On 9/18/2025 8:15 AM, Alan Grayson wrote:
On Thursday, September 18, 2025 at 7:23:39 AM UTC-6 John Clark wrote:
On Mon, Sep 15, 2025 at 4:16 AM Alan Grayson <[email protected]>
wrote:
/> If you claim there is no acceleration in GR, I'd like to
know how acceleration is defined, and how it can be calculated
to be zero. I tried using the 2nd derivative of dS, but it
doesn't yield zero. AG/
*That's how acceleration is determined in Newtonian physics but
that's not good enough for General Relativity. Newton assumed that
time proceeded at a constant rate for everybody, but Einstein knew
that wasn't true, you have to consider your motion not just
through space but also your motion through time. **And Einstein
realized that not only is it possible to accelerate through space
it is also possible to accelerate through time. *
*
*
*In General Relativity your "position" isn’t just defined by
(x,y,z), but by (t,x,y,z), it's called your "4-position" and
mathematically is expressed as a 4-vector, or as a rank-1 tensor.
The mathematical equation that defines it uses Christoffel symbols
that encode the 4-D non-Euclidean spacetime curvature you get in a
gravitational field.*
*Please elaborate on the role of the Christoffel symbols that enclode
the spacetime curvative in a gravitational field. What does "encode"
mean in this context? AG*
*It qualifies as being a tensor because when you change
coordinates, that is to say when you switch from one observer's
frame of reference to another, although its individual components
might change, the tensor as a whole does not change.*
*Can a tensor be defined as a mathematical entitity which is invariant
under a coordinate transformation. *
Not, invariant but covariant.
*What is the form of the entity that guarantees that invariance? *
It's not the form. For example you could write down the matrix defining
a 2nd order tensor in some particular coordinate system and then it
would be defined in every other coordinate system by its transformation
to that system. So it's not the same matrix in the new system, but it
is the same physical object...that's why it's call covariant, not
invariant. This often done in relative theory by choosing the
coordinate system in which it is relatively easy to write down the
tensor and then transforming that to the system of interest.
*Must the entity being measured in different frames have a particular
form, say as a matrix, for the invariance to make sense? AG *
**
*According to Einstein a body that is not experiencing a force
always follows the straightest possible line, and in curved 4-D
spacetime that would be a geodesic. You can use your 4-position to
calculate your "proper acceleration", the acceleration you feel
and that registers on an accelerometer. And you can also use it to
calculate your "coordinate acceleration", the change of spatial
coordinate you get when you change from one coordinate system to
another, or to put it another way, when you change from one frame
of reference to another. Proper acceleration is invariant, every
observer agrees whether or not you feel acceleration,*
*
*
*Brent says, I think, that proper accelation will not be felt. True or
not? AG *
No. Brent says no acceleration is felt when proper acceleration is
zero. Even though the Newtonian acceleration (the coordinate
3-acceleration) is non-zero, e.g. in orbiting. I thought you had bought
a text book so you could read this stuff.
Brent
*but coordinate acceleration is _NOT_ invariant.
*
*Is this Newtonian acceleration? AG *
*
*
*Einstein says thateverything is always moving through spacetime
at the speed of light, when you're just sitting under a tree and
not moving all your speed is in the time direction, but when you
get up and start walking a small part of your velocity vector is
in the spatial direction and so your speed through the time
dimension is reduced slightly. If you’re in a spaceship orbiting
Earth then the space-time curvature(mathematically expressed as
Christoffel symbols) bends your paththrough both 3-D space and 40
spacetime, but you haven't fired your rockets so your proper
acceleration is zero and you feel weightless.*
*
*
*John K Clark See what's on my new list at Extropolis
<https://groups.google.com/g/extropolis>*
ufw
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