Disagreement on the interpretation of the Andromeda paradox

[somanystylez]

What if one observer (the one that is moving with respect to Andromeda) is much closer to it then the one on Earth? Would it make a difference?

[Delta1212]

What if one observer (the one that is moving with respect to Andromeda) is much closer to it then the one on Earth? Would it make a difference?

If one observer was closer to Andromeda, the two observers are no longer co-located which means you can't determine which moment for each observer is simultaneous with which moment for the other any longer, and you have no basis left for comparing differences in simultaneity with Andromeda.

[xyzt]

The Lorentz time transformation (which you're about to try to use) transforms from one coordinate system to another. But, you attribute these coordinates, above, to both earth observers*The point of the Andromeda paradox is that the t coordinate of an event at a significant distance is different in two different frames.

 

*First Error: As explained [math](x_a,t_a)[/math] are the coordinates of the event in the Earth-Andromeda frame.

The three velocities** are [math]v_1[/math], [math]v_2[/math], and [math]v[/math]. If [math]v_2=v[/math] then it sounds like you're just being redundant.

**Second Error: There are only two velocities, not three, [math]v_1,v_2[/math]. [math]0,v[/math] are the values attributed to [math]v_1,v_2[/math] respectively.

 

 

 

Ok...

 

[math]t_{e2}-t_{e1}=\gamma(t_a+v_2 x_a/c^2)-\gamma(t_a+v_1 x_a/c^2)=\gamma vx_a/c^2[/math]

 

What you just said means...

 

[math]t_{e2} = \gamma(t_a+v_2 x_a/c^2)[/math]

 

and

 

[math]t_{e1} = \gamma(t_a+v_1 x_a/c^2)[/math]

 

This could only make sense if you have a third frame*** If [math]t_a[/math] is a coordinate in a third frame then [math]t_{e1}[/math] and [math]t_{e2}[/math] would be the same coordinate in two different frames. The third frame, however, would need to have its origin in the same place as the other two frames. So, now we have three earth observers. There is no need for that, I assure you.

***Third Error: There are only two frames, the Earth-Andromeda-Observer1 is one and Observer2 is the second one. The Lorentz transforms are quite trivial and self-explanatory.

 

If you evaluate your equation numerically, the only way it works is if [math]t_a[/math] is zero****.

****Fourth Error: This is outright false, the symbolic formulas work for any [math]t_a[/math]. These are the run of the mill Lorentz transforms, I do not understand why you have so much difficulty with the fact.

The further from zero, the less correct. In fact, the left hand side and right hand side would disagree by exactly the value of [math]t_a[/math]. *****

*****Fifth Error: You obviously missed the fact that [math]t_a[/math] is multiplied by [math]\gamma(v)-1[/math] and [math]\gamma(v) \approx 1[/math]. It is all in the original post.

So, if [math]t_a[/math] is half c (t_a = 150,000,000) then the equation would be off by 150,000,000 seconds. Nobody wants to be nearly 5 years off, so..******

******Sixth Error: There is no reason for [math]t_a[/math] to be either zero or an ungodly large number. All is needed is to observe that [math]x_a[/math] is a very large (cosmological) distance such that it dwarfs [math]t_a[/math]. Combined with the fact that [math]t_a[/math] is multiplied by [math]\gamma-1 \approx 0[/math], you should get the idea....

 

 

If [math]t_a[/math] has to be zero******** then:

 

[math]t_{e2}- {t_{e1}} = \gamma(t_a+v_2 x_a/c^2)- {\gamma(t_a+v_1 x_a/c^2)} =\gamma vx_a/c^2[/math]

*******Seventh Error: As explained, there is no reason for {math]t_a[/math] to be zero. It can be zero or it can be different from zero, the formalism I gave works either way.

Edited June 16, 2013 by xyzt

[Iggy]

*First Error: As explained [math](x_a,t_a)[/math] are the coordinates of the event in the Earth-Andromeda frame.

 

**Second Error: There are only two velocities, not three, [math]v_1,v_2[/math]. [math]0,v[/math] are the values attributed to [math]v_1,v_2[/math] respectively.

 

 

 

 

 

***Third Error: There are only two frames, the Earth-Andromeda-Observer1 is one and Observer2 is the second one. The Lorentz transforms are quite trivial and self-explanatory.

 

****Fourth Error: This is outright false, the symbolic formulas wort for any [math]t_a[/math]. These are the run of the mill Lorentz transforms, I do not understand why you have so much difficulty with the fact.

 

*****Fifth Error: You obviously missed the fact that [math]t_a[/math] is multiplied by [math]\gamma(v)-1[/math] and [math]\gamma(v) \approx 1[/math]. It is all in the original post.

 

******Sixth Error: There is no reason for [math]t_a[/math] to be either zero or an ungodly large number. All is needed is to observe that [math]x_a[/math] is a very large (cosmological) distance such that it dwarfs [math]t_a[/math]. Combined with the fact that [math]t_a[/math] is multiplied by [math]\gamma-1 \approx 0[/math], you should get the idea....

 

 

*******Seventh Error: As explained, there is no reason for {math]t_a[/math] to be zero. It can be zero or it can be different from zero, the formalism I gave works either way.

 

Well, that was boring to read.

 

 

So, if is half c (t_a = 150,000,000) then the equation would be off by 150,000,000 seconds. Nobody wants to be nearly 5 years off, so..******

****Fourth Error: This is outright false, the symbolic formulas wort for any [math]t_a[/math]. These are the run of the mill Lorentz transforms, I do not understand why you have so much difficulty with the fact.

 

Do it.

 

Let's see some numbers big boy.

[xyzt]

 

Well, that was boring to read.

 

 

Do it.

 

Let's see some numbers big boy.

I don't understand why you have so much difficulty with the elementary exercises:

 

[math]\gamma-1 =\frac{1}{\sqrt{1-(v/c)^2}}-1 \approx \frac{v^2}{2c^2}[/math]

 

For [math]v=0.1m/s[/math] it follows that [math]\gamma-1 \approx 0.5* 10^{-19}[/math]

 

so, for your chosen [math]t_a=1.5* 10^8[/math]

 

[math](\gamma-1)t_a \approx 0.75* 10^{-11}[/math]

Edited June 16, 2013 by xyzt

[D H]

Bzzzt, wrong.

 

You are computing time dilation, not the Lorentz transformation. Try again.

[xyzt]

Bzzzt, wrong.

 

You are computing time dilation, not the Lorentz transformation. Try again.

Actually, I am computing the correct extra term in the general formula explained in post 48. Nothing to do with any "time dilation". Let me tell you what I told Iggy, if you do not understand something, just ask, I will be more than happy to explain it to you.

Edited June 16, 2013 by xyzt

[D H]

You're right. It wasn't time dilation. You were calculating nonsense.

 

You had a week off. Why in the world didn't you use that time constructively to read up on what the Andromeda paradox is all about?

[Iggy]

I don't understand why you have so much difficulty with the elementary exercises:

 

[math]\gamma-1 =\frac{1}{\sqrt{1-(v/c)^2}}-1 \approx \frac{v^2}{2c^2}[/math]

 

For [math]v=0.1m/s[/math] it follows that [math]\gamma-1 \approx 0.5* 10^{-19}[/math]

 

so, for your chosen [math]t_a=1.5* 10^8[/math]

 

[math](\gamma-1)t_a \approx 0.75* 10^{-11}[/math]

Nope.

 

Try again.

 

This is your pitiful equation:

 

[math]t_{e2}-t_{e1}=\gamma(t_a+v_2 x_a/c^2)-\gamma(t_a+v_1 x_a/c^2)=\gamma vx_a/c^2[/math]

 

Plug some numbers into the left and right side and prove to me that t_a doesn't have to be zero for your pitiful excuse for nauseating and exhausting math is correct.

 

Go ahead and try to back up that claim.

 

edit:

 

by the way, please stop neg oneing all my posts. I've never given you a -1 no matter how much you deserve it (and frankly, no one deserves it more than you). Bother someone else with that crap, please!

Edited June 16, 2013 by Iggy

[hypervalent_iodine]

!

Moderator Note

Enough of the hostility.

 

Seriously, it's like moderating a kindergarten here lately. Play nice.

[xyzt]

 

Nope.

 

Try again.

 

This is your pitiful equation:

 

[math]t_{e2}-t_{e1}=\gamma(t_a+v_2 x_a/c^2)-\gamma(t_a+v_1 x_a/c^2)=\gamma vx_a/c^2[/math]

 

Plug some numbers into the left and right side and prove to me that t_a doesn't have to be zero for your pitiful excuse for nauseating and exhausting math is correct.********

 

 

******** Eighth Error: The correct equation, as already posted is:

 

[math]t_{e2}-t_{e1}=\gamma(v_2) v_2x_a/c^2+t_a (\gamma(v_2)-1)=\gamma(v)vx_a/c^2+t_a (\gamma(v)-1) \approx \gamma(v) vx_a/c^2 [/math]

 

As already explained, [math](\gamma(v)-1)t_a \approx \frac{v^2t_a}{2c^2} \approx 0[/math] so:

 

[math]t_{e2}-t_{e1} \approx vx_a/c^2[/math] as already correctly explained in post 2

Edited June 16, 2013 by xyzt

[Iggy]

The correct equation...

 

Try again.

 

The observers have speeds [math]v_1=0[/math] respectively [math]v_2=v[/math] wrt Andromeda, so:

 

[math]t_{e2}-t_{e1}=\gamma(t_a+v_2 x_a/c^2)-\gamma(t_a+v_1 x_a/c^2)=\gamma vx_a/c^2[/math]

 

Own your words.

 

[xyzt]

Try again.

 

 

Own your words.

 

Basic math says that the chain of equalities continues with [math]....=\gamma(t_a+v_2 x_a/c^2)-\gamma(t_a+v_1 x_a/c^2)=\gamma vx_a/c^2[/math]

 

The arithmetic in posts 48 and 55 is pretty basic.

Edited June 16, 2013 by xyzt

[Iggy]

 

If you evaluate your equation numerically, the only way it works is if [math]t_a[/math] is zero.

 

****Fourth Error: This is outright false, the symbolic formulas work for any [math]t_a[/math]. These are the run of the mill Lorentz transforms, I do not understand why you have so much difficulty with the fact.

 

Keep trying.

 

Do you need me to help?

[hypervalent_iodine]

!

Moderator Note

Since you all seem to have a problem in comprehending very simple requests in big red boxes, I am going to close this pending staff review.

[hypervalent_iodine]

!

Moderator Note

Staff have decided to reopen this thread.

 

However, the antagonizing jabs at one another are to stop. This goes for all parties equally.