# Will a spring extend more when going at relativistic speeds?

• I
Suppose Superman and Green Lantern are on an airless planet. Green Lantern whips up a spring and puts a mass on the end, and the spring extends some amount due to the gravitational pull on the mass. Superman then goes off with the spring and mass, and gets up to 0.9c and flies past Green Lantern, at the same height that the first spring measurement was made.

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

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

BvU
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Will the spring extend more now
No.

That better ?

Who is Green Lantern anyway ?

Orodruin
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"Relativistic mass" is not any form of inertia. It is just another name for the energy of an object divided by ##c^2##. Nothing in the object is actually changing just because you happen to use a coordinate system where it moves. You could just as well use a coordinate system where it is at rest.

Slightly better, however, a quick explanation as to why not would be nice, or a link that explains why not would be better. I understand your time and patience are not infinite, so thanks for so far.

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.

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

Orodruin
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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.
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.

• Dale and BvU
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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.
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.

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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BvU
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Second to that ! And idem Oro's suggestion !

russ_watters
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Slightly better, however, a quick explanation as to why not would be nice....
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!

This is resolved by The Principle of Relativity.

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pervect
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Suppose Superman and Green Lantern are on an airless planet. Green Lantern whips up a spring and puts a mass on the end, and the spring extends some amount due to the gravitational pull on the mass. Superman then goes off with the spring and mass, and gets up to 0.9c and flies past Green Lantern, at the same height that the first spring measurement was made.

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.

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.
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.

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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.

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).

pervect
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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. #### Attachments

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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.
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?

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 parallel to the direction of acceleration.

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 parallel to the direction of acceleration.
Oh, jeez, sorry for not being more specific, yes, I meant he is going perpendicular to the direction of acceleration.

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.

@pervect Huh? What's this all about? I was refering 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 refering 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?).

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

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

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?
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?

pervect
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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.