the midpoint between the two simultaneous events observes them without any trouble. And, to boot, the observer in cause observes the two events as being....simultaneous. You are hijacking this thread with your misconceptions.

You could do that only locally. You can never do it globally. You can only get local Cauchy patches, you cannot construct a full Cauchy surface out of those patches.

False, two events can be simultaneous, has nothing to do with "force" While forces need time to propagate this does not mean that two (or multiple ) events cannot be simultaneous. So, your claim has no basis. It is perhaps best if you opened a separate thread on your misconceptions about physics.

Close: it is the line parallel with the axis x'=0 for t'=constant.

No, it doesn't, time continues to run at its normal rate of 1 second per second. No, it doesn't. This is what the people who do not understand relativity say. This thread belongs in Trash. .

The flat spacetime formalism (Minkowski-Lorentz) applies only in a infinitesimal domain of the curved spacetime. There is no global notion of simultaneity in curved spacetime.

Yes, the thread is talking about the universe but you were babbling about "Minkowski diagrams" and "foliations", so I commented on your babbling <shrug>

I simply pointed out that you are posting prose without any mathematical backing. If you could follow the math (you obviously, couldn't) you would have learned that this is exactly what the math I have posted expresses. True. Except I am talking about flat spacetime. So, your post is a non-sequitur. There is no such thing. In GR , the notion of simultaneity is strictly local, there is no such thing as a global simultaneity, as in flat spacetime. You can stop posturing now.

I think you misunderstood, the AXIS [math]dt'=0[/math] represents all the simultaneous events in frame F'. I am not talking about [math]t'=0[/math]. I am talking about the set of events [math] (x'=arbitrary,t'=constant)[/math]. Do you understand the difference?

Yes, it does. You can ignore mdc's handwaving devoid of any math and look at the following: The Minkowski diagram presents two frames of reference, in motion with speed V wrt each other. Frame F has the x and t axis orthogonal on each other, with the t axis in the vertical direction. Frame F' has the x' and t' axes inclined wrt to x and t respectively. The angle made by x' with x is [math] arccos \frac{1}{\gamma}[/math]. Proof: [math]dx'=\gamma(dx-Vdt)[/math] [math]dt'=\gamma(dt-Vdx/c^2)[/math] The dx' axis represents the axis of "simultaneity" in F', i.e. [math]dt'=0[/math] This means [math]dt=Vdx/c^2[/math]. Substituting dt in [math]dx'=\gamma(dx-Vdt)[/math] gives [math]dx'=\frac{1}{\gamma}dx[/math], so the angle between dx' and dx is [math] arccos \frac{1}{\gamma}[/math]. In a similar manner, one can determine the angle made by dt' with dt. Here is how: dt' is the axis that represents all the points that have the same x', i.e. [math]dx'=0[/math]. This gives [math]dx=Vdt[/math]. Substituting into [math]dt'=\gamma(dt-Vdx/c^2)[/math] you get [math]dt'=\frac{1}{\gamma}dt[/math], The t',t axes make the same angle as the x',x axis in the Minkowski diagrams. As you can see above, there are two types of what is pretentiously called "foliation": [math]dt=0[/math] and [math]dx=0[/math], i.e. lines of equal time and lines of equal position.

Don't let the screendoor hit your backside on the way out.

I think you are mistaken because both u and v are functions of t, so one cannot separate the variables as you suggest. Think about it, Wolfram Alpha couldn't solve this system , the solution cannot be that simple.

[math]\frac{d}{dt}(\frac{u}{\sqrt{1-u^2-v^2}})=k_1[/math] produces [math]\frac{du}{\sqrt{1-u^2-v^2}}=k_1dt[/math] practically a variable separation which gives , by integrating wrt u in the LHS and t in the RHS the answers that I posted earlier. I think that you are claiming that one should get: [math]d\frac{u}{\sqrt{1-u^2-v^2}}=k_1dt[/math] resulting into [math]\frac{u}{\sqrt{1-u^2-v^2}}=\int{k_1dt}[/math] This doesn't seem right. Neither of them is right The right thing is to observe that [math]d(\frac{1}{\sqrt{1-u^2-v^2}})=\frac{udu+vdv}{(\sqrt{1-u^2-v^2})^3}[/math] So, the equation is [math]\frac{udu+vdv}{(\sqrt{1-u^2-v^2})^3}=k_1dt[/math] The problem is that the second equation produces: [math]\frac{udu+vdv}{(\sqrt{1-u^2-v^2})^3}=k_2dt[/math] so, the system is unsolvable if [math]k_1 \ne k_2[/math] and indeterminate if [math]k_1 = k_2[/math]. I think this is why Wolfram gave up on this problem. Unfortunately this is not an arbitrary exercise, it is rooted in particle physics.

Thank you, Both equations reduce to : [math]\sqrt{1-u^2-v^2}=k_1t+c_1[/math] [math]\sqrt{1-u^2-v^2}=k_2t+c_2[/math] This forces [math]k_1=k_2, c_1=c_2[/math], so, the system has a solution only for [math]k_1=k_2[/math]. It doesn't seem right, I don't think that it is allowable to integrate the equations as if v is a constant in the first equation while u is a constant for the second equation.

I am trying to solve the system of ODEs: [math]\frac{d}{dt}(\frac{u}{\sqrt{1-u^2-v^2}})=k_1[/math] [math]\frac{d}{dt}(\frac{v}{\sqrt{1-u^2-v^2}})=k_2[/math] where [math]k_1,k_2[/math] are two constants and [math]u(0)=u_0[/math], [math]v(0)=v_0[/math]. I tried using the symmetry by transforming into polar coordinates: [math]u=\rho(t) cos \theta (t)[/math] [math]v=\rho(t) sin \theta (t)[/math] but it did not help. Any ideas? This is not homework. PS: Wolfram Alpha couldn't solve it even after I made [math]k_1=k_2[/math]

It means that the sum of the partial derivatives of a function is zero. What does this have to do with relativity?

...and the trolling has restarted. Instead of wasting your time carping about things you do not understand why not invest in a book like "Gravitation" by Misner, Thorne, Wheeler? Then you too could learn the modern explanation of gravitational redshift (and more than that).

Thank you for clarifying your position. This is clearly a fringe thread, does not belong in the main forum.

it is brilliant, the work of genius, you should patent it.

did you make this up all by yourself or did you see it somewhere?

Err, you still don't get it, do you? From the page that you are citing: ”One can say that the energy levels of the absorber atoms are shifted towards the blue IN THE LABORATORY FRAME". Better phrased: "The em radiation emitted from the top of the tower is received blue shifted at the bottom, in the laboratory frame".

It's a bad book. Do yourself a favor and buy a better one. You've been told repeatedly that you are repeating errors, why not follow the advice?