General Relativity Basics: The Principle of Equivalence

[PeterDonis]

I do not believe the formula you use is correct.

First you should calculate the Schwarzschild radius rs using G and the mass of the Earth then solve the following equation for r ...

Technically this is correct, but since JDoolin was only considering things to the sixth significant figure, and the Schwarzschild radius is eight or nine orders of magnitude smaller than the radius of the Earth, the approximation he used was fine. Here are the numbers I come up with using your formula (ignoring the ln term since, as you say, it's way way smaller even than the small corrections we'll see below):

For g = 9.80664 m/s^2, r = 6375882.769 meters.

For g = 9.80665 m/s^2, r = 6375879.518 meters.

As you see, the approximation JDoolin used is fine to six figures; the differences with the more exact formula you gave only show up in the eighth figure.

(Edit: Actually it's the ninth figure.)

 

[JDoolin]

The usual quantitative measure of spacetime curvature is tidal gravity--the magnitude of tidal accelerations. For the spacetime outside a gravitating body like the Earth, tidal accelerations go like

$$\frac{2GM}{r^{3}} \Delta r$$

where r is the distance from the center of the Earth and $\Delta r$ is the size of the body being subjected to the tidal accelerations.

(Technically, according to GR, the quantitative measure of spacetime curvature is the Riemann curvature tensor, whose components are sort of a "limit" of tidal acceleration as the size of the object goes to zero--I say "sort of" because the limit of the above formula as $\Delta r$ goes to zero is zero, so obviously that isn't quite what the curvature tensor is. But its "radial" components outside a gravitating body like the Earth look just like the above formula, but with the $\Delta r$ taken out--i.e., just $\frac{2GM}{r^{3}}$. There's been plenty of discussion on these forums of the Riemann curvature tensor; I can try and find some good threads and post links to them.)



Actually, in a sense, yes, despite what I just said above--but it's still curvature of *spacetime*, not just space. There's an interesting discussion of this point in Misner, Thorne, and Wheeler, in Box 1.6 (pp. 32-33 in my edition). They compare the trajectories of a baseball, thrown over a distance of 10 meters, and a bullet fired from a rifle over the same distance. Obviously the spatial curvature of these two trajectories will be very different--they assume a slow thrown baseball, not a Nolan Ryan fastball, so the baseball's trajectory will be visibly curved in space, unlike the bullet's, whose spatial curvature is microscopic (but it still is there, as we'll see). But if we take time into account, we find that the curvatures of these two paths *are* the same (actually, we find that the tracks themselves are not "curved", being geodesics--but they reveal the curvature of the spacetime through which they travel, which has to be the same for both). Here's a quick back of the envelope calculation of that using their numbers:

Horizontal distance: 10 m for both baseball and bullet.

Vertical "height" of trajectory (highest point relative to start and end points): 5 m for baseball, 5 x 10^4 m (half a millimeter) for bullet.

Speed: 5 m/sec for baseball (about 10 mph, so a *very* slow pitch, hence the high trajectory), 500 m/sec for bullet.

Travel time: 2 sec for baseball, 2 x 10^-2 sec for bullet.

Spacetime interval: here MTW use the special relativistic formula $s^{2} = {ct}^{2} - x^{2}$; yes, this formula is only valid in an inertial frame, but since all the distances and times are very small compared to the radius of curvature we'll be calculating, we can treat all the numbers above as being measurements with respect to a momentarily comoving inertial frame. The intervals work out to be 6 x 10^8 meters for the baseball, and 6 x 10^6 meters for the bullet. (Note that the horizontal distance itself, 10 m, is negligible here, so the interval is basically ct; the horizontal distance only comes into play indirectly in determining t from the speed of each object.)

Curvature: here MTW use the formula (radius of curvature) = (spacetime interval)^2 / (8 * vertical rise). The derivation of this formula from the components of the Riemann curvature tensor is in the text; I won't go into it here. If you plug in the numbers above for the baseball and the bullet, you'll see that both of them result in a radius of curvature of about 1 light-year, which, as we've already seen, is indeed c^2 / g for a 1-g acceleration. And in fact, we can see that their formula for radius of curvature *must* give this result; if we use h for the vertical rise and t for the travel time, we have:

$$h = \frac{1}{2} g \left( \frac{1}{2} t \right)^{2} = \frac{1}{8} g t^{2}$$

$$r = \frac{\left( ct \right)^{2}}{8 h} = \frac{c^{2}}{g}$$

Note that we have to square 1/2 t in the formula for h because each object goes up and then comes down again, so the height h must be calculated using the time for half of the trajectory.

So this means that the actual "radius of curvature" of spacetime around the Earth *is* 1 light-year, even though the spatial radius of the Earth itself is only 6.3 million meters or so, or 9 orders of magnitude smaller.

It sounds like Misner Thorne Wheeler (MTW) might be an text on General Relativity, with comprehendible examples. I haven't seen it, so I'l see if I can locate it through the library system.


I'm curious/confused about the fomula

$$h = \frac{1}{2} g \left( \frac{1}{2} t \right)^{2} = \frac{1}{8} g t^{2}$$​

How is this time calculated; is it the time the ball take to go 10 meters? t=x/v ? Is this all done in a region of constant g?

I spent some time trying to collect some of the concepts in your posts--I haven't got to Passionflower's just yet.

$$\begin{center} \begin{tabular}{| c | c | c | c| } \hline Formula & ind. vars & name & derivation \\ \hline r_m=\frac {GM}{c^2} & M & Geometric Mass & 2*Schwartzchild Radius \\ \hline r_c=\frac{c^2}{a} & a & Rindler Horizon & position of stationary event in accelerating frame \\ \hline r_c= \frac {c^2 r^2} {G M} & M, r & spatial curvature & Rindler Horizon with Newtonian Gravity \\ \hline r_c=\frac{s^2}{8 h} & s,h & radius of curvature & Components of Riemann Curvature Tensor \\ \hline -(1+ c^2g00) & ? & time curvature & ? \\ \hline \end{tabular} \end{center}$$​

 

[PeterDonis]

It sounds like Misner Thorne Wheeler (MTW) might be an text on General Relativity, with comprehendible examples.

It is one of the classic GR texts--some would say *the* classic GR text, although, as the Usenet Physics FAQ says (I link to the page below), it "may not the best first book on GR for most students, in part because by offering so much it is liable to overwhelm a newcomer." Also, since it was published in 1973, it doesn't cover some important subject areas that were not really understood then (e.g., black hole evaporation). It's a weighty tome, so be advised that if you get it from the library, you may need a wheelbarrow to get it home. :-)

A more recent book is Wald's 1984 text "General Relativity", which covers a lot of the same ground as MTW but includes topics like black hole evaporation that are left out of MTW.

This page on the Usenet Physics FAQ is a pretty comprehensive guide to relativity texts:

http://www.edu-observatory.org/physics-faq/Administrivia/rel_booklist.html

There are texts listed there that may be better introductory ones; I'm not familiar enough with any of them to give an opinion (I have both MTW and Wald so I tend to draw examples from them).

I'm curious/confused about the fomula

$$h = \frac{1}{2} g \left( \frac{1}{2} t \right)^{2} = \frac{1}{8} g t^{2}$$​

How is this time calculated; is it the time the ball take to go 10 meters? t=x/v ? Is this all done in a region of constant g?

I just used the Newtonian formula for distance as a function of time, $d = \frac{1}{2} g t^{2}$, noting that the "t" in this formula will be the time for the ball or bullet to travel half the 10 meters, or 5 meters, since the formula applies to half the trajectory only (either half, since they're symmetric); since we were using the symbol "t" for the time to travel the whole 10 meters, I inserted "1/2 t" in the Newtonian formula.

Obviously this is an approximation, which assumes that g is constant in the spacetime region of interest, that the Earth's surface is flat to within the desired accuracy, and that the trajectories are parabolic. All of these assumptions are valid for the level of accuracy MTW were using (only 1 or 2 significant figures); they would *not* all be valid for the six-figure accuracy you were using, since, as you saw, g varies in the sixth figure over a height of 3 meters or so, less than the 5 meter height of the ball's trajectory. (I haven't calculated the error in the other assumptions, but I'm pretty sure the spatial curvature of the Earth, at least, shows up sooner than the sixth figure.)

 

[PeterDonis]

I spent some time trying to collect some of the concepts in your posts

A few comments/corrections:

(1) What I've been calling the "geometric mass" is usually just called "mass" or "mass in geometric units" and is usually just denoted M in the textbooks, which usually use geometric units (MTW does, for example; what you denote "M" in your table, MTW would denote $M_{conv}$ to make clear that it's in "conventional" rather than geometric units). It is *half* the Schwarzschild radius (which is usually written 2M).

(2) What I've been calling the "radius of curvature" $r_{c}$ is not one of the components of the Riemann curvature tensor; it's a scalar which can be calculated from those components, but it's an invariant (the same in all coordinate systems), which the curvature tensor components, individually, are not. (Actually, in 4-dimensional spacetime, there is in general not a single "radius of curvature", since that radius can be different along different dimensions--which is why we need a tensor to completely describe the curvature. For the specific case we've been considering, what we've been calling the "radius of curvature" should more properly be called something like the "radius of curvature in the r-t plane".)

(Edit: I should also note that the two items you've labeled as "spatial curvature" and "radius of curvature" are the same thing; the formula $s^{2} / 8 h$ from MTW for calculating the latter is just a shortcut way of calculating the former for that specific problem. Obviously you could set up problems with baseballs or bullets or other objects traveling at different speeds on different trajectories, but what you'll find is that no matter how you set the problems up, the interval s, the height h, and any other variables, will all vary together in such a way that the "radius of curvature" at a given radius r from the central mass is always the same, and is given by the formula you gave for "spatial curvature".)

(3) I haven't really discussed what you've labeled as "time curvature" yet, but for the case we've been considering, the metric component $g_{00}$, in Schwarzschild coordinates, is

$$g_{00} = g_{tt} = 1 - \frac{2 G M}{c^{2} r}$$

where I've used conventional units. The "time dilation factor", which is the amount of proper time elapsed for an observer hovering at radius r (and therefore accelerating to hold station), for a given amount of coordinate time t, is the square root of this. The expression Mentz114 gave in post #30 comes from the desire to show the correspondence between GR and Newtonian gravity; if we write the Newtonian "potential" as

$$U = - \frac{G M}{r}$$

we can see that

$$g_{tt} = 1 + \frac{2 U}{c^{2}}$$

which can be rearranged to look pretty much like what Mentz114 wrote down. (Note that I've used a "timelike" sign convention in the above, where the sign of the tt term in the metric is positive and the signs of the "spacelike" terms, rr and so forth, are negative; Mentz114's expression implicitly used the opposite convention, where the timelike terms are negative and the spacelike terms are positive, so his signs are switched around from mine.)

 

[PeterDonis]

Curvature: here MTW use the formula (radius of curvature) = (spacetime interval)^2 / (8 * vertical rise). The derivation of this formula from the components of the Riemann curvature tensor is in the text; I won't go into it here.

This statement has been nagging at me, so I went back and looked through that section of MTW more carefully (I was more or less going on memory before--I just briefly glanced at MTW to verify that the formula in Box 1.6 was the one I remembered). The statement I made in the quote above, as it stands, is false, as far as I can tell: I can't find any specific derivation of the "radius of curvature" formula from the Riemann tensor in MTW. In fact, no explicit derivation is given: there is simply a note in Box 1.6 of MTW about comparing the track to the "arc of a circle". That made me think of the following: suppose, in Euclidean geometry, you have a circle cut by a chord of length s, forming an arc of "height" (i.e., the length of the largest perpendicular from the arc to the chord) h. If r is the radius of the circle, the Pythagorean theorem gives:

$$\left( r - h \right)^{2} + \left( \frac{s}{2} \right)^{2} = r^{2}$$

We can rearrange this to (assuming I've done the algebra correctly):

$$r = \frac{s^{2}}{8 h} \left( 1 + 4 \frac{h^{2}}{s^{2}} \right)$$

which, assuming h << s, gives the formula MTW used.

Taken at face value, this means that MTW's "radius of curvature" calculation is basically hand-waving; they're just assuming that a formula from Euclidean geometry can be carried over to a calculation in a local inertial frame in GR. However, I don't think it's a coincidence that the result of the formula *exactly* equals the "Rindler horizon" value $c^{2} / g$. There probably is some connection with the curvature tensor that I haven't yet spotted that makes the "radius of curvature" formula make sense in this situation. However, it certainly isn't as simple as I had implied in my earlier post.

 

[JDoolin]

$$\begin{center} \begin{tabular}{| c | c | c | c| } \hline Formula & ind. vars & name & derivation \\ \hline r_m=\frac {GM_{conv}}{c^2} & M_{conv} & Geometric Mass & .5*Schwartzchild Radius \\ \hline r_c=\frac{c^2}{a} & a & Rindler Horizon & position of stationary event in accelerating frame \\ \hline r_c= \frac {c^2 r^2} {G M} & M, r & spatial curvature & Rindler Horizon with Newtonian Gravity \\ \hline r_c=\frac{s^2}{8 h} & s,h & radius of curvature & See attached diagram \\ \hline -(1+ c^2g00) & ? & time curvature & ? \\ \hline g_{tt}=1-\frac{2 G M_{conv}} {c^2 r} & M_{conv}, r & (grav time dilation)^2 & ? \\ \hline g_{tt}=1-\frac{2 U} {c^2} & U \in \lbrace - \infty , 0 \rbrace & (grav time dilation)^2 & ? \\ \hline \end{tabular} \end{center}$$​

I've ordered both Wald, and MTW through the library, so I'll get out the wheelbarrow soon.

I am just applying some corrections and additions to the table based on your responses. I fixed the Sch. radius and added equations for g_tt.

I also attached a diagram related to your calculation of s²/8h. I was thinking that the s was the space-time-interval, but that is definitely not the case in this diagram. Or if it is the case, it is only because $(ct)^2 - h^2 \approx (ct)^2+h^2$ for this low-velocity case.

 

[PeterDonis]

I...added equations for g_tt.

The second equation for $g_{tt}$ should have a plus sign in front of the U term; the term only becomes negative because the sign of U itself is negative (as you note in the table).

I also attached a diagram related to your calculation of s²/8h. I was thinking that the s was the space-time-interval, but that is definitely not the case in this diagram. Or if it is the case, it is only because $(ct)^2 - h^2 \approx (ct)^2+h^2$ for this low-velocity case.

The diagram as you've drawn it is backwards; the "radius" lines should go to the center of the circle.

In the diagram, of course, "s" is just a distance, as you note; but in MTW, "s" does refer to a spacetime interval (which is basically just the time interval, since it's so much larger than the space interval--here both "intervals" refer to the horizontal travel of the ball or the bullet, so the "time interval" is c times the travel time, 2 sec or 0.02 sec, and the "space interval" is just 10 meters in both cases; h only comes in when we apply the formula for radius of curvature). That's why I said the MTW argument was basically hand-waving.

 

[JDoolin]

The second equation for $g_{tt}$ should have a plus sign in front of the U term; the term only becomes negative because the sign of U itself is negative (as you note in the table).

Are you sure? The equation now has it so the gravitational time dilation is 1 when you are far away from any masses, and it increases toward infinity as you go into the black hole. The way you're saying, it would go to zero fairly quickly, (so time would be speed up in the gravitational well) and then tend toward negative infinity.

The diagram as you've drawn it is backwards; the "radius" lines should go to the center of the circle.

In the diagram, of course, "s" is just a distance, as you note; but in MTW, "s" does refer to a spacetime interval (which is basically just the time interval, since it's so much larger than the space interval--here both "intervals" refer to the horizontal travel of the ball or the bullet, so the "time interval" is c times the travel time, 2 sec or 0.02 sec, and the "space interval" is just 10 meters in both cases; h only comes in when we apply the formula for radius of curvature). That's why I said the MTW argument was basically hand-waving.

I made a new diagram.

c = 3*10^8; a = 9.8;
xmin = -c^2/a;
xmax = 5 c^2/a;
tbound = 3 c/a;
p1 = ContourPlot[(x + c^2/a)^2 - (c*t)^2 == c^4/a^2, {x, xmin, 
    xmax}, {t, -tbound, tbound}, Axes -> True];(*hyperbolic*)
p2 = ContourPlot[x == a/2 t^2, {x, xmin, xmax}, {t, -tbound, tbound}, 
   Axes -> True];(*quadratic*)
p3 = ContourPlot[
   x + c^2/a == c t, {x, xmin, xmax}, {t, -tbound, tbound}];(*linear*)
p4 = ContourPlot[
   x + c^2/a == -c t, {x, xmin, xmax}, {t, -tbound, tbound}];(*linear*)
p5 = ContourPlot[(x - c^2/a)^2 + (c t)^2 == (c^2/a)^2, {x, xmin, 
    xmax}, {t, -tbound, tbound}];(*circular*)
p6 = Graphics[{Thick, Blue, Line[{{-c^2/a, 0}, {c^2/a, 0}}]}];(*radius lines*)
Show[p1, p2, p3, p4, p5, p6]

 

[PeterDonis]

Are you sure? The equation now has it so the gravitational time dilation is 1 when you are far away from any masses, and it increases toward infinity as you go into the black hole. The way you're saying, it would go to zero fairly quickly, (so time would be speed up in the gravitational well) and then tend toward negative infinity.

Look again at post #39; the two formulas for $g_{tt}$ that I gave there have to agree, given the formula for U that I gave there (which agrees with the range you give for U). That means there needs to be a plus sign in front of the U term in the second formula, since U is negative.

You may be interpreting "time dilation" backwards from the way I was using the term. (It's kind of a difficult term since it can reasonably be interpreted either way; unfortunately, I'm not aware of any other handy colloquial term for what we're talking about.) The metric coefficient $g_{tt}$ (or more precisely its square root) tells you how much proper time elapses for each unit of coordinate time. So if you're deep in a gravity well, your $g_{tt}$ will be *less* than 1, meaning that, for example, *less* than 1 second of your proper time elapses for each second of Schwarzschild coordinate time t (the time that is experienced by observers very far away from the gravity well). (As I said, "time dilation" is probably not the best name for this effect, but I don't know of any other handy name for it.) This corresponds to U getting more and more *negative* as you get deeper into the gravity well; but again, the sign in front of U in the formula for $g_{tt}$ needs to be a plus sign for $g_{tt}$ to get smaller as U gets more and more negative.

It is true, by the way, that $g_{tt}$ goes to zero as the radius r goes to the Schwarzschild radius, and becomes negative (with the sign convention we're using) inside that radius, ultimately going to negative infinity as the radius goes to zero. This is one of the reasons that Schwarzschild coordinates are not good coordinates for doing physics close to or inside a black hole horizon. We've only been dealing with situations where the radius is much larger than the Schwarzschild radius, so we haven't had to deal with that problem.

I made a new diagram.

The circle part looks good.

Edit: OK, I see where the rest of the diagram is coming from. Basically, we have three curves that all look the same in the vicinity of the events in question: the hyperbola, the parabola, and the circle:

$$\left( x + r \right)^{2} - \left( c t \right)^{2} = r^{2}$$

$$x = \frac{ \left( c t \right)^{2} }{2 r}$$

$$\left( x - r \right)^{2} + \left( c t \right)^{2} = r^{2}$$

where I've written r, the "radius of curvature", in place of $c^2 / a$. These equations can be rearranged as follows:

$$r = \frac{ \left( c t \right)^{2} - x^{2} }{2 x}$$

$$r = \frac{ \left( c t \right)^{2} }{2 x}$$

$$r = \frac{ \left( c t \right)^{2} + x^{2} }{2 x}$$

When x is small enough, all three of these equations reduce to the second, which is basically MTW's formula (the only difference being, as I noted in a previous post, that MTW's time value "t" refers to a time twice as long as the "t" that appears in the above equations, so a factor of 8 appears in their denominator instead of 2).

 

[pervect]

What page of MTW is the bit about "radius of curvature"? If it is in conjunction with the Rindler space-time, I'm sure they weren't talking about the Riemann - because the Riemann tensor of that space-time is zero.

 

[PeterDonis]

What page of MTW is the bit about "radius of curvature"? If it is in conjunction with the Rindler space-time, I'm sure they weren't talking about the Riemann - because the Riemann tensor of that space-time is zero.

It's pp. 32-33 in my edition, Box 1.6, where they talk about the "radius of curvature" of the path of a ball and a bullet in the gravitational field of the Earth. I thought at first, based on my (obviously imperfect) memory of that section, that there was at least something there about relating the radius of curvature to the Riemann tensor, but as I noted in post #40, when I checked my memory by re-reading the section in detail, I couldn't find anything.

 

[JDoolin]

Look again at post #39; the two formulas for $g_{tt}$ that I gave there have to agree, given the formula for U that I gave there (which agrees with the range you give for U). That means there needs to be a plus sign in front of the U term in the second formula, since U is negative.

You may be interpreting "time dilation" backwards from the way I was using the term. (It's kind of a difficult term since it can reasonably be interpreted either way; unfortunately, I'm not aware of any other handy colloquial term for what we're talking about.)

Indeed I was interpreting "time dilation" backwards from the way you were using it. I think in Special Relativity, the time-dilation-factor is usually:

$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}= \frac {\Delta t_{observer(local)}}{\Delta t_{observed(distant)}}$$​

The passage of time for the observer is faster than the passage of time for the observed. The input variable is the relative velocity v, and the output has a range from $$(1,\infty)$$

while the gravitational based time dilation you are using, if you consider both observers, is

$$\frac {\Delta t_{observed(distant)}}{\Delta t_{observer(local)}}=\frac{c^2+2U_{observed}}{c^2+2U_{observer}}$$​

(This simplifies to the equation we were using before if you plug in $$U_{observer} = 0$$)

The input variables are $$U_{observer} \in (-\infty,0)$$ and $$U_{observed} \in (-\infty,0)$$ and the output has a range from $$(-\infty,\infty )$$.

 

[JDoolin]

$$1-\frac{2 G M_{conv}} {c^2 r}$$

G in N m^2 / kg^2 = m^3 / (s^2 kg)
M in kg
c in m/s
r in m

U=- G M / r
G in m^3/(s^2 kg)
M in kg
r in m
U in m^2/s^2 or Joules/kg

U is in Joule's per kilogram, which simplifies to m^2/s^2. Analtogous to voltage, V, in Joules per Coulomb, which doesn't simplify.

 

[PeterDonis]

I guess we're technically using potential energy per unit mass; sort of an equivalent to Voltage, but for mass.

Yes, the units of the "potential" U are velocity squared; that's true in Newtonian physics as well as relativity. The reason we can get away with this is the equivalence principle--more precisely, the version of it that says (in Newtonian terms) that all objects fall with the same acceleration in a gravitational field. That allows us to factor out the "test mass", the mass of the object we are using to probe the field, and do everything in units "per unit mass"--not just energy, but also, in more complicated problems, other quantities like angular momentum.

As you say, this is analogous to how voltage is used in problems dealing with electricity. However, there is no analogue of the equivalence principle for electricity; there are objects with different charge to mass ratios, which therefore "fall" with different accelerations in an electric field. It just happens that for many problems, we can ignore this because all the moving charge carriers have the same charge/mass ratio (typically they are all electrons), so we can factor it out and use voltage instead of the actual electric field.

 

[JDoolin]

Yes, the units of the "potential" U are velocity squared; that's true in Newtonian physics as well as relativity. The reason we can get away with this is the equivalence principle--more precisely, the version of it that says (in Newtonian terms) that all objects fall with the same acceleration in a gravitational field. That allows us to factor out the "test mass", the mass of the object we are using to probe the field, and do everything in units "per unit mass"--not just energy, but also, in more complicated problems, other quantities like angular momentum.

As you say, this is analogous to how voltage is used in problems dealing with electricity. However, there is no analogue of the equivalence principle for electricity; there are objects with different charge to mass ratios, which therefore "fall" with different accelerations in an electric field. It just happens that for many problems, we can ignore this because all the moving charge carriers have the same charge/mass ratio (typically they are all electrons), so we can factor it out and use voltage instead of the actual electric field.

I guess I've never been quite comfortable with the difference between "potential" and "potential energy"

"potential energy" has units of $kg \frac{m^2}{s^2}$ and, if I'm not mistaken, "potential" doesn't get introduced until it is introduced as "electric potential" with Coulomb's Law in most introductory physics books.

In any case, it's beginning to dawn on me how acceleration-based time dilation occurs in Rindler coordinates.

In the attached diagram, you see one observer traveling along arc AC, another along BD. Both observers view Events A&B to be simultaneous, and events C&D to be simultaneous. However, both also agree that BD takes much longer than AC.

Now I need to find the quantitative relationship between those two times.

 

[PeterDonis]

I guess I've never been quite comfortable with the difference between "potential" and "potential energy"

"potential energy" has units of $kg \frac{m^2}{s^2}$ and, if I'm not mistaken, "potential" doesn't get introduced until it is introduced as "electric potential" with Coulomb's Law in most introductory physics books.

I agree "potential" by itself is kind of a vague term, it gets used in a number of different ways. It would probably be better if relativity texts stuck to the term "energy per unit mass", which is what "potential" usually means when gravity is the only force under consideration, but which generalizes nicely to situations where other forces are in play without changing its meaning. Energy per unit mass also has the nice property that in "relativistic units", where the speed of light = 1, it is dimensionless.

Now I need to find the quantitative relationship between those two times.

One fairly easy way to do it is to transform to Rindler coordinates, as described in the Wikipedia page:

http://en.wikipedia.org/wiki/Rindler_coordinates

In these coordinates, the hyperbolas you've drawn become lines of constant x, and the lines AB and CD become lines of constant t. Once you write the metric in those coordinates, it's easy to calculate the times AC and BD, since each of them are measured along lines where only one coordinate changes (the Rindler t, in this case), so only one term in the metric is relevant.

 

[JDoolin]

I agree "potential" by itself is kind of a vague term, it gets used in a number of different ways. It would probably be better if relativity texts stuck to the term "energy per unit mass", which is what "potential" usually means when gravity is the only force under consideration, but which generalizes nicely to situations where other forces are in play without changing its meaning. Energy per unit mass also has the nice property that in "relativistic units", where the speed of light = 1, it is dimensionless.

Forgive me for nitpicking, but resolving to 1 does not mean "dimensionless." It just means that the correlation coefficient comes out to be 1.

You might have c=1 ly/y, or c= 1ls/s. The unfortunate fact is that we don't have any word that sounds intuitively like a distance that makes the speed of light = 1. The closest I've been able to come is 1 foot/nanosecond, which is close to the speed of light.

I do believe that it is nice when quantitities come out to be 1, but even when this happens, the units are still important.

Little things you notice after teaching introductory physics for a couple semesters: For instance, Kepler's Third Law is

$$\frac{T^2}{r^3}=\frac{4 \pi^2}{G M_{sun}}$$​

which becomes

$$\frac{T^2}{r^3}=1 year^2 / AU^3$$​

if you use units of Astronomical Units for r and Years for T.

This is getting off topic, but I don't really "believe" in dimensionless quantities, even when they come out dimensionless. The coefficient of friction is not unitless, but Newtons/Newton. i.e. Newtons of frictional force per Newton of parallel force.

On the other and if you take Torque, and divide by Work, you get a unitless quantity, but as far as I know, there's really no physical application of this at all. Work involves a dot-product of F*d and torque involves a cross-product of F*d. The directions of the force and distance make these two quantities completely conceptually different, even though they happen to have the same units.

Then there's the whole discussion of radians and degrees, which in some way measure the portion of a circle. They should have some kind of exact analogy in rapidities. If anything could convince me to "believe" (so to speak) in dimensionless quantities, it would be angles and rapidities.

One fairly easy way to do it is to transform to Rindler coordinates, as described in the Wikipedia page:

http://en.wikipedia.org/wiki/Rindler_coordinates

In these coordinates, the hyperbolas you've drawn become lines of constant x, and the lines AB and CD become lines of constant t. Once you write the metric in those coordinates, it's easy to calculate the times AC and BD, since each of them are measured along lines where only one coordinate changes (the Rindler t, in this case), so only one term in the metric is relevant.

I'm looking over the Wikipedia article. Now a couple of vaguely coherent thoughts are coming to mind.

they use t to represent the time according to one of the accelerated observers whose acceleration g=1. Off the top of my head, I'm thinking 1 ly/year^2 and 1 ls/s^2 are different quantities; but I'm not sure. I am calling this observer whose acceleration is g=1 the "reference observer"

I would prefer to see the equation relating tanh(t)=T/X with all of the variables, (a, c or r=c²/a) still present.

The value for t does not represent what any clocks would say, (except for the reference observer whose g=1). I'm not sure what notation, exactly, I'm looking for to find the relative speed of the clocks. (the issue might be entirely resolved if we have the equation with a and c or r=c²/a)

Finally, it may be interesting to note, precisely, the relationship between the reference-observer's passage of time, and his rapidity.

 

[PeterDonis]

they use t to represent the time according to one of the accelerated observers whose acceleration g=1. Off the top of my head, I'm thinking 1 ly/year^2 and 1 ls/s^2 are different quantities; but I'm not sure. I am calling this observer whose acceleration is g=1 the "reference observer"

Yes, 1 ly/y^2 and 1 ls/s^2 are different quantities; 1 ly/y and 1 ls/s are the same velocity, but when you divide that by 1 y you get a different acceleration than when you divide it by 1 s--the latter is a much larger acceleration (about 30 million times larger, since that's the number of seconds in a year).

I would prefer to see the equation relating tanh(t)=T/X with all of the variables, (a, c or r=c²/a) still present.

The "relativistic units" they're using express everything in terms of length, so converting back to conventional units just means replacing every time t with ct and every acceleration a with $a / c^{2}$, then canceling c's where possible (and rearranging so the c's don't appear in front of the t's, if desired). I come up with:

$$t = \frac{c}{g} arctanh \left( \frac{cT}{X} \right), x = \sqrt{X^{2} - c^{2}T^{2}$$

$$T = \frac{x}{c} sinh \left( \frac{gt}{c} \right), X = x cosh \left( \frac{gt}{c} \right)$$

$$ds^{2} = - \frac{g^{2} x^{2}}{c^{2}} dt^{2} + dx^{2} + dy^{2} + dz^{2}$$

The value for t does not represent what any clocks would say, (except for the reference observer whose g=1). I'm not sure what notation, exactly, I'm looking for to find the relative speed of the clocks. (the issue might be entirely resolved if we have the equation with a and c or r=c²/a)

That's what the metric, $ds^{2}$, tells you. The observers in question all have constant x, y, z, so the only non-zero coordinate differential along their paths is dt. That means their proper time, which is just $\sqrt{- ds^{2} / c^{2} }$, is given by

$$\tau = \frac{gx}{c^{2}} t$$

 

[JDoolin]

Yes, 1 ly/y^2 and 1 ls/s^2 are different quantities; 1 ly/y and 1 ls/s are the same velocity, but when you divide that by 1 y you get a different acceleration than when you divide it by 1 s--the latter is a much larger acceleration (about 30 million times larger, since that's the number of seconds in a year).



The "relativistic units" they're using express everything in terms of length, so converting back to conventional units just means replacing every time t with ct and every acceleration a with $a / c^{2}$, then canceling c's where possible (and rearranging so the c's don't appear in front of the t's, if desired). I come up with:

$$t = \frac{c}{g} arctanh \left( \frac{cT}{X} \right), x = \sqrt{X^{2} - c^{2}T^{2}$$

$$T = \frac{x}{c} sinh \left( \frac{gt}{c} \right), X = x cosh \left( \frac{gt}{c} \right)$$

$$ds^{2} = - \frac{g^{2} x^{2}}{c^{2}} dt^{2} + dx^{2} + dy^{2} + dz^{2}$$



That's what the metric, $ds^{2}$, tells you. The observers in question all have constant x, y, z, so the only non-zero coordinate differential along their paths is dt. That means their proper time, which is just $\sqrt{- ds^{2} / c^{2} }$, is given by


$$\tau = \frac{gx}{c^{2}} t$$

I feel compelled to make up another table:

$$\begin{center} \begin{tabular}{| c | c | c | }\hline Quantity & Circ & Hyperbola \\ \hline Unit Circle/Hyperbola & X^2+Y^2=1 & X^2-(cT)^2=\left(\frac{c^2}{a} \right)^2 \\ \hline Circle & X^2+Y^2=r^2 & X^2-(cT)^2=x^2 \\ \hline Vertical Component & Y=r sin(\theta) & c T=x sinh \left(\frac{a t}{c}\right) \\ \hline Horizontal Component & X=r cos(\theta) & X=x cosh \left(\frac{a t}{c}\right) \\ \hline Arc Length/Proper Time & \Delta s = r \theta & c \tau = x \left(\frac{a t}{c }\right) \\ \hline \end{tabular}\end{center}$$​

In this, it appears that t takes on the role of an angle (the time passed by the accelerated observer at x=c^2/a) , while x is analogous to the role of a radius (the distances from the rindler horizon as observed by the accelerated observers) A major difference, I think, is that while arc-length increases happily with both X and Y, proper-time goes up by cT and down by X.

I know $$\theta$$ represents the arc-length divided by the radius; which seems a perfect geometric description of the angle. However, I don't know if at/c has such an easy relationship*. I know of a geometric analogy between polar angle and hyperbolic angle which is shown http://en.wikipedia.org/wiki/File:Funhipgeom.png" and I went ahead and attached it. (*Never mind, Actually, it has basically the same relationship. at/c = c tau / x. theta = Delta s / r.)

 

[PeterDonis]

I know $$\theta$$ represents the arc-length divided by the radius; which seems a perfect geometric description of the angle. However, I don't know if at/c has such an easy relationship.

Obviously it does, since your last formula says that $at / c$ is the arc length $c \tau$ (which is the proper time converted to units of length) divided by the radius $x$.

Also, the slope of the line of simultaneity through a given point on the hyperbola (meaning the line through the origin that also passes through that point), which is just the boost velocity at that point, relative to the global inertial frame, is given by

$$\frac{c T}{X} = \beta = tanh \left( \frac{a t}{c} \right)$$

So I think we can equate $at / c$ with the rapidity.

 

[JDoolin]

$$\begin{center} \begin{tabular}{| c | c | c | }\hline Quantity & Circ & Hyperbola \\ \hline radius, proper distance & X^2+Y^2=r^2 & X^2-(cT)^2=x^2 \\ \hline Unit Circle/Hyperbola & X^2+Y^2=(l_u)^2 & X^2-(cT)^2=\left(\frac{c^2}{a_u} \right)^2 \\ \hline Arc Length/Proper Time & \Delta s = r \theta & \Delta\tau = \frac{x\phi}{c} =\frac{x a_u t}{c^2}\\ \hline Angle/Rapidity & \theta= \frac{\Delta s}{r}=arctan(Y/X) & \phi = \frac{a_u t}{c}=\frac{c\Delta\tau}{x} =arctanh(cT/X)\\ \hline Vertical Component & Y= r sin(\theta) & c T=x sinh (\phi) \\ \hline Horizontal Component & X= r cos(\theta) & X=x cosh (\phi) \\ \hline \end{tabular}\end{center}$$​

Some further observations:

• The passage of proper time is affected signifcantly, as a function of proper distance, x and [strike]rapidity[/strike] acceleration.
• The distances x are exactly the same as in the inertial frame. i.e. the rindler coordinate system does not have any effect on the measurement of proper length.
• I have replaced 1 with l_u, and a with a_u to denote a unit length and a unit acceleration. The unit circle is a different entity depending on whether you use feet, meters, etc. The unit hyperbola will also be different depending on what units you use.

I think I'm now ready to apply the equivalence principle to this system. From the perspective of the accelerated body, if I'm not mistaken, it would look like g->infinity as x->0, and g->0 as x->infinity. Potential energy would be determined as Work=Force*distance; "Plain Potential" would be Work/mass=acceleration * distance. Especially of interest is to determine g(x) , U(x), and $$d \tau(x)/d T$$.

Once I understand how the relationship between time-dilation and potential energy is found in this system, I may begin to understand the relationship between time-dilation and real gravitational potential energy.

 

[PeterDonis]

The distances x are exactly the same as in the inertial frame. i.e. the rindler coordinate system does not have any effect on the measurement of proper length.

I'm not sure this is completely correct. I agree that in the metric for Rindler coordinates that I quoted above, the metric coefficient for all the space coordinates is 1, meaning that observers at constant x, y, z coordinates will observe a constant spacelike interval ds between each other along curves of constant Rindler time t. However, the physical meaning of this is a little more complicated. The Wikipedia page I linked to in an earlier post discusses various physical notions of "distance" that do not all give the same answers in Rindler coordinates, whereas they do in, for example, global inertial coordinates such as the T, X we are using for the global inertial frame.

I think I'm now ready to apply the equivalence principle to this system. From the perspective of the accelerated body, if I'm not mistaken, it would look like g->infinity as x->0, and g->0 as x->infinity. Potential energy would be determined as Work=Force*distance; "Plain Potential" would be Work/mass=acceleration * distance. Especially of interest is to determine g(x) , U(x), and $$d \tau(x)/d T$$.

I haven't gone back and checked earlier posts in this thread, so I don't know if this has been brought up before, but what you're contemplating here goes beyond the equivalence principle. (Which doesn't mean that what you're contemplating is not of interest or worth doing.) The equivalence principle only applies locally; more precisely, it only applies over a small enough region of spacetime that tidal gravity can be neglected. Of course in Rindler coordinates, which are coordinates on flat spacetime, tidal gravity is zero everywhere; but the spacetime you'll be comparing them to, presumably Schwarzschild coordinates outside a gravitating body, is *not* flat, so you'll have to restrict your comparisons to a region of that spacetime that's small enough to set up a comparison with a region of Rindler coordinates. The exact size of that region will vary with the specific case; but certainly looking at what happens as any quantity of interest goes to infinity will *not* fall within a small region.

One example of how global comparisons break down between these two systems is the following: consider two observers accelerating at 1 g, one on a rocket out in deep space, the other standing on the surface of the Earth. Both observers launch an object straight up (i.e., in the direction they are accelerating) at "escape velocity", 11 km/s. The observer in the rocket ship will eventually catch up with the object and pass it. The observer on the Earth will see the object recede forever. Yet these observers are equivalent as far as the equivalence principle is concerned.

 

[JDoolin]

I'm not sure this is completely correct. I agree that in the metric for Rindler coordinates that I quoted above, the metric coefficient for all the space coordinates is 1, meaning that observers at constant x, y, z coordinates will observe a constant spacelike interval ds between each other along curves of constant Rindler time t. However, the physical meaning of this is a little more complicated. The Wikipedia page I linked to in an earlier post discusses various physical notions of "distance" that do not all give the same answers in Rindler coordinates, whereas they do in, for example, global inertial coordinates such as the T, X we are using for the global inertial frame.

The distances in x, and the values of X match at time T=t=tau=0. I think the wikipedia article refers to this as the "ruler distance"

These other notions of distance are interesting, too. I strongly suspect that if you were looking up, the same ruler-distances would look further than the same ruler-distances looking down. However, if you consider the radar-distance; timing a signal as it takes to reflect back. The radar distance would appear further going down then up than going up then down. Since the further apparent distance results in a shorter radar-distance, that seems a bit counter-intuitive; I may have to do some math to verify that.

I haven't gone back and checked earlier posts in this thread, so I don't know if this has been brought up before, but what you're contemplating here goes beyond the equivalence principle. (Which doesn't mean that what you're contemplating is not of interest or worth doing.) The equivalence principle only applies locally; more precisely, it only applies over a small enough region of spacetime that tidal gravity can be neglected.

Of course in Rindler coordinates, which are coordinates on flat spacetime, tidal gravity is zero everywhere; but the spacetime you'll be comparing them to, presumably Schwarzschild coordinates outside a gravitating body, is *not* flat, so you'll have to restrict your comparisons to a region of that spacetime that's small enough to set up a comparison with a region of Rindler coordinates. The exact size of that region will vary with the specific case; but certainly looking at what happens as any quantity of interest goes to infinity will *not* fall within a small region.

One example of how global comparisons break down between these two systems is the following: consider two observers accelerating at 1 g, one on a rocket out in deep space, the other standing on the surface of the Earth. Both observers launch an object straight up (i.e., in the direction they are accelerating) at "escape velocity", 11 km/s. The observer in the rocket ship will eventually catch up with the object and pass it. The observer on the Earth will see the object recede forever. Yet these observers are equivalent as far as the equivalence principle is concerned.

In "comparing" an infinite uniform gravitational field to a real gravitational field, I meant to highlight the differences; not just the similarities. You make an excellent point, bringing up tidal gravitation; if I'm not mistaken, this is dg/dr. However, notice that we have established that the rindler coordinates also has a nonzero dg/dx, since the born rigid system must have different accelerations at different ruler distances.

 

[JDoolin]

I think I can establish that in the inertial frame, the structure's acceleration at any given point X, at T=0 is given as:

$$g(X)=\frac{c^2}{X}$$​

I think this is the value of

$$g(X)=\frac{d^2 X}{dT^2}$$​

Where (X,T) represents the path of a particular accelerated object.

However, what is the appropriate value of the gravity g(x) in the accelerating frame? We want to use the local time tau, instead of the inertial coordinate time T.

$$g(x)=\frac{d^2 x_{obj}}{d \tau^2}$$​

We need to consider not dx, because in the accelerated frame, x is the ruler distance, which is not moving. Instead we need to consider the position, x_obj, of something that is stationary in the inertial frame, but it's position will change relative to x over time.

I'm kind of speculating that once we account for this, we'll find that the measured gravity will be the same for everybody on board the rocket, even though it looks like different accelerations to the inertial observer.

However, since tau gets faster and acceleration gets slower as you go up, I think it won't come out so nice. Or maybe it will. I'm not working through it real clearly just yet.

 

[PeterDonis]

You make an excellent point, bringing up tidal gravitation; if I'm not mistaken, this is dg/dr. However, notice that we have established that the rindler coordinates also has a nonzero dg/dx, since the born rigid system must have different accelerations at different ruler distances.

Not quite. It's true that, for the particular case we're considering, the Schwarzschild spacetime outside a gravitating body, the tidal gravity happens to equal dg/dr; but that's not true in general, and the fact that dg/dx is also nonzero in Rindler coordinates does not mean that there is nonzero tidal gravity in flat spacetime (which is what Rindler coordinates apply to).

The general rule is that "tidal gravity" is just another name for "spacetime curvature"; so tidal gravity is only nonzero if spacetime is curved, which it is outside a gravitating body, but is not (obviously) in flat spacetime, even for accelerating observers. Mathematically, tidal gravity/spacetime curvature is quantified by the Riemann tensor and various other tensors formed from it. In flat spacetime the Riemann tensor is zero everywhere; if we were to calculate its components in Rindler coordinates that's what we would find. In Schwarzschild spacetime, the particular component of the Riemann tensor that gives what we've been calling dg/dr is $R^{r}_{trt}$, which, when you work through the math, turns out to equal dg/dr. (Actually, it may only equal dg/dr for r much greater than the Schwarzschild radius, which of course it is for the cases we've been considering; for values of r close to the horizon, there is an extra factor in the denominator in the formula for g, which makes it diverge to infinity at the horizon, but all the Riemann tensor components are finite and well-behaved at the horizon. I'll have to go back and check the exact formulas.)

 

[JDoolin]

However, notice that we have established that the rindler coordinates also has a nonzero dg/dx, since the born rigid system must have different accelerations at different ruler distances.

Actually, what we established was that there was a nonzero dg/dX when g was based on the acceleration in the inertial frame. But because the rate of proper time $\tau$ is different at those different x values, this means that in the accelerated frame, the local value of g is also different at those different x values.

Instead of

$$g(x)=\frac{d^2 X_{rocketfloor}}{d T_{inertialclock}^2}$$​

we want:

$$g(x)=\frac{d^2 x_{inertialobject}}{d \tau_{rocketclock}^2}$$​

Not quite. It's true that, for the particular case we're considering, the Schwarzschild spacetime outside a gravitating body, the tidal gravity happens to equal dg/dr; but that's not true in general, and the fact that dg/dx is also nonzero in Rindler coordinates does not mean that there is nonzero tidal gravity in flat spacetime (which is what Rindler coordinates apply to).

Whether we claim zero tidal gravity, or not, I think it is important to explicitly define our variables. In the rindler coordinates, neither dg/dr, nor dg/dx are sufficient, because g is a function of the local rate of time.

The general rule is that "tidal gravity" is just another name for "spacetime curvature"; so tidal gravity is only nonzero if spacetime is curved, which it is outside a gravitating body, but is not (obviously) in flat spacetime, even for accelerating observers.

I thought it was obvous that spacetime was flat, even for accelerating observers, but now we have this nonzero dg/dX in the inertial frame which means it is no longer obvious that dg/dx is zero in the accelerated frame.

 

[PeterDonis]

I thought it was obvous that spacetime was flat, even for accelerating observers, but now we have this nonzero dg/dX in the inertial frame which means it is no longer obvious that dg/dx is zero in the accelerated frame.

I think it will indeed turn out when you do the computation you describe in post #59 that dg/dx will *not* be zero in the accelerated frame. The reason is simple: observers at different Rindler coordinates x certainly *feel* different accelerations, and that's what the function g(x) is supposed to capture--the acceleration that an observer at a given x *feels* (would measure with an accelerometer, etc.). So I expect g to vary with x.

But even assuming this is true, as I said before, it does *not* mean that the spacetime is curved. You can't change flat spacetime to curved spacetime by changing your coordinate system (or your state of motion); whether spacetime is flat or curved is a geometric invariant, just like the curvature of the Earth's surface. If you want, we can verify this by explicitly computing the components of the Riemann tensor in Rindler coordinates and verifying that they are all zero (or maybe a kind soul will point us to someplace where that computation has already been done).

 

[JDoolin]

I think it will indeed turn out when you do the computation you describe in post #59 that dg/dx will *not* be zero in the accelerated frame. The reason is simple: observers at different Rindler coordinates x certainly *feel* different accelerations, and that's what the function g(x) is supposed to capture--the acceleration that an observer at a given x *feels* (would measure with an accelerometer, etc.). So I expect g to vary with x.

I think I have a possible approach to the problem, somewhat inspired by your earlier reference to Misner Thorne Wheeler. To find the "gravity" measured by occupants of the rocket at any given x, we can use a diagram similar to the one attached, and use

$$(x_2 - x_1) = \frac{1}{2} a \tau ^2$$

I think that $\tau$ is fully determined by x₁ and $\Delta x$, so we should be able to find an explicit value of a for any given x.

But even assuming this is true, as I said before, it does *not* mean that the spacetime is curved. You can't change flat spacetime to curved spacetime by changing your coordinate system (or your state of motion); whether spacetime is flat or curved is a geometric invariant, just like the curvature of the Earth's surface. If you want, we can verify this by explicitly computing the components of the Riemann tensor in Rindler coordinates and verifying that they are all zero (or maybe a kind soul will point us to someplace where that computation has already been done).

That would be fine, I suppose, with the caveat that I'm effectively tensor illiterate. I got an MS degree in math and physics, but none of my professors ever talked about tensors. PS, I have looked at a few of the Susskind lectures on YouTube, and I am at least skimming MTW, and Wald now, but so far, it's all a little too abstract for me.

 

[SinghRP]

Newtonian equation motion in a gravitational field, written in full:

(Inertial mass) * (Acceleration) = (Intensity of gravitational field) * (Gravitational mass).

It is only when there is numerical equality between the inertial and gravitational mass that the acceleration is independent of the nature of the body. This is the Principle of Equivalence, which has double meanings.
Reference: The Meaning of Relativity, Albert Einstein.

 

[Mentz114]

That would be fine, I suppose, with the caveat that I'm effectively tensor illiterate.

You don't have to be. You understand the concept of spacetime and the geometric invariant ds². All that's needed is to make the leap to curved spacetime, where one *must* use covariant *and* contravariant components to define a geometric invariant.

I tried to explain this in post#9 and #11 in this thread https://www.physicsforums.com/showthread.php?t=431843

Do dimensional indexes throw you, i.e. writing t as x⁰, x as x¹, y as x² and so on ? Or the summation convention x^ax_a=x⁰x₀+x¹x₁+x²x₂+x³x₃ ?

If you hate all the indices try the diff. geom. approach.

 

[SinghRP]

I read discussions under General Physics, Classical Physics, and Specia Relativityl & General Relativity. I THINK there are several slightly varying interpretations of mass, force, accelaration, gravity, etc.

Gamow's One, Two, Three, Infinity raised my curiosity well enough to help me decide to become a physicist. Universities turned me into a marching physicist. Wigner's Symmetries and Reflections made me a humble physicist. I strongly recommend Wigner's essays. What an insigthful book!

 

[JDoolin]

I think it will indeed turn out when you do the computation you describe in post #59 that dg/dx will *not* be zero in the accelerated frame. The reason is simple: observers at different Rindler coordinates x certainly *feel* different accelerations, and that's what the function g(x) is supposed to capture--the acceleration that an observer at a given x *feels* (would measure with an accelerometer, etc.). So I expect g to vary with x.

Well, I think I've come to a conclusion; right now, the result seems surprising to me, but probably won't to you.

$$\begin{matrix} x cosh(\phi)=x+h \\ \phi=arccosh\left ( \frac{x+h}{x}\right )\\ \Delta \tau = \frac{x \phi}{c}\\ a = \frac{2 h}{\Delta \tau^2}\\ a=\lim_{h \to 0}\frac{2 h c^2}{x^2 ArcCosh^2(\frac{x+h}{x})}=c^2/x \end{matrix}$$​

(The attached diagram might help.)

It comes out so whether you do the calculation by (X and T) or by (x and $\tau$), you get the very same result for the acceleration.

 

[SinghRP]

General relativity has it that the spacetime continuum is curved. The physics of continuum is dealt with [stress] tensors.
My questions:
(1) The presence of a mass creates the curvature. By how?
(2) If the curvature due to matter is positive, is the sign due to antimatter negative?
(3) Do other fundamental interactions (the strong, the weak, and electromagnetic) create curvatures in their respective fields?

 

[Mentz114]

General relativity has it that the spacetime continuum is curved. The physics of continuum is dealt with [stress] tensors.
My questions:
(1) The presence of a mass creates the curvature. By how?
(2) If the curvature due to matter is positive, is the sign due to antimatter negative?
(3) Do other fundamental interactions (the strong, the weak, and electromagnetic) create curvatures in their respective fields?

Do you want to start another topic with these questions - you are hijacking someone else's topic.

 

[JDoolin]

You don't have to be. You understand the concept of spacetime and the geometric invariant ds². All that's needed is to make the leap to curved spacetime, where one *must* use covariant *and* contravariant components to define a geometric invariant.

I tried to explain this in post#9 and #11 in this thread https://www.physicsforums.com/showthread.php?t=431843

Do dimensional indexes throw you, i.e. writing t as x⁰, x as x¹, y as x² and so on ? Or the summation convention x^ax_a=x⁰x₀+x¹x₁+x²x₂+x³x₃ ?

If you hate all the indices try the diff. geom. approach.

I posted a reply (post 13) there in the other thread; yes I still want to understand tensors.