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The notion of weight in relativity

  1. Jul 13, 2013 #1
    The notion of "weight" in relativity

    I have recently been involved in a rather animated discussion on another forum on the subject of "weight". To make a long story short, the scenario involved a train moving straight and unaccelerated at relativistic speeds through a uniform gravitational field, as in on the surface of the earth. There is a lead ball of a given mass m on the train. In the cabin frame, the lead ball has a well defined and constant "weight", being the gravitational force acting on it. The question was - what would an observer at the embankment determine that weight to be ? In other words, how does the gravitational force acting on the lead ball transform if going from one inertial frame to another in this scenario ?

    The discussion involved a well-known crank who asserted that in the embankment frame the lead ball has an increased relativistic mass, and will thus crash through the cabin floor, destroying the train. Obviously this cannot be how things play out, but at the same time I couldn't come up with the appropriate maths to show otherwise.

    How can this be approached, mathematically ?
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  3. Jul 13, 2013 #2


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    That guy was obviously wrong. The train and its contents will become so massive at relativistic speeds that they will pull the earth up through the floor, destroying it (the earth).
  4. Jul 13, 2013 #3

    The lead ball will crash through the cabin floor, destroying the train. I just want to say that. :)

    Let's say, inside a large rocket, that is accelerating with constant proper acceleration, a train is moving at constant proper velocity relative to an observer standing on the rocket floor.

    The observer knows that a clock, inside the train, measuring how much time it takes for something to fall from the train ceiling to train floor, is time dilated, while the falling process is not time dilated.

    So the force of gravity has increased by the time dilation factor inside the train.
    Last edited: Jul 13, 2013
  5. Jul 13, 2013 #4
    Ha ha, yeah, you might be right :)
    Or the movement of the train might cause the earth to turn into a black hole...
  6. Jul 13, 2013 #5


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    This question often comes up. See the 1985 paper by Olson and Guarino here.
  7. Jul 13, 2013 #6


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    Nice :)
  8. Jul 13, 2013 #7
    What is the weight of the train in the embarkment frame?

    Let's again consider a spaceship with constant proper acceleration and a train that is moving inside the spacesip.

    It is not possible to reduce or increase the inertia of the cargo that the spaceship is carrying. Let's say the cargo is the train. So the moving train weighs the same as when it was standing still.

    Let's consider what happens when the train accelerates:

    Obviously the inertia of the fuel tank decreases, so the inertia of other parts increases. This redistribution of inertia will eventually cause the train to fall over, or break some other way.
  9. Jul 13, 2013 #8


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    Mathematically, first you need to define your notion of a uniform gravitational field. The Earth wont do, it's not uniform.

    The least error prone approach is probably to use Einstein's elevator to generate your uniform field. So you imagine a floor of an enormous elevator, accelerating upwards, and consider the fate of two observers, one intitally at rest on the floor, the other moving along it at some velocity.

    Replace the notion of "weight" with the 4-acceleration of each observer.

    You'll find a mathematical description of someone at rest on the elevator floor easily enough - just try http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken].

    Working out the description of someone moving at a constant velocity on the elevator floor will be trickier. I was reviewing some stuff I had written earlier, I wasn't really happy with how I did it and it started to get too involved when I started to redo it. So I'll leave that part up to you or some other poster for now - maybe I'll get some more time later.

    The tedious but non-argument inducing prone approach is to work everything out in inertial coordinates. This will naturally lead to a description in terms of the coordinate time t of the inertial observer. To calculate the 4-acceleration, though, you'll probably want to reparameterize the motion in terms of the proper time of each observer (which will be different than the coordinate time t of an inertial observer, obviously).

    What I expect you'll find if you work this all out is that the 4-accelerations of the two observers are not equal, and that the 4-acceleration required to keep the initially moving observer on the elevator floor is higher than the 4-acceleration required to keep the stationary observer on the floor. This is basically a consequence of relativistic time dilation.

    Another interesting consequence which I expect is is that due to the relativity of simultaneity, the floor that appears "flat" to the stationary observer will NOT appear flat to the moving observer. So part of the problem specification is a flat floor for some particular obserer (which we'll think of as the stationary one) and arguments based on the equivalence of all the observers will fail, for the floor will not be flat for all of them, but only for one of them.
    Last edited by a moderator: May 6, 2017
  10. Jul 14, 2013 #9


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    I'll sketch out a quick solution - using geometric units where c=1.

    We imagine the massive body sliding along the flat foor of an enormous, accelerating, elevator. We' will let the elevator accelerate in the z direction, and the body move in the x direction

    (t,x,z) will be inertial coordinates starting at some time t=0 when the elevator was at rest.
    We will also introduce tau, the proper time of the body.

    The motion of the massive object will be described, not by a constant velocity in the x direction, but by a constant momentum in the x direction. Having a constant velocity in the x direction is impossible - the velocity in the z direction soon approaches the speed of light, and the total velocity is v_tot^2 = vx^2 + vz^2. If vx remained constant, v_tot^2 would be greater than 1.

    We can thus write m*(dx/tau) = constant, and since m, the rest mass, is constant, we can write (dx/dtau) = K

    We also need the height of the elevator at time t - we can look this up at http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

    ##z = {\frac {\sqrt {1+{g}^{2}{t}^{2}}-1}{g}}##

    and it's derivative

    ##\frac{dz}{dt} = {\frac {gt}{\sqrt {1+{g}^{2}{t}^{2}}}}##

    Now we need to apply the relationship

    ##d\tau^2 = dt^2 - dx^2 - dz^2##

    We can rewrite this as

    [tex]1 = \left( \frac{dt}{d\tau} \right)^2 - \left( \frac{dx}{d\tau} \right) ^2 - \left( \frac{dz}{dt}\frac{dt}{d\tau} \right)^2 [/tex]

    We can then solve for ##dt/d\tau##:

    [tex]\frac{dt}{d\tau} = \sqrt{ \frac{1+K^2}{1-\frac{g^2t^2}{1+g^2 t^2 }} }[/tex]

    We already have ##dx/d\tau = K##, so all we need is ##dz/d\tau = (dz/dt)(dt/d\tau)##
    Ommiting the imetermediate algebraic steps
    ##\frac{dz}{d\tau} = g t \sqrt{1+K^2}##

    We could substitute for ##\tau ## at this point
    [tex] \tau = \int \frac{dt} { \left( dt / d\tau \right) } = \frac{1}{g \sqrt{1+K^2}} \sinh^{-1} gt[/tex]

    but it's easier (though perhaps not as instructive) to compute the 4-acceleration components

    [tex]a_t = \frac{d}{d\tau} \left( \frac{dt}{d\tau} \right) \quad a_x = 0 \quad a_z = \frac{d}{d\tau} \left( \frac{dz}{d\tau} \right) [/tex]

    via the chain rule ##\frac{d}{d\tau} = \left(dt / d\tau \right) \frac{d}{dt} ##, noting that since ##dx/d\tau = K## it's derivative is zero.

    The magnitude of the 4-aceleration ##\sqrt{|a_t ^2 - a_z ^2|}## is independent of time and is equal to

    ##A = g \left(1 + K^2 \right)##

    The value of K is the value of ##dx/d\tau##. If we let ##\beta_0 = dx/dt## at t=0, then ##K = \beta_0 / \sqrt{1-\beta_0 ^2}## and

    ##(1+K^2) = 1/(1-\beta_0^2) = \gamma_0^2##

    Thus we see the more or less expected result that the proper acceleration goes up as ##\gamma_0^2##, due to time dilation, and the fact that accleration is quadratic in time.
    Last edited by a moderator: May 6, 2017
  11. Jul 15, 2013 #10


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    Things you'll need to know to follow the derivation above (look them up if needed, or ask questions)

    0) In special relativity, space-time vectors transform by the Lorentz transformation. They're rather like normal 3-vectors, except that you need another component for the time.

    You can find a rather formal discussion on Wikki, it might be better to consult a modern SR textbook.

    1) A four vector, having components (t,x,y,z) has the property that (-t^2 + x^2 + y^2 + z^2) is invariant (does not change) under a Lorentz transformation. Here we have assumed geometric units, so that c=1, as I have in the rest of my derivation. Without this assumption, you need to stic a factor of "c^2" in front of the "t^2".

    2) Proper time has the property that it is invariant, because it does not change under a Lorentz transformation either. The invariant mass of an object (not to be confused with other sorts of mass) has the same quality, it also remains constant under a Lorentz transformation.

    2) The time and space interval between two events, (t2-t1), (x2-x1), (y2-y1), (z2-z1), is a 4-vector as above

    3) The rate of change of a position with respect to proper time is a 4-vector, as above, and is called a 4 velocity. You can basically prove that the 4-velocity is a 4-vector by considering the above- the displacement in the derivative is a 4 vector, and as long as we multiply it or divide it by something that remains invariant under a lorentz transform, it remains a 4-vector.

    4) Multiplying the 4-velocity by the invariant mass gives another 4-vector. This 4-vector is called the 4-momentum. The logic of why this is a 4-vector is similar to that above, mulitplying a 4 vector by a Lorentz invariant quantity yields another 4-vector.

    5) The rate of change of the 4-velocity with proper time is another 4-vector, called the 4-acceleration.

    6) If you calculate the 4-acceleration in the rest frame of an object, t=tau, d^2 t/ dtau^2 = 0, and hence the spatial components of the 4-accleration are just the 3-acceleration. Thus the magnitude of the 4-acceleration is the magnitude of the 3-acceleration in the objects own frame.

    Because of the nice properties of 4-vectors with respect to Lorentz transforms, the magnitude of the 4-acceleration will be the same in all frames related by the Lorentz transform. So we don't need to go to the object's frame to compute it's proper acceleration, we can use any frame we like that's convenient to solve the problem by use the 4-vector formalism.
  12. Jul 15, 2013 #11

    There must be some error, because I got proper acceleration going up as ##\gamma_0^1##.

    Maybe it would be a good idea to consider light beams sent from the middle of the train to various directions.

    A beam sent forwards hits the front wall at very low position, because "gravity" pulls the beam for a very long time.

    A beam sent backwards "falls" just for a very short time.

    A beam sent sideways hits the side wall at somewhat lower position, because of "gravity" pulling the beam for a time that is the time dilated time.

    (We have a moving train in an accelerating elevator here)

    For an observer standing on the elevator floor, the heights of the walls of the train are not contracted, therefore it's better to take the point of that observer.

    (For an inertial observer the walls are contracted and ... well all kinds of other changes)
  13. Jul 15, 2013 #12


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    I don't see the point of considering the light beams offhand.

    Perhaps we're calculating different things. I was basically calculating the accelerometer reading on the moving object. It might be worth computing the total force and 4-force required to keep the elevator in motion to figure out the floor loading. Because the transverse area doesn't change, the force/unit area should scale in the same manner as the total force.

    I wasn't focussed on the floor before, but in retrospect that seems to be the intent of the question.

    I think I'm getting a similar answer to yours when I answer that question, but there's something that bothers me about what I'm doing, so I need to think about it more before posting.
  14. Jul 16, 2013 #13


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    I'm not sure if we're totally justified in treating the slider (what I'll term the thing "sliding" on the floor of the elevator) as if it were an isolated system. It's interacting with the floor of the elevator. But if the pressure / c^2 is less than the density of the slider (which seems reasonable for reasonable accelerations), I suppose we can treat it as if it the slider had a mass rather than a stress-energy tensor (which is really the right way to work the problem, I think).

    Thus - it'd really be better to analyze the stress energy tensor of a perfect fluid blowing acrosss the elevator floor.

    But assuming that p/c^2 << rho, I think maybe we can agree that the floor loading (the pressure per unit area P on the floor) goes up as gamma^2 and not gamma. Correct me if I'm wrong, but you think the total force in the elevator frame is gamma, and I assume you'd agree that lorentz length contraction of the slider makes it's contact area 1/gamma of the slider's proper area, leading to a pressure increase of gamma^2 per unit area of the floor in the lab frame when it's moving compared to when it's at rest.

    I also get a pressure increase of gamma^2 in the frame of the slider, as that's what the proper acceleration is. In that frame, the pressure acts over the full contact area.

    The Z component of the pressure shouldn't be affected by a Lorentz boost in the x direction, the Lorentz boost should affect only the comonents in the x and t directions.
  15. Jul 16, 2013 #14

    This is a story about light beams from an inertial observers point of view:

    Let's say a light beam travels from one corner of an accelerating elevator to the diagonally opposite corner. It's a horizontal light beam in the elevator frame, at the beginning of its journey.

    An inertial observer decides that the same beam is a beam in a fast moving Einstein light clock. The velocity of the light clock is the x-component of the velocity of the light beam, where x-coordinate is aligned with two of the elevator walls. It's a rectangular elevator.

    The inertial observer knows that the beam descends two meters during the travel, which takes one second, for example. ("beam descends" means elevator moves upwards relative to beam)

    The inertial observer knows that an observer co-moving with the light clock will see the beam descending two meters during the travel, which according to him takes less than one second. A. Einsteins analysis of light clocks tells what the time is exactly. That answer is mathematically expressed in the time dilation formula.

    How rapidly light curves is a measure of acceleration.

    I see one problem: The light clock is moving "upwards" too, in the direction that the elevator is moving.
    Solution: Let's consider an elevator that has not gained much speed yet.
    Last edited: Jul 16, 2013
  16. Jul 16, 2013 #15


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    A horizontal beam should drop a distance of (1/2) a t^2. So if the light clock is 1 second long, the beam drops a distance of (1/2)*a, if it's 2 seconds long it drops 4 times as much, a distance of 2*a - it drops 4 times as much when you double the distance.

    If an observer in the elevator sets up a light clock along the floor, the sliding observer will agree on the drop distance (because it's transverse). But he won't agree on the length of the light clock.

    Suppose we have two light clocks along the floor of the elevator, in the x direction. The first light clock is 1 light second long in the elevator frame, the second light clock is 2 light seconds long in the elevator frame.

    The second light clock will have 4x the drop of the first as described above.

    Now suppose we have an observer moving along the floor of the elevator with a gamma factor of 2. He'll see the first light clock as being only half a light second long. He'll see the second light clock as having the correct length. Therefore he will compute his proper acceleration is being 4x that of the observer in the elevator frame.
  17. Jul 16, 2013 #16

    Well, I have a counter argument:

    When you are co-moving with a light clock, no part of the light clock is moving relative to you, except the light beam.

    But actually I have a correction to make:

    I said: "Curving rate of a light beam is a measure of acceleration"

    What the heck is "curving rate"?

    What I meant was:

    Turning rate of the velocity vector of a light beam is a measure of acceleration.

    Now distance is not involved, just angle and time.

    I guess radius of curvature of a light beam is also a measure of acceleration. Now time is not involved.
  18. Jul 16, 2013 #17


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    I'm afraid you've lost me again. Let me recap my argument, if it doesn't convince you, so be it. Regardless of whether you use a light clock to do the timing or some other sort of clock, falling objects that start at rest fall at a rate of s = (1/2) a t^2 (for a Newtonian fall) or s = (1/2) a t^2 + o(t^4) more generally. If we chose t small enough, we can ignore the order 4 terms.

    If you are measuring falling distance along the z axis and moving along the x-axis, relativistic effects don't measurably Lorentz contract distances in the direction that you fall , i.e. along the z axis, (not until your fall itself build up relativistic velocities, where the o(t^4) terms kick in).

    Therefore, if you have a time dilation factor of gamma compared to a coordinate observer, you will fall the same distance s in a time (t/gamma) that a coordinate observer whose clock is not time dilated considers to take a time of t.

    Thus (1/2) a_moving * t^2 / gamma^2 = (1/2) a_static * t^2, or a_moving / gamma^2 = a_static.
  19. Jul 17, 2013 #18

    I see. Doubling the acceleration does not halve the falling time, as I thought. And In post #3 there's the same error.

    Well maybe it's possible that proper acceleration goes up as gamma^2 then.
  20. Jul 17, 2013 #19
    Thank you very much pervect, this is much appreciated !
    So basically, the gist of the matter is that the weight of the object as determined by the two observers actually is different. So how does one then avoid the apparent paradox created by this ?
  21. Jul 17, 2013 #20


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    I'm not really sure of which apparent paradox you mean. Supplee's paradox is one common question/paradox we had a thread on that that here in PF, and there are some wiki articles and some literature on the topic as well.

    You might also ask "how can A be heavier than B, and B be heavier than A". A partial answer to that involves the fact that only one of A, B will see the floor of the spaceship as being flat, so they are not interchangable.
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