Relativistic formula for motion with constant acceleration

[Ibix]

Agreed. That is not consistent with what you wrote in #28, though.

 

[bobie]

Agreed. That is not consistent with what you wrote in #28, though.

That's the equation in #28:
$\sqrt( \frac{1-3*10^8/10^10 = .97}{1-3*10^8/ 10^12 = .9997})$
and this is #30
$\sqrt( \frac{1- R_s:3*10^8/B:10^10 = .97}{1-3*10^8/ A:10^12 = .9997})$

and the result is .985

 

[Ibix]

$$\frac{\gamma_r}{\gamma_R}=\sqrt{\frac{1-R_s/A}{1-R_s/B}}$$

$\sqrt( \frac{1- R_s:3*10^8/B:10^10 = .97}{1-3*10^8/ A:10^12 = .9997})$

Try reading them carefully.

 

[bobie]

Try reading them carefully.

$$\frac{\gamma_r}{\gamma_R}=\sqrt{\frac{1-R_s/A =10^{12}}{1-R_s/B=10^{10}}}$$.
even switching values in the right places we get
$$\frac{\gamma_r}{\gamma_R}=\sqrt{\frac{.9997}{.97}}$$.

I get an impossible result , grater than c : 1.0152

 

[Ibix]

Seriously? You can't even read back over the thread for the meaning of $\gamma$ without me holding your hand and tell you where to look?

 

[jbriggs444]

That's the equation in #28:
$\sqrt( \frac{1-3*10^8/10^10 = .97}{1-3*10^8/ 10^12 = .9997})$
and this is #30
$\sqrt( \frac{1- R_s:3*10^8/B:10^10 = .97}{1-3*10^8/ A:10^12 = .9997})$

and the result is .985

Note that in order to extend the vinculum in the square root sign over the entire expression, the expression should be enclosed in curly braces, not parentheses. By default the \sqrt takes as its argument the next symbol (in this case, just the leading parenthesis).

Similarly, the caret ("^") symbol uses the next symbol only by default. If you want to render 2¹⁰, you need to write 2^{10}.

e.g.

$\sqrt{ \frac{1-3*10^8/10^{10} = .97}{1-3*10^8/ 10^{12} = .9997}}$
and this is #30
$\sqrt{ \frac{1- R_s:3*10^8/B:10^{10} = .97}{1-3*10^8/ A:10^{12} = .9997}}$

The practice of mixing symbols, colons, the values of those symbols, equal signs and the values of sub-expressions in a single expression does not work for me. It ends up as pure gibberish.

 

[bobie]

Note that in order to extend the vinculum in the square root sign over the entire expression, the expression should be enclosed in curly braces, not parentheses. By default the \sqrt takes as its argument the next symbol (in this case, just the leading parenthesis)...

Thanks jbriggs, you are extremely kind

 

[bobie]

Note that in order t.

Hi jbriggs,
can you tell if there is a typo here (#7)?
$$\frac{\gamma_r}{\gamma_R}=\sqrt{\frac{1-R_s/R}{1-R_s/r}}$$.
shouldn't the ratio be inverted?

$$\frac{\gamma_R}{\gamma_r}=$$.

 

[jbriggs444]

The syntax looks fine. The semantics, I'm not going to throw out an opinion.

 

[bobie]

The syntax looks fine. The semantics, I'm not going to throw out an opinion.

what syntax?
I was referring to the fact that the dididend on the left points to r and the one on the right (under the squareroot sign) refers to R. It doesn't seem coherent, is it just a typo or it means something?

I am totally confused by this thread, I hope someone can help

 

[jbriggs444]

what syntax?

Syntax: The expression is well formed. It conforms to the rules for constructing well-formed formulas. That is the sort of "typo" that I am qualified to search for.

 

[PeterDonis]

shouldn't the ratio be inverted?

No.

 

[Ibix]

shouldn't the ratio be inverted?

Why would you think that?

 

[bobie]

Why would you think that?

well, normally, if on one side we get a ratio say between R and r, the ratio on the right side should follow the same order , that is ...R/ ...r.
In your formula , the order is surprisingly reversed, and I really can't see the logic of that, what is it?

 

[PeterDonis]

normally, if on one side we get a ratio say between R and r,

Which you don't have on the left side. You have a ratio between $\gamma_R$ and $\gamma_r$. You do realize that those $\gamma$ thingies vary inversely with $r$, right?

In your formula , the order is surprisingly reversed, and I really can't see the logic of that, what is it?

Um, that $\gamma$ varies inversely with $r$? Meaning that, as $r$ gets bigger, $\gamma$ gets smaller?

 

[bobie]

Which you don't have on the left side. You have a ratio between $\gamma_R$ and $\gamma_r$. You do realize that those $\gamma$ thingies vary inversely with $r$, right?
Um, that $\gamma$ varies inversely with $r$? Meaning that, as $r$ gets bigger, $\gamma$ gets smaller?

I meant:
the formula should give the ratio
$$\frac{\gamma_r^2}{\gamma_R^2}= \frac{1-R_s/A =10^{12}}{1-R_s/B=10^{10}}$$.
but, isn't .9997 the value of gamma at R (, = Ro, = A?

$$\frac{\gamma_r^2}{\gamma_R^2}= \frac{.9997}{.997}$$.
What do I get wrong?

 

[DrGreg]

$$\frac{1-R_s/A =10^{12}}{1-R_s/B=10^{10}}$$.

An expression like this makes no sense. It seems that you are saying that something equals 10¹², but it's not clear what. You should state this separately outside of the equation and not put equal signs within a fraction.

 

[bobie]

An expression like this makes no sense. It seems that you are saying that something equals 10¹², but it's not clear what. You should state this separately outside of the equation and not put equal signs within a fraction.

The post was a quick reminder that R = A = 10^12 cm,
$$\frac{\gamma_r^2}{\gamma_R^2}= \frac{1-R_s/(A =10^{12})}{1-R_s/B=10^{10}}$$.
therefore the factor on the dividend there , in my view, refers to the gamma factor at R and the ratio is therefore: $$\frac{\gamma_R^2}{\gamma_r^2}$$.

 

[DrGreg]

The post was a quick reminder that R = A = 10^12 cm,
$$\frac{\gamma_r^2}{\gamma_R^2}= \frac{1-R_s/(A =10^{12})}{1-R_s/B=10^{10}}$$.
therefore the factor on the dividend there , in my view, refers to the gamma factor at R and the ratio is therefore: $$\frac{\gamma_R^2}{\gamma_r^2}$$.

It would be better to write
$$\frac{\gamma_r^2}{\gamma_R^2}
= \frac{1-R_s/A}{1-R_s/B}
= \frac{1-R_s/10^{12}}{1-R_s/10^{10}}$$.

 

[bobie]

No.

I found on the web this formula:
$$ v = c \sqrt{1 - \frac{1}{\gamma^2}\left(1 - \frac{r_s}{r}\right)} $$

is it a good formula, in what differs from yours?

 

[PeterDonis]

I found on the web this formula:

Where? Please give a reference.

 

[Vanadium 50]

This has turned into a game of "guess and check". Is this the right equation? What about this one? What about this one?

 

[Ibix]

I found on the web this formula:
$$ v = c \sqrt{1 - \frac{1}{\gamma^2}\left(1 - \frac{r_s}{r}\right)} $$

is it a good formula, in what differs from yours?

So have you managed to get around to looking through this thread for the definition of $\gamma$, as I implied you should in #35? I strongly suspect the answer is no given your subsequent posts, this one in particular. If not, why not? If so, what have you done with it and where are you stuck?

We're not really interested in doing high-school level algebra for you, especially if you give us no evidence you've tried it yourself. That's basically why the responses are becoming less helpful.

 

[bobie]

This has turned into a game of "guess and check".

Where? Please give a reference.

We're not really interested in doing high-school level algebra for you, especially if you give us no evidence you've tried it yourself. That's basically why the responses are becoming less helpful.

More than "guess and check" it looks like a "cat and mouse" game, Vanadium

Peter Donis, you are a mentor, over 50 posts ago I simply asked for the relativistic formula for gravity, and did my honest whack,(#1)

Do you know the formula we must use to find the relativistic increase of velocity? from other calcs the velocity of F should be .9902 or so-
Thanks

I didn't ask anyone to hold my hand (#35) or do cheap algebra (#53) for me.
You are only patronizing me, as usual, do you think I'll get a formula sometime?, eventually?

 

[Ibix]

I didn't ask anyone to hold my hand (#35) or do cheap algebra (#53) for me.
You are only patronizing me, as usual, do you think I'll get a formula sometime?, eventually?

You already have a formula. I gave it to you in #7. You used it to get a ratio of $\gamma$s but seem incapable of getting to the velocity from there. Why? What have you tried? Apart from insisting (incorrectly) that the formula must be wrong?

 

[PeterDonis]

over 50 posts ago I simply asked for the relativistic formula for gravity

do you think I'll get a formula sometime?, eventually?

As @Ibix has pointed out, you've already gotten what you asked for. Plus a lot more.

Thread closed.