- #1

Idunno

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Will the spring extend more now, or the same as before? Does the relativistic mass equation (which, I understand, is not liked amongst those with higher education in physics) apply to this?

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- #1

Idunno

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Will the spring extend more now, or the same as before? Does the relativistic mass equation (which, I understand, is not liked amongst those with higher education in physics) apply to this?

- #2

BvU

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Nope: spring and mass both move at 0.9 cDoes the relativistic mass equation apply to this?

- #3

Idunno

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Thank you for replying, but i don't think you are really answering the question.

- #4

BvU

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- #6

Idunno

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Obviously, if one were to take the relativistic mass equation seriously, green lantern now sees the spring and mass as having increased, thus gravity pulls on it more and the spring is more extended than before. Superman sees the spring mass as the same and the planet mass to have increased, thus also would say the spring is more extended. So why is this wrong?

If you want to know who green lantern is, a quick internet search will answer that question.

- #7

BvU

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And in my first reply I mentioned that spring and mass both move with the same speed

But: I didn't take into account that Supe experiences a different gravity -- that complicates things, so I'd rather leave it out but I do remember there's a SR transformation formula for force

On the other hand: Greene sees the other guys length contracted

- #8

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No. First of all, you cannot talk about gravity in SR at all. SR has no description of gravity and you should therefore not expect ”relativistic mass” to do either. Second, in GR mass is not the source of gravity, the stress-energy-momentum tensor is so while total energy (again, another abd better word for relativistic mass) does influence gravity, you cannot disregard other parts. Third, as has been explained in many posts and in one of my Insight articles, relativistic mass is also not directly the inertia so it does not directly have any of the properties mass in classical mechanics does. This is why it is typically no longer used as a concept apart from in popular texts intended to sell and dazzle rather than to teach physics. I suggest you erease it from your relativity vocabulary.Obviously, if one were to take the relativistic mass equation seriously, green lantern now sees the spring and mass as having increased, thus gravity pulls on it more and the spring is more extended than before.

- #9

Nugatory

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The ##m## that appears in the relativistic mass equation refers to something altogether different than the ##m## that appears in the various gravity formulas (##F=mgh##, ##F=Gm_1m_2/r^2##), so changing the relativistic mass (which you can do just by choosing to describe the situation using a different coordinate system) doesn't increase the gravitational force.Obviously, if one were to take the relativistic mass equation seriously, green lantern now sees the spring and mass as having increased, thus gravity pulls on it more and the spring is more extended than before.

Confusion of this sort is the reason why relativistic mass is seldom used these days; in hindsight it was a mistake to ever introduce the concept. We have a FAQ on this subject - give me a moment and I'll post the link.

[Edit, a moment later: https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/]

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- #10

BvU

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Second to that ! And idem Oro's suggestion !

- #11

russ_watters

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The spring and mass are stationary with respect to each other. if speed mattered, then because you have an infinite number of different speed, there'd be no way to decide which to use in the calculations!Slightly better, however, a quick explanation as to why not would be nice...

This is resolved by The Principle of Relativity.

- #12

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Will the spring extend more now, or the same as before? Does the relativistic mass equation (which, I understand, is not liked amongst those with higher education in physics) apply to this?

I *think* the intent of this is for a stationary observer to measure the weight of a test mass with a spring scale in an unspecified "gravitational field", while the moving observer does the same.

To make the problem convenient to calculate, we replace the unspecified gravitational field with a uniform gravitational field, and further clarify what we mean by a "uniform gravitational field" to say that it's the sort of "field" one gets on an accelerating elevator (Einstein's elevator).

This gets rid of a lot of unnecessary complications, among which are the fact that the field flying by a massive object will be changing rapidly as one flies by and one gets distracted by how fast the scale responds, and is just easier tro calculate.

Putting the green lantern into it is just one superhero too many. (I'd actually opt for getting rid of Superman, too, but that's just me. The rationale is that you'll get more serious answers if you leave out the comic book theme.)

In those terms, yes, the weight of the test mass increases, and the spring on the spring scale stretches more for Superman than it does for the stationary observer.

The rationale behind replacing the massive object with the "field" of the elevator is called the "principle of Equivalence". It tends to cause a fair number of questions on its own in how it should be interpreted, but hopefully it's somewhat clear how it works in this context.

- #13

dagmar

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Since when, and in what possible way the phenomena observed in a closed elevator that undergoes constant acceleration g with initial velocity=0, will be any different than those in another elevator that begins to accelerate constantly ( same g ) with initial velocity v=velocity of Superman? Because I think that's exactly what the OP is asking.I *think* the intent of this is for a stationary observer to measure the weight of a test mass with a spring scale in an unspecified "gravitational field", while the moving observer does the same.

To make the problem convenient to calculate, we replace the unspecified gravitational field with a uniform gravitational field, and further clarify what we mean by a "uniform gravitational field" to say that it's the sort of "field" one gets on an accelerating elevator (Einstein's elevator).

This gets rid of a lot of unnecessary complications, among which are the fact that the field flying by a massive object will be changing rapidly as one flies by and one gets distracted by how fast the scale responds, and is just easier tro calculate.

Putting the green lantern into it is just one superhero too many. (I'd actually opt for getting rid of Superman, too, but that's just me. The rationale is that you'll get more serious answers if you leave out the comic book theme.)

In those terms, yes, the weight of the test mass increases, and the spring on the spring scale stretches more for Superman than it does for the stationary observer.

The rationale behind replacing the massive object with the "field" of the elevator is called the "principle of Equivalence". It tends to cause a fair number of questions on its own in how it should be interpreted, but hopefully it's somewhat clear how it works in this context.

No, the elongation of the spring is the same for both observers ( in the uniform gravitational field in question. ) Period.

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- #14

PeterDonis

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I think that's exactly what the OP is asking.

I don't. See below.

the elongation of the spring is the same for both observers ( in the uniform gravitational field in question. )

Here is how I think you are understanding the OP's question: we have a spring and a mass sitting on a planet, at rest relative to the planet's surface. One observer (GL) is sitting next to the spring/mass on the planet. The other observer (SM) is flying past at 0.9c, in a trajectory that is horizontal (i.e., perpendicular to the direction of the spring/mass extension). Both of them measure the elongation of the spring.

In this scenario, I agree that both of them would measure the same elongation, since it's perpendicular to the direction of relative motion.

However, here is how I am understanding the OP's question: there are two separate parts to the scenario.

(1) SM and GL are standing on the planet, at rest on its surface, with the spring and mass. Both of them measure the elongation of the spring.

(2) SM takes the spring/mass with him while GL stays on the planet's surface. SM flies past GL at 0.9c, holding the spring/mass system. Both SM and GL again measure the elongation of the spring.

SM and GL then both compare the measurements they made in #2 with the measurement they both made in #1.

In this scenario, the #2 measurements will not be the same as the #1 measurement. In fact, in #2, the spring's direction of elongation won't even be constant, since the direction of the planet relative to the spring/mass system is changing, so I'm not sure what "measurement" would count as "the" elongation of the spring.

- #15

PeterDonis

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Here is how I think you are understanding the OP's question

Actually, I misstated this; what I described is how others might be understanding the OP's question. But I think you (@dagmar) are understanding the OP's question to be comparing these two scenarios:

a closed elevator that undergoes constant acceleration g with initial velocity=0

another elevator that begins to accelerate constantly ( same g ) with initial velocity v=velocity of Superman?

I agree that the elongation of the spring, relative to the observer at rest in the elevator, will be the same in both cases; but I don't think that's what the OP is asking about (I described how I interpret the OP's question in my previous post).

- #16

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Since when, and in what possible way the phenomena observed in a closed elevator that undergoes constant acceleration g with initial velocity=0, will be any different than those in another elevator that begins to accelerate constantly ( same g ) with initial velocity v=velocity of Superman? Because I think that's exactly what the OP is asking.

No, the elongation of the spring is the same for both observers ( in the uniform gravitational field in question. ) Period.

No, this doesn't follow.

It is true that one can construct a stationary elevator with flat floors in a stationary frame. "Stationary frame" is just a word here to distinguish it from "Superman's frame", there is no claim that it's stationary in any absolute sense. Similarly, one can construct another elevator instantaneously co-moving with Superman with flat floors. But, non-intuitively, it turns out that the elevator with flat floors in the stationary frame won't have flat floors in superman's frame. And vica-versa. The elevator with flat floors in the stationary frame is a different elevator than the one with flat floors in Superman's frame.

To appreciate this takes some math, in particular the Lorentz transforms. It's a consequence of the relativity of simultaneity.

Special relativity can only be easily applied to accelerating frames if we use the idea of instantaneously co-moving inertial frames. Fortunately, this is quite possible to do. To proceed, we need to distinguish between the time t in the stationary frame, and the time t' in the moving frame, they are related by the Lorentz transform

$$t' = \gamma(t - \beta x/c) \quad x' = \gamma(x - \beta c t)$$

Here ##\beta## is the normalized velocity ##v/c##, and ##\gamma = 1/\sqrt{1-\beta^2}##.

Saying the floor is flat at some instant t means that x=constant for all t. When we perform a Lorentz boost , though, we find that x' as transformed won't be constant for all t', we findthat the flat floor of the elevator will no longer be flat in the moving frame. The Lorentz transformations are linear, which is why one might expect a flat floor to map to a flat floor. But the elevator is accelerating, and the Lorentz transformation mixes together space and time, so while the Lorentz transform is linear, the accelerated motion is not linear.

See for instance figure 11 of the paper https://arxiv.org/abs/0708.2490v1

The paper is mainly about Thomas precession, but it also illustrates the effect I'm talking about, where the floor (rails in the paper) appear curved.

- #17

Idunno

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That is what I am asking, yes. thanks for your reply. To get this straight, this is a GR problem, not and SR problem, and you wouldn't use the SR equation to calculate how much the spring elongates? Yes?I don't. See below.

Here is how I think you are understanding the OP's question: we have a spring and a mass sitting on a planet, at rest relative to the planet's surface. One observer (GL) is sitting next to the spring/mass on the planet. The other observer (SM) is flying past at 0.9c, in a trajectory that is horizontal (i.e., perpendicular to the direction of the spring/mass extension). Both of them measure the elongation of the spring.

In this scenario, I agree that both of them would measure the same elongation, since it's perpendicular to the direction of relative motion.

However, here is how I am understanding the OP's question: there are two separate parts to the scenario.

(1) SM and GL are standing on the planet, at rest on its surface, with the spring and mass. Both of them measure the elongation of the spring.

(2) SM takes the spring/mass with him while GL stays on the planet's surface. SM flies past GL at 0.9c, holding the spring/mass system. Both SM and GL again measure the elongation of the spring.

SM and GL then both compare the measurements they made in #2 with the measurement they both made in #1.

In this scenario, the #2 measurements will not be the same as the #1 measurement. In fact, in #2, the spring's direction of elongation won't even be constant, since the direction of the planet relative to the spring/mass system is changing, so I'm not sure what "measurement" would count as "the" elongation of the spring.

Would it simplify matters to have a long, straight platform of rock thick enough to have a decent gravitational field, so there is no complication due to a curved planet? Then SM flies over this long piece of rock at a constant height?

- #18

PeterDonis

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the elevator with flat floors in the stationary frame won't have flat floors in superman's frame. And vica-versa.

You're interpreting the OP's question as having Superman moving perpendicular to the direction of acceleration. (So am I, and, I suspect, everyone else except @dagmar .) But I think @dagmar is interpreting Superman to be moving

- #19

Idunno

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- #20

PeterDonis

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To get this straight, this is a GR problem, not and SR problem, and you wouldn't use the SR equation to calculate how much the spring elongates? Yes?

The version of the problem as you state it, with a planet, is a GR problem, yes, because spacetime isn't flat. But as @pervect has pointed out, you can greatly simplify the problem by simulating "gravity" with acceleration in flat spacetime. That version can be analyzed using just SR; however, it still involves some complications (such as the surface of the "elevator" that is accelerating not being flat in Superman's frame).

Would it simplify matters to have a long, straight platform of rock thick enough to have a decent gravitational field, so there is no complication due to a curved planet? Then SM flies over this long piece of rock at a constant height?

In GR, even the gravitational field of a long, straight platform of rock is not perfectly "uniform" (as it would be in Newtonian gravity). Also, there are velocity-dependent effects in GR that add further complications. The simplest scenario to analyze that brings in the essential point is probably pervect's version, where Green Lantern is in an "elevator" accelerating in flat spacetime, and Superman's trajectory is a combination of that acceleration and constant-velocity motion perpendicular to the acceleration at 0.9c.

- #21

dagmar

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@pervect Huh? What's this all about? I was referring to your #12 post. The direction of motion of the second elevator is at right angles with its flat floor, and both floors of the elevators are parallel to each other. It's shooting upwards the surface of an imaginary planet past a stationary elevator on the surface of the planet, at constant speed ( and the acceleration of gravity is constant, no matter what height above the surface of the planet, at least that's what your example indicates).

Edit: Let's remove the planet in the above example and stick to elevators only.

@PeterDonis Are we in agreement with what I wrote in the above sentence? Or are you referring to a real planet with ever-changing g according to some radial coordinate ? Even so, what's the difference? Do you feel any extra weight on your legs when you're in a platform and shooting upwards with constant velocity (if you take g to be the same for small heights above the surface of the Earth?).

I didn't read the last 2 new messages in this thread. I'll read them now...

Edit: Let's remove the planet in the above example and stick to elevators only.

@PeterDonis Are we in agreement with what I wrote in the above sentence? Or are you referring to a real planet with ever-changing g according to some radial coordinate ? Even so, what's the difference? Do you feel any extra weight on your legs when you're in a platform and shooting upwards with constant velocity (if you take g to be the same for small heights above the surface of the Earth?).

I didn't read the last 2 new messages in this thread. I'll read them now...

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- #22

dagmar

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@Idunno Well do tell us, which scenario do you imply? Please read the answers carefully this time.

- #23

Idunno

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As PeterDonis says, SM is flying perpendicular to the planet's acceleration due to gravity, at about the same height as the original measurement of the spring's elongation.@Idunno Well do tell us, which scenario do you imply? Please read the answers carefully this time.

First they stand side by side on the planet and measure the elongation of the spring when they put a mass on it.

Second, SM takes the spring mass system and gets up to 0.9c, and flies past GL, still on the surface. They both measure the spring elongation as he passes by GL, and compare it to what they first measured it to be. Is it the same, or different?

- #24

dagmar

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Perpendicular to the planet's acceleration, hence tangentially to the surface of the planet... As measured by each in his FOR (frame of reference) right?As PeterDonis says, SM is flying perpendicular to the planet's acceleration due to gravity, at about the same height as the original measurement of the spring's elongation.

First they stand side by side on the planet and measure the elongation of the spring when they put a mass on it.

Second, SM takes the spring mass system and gets up to 0.9c, and flies past GL, still on the surface. They both measure the spring elongation as he passes by GL, and compare it to what they first measured it to be. Is it the same, or different?

- #25

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That is what I am asking, yes. thanks for your reply. To get this straight, this is a GR problem, not and SR problem, and you wouldn't use the SR equation to calculate how much the spring elongates? Yes?

Would it simplify matters to have a long, straight platform of rock thick enough to have a decent gravitational field, so there is no complication due to a curved planet? Then SM flies over this long piece of rock at a constant height?

Not really, no. The problem can be done with SR, if one uses "Einstein's elevator" to generate what one might call "artificial gravity", and make use of the eqivalence principle to argue that the results should apply to gravity from a "real massive object".

The tensor methods of GR can be applied to the "Einstein's elevator" case, or to the gravity of an actual massive object.

The challenge is basically how to encapsulate the notion of weight in tensor notation, as GR uses tensors, but the quantity of physical interest here, weight, isn't a tensor concept. How would we physically weigh a moving truck, for instance? We have commercial truck scales that do just that. I'd argue that the basic technique applies, we measure the pressure the truck exerts on the floor through it's tires, and multiply the pressure by the contact area of the tires, to get the total weight of the truck. This approach needs a bit of an adjustment though, it turns out not to make sense unless one assumes that ##g_{00} = 1##. The issue here is that force or weight can be regarded as the rate of transfer of momentum (part of the energy momentum 4-vector) per unit time, but the physical notion of weight basically assumes that coordinate clocks tick at the same rate as the proper clocks carried by "the observer" measuring the weight.

This is s undoubtedly not the only approach to the issue of a tensor description of weight, but it seems nicely physical to me. Others might not like it, though.

However, for the vast majority of readers who aren't familiar with tensors, saying to "work the problem with GR" won't help, because they don't know tensors and hence can't really know GR, at least not well enough to get anything quantitative out of it. That's why I recommend a treatment with Einstein's elevator and SR. It's the only I-level approach that I'm aware of.

- #26

Mister T

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Does the relativistic mass equation (which, I understand, is not liked amongst those with higher education in physics) apply to this?

If ##m## is the mass then ##\gamma m## is the relativistic mass, where ##\gamma=\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}##. That is just a definition so it applies to this case and any other. And there are some people with a higher education in physics who like it just fine. The thing is, whether you like it or not makes no difference to the physics. It comes out the same either way.

The biggest reason for not liking it is that it creates problems for those trying to learn and understand relativity. For example, thinking that the Newtonian expression for the gravitational force, ##G\frac{Mm}{r^2}##, is a perfectly valid expression provided you replace ##m## with ##\gamma m##. In other words, that ##\gamma m## is a genuine relativistic generalization of ##m##. It is not, and the responses in this thread explain some of the reasons why.

- #27

PeterDonis

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@PeterDonis Are we in agreement with what I wrote in the above sentence?

I agree that Superman's motion, as seen in the elevator frame (i.e, Green Lantern's frame), is perpendicular to the direction of the elevator's acceleration.

However, @pervect is correct that, in Superman's frame, the elevator floor will not be flat. More precisely, the spatial locus occupied by the elevator floor at some constant time ##t'## in Superman's frame will not be a flat plane. The reason is relativity of simultaneity; surfaces of constant time in the two frames "cut" the world tube of the elevator floor at different angles, and only one such "cut" results in flat planes.

- #28

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@pervect Huh? What's this all about? I was referring to your #12 post. The direction of motion of the second elevator is at right angles with its flat floor, and both floors of the elevators are parallel to each other. It's shooting upwards the surface of an imaginary planet past a stationary elevator on the surface of the planet, at constant speed ( and the acceleration of gravity is constant, no matter what height above the surface of the planet, at least that's what your example indicates).

Edit: Let's remove the planet in the above example and stick to elevators only.

I suspect we're analyzing the same problem, and you're just not getting the point of what I was trying to say. I can see where it might be difficult to follow, but I'm not sure how to resolve the issue. I'd normaly suggest drawing a space-time diagram, but aside from the usual resistance in getting people to either draw their own space-time diagrams or look at space-time diagrams that others have drawn, one would need a 3-d diagram. One needs a time axis, an acceleration axis (the z-axis in my own thoughts), and another axis for superman to fly along (the x-axis, in my thoughts) that's perpendicular to the acceleration axis, to fully depict the problem.

The end result, though, is that one does not wind up with two elevators with parallel floors. Your intuition that it does is leading you astray here.

- #29

Idunno

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Yes.Perpendicular to the planet's acceleration, hence tangentially to the surface of the planet... As measured by each in his FOR (frame of reference) right?

- #30

dagmar

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Ok, I misinterpreted your example. Just remember that in case Superman flies along the radius of planet at constant speed there is no difference in elongation of string.Yes.

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- #31

Idunno

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Thanks everybody! :) Is there an easy way to calculate the spring extension when you use Einstein's elevator? Or is that pretty involved?

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- #32

dagmar

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Ok, draw your diagrams. 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. Who cares in which direction your vectors point at? Who cares in which direction your elevator floor is pointing at, or if its shape is cubic or parallelepiped? (Put a wedge shaped object on the floor to make it level and stand on it, to do your experiments!). You have found constant and equal in magnitude acceleration in both frames and that's exactly the measurement of the extension of your spring and that's equal in both frames, and regardless of the direction of motion of your second elevator with respect to the first, up or down, to and fro in any frame...I suspect we're analyzing the same problem, and you're just not getting the point of what I was trying to say. I can see where it might be difficult to follow, but I'm not sure how to resolve the issue. I'd normaly suggest drawing a space-time diagram, but aside from the usual resistance in getting people to either draw their own space-time diagrams or look at space-time diagrams that others have drawn, one would need a 3-d diagram. One needs a time axis, an acceleration axis (the z-axis in my own thoughts), and another axis for superman to fly along (the x-axis, in my thoughts) that's perpendicular to the acceleration axis, to fully depict the problem.

The end result, though, is that one does not wind up with two elevators with parallel floors. Your intuition that it does is leading you astray here.

I am repeating what I have just said:

Please do not lead the OP astray, by altering the meanings of the words that you've just written firstly, and by passing from idealized examples to real world situations secondly, and this applies to @PeterDonis.

Cheers, be good!

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- #33

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Thanks everybody! :) Is there an easy way to calculate the spring extension when you use Einstein's elevator? Or is that pretty involved?

Probably the easiest thing to calculate the proper acceleration of superman. This turns out to be

##\gamma_0^2##g, where ##\gamma_0 = 1/\sqrt{1-v^2/c^2}##, v being Superman's velocity relative to the foor of the elevator, and g is the acceleration of the elevator (as measured by an accelerometer at rest on the elevator floor). This formula is only valid if Superman is flying at "ground level", by the way.

However, this isn't necessarily all that easy to calculate without 4-vectors. I'm reluctant to do a 4-vector treatment if it isn't of interest, I'm also reluctant to try to stumble through a 3-vector treatment.

I will point out that the velocity v, of Superman, in the above expression is relative to the floor of the elevator, and not the velocity he has relative to an inertial observer, which is different (and not even constant).

So the first problem is to calculate the motion of the floor of the accelerating elevator in some inertial frame of reference, then one needs to calculate the motion of Superman in some inertial frame of reference. Once one understands the motion, one can compute the proper acceleration, there's a 3-vector formulation of it in Wiki, though I'd use the 4-vector formalism myself.

Wiki offers a helpful defintion of the proper acceleration, to illustrate why it is of interest:

In infinitesimal small durations there is always one inertial frame, which momentarily has the same velocity as the accelerated body, and in which the Lorentz transformation holds. The corresponding three-acceleration ain these frames can be directly measured by an accelerometer, and is called proper acceleration

By phrasing the question in terms of proper acceleration, we avoid having to consider the details of the spring.

There are a fair number of therads on PF where these calculations are carried out , at various mathematical levels. I'm not sure there is one I'd particularly care to recommend at the I level, though. People unfamiliar with them resist using 4-vector formulations, but the 3-vector forumations are very awkwards.

- #34

dagmar

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Oh , yes please do so, for the elevator example that you have given us and please announce the results!I'm reluctant to do a 4-vector treatment if it isn't of interest, I'm also reluctant to try to stumble through a 3-vector treatment.

- #35

PeterDonis

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Next, measure the magnitude of your 3-acceleration vectors and find them to be constant and equal. Who cares in which direction your vectors point at?

If they're different, then that's an observable difference.

I am repeating what I have just said:Since when, and in what possible way the phenomena observed in a closed elevator that undergoes constant acceleration g with initial velocity=0, will be any different than those in another elevator that begins to accelerate constantly ( same g in magnitude and direction ) with initial velocity v?

If the directions of acceleration are different in the two cases, that is a difference in observed phenomena. It might not make a difference to the extension of the spring, but it

Please do not lead the OP astray, by altering the meanings of the words that you've just written firstly, and by passing from idealized examples to real world situations secondly

Please do not lead the OP astray by making claims that are (a) not true, and (b) stronger than what you actually need to claim to answer the OP's question.

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