With what velocity are we moving along the time dimension?

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Does the question make sense? Velocity along time axis means $v_t=mathrm dt/mathrm dt$? If it doesn't, please explain where the flaw is. Taking time as measure like length? Or do we need to differentiate time with respect to some other quantity?










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    Does the question make sense? Velocity along time axis means $v_t=mathrm dt/mathrm dt$? If it doesn't, please explain where the flaw is. Taking time as measure like length? Or do we need to differentiate time with respect to some other quantity?










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      Does the question make sense? Velocity along time axis means $v_t=mathrm dt/mathrm dt$? If it doesn't, please explain where the flaw is. Taking time as measure like length? Or do we need to differentiate time with respect to some other quantity?










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      Does the question make sense? Velocity along time axis means $v_t=mathrm dt/mathrm dt$? If it doesn't, please explain where the flaw is. Taking time as measure like length? Or do we need to differentiate time with respect to some other quantity?







      time relativity differentiation






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      edited 17 mins ago









      Kyle Kanos

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      asked 1 hour ago









      Krishna Deshmukh

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          There is a variety of different conventions for defining some of the details, but the most common way to describe this, among relativists, would be the following. We take units in which $c=1$. There is a velocity four-vector which is tangent to the world-line of a particle. The normalization of this four-vector is defined so that its norm is 1 (in $+---$ signature). All of this is coordinate-independent.



          If we now specialize to Minkowski coordinates $(t,x,y,z)$ in flat spacetime, then the components of the velocity four-vector become the derivative of the coordinates with respect to proper time $tau$ (not coordinate time $t$), and the normalization condition ends up causing the timelike component of the velocity vector to be the Lorentz factor $gamma$. This is the closest thing we have, in common professional notation, to a useful way of defining something that is useful and corresponds in some way to the notion of a "velocity along the time dimension." It's $gamma$.



          In the special case where the particle is at rest with respect to the Minkowski frame being used, we have $gamma=1$. This is the justification you see for the statement in popularizations that we "move through spacetime at the speed of light," since the speed of light is 1. However, most relativists cringe at this phraseology, which seems to have been propagated by Brian Greene.






          share|cite|improve this answer



























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            In non-relativistic mechanics, time $t$ is a (universal) parameter and a particle's coordinates (in some inertial coordinate system) can be expressed as three functions, $x(t),y(t),z(t)$ of this universal parameter $t$. The velocity of the particle (in these coordinates) is then the derivative of the position with respect to the parameter $t$:



            $$mathbfv = fracdxdthatmathbfx + fracdydthatmathbfy + fracdzdthatmathbfz$$



            However, in relativistic mechanics (SR for simplicity), time $t$ is a coordinate which is reference frame dependent. Still, the world line of a particle can be parameterized with the proper time $tau$ which is essentially the time of an ideal clock fixed to the particle ('wristwatch time').



            The coordinates of the particle (in some inertial coordinate system) can then be expressed as four functions, $t(tau),x(tau),y(tau),z(tau)$ of the particle's proper time $tau$. The four-velocity of the particle is then the derivative of the four-position with respect to the parameter $tau$:



            $$vecU = cfracdtdtauhatmathbft + fracdxdtauhatmathbfx + fracdydtauhatmathbfy + fracdzdtauhatmathbfz$$



            So, in this coordinate system, the component of the particle's four-velocity in the time direction is



            $$U^0 = cfracdtdtau$$



            Now, it can be shown that (time dilation)



            $$dt = gamma_v dtau$$



            where



            $$gamma_v equiv left(1 - fracv^2c^2right)^-1/2$$



            and



            $$ v = sqrtleft(fracdxdtright)^2 + left(fracdydtright)^2 + left(fracdzdtright)^2$$



            thus



            $$U^0 = cgamma_v$$



            This is, I believe, a reasonable answer to the question "With what velocity are we moving along the time dimension?" if, by velocity, one means the the derivative of the coordinates with respect to a time parameter.



            (note: as I was finishing typing this answer up, I noticed that Ben Crowell had posted essentially the same answer but I'll post this anyhow since it's already done.)






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              At exactly 1 second per second.






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              • One second per second does not have the dimension of speed and therefore is not velocity, unless you use geometrized nits where the speed of light is unity.
                – safesphere
                1 hour ago










              • safesphere-So,how do we define the rate at which we are moving in time?Do we lack a concept to understand this?
                – Krishna Deshmukh
                1 hour ago










              • @KrishnaDeshmukh No, the concept is fine. The rate of our own (proper) time is $1$ (as in one second per second), not $gamma$ as some answers here suggest. In other words, our own $gamma=1$. To convert rate to speed, you need to multiply it by the speed of light. So we are moving in time at the rate of $gamma=1$ with the speed of $c$.
                – safesphere
                14 mins ago










              • How can one talk about a constant speed and its rate of change along time at once?If its changing then ,its not constant.
                – Krishna Deshmukh
                5 mins ago

















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              I'd like to think of it as moving at an infinitely fast pace. Time is not exactly spatial, so light speed in a vaccuum, parallel to safesphere's answer, would not truly apply. You pass an infinite amount of probabilities and times in an infinitesmally small amount of time, so the velocity is most likely infinite. But since this infinite velocity is in a seperate dimension, it won't affect us.






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                In relativity the time coordinate is $x_o=ct$ and its time derivative is $c$. Therefore the time component of four-velocity is the speed of light in vacuum.






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                • In relativity the time coordinate is xo=ct Well, not really. There is a variety of possible conventions. The most common convention among relativists is to work in units where $c=1$, and therefore never write any factors of $c$ anywhere. Therefore the time component of four-velocity is the speed of light in vacuum. Not true. This holds only for a four-velocity describing a world-line that is at rest relative to a particular Minkowski frame.
                  – Ben Crowell
                  1 hour ago










                • @BenCrowell You are wrong again. (1) For geometrized units where $c=1$, my answer stands. I never stated the specific value of $c$. (2) The question is "With what velocity are we moving along the time dimension" thus referring to the proper time. So my answer is correct despite your downvote. There is no time dilation in the proper frame.
                  – safesphere
                  1 hour ago











                • safesphere, since $vec U equiv dvecX/dtau$, isn't the time component of $vecU$ equal to $gamma c$ rather than $c$?
                  – Alfred Centauri
                  33 mins ago










                • @AlfredCentauri Yes, but the question is about the proper frame, so $gamma=1$.
                  – safesphere
                  26 mins ago










                • @safesphere Actually, the OP's question never mentions Proper Time and neither does your answer.
                  – K7PEH
                  19 mins ago










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                There is a variety of different conventions for defining some of the details, but the most common way to describe this, among relativists, would be the following. We take units in which $c=1$. There is a velocity four-vector which is tangent to the world-line of a particle. The normalization of this four-vector is defined so that its norm is 1 (in $+---$ signature). All of this is coordinate-independent.



                If we now specialize to Minkowski coordinates $(t,x,y,z)$ in flat spacetime, then the components of the velocity four-vector become the derivative of the coordinates with respect to proper time $tau$ (not coordinate time $t$), and the normalization condition ends up causing the timelike component of the velocity vector to be the Lorentz factor $gamma$. This is the closest thing we have, in common professional notation, to a useful way of defining something that is useful and corresponds in some way to the notion of a "velocity along the time dimension." It's $gamma$.



                In the special case where the particle is at rest with respect to the Minkowski frame being used, we have $gamma=1$. This is the justification you see for the statement in popularizations that we "move through spacetime at the speed of light," since the speed of light is 1. However, most relativists cringe at this phraseology, which seems to have been propagated by Brian Greene.






                share|cite|improve this answer
























                  up vote
                  1
                  down vote













                  There is a variety of different conventions for defining some of the details, but the most common way to describe this, among relativists, would be the following. We take units in which $c=1$. There is a velocity four-vector which is tangent to the world-line of a particle. The normalization of this four-vector is defined so that its norm is 1 (in $+---$ signature). All of this is coordinate-independent.



                  If we now specialize to Minkowski coordinates $(t,x,y,z)$ in flat spacetime, then the components of the velocity four-vector become the derivative of the coordinates with respect to proper time $tau$ (not coordinate time $t$), and the normalization condition ends up causing the timelike component of the velocity vector to be the Lorentz factor $gamma$. This is the closest thing we have, in common professional notation, to a useful way of defining something that is useful and corresponds in some way to the notion of a "velocity along the time dimension." It's $gamma$.



                  In the special case where the particle is at rest with respect to the Minkowski frame being used, we have $gamma=1$. This is the justification you see for the statement in popularizations that we "move through spacetime at the speed of light," since the speed of light is 1. However, most relativists cringe at this phraseology, which seems to have been propagated by Brian Greene.






                  share|cite|improve this answer






















                    up vote
                    1
                    down vote










                    up vote
                    1
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                    There is a variety of different conventions for defining some of the details, but the most common way to describe this, among relativists, would be the following. We take units in which $c=1$. There is a velocity four-vector which is tangent to the world-line of a particle. The normalization of this four-vector is defined so that its norm is 1 (in $+---$ signature). All of this is coordinate-independent.



                    If we now specialize to Minkowski coordinates $(t,x,y,z)$ in flat spacetime, then the components of the velocity four-vector become the derivative of the coordinates with respect to proper time $tau$ (not coordinate time $t$), and the normalization condition ends up causing the timelike component of the velocity vector to be the Lorentz factor $gamma$. This is the closest thing we have, in common professional notation, to a useful way of defining something that is useful and corresponds in some way to the notion of a "velocity along the time dimension." It's $gamma$.



                    In the special case where the particle is at rest with respect to the Minkowski frame being used, we have $gamma=1$. This is the justification you see for the statement in popularizations that we "move through spacetime at the speed of light," since the speed of light is 1. However, most relativists cringe at this phraseology, which seems to have been propagated by Brian Greene.






                    share|cite|improve this answer












                    There is a variety of different conventions for defining some of the details, but the most common way to describe this, among relativists, would be the following. We take units in which $c=1$. There is a velocity four-vector which is tangent to the world-line of a particle. The normalization of this four-vector is defined so that its norm is 1 (in $+---$ signature). All of this is coordinate-independent.



                    If we now specialize to Minkowski coordinates $(t,x,y,z)$ in flat spacetime, then the components of the velocity four-vector become the derivative of the coordinates with respect to proper time $tau$ (not coordinate time $t$), and the normalization condition ends up causing the timelike component of the velocity vector to be the Lorentz factor $gamma$. This is the closest thing we have, in common professional notation, to a useful way of defining something that is useful and corresponds in some way to the notion of a "velocity along the time dimension." It's $gamma$.



                    In the special case where the particle is at rest with respect to the Minkowski frame being used, we have $gamma=1$. This is the justification you see for the statement in popularizations that we "move through spacetime at the speed of light," since the speed of light is 1. However, most relativists cringe at this phraseology, which seems to have been propagated by Brian Greene.







                    share|cite|improve this answer












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                    share|cite|improve this answer










                    answered 1 hour ago









                    Ben Crowell

                    46k3147278




                    46k3147278




















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                        1
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                        In non-relativistic mechanics, time $t$ is a (universal) parameter and a particle's coordinates (in some inertial coordinate system) can be expressed as three functions, $x(t),y(t),z(t)$ of this universal parameter $t$. The velocity of the particle (in these coordinates) is then the derivative of the position with respect to the parameter $t$:



                        $$mathbfv = fracdxdthatmathbfx + fracdydthatmathbfy + fracdzdthatmathbfz$$



                        However, in relativistic mechanics (SR for simplicity), time $t$ is a coordinate which is reference frame dependent. Still, the world line of a particle can be parameterized with the proper time $tau$ which is essentially the time of an ideal clock fixed to the particle ('wristwatch time').



                        The coordinates of the particle (in some inertial coordinate system) can then be expressed as four functions, $t(tau),x(tau),y(tau),z(tau)$ of the particle's proper time $tau$. The four-velocity of the particle is then the derivative of the four-position with respect to the parameter $tau$:



                        $$vecU = cfracdtdtauhatmathbft + fracdxdtauhatmathbfx + fracdydtauhatmathbfy + fracdzdtauhatmathbfz$$



                        So, in this coordinate system, the component of the particle's four-velocity in the time direction is



                        $$U^0 = cfracdtdtau$$



                        Now, it can be shown that (time dilation)



                        $$dt = gamma_v dtau$$



                        where



                        $$gamma_v equiv left(1 - fracv^2c^2right)^-1/2$$



                        and



                        $$ v = sqrtleft(fracdxdtright)^2 + left(fracdydtright)^2 + left(fracdzdtright)^2$$



                        thus



                        $$U^0 = cgamma_v$$



                        This is, I believe, a reasonable answer to the question "With what velocity are we moving along the time dimension?" if, by velocity, one means the the derivative of the coordinates with respect to a time parameter.



                        (note: as I was finishing typing this answer up, I noticed that Ben Crowell had posted essentially the same answer but I'll post this anyhow since it's already done.)






                        share|cite|improve this answer


























                          up vote
                          1
                          down vote













                          In non-relativistic mechanics, time $t$ is a (universal) parameter and a particle's coordinates (in some inertial coordinate system) can be expressed as three functions, $x(t),y(t),z(t)$ of this universal parameter $t$. The velocity of the particle (in these coordinates) is then the derivative of the position with respect to the parameter $t$:



                          $$mathbfv = fracdxdthatmathbfx + fracdydthatmathbfy + fracdzdthatmathbfz$$



                          However, in relativistic mechanics (SR for simplicity), time $t$ is a coordinate which is reference frame dependent. Still, the world line of a particle can be parameterized with the proper time $tau$ which is essentially the time of an ideal clock fixed to the particle ('wristwatch time').



                          The coordinates of the particle (in some inertial coordinate system) can then be expressed as four functions, $t(tau),x(tau),y(tau),z(tau)$ of the particle's proper time $tau$. The four-velocity of the particle is then the derivative of the four-position with respect to the parameter $tau$:



                          $$vecU = cfracdtdtauhatmathbft + fracdxdtauhatmathbfx + fracdydtauhatmathbfy + fracdzdtauhatmathbfz$$



                          So, in this coordinate system, the component of the particle's four-velocity in the time direction is



                          $$U^0 = cfracdtdtau$$



                          Now, it can be shown that (time dilation)



                          $$dt = gamma_v dtau$$



                          where



                          $$gamma_v equiv left(1 - fracv^2c^2right)^-1/2$$



                          and



                          $$ v = sqrtleft(fracdxdtright)^2 + left(fracdydtright)^2 + left(fracdzdtright)^2$$



                          thus



                          $$U^0 = cgamma_v$$



                          This is, I believe, a reasonable answer to the question "With what velocity are we moving along the time dimension?" if, by velocity, one means the the derivative of the coordinates with respect to a time parameter.



                          (note: as I was finishing typing this answer up, I noticed that Ben Crowell had posted essentially the same answer but I'll post this anyhow since it's already done.)






                          share|cite|improve this answer
























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            In non-relativistic mechanics, time $t$ is a (universal) parameter and a particle's coordinates (in some inertial coordinate system) can be expressed as three functions, $x(t),y(t),z(t)$ of this universal parameter $t$. The velocity of the particle (in these coordinates) is then the derivative of the position with respect to the parameter $t$:



                            $$mathbfv = fracdxdthatmathbfx + fracdydthatmathbfy + fracdzdthatmathbfz$$



                            However, in relativistic mechanics (SR for simplicity), time $t$ is a coordinate which is reference frame dependent. Still, the world line of a particle can be parameterized with the proper time $tau$ which is essentially the time of an ideal clock fixed to the particle ('wristwatch time').



                            The coordinates of the particle (in some inertial coordinate system) can then be expressed as four functions, $t(tau),x(tau),y(tau),z(tau)$ of the particle's proper time $tau$. The four-velocity of the particle is then the derivative of the four-position with respect to the parameter $tau$:



                            $$vecU = cfracdtdtauhatmathbft + fracdxdtauhatmathbfx + fracdydtauhatmathbfy + fracdzdtauhatmathbfz$$



                            So, in this coordinate system, the component of the particle's four-velocity in the time direction is



                            $$U^0 = cfracdtdtau$$



                            Now, it can be shown that (time dilation)



                            $$dt = gamma_v dtau$$



                            where



                            $$gamma_v equiv left(1 - fracv^2c^2right)^-1/2$$



                            and



                            $$ v = sqrtleft(fracdxdtright)^2 + left(fracdydtright)^2 + left(fracdzdtright)^2$$



                            thus



                            $$U^0 = cgamma_v$$



                            This is, I believe, a reasonable answer to the question "With what velocity are we moving along the time dimension?" if, by velocity, one means the the derivative of the coordinates with respect to a time parameter.



                            (note: as I was finishing typing this answer up, I noticed that Ben Crowell had posted essentially the same answer but I'll post this anyhow since it's already done.)






                            share|cite|improve this answer














                            In non-relativistic mechanics, time $t$ is a (universal) parameter and a particle's coordinates (in some inertial coordinate system) can be expressed as three functions, $x(t),y(t),z(t)$ of this universal parameter $t$. The velocity of the particle (in these coordinates) is then the derivative of the position with respect to the parameter $t$:



                            $$mathbfv = fracdxdthatmathbfx + fracdydthatmathbfy + fracdzdthatmathbfz$$



                            However, in relativistic mechanics (SR for simplicity), time $t$ is a coordinate which is reference frame dependent. Still, the world line of a particle can be parameterized with the proper time $tau$ which is essentially the time of an ideal clock fixed to the particle ('wristwatch time').



                            The coordinates of the particle (in some inertial coordinate system) can then be expressed as four functions, $t(tau),x(tau),y(tau),z(tau)$ of the particle's proper time $tau$. The four-velocity of the particle is then the derivative of the four-position with respect to the parameter $tau$:



                            $$vecU = cfracdtdtauhatmathbft + fracdxdtauhatmathbfx + fracdydtauhatmathbfy + fracdzdtauhatmathbfz$$



                            So, in this coordinate system, the component of the particle's four-velocity in the time direction is



                            $$U^0 = cfracdtdtau$$



                            Now, it can be shown that (time dilation)



                            $$dt = gamma_v dtau$$



                            where



                            $$gamma_v equiv left(1 - fracv^2c^2right)^-1/2$$



                            and



                            $$ v = sqrtleft(fracdxdtright)^2 + left(fracdydtright)^2 + left(fracdzdtright)^2$$



                            thus



                            $$U^0 = cgamma_v$$



                            This is, I believe, a reasonable answer to the question "With what velocity are we moving along the time dimension?" if, by velocity, one means the the derivative of the coordinates with respect to a time parameter.



                            (note: as I was finishing typing this answer up, I noticed that Ben Crowell had posted essentially the same answer but I'll post this anyhow since it's already done.)







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 40 mins ago

























                            answered 45 mins ago









                            Alfred Centauri

                            46.5k344139




                            46.5k344139




















                                up vote
                                0
                                down vote













                                At exactly 1 second per second.






                                share|cite|improve this answer




















                                • One second per second does not have the dimension of speed and therefore is not velocity, unless you use geometrized nits where the speed of light is unity.
                                  – safesphere
                                  1 hour ago










                                • safesphere-So,how do we define the rate at which we are moving in time?Do we lack a concept to understand this?
                                  – Krishna Deshmukh
                                  1 hour ago










                                • @KrishnaDeshmukh No, the concept is fine. The rate of our own (proper) time is $1$ (as in one second per second), not $gamma$ as some answers here suggest. In other words, our own $gamma=1$. To convert rate to speed, you need to multiply it by the speed of light. So we are moving in time at the rate of $gamma=1$ with the speed of $c$.
                                  – safesphere
                                  14 mins ago










                                • How can one talk about a constant speed and its rate of change along time at once?If its changing then ,its not constant.
                                  – Krishna Deshmukh
                                  5 mins ago














                                up vote
                                0
                                down vote













                                At exactly 1 second per second.






                                share|cite|improve this answer




















                                • One second per second does not have the dimension of speed and therefore is not velocity, unless you use geometrized nits where the speed of light is unity.
                                  – safesphere
                                  1 hour ago










                                • safesphere-So,how do we define the rate at which we are moving in time?Do we lack a concept to understand this?
                                  – Krishna Deshmukh
                                  1 hour ago










                                • @KrishnaDeshmukh No, the concept is fine. The rate of our own (proper) time is $1$ (as in one second per second), not $gamma$ as some answers here suggest. In other words, our own $gamma=1$. To convert rate to speed, you need to multiply it by the speed of light. So we are moving in time at the rate of $gamma=1$ with the speed of $c$.
                                  – safesphere
                                  14 mins ago










                                • How can one talk about a constant speed and its rate of change along time at once?If its changing then ,its not constant.
                                  – Krishna Deshmukh
                                  5 mins ago












                                up vote
                                0
                                down vote










                                up vote
                                0
                                down vote









                                At exactly 1 second per second.






                                share|cite|improve this answer












                                At exactly 1 second per second.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered 1 hour ago









                                Cer

                                262




                                262











                                • One second per second does not have the dimension of speed and therefore is not velocity, unless you use geometrized nits where the speed of light is unity.
                                  – safesphere
                                  1 hour ago










                                • safesphere-So,how do we define the rate at which we are moving in time?Do we lack a concept to understand this?
                                  – Krishna Deshmukh
                                  1 hour ago










                                • @KrishnaDeshmukh No, the concept is fine. The rate of our own (proper) time is $1$ (as in one second per second), not $gamma$ as some answers here suggest. In other words, our own $gamma=1$. To convert rate to speed, you need to multiply it by the speed of light. So we are moving in time at the rate of $gamma=1$ with the speed of $c$.
                                  – safesphere
                                  14 mins ago










                                • How can one talk about a constant speed and its rate of change along time at once?If its changing then ,its not constant.
                                  – Krishna Deshmukh
                                  5 mins ago
















                                • One second per second does not have the dimension of speed and therefore is not velocity, unless you use geometrized nits where the speed of light is unity.
                                  – safesphere
                                  1 hour ago










                                • safesphere-So,how do we define the rate at which we are moving in time?Do we lack a concept to understand this?
                                  – Krishna Deshmukh
                                  1 hour ago










                                • @KrishnaDeshmukh No, the concept is fine. The rate of our own (proper) time is $1$ (as in one second per second), not $gamma$ as some answers here suggest. In other words, our own $gamma=1$. To convert rate to speed, you need to multiply it by the speed of light. So we are moving in time at the rate of $gamma=1$ with the speed of $c$.
                                  – safesphere
                                  14 mins ago










                                • How can one talk about a constant speed and its rate of change along time at once?If its changing then ,its not constant.
                                  – Krishna Deshmukh
                                  5 mins ago















                                One second per second does not have the dimension of speed and therefore is not velocity, unless you use geometrized nits where the speed of light is unity.
                                – safesphere
                                1 hour ago




                                One second per second does not have the dimension of speed and therefore is not velocity, unless you use geometrized nits where the speed of light is unity.
                                – safesphere
                                1 hour ago












                                safesphere-So,how do we define the rate at which we are moving in time?Do we lack a concept to understand this?
                                – Krishna Deshmukh
                                1 hour ago




                                safesphere-So,how do we define the rate at which we are moving in time?Do we lack a concept to understand this?
                                – Krishna Deshmukh
                                1 hour ago












                                @KrishnaDeshmukh No, the concept is fine. The rate of our own (proper) time is $1$ (as in one second per second), not $gamma$ as some answers here suggest. In other words, our own $gamma=1$. To convert rate to speed, you need to multiply it by the speed of light. So we are moving in time at the rate of $gamma=1$ with the speed of $c$.
                                – safesphere
                                14 mins ago




                                @KrishnaDeshmukh No, the concept is fine. The rate of our own (proper) time is $1$ (as in one second per second), not $gamma$ as some answers here suggest. In other words, our own $gamma=1$. To convert rate to speed, you need to multiply it by the speed of light. So we are moving in time at the rate of $gamma=1$ with the speed of $c$.
                                – safesphere
                                14 mins ago












                                How can one talk about a constant speed and its rate of change along time at once?If its changing then ,its not constant.
                                – Krishna Deshmukh
                                5 mins ago




                                How can one talk about a constant speed and its rate of change along time at once?If its changing then ,its not constant.
                                – Krishna Deshmukh
                                5 mins ago










                                up vote
                                0
                                down vote













                                I'd like to think of it as moving at an infinitely fast pace. Time is not exactly spatial, so light speed in a vaccuum, parallel to safesphere's answer, would not truly apply. You pass an infinite amount of probabilities and times in an infinitesmally small amount of time, so the velocity is most likely infinite. But since this infinite velocity is in a seperate dimension, it won't affect us.






                                share|cite|improve this answer
























                                  up vote
                                  0
                                  down vote













                                  I'd like to think of it as moving at an infinitely fast pace. Time is not exactly spatial, so light speed in a vaccuum, parallel to safesphere's answer, would not truly apply. You pass an infinite amount of probabilities and times in an infinitesmally small amount of time, so the velocity is most likely infinite. But since this infinite velocity is in a seperate dimension, it won't affect us.






                                  share|cite|improve this answer






















                                    up vote
                                    0
                                    down vote










                                    up vote
                                    0
                                    down vote









                                    I'd like to think of it as moving at an infinitely fast pace. Time is not exactly spatial, so light speed in a vaccuum, parallel to safesphere's answer, would not truly apply. You pass an infinite amount of probabilities and times in an infinitesmally small amount of time, so the velocity is most likely infinite. But since this infinite velocity is in a seperate dimension, it won't affect us.






                                    share|cite|improve this answer












                                    I'd like to think of it as moving at an infinitely fast pace. Time is not exactly spatial, so light speed in a vaccuum, parallel to safesphere's answer, would not truly apply. You pass an infinite amount of probabilities and times in an infinitesmally small amount of time, so the velocity is most likely infinite. But since this infinite velocity is in a seperate dimension, it won't affect us.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 58 mins ago









                                    Curious Fish

                                    6910




                                    6910




















                                        up vote
                                        -1
                                        down vote













                                        In relativity the time coordinate is $x_o=ct$ and its time derivative is $c$. Therefore the time component of four-velocity is the speed of light in vacuum.






                                        share|cite|improve this answer




















                                        • In relativity the time coordinate is xo=ct Well, not really. There is a variety of possible conventions. The most common convention among relativists is to work in units where $c=1$, and therefore never write any factors of $c$ anywhere. Therefore the time component of four-velocity is the speed of light in vacuum. Not true. This holds only for a four-velocity describing a world-line that is at rest relative to a particular Minkowski frame.
                                          – Ben Crowell
                                          1 hour ago










                                        • @BenCrowell You are wrong again. (1) For geometrized units where $c=1$, my answer stands. I never stated the specific value of $c$. (2) The question is "With what velocity are we moving along the time dimension" thus referring to the proper time. So my answer is correct despite your downvote. There is no time dilation in the proper frame.
                                          – safesphere
                                          1 hour ago











                                        • safesphere, since $vec U equiv dvecX/dtau$, isn't the time component of $vecU$ equal to $gamma c$ rather than $c$?
                                          – Alfred Centauri
                                          33 mins ago










                                        • @AlfredCentauri Yes, but the question is about the proper frame, so $gamma=1$.
                                          – safesphere
                                          26 mins ago










                                        • @safesphere Actually, the OP's question never mentions Proper Time and neither does your answer.
                                          – K7PEH
                                          19 mins ago














                                        up vote
                                        -1
                                        down vote













                                        In relativity the time coordinate is $x_o=ct$ and its time derivative is $c$. Therefore the time component of four-velocity is the speed of light in vacuum.






                                        share|cite|improve this answer




















                                        • In relativity the time coordinate is xo=ct Well, not really. There is a variety of possible conventions. The most common convention among relativists is to work in units where $c=1$, and therefore never write any factors of $c$ anywhere. Therefore the time component of four-velocity is the speed of light in vacuum. Not true. This holds only for a four-velocity describing a world-line that is at rest relative to a particular Minkowski frame.
                                          – Ben Crowell
                                          1 hour ago










                                        • @BenCrowell You are wrong again. (1) For geometrized units where $c=1$, my answer stands. I never stated the specific value of $c$. (2) The question is "With what velocity are we moving along the time dimension" thus referring to the proper time. So my answer is correct despite your downvote. There is no time dilation in the proper frame.
                                          – safesphere
                                          1 hour ago











                                        • safesphere, since $vec U equiv dvecX/dtau$, isn't the time component of $vecU$ equal to $gamma c$ rather than $c$?
                                          – Alfred Centauri
                                          33 mins ago










                                        • @AlfredCentauri Yes, but the question is about the proper frame, so $gamma=1$.
                                          – safesphere
                                          26 mins ago










                                        • @safesphere Actually, the OP's question never mentions Proper Time and neither does your answer.
                                          – K7PEH
                                          19 mins ago












                                        up vote
                                        -1
                                        down vote










                                        up vote
                                        -1
                                        down vote









                                        In relativity the time coordinate is $x_o=ct$ and its time derivative is $c$. Therefore the time component of four-velocity is the speed of light in vacuum.






                                        share|cite|improve this answer












                                        In relativity the time coordinate is $x_o=ct$ and its time derivative is $c$. Therefore the time component of four-velocity is the speed of light in vacuum.







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered 1 hour ago









                                        safesphere

                                        6,86611339




                                        6,86611339











                                        • In relativity the time coordinate is xo=ct Well, not really. There is a variety of possible conventions. The most common convention among relativists is to work in units where $c=1$, and therefore never write any factors of $c$ anywhere. Therefore the time component of four-velocity is the speed of light in vacuum. Not true. This holds only for a four-velocity describing a world-line that is at rest relative to a particular Minkowski frame.
                                          – Ben Crowell
                                          1 hour ago










                                        • @BenCrowell You are wrong again. (1) For geometrized units where $c=1$, my answer stands. I never stated the specific value of $c$. (2) The question is "With what velocity are we moving along the time dimension" thus referring to the proper time. So my answer is correct despite your downvote. There is no time dilation in the proper frame.
                                          – safesphere
                                          1 hour ago











                                        • safesphere, since $vec U equiv dvecX/dtau$, isn't the time component of $vecU$ equal to $gamma c$ rather than $c$?
                                          – Alfred Centauri
                                          33 mins ago










                                        • @AlfredCentauri Yes, but the question is about the proper frame, so $gamma=1$.
                                          – safesphere
                                          26 mins ago










                                        • @safesphere Actually, the OP's question never mentions Proper Time and neither does your answer.
                                          – K7PEH
                                          19 mins ago
















                                        • In relativity the time coordinate is xo=ct Well, not really. There is a variety of possible conventions. The most common convention among relativists is to work in units where $c=1$, and therefore never write any factors of $c$ anywhere. Therefore the time component of four-velocity is the speed of light in vacuum. Not true. This holds only for a four-velocity describing a world-line that is at rest relative to a particular Minkowski frame.
                                          – Ben Crowell
                                          1 hour ago










                                        • @BenCrowell You are wrong again. (1) For geometrized units where $c=1$, my answer stands. I never stated the specific value of $c$. (2) The question is "With what velocity are we moving along the time dimension" thus referring to the proper time. So my answer is correct despite your downvote. There is no time dilation in the proper frame.
                                          – safesphere
                                          1 hour ago











                                        • safesphere, since $vec U equiv dvecX/dtau$, isn't the time component of $vecU$ equal to $gamma c$ rather than $c$?
                                          – Alfred Centauri
                                          33 mins ago










                                        • @AlfredCentauri Yes, but the question is about the proper frame, so $gamma=1$.
                                          – safesphere
                                          26 mins ago










                                        • @safesphere Actually, the OP's question never mentions Proper Time and neither does your answer.
                                          – K7PEH
                                          19 mins ago















                                        In relativity the time coordinate is xo=ct Well, not really. There is a variety of possible conventions. The most common convention among relativists is to work in units where $c=1$, and therefore never write any factors of $c$ anywhere. Therefore the time component of four-velocity is the speed of light in vacuum. Not true. This holds only for a four-velocity describing a world-line that is at rest relative to a particular Minkowski frame.
                                        – Ben Crowell
                                        1 hour ago




                                        In relativity the time coordinate is xo=ct Well, not really. There is a variety of possible conventions. The most common convention among relativists is to work in units where $c=1$, and therefore never write any factors of $c$ anywhere. Therefore the time component of four-velocity is the speed of light in vacuum. Not true. This holds only for a four-velocity describing a world-line that is at rest relative to a particular Minkowski frame.
                                        – Ben Crowell
                                        1 hour ago












                                        @BenCrowell You are wrong again. (1) For geometrized units where $c=1$, my answer stands. I never stated the specific value of $c$. (2) The question is "With what velocity are we moving along the time dimension" thus referring to the proper time. So my answer is correct despite your downvote. There is no time dilation in the proper frame.
                                        – safesphere
                                        1 hour ago





                                        @BenCrowell You are wrong again. (1) For geometrized units where $c=1$, my answer stands. I never stated the specific value of $c$. (2) The question is "With what velocity are we moving along the time dimension" thus referring to the proper time. So my answer is correct despite your downvote. There is no time dilation in the proper frame.
                                        – safesphere
                                        1 hour ago













                                        safesphere, since $vec U equiv dvecX/dtau$, isn't the time component of $vecU$ equal to $gamma c$ rather than $c$?
                                        – Alfred Centauri
                                        33 mins ago




                                        safesphere, since $vec U equiv dvecX/dtau$, isn't the time component of $vecU$ equal to $gamma c$ rather than $c$?
                                        – Alfred Centauri
                                        33 mins ago












                                        @AlfredCentauri Yes, but the question is about the proper frame, so $gamma=1$.
                                        – safesphere
                                        26 mins ago




                                        @AlfredCentauri Yes, but the question is about the proper frame, so $gamma=1$.
                                        – safesphere
                                        26 mins ago












                                        @safesphere Actually, the OP's question never mentions Proper Time and neither does your answer.
                                        – K7PEH
                                        19 mins ago




                                        @safesphere Actually, the OP's question never mentions Proper Time and neither does your answer.
                                        – K7PEH
                                        19 mins ago

















                                         

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