Will a spring extend more when going at relativistic speeds?

[pervect]

Oh , yes please do so, for the elevator example that you have given us and please announce the results!

See https://www.physicsforums.com/threads/the-notion-of-weight-in-relativity.701257/#post-4445483 where I worked it out in the past.

The problem is worked out by describing Superman's position in inertial coordinates (t,x,y,z) using geometric units where c=1. In the initial problem, though, Superman is replaced by a sliding block. Einstien's train is also a good choice for the moving object.

One thing that's a bit confusing about this post when I re-read it is that I introduced a parameter K to describe superman's velocity. The relation between K and superman's normalized velocity $\beta_0$ and K is that
$$K = \beta_0 / \sqrt{1-\beta_0^2} \quad \beta_0 = K / \sqrt{1+K^2}$$

also useful is the relationship $\gamma_0 = 1/\sqrt{1-\beta_0^2} = \sqrt{1+K^2}$

The relation between $\beta_0$ and K can be and is derived from the original post by computing $dx/dt =\frac{dx/d\tau}{dt/d\tau}$ at t=0.

I use the notation $\beta_0$ because Superman's velocity in the x direction in the inertial frame varies with time. Superman's momentum in the x-direction stays constant, and so does his velocity relative to the floor of the rocket. But the x-component of his velocity in an inertial frame of reference does not stay constant. However, at t=0, the elevator is at rest, and superman's velocity relative to the elevator floor is the same as his velocity in the inertial frame of reference.

Another thing that would be helpful to show is that Superman's motion is hyperbolic motion, but I didn't do that in that thread. Note that while it is hyperoblic motion, Superman's proper acceleration is different from the proper accleration of the elevator.

It seems to me there should be an easier way to do this, but I didn't find it, so I'll stick with what I've already done.

 

[PeterDonis]

It might not make a difference to the extension of the spring

To be clear, I only said it might not. As the previous analysis @pervect linked to just now shows, it actually does.

 

[PeterDonis]

What you get is an hyperbola in the t-x plane for the stationary observer and another one in 3D space not belonging in any of the t-x, t-y or x-y planes. Uniform acceleration didn't you say? Right, now measure the radius of curvature of your two hyperbolas and find them to be constant and equal. Next, measure the magnitude of your 3-acceleration vectors and find them to be constant and equal.

To briefly summarize the key point of the analysis in the previous thread that @pervect linked to: the radius of curvature in spacetime of the two hyperbolas (aka the path curvature, aka the proper acceleration) is in fact not equal. Both accelerations are uniform (the same everywhere on each worldline), but they are not equal; Superman's is higher.

This might seem counterintuitive given that the conditions of the problem are that Superman has to stay the same height above the floor of the elevator (Green Lantern's elevator) as he flies past. But that condition does not require that Superman's proper acceleration is the same as Green Lantern's; it requires that Superman's coordinate acceleration in Green Lantern's frame is the same as Green Lantern's. (This can be translated into a coordinate-independent criterion, but it won't end up being "both of their proper accelerations are equal".) Viewed geometrically, we have two hyperbolas, one of which is in the t-x plane, the other of which is at an "angle" to the t-x plane, so the plane in which it lies is not any of t-x, t-y, or x-y, as you say. But the projection of the second hyperbola (Superman's) into the t-x plane must be identical to that of the first hyperbola (Green Lantern's); that's what the conditions of the problem require. And since the second hyperbola's plane is "at an angle" to the t-x plane, in order to have the same projected curvature in that plane, its actual curvature must be higher.

 

[dagmar]

o briefly summarize the key point of the analysis in the previous thread that @pervect linked to: the radius of curvature in spacetime of the two hyperbolas (aka the path curvature, aka the proper acceleration) is in fact not equal. Both accelerations are uniform (the same everywhere on each worldline), but they are not equal; Superman's is higher.

They're equal..It's not a Rindler universe.

Constant acceleration everywhere we said.

Excuse me now, but I don't have time to read your last messages. I have a marathon to run and you've kept me awake last night. In the meantime re-think your examples and don't be so rash.

 

[PeterDonis]

It's not a Rindler universe.

I'm not sure what you mean by that.

Constant acceleration everywhere we said.

Not if by that you mean that Superman and Green Lantern both have the same proper acceleration.

In the meantime re-think your examples and don't be so rash.

You should take your own advice. I don't have to re-think anything; @pervect already posted a link to a previous thread where this problem was analyzed. Please read it when you get a chance.

 

[dagmar]

You should take your own advice. I don't have to re-think anything; @pervect already posted a link to a previous thread where this problem was analyzed. Please read it when you get a chance.

Good. Be happy. Your analysis is irrelevant to me and to anyone who understands Physics for that matter.

If you believe that the spacetime direction of the 4-acceleration vector matters it is your problem. If you believe that the orientation of the walls of an elevator in regard to the outcomes of an experiment matter, it is your problem. The spatial vector of the acceleration in @pervect's example points always at the same direction and that's exactly what your spring's extension is measuring.

Bye.

 

[pervect]

An outline of tensor approach which gives the answer when one writes the trajectory of Superman in Rindler coordinates, (T,X,Z). as an alternative method follows. The techniques used are from GR, not SR.

The line element defines the coordinates, there are several forms, I'd take one of the simplest (but not completely general) forms and write:

$$ds^2 = -Z^2\,dT^2 + dX^2 + dZ^2$$

I've used geometric units, so c=1. Next we write the trajectory of Superman in these coordinates.

$$T=f(\tau) \quad X = \beta f(\tau) \quad Z=1$$

which implies that $X = \beta T$, and T is some function of proper time $f(\tau)$. This makes the velocity of superman relative to the surface of the elevator equal to $\beta$ , it makes the 4-acceleration of a point on the elevators surface (at Z=1) have unit magnitude, and it makes $g_{tt} = 1$ at the surface of the elevator.

Having chosen our coordinates (which are defined by the line element), we then compute the 4-velocity u of Superman, which has the components

u = ($dT/d\tau, dX/d\tau, dZ/d\tau$)

The normalization condition:

$g_{\mu\nu} u^\mu u^\nu = -1$

gives us the value of $f(\tau)$. 4-velocities must have a magnitude of -1 with a -+++ metric signature. It turns out that $df/d\tau$ is (unsurprisingly) independent of time T, though it does depend on Z. The normalized 4-velocity has components

$$
u = ( \frac{1}{\sqrt{Z^2-\beta^2}}, \frac{\beta}{\sqrt{Z^2-\beta^2}},0 )
$$

Then the 4-acceleration can be calculated from the 4 velocity $u^\mu$ by

$$a^\nu = u^\mu \nabla_\mu u^\nu$$

the only non-zero component of the 4-acceleration is the z component, which has a value of $\frac{Z}{Z^2-\beta^2}$. Setting Z=1 gives us the value of the 4-acceleraton of Superman (assuming he's flying on or close to the floor of the elevator), which is just $\frac{1}{1-\beta^2}$. This is different from the unit magnitude 4-acceleration of a point on the surface of the elevator.

As a check, one can compute the 4-velocity v of a point "at rest" , i..e. X=Z=constant and find

$$v = (\frac{1}{Z},0,0)$$

and use the same formula to calculate the proper acceleration associated with v. This results also has only one non-zero component, the z-component as before, but it has a different value, namely 1/Z. Setting Z=1, gives us the expected unit value for the 4-accelration of an object at rest on the surface of the elevator.

 

[Orodruin]

If you believe that the spacetime direction of the 4-acceleration vector matters it is your problem. If you believe that the orientation of the walls of an elevator in regard to the outcomes of an experiment matter, it is your problem. The spatial vector of the acceleration in @pervect's example points always at the same direction and that's exactly what your spring's extension is measuring.

You are missing the point. The point is that, contrary to your assertion in #39, Superman's proper acceleration is different from the one of Green Lantern. This is easy to show just using basic 4-vector analysis, as done in @pervect's post, and the result is that they differ by a factor $\gamma^2$, where the $\gamma$ is computed using the velocity of Superman in Green Lantern's instantaneous rest frame. Both have constant proper acceleration but your assertion that these proper accelerations are equal is just wrong.

 

[dagmar]

And what do you think you're doing by using Rindler worldlines to describe motion in an everywhere perfectly uniform grav. field like that of the elevator example in question??

No, everywhere at each point around your elevator thought experiment, acceleration stays the same in magnitude and direction ( its spatial vector part ). Just draw
the worldlines in a simple spacetime diagram and forget about it. Like I said there are simple hyperbolas with constant curvature everywhere.

You are wrong if you are thinking that you're going to cover up your mistakes by twisting a simple elevator example into something so complex like the real grav. field of a planet. Sorry.

 

[Orodruin]

And what do you think you're doing by using Rindler worldlines to describe motion in an everywhere perfectly uniform grav. field like that of the elevator example in question??

No, everywhere at each point around your elevator thought experiment, acceleration stays the same in magnitude and direction ( its spatial vector part ). Just draw
the worldlines in a simple spacetime diagram and forget about it. Like I said there are simple hyperbolas with constant curvature everywhere.

You are wrong if you are thinking that you're going to cover up your mistakes by twisting a simple elevator example into something so complex like the real grav. field of a planet. Sorry.

I suggest you stop the ad hominem argumentation and show your maths to make it clear exactly what you are doing as your argumentation is unclear.

The accelerating elevator example is an example in flat spacetime, there is nothing wrong with that and it should be sufficient to analyse the situation locally as this really has nothing to do with curvature and only something to do with the local relation between the observers' 4-velocities and 4-accelerations. @pervect is therefore perfectly free to use whatever coordinates he wants.

Unless you want a completely futile discussion, you need to write down your maths. Your reference to "simple hyperbolas ... everywhere" is unclear. Having hyperbolae with the same curvature in different parts of Minkowski space is what Bell's spaceship paradox is all about. There is no separate thing such as the "Rindler universe", it is just a patch of Minkowski space.

 

[PeterDonis]

Your analysis is irrelevant to me

Then you obviously don't need to be posting further in this thread. So you are now banned from doing so.

 

[Idunno]

OK, I hope I have this straight, so, if one were to blindly apply the SR mass equation, one would get that $Weight =m_0g\gamma = \frac {m_0g} {\sqrt{1 -\frac{v^2}{c^2}}}$ This is wrong, of course.

However, Pervect is saying that, using Einstein's elevator, $Weight =m_0g\gamma^2 = \frac {m_0g} {1 -\frac{v^2}{c^2}}$ If I am interpreting his past posts correctly and getting his symbols right, Is this right?

 

[PeterDonis]

Is this right?

Yes, but it's important to understand why. It's not because the mass of the object on the end of the spring increases (by a factor $\gamma$). It's because the proper acceleration of the object on the end of the spring increases (by a factor $\gamma^2$). In other words, the moving object has to accelerate harder than the stationary object. (Note also that this all assumes that both objects stay at a constant height above the floor of Green Lantern's elevator, as seen in Green Lantern's frame.)

To put it another way, for both objects, $W = m_0 a$, where $W$ is weight, $m_0$ is rest mass, and $a$ is proper acceleration. But for the stationary object, $a = g$, while for the moving object, $a = \gamma^2 g$.

 

[pervect]

OK, I hope I have this straight, so, if one were to blindly apply the SR mass equation, one would get that $Weight =m_0g\gamma = \frac {m_0g} {\sqrt{1 -\frac{v^2}{c^2}}}$ This is wrong, of course.

However, Pervect is saying that, using Einstein's elevator, $Weight =m_0g\gamma^2 = \frac {m_0g} {1 -\frac{v^2}{c^2}}$ If I am interpreting his past posts correctly and getting his symbols right, Is this right?

It's a bit more complicated. When I think of the weight of a moving block (which I prefer over Superman), I think of weighing it with a truck scale.
With this sort of "weight", the weight is gamma*m*g, m being the rest or invariant mass. The test mass (the block) is moving in this case, but not the spring scale that's doing the weighing.

If the scale moves along with the test mass, the "weight" measured is the magnitude of the 4-acceleration of the block. Physically, this could be measured by any sort of accelerometer. A common technique for a gravimeter / accelerometer measures the time it takes a body to fall in a vacuum, this could profitably be used to replace the spring scale in the thought experiment. The sort of weight" measured in that manner is proportional to gamma^2*m*g.

 

[pervect]

OK, here's the super simple version involving the notion of a "dropped mass" style of accelerometer. These are commonly used as "absolute" gravimeters.. It gets the right answer, but it's not very rigorous, I have some concenrs in that regard.

Before we get into this, though, I want to point out that the "dropped mass" in this case is just an instantaneously co-moving inertial observer. The observer on the elevator, and also Superman, are accelerating. The "dropped test mass" is moving inertially.

If we consider an instantaneously co-moving observer who starts out "at rest" with respect to the elevator floor, they will see the elevator accelerate away from them , to a high degree of precision obeying the Newtonian law s = 1/2 a t^2, where a is the acceleration of the elevator. If they watch long enough, eventually the elevator will reach relativistic velocities and we'd need to do a relativistic analysis, but if we consider short time periods this is not necessary.

So if the acceleration, a, of the elevevator is 32 feet/sec^2 (1 gravity), at t = 1 second, s = 16 feet.

The accelerating observer at rest on the elevator floor will see essentially the same thing as their associated instantaneously co-moving inertial observer.

Now let's consider Superman's point of view. Superman also has an instantaneously co-moving observer. Superman and that observer both share essentially the same inertial frame of reference, which is different from the frame of reference of the elevator floor. Due to time dilation, the moving observer's clocks will run slower by the factor gamma. So 1 second on the stationary observers clock is only 1/2 a second on the moving clock, if we assume the gamma factor is 2. This means the elevator accelerates away from the moving observer by 16 feet in 1/2 second. Plugging that into s=1/2 a t^2, the moving observer computes that the elevator is accelerating away at 4 gravities. When we generalize this to a general factor of gamma , we see that the acceleration went up by a factor of gamma^2.

Thus superman (and the instantaneously inertial observer co-moving with him) sees the elevator's acceleration as $\gamma^2 a$.

Lorentz contraction doesn't have any effect on the analysis, the distances are measured perpendicularly to the direction of motion.

The issue that I'm concerned with is to justify how we can ignore the relativity of simultaneity.

 

[jartsa]

OK, here's the super simple version involving the notion of a "dropped mass" style of accelerometer. These are commonly used as "absolute" gravimeters.. It gets the right answer, but it's not very rigorous, I have some concenrs in that regard.

Before we get into this, though, I want to point out that the "dropped mass" in this case is just an instantaneously co-moving inertial observer. The observer on the elevator, and also Superman, are accelerating. The "dropped test mass" is moving inertially.

If we consider an instantaneously co-moving observer who starts out "at rest" with respect to the elevator floor, they will see the elevator accelerate away from them , to a high degree of precision obeying the Newtonian law s = 1/2 a t^2, where a is the acceleration of the elevator. If they watch long enough, eventually the elevator will reach relativistic velocities and we'd need to do a relativistic analysis, but if we consider short time periods this is not necessary.

So if the acceleration, a, of the elevevator is 32 feet/sec^2 (1 gravity), at t = 1 second, s = 16 feet.

The accelerating observer at rest on the elevator floor will see essentially the same thing as their associated instantaneously co-moving inertial observer.

Now let's consider Superman's point of view. Superman also has an instantaneously co-moving observer. Superman and that observer both share essentially the same inertial frame of reference, which is different from the frame of reference of the elevator floor. Due to time dilation, the moving observer's clocks will run slower by the factor gamma. So 1 second on the stationary observers clock is only 1/2 a second on the moving clock, if we assume the gamma factor is 2. This means the elevator accelerates away from the moving observer by 16 feet in 1/2 second. Plugging that into s=1/2 a t^2, the moving observer computes that the elevator is accelerating away at 4 gravities. When we generalize this to a general factor of gamma , we see that the acceleration went up by a factor of gamma^2.

Thus superman (and the instantaneously inertial observer co-moving with him) sees the elevator's acceleration as $\gamma^2 a$.

Lorentz contraction doesn't have any effect on the analysis, the distances are measured perpendicularly to the direction of motion.

The issue that I'm concerned with is to justify how we can ignore the relativity of simultaneity.

Shouldn't the vertical motion of the elevator be slowed down according an observer that observes an elevator that has lot of horizontal speed? (Compared to an observer that is glued on the elevator floor)

And shouldn't the elevator floor still be curved, according to that observer?(I'm not sure what is discussed here, but I'm guessing that we want to know what superman flying through an accelerating elevator observes )

 

[PeterDonis]

The issue that I'm concerned with is to justify how we can ignore the relativity of simultaneity.

If the acceleration is independent of time, as it is in this scenario, then relativity of simultaneity is not an issue because it doesn't matter where the different spacelike slices of the two observers (Green Lantern and Superman) intersect the world tube of the elevator; the curvature of the world tube is the same everywhere, and so are the curvatures of the worldlines of the two observers.

If the acceleration is not independent of time, then you will indeed have to take relativity of simultaneity into account.

 

[Idunno]

Yes, but it's important to understand why. It's not because the mass of the object on the end of the spring increases (by a factor $\gamma$). It's because the proper acceleration of the object on the end of the spring increases (by a factor $\gamma^2$). In other words, the moving object has to accelerate harder than the stationary object. (Note also that this all assumes that both objects stay at a constant height above the floor of Green Lantern's elevator, as seen in Green Lantern's frame.)

To put it another way, for both objects, $W = m_0 a$, where $W$ is weight, $m_0$ is rest mass, and $a$ is proper acceleration. But for the stationary object, $a = g$, while for the moving object, $a = \gamma^2 g$.

Thanks a bunch everyone, I think I understand all this much better now, for whatever that's worth. :)

 

[pervect]

Something I'd like to clear up. Calculating the 4-acceleration from the 4-velocity is a very straightforwards procedure that can be found in most texts, in the context of either special relativity or general relativity.

The tricky part of the problem problem (IMO) is finding the 4-velocity of Superman, aka the sliding block. I do this in https://www.physicsforums.com/threads/the-notion-of-weight-in-relativity.701257/#post-4445483, but it's helpful to eliminate the factor K I used in the derivation, and replace it with $\beta_0$, the block's normalized velocity, and $\gamma_0 = 1/\sqrt{1-\beta_0^2}$

The formulae to do the conversion are in the original post, but the converted results are not written down explicitly. Doing so now, we find

$$\frac{dt}{d\tau} = \gamma_0 \sqrt{1+g^2\,t^2} \quad \frac{dx}{d\tau} = \beta_0 \, \gamma_0 \quad \frac{dz}{d\tau} = \gamma_0 \, g \, t$$

This express the 4-velocity in terms of t, not $\tau$. My original intent was to stop here, but as I wrote this I realized it'd be helpful to go further and additionaly re-write the proper velocity as a function of $\tau$ rather than t. To do this we make use of the relation between t and $\tau$ derived in the original post:

$$\tau = \frac{\sinh^{-1} gt}{ \gamma_0 g} \quad t = \frac{1}{g} \sinh \gamma_0 g \tau $$

We can then substitute to find the 4-velocity as a function of $\tau$

$$\frac{dt}{d\tau} = \gamma_0 \sqrt{1+\sinh^2 \gamma_0 g \tau} = \gamma_0 \cosh \gamma_0 g \tau \quad \frac{dx}{d\tau} = \beta_0 \gamma_0 \quad \frac{dz}{d\tau} = \gamma_0 \, \sinh \gamma_0 g \tau $$

While I'm at, I realized I can write the 3-velocity of Superman as a function of time:

$$\frac{dx}{dt} = \frac{\beta_0}{\sqrt{1+g^2\,t^2 }} \quad \frac{dz}{dt}= \frac{g\,t}{ \sqrt{1+g^2t^2}} $$

We can see that the velocity $dz/dt$ is the same function of time t as for the relativistic rocket, as it should be.

 

[PeterDonis]

Thread closed for moderation, since it looks like the OP question has been answered and the thread is now bogged down in an off topic tangent.

 

[PeterDonis]

A number of off topic posts have been deleted. Since the thread has run its course and the OP has expressed satisfaction with the answers given, the thread will remain closed.