is space expansion the same as time dilation ?

aanpalendmw
2022-07-16
Answered

is space expansion the same as time dilation ?

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Abraham Norris

Answered 2022-07-17
Author has **16** answers

Not at all, because the time-dilation is a local thing, it happens in small regions that don't care about the global expansion of the universe, and it has to do with the speed with which you traverse a closed loop in space, just like the length of a spiral along one turn is different if the spiral is stretched or smooshed. If you have a muon going in a circle fast in a magnetic field, the time-dilation is only a function of the speed, as can be seen by how fast the muon decays. It has no relation to the expansion of the universe, which is only visible on galactic distance scales.

asked 2022-08-13

In string theory, If an open string obeys the Neumann boundary condition, then in the static gauge, one can show that the end points move at the speed of light. The derivation is straightforward, but how can this apply to the massive string?

asked 2022-07-21

Let $A$ be a rocket moving with velocity $v$.

Then the slope of its worldline in a spacetime diagram is given by $c/v$.

Since it is a slope, $c/v=\mathrm{tan}(\theta )$ for some $\theta $ $>45$ and theta $<90$.

Does this impose a mathematical limit on $v$?

If so what is it?

As in, we know $\mathrm{tan}(89.9999999999)=572957795131.$

And $c=299792458.$

Using $\mathrm{tan}(89.9999999999)$ as our limit of precision, the smallest $v$ we can use is:

$c/v=\mathrm{tan}(89.9999999999)$

$\Rightarrow 299792458/v=572957795131$

Therefore, $v=1911.18\text{}m/s$

What is the smallest non zero value of $v$?

Then the slope of its worldline in a spacetime diagram is given by $c/v$.

Since it is a slope, $c/v=\mathrm{tan}(\theta )$ for some $\theta $ $>45$ and theta $<90$.

Does this impose a mathematical limit on $v$?

If so what is it?

As in, we know $\mathrm{tan}(89.9999999999)=572957795131.$

And $c=299792458.$

Using $\mathrm{tan}(89.9999999999)$ as our limit of precision, the smallest $v$ we can use is:

$c/v=\mathrm{tan}(89.9999999999)$

$\Rightarrow 299792458/v=572957795131$

Therefore, $v=1911.18\text{}m/s$

What is the smallest non zero value of $v$?

asked 2022-08-10

$\rho ={\varphi}^{\ast}\varphi \phantom{\rule{thinmathspace}{0ex}}$

$\rho ={\varphi}^{\ast}\varphi \phantom{\rule{thinmathspace}{0ex}}$

......

$J=-\frac{i\hslash}{2m}({\varphi}^{\ast}\mathrm{\nabla}\varphi -\varphi \mathrm{\nabla}{\varphi}^{\ast})$

with the conservation of probability current and density following from the Schrödinger equation:

$\mathrm{\nabla}\cdot J+\frac{\mathrm{\partial}\rho}{\mathrm{\partial}t}=0.$

The fact that the density is positive definite and convected according to this continuity equation, implies that we may integrate the density over a certain domain and set the total to 1, and this condition will be maintained by the conservation law. A proper relativistic theory with a probability density current must also share this feature. Now, if we wish to maintain the notion of a convected density, then we must generalize the Schrödinger expression of the density and current so that the space and time derivatives again enter symmetrically in relation to the scalar wave function. We are allowed to keep the Schrödinger expression for the current, but must replace by probability density by the symmetrically formed expression

$\rho =\frac{i\hslash}{2m}({\psi}^{\ast}{\mathrm{\partial}}_{t}\psi -\psi {\mathrm{\partial}}_{t}{\psi}^{\ast}).$

which now becomes the 4th component of a space-time vector, and the entire 4-current density has the relativistically covariant expression

${J}^{\mu}=\frac{i\hslash}{2m}({\psi}^{\ast}{\mathrm{\partial}}^{\mu}\psi -\psi {\mathrm{\partial}}^{\mu}{\psi}^{\ast})$

1. What exactly are ${\mathrm{\partial}}_{t}$ and ${\mathrm{\partial}}^{\mu}$?

2. Are they tensors?

3. If they are, how are they defined?

$\rho ={\varphi}^{\ast}\varphi \phantom{\rule{thinmathspace}{0ex}}$

......

$J=-\frac{i\hslash}{2m}({\varphi}^{\ast}\mathrm{\nabla}\varphi -\varphi \mathrm{\nabla}{\varphi}^{\ast})$

with the conservation of probability current and density following from the Schrödinger equation:

$\mathrm{\nabla}\cdot J+\frac{\mathrm{\partial}\rho}{\mathrm{\partial}t}=0.$

The fact that the density is positive definite and convected according to this continuity equation, implies that we may integrate the density over a certain domain and set the total to 1, and this condition will be maintained by the conservation law. A proper relativistic theory with a probability density current must also share this feature. Now, if we wish to maintain the notion of a convected density, then we must generalize the Schrödinger expression of the density and current so that the space and time derivatives again enter symmetrically in relation to the scalar wave function. We are allowed to keep the Schrödinger expression for the current, but must replace by probability density by the symmetrically formed expression

$\rho =\frac{i\hslash}{2m}({\psi}^{\ast}{\mathrm{\partial}}_{t}\psi -\psi {\mathrm{\partial}}_{t}{\psi}^{\ast}).$

which now becomes the 4th component of a space-time vector, and the entire 4-current density has the relativistically covariant expression

${J}^{\mu}=\frac{i\hslash}{2m}({\psi}^{\ast}{\mathrm{\partial}}^{\mu}\psi -\psi {\mathrm{\partial}}^{\mu}{\psi}^{\ast})$

1. What exactly are ${\mathrm{\partial}}_{t}$ and ${\mathrm{\partial}}^{\mu}$?

2. Are they tensors?

3. If they are, how are they defined?

asked 2022-05-01

Under the Lorentz transformations, quantities are classed as four-vectors, Lorentz scalars etc depending upon how their measurement in one coordinate system transforms as a measurement in another coordinate system.

The proper length and proper time measured in one coordinate system will be a calculated, but not measured, invariant for all other coordinate systems.

So what kind of invariants are proper time and proper length?

The proper length and proper time measured in one coordinate system will be a calculated, but not measured, invariant for all other coordinate systems.

So what kind of invariants are proper time and proper length?

asked 2022-05-10

Take the following gedankenexperiment in which two astronauts meet each other again and again in a perfectly symmetrical setting - a hyperspherical (3-manifold) universe in which the 3 dimensions are curved into the 4. dimension so that they can travel without acceleration in straight opposite directions and yet meet each other time after time.

On the one hand this situation is perfectly symmetrical - even in terms of homotopy and winding number. On the other hand the Lorentz invariance should break down according to GRT, so that one frame is preferred - but which one?

So the question is: Who will be older? And why?

And even if there is one prefered inertial frame - the frame of the other astronaut should be identical with respect to all relevant parameters so that both get older at the same rate. Which again seems to be a violation of SRT in which the other twin seems to be getting older faster/slower...

How should one find out what the preferred frame is when everything is symmetrical - even in terms of GRT

On the one hand this situation is perfectly symmetrical - even in terms of homotopy and winding number. On the other hand the Lorentz invariance should break down according to GRT, so that one frame is preferred - but which one?

So the question is: Who will be older? And why?

And even if there is one prefered inertial frame - the frame of the other astronaut should be identical with respect to all relevant parameters so that both get older at the same rate. Which again seems to be a violation of SRT in which the other twin seems to be getting older faster/slower...

How should one find out what the preferred frame is when everything is symmetrical - even in terms of GRT

asked 2022-08-16

Let us assume that we placed lot's of ball touching each other in a hollow cylindrical tube, now if we push one ball at the end the ball at the other end move's instantly. So how do the information from one side of the tube travel's to the other side of the tube instantly. and at what speed at it travel's. If it travels at the speed of light so if make a long tube of the length of about $3\ast {10}^{8}$ m it would take 1 second to see the effect at the other end assuming that nothing will bend or compress all thing's are ideal?

asked 2022-09-17

How do Vectors transform from one inertial reference frame to another inertial reference frame in special relativity.

A bound vector in an inertial reference frame ($x$,$ct$) has its line of action as one of the space axis in that frame and is described by $x$*$i$*,then what would it be in form of new base vectors ($\mathbf{a}$) and ($\mathbf{b}$) in a different inertial system ($x\u2018$,$ct\u2018$) moving with respect to the former inertial system with v*i* velocity.Let (i) and (j) be the two bounded unit vectors with the line of action as co-ordinate axis($x$) and ($ct$) respectively and senses in the positive side of co-ordinates and similarly ($\mathbf{a}$) and ($\mathbf{b}$) are defined for co-ordinates ($x\u2018$) and ($ct\u2018$) respectively.

A bound vector in an inertial reference frame ($x$,$ct$) has its line of action as one of the space axis in that frame and is described by $x$*$i$*,then what would it be in form of new base vectors ($\mathbf{a}$) and ($\mathbf{b}$) in a different inertial system ($x\u2018$,$ct\u2018$) moving with respect to the former inertial system with v*i* velocity.Let (i) and (j) be the two bounded unit vectors with the line of action as co-ordinate axis($x$) and ($ct$) respectively and senses in the positive side of co-ordinates and similarly ($\mathbf{a}$) and ($\mathbf{b}$) are defined for co-ordinates ($x\u2018$) and ($ct\u2018$) respectively.