Misunderstanding of "proper velocity"

[etotheipi]

I've been looking around and have gotten quite muddled about some concepts in special relativity. In the following I won't use four-vectors since I'd like to clear up the confusion first before adding more complexity!

The proper velocity/celerity is said to be $\frac{dx}{d \tau} = \frac{dx}{dt} \frac{dt}{d\tau} = v \gamma$, and this is supposedly frame invariant. If a rocket accelerates away from a planet at $v$ (in one dimension), in the planet frame the proper velocity of the rocket is then $v\gamma(v)$ whilst the proper velocity of the rocket in the rocket frame appears to be zero? However, this can't be right as this quantity is supposed to be invariant.

I know I must be doing something horrifically wrong but I don't know whereabouts to go from here!

 

[Orodruin]

The proper velocity/celerity is said to be dxdτ=dxdtdtdτ=vγdxdτ=dxdtdtdτ=vγ\frac{dx}{d \tau} = \frac{dx}{dt} \frac{dt}{d\tau} = v \gamma, and this is supposedly frame invariant.

This object is clearly frame dependent as it is zero if $v = 0$ and non-zero otherwise.

While $\tau$ is frame invariant, $x$ is not.

I won't use four-vectors since I'd like to clear up the confusion first before adding more complexity!

I would say using 4-vectors generally decreases the complexity, not the other way around.

 

[Orodruin]

Addendum: Note that what you have defined is the x-component of the 4-velocity, which is an invariant object in itself. However, its components are going to differ between frames, just as a normal 3-vector is an invariant geometrical object, but its components depend on the coordinate system.

 

[etotheipi]

This object is clearly frame dependent as it is zero if $v = 0$ and non-zero otherwise.

While $\tau$ is frame invariant, $x$ is not.I would say using 4-vectors generally decreases the complexity, not the other way around.

Thank you, that's reassuring, though I still don't fully understand a few points. If the proper acceleration (which hopefully this time I'm right in saying is frame invariant) were to be determined from both frames, both values should agree upon whatever the value on an accelerometer on the rocket would measure. I'm now supposing that the rocket is speeding away at some acceleration $a$ as measured by the accelerometer.

If the proper acceleration equals $\frac{du}{dt}$ where $u$ is the proper velocity and $t$ is the coordinate time, then wouldn't this turn out to be zero for the frame traveling with the rocket (in which the proper velocity is always zero) but non-zero for the planet frame?

 

[Orodruin]

Thank you, that's reassuring, though I still don't fully understand a few points. If the proper acceleration (which hopefully this time I'm right in saying is frame invariant) were to be determined from both frames, both values should agree upon whatever the value on an accelerometer on the rocket would measure. I'm now supposing that the rocket is speeding away at some acceleration $a$ as measured by the accelerometer.

If the proper acceleration equals $\frac{du}{dt}$ where $u$ is the proper velocity and $t$ is the coordinate time, then wouldn't this turn out to be zero for the frame traveling with the rocket (in which the proper velocity is always zero) but non-zero for the planet frame?

No, this is not the definition of the proper acceleration. The proper acceleration is equal to the acceleration in the instantaneous rest frame (i.e., the frame where the velocity is zero at the moment of consideration). This does not mean that the derivative of the velocity in that frame is zero.

It is also generally easier to compute the proper acceleration based on the derivative of the 4-velocity with respect to proper time.

 

[Orodruin]

then wouldn't this turn out to be zero for the frame traveling with the rocket

The frame traveling with the rocket is non-inertial (because the rocket is non-inertial). When we talk about the instantaneous rest frame, we refer to the inertial frame where the rocket is momentarily at rest, not the accelerated frame where the rocket is always at rest.

 

[etotheipi]

The frame traveling with the rocket is non-inertial (because the rocket is non-inertial). When we talk about the instantaneous rest frame, we refer to the inertial frame where the rocket is momentarily at rest, not the accelerated frame where the rocket is always at rest.

Thank you for your help, for context the question I was trying to solve when I got confused went as follows:

Suppose a spacecraft accelerates with constant acceleration $a$ (as measured by the spacecraft ’s onboard accelerometers). At $t=0$ it is at rest with respect to a planet. Work out its speed relative to the planet as a function of time (a) as measured by clocks on the spacecraft , and (b) as measured by clocks on the planet. Note that the instantaneous speed of the craft relative to the planet will be agreed upon by spacecraft and planet

I didn't really know where to start so I found a relation which stated something along the lines of $a_{proper} = \gamma^{3} a_{coord}$. So in the planet frame the coordinate acceleration would be measured to be $\frac{a}{\gamma^{3}}$ (resulting in $v = \frac{a}{\gamma^{3}} t$). I don't know if this is correct though!

 

[Orodruin]

I didn't really know where to start so I found a relation which stated something along the lines of $a_{proper} = \gamma^{3} a_{coord}$. So in the planet frame the coordinate acceleration would be measured to be $\frac{a}{\gamma^{3}}$ (resulting in $v = \frac{a}{\gamma^{3}} t$). I don't know if this is correct though!

This is not correct because $\gamma$ depends on $v$ and is therefore not a constant. You therefore cannot integrate just by multiplying by $t$. (a) is most easily solved by deriving a relation between $dx$ in the original rest frame and $d\tau$. This can be done by using the formula for velocity addition.

 

[etotheipi]

This is not correct because $\gamma$ depends on $v$ and is therefore not a constant. You therefore cannot integrate just by multiplying by $t$. (a) is most easily solved by deriving a relation between $dx$ in the original rest frame and $d\tau$. This can be done by using the formula for velocity addition.

I wonder if I'm going in the right direction: In the inertial planet frame,

$dv = adt \implies v=at$

whilst we also have $dt = \gamma(v) d\tau$ which seems to suggest

$\sqrt{1-\frac{v^{2}}{c^{2}}} dv = a d\tau$, which might then be integrated? Though this doesn't seem quite right and doesn't use velocity addition as you suggested... I'll keep looking!

 

[mfb]

$dv = adt \implies v=at$

Only for constant a. The spacecraft won't maintain that for a long time.

Have you heard of rapidity? Rapidity can be added easily.

 

[etotheipi]

Only for constant a. The spacecraft won't maintain that for a long time.

Have you heard of rapidity? Rapidity can be added easily.

I suppose you're right that the assumption of constant acceleration is questionable, though it is asserted in the question so I'm happy to take it to be so. I am not familiar with rapidity, though I can have a look!

 

[PeterDonis]

Only for constant a. The spacecraft won't maintain that for a long time.

That's not the problem since we can always consider an idealized spacecraft that can maintain constant proper acceleration indefinitely. The problem is that "constant a" means constant proper acceleration, which is not the same as constant coordinate acceleration in a fixed inertial frame. But in post #9 the equations are written using an assumption of constant coordinate acceleration.

 

[PeroK]

I didn't really know where to start so I found a relation which stated something along the lines of $a_{proper} = \gamma^{3} a_{coord}$. So in the planet frame the coordinate acceleration would be measured to be $\frac{a}{\gamma^{3}}$ (resulting in $v = \frac{a}{\gamma^{3}} t$). I don't know if this is correct though!

If you expand $\gamma$ here you get:

$\frac{dv}{dt} = a(1- \frac{v^2}{c^2})^{3/2}$

If you are stuck for a better idea you can always integrate that!

Note: if you are going to use the shorthand $\gamma$ for a particle, then you must remember it is a function of $v$, hence $t$ etc. Unlike the $\gamma$ factor for the transformation between inertial reference frames, which is a constant.

 

[mfb]

That's not the problem since we can always consider an idealized spacecraft that can maintain constant proper acceleration indefinitely. The problem is that "constant a" means constant proper acceleration, which is not the same as constant coordinate acceleration in a fixed inertial frame. But in post #9 the equations are written using an assumption of constant coordinate acceleration.

I assumed the acceleration a to be in the reference frame of Earth, because that's what the equations used. In that case you can't maintain it indefinitely as the speed of light is the limit (and before that the spacecraft needs implausible proper acceleration).

 

[Orodruin]

I suppose you're right that the assumption of constant acceleration is questionable, though it is asserted in the question so I'm happy to take it to be so. I am not familiar with rapidity, though I can have a look!

The question states that the proper acceleration is constant, not that the acceleration in the lab frame is constant. As $\gamma$ depends on the velocity and there is a factor $\gamma^3$ between the proper acceleration and the acceleration in the lab frame, the acceleration in the lab frame will certainly not be constant.

 

[pervect]

I've been looking around and have gotten quite muddled about some concepts in special relativity. In the following I won't use four-vectors since I'd like to clear up the confusion first before adding more complexity!

The proper velocity/celerity is said to be $\frac{dx}{d \tau} = \frac{dx}{dt} \frac{dt}{d\tau} = v \gamma$, and this is supposedly frame invariant. If a rocket accelerates away from a planet at $v$ (in one dimension), in the planet frame the proper velocity of the rocket is then $v\gamma(v)$ whilst the proper velocity of the rocket in the rocket frame appears to be zero? However, this can't be right as this quantity is supposed to be invariant.

I know I must be doing something horrifically wrong but I don't know whereabouts to go from here!

I'll introduce by making a note that this response is written in the context of special relativity, not general relativity.

The components of the proper velocity are not frame independent. The object as a whole can be regarded as frame independent, though, even though it's individual components are not, because it is a rank 1 tensor, i.e. a 4-vector. So it is a a "geometric object". The components of this object transform in a standard manner, i.e via the Lorentz transform. When all objects transform via the same transforms, we can see that the choice of frame doesn't matter, because all the objects transform according to the same consistent scheme.

The scheme includes tensor of all ranks, scalars (rank 0 tensors that are numbers that are frame independent), vectors, rank 1 tensors that transform via the Lorentz transform, and higher rank tensors (rank 2 and above), that you may not need at this point but eventually become important and useful.

 

[PeterDonis]

I assumed the acceleration a to be in the reference frame of Earth, because that's what the equations used.

It can't be, because the question specifies a as the constant proper acceleration, which will not be the same as coordinate acceleration for more than an instant in any frame. The OP's equations are confusing the two, which is why they are giving him incorrect answers.

 

[SiennaTheGr8]

The word proper is used inconsistently in the literature, and indeed even on this thread!