[continued from previous message]
>> That is NOT how it works. Unless you divide the metric by a constant value,
>> it is *always* _ds²_ LHS (or RHS) *regardless* what kind of interval it is
>> [possibilities are ds² < 0, ds² = 0, and ds² > 0. In the sign convention
>> (+−−−), ds² > 0 means timelike; in (−+++), spacelike]. So you can
write, if
>> you want
>>
>> c²ds² = -(1 − 2GM/[c²r]) c⁴dt² + 1/(1 − 2GM/[c²r]) c²dr²
>> + c²r² [dθ² + sin²(θ) dφ²),
>
> Which would be pointless.
That is what I said:
>> i.e. you can multiply the whole equation by c². It does not change
^^^^^^^^^^^^^^^^^^
>> anything, but it is also not particularly helpful.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
>>> No, it is the metric with signature (-+++) because:
>>
>> Nonsense. In your paper you clearly use the sign convention (+−−−)
[which
>> is OK, just less common in GR], and that is what I was referring to.
>>
>>> c²ds² = - c²dτ² = -(1 − 2GM/[c²r])c²dt² + 1/(1 −
2GM/[c²r])dr²
>>> + r² (dθ² + sin² θ dφ)
>
> Typo again.
> I meant to write:
>
> ds² = - c²dτ² = -(1 − 2GM/[c²r])c²dt² + 1/(1 − 2GM/[c²r])dr²
> + r² (dθ² + sin² θ dφ)
> Which is correct.
Yes.
> When s² < 0 and thus time like, then (-c²τ²) < 0 and thus time like.
[_timelike_ or _time-like_: ]
Correct for _ds²_ and _-c²dτ²_.
>> That is nonsense (see above), and not how you wrote it in your paper (in
>> multiple respects).
>
> In the paper I wrote:
> c²dτ² = (1 − 2GM/[c²r])c²dt² - 1/(1 − 2GM/[c²r])dr²
> - r² (dθ² + sin² θ dφ)
>
> You can argue that the signature in this case is (+---)
Yes.
> But then the same metric as above would be:
> -ds² = (1 − 2GM/[c²r])c²dt² - 1/(1 − 2GM/[c²r])dr²
No, it is always ds² (with a positive sign).
>> In particular, while you *can* use the Schwarzschild metric (in whatever
>> coordinates you want) to describe the "gravitational field" of Earth, the
>> metric is NOT *defined* based on the gravitational field of Earth, but
>> describes a spacetime corresponding to *any* spherically-symmetric
>> mass/energy distribution. That distinction is missing from your paper, and
>> that is misleading.
>
> This is nonsense.
No, it is not. You fail(ed) to make the distinction between the
Schwarzschild metric and the Schwarzschild metric as applied to your
use-case: You wrote that the M in the *metric* *is* the mass of Earth, and
that is simply not so.
> Of course I can use the Schwarzschild metric to find the proper
> times of satellites orbiting the Earth without explaining that
> it also could be used on othe massive bodies than the Earth.
Yes, but, at best, you are confusing the reader then.
>>>> [A few months before I registered for the GR course, motivated by a
need
>>>> for a better explanation of "time dilation", I did a similar, first
more
>>>> general and then in its application more detailed calculation,
published
>>>> on Quora (currently unavailable; they blocked me without good
reason);
>>>> when previously for the Schwarzschild spacetime I had considered
>>>> "kinetic time dilation" and "gravitational time dilation"
separately,
>>>> simply arguing conceptually that one offsets the other.]
>>>>
>>>> "The difference between (4) and (5) is less than 10^-25 for all r"
>>>>
>>>> That is a questionable claim (for *all* r?); you should
substantiate it.
>>>
>>> It is true, tough.
>> Prove it, then.
>
> Is it really necessary?
>
> sqrt(1 - x) ~ (1 - x/2) when x << 1
>
> The approximation gets worse the higher x is.
> In (4) and (5) the highest value x can have is when
> r is smallest, which is the radius of the Earth.
The smallest value for r is 0. The smallest physically meaningful value for
r in eq. (4) and (5) is when 2μ/(c^2 r) + v^2/c^2 = 1, i.e. 2μ/(c^2 - v^2).
> So just compare (4) and (5) with r = Earth's radius
That does not prove your claim.
>> Please trim your quotes to the relevant minimum next time.
Why did you not? :-(
--
PointedEars
Twitter: @PointedEars2
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