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Space-Time Questions

  1. Nov 19, 2013 #1
    Hi.
    I know that as an object travels faster, its mass increases. Would this increase create a larger curvature in space-time?
     
  2. jcsd
  3. Nov 19, 2013 #2
    That'sa good question.

    I'm pretty sure that.... yes. It would. If you travel AT the speed of light (which is impossible) you would have infinite mass and would suck the whole universe :p

    Correct me if I'm wrong, PFers
     
  4. Nov 19, 2013 #3

    PAllen

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    Turn this question around to see how problematic it is:

    If someone else somewhere travels fast relative to me, my mass and the curvature around me increase. Say what?

    More correct:

    If you are traveling faster relative to A than to B, then A will measure you as having more momentum.

    As for curvature, for a body or system in reasonable isolation from other bodies, it's curvature contribution is determined by its energy/pressure measured in its COM frame.
     
    Last edited: Nov 19, 2013
  5. Nov 19, 2013 #4

    PeterDonis

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    No. Spacetime curvature is generated by the stress-energy tensor, which is covariant; the individual components of the SET (of which energy, i.e., "mass" in the sense you're using the term, is one) change, but they change in such a way that the spacetime curvature generated stays the same.
     
  6. Nov 19, 2013 #5

    Nugatory

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    That's not right, although there are enough oversimplified popularizations out that you can be forgiven for thinking that it is.

    But think about it for a moment.... Is your mass increasing by outrageous amounts right now? Why not? After all, you are moving at 99.999% of the speed of light relative to the ravenous bug-bladder beasts of the planet Traal (that's a literary allusion - If you don't recognize it, google for it! :smile:). Of course, you're also pretty much at rest relative to the surface of the earth, so maybe your mass shouldn't be increasing at all?

    What relativity actually says is that an observer will measure the mass of an object that is moving relative to the observer to be greater than the mass of the same object as measured by an observer at rest relative to the object. Furthermore, this effect is completely symmetrical. If I fly past at .6c you will measure my mass to be 1.25 times what it would be if I were at rest relative to you - but as far as I am concerned, I am at rest, you are moving backwards at .6c, there's nothing wrong with my mass, and you're the one who is 25% heavier.
     
  7. Nov 19, 2013 #6
    @PeterDonis your answer makes sense. If someone else was travelling almost at the speed of light, would it affect my mass? No, that wouldn't make sense :p

    The theory of relativity is quite complex, when you think you fully understand you, you find out you don't.

    @Nugatory that doesn't answer his main question. Would the space time curvature increase?
     
  8. Nov 19, 2013 #7
    The modern understanding is that mass is invariant and does not increase with relative velocity. One of the reasons that mass is considered invariant is exactly the one you raise. An object with relative motion does not have an increased effective gravitational mass. As PAllen alluded to, an observer with a high velocity relative to the Earth does not see the Earth turn into a black hole.

    The concept of relativistic mass is outdated. It requires that we have one one type of relativistic mass for how an an objects responds to a parallel force and another type of relativistic mass for how an object responds to a transverse force, another type of mass in the kinetic energy equation and yet another type of relativistic mass for the gravitational mass of an object.

    The modern usage is that there is only one type of mass, the invariant rest mass. We simply have to accept that the equations of relativity are different to those of Newtonian physics and we do not obtain the relativistic equations simply by substituting relativistic mass into the Newtonian equations.
     
    Last edited: Nov 19, 2013
  9. Nov 19, 2013 #8

    Nugatory

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    No. It can't because different observers will see different mass increases, so if the perceived mass increase affected the curvature of space-time then different observers would get different results for the curvature of space-time at the same point. You'll find similar arguments in some of the traditional "paradoxes" of special relativity: for example, two different observers may disagree about the simultaneity of various events occurring inside the triggering mechanism for a bomb, but they all have to come to the same conclusion about whether the bomb explodes or not.
     
  10. Nov 19, 2013 #9

    WannabeNewton

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    Here's an analogy. Say we are in a background global inertial frame observing a single charged particle moving in some arbitrary motion (in free space). Using the definitions ##\vec{E} = -\vec{\nabla}\varphi - \partial_t \vec{A}## and ##\vec{B} = \vec{\nabla}\times \vec{A}##, we can derive the equations ##\vec{\nabla}^2 \vec{A} - \frac{1}{c^2}\partial_t^2 \vec{A} = -\frac{4\pi}{c} \vec{j}## and ##\vec{\nabla}^2 \varphi - \frac{1}{c^2}\partial_t^2 \varphi= -4\pi \rho## from Maxwell's equations. Solving these for the single charged particle gives us the Lienard-Wiechert potentials. At this point this is all just basic review.

    The point is that in this particular background global inertial frame we have some charged particle moving along some arbitrary (but predetermined) trajectory and based on the charge density ##\rho## and 3-current density ##\vec{j}## that we see in this frame, we attribute to this charged particle an electric field and magnetic field relative to this frame that will, through the Lienard-Wiechert potentials, depend on both the velocity ("Coulomb-like" field) and acceleration ("radiation" field) of the particle. The electric field and magnetic field together determine an electromagnetic field and the important detail is that the electromagnetic field is a covariant object.

    The question to then ask is: if I go to the instantaneous rest frame of the charged particle, where ##\vec{j} = 0##, will the electromagnetic field change? I mean if I go to this instantaneous rest frame then for that instant all I see is a Coulomb field which is clearly not the same as the electric and magnetic fields we get from the Lienard-Wiechert potentials in the original background global inertial frame so what happens to the electromagnetic field? The point is that the electromagnetic field doesn't care about what frame you use to represent it; it's characteristics are purely geometric (in the sense that it is a 4-tensor). The electric and magnetic fields transform from frame to frame so as to keep the electromagnetic field covariant -the relative nature of motion is already accounted for in the covariant framework. I mean this is all very intuitive when it comes to relativistic electrodynamics right?

    Now just replace the electromagnetic field with space-time curvature and the 4-current density with the energy-momentum tensor and the answer is the exact same.

    Just as a side note, you shouldn't talk about the space-time curvature due to a point particle in the general relativity setting. Point particles work fine for Poisson's equation for the Newtonian potential and for Maxwell's equations because point particles are given by dirac delta functions (which are distributions) and distributional sources are well defined for linear field equations. The EFEs are non-linear and as such products of distributions have no general meaning.
     
  11. Nov 19, 2013 #10

    pervect

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    Space time curvature is a rank 4 tensor

    Because of its a complex structure, one can't directly compare curvatures to decide if one is greater or lesser than another, so the question doesn't really have an answer as it is asked. This is similar to the way that one cannot compare complex numbers to determine which is "greater" and which is "lesser".

    I assume the OP is assuming the curvature tensor is a single real number - this is not the case. Because it is not the case, there's no unique way to rank-order curvature tensors.

    The details below are overkill, but it may be helpful to describe what the structure of the (Riemann) curvature tensor is. (There are several curvature tensors, the Riemann is the most complete).

    One can think of the Riemann , a rank 4 tensor, as a 4x4x4x4 hypercube which has 256 cells , each cell which has a place for a number. Each number measures one "component" of the curvature.

    The Riemann curvature tensor has a high degree of symmetry. Because of this symmetry (google for Biancci identities if you want the exact details), only 36 of these 256 cells have non-zero numbers, the remaning cells contain zeroes.

    There are only 21 unique values, and only 20 degrees of freedom. (There are less degrees of freedom than unique values because of a constraint equation).
     
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