Is time travel... travelling from one 3 dimensional space time to another 3 dimensional space time?

[Judy]

If time is a 4th dimensional thing but still exist in our 3 dimensional world..., time exist as a way or medium.. as we actually do time travel in 'x' time every second, every minute and it's the rate of time our universe is going through and to time travel.., we mean to change that rate of time

 

[Nugatory]

It’s tempting to think of the passage of time as moving through the fourth dimension from one three-dimensional space to another, but it turns out not to work.

However I don’t know of any easy way of explaining why it doesn’t work, except to say “learn special relativity and then you’ll know”. For some things there’s no substitute for beginning at the beginning with a good textbook. One of my favorites is Taylor and Wheeler’s “Spacetime Physics” which is available free on the Internet and (just barely) within the grasp of a motivated high school student.

 

[thomsj4]

Consider a 4-dimensional manifold with metric signature $- + + +$, and let $(t,x,y,z)$ be coordinates such that the spacetime interval is given by $ds^2 = -c^2,dt^2 + dx^2 + dy^2 + dz^2$. Define a scalar field $\Phi(t,x,y,z)$ that vanishes everywhere except along two distinct 3-dimensional spatial hypersurfaces: one at $t = t_0$ and one at $t = t_1$. Suppose we construct a continuous mapping $F : \mathbb{R}^4 \to \mathbb{R}^4$ such that $F$ restricts to the identity on each hypersurface but imposes a phase shift in the metric over an intermediate time interval. Formally, let $F^\mu(\tau,\sigma,\rho,\chi)$ satisfy $F^0(\tau,\sigma,\rho,\chi) = t_0 + \gamma(\tau)$ for some smooth function $\gamma$ with $\gamma(\tau_0) = 0$ and $\gamma(\tau_f) = t_1 - t_0$. Impose a boundary condition so that $F$ preserves proper lengths on each 3D slice by requiring that the induced metric $g’{\alpha\beta} = \frac{\partial F^\mu}{\partial \xi^\alpha} \frac{\partial F^\nu}{\partial \xi^\beta} g{\mu\nu}$ reduces to $g_{\alpha\beta}$ when $\tau = \tau_0$ or $\tau = \tau_f$.

We now define a 1-form $\omega = \Phi,d\eta$ for a new coordinate $\eta$, where $\Phi$ localizes $\omega$ in the time region $t_0 \le t \le t_1$. This 1-form modifies the geodesic equations by adding a term proportional to $\omega_\mu u^\mu$, where $u^\mu$ is the 4-velocity. The resulting geodesic equation can be written as

$$
\frac{d^2 x^\mu}{d\lambda^2} + \Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\lambda} \frac{dx^\beta}{d\lambda} = \kappa,\omega_\nu,g^{\mu\nu},

$$

where $\kappa$ is a coupling constant. Because $\omega$ is only active between $t_0$ and $t_1$, any observer crossing this region evolves along a timelike curve that smoothly lifts them from the hypersurface $t = t_0$ to $t = t_1$. The constant-time slices $t = t_0$ and $t = t_1$ thus appear as distinct 3-dimensional “copies” of space.

 

[Nugatory]

@thomsj4 please respect the thread level prefixes
@Judy It's really not THAT complicated

 

[Judy]

@thomsj4 please respect the thread level prefixes
@Judy It's really not THAT complica

@thomsj4 please respect the thread level prefixes
@Judy It's really not THAT complicated

Thankyou very much for your guidance.. (from my whole heart)

 

[Judy]

Consider a 4-dimensional manifold with metric signature $- + + +$, and let $(t,x,y,z)$ be coordinates such that the spacetime interval is given by $ds^2 = -c^2,dt^2 + dx^2 + dy^2 + dz^2$. Define a scalar field $\Phi(t,x,y,z)$ that vanishes everywhere except along two distinct 3-dimensional spatial hypersurfaces: one at $t = t_0$ and one at $t = t_1$. Suppose we construct a continuous mapping $F : \mathbb{R}^4 \to \mathbb{R}^4$ such that $F$ restricts to the identity on each hypersurface but imposes a phase shift in the metric over an intermediate time interval. Formally, let $F^\mu(\tau,\sigma,\rho,\chi)$ satisfy $F^0(\tau,\sigma,\rho,\chi) = t_0 + \gamma(\tau)$ for some smooth function $\gamma$ with $\gamma(\tau_0) = 0$ and $\gamma(\tau_f) = t_1 - t_0$. Impose a boundary condition so that $F$ preserves proper lengths on each 3D slice by requiring that the induced metric $g’{\alpha\beta} = \frac{\partial F^\mu}{\partial \xi^\alpha} \frac{\partial F^\nu}{\partial \xi^\beta} g{\mu\nu}$ reduces to $g_{\alpha\beta}$ when $\tau = \tau_0$ or $\tau = \tau_f$.

We now define a 1-form $\omega = \Phi,d\eta$ for a new coordinate $\eta$, where $\Phi$ localizes $\omega$ in the time region $t_0 \le t \le t_1$. This 1-form modifies the geodesic equations by adding a term proportional to $\omega_\mu u^\mu$, where $u^\mu$ is the 4-velocity. The resulting geodesic equation can be written as

$$
\frac{d^2 x^\mu}{d\lambda^2} + \Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\lambda} \frac{dx^\beta}{d\lambda} = \kappa,\omega_\nu,g^{\mu\nu},

$$

where $\kappa$ is a coupling constant. Because $\omega$ is only active between $t_0$ and $t_1$, any observer crossing this region evolves along a timelike curve that smoothly lifts them from the hypersurface $t = t_0$ to $t = t_1$. The constant-time slices $t = t_0$ and $t = t_1$ thus appear as distinct 3-dimensional “copies” of space.

Thankyou very much...
What would I conclude with it?

 

[L Drago]

It’s tempting to think of the passage of time as moving through the fourth dimension from one three-dimensional space to another, but it turns out not to work.

However I don’t know of any easy way of explaining why it doesn’t work, except to say “learn special relativity and then you’ll know”. For some things there’s no substitute for beginning at the beginning with a good textbook. One of my favorites is Taylor and Wheeler’s “Spacetime Physics” which is available free on the Internet and (just barely) within the grasp of a motivated high school student.

Time is the fourth dimension. We travel in the arrow of time and that is in forward Direction and through passage of time. I also don't think we travel from one three dimensional space to another while travelling in passage of time as when I am sitting on Earth time is moving forward I am moving in the passage of time but I am at rest relative to Earth and I am sitting that is I am not travelling from one three dimensional place to another although according to high speed moving dark matter besides me, I am travelling in very high speeds. Am I right here ?

 

[Nugatory]

Am I right here ?

You're not exactly wrong, but you are doing a pretty good job of proving my point: "For some things there’s no substitute for beginning at the beginning with a good textbook."

You should be cautious about jumping into threads when you're not confident in your own understanding of the subject.

 

[L Drago]

You're not exactly wrong, but you are doing a pretty good job of proving my point: "For some things there’s no substitute for beginning at the beginning with a good textbook."

You should be cautious about jumping into threads when you're not confident in your own understanding of the subject.

Thank you sir.

 

[Judy]

Time is the fourth dimension. We travel in the arrow of time and that is in forward Direction and through passage of time. I also don't think we travel from one three dimensional space to another while travelling in passage of time as when I am sitting on Earth time is moving forward I am moving in the passage of time but I am at rest relative to Earth and I am sitting that is I am not travelling from one three dimensional place to another although according to high speed moving dark matter besides me, I am travelling in very high speeds. Am I right here ?

No, but I meant like our 3 dimensional universe is moving this way not us (I'm unsure if i should post this or not, please correct me, I'm not clear with many physics related theories)

 

[Ivan Nikiforov]

If time is a 4th dimensional thing but still exist in our 3 dimensional world..., time exist as a way or medium.. as we actually do time travel in 'x' time every second, every minute and it's the rate of time our universe is going through and to time travel.., we mean to change that rate of time

The passage of time can be slowed down, but it cannot be turned back. Time travel to the past is impossible due to the enthalpy of the systems. Simply put, firewood that has burned in the fireplace and dissipated by heat, light and smoke cannot be returned to its original state

 

[phinds]

The passage of time can be slowed down, but it cannot be turned back.

No, it cannot. Time always moves forward at one second per second.

 

[Ivan Nikiforov]

No, it cannot. Time always moves forward at one second per second.

I was referring to the relativistic time dilation during space travel. Yes, so far it is technically impossible to realize movement in space at high speeds. However, when it is possible, the space traveler will be able to use time dilation and, in fact, will be able to travel to the future on his return. This is very well demonstrated in the wonderful film Interstellar.

 

[phinds]

I was referring to the relativistic time dilation during space travel. Yes, so far it is technically impossible to realize movement in space at high speeds. However, when it is possible, the space traveler will be able to use time dilation and, in fact, will be able to travel to the future on his return. This is very well demonstrated in the wonderful film Interstellar.

You are correct about differential aging (not time dilation) being able to take you into the future even while time progresses for you at one second per second, but the "wonderful" film Interstellar has the concept of differential aging correct qualitatively but actual use of it that is quite wrong quantitatively (and of course that says nothing about its ridiculous treatment of a black hole).

 

[Ivan Nikiforov]

You are correct about differential aging (not time dilation) being able to take you into the future even while time progresses for you at one second per second, but the "wonderful" film Interstellar has the concept of differential aging correct qualitatively but actual use of it that is quite wrong quantitatively (and of course that says nothing about its ridiculous treatment of a black hole).

I agree with you. The plot of the black hole is absolutely fantastic.