Single equation for space-time curvature?

[PhysicsStuff]

Is there a single equation that can model both spatial and temporal metric contraction simultaneously? And also what's that equation that can model the actual degree of curvature n space-time that uses trig functions and how do you use that in combination with the two previously mentioned equations to model the dimensional metric contraction caused by curvature?

 

[PeterDonis]

Can you give a specific example of a situation in which "spatial and temporal metric contraction" occurs?

 

[PhysicsStuff]

Can you give a specific example of a situation in which "spatial and temporal metric contraction" occurs?

Uh...ok...gravity? Traveling over 1% the speed of light (observably, anyway)? You put a ruler stick in a higher gravity field and from an outside frame of reference it looks smaller while simultaneously clocks run slower. You travel near the speed of light you are in a 4-dimension-ally rotated frame such that your ruler appears smaller and your clock is slower to an outside observer that measures you traveling near the speed of light.

 

[K^2]

The thing about General Relativity is that you can do a general coordinate system transformation, and still be able to get a metric out of it. There isn't really a formula for a specific transform you can do, because it's just whatever you need. You do have to figure out the new metric, of course. But the way you do that is by making sure that all possible scalar products are preserved.

 

[PhysicsStuff]

The thing about General Relativity is that you can do a general coordinate system transformation, and still be able to get a metric out of it. There isn't really a formula for a specific transform you can do, because it's just whatever you need. You do have to figure out the new metric, of course. But the way you do that is by making sure that all possible scalar products are preserved.

Well I mean you rotate a space-time coordinate system a certain amount to observe a certain specific metric contraction though right? Let's say both axis rotate 10 degrees from 20 miles away. What's are the new metric contractions from an observer 20 miles away?

 

[Nugatory]

Uh...ok...gravity? Traveling over 1% the speed of light (observably, anyway)? You put a ruler stick in a higher gravity field and from an outside frame of reference it looks smaller while simultaneously clocks run slower. You travel near the speed of light you are in a 4-dimension-ally rotated frame such that your ruler appears smaller and your clock is slower to an outside observer that measures you traveling near the speed of light.

You are describing two different effects governed by two different sets of equations.

The gravitational case is described by the Einstein field equations, which specify how spacetime is curved and how this curvature is represented in the metric. This is general relativity.

The length contraction and time dilation effect has no curvature effects and the metric doesn't change, and is described by the Lorentz transformations of special relativity.

Neither can be described in a single equation in the way that (I think) you're asking for.

The Einstein field equations can be written as $G=8\pi{T}$, but neither $G$ nor $T$ are simple variables. They're multi-component tensors and that innocent-looking equality is actually a compact representation of a collection of shriekingly difficult non-linear differential equations.

The Lorentz transforms can be written as a single matrix equation, but that matrix equation is a compact representation of the four transformation equations, one for each coordinate, that you see more often in introductory texts. By choosing the relative velocity in the direction of one of the axes, you can get it down to two equations, but that's about it.

 

[Nugatory]

Well I mean you rotate a space-time coordinate system a certain amount to observe a certain specific metric contraction though right? Let's say both axis rotate 10 degrees from 20 miles away.

Interesting that you're talking in terms of "rotation"... The Lorentz transformations are not rotations.

Long ago SR was sometimes taught using a formalism in which the $t$ coordinate was imaginary, an $ict$ term appeared in various equations, and the Lorentz transforms in Minkowski space had the same formal appearance as a rotation in Euclidean space. This approach has pretty much fallen out of favor because spacetime is not Euclidean and trying to understand SR in terms of Euclidean rotations is a dead end, something you have to unlearn, before you can tackle GR.

 

[PhysicsStuff]

You are describing two different effects governed by two different sets of equations.

The gravitational case is described by the Einstein field equations, which specify how spacetime is curved and how this curvature is represented in the metric. This is general relativity.

The length contraction and time dilation effect has no curvature effects and the metric doesn't change, and is described by the Lorentz transformations of special relativity.

Neither can be described in a single equation in the way that (I think) you're asking for.

The Einstein field equations can be written as $G=8\pi{T}$, but neither $G$ nor $T$ are simple variables. They're multi-component tensors and that innocent-looking equality is actually a compact representation of a collection of shriekingly difficult non-linear differential equations.

The Lorentz transforms can be written as a single matrix equation, but that matrix equation is a compact representation of the four transformation equations, one for each coordinate, that you see more often in introductory texts. By choosing the relative velocity in the direction of one of the axes, you can get it down to two equations, but that's about it.

Well I mean I was looking up stuff like this http://www.phy.olemiss.edu/HEP/QuarkNet/time.html
http://www.phy.olemiss.edu/HEP/QuarkNet/length.html
and while I was looking something up I ound more complicated equations to model the angle o curvature, literally the actual "angle", theta, modeled using a sine function. And so, I was wondering how exactly can you combine the two different views to say "length and time metrics contract because space-time 4 dimensiontally rotates. Rotating space-time at a certain angle will give you a certain metric contraction".

Interesting that you're talking in terms of "rotation"... The Lorentz transformations are not rotations.

Long ago SR was sometimes taught using a formalism in which the $t$ coordinate was imaginary, an $ict$ term appeared in various equations, and the Lorentz transforms in Minkowski space had the same formal appearance as a rotation in Euclidean space. This approach has pretty much fallen out of favor because spacetime is not Euclidean and trying to understand SR in terms of Euclidean rotations is a dead end, something you have to unlearn, before you can tackle GR.

Yeah it's not a Euclidean rotation, but it's a 4-dimensional rotation...you can't just use polar coordinates because both axis of the coordinate system are rotating...and those axis represent a total of 4 dimensions...
And I think that's how Einstein described it as one point, he said gravity was a fictitious force and that actually a gravitational field is an accelerated frame of reference caused by the rotation of space-time.

 

[PeterDonis]

Uh...ok...gravity? Traveling over 1% the speed of light (observably, anyway)?

These are two different kinds of phenomena; traveling fast is not the same as being in a gravity field. See below.

You put a ruler stick in a higher gravity field and from an outside frame of reference it looks smaller

Does it? How would you measure this? There is a difference between this and the situation with clocks--see next comment.

while simultaneously clocks run slower.

This can be directly measured: have two observers exchange round-trip light signals, one far out in empty space and another deep in a gravity well. The second observer will see fewer ticks of his clock elapse between successive round-trip light signals than the first observer.

You travel near the speed of light you are in a 4-dimension-ally rotated frame such that your ruler appears smaller and your clock is slower to an outside observer that measures you traveling near the speed of light.

This is not the same as being in a gravity well. First, with respect to clocks, there is no way to directly measure the clock effect the way I described it in the gravity case above. (If you disagree, please describe explicitly how you would measure it.) But that's not as important as the next point.

The more important point is, unlike the case with gravity, both effects, even if we can settle on how they are to be physically measured, are symmetric: if you and I are in relative motion, you see my clocks run slow and my rulers contracted, and I see your clocks run slow and your rulers contracted. But the two observers I described above in the gravity case are *not* symmetric with respect to their clocks: the observer far out in empty space sees the clock of the observer in the gravity well run slow, but the observer in the gravity well sees the clock of the observer far out in empty space run *fast*. So the analogy you are trying to draw between the two kinds of phenomena fails.

 

[K^2]

Boosts are hyper-rotations. A lot of the algebra is similar, but distinction is important.

 

[PeterDonis]

And I think that's how Einstein described it as one point, he said gravity was a fictitious force

Yes.

and that actually a gravitational field is an accelerated frame of reference caused by the rotation of space-time.

No. If nothing else, a real gravitational field is not uniform: for example, the "acceleration due to gravity" of the Earth points in different directions at different points. But a "rotation of spacetime", at least as you're using the term, must be in the same direction everywhere.

What you *can* say is that, in a sufficiently small region of spacetime (small enough that tidal effects--the effects of changes in the magnitude or direction of the gravity field--are negligible), being at rest in an accelerated frame is equivalent to being at rest in a gravity field. That's what Einstein meant when he said that gravity was a fictitious force. But this equivalence only holds locally, not globally.

 

[pervect]

Is there a single equation that can model both spatial and temporal metric contraction simultaneously? And also what's that equation that can model the actual degree of curvature n space-time that uses trig functions and how do you use that in combination with the two previously mentioned equations to model the dimensional metric contraction caused by curvature?

There are 10 simultaneous partial differential equations involving the 10 metric coefficients g_ij, and their first and second partial derviatves with respect to time and space. This is the "left side" of Einstein's equation,

G_ij = 8 pi T_ij

The capital G_ij is not the same as the lower case g_ij G_ij is the "Einstein tensor". The equation above becomes a partial differential tensor equation when one expands each of the G_ij in some basis in terms of the g_ij and their first and second partial derivatives.

In a vacuum, the right hand side of the equation, i.e. all the T_ij are zero. More generally the T_ij represent the matter distribution.

By some counts there would be 16 g_ij, as i and j can each take on the values 0,1,2,3. But because of the symmetry condition g_ij = g_ji, there are only 10 "different" metric coefficients.

While there are 10 equations, they're not independent, there is one "constraint equation", so there are actually 9 indepenent equations and 9 variables. Similar remarks can be made about the right hand side, T_ij.

The 10 g_ij are the metric coefficients - I'm not sure why you refer to them as "metric contractions", that's not a standard term. Possibly you are thinking of something else when you ask about "metric contraction" (perhaps related to time dilation?), but it's not clear to me what you might be thinking of. I'm concerned that you may have intended to asked a different question than the one I answered due to semantic issues.

The metric coefficients are a quadratic form that gives the square of the "distance" (more precisely the Lorentz interval) between two points.

Thus, if you have (t,x,y,z) as coordinates, there will be some expression for the square of the lorentz interval, ds^2

ds^2 = g_00 dt^2 + 2 g_01 dt dx + 2 g_02 dt dy + 2 g_03 dt dz + g_11 dx^2 + 2 g_12 dx dy + 2 g_13 dx dz + g_22 dy^2 + 2 g_23 dy dz + g_33 dz^2

the 10 numbers (g_00, g_01, g_02, g_03, g_11, g_12, g_13, g_22, g_23, g_33) are the metric coefficients.

ds^2 could be the square of a distance, or the square of a proper time, depending on your sign conventions.

 

[K^2]

That's what Einstein meant when he said that gravity was a fictitious force. But this equivalence only holds locally, not globally.

Local equivalence is all that matters for there to be a fictitious force, though. That might be what you meant, but I wanted to clarify.

It also just occurred to me that the confusion with rotation and gravity might have come from a centrifuge analogy. If you write Minkowski metric in polar coordinates, and then go to a rotating frame of reference, you will get "gravity" terms that are equivalent to Centrifugal and Coriolis forces.

 

[PhysicsStuff]

These are two different kinds of phenomena; traveling fast is not the same as being in a gravity field. See below.

it's very strange you say that because I thought you're suppose to be able to treat acceleration and gravitational field strength the same because of the whole gravity being fictitious force thing...
http://en.wikipedia.org/wiki/Equivalence_principle

(Inertial mass) \ (Acceleration)) = (Intensity of the gravitational field) \ (Gravitational mass))

Does it? How would you measure this? There is a difference between this and the situation with clocks--see next comment.

Uh yeah I'm pretty sure that's how it works, I've had more than one person tell me that. You measure it simply by seeing that a ruler is smaller than it should be and that a clock is slower than it should be. They've measured that clocks go faster with very precise clocks in above the Earth...

This can be directly measured: have two observers exchange round-trip light signals, one far out in empty space and another deep in a gravity well. The second observer will see fewer ticks of his clock elapse between successive round-trip light signals than the first observer.

But the point is, if I accelerate near the speed of light, an outside observer would observe the same time dilation as if I were in an intense gravitational field...

This is not the same as being in a gravity well. First, with respect to clocks, there is no way to directly measure the clock effect the way I described it in the gravity case above. (If you disagree, please describe explicitly how you would measure it.) But that's not as important as the next point.

Ok, I do disgree, http://en.wikipedia.org/wiki/Time_dilation#Experimental_confirmation
read the "experimental confirmation" section.

The more important point is, unlike the case with gravity, both effects, even if we can settle on how they are to be physically measured, are symmetric: if you and I are in relative motion, you see my clocks run slow and my rulers contracted, and I see your clocks run slow and your rulers contracted. But the two observers I described above in the gravity case are *not* symmetric with respect to their clocks: the observer far out in empty space sees the clock of the observer in the gravity well run slow, but the observer in the gravity well sees the clock of the observer far out in empty space run *fast*. So the analogy you are trying to draw between the two kinds of phenomena fails.

But...if both people are moving at the same speed, their clocks and rulers will agree with each other, but disagree with an outside frame of reference, and the same is true of them being in a gravitational field...

There are 10 simultaneous partial differential equations involving the 10 metric coefficients g_ij, and their first and second partial derviatves with respect to time and space - and the distribution of mass-energy, as expressed by the stress-energy tensor, T_ij, which appears on the right hand side.

By some counts there would be 16 g_ij, as i and j can each take on the values 0,1,2,3. But because of the symmetry condition g_ij = g_ji, there are only 10 "different" metric coefficients.

While there are 10 equations, they're not independent, there is one "constraint equation", so there are actually 9 indepenent equations and 9 variables. There is also one constraint equation on the stress-energy tensor T_ij.

The 10 g_ij are the metric coefficients - I'm not sure why you refer to them as "metric contractions", that's not a standard term. Possibly you are thinking of something else when you ask about "metric contraction" (perhaps related to time dilation?), but it's not clear to me what you might be thinking of. I'm concerned that you may have intended to asked a different question than the one I answered due to semantic issues.

The metric coefficients are a quadratic form that gives the square of the "distance" (more precisely the Lorentz interval) between two points.

Thus, if you have (t,x,y,z) as coordinates, there will be some expression for the square of the lorentz interval, ds^2

ds^2 = g_00 dt^2 + 2 g_01 dt dx + 2 g_02 dt dy + 2 g_03 dt dz + g_11 dx^2 + 2 g_12 dx dy + 2 g_13 dx dz + g_22 dy^2 + 2 g_23 dy dz + g_33 dz^2

the 10 numbers (g_00, g_01, g_02, g_03, g_11, g_12, g_13, g_22, g_23, g_33) are the metric coefficients.

ds^2 could be the square of a distance, or the square of a proper time, depending on your sign conventions.

Alright do you have a link that can more precisely and in a more organized way display the equations you are trying to show? When I say metric contraction I am trying to generalize length contraction and time dilation simultaneously, since you would measure time passing slower via a meter of time appearing to shrink.

 

[WannabeNewton]

Yeah it's not a Euclidean rotation, but it's a 4-dimensional rotation...you can't just use polar coordinates because both axis of the coordinate system are rotating...and those axis represent a total of 4 dimensions...

There is some merit to what you're trying to say but what you have actually stated is inaccurate. Allow me to make it more rigorous. In Minkowski space-time, an arbitrary boost can be represented by the vector field $\xi^{\mu} = 2E^{[\nu}u^{\mu]}x_{\nu}$ where $E^{\mu}$ is a space-like vector field, $u^{\mu}$ is the 4-velocity of an inertial observer, and $x^{\mu}$ is the position vector field relative to this inertial observer. If $\eta^{\mu}$ is a constant vector field on Minkowski space-time then it can be shown (with little effort) that $\mathcal{L}_ {\xi} \eta^{\mu} = 0$ if and only if $\eta^{\mu}$ is orthogonal to both $u^{\mu}$ and $E^{\mu}$, where $\mathcal{L}_{\xi}$ is the Lie derivative along $\xi^{\mu}$. What this means is that boosts generated by $\xi^{\mu}$ rotate all codimension 2 submanifolds to which $u^{\mu}$ and $E^{\mu}$ are tangent. In this sense the boost is within the plane determined by $u^{\mu}$ and $E^{\mu}$ and rotates this plane. If we had a simple boost along the $x$-axis of this inertial observer's frame then this plane would correspond to the $t$-$x$ plane.

In contrast, an arbitrary rotation can be represented by $\xi^{\nu} = \epsilon^{\mu\nu\alpha\beta}x_{\mu}u_{\alpha}B_{\beta}$ where $B^{\mu}$ is also a space-like vector field. It can similarly be shown with little effort that $\mathcal{L}_{\xi}\eta^{\mu} = 0$ if and only if $\eta^{\mu} = \alpha u^{\mu} + \beta B^{\mu}$ hence the rotations generated by $\xi^{\mu}$ rotate all codimension 2 submanifolds to which $u^{\mu}$ and $B^{\mu}$ are orthogonal. And in this sense $B^{\mu}$ determines the axis of rotation.

And I think that's how Einstein described it as one point, he said gravity was a fictitious force and that actually a gravitational field is an accelerated frame of reference caused by the rotation of space-time.

This is a nonsensical statement.

 

[Nugatory]

traveling fast is not the same as being in a gravity field.

it's very strange you say that because I thought you're suppose to be able to treat acceleration and gravitational field strength the same because of the whole gravity being fictitious force thing...

Traveling fast at a constant velocity is a very different thing than acceleration, so the equivalence principle between gravitation and acceleration doesn't contradict PeterDonis.

 

[PhysicsStuff]

There is some merit to what you're trying to say but what you have actually stated is inaccurate. Allow me to make it more rigorous. In Minkowski space-time, an arbitrary boost can be represented by the vector field $\xi^{\mu} = 2E^{[\nu}u^{\mu]}x_{\nu}$ where $E^{\mu}$ is a space-like vector field, $u^{\mu}$ is the 4-velocity of an inertial observer, and $x^{\mu}$ is the position vector field relative to this inertial observer. If $\eta^{\mu}$ is a constant vector field on Minkowski space-time then it can be shown (with little effort) that $\mathcal{L}_ {\xi} \eta^{\mu} = 0$ if and only if $\eta^{\mu}$ is orthogonal to both $u^{\mu}$ and $E^{\mu}$, where $\mathcal{L}_{\xi}$ is the Lie derivative along $\xi^{\mu}$. What this means is that boosts generated by $\xi^{\mu}$ rotate all codimension 2 submanifolds to which $u^{\mu}$ and $E^{\mu}$ are tangent. In this sense the boost is within the plane determined by $u^{\mu}$ and $E^{\mu}$ and rotates this plane. If we had a simple boost along the $x$-axis of this inertial observer's frame then this plane would correspond to the $t$-$x$ plane.

In contrast, an arbitrary rotation can be represented by $\xi^{\nu} = \epsilon^{\mu\nu\alpha\beta}x_{\mu}u_{\alpha}B_{\beta}$ where $B^{\mu}$ is also a space-like vector field. It can similarly be shown with little effort that $\mathcal{L}_{\xi}\eta^{\mu} = 0$ if and only if $\eta^{\mu} = \alpha u^{\mu} + \beta B^{\mu}$ hence the rotations generated by $\xi^{\mu}$ rotate all codimension 2 submanifolds to which $u^{\mu}$ and $B^{\mu}$ are orthogonal. And in this sense $B^{\mu}$ determines the axis of rotation.

Ok so there you use the word rotation, a different type of rotation though. How do I use this to described metric contraction in an x,y,z,t coordinate system?

This is a nonsensical statement.

No it's completely sensical you just obviously don't agree that it's true.

Traveling fast at a constant velocity is a very different thing than acceleration, so the equivalence principle between gravitation and acceleration doesn't contradict PeterDonis.

So what if you indefinitely "accelerated" to approach the speed of light? And what about the equivalence between the time dilation and length contraction? Does it mean nothing? Just pure coincidence?

 

[WannabeNewton]

Alright do you have a link that can more precisely and in a more organized way display the equations you are trying to show? When I say metric contraction I am trying to generalize length contraction and time dilation simultaneously, since you would measure time passing slower via a meter of time appearing to shrink.

Yeah, a textbook on general relativity would be a logical place to start. It would also clear up the numerous other confusions you seem to be having about general relativity.

 

[WannabeNewton]

No it's completely sensical you just obviously don't agree that it's true.

Yes, you're clearly right. I know nothing about general relativity, sorry for trying to correct you.

 

[PhysicsStuff]

Yeah, a textbook on general relativity would be a logical place to start. It would also clear up the numerous other confusions you seem to be having about general relativity.

Like...measuring time in meters? Did Einstein not postulate that time was a dimension? You think time has never been measured in meters?

Yes, you're clearly right. I know nothing about general relativity, sorry for trying to correct you.

No idea why you're getting sarcastically defensive, just take a break. Maybe it's because Einstein's models were more accurate than Newton's.

 

[Nugatory]

And what about the equivalence between the time dilation and length contraction? Does it mean nothing? Just pure coincidence?

It's not a coincidence, but it has nothing to do with the equivalence principle or acceleration. The relationship between time dilation and length contraction is derived from the Lorentz transforms. Relative velocity, but not acceleration, appears in these transforms and in the derivation of the length contraction and time dilation formulas from them.

 

[PhysicsStuff]

It's not a coincidence, but it has nothing to do with the equivalence principle or acceleration. The relationship between time dilation and length contraction is derived from the Lorentz transforms. Relative velocity, but not acceleration, appears in these transforms and in the derivation of the length contraction and time dilation formulas from them.

But that's more or less what I'm trying to get at, a way to derive both length contraction and time dilation from some single thing, like the Loretnz transformation, which from my point of view seems to show a higher dimensional relative rotation of both the axis of space and time to create a smaller metric of both time and space in Minkowski space, and I know I've seen trigonometric functions used to describe space-time curvature somewhere.

 

[PeterDonis]

it's very strange you say that because I thought you're suppose to be able to treat acceleration and gravitational field strength the same because of the whole gravity being fictitious force thing...
http://en.wikipedia.org/wiki/Equivalence_principle

Read my next post, in response to your response to Nugatory.

Uh yeah I'm pretty sure that's how it works, I've had more than one person tell me that. You measure it simply by seeing that a ruler is smaller than it should be

And how do you "see" this? I'm asking for an explicit measurement procedure, like the one I gave that shows how to directly measure the difference in clock rates by exchanging round-trip light signals.

But the point is, if I accelerate near the speed of light, an outside observer would observe the same time dilation as if I were in an intense gravitational field...

Ok, I do disgree, http://en.wikipedia.org/wiki/Time_dilation#Experimental_confirmation
read the "experimental confirmation" section.

These experiments do illustrate a crucial point that I should have been more explicit about. To see it, let's compare two different kinds of experiments with muons:

(1) Muons are created in the laboratory and put into a high-energy accelerator, where they go around in a tight circle at close to the speed of light. Their half-lives are measured to be significantly longer, by laboratory clocks, than the half-lives of muons at rest.

(2) Muons are created high in the Earth's atmosphere by cosmic rays, and manage to reach the surface of the Earth to be measured before they decay. In the Earth ("lab") frame, the time it takes the muons to travel the distance involved is much longer than their half-life; yet significant numbers of muons are detected at the surface, indicating that their half-lives are significantly longer than the half-lives of muons at rest.

Did you notice a critical fact about the first experiment, as compared to the second? The first experiment involves *periodic* motion--i.e., one object stays put (the "laboratory" clock), while another object (the muon) moves in a periodic orbit that returns to the same place. The two clocks get compared each time the moving object and the stationary object are co-located, and the moving object's clock shows less elapsed time. The key is that the clock comparison is direct, because the clocks are co-located when the comparison is done. So the effect is asymmetric: the muons "agree" that the laboratory clock shows more elapsed time--the lab clock is running fast.

In the second experiment, the motion is linear: and that means the effect is symmetric. In other words, according to the muons, the *Earth* clocks are running *slow*, not fast. (Also, the distance the Earth has to travel to reach the muons, in the muons' rest frame, is much shorter due to length contraction--and according to the muons, *that*, not time dilation, is why they can reach the surface before decaying.) So the two situations are not really the same.

If you are trying to make an analogy with gravity, the first situation (muons in an accelerator) does work better because it is asymmetric, as time dilation due to gravity is. However, the fact that the muons have to move very fast is still a big difference, because it means length contraction due to relative motion comes into play. I think this may have misled you into thinking that there should be a similar "length contraction due to gravity" to go with time dilation due to gravity; but there isn't.

But...if both people are moving at the same speed, their clocks and rulers will agree with each other, but disagree with an outside frame of reference, and the same is true of them being in a gravitational field...

So what? I didn't say that nothing was the same between the two situations; I said that a particularly important thing is *not* the same (symmetry vs. asymmetry of the time dilation). You haven't addressed that at all.

 

[PeterDonis]

Local equivalence is all that matters for there to be a fictitious force, though. That might be what you meant, but I wanted to clarify.

Yes, that's what I meant.

It also just occurred to me that the confusion with rotation and gravity might have come from a centrifuge analogy. If you write Minkowski metric in polar coordinates, and then go to a rotating frame of reference, you will get "gravity" terms that are equivalent to Centrifugal and Coriolis forces.

Yes, and you can do the same with any metric by adopting an appropriate chart. That's why it's important to understand the difference between what I would call "completely fictitious forces" like centrifugal and coriolis, and "partially fictitious forces" like gravity, which has a part that can be transformed away by a change of frame (the connection--the analogue to the centrifugal and coriolis forces), and a part that can't (the curvature--tidal effects). I suspect the OP is confusing the two.

 

[PeterDonis]

But that's more or less what I'm trying to get at, a way to derive both length contraction and time dilation from some single thing, like the Loretnz transformation, which from my point of view seems to show a higher dimensional relative rotation of both the axis of space and time to create a smaller metric of both time and space in Minkowski space

This is fine as a way of describing *relative motion*. However, it does not work as a way of describing gravity. There are some similarities, but also some key differences, which you do not appear to be taking into account. So I don't think what you are looking for exists.

I know I've seen trigonometric functions used to describe space-time curvature somewhere.

Trig functions can be used to describe Lorentz boosts (i.e., "4-dimensional rotations" in spacetime)--more precisely, hyperbolic trig functions can. I'm not aware of any special role that trig functions play in describing curvature, though. If you can find a reference that uses them in this way, that would be helpful.

 

[PhysicsStuff]

And how do you "see" this? I'm asking for an explicit measurement procedure, like the one I gave that shows how to directly measure the difference in clock rates by exchanging round-trip light signals.

I don't know how to make it any simpler. You put a really HUGE clock near a black hole so that you can see how fast it ticks from miles away, and you see it's slower. In the wikipedia example I gave you, instead of doing something like that, they used some kind of super precise atomic clock which measured that when accelerating relative to an observer, than the accelerated clock will appear slower to the observer. When not accelerating and simply in a lower gravitational field, the clock further away from Earth was faster.
In the instance of a ruler, stick a huge ruler near a black hole and watch it shrink a lot more than it should because space in a way becomes more "compact", that the metric of space shrinks.

(2) Muons are created high in the Earth's atmosphere by cosmic rays, and manage to reach the surface of the Earth to be measured before they decay. In the Earth ("lab") frame, the time it takes the muons to travel the distance involved is much longer than their half-life; yet significant numbers of muons are detected at the surface, indicating that their half-lives are significantly longer than the half-lives of muons at rest.

Did you notice a critical fact about the first experiment, as compared to the second? The first experiment involves *periodic* motion--i.e., one object stays put (the "laboratory" clock), while another object (the muon) moves in a periodic orbit that returns to the same place. The two clocks get compared each time the moving object and the stationary object are co-located, and the moving object's clock shows less elapsed time. The key is that the clock comparison is direct, because the clocks are co-located when the comparison is done. So the effect is asymmetric: the muons "agree" that the laboratory clock shows more elapsed time--the lab clock is running fast.

In the second experiment, the motion is linear: and that means the effect is symmetric. In other words, according to the muons, the *Earth* clocks are running *slow*, not fast. (Also, the distance the Earth has to travel to reach the muons, in the muons' rest frame, is much shorter due to length contraction--and according to the muons, *that*, not time dilation, is why they can reach the surface before decaying.) So the two situations are not really the same.

If you are trying to make an analogy with gravity, the first situation (muons in an accelerator) does work better because it is asymmetric, as time dilation due to gravity is. However, the fact that the muons have to move very fast is still a big difference, because it means length contraction due to relative motion comes into play. I think this may have misled you into thinking that there should be a similar "length contraction due to gravity" to go with time dilation due to gravity; but there isn't.
So what? I didn't say that nothing was the same between the two situations; I said that a particularly important thing is *not* the same (symmetry vs. asymmetry of the time dilation). You haven't addressed that at all.

Yeah I see what you're trying to say just fine, someone in one location moving linearly will observe the same linear time dilation as another observer with the same linear speed at a greater distance and their clocks will agree, whereas someone separated by distance in a gravitational field (assuming the distance is vertical) will not be able to agree on the same time. That's fine, but it still does not show there isn't equivalence in metric contraction with accelerating and being in a higher gravitational field. I think it's just that in one instance space-time is rotating, and in another, the relative coordinate system is rotating.

Trig functions can be used to describe Lorentz boosts (i.e., "4-dimensional rotations" in spacetime)--more precisely, hyperbolic trig functions can. I'm not aware of any special role that trig functions play in describing curvature, though. If you can find a reference that uses them in this way, that would be helpful.

Yeah I don't think what I'm looking for exists either now, but I still think that in both instances, some type of rotation or curvature is happening, even if they aren't the same types, and the same types of metric contractions are happening.

 

[Nugatory]

But that's more or less what I'm trying to get at, a way to derive both length contraction and time dilation from some single thing, like the Loretnz transformation, which from my point of view seems to show a higher dimensional relative rotation of both the axis of space and time to create a smaller metric of both time and space in Minkowski space, and I know I've seen trigonometric functions used to describe space-time curvature somewhere.

If you've seen trig functions used to describe a rotation to get time dilation and length contraction, you've likely been exposed to the dead end I described in post #7 of this thread: https://www.physicsforums.com/showpost.php?p=4604968&postcount=7
If that's what it is, the sooner you can unlearn it, the better.

Although the Lorentz transforms in their simplest form are two equations not one, you always use the two of them together (which is why they can be written as a single matrix equation: $X' = \Lambda{X}$) where $X$ and $X'$ are vectors formed from the x, y, z, and t coordinates and $\Lambda$ is a 4x4 matrix).

Learn how to use them, get them down cold, before you try to take on the enormously more hairy problems that come with curved spacetime.

 

[PhysicsStuff]

If you've seen trig functions used to describe a rotation to get time dilation and length contraction, you've likely been exposed to the dead end I described in post #7 of this thread: https://www.physicsforums.com/showpost.php?p=4604968&postcount=7
If that's what it is, the sooner you can unlearn it, the better.

Although the Lorentz transforms in their simplest form are two equations not one, you always use the two of them together (which is why they can be written as a single matrix equation: $X' = \Lambda{X}$) where $X$ and $X'$ are vectors formed from the x, y, z, and t coordinates and $\Lambda$ is a 4x4 matrix).

Learn how to use them, get them down cold, before you try to take on the enormously more hairy problems that come with curved spacetime.

But you can still say it's a rotation of the "coordinate system", can't you? And that rotation can create the same type of time dilation that you see with gravity right? There can't be 0 connection.

 

[WannabeNewton]

No idea why you're getting sarcastically defensive, just take a break. Maybe it's because Einstein's models were more accurate than Newton's.

No it's because you're trying to argue steadfastly using non-mathematical handwavy statements when you don't even have a basic understanding of general relativity. That's not helping anyone.

 

[PhysicsStuff]

No it's because you're trying to argue and stick to non-mathematical handwavy statements when you don't even have a basic understanding of general relativity. That's not helping anyone.

If I didn't have any understanding how could I even ask these questions? My suggestion is let the people who are already Newtons handle the situation as you are obviously no longer trying to argue for an understanding for something in physics.

 

[berkeman]

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