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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?
time relativity differentiation
edited 17 mins ago
Kyle Kanos
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asked 1 hour ago
Krishna Deshmukh
595
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up vote
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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
edited 17 mins ago
Kyle Kanos
21.4k114792
asked 1 hour ago
Krishna Deshmukh
595
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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
edited 17 mins ago
Kyle Kanos
21.4k114792
asked 1 hour ago
Krishna Deshmukh
595
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
time relativity differentiation
edited 17 mins ago
Kyle Kanos
21.4k114792
asked 1 hour ago
Krishna Deshmukh
595
edited 17 mins ago
Kyle Kanos
21.4k114792
asked 1 hour ago
Krishna Deshmukh
595
edited 17 mins ago
Kyle Kanos
21.4k114792
edited 17 mins ago
Kyle Kanos
21.4k114792
edited 17 mins ago
Kyle Kanos
21.4k114792
21.4k114792
asked 1 hour ago
Krishna Deshmukh
595
asked 1 hour ago
Krishna Deshmukh
595
asked 1 hour ago
Krishna Deshmukh
595
595
add a comment |Â
add a comment |Â
5 Answers
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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.
answered 1 hour ago
Ben Crowell
46k3147278
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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.)
answered 45 mins ago
Alfred Centauri
46.5k344139
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At exactly 1 second per second.
answered 1 hour ago
Cer
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
add a comment |Â
up vote
0
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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.
answered 58 mins ago
Curious Fish
6910
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-1
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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.
answered 1 hour ago
safesphere
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
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5 Answers
5
active
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5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered 1 hour ago
Ben Crowell
46k3147278
add a comment |Â
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.
answered 1 hour ago
Ben Crowell
46k3147278
add a comment |Â
up vote
1
down vote
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.
answered 1 hour ago
Ben Crowell
46k3147278
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.
answered 1 hour ago
Ben Crowell
46k3147278
answered 1 hour ago
Ben Crowell
46k3147278
answered 1 hour ago
Ben Crowell
46k3147278
answered 1 hour ago
Ben Crowell
46k3147278
46k3147278
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add a comment |Â
up vote
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.)
answered 45 mins ago
Alfred Centauri
46.5k344139
add a comment |Â
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.)
answered 45 mins ago
Alfred Centauri
46.5k344139
add a comment |Â
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.)
answered 45 mins ago
Alfred Centauri
46.5k344139
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.)
answered 45 mins ago
Alfred Centauri
46.5k344139
edited 40 mins ago
answered 45 mins ago
Alfred Centauri
46.5k344139
answered 45 mins ago
Alfred Centauri
46.5k344139
answered 45 mins ago
Alfred Centauri
46.5k344139
46.5k344139
add a comment |Â
add a comment |Â
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0
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At exactly 1 second per second.
answered 1 hour ago
Cer
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
add a comment |Â
up vote
0
down vote
At exactly 1 second per second.
answered 1 hour ago
Cer
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
add a comment |Â
up vote
0
down vote
up vote
0
down vote
At exactly 1 second per second.
answered 1 hour ago
Cer
262
At exactly 1 second per second.
answered 1 hour ago
Cer
262
answered 1 hour ago
Cer
262
answered 1 hour ago
Cer
262
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
add a comment |Â
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
add a comment |Â
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.
answered 58 mins ago
Curious Fish
6910
add a comment |Â
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.
answered 58 mins ago
Curious Fish
6910
add a comment |Â
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.
answered 58 mins ago
Curious Fish
6910
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.
answered 58 mins ago
Curious Fish
6910
answered 58 mins ago
Curious Fish
6910
answered 58 mins ago
Curious Fish
6910
answered 58 mins ago
Curious Fish
6910
6910
add a comment |Â
add a comment |Â
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.
answered 1 hour ago
safesphere
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
 |Â
show 1 more comment
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.
answered 1 hour ago
safesphere
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
 |Â
show 1 more comment
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.
answered 1 hour ago
safesphere
6,86611339
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.
answered 1 hour ago
safesphere
6,86611339
answered 1 hour ago
safesphere
6,86611339
answered 1 hour ago
safesphere
6,86611339
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
 |Â
show 1 more comment
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
 |Â
show 1 more comment
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