Can the Lorentz factor be modified for non-inertial frames?

[bwana]

The term "Lorentz transformations" only refers to transformations between inertial frames.
However, if we differentiate velocity with respect to time, we obtain acceleration.
The Lorentz factor says:
t(0 reference frame observer at rest) / t(moving) = 1 / Sqrt[1–(v/c)^2]
t₀ / t_m = 1/[(1-(v/c)²]^1/2
If we want t(moving) to be changing then we need to differentiate the lorentz factor with respect to t(moving)
v is a function of t so
t₀ / t_accelerating = ∂ (1/ [(1-(v(t) / c)²]^1/2) ∂t
t₀ / t_accelerating = (v[t] ∂v[t]/∂t) / (c² (1 - v[t]²/c²])^3/2]

It looks better here:
https://www.wolframalpha.com/input/?i=differentiate+1/(1-(v(t)/c)^2)^1/2
so the calculation would require we know the acceleration AND velocity at the moment we want to know what the time dilation is doing.

Although we can do the mathematical gymnastics, I do not know if there are any assumptions in the derivation of the Lorentz transformation that REQUIRE an inertial frame.
Or is the equation above really valid?

 

[PeterDonis]

The Lorentz factor says:
t(0 reference frame observer at rest) / t(moving) = 1 / Sqrt[1–(v/c)^2]
t0 / tm = 1/[(1-(v/c)2]1/2

No, that's not what the Lorentz transformation says. You can't just transform $t$. You have to transform the spatial coordinates as well. Just looking at $t$ alone will give you wrong answers.

I do not know if there are any assumptions in the derivation of the Lorentz transformation that REQUIRE an inertial frame.

Yes, there are. As you said yourself, the Lorentz transformation only works between inertial frames.

s the equation above really valid?

Not as an equation that tells you anything about Lorentz transformations.

What do you want to do with such an equation?

 

[bhobba]

Although we can do the mathematical gymnastics, I do not know if there are any assumptions in the derivation of the Lorentz transformation that REQUIRE an inertial frame.

The derivation depends critically on being in an inertial frame:
http://www2.physics.umd.edu/~yakovenk/teaching/Lorentz.pdf

A point needs to be made and books on classical mechanics often do not make it - they say an inertial frame is one in which Newtons first law holds. That's true - but everything is much clearer, and you see why the first law is true, if you look at the symmetry properties of an inertial frame. Using this by definition it is a frame in which the laws of physics are the same at all points, all directions, and all instants of time. It can easily be shown all inertial frames travel at constant velocity with respect to each other. What it doesn't tell us however is about a frame that travels at constant velocity relative to an inertial frame. That is the real content of the first law - all frames traveling at constant velocity to an inertial frame is also inertial. Then the first law is easy. Have a particle at rest in an inertial frame. If it accelerates in some direction then we have this law particles at rest accelerate in such and such direction. This direction is then special. Rotate the frame - then in that frame it must accelerate in the same direction - but it will be in a different direction in the first frame - contradiction. Hence a particle at rest remains at rest. If we have a particle moving at constant velocity in an inertial frame simply go to a frame where it is at rest - it must remain at rest in that frame - so in the original frame continues to move at constant velocity.

Accelerations can be handled in SR (ie you do not have to go to GR) by means of a simple trick. You break the acceleration into a large number of constant velocities with a jump in between. This is how the twin paradox is resolved in SR:
http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_spacetime.html

Those dx and dt's are the breaking it up bit and you then do the usual calculus thing and take them to zero or if you are lazy like me, despite my love of analysis, just think of them as actual numbers so small you do not have to worry about them.

Thanks
Bill

 

[bwana]

Differentiating the Lorentz transformation would let me calculate the time dilation occurring in an accelerating frame. Another formula also let's me calculate time dilation in a gravitationally accelerated frame - the Schwarzschild equation. The equivalence principle says these two accelerations are equivalent. This gives an interesting equivalence with perplexing conclusions.

Although I appreciate your replies, they seem not to address my question - is the derivative of the Lorentz equation correct? And to keep things simple, just consider motion in one direction and just consider gravity in the same direction as well.

 

[PeterDonis]

Differentiating the Lorentz transformation would let me calculate the time dilation occurring in an accelerating frame.

How? What does the Lorentz transformation or its derivative have to do with an accelerating frame? Why do you think this has anything to do with an accelerating frame?

(Note also that the thing you've taken a derivative of is not the Lorentz transformation, as I've already pointed out. It's the gamma factor that appears in the transformation, but that's not the same thing as the transformation itself. I'm not sure what it would even mean to take the derivative of a transformation.)

If you want to know how things look in an accelerating frame, you need to actually look at a valid one. I suggest looking up Rindler coordinates.

Another formula also let's me calculate time dilation in a gravitationally accelerated frame - the Schwarzschild equation.

What formula are you talking about?

The equivalence principle says these two accelerations are equivalent.

Sort of.

This gives an interesting equivalence with perplexing conclusions.

I'm not sure what you mean by this.

is the derivative of the Lorentz equation correct?

"Correct" in what sense? If you mean, did you do the math right, it looks right to me (at least as taking a derivative of the Lorentz gamma factor--see my comment above). But doing the math right does not mean the math you've done means anything physically.

 

[Ibix]

Differentiating the Lorentz transformation would let me calculate the time dilation occurring in an accelerating frame.

The Lorentz transforms assume an array of inertial clocks at rest with respect to one another, and another array of inertial clocks at rest with respect to one another and moving with velocity v with respect to the first set. The Lorentz transforms then tell you that position x and time t according to one set of clocks is position x' and time t' according to the other.

You need an equivalent process for a non-inertial frame, starting with deciding on the state of motion of your clocks and specifying a clock synchronisation procedure. Or to derive a transformation between an inertial frame and your chosen non-inertial system, which is equivalent. Simply differentiating the gamma factor does not do all this.

It seems to me that what you actually want to know is the tick rate of an accelerating clock compared to that of an inertial one. That's a rather simpler problem. Is that right?

 

[bwana]

thank you ibix. you get where I'm going with this.
the fundamental conundrum that bothers me is the following:
consider an elevator at rest on the ground ( in a gravity well) compared to another at rest in deep space. We can calculate how much slower the a clock in the elevator on Earth ticks based on gravitic time dilation. a person on the ground can call the person in the deep space elevator after a while and confirm that their clocks have ticked at different rates even tho the clocks are identical.
T_gravity>T_space

now consider a third elevator in deep space moving relative to the elevator at rest in deep space. We can calculate the time dilation of a clock in this moving elevator as well. T_moving
And the person riding in this elevator can call the person in the elevator not moving in deep space and confirm that clocks have ticked at different rates.
T_moving>T_space
in fact, this elevator can move at a specific velocity, v, such that his time dilation is the same as the dilation experienced by the person on the planet relative to the stationary elevator in deep space.
T_moving=T_gravity

now consider a fourth elevator in deep space moving at velocity v, which then turns on its rocket motor to accelerate at 1 g.
T_accel
The time dilation in this elevator relative to the stationary elevator in deep space has to be more than the elevator moving at a steady speed,v.
T_accel>T_moving=T_gravity
And by extension, the time dilation in this accelerating elevator has to be greater than the time dilation experienced by the person on the ground.
The problem I have is that T_accel>T_gravity
The equivalence principle says that the same physics happens in this accelerating elevator as the elevator on the ground. There is no experiment that can be done to distinguish between the two environments. Yet by virtue of the logic above, the time dilation in the accelerating elevator is greater than the time dilation of elevator in the gravity well.

You might argue that special relativity does not apply to the accelerating elevator because it is not an inertial frame of reference. But we can approximate the accelerating elevator with many little intervals of different speeds within each of which the elevator speed is constant.

The conclusion I would draw is that I can tell whether I am on a planet or in an accelerating elevator by measuring my time dilation. On the planet time dilation has the same rate. In an accelerating elevator, my rate of time dilation should be increasing. But the equivalence principle says I shouldn't be able to tell the difference.

 

[Nugatory]

You might argue that special relativity does not apply to the accelerating elevator because it is not an inertial frame of reference.

That argument would be incorrect. Special relativity works just fine for accelerated frames in flat spacetime (it has to, because the difference between an accelerated and an inertial frame is just a coordinate transformation). General relativity is only required when spacetime is not flat, meaning that tidal gravitational effects are present.

now consider a third elevator in deep space moving relative to the elevator at rest in deep space. We can calculate the time dilation of a clock in this moving elevator as well.
And the person riding in this elevator can call the person in the elevator not moving in deep space and confirm that clocks have ticked at different rates.

Yes, but they won't hear what you're expecting them to hear. Both elevator people will discover that the other's clock is slower than their own. You've overlooked the relativity of simultaneity, which makes trying to compare "rates" of time dilation treacherous (so treacherous that I put scare-quotes around the word "rates").

Also, you must remember that the equivalence principle only applies locally.

 

[bhobba]

The equivalence principle says that the same physics happens in this accelerating elevator as the elevator on the ground. There is no experiment that can be done to distinguish between the two environments.

That's an over simplification. I suggest you do a bit of reading on the equivalence principle - in particular the strong and weak versions

Yet by virtue of the logic above, the time dilation in the accelerating elevator is greater than the time dilation of elevator in the gravity well.

You really have got yourself into a logical knot. You should not be comparing clocks until they are bought together by slow transport. I suggest you read the following interesting book:
https://www.amazon.com/dp/0393337685/?tag=pfamazon01-20

Einstein made that and similar mistakes that were not picked up in peer review. In fact his early papers were so riddled with them when they issued a book collecting his early papers the commentary correcting the mistakes was more than the papers themselves. I must also add this in no way diminishes Einsteins genius, but is simply a reflection on how subtle the material is and it took a while to fully flesh out what was going on.

You might argue that special relativity does not apply to the accelerating elevator because it is not an inertial frame of reference. But we can approximate the accelerating elevator with many little intervals of different speeds within each of which the elevator speed is constant.

Yes, as I explained you can do that, but that is not the mistake you are making. Try rewriting your objection where clock comparisons are done when the clocks are bought together. As it stands its like one of those logical conundrums where you know there has to be a mistake, but groan at finding it - 1=0 proofs are often like that. Nugatory already above posted one issue because you are not comparing clocks that have been bought together - you constantly run into issues like that trying to do what you are doing.

As an aside what is necessary to derive GR these days is simpler than when Einstein developed it. This is due to a theorem called Lovelock's Theorem:
https://en.wikipedia.org/wiki/Lovelock's_theorem

It almost seems to fall out of SR and its tensor formulation, plus a few generally accepted physical principles such as a field has a Lagrangian. It's an interesting exercise seeing just what is necessary to assume so the theorem can be applied, but that would be a thread in itself.

Using the equivalence principle you run into a long standing issue that has created considerable, sometimes heated, debate, even here, eg:
https://www.physicsforums.com/threads/do-tidal-forces-mean-the-equivalence-principle-is-bs.505084/

It's to do with the issue of tidal forces. Gravity is space-time curvature so you can't entirely get rid of it using accelerating frames. Personally, because of that I am not a fan of the equivalence principle as a justification for GR, but opinions vary. It is however suggestive of space-time curvature, so does have heuristic value in why you would want to investigate curved space-time. But as I said - opinions vary. I would say my view is a minority one. The technically correct answer was posted by Atty at the end of the link above.

Thanks
Bill

 

[Cryo]

If you are only interested in time-dilation, I see no problem with working with accelerating observer.

Lets say we have an accelerating observer $A$ and a stationary one $B$. I will describe everything the the frame of observer $B$, who uses $t$ for time and $\mathbf{r}=\left(x,y,z\right)$ for space.

Now, the world-line of observer $A$, as seen by $B$ is: $\bar{x}^\alpha=\bar{x}^\alpha\left(\tau\right)=\left(c\bar{t},\bar{\mathbf{r}}\right)^\alpha$, where $\tau$ is $A$'s proper time. Basically, $\tau$ is proportional to the length of $A$'s world-line from the point labeled by $\tau=0$ (or any other point we choose). Function $\bar{t}=\bar{t}\left(\tau\right)$ is the temporal position of $A$ as seen by $B$ ($\bar{\mathbf{r}}$ is the spatial position of $A$)

Now let's look at the four-velocity: $u^\alpha\left(\tau\right)=\frac{d \bar{x}^\alpha}{d\tau}=\left(c, \mathbf{v}\right)^\alpha \frac{d\bar{t}}{d\tau}$, where $\mathbf{v}=\frac{d\bar{\mathbf{r}}}{d\bar{t}}$ is the velocity of $A$ relative to $B$. Note the velocity need not be constant! Also note that the magnitude squared of four-velocity is $u^\alpha u_\alpha = \left(c^2 - v^2\right)\left(\frac{d\bar{t}}{d\tau}\right)^2$

Next, consider a small displacement along the world-line of $A$: $\delta \bar{x}^\alpha$. We know that since proper time measures length along the world-line. It follows that the length of the displacement $\delta \bar{x}^\alpha$is proportional to the difference in proper time across this displacement: $\delta \bar{x}^\alpha \delta \bar{x}_\alpha = c^2 (\delta \tau)^2$, it follows that $\frac{\delta \bar{x}^\alpha}{\delta\tau}\frac{\delta \bar{x}_\alpha}{\delta\tau}=c^2$, so $u^\alpha u_\alpha = \frac{d \bar{x}^\alpha}{d\tau}\frac{d \bar{x}_\alpha}{d\tau}=c^2$

Putting this together we get: $c^2 = \left(c^2 - v^2\right)\left(\frac{d\bar{t}}{d\tau}\right)^2$, so $\gamma=\frac{d\bar{t}}{d\tau}=1/\sqrt{1-v^2/c^2}$ (we always choose the positive solution). At no point did I assume that speed or velocity are constant.

So the equation for time-dilation, i.e. the difference in the rate of flow of lab-time relative to proper time, is the same for accelerated or not accelerated observers. The way I think about this, is that time-dilation is like the angle between the temporal axis of the stationary observer, and the tangent to the world-line of the moving observer. All that changes in case of acceleration, is that tangent now changes along the world-line. That's ok. You can still define the angle the same way at any proper time.

 

[Ibix]

thank you ibix. you get where I'm going with this.
the fundamental conundrum that bothers me is the following:

There are a number of fundamental misconceptions in this.

First, kinematic and gravitational time dilation are different. In this case, the important difference is that kinematic time dilation is reciprocal - if I see your clock ticking slow, you will also see my clock ticking slow. This follows from the principle of relativity - we can both regard ourselves as at rest and the other as moving. Gravitational time dilation is not reciprocal - if I see your clock ticking slow you will see mine ticking fast. So your second and third lifts will both see regard the other's clock as ticking slow. Both will agree that the first lift's clock ticks slower than the second one, but the relationship between the first and third depends on details you haven't specified, and are frame dependent.

Secondly, it's not acceleration that leads to gravitational time dilation. It's the difference in gravitational potential between the two clocks being observed. So the fourth lift turning on its engine has no effect on the clock rate, as Cryo points out, except that its velocity changes.

The conclusion I would draw is that I can tell whether I am on a planet or in an accelerating elevator by measuring my time dilation. On the planet time dilation has the same rate. In an accelerating elevator, my rate of time dilation should be increasing. But the equivalence principle says I shouldn't be able to tell the difference.

And here is one more problem. You cannot "measure your time dilation". You can only compare your clock rate to a different clock. If you do that locally in an accelerating frame you need to compare a clock hung on the wall to a falling clock. Inevitably the falling clock is changing speed compared to the wall one, whether due to curved spacetime or an accelerating room.

If you do large scale experiments you certainly can tell the difference between acceleration and gravity. The ancient Greeks measured the curvature of the Earth, so they could have determined if they were in a lift. The equivalence principle only applies to local measurements made in a small volume.

 

[sweet springs]

The term "Lorentz transformations" only refers to transformations between inertial frames.

I found the formula (99) in Moller describing clock proper time moving in gravitational field.
$$d\tau=dt\sqrt{1+\frac{2\chi}{c^2}-\frac{u^2}{c^2}}$$
It is not a direct answer to your problem but it may suggest the way to learn.

 

[pervect]

There is a derivation of the transform equation from a set of coordinates that represents an accelerated frame to inertial coordinates in MTW's textbook "Gravitation".

The basic technique used is to transform the basis vectors from an instantaneously co-moving inertial frame to the stationary inertial frame. Knowing how the basis vectors transform, and also knowing the equations of motion of the worldline of the accelerated observer, one can find a set of coordinates that represents an transformation from an "accelerated frame" to a static inertial frame, or vica versa. By some definitions, a frame is defined by it's basis vectors, so the coordinate transformations mentioned above are a bit of an extra.

It's a bit long to type out, but one can borrow the textbook from the library. It does use tensor notation, however.